# Heat and gravity.

PRAJWAL NEUPANE

Geodynamics

First published in 1982, Don Turcotte and Jerry Schubert’s Geodynamics be-

came a classic textbook for several generations of students of geophysics and

geology. In this second edition, the authors bring this classic text completely

up-to-date. Important additions include a chapter on chemical geodynamics,

an updated coverage of comparative planetology based on recent planetary

missions, and a variety of other new topics.

Geodynamics provides the fundamentals necessary for an understanding of

the workings of the solid Earth. The Earth is a heat engine, with the source

of the heat the decay of radioactive elements and the cooling of the Earth

from its initial accretion. The work output includes earthquakes, volcanic

eruptions, and mountain building. Geodynamics comprehensively explains

these concepts in the context of the role of mantle convection and plate

tectonics. Observations such as the Earth’s gravity field, surface heat flow,

distribution of earthquakes, surface stresses and strains, and distribution of

elements are discussed. The rheological behavior of the solid Earth, from an

elastic solid to fracture to plastic deformation to fluid flow, is considered.

Important inputs come from a comparison of the similarities and differences

between the Earth, Venus, Mars, Mercury, and the Moon. An extensive set

of student exercises is included.

This new edition of Geodynamics will once again prove to be a classic

textbook for intermediate to advanced undergraduates and graduate stu-

dents in geology, geophysics, and Earth science.

Donald L. Turcotte is Maxwell Upson Professor of Engineering, Depart-

ment of Geological Sciences, Cornell University. In addition to this book, he

is author or co-author of 3 books and 276 research papers, including Fractals

and Chaos in Geology and Geophysics (Cambridge University Press, 1992

and 1997) and Mantle Convection in the Earth and Planets (with Gerald

Schubert and Peter Olson; Cambridge University Press, 2001). Professor

Turcotte is a Fellow of the American Geophysical Union, Honorary Fellow

of the European Union of Geosciences, and Fellow of the Geological So-

ciety of America. He is the recipient of several medals, including the Day

Medal of the Geological Society of America, the Wegener Medal of the Euro-

pean Union of Geosciences, the Whitten Medal of the American Geophysical

Union, the Regents (New York State) Medal of Excellence, and Caltech’s

Distinguished Alumnus Award. Professor Turcotte is a member of the Na-

tional Academy of Sciences and the American Academy of Arts and Sciences.

Gerald Schubert is a Professor in the Department of Earth and Space

Sciences and the Institute of Geophysics and Planetary Physics at the Uni-

iv

versity of California, Los Angeles. He is co-author with Donald Turcotte

and Peter Olson of Mantle Convection in the Earth and Planets (Cambridge

University Press, 2001), and author of over 400 research papers. He has par-

ticipated in a number of NASA’s planetary missions and has been on the

editorial boards of many journals, including Icarus, Journal of Geophysical

Research, Geophysical Research Letters, and Annual Reviews of Earth and

Planetary Sciences. Professor Schubert is a Fellow of the American Geo-

physical Union and a recipient of the Union’s James B. MacElwane medal.

He is a member of the American Academy of Arts and Sciences.

Contents

Preface page x

Preface to the Second Edition xiii

1 Plate Tectonics 1

1.1 Introduction 1

1.2 The Lithosphere 9

1.3 Accreting Plate Boundaries 10

1.4 Subduction 15

1.5 Transform Faults 23

1.6 Hotspots and Mantle Plumes 25

1.7 Continents 30

1.8 Paleomagnetism and the Motion of the Plates 36

1.9 Triple Junctions 59

1.10 The Wilson Cycle 65

1.11 Continental Collisions 70

1.12 Volcanism and Heat Flow 76

1.13 Seismicity and the State of Stress in the Lithosphere 85

1.14 The Driving Mechanism 90

1.15 Comparative Planetology 91

1.16 The Moon 92

1.17 Mercury 97

1.18 Mars 99

1.19 Phobos and Deimos 105

1.20 Venus 105

1.21 The Galilean Satellites 107

2 Stress and Strain in Solids 127

2.1 Introduction 127

2.2 Body Forces and Surface Forces 128

vi Contents

2.3 Stress in Two Dimensions 140

2.4 Stress in Three Dimensions 146

2.5 Pressures in the Deep Interiors of Planets 148

2.6 Stress Measurement 151

2.7 Basic Ideas about Strain 154

2.8 Strain Measurements 167

3 Elasticity and Flexure 185

3.1 Introduction 185

3.2 Linear Elasticity 187

3.3 Uniaxial Stress 189

3.4 Uniaxial Strain 191

3.5 Plane Stress 193

3.6 Plane Strain 196

3.7 Pure Shear and Simple Shear 197

3.8 Isotropic Stress 198

3.9 Two-Dimensional Bending or Flexure of Plates 199

3.10 Bending of Plates under Applied Moments and Vertical

Loads 205

3.11 Buckling of a Plate under a Horizontal Load 210

3.12 Deformation of Strata Overlying an Igneous Intrusion 212

3.13 Application to the Earth’s Lithosphere 216

3.14 Periodic Loading 217

3.15 Stability of the Earth’s Lithosphere Under an End Load 220

3.16 Bending of the Elastic Lithosphere under the Loads of

Island Chains 222

3.17 Bending of the Elastic Lithosphere at an Ocean Trench 227

3.18 Flexure and the Structure of Sedimentary Basins 230

4 Heat Transfer 237

4.1 Introduction 237

4.2 Fourier’s Law of Heat Conduction 238

4.3 Measuring the Earth’s Surface Heat Flux 240

4.4 The Earth’s Surface Heat Flow 242

4.5 Heat Generation by the Decay of Radioactive Elements 244

4.6 One-Dimensional Steady Heat Conduction 249

4.7 A Conduction Temperature Profile for the Mantle 253

4.8 Continental Geotherms 254

4.9 Radial Heat Conduction in a Sphere or Spherical Shell 260

4.10 Temperatures in the Moon 263

4.11 Steady Two- and Three-Dimensional Heat Conduction 264

Contents vii

4.12 Subsurface Temperature 266

4.13 One-Dimensional, Time-Dependent Heat Conduction 269

4.14 Periodic Heating of a Semi-Infinite Half-Space 271

4.15 Instantaneous Heating or Cooling of a Semi-Infinite

Half-Space 276

4.16 Cooling of the Oceanic Lithosphere 285

4.17 Plate Cooling Model of the Lithosphere 290

4.18 The Stefan Problem 294

4.19 Solidification of a Dike or Sill 300

4.20 The Heat Conduction Equation in a Moving Medium 304

4.21 One-Dimensional, Unsteady Heat Conduction in an

Infinite Region 307

4.22 Thermal Stresses 310

4.23 Ocean Floor Topography 317

4.24 Changes in Sea Level 323

4.25 Thermal and Subsidence History of Sedimentary Basins 325

4.26 Heating or Cooling a Semi-Infinite Half-Space 333

4.27 Frictional Heating on Faults 335

4.28 Mantle Geotherms and Adiabats 337

4.29 Thermal Structure of the Subducted Lithosphere 345

4.30 Culling Model for the Erosion and Deposition of Sediments 348

5 Gravity 354

5.1 Introduction 354

5.2 Gravitational Acceleration 355

5.3 Centrifugal Acceleration and the Acceleration of Gravity 365

5.4 The Gravitational Potential and the Geoid 366

5.5 Moments of Inertia 373

5.6 Surface Gravity Anomalies 378

5.7 Bouguer Gravity Formula 383

5.8 Reductions of Gravity Data 385

5.9 Compensation 387

5.10 The Gravity Field of a Periodic Mass Distribution on a

Surface 389

5.11 Compensation Due to Lithospheric Flexure 391

5.12 Isostatic Geoid Anomalies 394

5.13 Compensation Models and Observed Geoid Anomalies 397

5.14 Forces Required to Maintain Topography and the Geoid 405

6 Fluid Mechanics 411

6.1 Introduction 411

viii Contents

6.2 One-Dimensional Channel Flows 412

6.3 Asthenospheric Counterflow 418

6.4 Pipe Flow 421

6.5 Artesian Aquifer Flows 425

6.6 Flow Through Volcanic Pipes 426

6.7 Conservation of Fluid in Two Dimensions 427

6.8 Elemental Force Balance in Two Dimensions 428

6.9 The Stream Function 432

6.10 Postglacial Rebound 434

6.11 Angle of Subduction 442

6.12 Diapirism 447

6.13 Folding 456

6.14 Stokes Flow 467

6.15 Plume Heads and Tails 476

6.16 Pipe Flow with Heat Addition 481

6.17 Aquifer Model for Hot Springs 485

6.18 Thermal Convection 488

6.19 Linear Stability Analysis for the Onset of Thermal

Convection 492

6.20 A Transient Boundary-Layer Theory 500

6.21 A Steady-State Boundary-Layer Theory 505

6.22 The Forces that Drive Plate Tectonics 516

6.23 Heating by Viscous Dissipation 521

6.24 Mantle Recycling and Mixing 525

7 Rock rheology 538

7.1 Introduction 538

7.2 Elasticity 540

7.3 Diffusion Creep 553

7.4 Dislocation Creep 568

7.5 Shear Flows of Fluids 574

7.6 Mantle Rheology 588

7.7 Rheological Effects on Mantle Convection 597

7.8 Mantle Convection and the Cooling of the Earth 599

7.9 Crustal Rheology 605

7.10 Viscoelasticity 609

7.11 Elastic–Perfectly Plastic Behavior 615

8 Faulting 627

8.1 Introduction 627

8.2 Classification of Faults 628

Contents ix

8.3 Friction on Faults 632

8.4 Anderson Theory of Faulting 637

8.5 Strength Envelope 642

8.6 Thrust Sheets and Gravity Sliding 643

8.7 Earthquakes 647

8.8 San Andreas Fault 659

8.9 North Anatolian Fault 664

8.10 Some Elastic Solutions for Strike–Slip Faulting 667

8.11 Stress Diffusion 679

8.12 Thermally Activated Creep on Faults 682

9 Flows in Porous Media 692

9.1 Introduction 692

9.2 Darcy’s Law 693

9.3 Permeability Models 695

9.4 Flow in Confined Aquifers 697

9.5 Flow in Unconfined Aquifers 700

9.6 Geometrical Form of Volcanoes 717

9.7 Equations of Conservation of Mass, Momentum, and

Energy for Flow in Porous Media 722

9.8 One-Dimensional Advection of Heat in a Porous Medium 725

9.9 Thermal Convection in a Porous Layer 729

9.10 Thermal Plumes in Fluid-Saturated Porous Media 735

9.11 Porous Flow Model for Magma Migration 746

9.12 Two-Phase Convection 752

10 Chemical Geodynamics 761

10.1 Introduction 761

10.2 Radioactivity and Geochronology 763

10.3 Geochemical Reservoirs 771

10.4 A Two-Reservoir Model with Instantaneous Crustal

Differentiation 776

10.5 Noble Gas Systems 786

10.6 Isotope Systematics of OIB 788

Appendix A Symbols and Units 795

Appendix B Physical Constants and Properties 806

Appendix C Answers to Selected Problems 815

Index 828

Preface

This textbook deals with the fundamental physical processes necessary for

an understanding of plate tectonics and a variety of geological phenomena.

We believe that the appropriate title for this material is geodynamics. The

contents of this textbook evolved from a series of courses given at Cornell

University and UCLA to students with a wide range of backgrounds in

geology, geophysics, physics, mathematics, chemistry, and engineering. The

level of the students ranged from advanced undergraduate to graduate.

In all cases we present the material with a minimum of mathematical

complexity. We have not introduced mathematical concepts unless they are

essential to the understanding of physical principles. For example, our treat-

ment of elasticity and fluid mechanics avoids the introduction or use of

tensors. We do not believe that tensor notation is necessary for the under-

standing of these subjects or for most applications to geological problems.

However, solving partial differential equations is an essential part of this

textbook. Many geological problems involving heat conduction and solid and

fluid mechanics require solutions of such classic partial differential equations

as Laplace’s equation, Poisson’s equation, the biharmonic equation, and the

diffusion equation. All these equations are derived from first principles in the

geological contexts in which they are used. We provide elementary explana-

tions for such important physical properties of matter as solid-state viscosity,

thermal coefficient of expansion, specific heat, and permeability. Basic con-

cepts involved in the studies of heat transfer, Newtonian and non-Newtonian

fluid behavior, the bending of thin elastic plates, the mechanical behavior of

faults, and the interpretation of gravity anomalies are emphasized. Thus it

is expected that the student will develop a thorough understanding of such

fundamental physical laws as Hooke’s law of elasticity, Fourier’s law of heat

conduction, and Darcy’s law for fluid flow in porous media.

The problems are an integral part of this textbook. It is only through

Preface xi

solving a substantial number of exercises that an adequate understanding

of the underlying physical principles can be developed. Answers to selected

problems are provided.

The first chapter reviews plate tectonics; its main purpose is to provide

physics, chemistry, and engineering students with the geological background

necessary to understand the applications considered throughout the rest of

the textbook. We hope that the geology student can also benefit from this

summary of numerous geological, seismological, and paleomagnetic observa-

tions. Since plate tectonics is a continuously evolving subject, this material

may be subject to revision. Chapter 1 also briefly summarizes the geologi-

cal and geophysical characteristics of the other planets and satellites of the

solar system. Chapter 2 introduces the concepts of stress and strain and dis-

cusses the measurements of these quantities in the Earth’s crust. Chapter 3

presents the basic principles of linear elasticity. The bending of thin elastic

plates is emphasized and is applied to problems involving the bending of the

Earth’s lithosphere. Chapter 4 deals mainly with heat conduction and the

application of this theory to temperatures in the continental crust and the

continental and oceanic lithospheres. Heat transfer by convection is briefly

discussed and applied to a determination of temperature in the Earth’s man-

tle. Surface heat flow measurements are reviewed and interpreted in terms

of the theory. The sources of the Earth’s surface heat flow are discussed.

Problems involving the solidification of magmas and extrusive lava flows are

also treated. The basic principles involved in the interpretation of gravity

measurements are given in Chapter 5. Fluid mechanics is studied in Chapter

6; problems involving mantle convection and postglacial rebound are empha-

sized. Chapter 7 deals with the rheology of rock or the manner in which it

deforms or flows under applied forces. Fundamental processes are discussed

from a microscopic point of view. The mechanical behavior of faults is dis-

cussed in Chapter 8 with particular attention being paid to observations of

displacements along the San Andreas fault. Finally, Chapter 9 discusses the

principles of fluid flow in porous media, a subject that finds application to

hydrothermal circulations in the oceanic crust and in continental geothermal

areas.

The contents of this textbook are intended to provide the material for a

coherent one-year course. In order to accomplish this goal, some important

aspects of geodynamics have had to be omitted. In particular, the fundamen-

tals of seismology are not included. Thus the wave equation and its solutions

are not discussed. Many seismic studies have provided important data rele-

vant to geodynamic processes. Examples include (1) the radial distribution

of density in the Earth as inferred from the radial profiles of seismic veloci-

xii Preface

ties, (2) important information on the locations of plate boundaries and the

locations of descending plates at ocean trenches provided by accurate deter-

minations of the epicenters of earthquakes, and (3) details of the structure

of the continental crust obtained by seismic reflection profiling using arti-

ficially generated waves. An adequate treatment of seismology would have

required a very considerable expansion of this textbook. Fortunately, there

are a number of excellent textbooks on this subject.

A comprehensive study of the spatial and temporal variations of the

Earth’s magnetic field is also considered to be outside the scope of this

textbook. A short discussion of the Earth’s magnetic field relevant to pale-

omagnetic observations is given in Chapter 1. However, mechanisms for the

generation of the Earth’s magnetic field are not considered.

In writing this textbook, several difficult decisions had to be made. One

was the choice of units; we use SI units throughout. This system of units

is defined in Appendix 1. We feel there is a strong trend toward the use of

SI units in both geology and geophysics. We recognize, however, that many

cgs units are widely used. Examples include µcal cm−2 s−1 for heat flow,

kilobar for stress, and milligal for gravity anomalies. For this reason we have

often included the equivalent cgs unit in parentheses after the SI unit, for

example, MPa (kbar). Another decision involved the referencing of original

work. We do not believe that it is appropriate to include a large number of

references in a basic textbook. We have credited those individuals making

major contributions to the development of the theory of plate tectonics and

continental drift in our brief discussion of the history of this subject in

Chapter 1. We also provide references to data. At the end of each chapter a

list of recommended reading is given. In many instances these are textbooks

and reference books, but in some cases review papers are included. In each

case the objective is to provide background material for the chapter or to

extend its content.

Many of our colleagues have read all or parts of various drafts of this

textbook. We acknowledge the contributions made by Jack Bird, Peter Bird,

Muawia Barazangi, Allan Cox, Walter Elsasser, Robert Kay, Suzanne Kay,

Mark Langseth, Bruce Marsh, Jay Melosh, John Rundle, Sean Solomon,

David Stevenson, Ken Torrance, and David Yuen. We particularly wish to

acknowledge the many contributions to our work made by Ron Oxburgh

and the excellent manuscript preparation by Tanya Harter.

Preface to the Second Edition

As we prepared our revisions for this second edition of Geodynamics we were

struck by the relatively few changes and additions that were required. The

reason is clear: this textbook deals with fundamental physical processes that

do not change. However, a number of new ideas and concepts have evolved

and have been included where appropriate.

In revising the first chapter on plate tectonics we placed added emphasis

on the concept of mantle plumes. In particular we discussed the association

of plume heads with continental flood basalts. We extensively revised the

sections on comparative planetology. We have learned new things about the

Moon, and the giant impact hypothesis for its origin has won wide accep-

tance. For Venus, the Magellan mission has revolutionized our information

about the planet. The high-resolution radar images, topography, and grav-

ity data have provided new insights that emphasize the tremendous differ-

ences in structure and evolution between Venus and the Earth. Similarly,

the Galileo mission has greatly enhanced our understanding of the Galilean

satellites of Jupiter.

In Chapter 2 we introduce the crustal stretching model for the isostatic

subsidence of sedimentary basins. This model provides a simple explanation

for the formation of sedimentary basins. Space-based geodetic observations

have revolutionized our understanding of surface strain fields associated with

tectonics. We introduce the reader to satellite data obtained from the global

positioning system (GPS) and synthetic aperture radar interferometry (IN-

SAR). In Chapter 4 we introduce the plate cooling model for the thermal

structure of the oceanic lithosphere as a complement to the half-space cool-

ing model. We also present in this chapter the Culling model for the diffu-

sive erosion and deposition of sediments. In Chapter 5 we show how geoid

anomalies are directly related to the forces required to maintain topography.

In Chapter 6 we combine a pipe-flow model with a Stokes-flow model in

xiv Preface to the Second Edition

order to determine the structure and strength of plume heads and plume

tails. The relationship between hotspot swells and the associated plume flux

is also introduced. In addition to the steady-state boundary-layer model for

the structure of mantle convection cells, we introduce a transient boundary-

layer model for the stability of the lithosphere.

Finally, we conclude the book with a new Chapter 10 on chemical geo-

dynamics. The concept of chemical geodynamics has evolved since the first

edition was written. The object is to utilize geochemical data, particularly

the isotope systematics of basalts, to infer mantle dynamics. Questions ad-

dressed include the homogeneity of the mantle, the fate of subducted litho-

sphere, and whether whole mantle convection or layered mantle convection

is occurring.

The use of SI units is now firmly entrenched in geology and geophysics,

and we use these units throughout the book. Since Geodynamics is meant to

be a textbook, large numbers of references are inappropriate. However, we

have included key references and references to sources of data in addition to

recommended collateral reading.

In addition to the colleagues who we acknowledge in the preface to the

first edition, we would like to add Claude Allègre, Louise Kellogg, David

Kohlstedt, Bruce Malamud, Mark Parmentier, and David Sandwell. We also

acknowledge the excellent manuscript preparation by Stacey Shirk and Ju-

dith Hohl, and figure preparation by Richard Sadakane.

1

Plate Tectonics

1.1 Introduction

Plate tectonics is a model in which the outer shell of the Earth is divided into

a number of thin, rigid plates that are in relative motion with respect to one

another. The relative velocities of the plates are of the order of a few tens of

millimeters per year. A large fraction of all earthquakes, volcanic eruptions,

and mountain building occurs at plate boundaries. The distribution of the

major surface plates is illustrated in Figure 1–1.

The plates are made up of relatively cool rocks and have an average

thickness of about 100 km. The plates are being continually created and

consumed. At ocean ridges adjacent plates diverge from each other in a pro-

cess known as seafloor spreading. As the adjacent plates diverge, hot mantle

rock ascends to fill the gap. The hot, solid mantle rock behaves like a fluid

because of solid-state creep processes. As the hot mantle rock cools, it be-

comes rigid and accretes to the plates, creating new plate area. For this

reason ocean ridges are also known as accreting plate boundaries. The ac-

cretionary process is symmetric to a first approximation so that the rates

of plate formation on the two sides of a ridge are approximately equal. The

rate of plate formation on one side of an ocean ridge defines a half-spreading

velocity u. The two plates spread with a relative velocity of 2u. The global

system of ocean ridges is denoted by the heavy dark lines in Figure 1–1.

Because the surface area of the Earth is essentially constant, there must

be a complementary process of plate consumption. This occurs at ocean

trenches. The surface plates bend and descend into the interior of the Earth

in a process known as subduction. At an ocean trench the two adjacent plates

converge, and one descends beneath the other. For this reason ocean trenches

are also known as convergent plate boundaries. The worldwide distribution

2 Plate Tectonics

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1.1 Introduction 3

Figure 1.2 Accretion of a lithospheric plate at an ocean ridge and its sub- duction at an ocean trench. The asthenosphere, which lies beneath the lithosphere, is shown along with the line of volcanic centers associated with subduction.

of trenches is shown in Figure 1–1 by the lines with triangular symbols,

which point in the direction of subduction.

A cross-sectional view of the creation and consumption of a typical plate

is illustrated in Figure 1–2. That part of the Earth’s interior that comprises

the plates is referred to as the lithosphere. The rocks that make up the

lithosphere are relatively cool and rigid; as a result the interiors of the plates

do not deform significantly as they move about the surface of the Earth. As

the plates move away from ocean ridges, they cool and thicken. The solid

rocks beneath the lithosphere are sufficiently hot to be able to deform freely;

these rocks comprise the asthenosphere, which lies below the lithosphere. The

lithosphere slides over the asthenosphere with relatively little resistance.

As the rocks of the lithosphere become cooler, their density increases

because of thermal contraction. As a result the lithosphere becomes gravi-

tationally unstable with respect to the hot asthenosphere beneath. At the

ocean trench the lithosphere bends and sinks into the interior of the Earth

because of this negative buoyancy. The downward gravitational body force

on the descending lithosphere plays an important role in driving plate tec-

tonics. The lithosphere acts as an elastic plate that transmits large elas-

tic stresses without significant deformation. Thus the gravitational body

force can be transmitted directly to the surface plate and this force pulls

the plate toward the trench. This body force is known as trench pull. Ma-

jor faults separate descending lithospheres from adjacent overlying litho-

spheres. These faults are the sites of most great earthquakes. Examples are

the Chilean earthquake in 1960 and the Alaskan earthquake in 1964. These

4 Plate Tectonics

Figure 1.3 Izalco volcano in El Salvador, an example of a subduction zone volcano (NOAA—NGDC Howell Williams).

are the largest earthquakes that have occurred since modern seismographs

have been available. The locations of the descending lithospheres can be

accurately determined from the earthquakes occurring in the cold, brittle

rocks of the lithospheres. These planar zones of earthquakes associated with

subduction are known as Wadati–Benioff zones.

Lines of active volcanoes lie parallel to almost all ocean trenches. These

volcanoes occur about 125 km above the descending lithosphere. At least

a fraction of the magmas that form these volcanoes are produced near the

upper boundary of the descending lithosphere and rise some 125 km to the

surface. If these volcanoes stand on the seafloor, they form an island arc,

as typified by the Aleutian Islands in the North Pacific. If the trench lies

adjacent to a continent, the volcanoes grow from the land surface. This is

the case in the western United States, where a volcanic line extends from

Mt. Baker in the north to Mt. Shasta in the south. Mt. St. Helens, the

site of a violent eruption in 1980, forms a part of this volcanic line. These

volcanoes are the sites of a large fraction of the most explosive and violent

volcanic eruptions. The eruption of Mt. Pinatubo in the Philippines in 1991,

the most violent eruption of the 20th century, is another example. A typical

subduction zone volcano is illustrated in Figure 1–3.

The Earth’s surface is divided into continents and oceans. The oceans have

an average depth of about 4 km, and the continents rise above sea level. The

reason for this difference in elevation is the difference in the thickness of the

crust. Crustal rocks have a different composition from that of the mantle

rocks beneath and are less dense. The crustal rocks are therefore gravita-

1.1 Introduction 5

tionally stable with respect to the heavier mantle rocks. There is usually a

well-defined boundary, the Moho or Mohorovičić discontinuity, between the

crust and mantle. A typical thickness for oceanic crust is 6 km; continental

crust is about 35 km thick. Although oceanic crust is gravitationally stable,

it is sufficiently thin so that it does not significantly impede the subduction

of the gravitationally unstable oceanic lithosphere. The oceanic lithosphere

is continually cycled as it is accreted at ocean ridges and subducted at ocean

trenches. Because of this cycling the average age of the ocean floor is about

108 years (100 Ma).

On the other hand, the continental crust is sufficiently thick and gravita-

tionally stable so that it is not subducted at an ocean trench. In some cases

the denser lower continental crust, along with the underlying gravitationally

unstable continental mantle lithosphere, can be recycled into the Earth’s in-

terior in a process known as delamination. However, the light rocks of the

upper continental crust remain in the continents. For this reason the rocks

of the continental crust, with an average age of about 109 years (1 Ga), are

much older than the rocks of the oceanic crust. As the lithospheric plates

move across the surface of the Earth, they carry the continents with them.

The relative motion of continents is referred to as continental drift.

Much of the historical development leading to plate tectonics concerned

the validity of the hypothesis of continental drift: that the relative positions

of continents change during geologic time. The similarity in shape between

the west coast of Africa and the east coast of South America was noted as

early as 1620 by Francis Bacon. This “fit” has led many authors to spec-

ulate on how these two continents might have been attached. A detailed

exposition of the hypothesis of continental drift was put forward by Frank

B. Taylor (1910). The hypothesis was further developed by Alfred Wegener

beginning in 1912 and summarized in his book The Origin of Continents

and Oceans (Wegener, 1946). As a meteorologist, Wegener was particularly

interested in the observation that glaciation had occurred in equatorial re-

gions at the same time that tropical conditions prevailed at high latitudes.

This observation in itself could be explained by polar wander, a shift of the

rotational axis without other surface deformation. However, Wegener also

set forth many of the qualitative arguments that the continents had formerly

been attached. In addition to the observed fit of continental margins, these

arguments included the correspondence of geological provinces, continuity of

structural features such as relict mountain ranges, and the correspondence

of fossil types. Wegener argued that a single supercontinent, Pangaea, had

formerly existed. He suggested that tidal forces or forces associated with the

6 Plate Tectonics

rotation of the Earth were responsible for the breakup of this continent and

the subsequent continental drift.

Further and more detailed qualitative arguments favoring continental drift

were presented by Alexander du Toit, particularly in his book Our Wan-

dering Continents (du Toit, 1937). Du Toit argued that instead of a single

supercontinent, there had formerly been a northern continent, Laurasia, and

a southern continent, Gondwanaland, separated by the Tethys Ocean.

During the 1950s extensive exploration of the seafloor led to an improved

understanding of the worldwide range of mountains on the seafloor known

as mid-ocean ridges. Harry Hess (1962) hypothesized that the seafloor was

created at the axis of a ridge and moved away from the ridge to form an ocean

in a process now referred to as seafloor spreading. This process explains the

similarity in shape between continental margins. As a continent breaks apart,

a new ocean ridge forms. The ocean floor created is formed symmetrically

at this ocean ridge, creating a new ocean. This is how the Atlantic Ocean

was formed; the mid-Atlantic ridge where the ocean formed now bisects the

ocean.

It should be realized, however, that the concept of continental drift won

general acceptance by Earth scientists only in the period between 1967 and

1970. Although convincing qualitative, primarily geological, arguments had

been put forward to support continental drift, almost all Earth scientists

and, in particular, almost all geophysicists had opposed the hypothesis.

Their opposition was mainly based on arguments concerning the rigidity

of the mantle and the lack of an adequate driving mechanism.

The propagation of seismic shear waves showed beyond any doubt that

the mantle was a solid. An essential question was how horizontal displace-

ments of thousands of kilometers could be accommodated by solid rock. The

fluidlike behavior of the Earth’s mantle had been established in a general

way by gravity studies carried out in the latter part of the nineteenth cen-

tury. Measurements showed that mountain ranges had low-density roots.

The lower density of the roots provides a negative relative mass that nearly

equals the positive mass of the mountains. This behavior could be explained

by the principle of hydrostatic equilibrium if the mantle behaved as a fluid.

Mountain ranges appear to behave similarly to blocks of wood floating on

water.

The fluid behavior of the mantle was established quantitatively by N. A.

Haskell (1935). Studies of the elevation of beach terraces in Scandinavia

showed that the Earth’s surface was still rebounding from the load of the

ice during the last ice age. By treating the mantle as a viscous fluid with

a viscosity of 1020 Pa s, Haskell was able to explain the present uplift of

1.1 Introduction 7

Scandinavia. Although this is a very large viscosity (water has a viscosity of

10−3 Pa s), it leads to a fluid behavior for the mantle during long intervals

of geologic time.

In the 1950s theoretical studies had established several mechanisms for

the very slow creep of crystalline materials. This creep results in a fluid

behavior. Robert B. Gordon (1965) showed that solid-state creep quantita-

tively explained the viscosity determined from observations of postglacial

rebound. At temperatures that are a substantial fraction of the melt tem-

perature, thermally activated creep processes allow mantle rock to flow at

low stress levels on time scales greater than 104 years. The rigid lithosphere

includes rock that is sufficiently cold to preclude creep on these long time

scales.

The creep of mantle rock was not a surprise to scientists who had studied

the widely recognized flow of ice in glaciers. Ice is also a crystalline solid, and

gravitational body forces in glaciers cause ice to flow because its temperature

is near its melt temperature. Similarly, mantle rocks in the Earth’s interior

are near their melt temperatures and flow in response to gravitational body

forces.

Forces must act on the lithosphere in order to make the plates move. We-

gener suggested that either tidal forces or forces associated with the rotation

of the Earth caused the motion responsible for continental drift. However, in

the 1920s Sir Harold Jeffreys, as summarized in his book The Earth (Jeffreys,

1924), showed that these forces were insufficient. Some other mechanism had

to be found to drive the motion of the plates. Any reasonable mechanism

must also have sufficient energy available to provide the energy being dis-

sipated in earthquakes, volcanoes, and mountain building. Arthur Holmes

(1931) hypothesized that thermal convection was capable of driving mantle

convection and continental drift. If a fluid is heated from below, or from

within, and is cooled from above in the presence of a gravitational field, it

becomes gravitationally unstable, and thermal convection can occur. The

hot mantle rocks at depth are gravitationally unstable with respect to the

colder, more dense rocks in the lithosphere. The result is thermal convec-

tion in which the colder rocks descend into the mantle and the hotter rocks

ascend toward the surface. The ascent of mantle material at ocean ridges

and the descent of the lithosphere into the mantle at ocean trenches are

parts of this process. The Earth’s mantle is being heated by the decay of

the radioactive isotopes uranium 235 (235U), uranium 238 (238U), thorium

232 (232Th), and potassium 40 (40K). The volumetric heating from these

isotopes and the secular cooling of the Earth drive mantle convection. The

heat generated by the radioactive isotopes decreases with time as they de-

8 Plate Tectonics

cay. Two billion years ago the heat generated was about twice the present

value. Because the amount of heat generated is less today, the vigor of the

mantle convection required today to extract the heat is also less. The vigor

of mantle convection depends on the mantle viscosity. Less vigorous mantle

convection implies a lower viscosity. But the mantle viscosity is a strong

function of mantle temperature; a lower mantle viscosity implies a cooler

mantle. Thus as mantle convection becomes less vigorous, the mantle cools;

this is secular cooling. As a result, about 80% of the heat lost from the in-

terior of the Earth is from the decay of the radioactive isotopes and about

20% is due to the cooling of the Earth (secular cooling).

During the 1960s independent observations supporting continental drift

came from paleomagnetic studies. When magmas solidify and cool, their

iron component is magnetized by the Earth’s magnetic field. This remanent

magnetization provides a fossil record of the orientation of the magnetic

field at that time. Studies of the orientation of this field can be used to

determine the movement of the rock relative to the Earth’s magnetic poles

since the rock’s formation. Rocks in a single surface plate that have not

been deformed locally show the same position for the Earth’s magnetic poles.

Keith Runcorn (1956) showed that rocks in North America and Europe gave

different positions for the magnetic poles. He concluded that the differences

were the result of continental drift between the two continents.

Paleomagnetic studies also showed that the Earth’s magnetic field has

been subject to episodic reversals. Observations of the magnetic field over

the oceans indicated a regular striped pattern of magnetic anomalies (regions

of magnetic field above and below the average field value) lying parallel to

the ocean ridges. Frederick Vine and Drummond Matthews (1963) correlated

the locations of the edges of the striped pattern of magnetic anomalies with

the times of magnetic field reversals and were able to obtain quantitative

values for the rate of seafloor spreading. These observations have provided

the basis for accurately determining the relative velocities at which adjacent

plates move with respect to each other.

By the late 1960s the framework for a comprehensive understanding of

the geological phenomena and processes of continental drift had been built.

The basic hypothesis of plate tectonics was given by Jason Morgan (1968).

The concept of a mosaic of rigid plates in relative motion with respect to

one another was a natural consequence of thermal convection in the mantle.

A substantial fraction of all earthquakes, volcanoes, and mountain building

can be attributed to the interactions among the lithospheric plates at their

boundaries (Isacks et al., 1968). Continental drift is an inherent part of plate

1.2 The Lithosphere 9

tectonics. The continents are carried with the plates as they move about the

surface of the Earth.

Problem 1.1 If the area of the oceanic crust is 3.2 × 108 km2 and new

seafloor is now being created at the rate of 2.8 km2 yr−1, what is the mean

age of the oceanic crust? Assume that the rate of seafloor creation has been

constant in the past.

1.2 The Lithosphere

An essential feature of plate tectonics is that only the outer shell of the

Earth, the lithosphere, remains rigid during intervals of geologic time. Be-

cause of their low temperature, rocks in the lithosphere do not significantly

deform on time scales of up to 109 years. The rocks beneath the lithosphere

are sufficiently hot so that solid-state creep can occur. This creep leads to

a fluidlike behavior on geologic time scales. In response to forces, the rock

beneath the lithosphere flows like a fluid.

The lower boundary of the lithosphere is defined to be an isotherm (surface

of constant temperature). A typical value is approximately 1600 K. Rocks

lying above this isotherm are sufficiently cool to behave rigidly, whereas

rocks below this isotherm are sufficiently hot to readily deform. Beneath the

ocean basins the lithosphere has a thickness of about 100 km; beneath the

continents the thickness is about twice this value. Because the thickness of

the lithosphere is only 2 to 4% of the radius of the Earth, the lithosphere

is a thin shell. This shell is broken up into a number of plates that are in

relative motion with respect to one another. The rigidity of the lithosphere

ensures, however, that the interiors of the plates do not deform significantly.

The rigidity of the lithosphere allows the plates to transmit elastic stresses

during geologic intervals. The plates act as stress guides. Stresses that are

applied at the boundaries of a plate can be transmitted throughout the

interior of the plate. The ability of the plates to transmit stress over large

distances has important implications with regard to the driving mechanism

of plate tectonics.

The rigidity of the lithosphere also allows it to bend when subjected to

a load. An example is the load applied by a volcanic island. The load of

the Hawaiian Islands causes the lithosphere to bend downward around the

load, resulting in a region of deeper water around the islands. The elastic

bending of the lithosphere under vertical loads can also explain the structure

of ocean trenches and some sedimentary basins.

However, the entire lithosphere is not effective in transmitting elastic

10 Plate Tectonics

Figure 1.4 An accreting plate margin at an ocean ridge.

stresses. Only about the upper half of it is sufficiently rigid so that elas-

tic stresses are not relaxed on time scales of 109 years. This fraction of the

lithosphere is referred to as the elastic lithosphere. Solid-state creep pro-

cesses relax stresses in the lower, hotter part of the lithosphere. However,

this part of the lithosphere remains a coherent part of the plates. A detailed

discussion of the difference between the thermal and elastic lithospheres is

given in Section 7–10.

1.3 Accreting Plate Boundaries

Lithospheric plates are created at ocean ridges. The two plates on either side

of an ocean ridge move away from each other with near constant velocities of

a few tens of millimeters per year. As the two plates diverge, hot mantle rock

flows upward to fill the gap. The upwelling mantle rock cools by conductive

heat loss to the surface. The cooling rock accretes to the base of the spreading

plates, becoming part of them; the structure of an accreting plate boundary

is illustrated in Figure 1–4.

As the plates move away from the ocean ridge, they continue to cool and

the lithosphere thickens. The elevation of the ocean ridge as a function of

distance from the ridge crest can be explained in terms of the temperature

distribution in the lithosphere. As the lithosphere cools, it becomes more

dense; as a result it sinks downward into the underlying mantle rock. The

topographic elevation of the ridge is due to the greater buoyancy of the

1.3 Accreting Plate Boundaries 11

thinner, hotter lithosphere near the axis of accretion at the ridge crest. The

elevation of the ocean ridge also provides a body force that causes the plates

to move away from the ridge crest. A component of the gravitational body

force on the elevated lithosphere drives the lithosphere away from the accre-

tional boundary; it is one of the important forces driving the plates. This

force on the lithosphere is known as ridge push and is a form of gravitational

sliding.

The volume occupied by the ocean ridge displaces seawater. Rates of

seafloor spreading vary in time. When rates of seafloor spreading are high,

ridge volume is high, and seawater is displaced. The result is an increase

in the global sea level. Variations in the rates of seafloor spreading are the

primary cause for changes in sea level on geological time scales. In the Creta-

ceous (≈80 Ma) the rate of seafloor spreading was about 30% greater than

at present and sea level was about 200 m higher than today. One result

was that a substantial fraction of the continental interiors was covered by

shallow seas.

Ocean ridges are the sites of a large fraction of the Earth’s volcanism.

Because almost all the ridge system is under water, only a small part of this

volcanism can be readily observed. The details of the volcanic processes at

ocean ridges have been revealed by exploration using submersible vehicles.

Ridge volcanism can also be seen in Iceland, where the oceanic crust is

sufficiently thick so that the ridge crest rises above sea level. The volcanism

at ocean ridges is caused by pressure-release melting. As the two adjacent

plates move apart, hot mantle rock ascends to fill the gap. The temperature

of the ascending rock is nearly constant, but its pressure decreases. The

pressure p of rock in the mantle is given by the simple hydrostatic equation

p = ρgy, (1.1)

where ρ is the density of the mantle rock, g is the acceleration of gravity, and

y is the depth. The solidus temperature (the temperature at which the rock

first melts) decreases with decreasing pressure. When the temperature of

the ascending mantle rock equals the solidus temperature, melting occurs, as

illustrated in Figure 1–5. The ascending mantle rock contains a low-melting-

point, basaltic component. This component melts to form the oceanic crust.

Problem 1.2 At what depth will ascending mantle rock with a tempera-

ture of 1600 K melt if the equation for the solidus temperature T is

T (K) = 1500 + 0.12p (MPa).

12 Plate Tectonics

Figure 1.5 The process of pressure-release melting is illustrated. Melting occurs because the nearly isothermal ascending mantle rock encounters pressures low enough so that the associated solidus temperatures are below the rock temperatures.

Figure 1.6 Typical structure of the oceanic crust, overlying ocean basin, and underlying depleted mantle rock.

Assume ρ = 3300 kg m−3, g = 10 m s−2, and the mantle rock ascends at

constant temperature.

The magma (melted rock) produced by partial melting beneath an ocean

ridge is lighter than the residual mantle rock, and buoyancy forces drive

1.3 Accreting Plate Boundaries 13

Table 1.1 Typical Compositions of Important Rock Types

Clastic Continental Granite Diorite Sediments Crust Basalt Harzburgite “Pyrolite” Chondrite

SiO2 70.8 57.6 70.4 61.7 50.3 45.3 46.1 33.3 Al2O3 14.6 16.9 14.3 15.8 16.5 1.8 4.3 2.4 Fe2O3 1.6 3.2 —— —— —— —— —— —— FeO 1.8 4.5 5.3 6.4 8.5 8.1 8.2 35.5 MgO 0.9 4.2 2.3 3.6 8.3 43.6 37.6 23.5 CaO 2.0 6.8 2.0 5.4 12.3 1.2 3.1 2.3 Na2O 3.5 3.4 1.8 3.3 2.6 —— 0.4 1.1 K2O 4.2 3.4 3.0 2.5 0.2 —— 0.03 —— TiO2 0.4 0.9 0.7 0.8 1.2 —— 0.2 ——

it upward to the surface in the vicinity of the ridge crest. Magma cham-

bers form, heat is lost to the seafloor, and this magma solidifies to form the

oceanic crust. In some localities slices of oceanic crust and underlying man-

tle have been brought to the surface. These are known as ophiolites; they

occur in such locations as Cyprus, Newfoundland, Oman, and New Guinea.

Field studies of ophiolites have provided a detailed understanding of the

oceanic crust and underlying mantle. Typical oceanic crust is illustrated in

Figure 1–6. The crust is divided into layers 1, 2, and 3, which were origi-

nally associated with different seismic velocities but subsequently identified

compositionally. Layer 1 is composed of sediments that are deposited on the

volcanic rocks of layers 2 and 3. The thickness of sediments increases with

distance from the ridge crest; a typical thickness is 1 km. Layers 2 and 3 are

composed of basaltic rocks of nearly uniform composition. A typical com-

position of an ocean basalt is given in Table 1–1. The basalt is composed

primarily of two rock-forming minerals, plagioclase feldspar and pyroxene.

The plagioclase feldspar is 50 to 85% anorthite (CaAl2Si2O8) component

and 15 to 50% albite (NaAlSi3O8) component. The principal pyroxene is

rich in the diopside (CaMgSi2O6) component. Layer 2 of the oceanic crust is

composed of extrusive volcanic flows that have interacted with the seawater

to form pillow lavas and intrusive flows primarily in the form of sheeted

dikes. A typical thickness for layer 2 is 1.5 km. Layer 3 is made up of gab-

bros and related cumulate rocks that crystallized directly from the magma

chamber. Gabbros are coarse-grained basalts; the larger grain size is due to

slower cooling rates at greater depths. The thickness of layer 3 is typically

4.5 km.

Studies of ophiolites show that oceanic crust is underlain primarily by a

14 Plate Tectonics

peridotite called harzburgite. A typical composition of a harzburgite is given

in Table 1–1. This peridotite is primarily composed of olivine and orthopy-

roxene. The olivine consists of about 90% forsterite component (Mg2SiO4)

and about 10% fayalite component (Fe2SiO4). The orthopyroxene is less

abundant and consists primarily of the enstatite component (MgSiO3). Rel-

ative to basalt, harzburgite contains lower concentrations of calcium and

aluminum and much higher concentrations of magnesium. The basalt of the

oceanic crust with a density of 2900 kg m−3 is gravitationally stable with

respect to the underlying peridotite with a density of 3300 kg m−3. The

harzburgite has a greater melting temperature (≃500 K higher) than basalt

and is therefore more refractory.

Field studies of ophiolites indicate that the harzburgite did not crystallize

from a melt. Instead, it is the crystalline residue left after partial melting

produced the basalt. The process by which partial melting produces the

basaltic oceanic crust, leaving a refractory residuum of peridotite, is an

example of igneous fractionation.

Molten basalts are less dense than the solid, refractory harzburgite and

ascend to the base of the oceanic crust because of their buoyancy. At the

base of the crust they form a magma chamber. Since the forces driving plate

tectonics act on the oceanic lithosphere, they produce a fluid-driven fracture

at the ridge crest. The molten basalt flows through this fracture, draining the

magma chamber and resulting in surface flows. These surface flows interact

with the seawater to generate pillow basalts. When the magma chamber is

drained, the residual molten basalt in the fracture solidifies to form a dike.

The solidified rock in the dike prevents further migration of molten basalt,

the magma chamber refills, and the process repeats. A typical thickness of

a dike in the vertical sheeted dike complex is 1 m.

Other direct evidence for the composition of the mantle comes from xeno-

liths that are carried to the surface in various volcanic flows. Xenoliths are

solid rocks that are entrained in erupting magmas. Xenoliths of mantle peri-

dotites are found in some basaltic flows in Hawaii and elsewhere. Mantle

xenoliths are also carried to the Earth’s surface in kimberlitic eruptions.

These are violent eruptions that form the kimberlite pipes where diamonds

are found.

It is concluded that the composition of the upper mantle is such that

basalts can be fractionated leaving harzburgite as a residuum. One model

composition for the parent undepleted mantle rock is called pyrolite and its

chemical composition is given in Table 1–1. In order to produce the basaltic

oceanic crust, about 20% partial melting of pyrolite must occur. Incompat-

ible elements such as the heat-producing elements uranium, thorium, and

1.4 Subduction 15

potassium do not fit into the crystal structures of the principal minerals

of the residual harzburgite; they are therefore partitioned into the basaltic

magma during partial melting.

Support for a pyrolite composition of the mantle also comes from studies of

meteorites. A pyrolite composition of the mantle follows if it is hypothesized

that the Earth was formed by the accretion of parental material similar to

Type 1 carbonaceous chondritic meteorites. An average composition for a

Type 1 carbonaceous chondrite is given in Table 1–1. In order to generate a

pyrolite composition for the mantle, it is necessary to remove an appropriate

amount of iron to form the core as well as some volatile elements such as

potassium.

A 20% fractionation of pyrolite to form the basaltic ocean crust and a

residual harzburgite mantle explains the major element chemistry of these

components. The basalts generated over a large fraction of the ocean ridge

system have near-uniform compositions in both major and trace elements.

This is evidence that the parental mantle rock from which the basalt is frac-

tionated also has a near-uniform composition. However, both the basalts of

normal ocean crust and their parental mantle rock are systematically de-

pleted in incompatible elements compared with the model chondritic abun-

dances. The missing incompatible elements are found to reside in the conti-

nental crust.

Seismic studies have been used to determine the thickness of the oceanic

crust on a worldwide basis. The thickness of the basaltic oceanic crust has

a nearly constant value of about 6 km throughout much of the area of the

oceans. Exceptions are regions of abnormally shallow bathymetry such as

the North Atlantic near Iceland, where the oceanic crust may be as thick

as 25 km. The near-constant thickness of the basaltic oceanic crust places

an important constraint on mechanisms of partial melting beneath the ridge

crest. If the basalt of the oceanic crust represents a 20% partial melt, the

thickness of depleted mantle beneath the oceanic crust is about 24 km.

However, this depletion is gradational so the degree of depletion decreases

with depth.

1.4 Subduction

As the oceanic lithosphere moves away from an ocean ridge, it cools, thick-

ens, and becomes more dense because of thermal contraction. Even though

the basaltic rocks of the oceanic crust are lighter than the underlying mantle

rocks, the colder subcrustal rocks in the lithosphere become sufficiently dense

to make old oceanic lithosphere heavy enough to be gravitationally unstable

16 Plate Tectonics

with respect to the hot mantle rocks immediately underlying the lithosphere.

As a result of this gravitational instability the oceanic lithosphere founders

and begins to sink into the interior of the Earth at ocean trenches. As the

lithosphere descends into the mantle, it encounters increasingly dense rocks.

However, the rocks of the lithosphere also become increasingly dense as a

result of the increase of pressure with depth (mantle rocks are compressible),

and they continue to be heavier than the adjacent mantle rocks as they de-

scend into the mantle so long as they remain colder than the surrounding

mantle rocks at any depth. Phase changes in the descending lithosphere and

adjacent mantle and compositional variations with depth in the ambient

mantle may complicate this simple picture of thermally induced gravita-

tional instability. Generally speaking, however, the descending lithosphere

continues to subduct as long as it remains denser than the immediately ad-

jacent mantle rocks at any depth. The subduction of the oceanic lithosphere

at an ocean trench is illustrated schematically in Figure 1–7.

The negative buoyancy of the dense rocks of the descending lithosphere

results in a downward body force. Because the lithosphere behaves elasti-

cally, it can transmit stresses and acts as a stress guide. The body force

acting on the descending plate is transmitted to the surface plate, which is

pulled toward the ocean trench. This is one of the important forces driving

plate tectonics and continental drift. It is known as slab pull.

Prior to subduction the lithosphere begins to bend downward. The con-

vex curvature of the seafloor defines the seaward side of the ocean trench.

The oceanic lithosphere bends continuously and maintains its structural in-

tegrity as it passes through the subduction zone. Studies of elastic bending

at subduction zones are in good agreement with the morphology of some

subduction zones seaward of the trench axis (see Section 3–17). However,

there are clearly significant deviations from a simple elastic rheology. Some

trenches exhibit a sharp “hinge” near the trench axis and this has been

attributed to an elastic–perfectly plastic rheology (see Section 7–11).

As a result of the bending of the lithosphere, the near-surface rocks are

placed in tension, and block faulting often results. This block faulting allows

some of the overlying sediments to be entrained in the upper part of the

basaltic crust. Some of these sediments are then subducted along with the

basaltic rocks of the oceanic crust, but the remainder of the sediments are

scraped off at the base of the trench. These sediments form an accretionary

prism (Figure 1–7) that defines the landward side of many ocean trenches.

Mass balances show that only a fraction of the sediments that make up layer

1 of the oceanic crust are incorporated into accretionary prisms. Since these

sediments are derived by the erosion of the continents, the subduction of

1.4 Subduction 17

Figure 1.7 Subduction of oceanic lithosphere at an ocean trench. Sediments forming layer 1 of the oceanic crust are scraped off at the ocean trench to form the accretionary prism of sediments. The volcanic line associated with subduction and the marginal basin sometimes associated with subduction are also illustrated.

sediments is a mechanism for subducting continental crust and returning it

to the mantle.

The arclike structure of many ocean trenches (see Figure 1–1) can be

qualitatively understood by the ping-pong ball analogy. If a ping-pong ball

is indented, the indented portion will have the same curvature as the original

ball, that is, it will lie on the surface of an imaginary sphere with the same

radius as the ball, as illustrated in Figure 1–8. The lithosphere as it bends

downward might also be expected to behave as a flexible but inextensible

thin spherical shell. In this case the angle of dip α of the lithosphere at the

trench can be related to the radius of curvature of the island arc. A cross

section of the subduction zone is shown in Figure 1–8b. The triangles OAB,

BAC, and BAD are similar right triangles so that the angle subtended by the

indented section of the sphere at the center of the Earth is equal to the angle

of dip. The radius of curvature of the indented section, defined as the great

circle distance BQ, is thus aα/2, where a is the radius of the Earth. The

radius of curvature of the arc of the Aleutian trench is about 2200 km. Taking

a = 6371 km, we find that α = 39.6◦. The angle of dip of the descending

lithosphere along much of the Aleutian trench is near 45◦. Although the

18 Plate Tectonics

Figure 1.8 The ping-pong ball analogy for the arc structure of an ocean trench. (a) Top view showing subduction along a trench extending from S to T. The trench is part of a small circle centered at Q. (b) Cross section of indented section. BQR is the original sphere, that is, the surface of the Earth. BPR is the indented sphere, that is, the subducted lithosphere. The angle of subduction α is CBD. O is the center of the Earth.

ping-pong ball analogy provides a framework for understanding the arclike

structure of some trenches, it should be emphasized that other trenches

do not have an arclike form and have radii of curvature that are in poor

agreement with this relationship. Interactions of the descending lithosphere

with an adjacent continent may cause the descending lithosphere to deform

so that the ping-pong ball analogy would not be valid.

Ocean trenches are the sites of many of the largest earthquakes. These

earthquakes occur on the fault zone separating the descending lithosphere

1.4 Subduction 19

from the overlying lithosphere. Great earthquakes, such as the 1960 Chilean

earthquake and the 1964 Alaskan earthquake, accommodate about 20 m of

downdip motion of the oceanic lithosphere and have lengths of about 350

km along the trench. A large fraction of the relative displacement between

the descending lithosphere and the overlying mantle wedge appears to be

accommodated by great earthquakes of this type. A typical velocity of sub-

duction is 0.1 m yr−1 so that a great earthquake with a displacement of 20

m would be expected to occur at intervals of about 200 years.

Earthquakes within the cold subducted lithosphere extend to depths of

about 660 km. The locations of these earthquakes delineate the structure

of the descending plate and are known as the Wadati-Benioff zone. The

shapes of the upper boundaries of several descending lithospheres are given

in Figure 1–9. The positions of the trenches and the volcanic lines are also

shown. Many subducted lithospheres have an angle of dip near 45◦. In the

New Hebrides the dip is significantly larger, and in Peru and North Chile

the angle of dip is small.

The lithosphere appears to bend continuously as it enters an ocean trench

and then appears to straighten out and descend at a near-constant dip angle.

A feature of some subduction zones is paired belts of deep seismicity. The

earthquakes in the upper seismic zone, near the upper boundary of the

descending lithosphere, are associated with compression. The earthquakes

within the descending lithosphere are associated with tension. These double

seismic zones are attributed to the “unbending,” i.e., straightening out, of

the descending lithosphere. The double seismic zones are further evidence

of the rigidity of the subducted lithosphere. They are also indicative of the

forces on the subducted lithosphere that are straightening it out so that it

descends at a typical angle of 45◦.

Since the gravitational body force on the subducted lithosphere is down-

ward, it would be expected that the subduction dip angle would be 90◦. In

fact, as shown in Figure 1–9, the typical dip angle for a subduction zone

is near 45◦. One explanation is that the oceanic lithosphere is “foundering”

and the trench is migrating oceanward. In this case the dip angle is deter-

mined by the flow kinematics. While this explanation is satisfactory in some

cases, it has not been established that all slab dips can be explained by the

kinematics of mantle flows. An alternative explanation is that the subducted

slab is supported by the induced flow above the slab. The descending litho-

sphere induces a corner flow in the mantle wedge above it, and the pressure

forces associated with this corner flow result in a dip angle near 45◦ (see

Section 6–11).

One of the key questions in plate tectonics is the fate of the descending

20 Plate Tectonics

plates. Earthquakes terminate at a depth of about 660 km, but termination

of seismicity does not imply cessation of subduction. This is the depth of

a major seismic discontinuity associated with the solid–solid phase change

from spinel to perovskite and magnesiowüstite; this phase change could act

to deter penetration of the descending lithosphere. In some cases seismic

activity spreads out at this depth, and in some cases it does not. Shallow

subduction earthquakes generally indicate extensional stresses where as the

deeper earthquakes indicate compressional stresses. This is also an indica-

tion of a resistance to subduction. Seismic velocities in the cold descending

lithosphere are significantly higher than in the surrounding hot mantle. Sys-

tematic studies of the distribution of seismic velocities in the mantle are

known as mantle tomography. These studies have provided examples of the

descending plate penetrating the 660-km depth.

The fate of the descending plate has important implications regarding

mantle convection. Since plates descend into the lower mantle, beneath a

depth of 660 km, some form of whole mantle convection is required. The

entire upper and at least a significant fraction of the lower mantle must

take part in the plate tectonic cycle. Although there may be a resistance to

convection at a depth of 660 km, it is clear that the plate tectonic cycle is

not restricted to the upper mantle above 660 km.

Volcanism is also associated with subduction. A line of regularly spaced

volcanoes closely parallels the trend of the ocean trench in almost all cases.

These volcanics may result in an island arc or they may occur on the con-

tinental crust (Figure 1–10). The volcanoes lie 125 to 175 km above the

descending plate, as illustrated in Figure 1–9.

It is far from obvious why volcanism is associated with subduction. The

descending lithosphere is cold compared with the surrounding mantle, and

thus it should act as a heat sink rather than as a heat source. Because the

flow is downward, magma cannot be produced by pressure-release melting.

One source of heat is frictional dissipation on the fault zone between the

descending lithosphere and the overlying mantle. However, there are several

problems with generating island-arc magmas by frictional heating. When

rocks are cold, frictional stresses can be high, and significant heating can

occur. However, when the rocks become hot, the stresses are small, and it

appears to be impossible to produce significant melting simply by frictional

heating.

It has been suggested that interactions between the descending slab and

the induced flow in the overlying mantle wedge can result in sufficient heating

of the descending oceanic crust to produce melting. However, thermal models

of the subduction zone show that there is great difficulty in producing enough

1.4 Subduction 21

F ig

u re

1. 9

T h e

sh a pe

s o f th

e u p pe

r bo

u n d a ri

es o f d es

ce n d in

g li th

o sp

h er

es a t se

ve ra

l oc

ea n ic

tr en

ch es

ba se

d o n

th e

d is

tr ib

u ti o n s

o f ea

rt h qu

a ke

s. T

h e

n a m

es o f th

e tr en

ch es

a re

a bb

re vi

a te

d fo

r cl

a ri

ty (N

H =

N ew

H eb

ri d es

, C

A =

C en

tr a l A

m er

ic a , A

L T

= A

le u ti a n , A

L K

= A

la sk

a , M

= M

a ri

a n a , IB

= Iz

u – B o n in

, K

E R

= K

er m

a d ec

, N

Z =

N ew

Z ea

la n d , T

= T o n ga

, K

K =

K u ri

le – K

a m

ch a tk

a , N

C =

N o rt

h C

h il e,

P =

P er

u ).

T h e

lo ca

ti o n s

o f th

e vo

lc a n ic

li n es

a re

sh o w n

by th

e so

li d

tr ia

n gl

es . T

h e

lo ca

ti o n s

o f th

e tr en

ch es

a re

sh o w n

ei th

er a s

a ve

rt ic

a l li n e

o r

a s

a h o ri

zo n ta

l

li n e

if th

e tr en

ch – vo

lc a n ic

li n e

se pa

ra ti o n

is va

ri a bl

e (I

sa ck

s a n d

B a ra

za n gi

, 1 9 7 7 ).

22 Plate Tectonics

Figure 1.10 Eruption of ash and steam from Mount St. Helens, Washington, on April 3, 1980. Mount St. Helens is part of a volcanic chain, the Cascades, produced by subduction of the Juan de Fuca plate beneath the western margin of the North American plate (Washington Department of Natural Resources).

heat to generate the observed volcanism. The subducted cold lithospheric

slab is a very large heat sink and strongly depresses the isotherms above the

slab. It has also been argued that water released from the heating of hydrated

minerals in the subducted oceanic crust can contribute to melting by de-

pressing the solidus of the crustal rocks and adjacent mantle wedge rocks.

However, the bulk of the volcanic rocks at island arcs have near-basaltic com-

positions and erupt at temperatures very similar to eruption temperatures

at accretional margins. Studies of the petrology of island-arc magmas indi-

cate that they are primarily the result of the partial melting of rocks in the

mantle wedge above the descending lithosphere. Nevertheless, geochemical

evidence indicates that partial melting of subducted sediments and oceanic

crust does play an important role in island-arc volcanism. Isotopic studies

have shown conclusively that subducted sediments participate in the melt-

ing process. Also, the locations of the surface volcanic lines have a direct

geometrical relationship to the geometry of subduction. In some cases two

adjacent slab segments subduct at different angles, and an offset occurs in

the volcanic line; for the shallower dipping slab, the volcanic line is farther

from the trench keeping the depth to the slab beneath the volcanic line

nearly constant.

Processes associated with the subducted oceanic crust clearly trigger sub-

duction zone volcanism. However, the bulk of the volcanism is directly asso-

ciated with the melting of the mantle wedge in a way similar to the melting

1.5 Transform Faults 23

beneath an accretional plate margin. A possible explanation is that “flu-

ids” from the descending oceanic crust induce melting and create sufficient

buoyancy in the partially melted mantle wedge rock to generate an ascend-

ing flow and enhance melting through pressure release. This process may be

three-dimensional with ascending diapirs associated with individual volcanic

centers.

In some trench systems a secondary accretionary plate margin lies be-

hind the volcanic line, as illustrated in Figure 1–7. This back-arc spreading

is very similar to the seafloor spreading that is occurring at ocean ridges.

The composition and structure of the ocean crust that is being created are

nearly identical. Back-arc spreading creates marginal basins such as the Sea

of Japan. A number of explanations have been given for back-arc spreading.

One hypothesis is that the descending lithosphere induces a secondary con-

vection cell, as illustrated in Figure 1–11a. An alternative hypothesis is that

the ocean trench migrates away from an adjacent continent because of the

“foundering” of the descending lithosphere. Back-arc spreading is required

to fill the gap, as illustrated in Figure 1–11b. If the adjacent continent is

being driven up against the trench, as in South America, marginal basins

do not develop. If the adjacent continent is stationary, as in the western Pa-

cific, the foundering of the lithosphere leads to a series of marginal basins as

the trench migrates seaward. There is observational evidence that back-arc

spreading centers are initiated at volcanic lines. Heating of the lithosphere

at the volcanic line apparently weakens it sufficiently so that it fails under

tensional stresses.

Problem 1.3 If we assume that the current rate of subduction, 0.09 m2

s−1, has been applicable in the past, what thickness of sediments would have

to have been subducted in the last 3 Gyr if the mass of subducted sediments

is equal to one-half the present mass of the continents? Assume the density

of the continents ρc is 2700 kg m−3, the density of the sediments ρs is 2400

kg m−3, the continental area Ac is 1.9× 108 km2, and the mean continental

thickness hc is 35 km.

1.5 Transform Faults

In some cases the rigid plates slide past each other along transform faults.

The ocean ridge system is not a continuous accretional margin; rather, it is

a series of ridge segments offset by transform faults. The ridge segments lie

nearly perpendicular to the spreading direction, whereas the transform faults

lie parallel to the spreading direction. This structure is illustrated in Figure

24 Plate Tectonics

Figure 1.11 Models for the formation of marginal basins. (a) Secondary mantle convection induced by the descending lithosphere. (b) Ascending convection generated by the foundering of the descending lithosphere and the seaward migration of the trench.

Figure 1.12 (a) Segments of an ocean ridge offset by a transform fault. (b) Cross section along a transform fault.

1–12a. The orthogonal ridge–transform system has been reproduced in the

laboratory using wax that solidifies at the surface. Even with this analogy,

the basic physics generating the orthogonal pattern is not understood. The

relative velocity across a transform fault is twice the spreading velocity.

1.6 Hotspots and Mantle Plumes 25

This relative velocity results in seismicity (earthquakes) on the transform

fault between the adjacent ridge sections. There is also differential vertical

motion on transform faults. As the seafloor spreads away from a ridge crest,

it also subsides. Since the adjacent points on each side of a transform fault

usually lie at different distances from the ridge crest where the crust was

formed, the rates of subsidence on the two sides differ. A cross section along

a transform fault is given in Figure 1–12b. The extensions of the transform

faults into the adjacent plates are known as fracture zones. These fracture

zones are often deep valleys in the seafloor. An ocean ridge segment that

is not perpendicular to the spreading direction appears to be unstable and

transforms to the orthogonal pattern.

A transform fault that connects two ridge segments is known as a ridge–

ridge transform. Transform faults can also connect two segments of an ocean

trench. In some cases one end of a transform fault terminates in a triple junc-

tion of three surface plates. An example is the San Andreas fault in Califor-

nia; the San Andreas accommodates lateral sliding between the Pacific and

North American plates.

1.6 Hotspots and Mantle Plumes

Hotspots are anomalous areas of surface volcanism that cannot be directly

associated with plate tectonic processes. Many hotspots lie well within the

interiors of plates; an example is the volcanism of the Hawaiian Islands

(Figure 1–13). Other hotspots lie at or near an ocean ridge, an example

is the volcanism that forms Iceland. Much more voluminous than normal

ocean ridge volcanism; this volcanism resulted in a thick oceanic crust and

the elevation of Iceland above sea level.

In many cases hotspots lie at the end of well-defined lines of volcanic

edifices or volcanic ridges. These are known as hotspot tracks. The hotspot

track associated with the Hawaiian hotspot is the Hawaiian–Emperor island–

seamount chain that extends across the Pacific plate to the Aleutian Islands.

There is little agreement on the total number of hotspots. The positions

of thirty hotspots are given in Table 1–2, and twenty of the most prominent

hotspots are shown in Figure 1–14. Also shown in this figure are some of

the hotspot tracks. Some compilations of hotspots list as many as 120 (see

Figure 1–15). The definition of a hotspot tends to be quite subjective, partic-

ularly with regard to volcanism on or adjacent to plate boundaries. Hotspots

occur both in the oceans and on the continents. They do not appear to be

uniformly distributed over the Earth’s surface. There are numerous hotspots

26 Plate Tectonics

Figure 1.13 Satellite photograph of the island of Hawaii. The island is dom- inated by the active volcano Mauna Loa near its center (NASA STS61A- 50-0057).

in Africa and relatively few in South America, North America, Europe, and

Asia.

Jason Morgan (1971) attributed hotspot volcanism to a global array of

deep mantle plumes. Mantle plumes are quasi-cylindrical concentrated up-

wellings of hot mantle rock and they represent a basic form of mantle con-

vection. Pressure-release melting in the hot ascending plume rock produces

the basaltic volcanism associated with most hotspots. The hypothesis of

fixed mantle plumes impinging on the base of the moving lithospheric plates

explains the origin of hotspot tracks (see Figure 1–16).

The prototype example of a hotspot track is the Hawaiian–Emperor chain

of volcanic islands and sea-mounts illustrated in Figure 1–17. The associated

hot-spot volcanism has resulted in a nearly continuous volcanic ridge that

extends some 4000 km from near the Aleutian Islands to the very active

1.6 Hotspots and Mantle Plumes 27

Table 1.2 Hotspot Locations

Overlying Latitude Longitude Hotspot Plate (Degrees) (Degrees)

Hawaii Pacific 20 −157 Samoa Pacific −13 −173 St. Helena Africa −14 −6 Bermuda N. America 33 −67 Cape Verde Africa 14 −20 Pitcairn Pacific −26 −132 MacDonald Pacific −30 −140 Marquesas Pacific −10 −138 Tahiti Pacific −17 −151 Easter Pac-Naz −27 −110 Reunion Indian −20 55 Yellowstone N. America 43 −111 Galapagos Nazca 0 −92 Juan Fernandez Nazca −34 −83 Ethiopia Africa 8 37 Ascencion S. Am–Afr −8 −14 Afar Africa 10 43 Azores Eurasia 39 −28 Iceland N. Am–Eur 65 −20 Madeira Africa 32 −18 Canary Africa 28 −17 Hoggar Ind–Ant −49 69 Bouvet Afr–Ant −54 2 Pr. Edward Afr–Ant −45 50 Eifel Eurasia 48 8 San Felix Nazca −24 −82 Tibesti Africa 18 22 Trinadade S. America −20 −30 Tristan S. Am–Afr −36 −13

Source: After Crough and Jurdy (1980).

Kilauea volcano on the island of Hawaii. There is a remarkably uniform age

progression, with the age of each volcanic shield increasing systematically

with distance from Kilauea. Directly measured ages and ages inferred from

seafloor magnetic anomalies are given in Figure 1–17. These ages are given

as a function of distance from Kilauea in Figure 1–18 and they correlate

very well with a propagation rate of 90 mm yr−1across thePacific plate.

A striking feature of this track is the bend that separates the near-linear

trend of the Emperor chain from the near-linear trend of the Hawaiian chain.

The bend in the track occurred at about 43 Ma when there was an abrupt

shift in the motion of the Pacific plate. This shift was part of a global

28 Plate Tectonics

reorientation of plate motions over a span of a few million years. This shift

has been attributed to the continental collision between India and Asia,

which impeded the northward motion of the Indian plate.

Many hotspots are associated with linear tracks as indicated in Figure 1–

14. When the relative motions of the plates are removed the hotspots appear

to be nearly fixed with respect to each other. However, they are certainly

not precisely fixed. Systematic studies have shown that the relative motion

among hotspots amounts to a few mm yr−1. These results are consistent

with plumes that ascend through a mantle in which horizontal velocities are

about an order of magnitude smaller than the plate velocities.

Many hotspots are also associated with topographic swells. Hotspot swells

are regional topographic highs with widths of about 1000 km and anomalous

elevations of up to 3 km. The hotspot swell associated with the Hawaiian

hotspot is illustrated in Figure 1–19. The swell is roughly parabolic in form

and extends upstream from the active hotspot. The excess elevation asso-

ciated with the swell decays rather slowly down the track of the hotspot.

Hotspot swells are attributed to the interaction between the ascending hot

mantle rock in the plume and the lithospheric plate upon which the plume

impinges.

The volcanic rocks produced at most hotspots are primarily basalt. In

terms of overall composition, the rocks are generally similar to the basaltic

rocks produced at ocean ridges. It appears that these volcanic rocks are also

produced by about 20% partial melting of mantle rocks with a pyrolite com-

position. However, the concentrations of incompatible elements and isotopic

ratios differ from those of normal mid-ocean ridge basalts. Whereas the mid-

ocean ridge basalts are nearly uniformly depleted in incompatible elements,

the concentrations of these elements in hotspot basalts have considerable

variation. Some volcanoes produce basalts that are depleted, some produce

basalts that have near chondritic ratios, and some volcanoes produce basalts

that are enriched in the incompatible elements. These differences will be dis-

cussed in some detail in Chapter 10.

The earthquakes of the Wadati–Benioff zone define the geometry of the

subducted oceanic lithosphere. No seismicity is associated with mantle plumes,

and little direct observational evidence exists of their structure and origin.

Thus we must depend on analytical, numerical, and laboratory studies for

information. These studies indicate that plumes originate in a lower hot

thermal boundary layer either at the base of the mantle (the D′′-layer of

seismology) or at an interface in the lower mantle between an upper con-

vecting mantle layer and an isolated lower mantle layer. Plumes result from

the gravitational instability of the hot lower thermal boundary layer just as

1.6 Hotspots and Mantle Plumes 29

Figure 1.14 Hotspot and hotspot track locations: 1, Hawaii (Hawaiian– Emperor Seamount Chain); 2, Easter (Tuomoto–Line Island Chain); 3, MacDonald Seamount (Austral–Gilbert–Marshall Island Chain); 4, Bellany Island; 5, Cobb Seamount (Juan de Fuca Ridge); 6, Yellowstone (Snake River Plain–Columbia Plateau); 7, Galapagos Islands; 8, Bermuda; 9, Ice- land; 10, Azores; 11, Canary Islands; 12, Cape Verde Islands; 13, St. He- lena; 14, Tristan de Cunha (Rio Grande Ridge (w), Walvis Ridge (e)); 15, Bouvet Island; 16, Prince Edward Island; 17, Reunion Island (Mauri- tius Plateau, Chagos–Lacadive Ridge); 18, Afar; 19, Eifel; 20, Kerguelen Plateau (Ninety-East Ridge).

the subducted lithosphere results from the gravitational instability of the

cold, surface thermal boundary layer, the lithosphere.

Numerical and laboratory studies of the initiation of plumes show a lead-

ing diapir or plume head followed by a thin cylindrical conduit or plume tail

that connects the diapir to the source region. An example from a laboratory

experiment is given in Figure 1–20. Confirmation of this basic model comes

from the association of massive flood basalts with plume heads. There is

convincing observational evidence that flood basalt eruptions mark the ini-

tiation of hotspot tracks. As specific examples, the hotspot tracks of the

currently active Reunion, Iceland, Tristan da Cunha, and Prince Edward

hotspots originate, respectively, in the Deccan, Tertiary North Atlantic,

Parana, and Karoo flood basalt provinces.

The association of the Reunion hotspot with the Deccan flood basalt

province is illustrated in Figure 1–21. Pressure-release melting in the plume

head as it approached and impinged on the lithosphere can explain the

30 Plate Tectonics

eruption of the Deccan traps in India with a volume of basaltic magma in

excess of 1.5 × 106 km3 in a time interval of less than 1 Myr. Since then,

Reunion hotspot volcanism has been nearly continuous for 60 Myr with an

average eruption rate of 0.02 km3 yr−1. As the Indian plate moved northward

the hotspot track formed the Chagos–Laccadive Ridge. The hotspot track is

then offset by seafloor spreading on the central Indian Ridge and forms the

Mascarene Ridge on the Indian plate that connects to the currently active

volcanism of the Reunion Islands.

1.7 Continents

As described in the previous sections, the development of plate tectonics

primarily involves the ocean basins, yet the vast majority of geological data

comes from the continents. There is essentially no evidence for plate tectonics

in the continents, and this is certainly one reason why few geologists were

willing to accept the arguments in favor of continental drift and mantle

convection for so long. The near surface rocks of the continental crust are

much older than the rocks of the oceanic crust. They also have a more silicic

composition. The continents include not only the area above sea level but

also the continental shelves. It is difficult to provide an absolute definition

of the division between oceanic and continental crust. In most cases it is

appropriate to define the transition as occurring at an ocean depth of 3 km.

The area of the continents, including the margins, is about 1.9 × 108 km2,

or 37% of the surface of the Earth.

The rocks that make up the continental crust are, in bulk, more silicic

and therefore less dense than the basaltic rocks of the oceanic crust. Also,

the continental crust with a mean thickness of about 40 km is considerably

thicker than the oceanic crust. These two effects make the continental litho-

sphere gravitationally stable and prevent it from being subducted. Although

continental crust cannot be destroyed by subduction, it can be recycled in-

directly by delamination. The mantle portion of the continental lithosphere

is sufficiently cold and dense to be gravitationally unstable. Thus it is pos-

sible for the lower part of the continental lithosphere, including the lower

continental crust, to delaminate and sink into the lower mantle. This is par-

tial subduction or delamination. It has been suggested that delamination is

occurring in continental collision zones such as the Himalayas and the Alps

and behind subduction zones such as in the Altiplano in Peru. There are

a number of continental areas in which the mantle lithosphere is absent.

One example is the western United States. Crustal doubling such as in Ti-

bet has also been attributed to the absence of mantle lithosphere beneath

1.7 Continents 31

F ig

u re

1. 15

D is

tr ib

u ti o n

o f su

rf a ce

h o ts

po ts

, ce

n te

rs o f in

tr a p la

te vo

lc a n is

m , a n d

a n o m

a lo

u s

p la

te m

a rg

in vo

lc a n is

m .

32 Plate Tectonics

Figure 1.16 Formation of an island–seamount chain by the movement of a lithospheric plate over a melting anomaly in the upper mantle.

Figure 1.17 Age progression of the Hawaiian–Emperor seamount and island chain (Molnar and Stock, 1987). Dated seamounts and islands are shown in normal print and dates inferred from magnetic anomalies are shown in bold print.

Asia. Delamination is an efficient mechanism for the removal of continental

lithosphere. Continental crust can also be recycled into the mantle by the

subduction of sediments. Although there is evidence of the recycling of the

continental crust, it is much less efficient than the recycling of oceanic crust

by the plate tectonic cycle. The result is that the continental crust is nearly

a factor of 10 older than oceanic crust. Continental crust older than 1 billion

years is common, and some is older than 3 billion years.

Determining the relative age of continental rocks has been an important

1.7 Continents 33

Figure 1.18 Ages of islands and seamounts in the Hawaiian–Emperor chain as a function of distance from the currently active Kilauea volcano. The straight line gives a constant rate of propagation across the Pacific plate of 90 mm yr−1.

Figure 1.19 Bathymetric profile across the Hawaiian ridge at Oahu (Watts, 1976).

aspect of the historical development of geology. The early classification of the

age of rocks was based on the fossils found in sedimentary rocks. By studying

the evolution of the species involved, and their relative positions in the

stratigraphic column, an uncalibrated, relative time scale was developed. The

divisions of the time scale were associated with sedimentary unconformities.

These are recognized as discontinuities in the sedimentation process, where

adjacent strata often contain dissimilar fossils. These unconformities occur

worldwide.

It is now recognized that major unconformities correspond with times of

low sea level. During these periods erosion occurred over a large fraction of

the continents, causing gaps in the sedimentary record. During periods of

high sea level much of the area of the continents was covered with shallow

34 Plate Tectonics

Figure 1.20 Photograph of a low-density, low-viscosity glucose fluid plume ascending in a high-density, high-viscosity glucose fluid (Olson and Singer, 1985).

seas, and sediments were deposited. The causes of the periods of high and

low sea levels are not fully understood. Clearly, ice ages can cause periods

of low sea level. Also, on a longer time scale, variations in the volume of the

ocean ridge system can change the sea level.

Quantitative measurements of the concentrations of radioactive isotopes

and their daughter products in rocks have provided an absolute geological

time scale. The science of dating rocks by radioisotopic techniques is known

as geochronology. Geochronological methods will be discussed in Section 10–

2.

The radiometrically calibrated geological time scale is given in Table 1–3.

Note that the Precambrian period, during which fossils were not available

for classification purposes, represents 88% of the Earth’s history.

Erosion and sedimentation play an important role in shaping the surface

of the continents. Mountain ranges that are built by plate tectonic processes

are eroded to near sea level in a few million years. Any areas of the con-

tinents that are depressed below sea level are filled with these sediments

to form sedimentary basins. The base of a sedimentary basin is referred to

as the basement. Because the ages of basement rocks are not well known

on a worldwide basis, it is difficult to specify a mean age for continental

1.7 Continents 35

Figure 1.21 The relationship between the Reunion hotspot track and the Deccan flood basalts (White and McKenzie, 1989).

rocks. Regions of the continents where Precambrian metamorphic and ig-

neous rocks are exposed are known as continental shields. Detailed studies

of Precambrian terrains indicate that the plate tectonic processes that are

occurring today have been going on for at least 3 billion years.

It is relatively easy to estimate the composition of the upper continental

crust, but it is difficult to estimate the composition of the crust as a whole.

Direct evidence for the composition of the lower continental crust comes

from surface exposures of high-grade metamorphic rocks and lower crustal

36 Plate Tectonics

xenoliths transported to the surface in diatremes and magma flows. Indirect

evidence of the composition

of the lower crust comes from comparisons between seismic velocities and

laboratory studies of relevant minerals. An estimate of the bulk composi-

tion of the continental crust is given in Table 1–1. It is compared with the

mean composition of clastic sediments (representative of the upper conti-

nental crust) and with a typical basalt composition. Estimates of the mean

composition of the continental crust are clearly more basic (less silicic) than

the composition of the upper continental crust, but they do not approach a

basaltic composition.

An important question is: How is continental crust formed? One hypoth-

esis is that it is formed from partial melts of the mantle. But as we have dis-

cussed, mantle melts have near-basaltic compositions. Thus, if this were the

case, the mean composition of the continental crust would also be basaltic.

However, as seen in Table 1–1, the mean continental crust is considerably

more silicic than the composition of basalts. A preferred hypothesis for the

generation of the continental crust consists of three steps: 1) Basaltic volcan-

ism from the mantle associated with island-arc volcanics, continental rifts,

and hotspots is responsible for the formation of the continental crust. 2) In-

tracrustal melting and high-temperature metamorphism are responsible for

the differentiation of the continental crust so that the upper crust is more

silicic and the lower crust is more basic. Basaltic magmas from the mantle

that intrude into a basaltic continental crust in the presence of water can

produce the granitic rocks associated with the bulk continental crust. 3)

Delamination of substantial quantities of continental lithosphere including

the mantle and lower crust returns a fraction of the more basic lower crust

to the mantle. The residuum, composed primarily of the upper crust, thus

becomes more silicic and forms the present continental crust.

1.8 Paleomagnetism and the Motion of the Plates

Although qualitative geological arguments had long favored the continental

drift theory, it remained for paleomagnetic studies to provide quantitative

confirmation. Paleomagnetism is the study of the Earth’s past magnetic field

from the records preserved in magnetized rocks. The silicate minerals mak-

ing up the bulk of a rock are either paramagnetic (olivine, pyroxene, garnet,

amphiboles) or diamagnetic (quartz, feldspar) and are incapable of acquir-

ing a permanent magnetization. However, rocks containing small amounts

of ferromagnetic, or more accurately ferrimagnetic, minerals, that is, iron

oxides such as magnetite Fe3O4 and hematite Fe2O3 and iron sulfides such

1.8 Paleomagnetism and the Motion of the Plates 37

Table 1.3 Geologic Time Scale

Age (Ma) Period Era Eon

0 Holocene0.01

Upper 0.13

Middle Pleistocene 0.8

Lower 1.8

Upper 3.6 Pliocene

Lower 5.3

Upper 11.2

Middle Miocene 16.4

Lower 23.8

Upper Cenozoic 28.5 Oligocene

Lower 33.7

Upper 37.0

Middle Eocene 49.0

Lower 54.8

Upper 61.0 Paleocene

Lower 65.0

Upper 98.9 Cretaceous

Lower 142.0

Upper 159.4

Middle Jurassic Mesozoic 180.1

Lower

Ph an

er oz

oi c

205.7 Upper

227.4 Middle Triassic

241.7 Lower

248.2 Upper

256.0 Permian Lower

290.0 Upper

323.0 Carboniferous Lower

354.0 Upper

370.0 Middle Devonian

391.0 Lower Paleozoic

417.0 Silurian

443.0 Upper

470.0 Ordovician Lower

495.0 Upper

505.0 Middle Cambrian

518.0 Lower

545.0 Proterozoic

2500 Precambrian Archean 4550

N eo

ge ne

Pa le

og en

e

Te rt

ia ry

Q ua

te rn

ar y

38 Plate Tectonics

Figure 1.22 Declination and inclination of the magnetic field.

as pyrrhotite Fe1−yS, can acquire a weak permanent magnetism when they

are formed. The fossil magnetism in a rock is referred to as natural remanent

magnetism (NRM).

A rock can acquire NRM in several ways. When a mineral is heated above

its Curie temperature, all magnetism is lost. For magnetite the Curie tem-

perature is 851 K. When a rock containing ferromagnetic minerals is cooled

to a temperature below the Curie temperature, known as the blocking tem-

perature, in the presence of a magnetic field, it can acquire a remanent

magnetism. This is known as thermoremanent magnetism (TRM). In some

cases magnetic minerals are formed by chemical processes at low tempera-

tures. As a grain of a ferromagnetic mineral grows, it reaches a size where

it becomes magnetically stable. If this occurs in an applied magnetic field,

a chemical remanent magnetism (CRM) may be acquired. A sedimentary

rock may also acquire a remanent magnetism during its formation. As small

particles of ferromagnetic minerals fall through water in the presence of a

magnetic field, their magnetic moments become partially aligned with the

ambient magnetic field; the result is that the sedimentary rock that is formed

with these particles present has a depositional remanent magnetism (DRM).

Rocks may also acquire magnetism after they are formed. This type of

magnetism may usually be removed by subjecting the rock to alternating

magnetic fields or by heating the rock to a substantial fraction of the Curie

temperature. After it has been confirmed that the magnetism in a rock is

in fact the remanent magnetism acquired at the time of its formation, the

orientation or direction of the remanent field is determined. This is normally

expressed in terms of the declination D or magnetic azimuth, which is the

angle between geographic north and the magnetic field direction measured

1.8 Paleomagnetism and the Motion of the Plates 39

Figure 1.23 The Earth’s dipole magnetic field.

positive clockwise (0 to 360◦), and the inclination I, which is the angle

between the horizontal and the field direction measured positive downward

(−90 to +90◦) (Figure 1–22).

In addition to declination and inclination, the complete specification of a

remanent magnetic field requires the determination of its magnitude B. The

SI unit of B is the tesla or weber m−2. Figure 1–22 clearly shows that the

horizontal BH and vertical BV components of the magnetic field are related

to the magnitude of the field and the inclination by

BH = B cos I (1.2)

BV = B sin I. (1.3)

The horizontal field can be further resolved into a northward component

BHN and an eastward component BHE given by

BHN = B cos I cosD (1.4)

BHE = B cos I sinD. (1.5)

The present-day magnetic field of the Earth can be reasonably approx-

imated as a dipole magnetic field, the form of which is sketched in Figure

1–23. The horizontal and vertical components of the Earth’s dipole magnetic

field, Bθ and Br, at its surface, assuming that the Earth is a sphere of radius

a, are given by

Bθ = µ0m

4πa3 sin θm (1.6)

40 Plate Tectonics

Br = µ0m

2πa3 cos θm, (1.7)

where µ0 is the permeability of free space (µ0 = 4π× 10−7 tesla m A−1), m

is the dipole moment (A m2), and θm is the magnetic colatitude (magnetic

latitude φm = π/2 − θm) (see Figure 1–23). The magnetic poles are the

positions where the dipole field lines are vertical. At the north magnetic pole

(θm = 0, φm = π/2), Bθ = 0, Br = µ0m/2πa 3, the inclination is π/2 rad

or 90◦, and the field is directed into the Earth. At the south magnetic pole

(θm = π, φm = −π/2), Bθ = 0, Br =−µ0m/2πa 3, the inclination is −π/2

rad or −90◦, and the field is directed out from the Earth. The magnetic field

lines of the Earth’s present dipole magnetic field leave at the south magnetic

pole and enter at the north magnetic pole (Figure 1–23). At the magnetic

equator (θm = π/2, φm = 0), Br = 0, Bθ = µ0m/4πa 3, the field lines are

horizontal, and the inclination is zero. The angle of inclination of the dipole

magnetic field is given by

tan I = Br Bθ , (1.8)

and its magnitude B can be written

B = (B2 r +B2

θ ) 1/2. (1.9)

By substituting for Br and Bθ from Equations (1–6) and (1–7), we can

rewrite these expressions for I and B as

tan I = 2cot θm = 2 tan φm (1.10)

B = µ0m

4πa3 (sin2 θm + 4cos2 θm)1/2

= µ0m

4πa3 (1 + 3 cos2 θm)1/2

= µ0m

4πa3 (1 + 3 sin2 φm)1/2. (1.11)

The Earth’s magnetic field is only approximately a dipole. The present

locations (latitude and longitude) of the magnetic poles are 73◦N, 100◦W

and 68◦S, 143◦E. The magnetic poles of the dipole field that is the best fit

to the Earth’s field are at 79◦N, 70◦W and 79◦S, 110◦E. Thus the axis of the

dipole field makes an angle of about 11◦ with the Earth’s rotational axis.

The moment of the dipole field is m = 7.94 × 1022 A m2, and the surface

magnetic field at the magnetic equator is Bθ = 3.07 × 10−5 teslas. Maps of

the magnitude, declination, and inclination of the present magnetic field of

the Earth are presented in Figure 1–24.

1.8 Paleomagnetism and the Motion of the Plates 41

Problem 1.4 Assume that the Earth’s magnetic field is a dipole. What is

the maximum intensity of the field at the core–mantle boundary?

Problem 1.5 Assume that the Earth’s magnetic field is a dipole. At what

distance above the Earth’s surface is the magnitude of the field one-half of

its value at the surface?

If a dipole field is a reasonable approximation of the Earth’s magnetic

field throughout geologic time, a paleomagnetic measurement of declination

and inclination can be used to locate the magnetic pole position at the time

the rock acquired its magnetization. Suppose that the paleomagnetic mea-

surement is carried out at a north latitude φ and an east longitude ψ, as

in Figure 1–25. From the definition of declination it is clear that the paleo-

magnetic north pole lies an angular distance θm along a great circle making

an angle D with the meridian through the measurement point. Geographic

north, paleomagnetic north, and the measurement point define a spherical

triangle with sides π/2 − φ, θm, and π/2 − φp, where φp is the latitude of

the paleomagnetic pole. The triangle contains the included angle D. Using

a result from spherical trigonometry, we can write

cos

(

π

2 − φp

)

= cos

(

π

2 − φ

)

cos θm

+ sin

(

π

2 − φ

)

sin θm cosD.

(1.12)

This can be simplified by noting that cos(π/2−φp) = sinφp, cos(π/2−φ) =

sinφ, and sin(π/2 − φ) = cosφ. The result is

sinφp = sinφ cos θm + cosφ sin θm cosD. (1.13)

The magnetic colatitude θm can be determined from Equations (1–8) and

(1–10). The angle between the meridians passing through the measurement

point and the paleomagnetic north pole is ψp − ψ, where ψp is the east lon-

gitude of the paleomagnetic pole. A second spherical trigonometric formula

allows us to write

sin(ψp − ψ)

sin θm =

sinD

sin(π/2 − φp) =

sinD

cosφp (1.14)

or

sin(ψp − ψ) = sin θm sinD

cosφp , (1.15)

42 Plate Tectonics

F igu

re 1.24

P resen

t-d a y

m a gn

etic fi eld

o f th

e E a rth

. (a

) M

a gn

itu d e, µ T

. C

o n tin

u ed

o n

p p . 2 6 – 7 .

1.8 Paleomagnetism and the Motion of the Plates 43

F ig

u re

1. 24

(c o n t. )

(b )

D ec

li n a ti o n , d eg

.

44 Plate Tectonics

F igu

re 1.24

(co n t.)

(c) In

clin a tio

n , d eg.

1.8 Paleomagnetism and the Motion of the Plates 45

if cos θm > sinφ sinφp. If cos θm < sinφ sinφp, Equation (1–15) must be

replaced by

sin(π + ψ − ψp) = sin θm sinD

cosφp . (1.16)

Paleomagnetic measurements are useful only if the orientation of the sam-

ple has remained fixed with respect to the rest of the geological province,

since the sample was magnetized. Usually the absence of subsequent de-

formation can be established with some certainty and the reliability of the

measurement established.

Problem 1.6 The measured declination and inclination of the paleomag-

netic field in Upper Triassic rocks at 41.5◦N and 72.7◦W are D = 18◦ and

I = 12◦. Determine the paleomagnetic pole position.

Problem 1.7 The measured declination and inclination of the paleo-

magnetic field in Oligocene rocks at 51◦N and 14.7◦E are D = 200◦ and

I = −63◦. Determine the paleomagnetic pole position.

Problem 1.8 The measured declination and inclination of the paleomag-

netic field in Lower Cretaceous rocks at 45.5◦N and 73◦W are D = 154◦ and

I = −58◦. Determine the paleomagnetic pole position.

Paleomagnetic measurements can indicate the position of the magnetic

pole as a function of time for rocks of different ages. However, before dis-

cussing these results, we should note that one of the early conclusions of

paleomagnetic measurements was that the Earth’s magnetic field has been

subject to periodic reversals in which the north magnetic pole became the

south magnetic pole and vice versa. This was apparent from the reversed

orientations of the remanent magnetic field in a series of rocks of different

ages from the same locality. A summary of dated rocks with normal and

reversed polarities for the last 5 Ma is given in Figure 1–26. Measurements

indicate that for the past 720,000 years the magnetic field has been in its

present (normal) orientation; this magnetic time period is referred to as the

Brunhes epoch. Between 0.72 and 2.45 Ma, there was a period known as

the Matuyama epoch during which the orientation of the field was predom-

inantly reversed. Periods of normal polarity for the last 170 Ma are given in

Table 1–4.

The mechanism for magnetic field reversals is not known. In fact, the

way in which the Earth’s magnetic field is generated is only qualitatively

understood. It is well established from seismology that the outer core of the

Earth is primarily composed of liquid iron. Presumably, electric currents in

46 Plate Tectonics

the highly electrically conducting liquid iron generate the Earth’s magnetic

field. However, the currents that create the magnetic field are themselves

driven by motions of the conducting liquid in the presence of the magnetic

field. The field generation mechanism requires the presence of the field itself.

The process by which fluid motions maintain the magnetic field against its

tendency to decay because of ohmic dissipation is known as regenerative

dynamo action.

An energy source is required to overcome the resistive losses. Possible

energy sources are the decay of radioactive elements in the core, the cooling

of the core, the latent heat release upon solidification of the inner core,

and the gravitational energy release that accompanies solidification of the

inner core. The last energy source exists because the outer core contains an

alloying element lighter than iron. The light element does not enter the inner

core when solidification occurs at the inner core–outer core boundary. As a

result, growth of the inner core concentrates the light element in the outer

core, causing outer-core liquid to become increasingly lighter with time. This

releases gravitational potential energy in the same way that separation of

the entire core did early in the Earth’s evolution.

One or more of these energy sources drives the thermal or chemical con-

vective motions of the highly conducting liquid iron that result in a self-

excited dynamo; however, detailed theories of the process are not available.

Self-excited mechanical dynamos built in the laboratory exhibit random re-

versals of the resulting field. Presumably, the dynamo in the Earth’s core is

subject to random fluctuations that aperiodically lead to field reversals.

It is believed that the rotation of the Earth has an important influence on

the generation of the field. We have already noted that the Earth’s present

dipole axis is nearly aligned with its axis of rotation. It is implicitly assumed

in the use of paleomagnetic measurements that the magnetic poles and the

geographic poles coincide. A measurement of a paleomagnetic pole can then

be used to deduce the motion of the plate on which the measurement was

made.

Many paleomagnetic measurements have been made. Data are divided into

geological periods and into continental areas that appear to have remained

a single unit over the periods considered. Average pole positions are given in

Table 1–5. If no relative motion occurred among the continental blocks, all

measurements during a particular period should give the same pole position.

Clearly, as can be seen from Table 1–5, this is not the case. If a sequence

of pole positions for a particular continental area is plotted, it should form

a continuous path terminating close to the present position of the magnetic

pole; this is known as the polar wandering path for the magnetic pole. A polar

Figure 1.25 Geometry for determining the latitude and longitude of a pa- leomagnetic field.

wandering path of a plate can be used to determine the absolute position of

that plate relative to the geographic poles. The relation between the polar

wandering paths of two adjacent plates can be used to determine relative

velocities between the plates. The polar wandering paths for North America

and Europe are shown in Figure 1–27. The systematic divergence of the

paths over the past several hundred million years was one of the first pieces

of quantitative evidence that continental drift was occurring. Unfortunately

the considerable scatter in paleomagnetic measurements makes it difficult

to obtain reliable data. Much of this scatter can be attributed to deviations

of the magnetic poles from the geographic poles.

The magnitude of the magnetic field at the Earth’s surface varies both in

space and in time. The spatial variations are known as magnetic anomalies.

In the continents, regions of high magnetic field, that is, positive magnetic

anomalies, are usually associated with concentrations of magnetic minerals

in the Earth’s crust. Regional surveys of the magnetic field are an important

method of exploration for economic deposits of minerals.

Similar magnetic surveys over the oceans have shown a pattern of striped

magnetic anomalies, that is, elongated continuous zones of positive mag-

netic anomalies some tens of kilometers wide separated from one another

by zones of negative magnetic anomalies. The zones of striped magnetic

48 Plate Tectonics

Figure 1.26 Measurements of the polarity of the Earth’s magnetic field for the last 5 million years. Each short line indicates a dated polarity determi- nation from a volcanic rock. The shaded periods are intervals of predomi- nantly normal polarity.

1.8 Paleomagnetism and the Motion of the Plates 49

Table 1.4 Ages in Ma of Periods of Normal Polarity of the Earth’s

Magnetic Field for the Last 170 Ma

Normal Polarity Normal Polarity Normal Polarity Anomaly Interval Anomaly Interval Anomaly Interval

1.1 0.00 0.72 6.7 22.90 23.05 M1 118.70 121.81 1.2 0.91 0.97 6.8 23.25 23.38 M2 122.25 123.03 2.1 1.65 1.88 6.9 23.62 23.78 M4 125.36 126.46 2.2 2.06 2.09 7.1 25.01 25.11 M6 127.05 127.21 2.3 2.45 2.91 7.2 25.17 25.45 M7 127.34 127.52 2.4 2.98 3.07 7.3 25.84 26.01 M8 127.97 128.33 2.5 3.17 3.40 8.1 26.29 26.37 M9 128.60 128.91 3.1 3.87 3.99 8.2 26.44 27.13 M10.1 129.43 129.82 3.2 4.12 4.26 9.1 27.52 28.07 M10.2 130.19 130.57 3.3 4.41 4.48 9.2 28.12 28.51 M10.3 130.63 131.00 3.4 4.79 5.08 10.1 29.00 29.29 M10.4 131.02 131.36 3.5 5.69 5.96 10.2 29.35 29.58 M11.1 131.65 132.53 3.6 6.04 6.33 11.1 30.42 30.77 M11.2 133.03 133.08 3.7 6.66 6.79 11.2 30.82 31.21 M11.3 133.50 134.31 4.1 7.01 7.10 12 31.60 32.01 M12.1 134.42 134.75 4.2 7.17 7.56 13.1 34.26 34.44 M12.2 135.56 135.66 4.3 7.62 7.66 13.2 34.50 34.82 M12.3 135.88 136.24 4.4 8.02 8.29 15.1 36.12 36.32 M13 136.37 136.64 4.5 8.48 8.54 15.2 36.35 36.54 M14 137.10 137.39 4.6 8.78 8.83 15.3 36.93 37.16 M15 138.30 139.01 5.1 8.91 9.09 16.1 37.31 37.58 M16 139.58 141.20 5.2 9.14 9.48 16.2 37.63 38.01 M17 141.85 142.27 5.3 9.49 9.80 17.1 38.28 39.13 M18 143.76 144.33 5.4 9.83 10.13 17.2 39.20 39.39 M19.1 144.75 144.88 5.5 10.15 10.43 17.3 39.45 39.77 M19.2 144.96 145.98 5.6 10.57 10.63 18.1 39.94 40.36 M20.1 146.44 146.75 5.7 11.11 11.18 18.2 40.43 40.83 M20.2 146.81 147.47 5.8 11.71 11.90 18.3 40.90 41.31 M21 148.33 149.42 5.9 12.05 12.34 19 42.14 42.57 M22.1 149.89 151.46 5.10 12.68 12.71 20 43.13 44.57 M22.2 151.51 151.56 5.11 12.79 12.84 21 47.01 48.51 M22.3 151.61 151.69 5.12 13.04 13.21 22 50.03 50.66 M22.4 152.53 152.66 5.13 13.40 13.64 23.1 51.85 52.08 M23.1 152.84 153.21 5.14 13.87 14.24 23.2 52.13 52.83 M23.2 153.49 153.52 5.15 14.35 14.79 23.3 53.15 53.20 M24.1 154.15 154.48 5.16 14.98 15.07 24.1 53.39 53.69 M24.2 154.85 154.88 5.17 15.23 15.35 24.2 54.05 54.65 M24.3 155.08 155.21 5.18 16.27 16.55 25 57.19 57.80 M24.4 155.48 155.84 5.19 16.59 16.75 26 58.78 59.33 M25.1 156.00 156.29 5.20 16.82 16.99 27 61.65 62.17 M25.2 156.55 156.70 5.21 17.55 17.87 28 62.94 63.78 M25.3 156.78 156.88 5.22 18.07 18.09 29 64.16 64.85 M25.4 156.96 157.10 5.23 18.50 19.00 30 65.43 67.14 M26.1 157.20 157.30 6.1 19.26 20.23 31 67.23 68.13 M26.2 157.38 157.46 6.2 20.52 20.74 32.1 70.14 70.42 M26.3 157.53 157.61 6.3 20.97 21.37 32.2 70.69 72.35 M26.4 157.66 157.85 6.4 21.60 21.75 32.3 72.77 72.82 M27 158.01 158.21 6.5 21.93 22.03 33 73.12 79.09 M28 158.37 158.66 6.6 22.23 22.60 34 84.00 118.00 M29 158.87 159.80

J-QZ 160.33 169.00

Source: Harland et al. (1990).

anomalies generally lie parallel to ocean ridges and are symmetric with re-

spect to the ridge crest. A typical pattern adjacent to the mid-Atlantic ridge

is shown in Figure 1–28. A typical magnetic anomaly profile perpendicular

to the East Pacific Rise is given in Figure 1–29. The magnitude of any in-

dividual anomaly is a few hundred nanoteslas, or about 1% of the Earth’s

dipole field at the surface. The magnetic anomalies are attributed to ther-

mal remanent magnetism in the basaltic oceanic crust. As the volcanic rocks

of the oceanic crust cool through the magnetic blocking temperature near

the ocean ridge, a thermal remanent magnetism is acquired in the direc-

tion of the Earth’s magnetic field. This magnetization of the oceanic crust

produces the magnetic anomalies as a consequence of the episodic reversals

50 Plate Tectonics

T ab

le 1.5

P o sitio

n o f th

e N

o rth

M a gn

etic P o le

in D

iff eren

t G

eo logica

l P eriod

s a s

D eterm

in ed

by

P a leo

m a gn

etic S tu

d ies

N o rth

R u

ssian Sib

erian So

u th

A m

erica Eu

ro p e

P latfo

rm P

latfo rm

A frica

A m

erica A

u stralia

In d

ia

U 87N

,140E 80N

,157E 78N

,191E 66N

, 234E 87N

,152E 77N

,275E 82N

,62E Tertiary

L 85N

,197E 75N

,151E 68N

,192E 57N

,152E 85N

,186E 70N

,306E

C retaceous

64N ,187E

86N ,0E

66N ,166E

77N ,176E

61N ,260E

78N ,236E

53N ,329E

22N ,295E

Jurassic 76N

,142E 36N

,50E 65N

,138E 84N

,256E 65N

,262E 48N

,331E Triassic

62N ,100E

45N ,143E

51N ,154E

47N ,151E

80N ,71E

20N ,308E

Perm ian

46N ,117E

45N ,160E

44N ,162E

27N ,269E

7S,304E

U 38N

,161E 43N

,168E 34N

,144E 46N

,220E 60N

,180E 46N

,315E 26S,312E

C arboniferous

L 37N

,126E 22N

,168E 26N

,206E 73N

,34E 43N

,151E D

evonian 36N

,162E 28N

,151E 72N

,174E

Silurian 29N

,123E O

N ,136E

28N ,149E

24N ,139E

54N ,91E

O rdovician

28N ,192E

10N ,176E

25S,131E 24S,165E

11S,143E 2N

,188E C

am brian

7N ,140E

22N ,167E

8N ,189E

36S,127E 28N

,212E

S o u rce:

A fter

M . W

. M

cE lh

in n y

(1 9 7 3 ).

1.8 Paleomagnetism and the Motion of the Plates 51

Figure 1.27 Polar wandering paths for North America and Europe. Num- bers give time before present in millions of years.

in the Earth’s magnetic field. Ocean floor created in the last 720,000 years

has been magnetized in the direction of the Earth’s present magnetic field,

leading to a positive magnetic anomaly (see Figure 1–26). However, ocean

floor created between 2.45 and 0.72 Ma was primarily magnetized in the

direction of the reversed field. This magnetization is opposite to the present

Earth’s field and therefore subtracts from it, leading to a zone of low field or

a negative magnetic anomaly, as illustrated in Figure 1–29b. The conclusion

is that the stripes of seafloor with positive magnetic anomalies were created

during periods of normal polarity of the Earth’s magnetic field and stripes of

the seafloor with negative magnetic anomalies were created during periods

of reversed polarity of the Earth’s magnetic field.

Since the dates of the field reversals are known independently from geochrono-

logical studies, the widths of the magnetic stripes can be used to determine

the velocity of seafloor spreading. For the example given in Figure 1–29a,

the distance from the ridge crest to the edge of each anomaly is plotted

against the time of known field reversal in Figure 1–29c. The result is nearly

a straight line, the slope of which is the velocity of seafloor spreading, 45

mm yr−1 in this case. Velocities of seafloor spreading (half-spreading rates)

range upward to about 100 mm yr−1.

52 Plate Tectonics

Figure 1.28 Striped pattern of magnetic anomalies parallel to the Mid- Atlantic ridge (Heirtzler et al., 1966).

Problem 1.9 Determine the velocity of seafloor spreading on the East

Pacific Rise from the magnetic anomaly profile given in Figure 1–30a.

Problem 1.10 Determine the velocity of seafloor spreading on the

South East Indian Rise from the magnetic anomaly profile given in Figure

1–30b.

Shipboard magnetometers have been used to obtain maps of magnetic

anomalies over a large fraction of the world’s oceans. Striped patterns of

magnetic anomalies have not been obtained near the paleomagnetic equator.

At the magnetic equator the magnetic field is horizontal, and the magneti-

zation of the ferromagnetic minerals in the oceanic crust does not produce

a significant surface magnetic anomaly. The maps of magnetic anomalies

have been used to determine the age of a large fraction of the ocean floor

(Figure 1–31). This distribution of ages has been confirmed by the Deep Sea

Drilling Project (DSDP). The deep-sea capability of the drilling ship Glomar

Challenger made it possible to drill a large number of cored holes through

1.8 Paleomagnetism and the Motion of the Plates 53

Figure 1.29 (a) Magnetic anomaly profile perpendicular to the East Pa- cific Rise (52◦S, 118◦W). (b) Induced magnetization in the oceanic crust due to episodic reversals of the Earth’s magnetic field. (c) Correlation of the positions x of the magnetic anomalies with t of field reversals to give the velocity u of seafloor spreading.

the sedimentary cover and into the underlying basaltic oceanic crust. If we

hypothesize that the age of the oldest sediments in the sedimentary sequence

adjacent to the volcanic crust, as determined from studies of fossils, corre-

sponds to the age of the volcanic rocks, then we can determine the age of the

seafloor. This has been done for a number of DSDP holes, and the results

have been compared with the age of the seafloor inferred from studies of

the magnetic anomalies in Figure 1–32. The excellent agreement is striking

confirmation of the magnetic method for determining the age of the seafloor.

Because the surface area of the Earth remains essentially constant, the

velocities of seafloor spreading at ocean ridges can be related to velocities

of subduction at ocean trenches. As a result the relative velocities among

the rigid plates can be determined. The ten major plates are illustrated in

Figure 1–1. The relative motion between two adjacent plates can be obtained

using Euler’s theorem. This theorem states that any line on the surface of a

sphere can be translated to any other position and orientation on the sphere

by a single rotation about a suitably chosen axis passing through the center

of the sphere. In terms of the Earth this means that a rigid surface plate

can be shifted to a new position by a rotation about a uniquely defined

axis. The point where this axis intersects the surface of the Earth is known

54 Plate Tectonics

Figure 1.30 Typical profiles of the magnetic anomaly pattern (a) perpen- dicular to the East Pacific Rise at 61◦S and 151◦W and (b) perpendicular to the South East Indian Rise at 54◦S and 142◦E.

as a pole of rotation. This is illustrated in Figure 1–33, where plate B is

rotating counterclockwise with respect to plate A. Ridge segments lie on

lines of longitude emanating from the pole of rotation P. Transform faults

lie on small circles with their centers at the pole of rotation.

The relative motion between two adjacent plates is completely specified

when the latitude and longitude of the pole of rotation together with the

angular velocity of rotation ω are given. The location of the pole of rotation

can be determined from the orientations of ridge crests, magnetic lineaments,

and transform faults. The angular velocity of rotation can be obtained from

the seafloor-spreading velocities determined from widths of the magnetic

lineaments and the requirement that surface area must be preserved.

The latitudes and longitudes of the poles of rotation for relative motions

among ten plates are given in Table 1–6. The angular velocities of rotation

are also given. The relative velocity u between plates at any plate boundary

is

u = ωa sin ∆, (1.17)

where a is the radius of the Earth and ∆ is the angle subtended at the center

of the Earth by the pole of rotation P and point A on the plate boundary

(see Figure 1–34). Note that ω is in radians per unit time. The angle ∆ can

1.8 Paleomagnetism and the Motion of the Plates 55

F ig

u re

1. 31

M a p

o f se

a fl oo

r a ge

s (M

u ll er

et a l. , 1 9 9 7 ).

56 Plate Tectonics

Figure 1.32 Correlation of the ages of the oldest sediments in DSDP holes with the predicted ages of the oceanic crust based on seafloor magnetic anomalies.

Figure 1.33 Plate B is moving counterclockwise relative to plate A. The motion is defined by the angular velocity ω about the pole of rotation P. Double lines are ridge segments, and arrows denote directions of motion on transform faults.

be related to the colatitude θ and east longitude ψ of the pole of rotation and

the colatitude θ′ and east longitude ψ′ of the point on the plate boundary

by the same spherical trigonometry formula used in Equation (1–12). By

referring to Figure 1–35 we can write

cos ∆ = cos θ cos θ′ + sin θ sin θ′ cos(ψ − ψ′).

(1.18)

1.8 Paleomagnetism and the Motion of the Plates 57

Table 1.6 Pole Positions and Rates of Rotation for Relative Motion

Between Adjacent Surface Plates∗

Plates Lat. (N) Long. (E) ω (deg/Myr)

EU–NA 62.4 135.8 0.21 AF–NA 78.8 38.3 0.24 AF–EU 21.0 −20.6 0.12 NA–SA 16.3 −58.1 0.15 AF–SA 62.5 −39.4 0.31 AN–SA 86.4 −40.7 0.26 NA–CA −74.3 −26.1 0.10 CA–SA 50.0 −65.3 0.18 NA–PA 48.7 −78.2 0.75 CO–PA 36.8 −108.6 2.00 CO–NA 27.9 −120.7 1.36 CO–NZ 4.8 −124.3 0.91 NZ–PA 55.6 −90.1 1.36 NZ–AN 40.5 −95.9 0.52 NZ–SA 56.0 −94.0 0.72 AN–PA 64.3 −84.0 0.87 PA–AU −60.1 −178.3 1.07 EU–PA 61.1 −85.8 0.86 CO–CA 24.1 −119.4 1.31 NZ–CA 56.2 −104.6 0.55 AU–AN 13.2 38.2 0.65 AF–AN 5.6 −39.2 0.13 AU–AF 12.4 49.8 0.63 AU–IN −5.6 77.1 0.30 IN–AF 23.6 28.5 0.41 AR–AF 24.1 24.0 0.40 IN–EU 24.4 17.7 0.51 AR–EU 24.6 13.7 0.50 AU–EU 15.1 40.5 0.69 IN–AR 3.0 91.5 0.03

∗ Global plate motion model NUVEL-1A. The first plate moves counterclockwise relative to the second plate. Abbreviations: PA, Pacific; NA, North America; SA,

South America; AF, Africa; CO, Cocos; NZ, Nazca; EU, Eurasia; AN, Antarctica; AR, Arabia; IN, India; AU, Australia; CA, Caribbean. See Figure

1–1 for plate geometries. Source: DeMets et al. (1994).

The surface distance s between points A and P is

s = a∆, (1.19)

with ∆ in radians. This relation along with Equation (1–18) can be used

to determine the distance between two points on the surface of the Earth

58 Plate Tectonics

Figure 1.34 Geometry for the determination of the relative plate velocity at point A on the boundary between two plates in terms of the rate of rotation ω about pole P.

Figure 1.35 Geometry for determining the angle between point A on a plate boundary and a pole of rotation.

given the latitudes and longitudes of the points. Using Equations (1–17) and

(1–18), one can find the relative velocity between two plates, at any point

on the boundary between the plates, once the latitude and longitude of the

point on the boundary have been specified.

As a specific example let us determine the relative velocity across the San

Andreas fault at San Francisco (37.8◦N, 122◦W). We assume that the entire

relative velocity between the rigid Pacific and North American plates is

accommodated on this fault. From Table 1–6 we find θ = 90◦ − 48.7◦ = 41.3◦

and ψ = −78.2◦. Since θ′ = 52.2◦ and ψ′ = 238◦, we find from Equation

1.9 Triple Junctions 59

(1–18) that ∆ = 33.6◦; with ω = 0.75◦ Myr−1, we find from Equation (1–17)

that the relative velocity across the fault is 46 mm yr−1.

Problem 1.11 Determine the declination and inclination of the Earth’s

magnetic field at Boston (φ = 42.5◦, ψ = −71◦). Use the dipole approxima-

tion to the field, but do not assume that the geographic and magnetic poles

coincide.

Problem 1.12 Determine the declination and inclination of the Earth’s

magnetic field at Chicago (φ = 41.8◦, ψ = −87.5◦). Use the dipole approx-

imation to the field, but do not assume that the geographic and magnetic

poles coincide.

Problem 1.13 What are the surface distances between the Earth’s mag-

netic poles and geographic poles?

Problem 1.14 What is the surface distance between the Earth’s mag-

netic poles and the best-fit dipole poles?

Problem 1.15 Plot the distance between the paleomagnetic poles ob-

tained from North American and European rocks as a function of time, and

discuss the results.

Problem 1.16 Plot the distance between the paleomagnetic poles ob-

tained from the Russian and Siberian Platform rocks as a function of time,

and discuss the results.

Problem 1.17 What is the spreading rate between the North American

and Eurasian plates in Iceland (65◦N, 20◦W)?

Problem 1.18 What is the relative plate velocity between the Nazca

and South American plates at Lima, Peru (12◦S, 77◦W)?

Problem 1.19 What is the relative plate velocity between the Indian

and Eurasian plates in the Himalayas (30◦N, 81◦E)?

1.9 Triple Junctions

A plate boundary can end only by intersecting another plate boundary;

this intersection is a triple junction. Since there are three types of plate

boundaries – ridges, trenches, and transform faults – there are in princi-

ple ten types of triple junctions. However, some of these triple junction

cannot exist. An example is a triple junction of three transform faults. The

required condition for the existence of a triple junction is that the three

vector velocities defining relative motions between plate pairs at a triple

60 Plate Tectonics

Figure 1.36 (a) Schematic of a ridge–ridge–ridge (RRR) triple junction of plates A, B, and C. (b) Vector velocities for relative motion between the plates.

junction must form a closed triangle. For many types of triple junctions this

condition requires a particular orientation of the plate boundaries.

As a specific example let us consider the ridge– ridge–ridge (RRR) triple

junction illustrated in Figure 1–36a. The ridge between plates A and B lies in

the north-south direction (an azimuth with respect to the triple junction of

0◦). Since the relative velocity across a ridge is perpendicular to the ridge,

the vector velocity of plate B relative to plate A, uBA, has an azimuth,

measured clockwise from north, of 90◦; we assume that the magnitude is

uBA = 100 mm yr−1. The ridge between plates B and C has an azimuth of

110◦ relative to the triple junction. The vector velocity of plate C relative

to plate B, uCB, therefore has an azimuth of 200◦; we assume that the

magnitude uCB = 80 mm yr−1. The problem is to find the azimuth of the

ridge between plates A and C, α, and the azimuth and magnitude of the

relative velocity uAC.

The velocity condition for all triple junctions requires that

uBA + uCB + uAC = 0. (1.20)

This is illustrated in Figure 1–36b. In order to determine the magnitude of

the velocity uAC we use the law of cosines:

uAC = (1002 + 802 − 2 · 100 · 80 · cos 70◦)1/2

1.9 Triple Junctions 61

= 104.5 mm yr−1. (1.21)

The angle α is then determined using the law of sines:

sin(α− 180◦) = 80

104.5 sin 70◦ = 0.7518 = −sinα,

α = 228.7◦. (1.22)

The azimuth of the ridge is 228.7◦, and the azimuth of uAC is 318.7◦. An

example of an RRR triple junction is the intersection of the Nazca, Cocos,

and Pacific plates (see Figure 1–1).

Problem 1.20 Consider an RRR triple junction of plates A, B, and

C. The ridge between plates A and B lies in a north–south direction (an

azimuth of 0◦ with respect to the triple junction) and has a relative velocity

of 60 mm yr−1. The ridge between plates B and C has an azimuth of 120◦

with respect to the triple junction, and the ridge between plates A and C

has an azimuth of 270◦ with respect to the triple junction. Determine the

azimuths and magnitudes of the relative velocities between plates B and C

and C and A.

We next consider a trench–trench–trench (TTT) triple junction. In gen-

eral this type of triple junction cannot exist. A geometry that is acceptable

is illustrated in Figure 1–37a. Both plates A and B are being subducted

beneath plate C along a single north–south trench. Plate A is also being

subducted beneath plate B along a trench that has an azimuth of 135◦ with

respect to the triple junction. Since oblique subduction can occur, the rel-

ative velocities between plates where subduction is occurring need not be

perpendicular to the trench. We assume that the velocity of plate A relative

to plate B has a magnitude uAB = 50 mm yr−1 and an azimuth of 225◦. We

also assume that the relative velocity of plate B with respect to plate C has

a magnitude uBC = 50 mm yr−1 and an azimuth of 270◦. Applying the law

of cosines to the velocity triangle of Figure 1–37b, we find

uAC = (502 + 502 − 2 · 50 · 50 · cos 135◦)1/2

= 92.4 mm yr−1. (1.23)

The angle α in Figure 1–37b is determined from the law of sines:

sinα = 50

92.4 sin 135◦ = 0.383, α = 22.5◦, (1.24)

so that the azimuth of uAC is 247.5◦. The velocity at which subduction is

occurring is uAC cosα = 85.4 mm yr−1, and the velocity of migration of the

triple junction along the north–south trench is uAC sinα = 35.4 mm yr−1.

62 Plate Tectonics

Figure 1.37 (a) Illustration of a trench–trench–trench (TTT) triple junc- tion of plates A, B, and C. (b) Vector velocities for relative motion between the plates.

Figure 1.38 Another TTT triple junction.

An example of a TTT triple junction is the intersection of the Eurasian,

Pacific, and Philippine plates (see Figure 1–1).

Problem 1.21 Show that a triple junction of three transform faults

cannot exist.

Problem 1.22 Consider the TTT triple junction illustrated in Figure 1–

38. This triple junction is acceptable because the relative velocity between

plates C and A, uCA, is parallel to the trench in which plate B is being

subducted beneath plate C. The trench between plates C and B has an

azimuth of 180◦ so that uCA has an azimuth of 0◦; assume that uCA = 50

mm yr−1. Also assume that the azimuth and magnitude of uBA are 315◦

and 60 mm yr−1. Determine the azimuth and magnitude of uBC.

1.9 Triple Junctions 63

Figure 1.39 (a) A trench–ridge–fault (TRF) triple junction of plates A, B, and C. (b) Vector velocities for the relative motions between the plates.

As our final example we consider a ridge–trench–fault (RTF) triple junc-

tion. This is another type of triple junction that cannot generally exist.

An acceptable geometry is illustrated in Figure 1–39a; the trench and the

transform fault are aligned in the north–south direction. Plate C is being

subducted beneath plate B; plate A is sliding past plate B on a transform

fault. The velocity of plate B relative to plate A has a magnitude uBA = 50

mm yr−1 and an azimuth of 180◦ (the orientation of the fault requires an

azimuth of either 0◦ or 180◦). The ridge has an azimuth of 225◦ with respect

to the triple junction. This constrains the relative velocity between plates

A and C to have an azimuth of 315◦; we assume that uAC = 40 mm yr−1.

Applying the law of cosines to the velocity triangle in Figure 1–39b we get

uCB = (502 + 402 − 2 · 40 · 50 cos 45◦)1/2

= 35.7 mm yr−1, (1.25)

and from the law of sines we find

sinα = 40

35.7 sin 45◦ = 0.79, α = 52.4◦. (1.26)

The rate at which the ridge is migrating northward along the trench–transform

boundary is uCB cosα+uAC cos 45◦ = 50.1 mm yr−1. An example of an RTF

triple junction is the intersection of the Pacific, North American, and Cocos

plates (see Figure 1–1).

It should be emphasized that the relative plate motions given in Table

1–6 are only instantaneously valid. As plates evolve, their poles of rotation

migrate, and their angular velocities change. Plate boundaries and triple

64 Plate Tectonics

Figure 1.40 A TTR triple junction.

junctions must also evolve. One result is that a plate boundary may cease to

be active or new plate boundaries and triple junctions may form. Another

consequence is that plate boundaries may become broad zones of diffuse

deformation. The western United States is an example of such a zone; the

deformation associated with the interaction of the Pacific, Juan de Fuca,

and North American plates extends from the Colorado Front in Wyoming,

Colorado, and New Mexico, to the Pacific Coast (see Section 1–13).

Problem 1.23 Consider the TTR triple junction illustrated in Figure

1–40. A ridge with an azimuth of 135◦ relative to the triple junction is

migrating along a north–south trench. If the azimuth and magnitude of uBA

are 270◦ and 50 mm yr−1 and uCB = 40 mm yr−1, determine the azimuth

and magnitude of uCA. Also determine the direction and rate of migration

of the ridge relative to plate A.

Problem 1.24 Consider the TTF triple junction illustrated in Figure 1–

41a. A right-lateral transform fault has an azimuth of 45◦ with respect to the

triple junction that is migrating along a north–south trench. If the azimuth

and magnitude of uBA are 270◦ and 50 mm yr−1 and uCB = 50 mm yr−1,

determine the azimuth and magnitude of uCA. Also determine the direction

and rate of migration of the fault along the trench.

Problem 1.25 Consider the TTF triple junction illustrated in Figure 1–

41b. A left-lateral transform fault has an azimuth of 0◦, and two trenches

have azimuths of 180◦ and 225◦. If the azimuth and magnitude of uCB are

90◦ and 10 mm yr−1 and uAB = 50 mm yr−1, determine the azimuth and

magnitude of uAC.

1.10 The Wilson Cycle 65

Figure 1.41 Two TTF triple junctions.

1.10 The Wilson Cycle

J. Tuzo Wilson (1966) proposed that continental drift is cyclic. In particular

he proposed that oceans open and close cyclically; this concept is now known

as the Wilson cycle and was based on the opening and closing of the Atlantic

Ocean. The Wilson cycle, in its simplest form, is illustrated in Figure 1–42.

The first step in the Wilson cycle, illustrated in Figure 1–42a, is the

breakup of a continent. This occurs on continental rift zones. The first stage

of the splitting process is the formation of a rift valley. When a continent

starts to fracture under tensional stresses, a rift valley is formed. The cen-

tral block of the rift valley, known as a graben, subsides, as shown in Figure

1–42a, and the edges of the adjacent blocks are uplifted. The faults that

occur on the sides of the down-dropped central graben are known as normal

faults. Displacements on the normal faults accommodate horizontal exten-

sion. Examples of rift valleys that may be in the first stage of continental

splitting include the East African rift system and the Rio Grande (river

valley) rift. There is ample evidence in the geological record, however, that

some rift valleys never evolve into an ocean. The splitting process may be

aborted. Once the formation of the rift valley relieves the tensional stresses,

no further horizontal extension may occur.

The Red Sea and the Gulf of Aden are rift valleys that have progressed to

the formation of accreting plate margins. Together with the East African rift

they define a three-armed pattern that can be seen in the satellite photograph

in Figure 1–43. If all the rifts of a three-armed system develop into accreting

plate margins, an RRR triple junction is formed. In many cases, however,

66 Plate Tectonics

only two arms develop into accreting margins, and the third becomes a relict

rift zone in the continent. This third arm is referred to as a failed arm. An

example of a failed arm is the Benue rift on the western margin of Africa

shown in Figure 1–53. The other two arms of this system became part of the

early mid-Atlantic ridge at which the Atlantic Ocean formed. The failed arm

of the system eventually became filled with sediments; the sediment-filled

fossil rift is known as an aulacogen.

The second stage of continent splitting is the formation of a seafloor-

spreading center, or ocean ridge. This is illustrated in Figure 1–42b. The

normal faults associated with the margins of the rift valley now form the

margins of a new ocean. Upwelling hot mantle rock partially melts to form

new ocean crust and the first stages of an ocean ridge. An example of an

ocean at this early stage of development is the Red Sea (Figure 1–43). As

seafloor spreading continues at the spreading center, an ocean is formed.

Because the creation of new seafloor at an ocean ridge is very nearly a

symmetric process, the ocean ridge bisects the newly created ocean. This is

illustrated in Figure 1–42c. An example is the Atlantic Ocean. The margins

of the opening ocean are known as passive continental margins in contrast

to the active continental margins where subduction is occurring.

As the seafloor at the continental margin grows older, the lithosphere

becomes thicker and more dense. Eventually the lithosphere becomes suf-

ficiently unstable so that it founders and an ocean trench develops and

subduction begins. This is illustrated in Figure 1–42d. Trenches apparently

form immediately adjacent to one of the continents. This is the site of the

oldest, coldest, and most unstable oceanic lithosphere. Also, the continental

margin is inherently a zone of weakness. As the ocean basin adjacent to a

continent grows older, it continues to subside relative to the continent. This

differential subsidence is accommodated on the normal faults associated with

the continental margin. These normal faults are zones of weakness, and they

may play a key role in the formation of new ocean trenches, when a passive

continental margin is converted to an active continental margin.

If the rate of subduction is greater than the rate of seafloor spreading,

the size of the ocean will decrease. Eventually the ocean ridge itself will be

subducted (see Figure 1–42e). Ridge subduction is occurring along the west

coast of North America. The remanents of the Juan de Fuca ridge form the

boundary between the Juan de Fuca plate and the Pacific plate (Figure 1–1).

The northern part of this ridge was subducted beneath the Aleutian trench.

Other parts of the ridge were subducted off the west coast of California. In

these cases, the subduction led to the transformation of the convergent plate

boundaries between the North American plate and the Juan de Fuca plate

1.10 The Wilson Cycle 67

F ig

u re

1. 42

T h e

W il so

n cy

cl e.

68 Plate Tectonics

Figure 1.43 Satellite photograph of the Red Sea (NASA STS040-078-088).

(also known as the Farallon plate) to the present transform fault boundaries

between the North American and Pacific plates.

After ridge subduction, the remainder of the oceanic plate will be sub-

ducted and the continents will collide (Figure 1–42f). The implications of a

continental collision are discussed in the next section.

Evidence of the past motions of the continents comes from many sources.

The distribution of magnetic lineations on the seafloor can be used to re-

construct the positions of the continents for about the last 150 Ma. Because

there is very little seafloor older than this, reconstructions prior to about

150 Ma are primarily based on paleomagnetic measurements in continental

rocks. Many other sources of information contribute to paleoreconstructions.

Dated orogenic events provide information on the locations of ocean trenches

and continental collision zones. The spatial distributions of fossils, glacia-

tions, and morphological features provide additional latitude control.

Continental reconstructions for the last 170 Ma are given in Figure 1–44.

1.10 The Wilson Cycle 69

Figure 1.44 Continental reconstructions. (a) 170 Ma. Continued on pp. 42–3. (Smith et al., 1974)

Reassembly of the continents clearly resembles the construction of a jigsaw

puzzle. Not only does South America fit against Africa, but Australia can be

fit together with Southeast Asia and Antarctica as well as Greenland with

North America and Europe. Continental reconstructions can be extended

even farther back in time, but the uncertainties become large.

At 170 Ma the supercontinent Pangaea was being rifted to form the north-

ern continent Laurasia (composed of North America, Europe, and Asia) and

the southern continent Gondwanaland composed of South America, Africa,

India, Australia, and parts of Antarctica and Southeast Asia. Between these

continents the Tethys Ocean was being formed. Between 170 and 100 Ma

the central Atlantic Ocean began to form as North America rotated away

from Africa. Simultaneously the Tethys Ocean was closing. Between 100 and

50 Ma the Atlantic Ocean continued to form and the Indian Ocean formed

70 Plate Tectonics

Figure 1.44 (cont.) (b) 100 Ma. (Smith et al., 1974)

as Australia and Antarctica rotated away from Africa. The Tethys Ocean

continued to close. In the last 50 Ma the Atlantic has continued to open,

India has collided with Eurasia, and Australia has moved northward relative

to Antarctica. Today the Mediterranean Sea, Black Sea, and Caspian Sea

are the only relics of the Tethys Ocean.

1.11 Continental Collisions

The collision of two continents when an ocean closes is a major cause of

mountain building. At present a continental collision is occurring along a

large fraction of the southern boundary of the Eurasian plate. The style of

this collision varies considerably from west to east. The mountain building

associated with a continental collision is referred to as an orogeny. The re-

gion where mountain building is occurring is referred to as an orogenic zone.

1.11 Continental Collisions 71

F ig

u re

1. 44

(c o n t. )

(c )

72 Plate Tectonics

The collision between the Eurasian and African plates has resulted in the

formation of the Alps; this is a relatively subdued continental collision and

the Alpine orogenic zone is relatively narrow. One model for this collision

is illustrated in the cross section given in Figure 1–45. A flake of the upper

continental crust of the Eurasian plate has overridden the continental crust

of the African plate. The forces associated with the southward dipping sub-

duction of the Eurasian plate has driven the upper Eurasian crust several

hundred kilometers over the African crust. The lower Eurasian crust has

been delaminated and is being subducted into the mantle with the Eurasian

lithosphere. The underlying African crust is exposed through the overly-

ing upper Eurasian crust at several points in the Alps. The splitting of the

Eurasian crust at a depth of about 15 km requires an intracrustal decolle-

ment. This type of splitting is often observed in the geological record and is

attributed to a soft crustal rheology at intermediate crustal depths.

The continental collision between the Eurasian and the Indian plates has

resulted in a much broader orogenic zone that extends throughout much of

China. This orogenic zone is illustrated in Figure 1–46. The collision has

resulted in the formation of the Himalayas, the highest and most extensive

mountain range in the world. A satellite photograph of the Himalayas look-

ing to the northwest is given in Figure 1–47. Imbedded within the Himalayas

is the Indus suture, the actual boundary between the Indian plate and the

Eurasian plate. The Tibetan plateau is a broad region of elevated topogra-

phy with extensive faulting but very little recent volcanism. Faulting extends

throughout much of China. A substantial fraction of the largest historical

earthquakes has occurred on these faults and in many cases the death toll

has been very high. Reports claimed that there were 655,000 deaths during

the Tangshan earthquake of July 28, 1976. The largest reported death toll in

an earthquake was the 800,000 deaths attributed to the Shensi earthquake

on January 23, 1556.

It is necessary to explain why this orogenic zone is so broad and why the

orogeny is principally on the Eurasian plate with relatively little deformation

on the Indian plate. One explanation for the asymmetric deformation is that

the Eurasian lithosphere in Tibet and China was thin and weak prior to the

collision. This area may have resembled the present western United States

which has a weak and easily deformable lithosphere. A simplified model for

this continental collision is given in Figure 1–48. The Indian continental

crust and lithosphere have been thrust beneath the Eurasian crust across

the entire width of the Tibetan plateau.

Continental collisions can produce large amounts of horizontal strain. It

is estimated that the original continental crust in the Himalayas has been

1.11 Continental Collisions 73

Figure 1.45 Cross section of the Alpine orogenic zone (after Schmid et al., 1997).

shortened by 300 km or more. Strain in the crust is accommodated by both

brittle and ductile mechanisms. The brittle upper crust can be compressed

and thickened by displacements on a series of thrust faults that form a thrust

belt such as that illustrated in Figure 1–49; each of the upthrust blocks

forms a mountain range. Sedimentary basins often form over the downthrust

blocks. In the Wyoming thrust belt these sedimentary basins are the sites of

major oil fields. Crustal thickening and shortening resulting from thrusting

during the collision of India and Asia are illustrated in Figure 1–48. In

some cases the entire brittle part of the continental crust is thrust over

the adjacent continental crust as a thrust sheet. Evidence indicates that a

thrust sheet in the southern Appalachian Mountains extends over hundreds

of kilometers. This structure is associated with the continental collision that

occurred when the proto-Atlantic ocean closed at about 250 Ma.

The crust can also be compressed by ductile deformation, one result of

which is folding. The convex upward or top of a fold is known as an anticline,

and the concave upward or bottom of a fold is known as a syncline. On a large

scale these are known as anticlinoria and synclinoria. Folding is illustrated

in Figure 1–50. When a region of large-scale folding is eroded, the easily

eroded strata form valleys, and the resistant stata form ridges. This type

of valley and ridge topography is found in Pennsylvania and West Virginia

(Figure 1–51). The ridges are primarily sandstone, and the valleys are the

74 Plate Tectonics

F igu

re 1.46

Illu stra

tio n

o f th

e o rogen

ic zo

n e

resu ltin

g fro

m th

e co

n tin

en ta

l co

llisio n

betw een

th e

E u ra

sia n

a n d

In d ia

n p la

tes. T

h e

m a jo

r fa

u lts

a n d

zo n es

o f vo

lca n ism

a re

sh o w n . T

h e

In d u s

su tu

re is

th e

p ro

ba ble

bo u n d a ry

betw een

th e

p la

tes (a

fter T a p po

n ier

a n d

M o ln

a r,

1 9 7 7 ).

1.11 Continental Collisions 75

Figure 1.47 Satellite photograph of the Himalayas and the Tibetan plateau (NASA STS 41G-120-0022).

result of the erosion of shales. The large-scale folding in this area was also

the result of the continental collision that occurred at about 250 Ma. An

extreme amount of deformation occurs in the formation of nappes. A nappe

may be either a thrust sheet or a recumbent fold, that is, a fold whose limbs

are almost parallel and roughly horizontal (Figure 1–45).

The effects of continental collisions appear to vary widely. The collision

between India and Asia is responsible not only for the Himalaya Mountains

but also for tectonics and seismicity throughout China. In contrast, the

Alpine orogeny in Europe is narrowly confined. There is also observational

evidence that the continental collision that resulted in the formation of the

Appalachian Mountains was relatively mild. This difference in collisional

orogenies may be attributed to the characteristics of the orogenic zone prior

to collision. China may have resembled the western United States; that is, its

lithosphere may have been very thin prior to collision, and it may therefore

have been easily deformed during the collision.

76 Plate Tectonics

Figure 1.48 A schematic cross section of the Himalaya and the southern Tibetan plateau showing underthrusting of the Indian continental crust and lithosphere beneath the Eurasian crust. MBT, Main Boundary Thrust; MCT, Main Central Thrust; STD, South Tibetan Detachment (after Zhao and Nelson, 1993).

Figure 1.49 Horizontal compression resulting in continental collision and a series of thrust faults. Each uplifted block creates a mountain range and each downdropped block a sedimentary basin.

1.12 Volcanism and Heat Flow

As we have previously discussed, volcanism is associated with both accre-

tionary plate margins and subduction zones. The worldwide distribution of

active volcanoes is shown in Figure 1–52. Clearly, most volcanoes are associ-

ated with plate margins. Continuous volcanism occurs along the worldwide

ocean ridge system, where it forms the 6-km-thick basaltic oceanic crust.

Only a small fraction of this volcanism is included in the distribution of

active volcanoes given in Figure 1–52; the remainder occurs on the seafloor,

where it cannot be readily observed. The linear chains of active volcanoes

associated with ocean trenches are clearly illustrated in Figure 1–52. How-

ever, significant gaps in the chains do occur even when active subduction is

going on. Examples are in Peru and central Chile.

1.12 Volcanism and Heat Flow 77

Figure 1.50 Large-scale folding resulting from horizontal compression. The easily eroded strata form valleys, and the resistive strata form ridges.

While a large fraction of the Earth’s volcanism can be directly attributed

to plate boundaries, there are many exceptions. The most obvious example

is the volcanism of the Hawaiian Islands. This intraplate volcanism occurs

near the center of the Pacific plate. As discussed in Section 1–6 centers

of intraplate volcanism are referred to as hotspots. The locations of thirty

hotspots are given in Table 1–2 and the distribution of twenty hotspots

is shown on the map in Figure 1–14. These are both intraplate hotspots

and hotspots located on or near oceanic ridges. One example of a hotspot

on an oceanic ridge is Iceland, where very high rates of volcanism have

produced anomalously thick oceanic crust. Other examples are the Azores

and Galapagos Islands, where anomalous volcanism has produced groups of

islands near an ocean ridge.

In many ways hotspot volcanics are notable for their differences rather

than their similarities. We will now discuss in somewhat more detail the

hotspot volcanics of Africa and western North America. The distribution

of volcanic rocks in Africa that are younger than 26 million years is shown

in Figure 1–53. Active volcanism is occurring throughout much of Africa.

The East African rift system is a nearly linear feature extending southwest

several thousand kilometers from its junction with the Red Sea and the

Gulf of Aden. The rift is characterized by tensional tectonics and horizontal

extension. As discussed previously, the rift may represent the first stage of

continental rupture.

The East African rift is also characterized by near-circular regions of el-

evated topography referred to as swells. The relationship of these features

to the rifting process is uncertain. They may be associated with convec-

tive plumes in the mantle. An alternative hypothesis for continental rifts is

that they are the direct result of tensional stresses in the lithosphere. The

78 Plate Tectonics

Figure 1.51 Space shuttle radar image of the Appalachian foldbelt in Penn- sylvania. The more resistant strata such as sandstones form the narrow ridges (NASA PIA01306).

tensional stresses cause the continental lithosphere to rupture, leading to

volcanism and uplift.

Other areas of extensive volcanism in Africa are the Tibesti area in north-

east Chad and Hoggar to the west. In addition to volcanism these areas are

associated with crustal swells; however, they do not appear to be associated

with any linear features. The Haruj volcanics north of Tibesti are not as-

sociated with any apparent crustal elevation. Adjacent to the continental

margin in the Gulf of Guinea is the Cameroon line of recent volcanics. This

is a series of active volcanic centers that pass from oceanic onto continental

crust. Although this is a linear chain of volcanic centers, the linear progres-

sion in time is not well defined. Farther to the north on the continental

margin of Africa lie the Canary Islands. In this group of volcanic islands,

volcanism has been centered for a long period.

It is evident that very diverse types of intraplate volcanism occur in Africa.

1.12 Volcanism and Heat Flow 79

Figure 1.52 Distribution of active volcanoes in the Quaternary.

It is not clear whether all this volcanism can be attributed to a single mech-

anism. If mantle plumes are responsible for this volcanism, how many are

required? Why does the volcanism in Africa have such a variety of forms?

The western United States is another area of extensive volcanism. The

distribution of recent surface volcanic rocks (with ages of less than 7 million

years) is given in Figure 1–54. Because the San Andreas fault in California

is recognized as a major plate boundary between the Pacific and North

American plates, the volcanism of this area may be classified as being plate-

margin-related; however, the volcanism extends more than 1500 km from

the plate margin.

Yellowstone National Park in the northwest corner of Wyoming is the

center of extensive recent volcanism that occurs at the end of a track of

volcanism that extends along the Snake River plain. For this reason the

Yellowstone area is classified as a hotspot (see Figure 1–14), and it is thus

a possible site of a mantle plume. The ages of surface volcanic rocks on the

Snake River plain are given as a function of the distance from Yellowstone

in Figure 1–55. The ages of the oldest volcanic rocks tend to increase with

distance from Yellowstone; however, young volcanic rocks occur along much

of the length of the Snake River plain. It is difficult to associate these young

volcanics, which extend over a distance of some 500 km, with a single melting

80 Plate Tectonics

anomaly beneath Yellowstone. Also, it is clear from Figure 1–54 that very

young volcanics extend throughout the western United States. Some of these

volcanics form the volcanic line associated with subduction of the Juan

de Fuca plate beneath Washington, Oregon, and northern California. But

much volcanism remains unexplained. Small amounts of volcanism appear

to be associated with the tensional tectonics of the Rio Grande rift in New

Mexico and the Basin and Range province in Nevada and Arizona. Further

discussion of this is given in the next section.

Variations in surface heat flow can also be correlated with the distribution

of plates. On a worldwide basis the average surface heat flow is about 87

mW m−2. For rocks with an average thermal conductivity this corresponds

to an increase in temperature with depth of 25 K km−1. The heat being lost

to the surface of the Earth can be attributed to the heat produced by the

decay of the radioactive isotopes and to the cooling of the Earth.

Plate margins and other areas where volcanism occurs are generally char-

acterized by surface heat flows much higher than the average value just

given. The high heat flows may be the result of a thin lithosphere or, in

some cases, the migration of magma through a relatively thick lithosphere.

The cooling of the oceanic lithosphere as it spreads from an oceanic ridge

leads to a systematic decrease with age in the heat flux to the seafloor.

1.12 Volcanism and Heat Flow 81

Figure 1.53 Distribution of volcanic rocks in Africa. Dark areas are surface volcanic rocks with ages less than 26 Ma. Also shown are some of the tectonic provinces and areas of crustal doming.

82 Plate Tectonics

Figure 1.54 Distribution of recent volcanic rocks in the western United States. Dark areas are rocks younger than 1.5 Ma; shaded areas are rocks with ages between 1.5 and 7 Ma.

1.12 Volcanism and Heat Flow 83

Figure 1.55 Ages of volcanic rocks in the Snake River plain as a function of the distance from Yellowstone Caldera.

Figure 1.56 Old Faithful Geyser in Yellowstone National Park, Wyoming (J. R. Stacy 692, U.S. Geological Survey).

84 Plate Tectonics

Figure 1.57 Global distribution of earthquakes with magnitudes greater than 5.1 for the period 1964–1995 (Engdahl et al., 1998).

The occurrence of hot springs is also strongly correlated with volcanism.

In continental areas with no volcanism, the temperatures of springs seldom

exceed 293 K. Most boiling hot springs and geysers (Figure 1–56) are di-

rectly associated with the cooling of a magma body at a relatively shallow

depth. The circulation of heated ground water near a cooling intrusion ac-

celerates the solidification of the intrusion and plays an important role in

the emplacement of ore deposits. Minerals dissolve in the hot water; when

the water boils or is cooled, the minerals precipitate to form ore deposits.

Hydrothermal circulation of seawater in the oceanic crust also is believed to

play a significant role in removing the heat at ocean ridges and in concen-

trating minerals on the seafloor and in the oceanic crust. Exploration with

the deep submersible Alvin has provided actual observations of hot water

from the crust venting directly to the ocean. Submarine thermal springs

have been discovered on the Galápagos rift and the East Pacific rise crest.

Spectacular submarine hot springs with exit water temperatures near 700

K have also been discovered at the latter location.

1.13 Seismicity and the State of Stress in the Lithosphere 85

Figure 1.58 View along the San Andreas fault in Choia Valley. Note the streams have been offset by the right-lateral displacement on the fault (R. E. Wallace 304, U.S. Geological Survey).

1.13 Seismicity and the State of Stress in the Lithosphere

Just as in the case of volcanoes, the occurrences of earthquakes strongly

correlate with plate boundaries. The worldwide distribution of seismicity is

illustrated in Figure 1–57. Earthquakes occur on all types of plate bound-

aries; however, subduction zones and areas of continental collision are noted

for their very large earthquakes. Large earthquakes also occur in plate inte-

riors but with a much lower frequency.

Earthquakes are associated with displacements on preexisting faults. A

typical displacement in a very large earthquake is 10 m. If the relative ve-

locity across a plate boundary is 50 mm yr−1, it would take 200 years to

accumulate this displacement. Large earthquakes at subduction zones and

major transform faults such as the San Andreas reoccur in about this pe-

riod. Since regular displacements do not have to be accommodated in plate

interiors, the period between major intraplate earthquakes is much longer.

The near-surface expressions of major faults are broad zones of fractured

rock with a width of a kilometer or more (Figure 1–58). Smaller faults may

have zones of fault gouge with widths of a few centimeters or less. Small

86 Plate Tectonics

Figure 1.59 Cross sections of (a) a normal fault and (b) a thrust fault and top views of (c) right-lateral and (d) left-lateral strike-slip faults.

faults grade down to rock fractures across which there is no offset displace-

ment. The total offset across major faults may be hundreds of kilometers.

A fault zone is a zone of weakness. When the regional stress level becomes

sufficiently large, the fault ruptures and an earthquake occurs. There is ex-

tensive geological evidence that faults become reactivated. Large stresses

can reactivate faults that have been inactive for tens or hundreds of millions

of years.

The direction of the displacement on a fault can be used to determine the

state of stress responsible for the displacement. Since voids cannot be created

in the Earth’s deep interior, displacements on faults are parallel to the fault

surface. If a region is in a state of tensional stress, normal faulting will occur,

as illustrated in Figure 1–59a. If a region is in a state of compressional stress,

thrust faulting will occur, as illustrated in Figure 1–59b.

If a region is in a state of shear, strike–slip faulting will occur, as illus-

trated in Figures 1–59c and 1–59d. If, to an observer standing on one side of

the fault, the motion on the other side of the fault is to the left, the fault is a

left-lateral or sinistral strike–slip fault. If the motion on the other side of the

fault is to the right, it is a right-lateral or dextral strike–slip fault. The dis-

placement during many earthquakes combines the horizontal displacement

associated with strike–slip faulting and the vertical displacement associated

with either normal or thrust faulting.

As discussed previously, the lithosphere can transmit stress over large dis-

tances. There are several sources for the stress in the lithosphere. One source

is the body forces that drive the motion of the surface plates. These include

the negative buoyancy on the descending plate at a subduction zone and the

1.13 Seismicity and the State of Stress in the Lithosphere 87

gravitational sliding of a plate off an ocean ridge. Changes of temperature

lead to thermal stresses. When the temperature increases or decreases, rock

expands or contracts. The expansion or contraction can cause very large

stresses. Erosion and sedimentation also cause a buildup of stress through

the addition or removal of surface loads. Glaciation and deglaciation act sim-

ilarly. Because the Earth is not a perfect sphere but rather a spheroid with

polar flattening and an equatorial bulge, plates must deform as they change

latitude. This deformation leads to membrane stresses in the lithosphere.

Plate interactions such as continental collisions are an important source of

stress. Large displacements of the cool, near-surface rocks often occur in

these zones. If these deformations occur on faults, high stress levels and ma-

jor earthquakes can be expected. The state of stress in the lithosphere is the

result of all these factors and other contributions.

As a specific example of seismicity and stress we again turn to the west-

ern United States. The distribution of seismicity in this region is given in

Figure 1–60. Also included in the figure are the relative velocities between

plates and the directions of lithospheric stress inferred from displacements

on faults. The Juan de Fuca plate is being formed on the Juan de Fuca

ridge with a half-spreading rate of 29 mm yr−1. The seismicity on a trans-

form fault offsetting two segments of the ridge is clearly illustrated. Because

the lithosphere is thin at the ridge and the rocks are hot and weak, rela-

tively little seismicity is associated with the spreading center at the ridge

crest. The Juan de Fuca plate is being subducted at a rate of about 15

mm yr−1 at a trench along the Oregon and Washington coasts. The seis-

micity in Oregon and Washington associated with this subduction is also

relatively weak. This may be due to aseismic slip on the fault zone between

the descending oceanic plate and the overlying continental lithosphere. In

this case the relative displacement is accommodated without the buildup of

the large stresses required for extensive seismicity. An alternative explana-

tion is that the accumulated displacement was relieved in a great earthquake

several hundred years ago and insufficient displacement has accumulated to

cause high stresses. The historic record of earthquakes in the western United

States is relatively short, and since the subduction velocity is quite small,

the recurrence period would be expected to be of the order of 500 years.

Although the extensive seismicity usually associated with active subduction

is absent in the Pacific Northwest, a well-defined line of active volcanoes lies

parallel to the trench. The volcanoes extend from Mount Baker in Wash-

ington to Mount Shasta in northern California. These volcanoes have had

violent eruptions throughout the recent geological past. An eruption about

6000 years ago removed the upper 2 km of Mount Mazama, creating Crater

88 Plate Tectonics

Figure 1.60 Distribution of seismicity in the geological provinces of the western United States (stars). Solid arrows give relative plate velocities; open arrows give stress directions inferred from seismic focal mechanism studies.

Lake in Oregon. The spectacular eruption of Mount St. Helens, Washington

(Figure 1–10), on May 18, 1980, blew out its entire north flank, a volume of

about 6 km3.

The velocity between the Pacific and North American plates is 47 mm yr−1

in California. A large fraction of this is accommodated by displacements on

the San Andreas fault. In the north the fault terminates in a fault–fault–

trench (FFT) triple junction with the Juan de Fuca plate. In the south the

fault terminates in a series of small spreading centers (ocean ridges) extend-

ing down the Gulf of California. Along much of the fault, displacements are

almost entirely right-lateral strike–slip. However, north of Los Angeles the

fault bends, introducing a thrusting component. Motion on thrust faults in

1.13 Seismicity and the State of Stress in the Lithosphere 89

this region is responsible for the formation of a series of mountain ranges

known as the Transverse Ranges.

A great earthquake occurred on the northern section of the San Andreas

fault in 1906; the average displacement was 4 m. A great earthquake oc-

curred on the southern section of the San Andreas fault in 1857; the average

displacement was 7 m. A detailed discussion of the San Andreas fault is

given in Section 8–8.

It is clear that the displacements on accreting plate margins, subduction

zones, and transform faults cannot explain the entire distribution of seismic-

ity in the western United States. Major earthquakes occur throughout the

region. Rapid mountain building is associated with the Rocky Mountains

and the Sierra Nevada. The Basin and Range province is a region of exten-

sive normal faulting. The presence of many graben structures is evidence

of crustal extension due to tensional stresses. The asthenosphere rises to

the base of the continental crust in this region, and the lithosphere is thin

and weak. Considerable volcanism occurs throughout the province. The Rio

Grande rift, which marks the eastern boundary of this area of volcanism,

seismicity, and mountain building, is also an extensional feature. The stress

directions shown in Figure 1–60 indicate the entire western United States

appears to be extending because of tensional stresses. Although there is no

comprehensive understanding of this area, it is likely that the seismicity, vol-

canism, and mountain building are the result of complex interactions of the

Pacific, North American, and Juan de Fuca plates that are deforming the

entire region. It is likely that there is a geometrical incompatibility between

the strike–slip motion on the San Andreas fault and the time-dependent

relative displacement between the Pacific and North American plates. As a

result the western part of the North American plate is being deformed.

China is another region of extensive tectonics. It is the site of extensive

seismicity and mountain building. Deformation associated with the continen-

tal collision between India and Asia extends several thousands of kilometers

north of the suture zone.

Seismicity can also occur in plate interiors. An example is New Madrid,

Missouri, where three very large earthquakes struck in 1811 and 1812. A

significant number of small earthquakes occur in this region at the present

time. It should not be surprising that earthquakes occur in plate interiors,

since the elastic lithosphere can transmit large stresses. These intraplate

earthquakes are likely to occur where the elastic properties of the plate

change and stress concentrates.

90 Plate Tectonics

1.14 The Driving Mechanism

Plate tectonics provides a general framework for understanding the world-

wide distribution of seismicity, volcanism, and mountain building. These

phenomena are largely associated with plate interactions at plate margins.

However, an explanation must also be given for the relative motions of the

plates. The basic mechanism responsible for plate tectonics must provide

the energy for the earthquakes, volcanism, and mountain building. The only

source of energy of sufficient magnitude is heat from the interior of the

Earth. This heat is the result of the radioactive decay of the uranium iso-

topes 238U and 235U, the thorium isotope 232Th, and the potassium isotope 40K as well as the cooling of the Earth. An accurate estimate of the heat

lost from the interior of the Earth can be obtained from measurements of

the surface heat flow. The energy associated with seismicity, volcanism, and

mountain building is about 1% of the heat flow to the surface.

Heat can be converted to directed motion by thermal convection. Consider

a horizontal fluid layer in a gravitational field that is heated from within and

cooled from above. The cool fluid near the upper boundary is heavier than

the hotter fluid in the layer. Buoyancy forces cause the cool fluid to sink,

and it is replaced by hot rising fluid. Laboratory experiments show that

under appropriate conditions two-dimensional convection cells develop, as

illustrated in Figure 1–61. A thin thermal boundary layer of cool fluid forms

adjacent to the upper boundary of the layer. Thermal boundary layers from

two adjacent cells merge and separate from the upper boundary to form a

cool descending plume. The negative buoyancy of the cool descending plume

drives the flow. The thin thermal boundary layer is directly analogous to the

lithosphere. The separation of the thermal boundary layers to form the cool

descending plume is analogous to subduction. The buoyancy body force on

the cool descending plume is analogous to the body force on the descending

lithosphere.

Ascending mantle plumes can also be associated with thermal convection.

There is expected to be a hot thermal boundary layer at the base of the

convecting mantle. Buoyancy forces on the low-density mantle rock would

be expected to form hot ascending plumes either at the base of the mantle

(the D′′-layer of seismology) or at an interface in the lower mantle between

an upper convecting layer and an isolated lower layer.

The fluidlike behavior of the Earth’s crystalline mantle can be explained

by solid-state creep processes. At low stress levels and temperatures ap-

proaching the rock solidus, the dominant creep process is the diffusion of

ions and vacancies through the crystal lattice. This diffusion relieves an

1.15 Comparative Planetology 91

Figure 1.61 Boundary layer model for two-dimensional thermal convection in a fluid layer heated from within and cooled from above.

applied stress and results in strain. The strain rate is proportional to the

stress, resulting in a Newtonian fluid behavior. At higher stress levels creep

is the result of the movement of dislocations through the crystal lattice. In

this case, the strain rate is proportional to the third power of the stress,

resulting in a non-Newtonian fluid behavior. Both diffusion creep and dislo-

cation creep are thermally activated; that is, the strain rates are inversely

proportional to the exponential of the inverse absolute temperature.

The strain rate in the mantle is so small that it cannot be reproduced

in the laboratory. However, extrapolations of laboratory measurements give

fluid properties consistent with thermal convection in the mantle. Indepen-

dent information on the fluid behavior of the mantle comes from studies

of glacial loading and unloading. When an ice sheet forms, its weight forces

mantle rock to flow from beneath it, and the Earth’s surface subsides. When

the ice sheet melts, the mantle rock returns, and the Earth’s surface re-

bounds. It takes about 10,000 years for the rebound to take place. Data on

the rate of rebound come from dated, elevated beaches. These data have

been used to obtain a viscosity for the mantle of about 1021 Pa s. Although

this viscosity is large, it allows a fluid behavior of the mantle on geological

time scales.

1.15 Comparative Planetology

Space missions have provided extensive information on the other planets and

the planetary satellites of the solar system. It appears that plate tectonics is

92 Plate Tectonics

unique to the Earth. The Moon and Mercury have continuous lithospheres

whose surfaces have been shaped largely by impacts and volcanic processes.

Although impact cratering and volcanism have also been prevalent on Mars,

its surface has also been modified by its atmosphere and the flow of a surface

fluid, presumably water. Cloud-covered Venus has yielded its secrets to the

eyes of Earth-based and spacecraft radar systems. Cratering and volcanism

have extensively modified its surface, but there is no direct evidence of plate

tectonic features such as extensive ridge or trench systems. The Galilean

satellites of Jupiter have been shown to have very diverse features including

very active volcanism on Io. The surface of Ganymede shows impact craters

and tectonic structures resulting from dynamical processes in an underlying

predominantly ice lithosphere. Callisto is a heavily cratered object about the

same size as Ganymede, but there is no sign that its surface has been altered

by internal activity. Europa is mainly a rocky object, somewhat smaller than

the Moon, with a relatively thin outer shell of water that is ice at the surface

but may be liquid at depth.

The surface of Io has been recently formed by a style of volcanism ap-

parently unique to that body. Io is the only body in the solar system, other

than the Earth, on which we have observed active volcanism; this satellite

is the most volcanically active body in the solar system. Lithospheric plate

evolution has destroyed much of the evidence of the early evolution of the

Earth by continuously creating new surface rocks and returning old surface

rocks to the mantle. The pervasive volcanism of Io has had the same effect

by blanketing the surface with recently formed lavas. However, bodies such

as Mercury, the Moon, Callisto, and the satellites of Mars preserve the early

records of their evolutions in their cratered surfaces and thereby provide

information on the early history of the solar system. Some of the physical

properties of the terrestrial-like bodies of the solar system are summarized

in Appendix 2, Section C.

1.16 The Moon

Early telescopic observations showed that the near side of the Moon is com-

posed of two types of surfaces: topographically low, dark areas-referred to as

maria (or seas) and topographically elevated, light areas known as highlands

(Figure 1–62). The highlands are more heavily cratered and are therefore

presumed to be older because the flux of meteorities is known to have de-

creased with time. Because of its synchronous rotation with respect to the

Earth, the far side of the Moon was first observed from spacecraft in lunar

1.16 The Moon 93

Figure 1.62 Full hemispheric image of the Moon taken by the Galileo space- craft on December 7, 1992, on its way to explore the Jupiter system in 1995–97. The distinct bright ray crater at the bottom of the image is the Tycho impact basin. The dark areas are basaltic-rock-filled impact basins: Oceanus Procellarum (on the left), Mare Imbrium (center left), Mare Seren- itatis and Mare Tranquillitatis (center), and Mare Crisium (near the right edge) (NASA Image PIA00405).

orbit. Highland terrain dominates the far side of the Moon; there are no

extensive maria on the farside lunar surface.

The first manned landing on the Moon took place on July 20, 1969. This

Apollo 11 mission as well as the subsequent Apollo 12, 15, and 17 missions

landed on the lunar maria. Chemical studies of the rocks returned on these

missions showed that the maria are composed of basaltic rocks similar in

major element chemistry to the basalts of the oceanic crust. Radioactive

dating of these rocks gives ages of 3.16 to 3.9 Ga.

The Apollo 14 and 16 missions returned samples from the lunar high-

lands. These rocks have a much more complex chemical history than the

mare rocks. They have been extensively shocked and melted by meteorite

bombardment. Detailed chemical studies have shown, however, that these

rocks are highly fractionated igneous rocks. Radioactive dating of the high-

land rocks indicates that they crystallized about 4.5 Ga ago, close to the

estimated age of the solar system.

The evolution of the Moon can be divided into three phases: (1) highlands

94 Plate Tectonics

formation, (2) mare formation, and (3) surface quiescence. The highlands of

the Moon formed early in its evolution, one hypothesis being that they crys-

tallized from a global magma ocean. It is difficult to date the crystallization

of the highlands exactly, but it certainly terminated by 4.0 Ga and probably

before 4.4 Ga. The terminal bombardment between 3.8 and 4.0 Ga resulted

in the excavation of many large, deep basins. These basins, particularly on

the nearside, were subsequently filled by mare basaltic volcanism. The lunar

maria constitute some 17% of the surface area.

One of the major discoveries of the Apollo missions was that the Moon

is made up of a variety of igneous rock types that differ widely in both

their chemistry and mineral composition. The major differences between

the lunar maria and highlands indicate large-scale chemical differentiation

of the Moon. Early recognition of the fact that the highlands are composed

mostly of plagioclase, a relatively light mineral, led to the suggestion that

this mineral represents crystal flotation at the top of a deep magma ocean.

The basic argument for a “magma ocean” is the need for a mechanism

to float a plagioclase-rich crust, while denser minerals such as olivine and

pyroxene sink. As the Moon formed, its outer portions consisted of a layer

of molten silicate magma in which plagioclase floated and accumulated into

the first stable lunar crust.

The solidification of the magma ocean must have occurred in about 100

Myr after the formation of the solar system because of the age of the re-

turned lunar samples. Seismic studies carried out on the Apollo missions

showed that the lunar crust has a thickness between 60 and 100 km. Seis-

mic velocities and the mean density of the Moon indicate that the lunar

mantle is composed primarily of peridotite similar to the Earth’s mantle.

It is hypothesized that the lunar crust represents about a 20% partial melt

fraction of a primitive lunar mantle with a composition similar to pyrolite.

It is expected that there is a layer of depleted mantle rock beneath the lunar

crust with a thickness of about 300–500 km.

Subsequent to the solidification of the magma ocean, the morphology

of the lunar surface was strongly affected by collisions with the remaining

planetesimals and large meteorites. These collisions created large basins; the

largest of the colliding bodies created the Imbrium basin, an event that has

been dated at 3.86 Ga. A period of volcanism lasting 1 Gyr then filled the

floors of these preexisting impact basins with the dark basaltic rocks that

form the lunar maria. This volcanism terminated at about 3 Ga. Since then

the lunar surface has remained virtually unaltered.

All of the smooth dark regions visible on the Moon’s nearside consist

of basaltic rocks that partly or completely fill the multiring mare basins.

1.16 The Moon 95

Nearly all of the basalts occur on the nearside. A significant time interval

elapsed between the formation of a large mare basin by impact and its

subsequent filling with basaltic magma flows to form a dark lunar maria.

Current information dates the Imbrium basin at 3.86 Ga, but the lavas that

fill it date at about 3.3 Ga. The primary landforms resulting from lunar

basaltic volcanism are vast, smooth plains, indicating low lava viscosities

and high eruption rates. Major basaltic eruptions lasted a minimum of 800

million years, i.e., from 3.9 to 3.1 Ga. On the basis of low crater densities

on some formations, minor eruptions could have continued until as recently

as 2 Ga.

Although lunar rocks are similar to igneous rocks on the Earth, there are

significant differences between the two bodies. Unlike the Earth, the Moon

does not have a large iron core. The Moon may have a small iron core,

but its radius is constrained by the measured values of lunar mass, radius,

and moment of inertia to have a value less than about 350 km. Since the

mean density of the Moon is only 3340 kg m−3, the missing iron cannot

be distributed through the lunar mantle. It is therefore concluded that the

Moon is deficient in metallic iron relative to the Earth. The Moon also has

fewer volatile elements than the Earth; there is no evidence of a significant

presence of water during the evolution of the Moon.

Magnetic field measurements were made by small satellites left in lunar

orbit by the Apollo 15 and 16 missions. Although localized regions of mag-

netized rock were detected by the subsatellites, no global lunar magnetic

field could be measured. A lunar magnetic dipole moment can be no larger

than 1016 A m2. This is nearly seven orders of magnitude smaller than the

Earth’s dipole moment. The absence of a present-day global lunar magnetic

field is presumably due to the absence of an active dynamo in the Moon.

This may indicate that the Moon has no core; on the other hand, a small

lunar core could have cooled, or solidified, sufficiently so that convective

motions in it are no longer possible. It has been suggested that the localized

areas of remanent lunar magnetism were magnetized in the ambient field of

an ancient lunar dynamo.

The Moon is the only body other than the Earth for which we have in situ

determinations of the surface heat flux. Two lunar heat flow measurements

have been made, one on the Apollo 15 mission and the other on Apollo 17.

The measured heat flow values are 20 mW m−2 and 16 mW m−2. Although

these two determinations may not be representative of the average lunar

heat flow, the values are consistent with the Earth’s surface heat loss if the

differences in the sizes of the planets are accounted for.

The lunar gravity field is known quite well from the radio tracking of

96 Plate Tectonics

the many spacecraft that have been placed in lunar orbit. A map of the

Moon’s gravity field is shown in a later chapter (Figure 5–13). The lunar

maria are sites of positive gravity anomalies, or excess concentrations of

mass known as mascons. These surface loads appear to be supported by the

lunar lithosphere, an observation that implies that the Moon’s lithosphere

is thicker and therefore stronger than the Earth’s. The Earth’s lithosphere

is not thick enough to support large excess loads – mountains, for example

– with the consequence that such loads tend to depress the lithosphere and

subside. Since the maria were formed by 3 Ga, the Moon’s lithosphere must

have thickened sufficiently by then to support the mascons.

The Moon’s motion about the Earth is prograde; that is, it rotates in

the same sense as the rotation of the planets about the sun. In its present

prograde orbit the tidal interactions between the Earth and the Moon cause

the separation between the bodies to increase; in the past the Moon was

closer to the Earth. Extrapolation of the present rate of tidal dissipation

back in time would bring the Moon to within a few Earth radii of the

Earth between 1.5 and 3 Ga. Since there is little evidence to support a close

approach of the two bodies during this period, it is presumed that the rate

of tidal dissipation in the past has been lower than at present. Nevertheless,

it is highly likely that the Moon has been considerably closer to the Earth

than it is today.

Theories for the origin of the Moon have been debated for more than a

century. The classic theories claim (1) that the Moon formed as a separate

planet and was captured by the Earth, (2) that the Moon was originally

part of the Earth and that the Earth broke into two parts, and (3) that

the Earth and moon formed as a binary planet. None of these theories has

been able to satisfy all the major constraints on lunar origin, which include

the large prograde angular momentum of the Earth–Moon system relative

to the other planets and the Moon’s depletion in volatile elements and iron

compared with the cosmic (chondritic) abundances. Another theory pro-

poses that the Moon formed by accreting from a disc of ejecta orbiting the

Earth after the impact of a Mars-size body with the Earth. The giant im-

pact origin of the Moon has gained widespread support because it does not

violate any of the major observational constraints on lunar origin. One of

the major consequences of the giant impact hypothesis of lunar origin is a

hot, partially molten (or perhaps completely molten) Moon upon accretion

from the circumterrestrial ejecta disk.

1.17 Mercury 97

Figure 1.63 Hemispheric image of Mercury acquired by the Mariner 10 spacecraft on March 24, 1974 (NASA Image PIA00437).

1.17 Mercury

Although it is the smallest of the terrestrial planets, Mercury is the densest

(Appendix 2, Section C). If the planet has the cosmic abundance of heavy

elements, then its large density requires that Mercury is 60% to 70% Fe by

mass. With the iron concentrated in a central core, Mercury could best be

described as a ball of iron surrounded by a thin silicate shell.

In photographs obtained by the Mariner 10 spacecraft during 1974 and

1975 (Figure 1–63), portions of Mercury’s surface strongly resemble the

heavily cratered lunar highlands. In addition, there are large areas of rel-

atively smooth terrain and a number of ringed basins believed to be im-

pact structures. The largest of these is the 1300-km-diameter Caloris basin,

which is similar to the Imbrium and Orientale basins on the Moon. The

Caloris basin is covered with a relatively smooth plains type of material,

perhaps similar to the lunar maria, having many fewer craters than the

heavily cratered terrain. Areas of relatively smooth terrain known as in-

tercrater plains are also found interspaced between the basins and craters.

Lobate scarps, probably curved fault scarps, which are several kilometers

high and extend for hundreds of kilometers across Mercury’s surface, have

98 Plate Tectonics

Figure 1.64 Photomosaic of Mariner 10 images of the Michelangelo Quad- rangle H-12 on Mercury. In addition to the numerous impact craters, ejecta deposits are seen as bright lines or rays radiating outward from some young craters. Several large lobate scarps are visible in the lower left side of the image (NASA Image PIA02237).

no lunar counterpart (Figure 1–64). These scarps are suggestive of thrust

faults resulting from crustal shortening and compression.

Several hypotheses have been advanced to explain the compressional sur-

face features on Mercury. The first hypothesis concerns tidal despining. Early

in its evolution Mercury may have had a rapid rotation. If the planet was

hot it would have had a near hydrostatic shape with considerable polar flat-

tening and an equatorial bulge. As the planet cooled, a global lithosphere

developed with considerable rigidity and ellipticity. However, tidal interac-

tions with the sun gradually slowed the rotation of the planet. The rigidity

of the lithosphere preserved a fossil ellipticity associated with the early rapid

rotation but as a result large lithospheric stresses developed. The resultant

compressional stresses in the equatorial region are one explanation for the

observed compressional features. An alterative explanation is that they were

caused by the formation and/or solidification of the large iron core on Mer-

cury. Mercury’s high mean density of 5440 kg m−3, almost equal to the

Earth’s, is attributed to a large iron core with a 500 to 600 km thick cover

of silicate rocks. One explanation for the high mean density is that a massive

collision blasted off a large fraction of an early mantle of larger size.

Magnetic field measurements by Mariner 10 showed that Mercury has an

intrinsic global magnetic field. Because of the limited amount of data, there

are large uncertainties in the inferred value of Mercury’s magnetic dipole

moment. Most estimates lie in the range of 2 to 5 × 1019 A m2, or about

1.18 Mars 99

5 × 10−4 of the Earth’s magnetic field strength. Although a magnetized

crust cannot be ruled out as a source of this field, it seems more likely that

it originates by dynamo action in a liquid part of Mercury’s core.

Because of the similarities in the surfaces of Mercury and the Moon, their

evolutions must have been similar in several respects. Separation of the

iron and silicates in Mercury and crustal differentiation must have occurred

very early in its history because the planet’s surface preserves an ancient

record of heavy bombardment similar to the lunar highlands. The filling of

the Caloris basin must have occurred subsequent to the termination of this

severe cratering phase because the basin material is relatively free of craters.

The lobate scarps must also have formed at the end of or subsequent to the

early phase of severe bombardment because they sometimes pass through

and deform old craters (Figure 1–64). The scarps may be a consequence of

the cooling and contraction of the core, and if so, they are the only surface

features that distinguish Mercury with its large core from the Moon with

only a very small core or none at all.

1.18 Mars

The first detailed photographs of the Martian surface were returned from the

flybys of Mariner 4 (1965) and Mariners 6 and 7 (1969). These photographs

showed a heavily cratered surface much like that of the Moon. However,

the Mariner 9 (1971) photographs revealed that the earlier spacecraft had

photographed only a single type of terrain on a planet of great geologi-

cal diversity. There are volcanoes that dwarf the largest volcanic structures

on Earth, a huge canyon complex comparable to the East African rift sys-

tem, meandering channels with multiple braided features and stream-lined

islands, sand dunes, and polar caps. The richness and variety of Martian

geologic forms was not fully realized prior to the pictures returned by the

Viking 1 and 2 orbiters and landers (1976) and the Mars orbiter camera

aboard the Mars Global Surveyor (1999). The surface of Mars is character-

ized by a wide variety of volcanic and tectonic landforms. However, there is

no evidence of the global system of ridges and trenches that are character-

istic of active plate tectonics. Thus, it is concluded that Mars does not have

active plate tectonics.

The most striking global feature of the Martian surface is its hemispheric

asymmetry. Much of the southern hemisphere of Mars is covered by densely

cratered highlands, whereas most of the northern hemisphere is made up of

lightly cratered plains. The heavily cratered terrain in the southern hemi-

100 Plate Tectonics

sphere is probably the remnant of the postaccretionary surface of the planet,

and the younger northern plains are probably volcanic in origin.

The southern highlands cover more than 60% of the surface of Mars. Most

of the highland terrain consists of ancient densely cratered rock (largely

impact breccias) formed early in the planet’s history when impact rates were

high. Extensive lava flows have covered large areas within the highlands. The

large, roughly circular basins of Argyre and Hellas (Figure 1–65) are located

in the southern hemisphere and are generally believed to be impact basins

similar to the mare basins on the Moon. The Hellas basin has a rim diameter

of ∼2300 km and is one of the largest impact structures in the solar system.

It is the dominant surface feature of the Martian southern highlands. The

Argyre basin has a diameter in excess of 1500 km. Volcanic plains cover

much of the northern lowlands of Mars. These volcanic plains are similar to

the volcanic plains that dominate other planetary surfaces, e.g., Venus, and

they are much less cratered than the southern highlands.

The approximately hemispherical dichotomy is generally held to be an

ancient, first-order feature of the Martian crust. The dichotomy has been

ascribed variously to a very long-wavelength mantle convective planform,

to subcrustal erosion due to mantle convection, to post-accretional core for-

mation, to one large impact, and to several impact events. Sleep (1994)

has proposed that the lowland crust was formed in an episode of seafloor

spreading on Mars. He hypothesized a hemispheric subduction event that

destroyed the original primitive crust in the northern hemisphere, and pro-

posed a well-defined sequence of seafloor-spreading events that created the

northern volcanic plains.

One of the major volcanotectonic features on Mars is the Tharsis region,

which is a large elevated region composed of relatively young volcanics. The

horizontal scale is some 3000 km with the elevation rising about 10 km above

the mean surface elevation. The region exhibits a complex history of episodic

tectonism, closely associated with volcanism on local and regional scales. The

entire Tharsis uplift appears to be the result of extensive volcanism.

Three immense volcanic shields (Arsia, Pavonis, and Ascraeus Montes)

form the Tharsis Montes, a linear chain of volcanoes extending northeast-

ward across the Tharsis rise (Figures 1–65 and 1–66). These three shields

have gentle slopes of a few degrees (the upper slopes are commonly steeper

than the lower slopes), wide calderas, and flank vents. The shields appear

to be the result of basaltic flows and are similar to the intraplate shield vol-

canoes of the Hawaiian Islands. The Martian shield volcanoes rise 10 to 18

km above the Tharsis rise and attain elevations of 18 to 26 km. Along the

Tharsis axial trend, volcanoes stretch from Arsia Mons to near Tempe Pat-

1.18 Mars 101

Figure 1.65 Composite images of the two hemispheres of Mars. Upper left is the “eastern” hemisphere. The hemispheric dichotomy between the young, smooth, low-lying northern plains and the heavily cratered, old, south- ern highlands is clearly illustrated. The dark circular Hellas basin in the south is accepted to be an impact structure. Lower right is the “western” hemisphere. The three giant shield volcanoes that form the linear Tharsis Montes chain lie near the equator. Olympus Mons, the tallest mountain in the solar system, lies to the northwest of this chain. To the east the Valles Marineris canyon system is seen (NASA Image PIA02040).

era, some 4000 km. Lava flows that were erupted from the Tharsis Montes

and surrounding vents cover nearly 7 × 106 km2.

Olympus Mons (Figures 1–65 and 1–66) is a shield volcano nearly 600 km

102 Plate Tectonics

in diameter and over 26 km high, the tallest mountain in the solar system.

Flows on the flanks of the volcano and adjacent volcanic plains that were

erupted from fissures east of the volcano are among the youngest flows on

Mars. The extreme height of the Martian volcanoes can be attributed to the

low surface gravity and the lack of relative motion between the lithosphere

and the magma source. The presence of shield volcanoes on Mars and their

absence on the Moon can be attributed to differences in the viscosities of the

erupted lavas. A significant gravity anomaly is associated with the Tharsis

uplift. This gravity anomaly can be explained if the volcanic construct is

partially supported by the elastic lithosphere on Mars. Because Mars is

smaller than the Earth, it would be expected to cool more efficiently – it has

a larger surface area to volume ratio – and has a thicker lithosphere, other

factors being the same. This additional thickness and the smaller radius give

the elastic lithosphere on Mars a much greater rigidity.

Another major tectonic feature on Mars is an enormous canyon system,

Valles Marineris, extending eastward from Tharsis for about 4500 km (Figure

1–67; see also Figure 1–68). Individual canyons are up to 200 km wide and

several kilometers deep. In the central section (Figure 1–68), the system

is about 600 km wide and over 7 km deep. The Valles Marineris system

might be a complex set of fractures in the Martian crust caused by the

large topographic bulge containing the Tharsis volcanic region. The system

is roughly radial to this bulge, as are other prominent fractures.

Numerous channels are widely distributed over the Martian surface. They

display a variety of morphologic forms, including braiding and stream-lined

islands, strongly suggestive of formation by flowing water (Figure 1–69). If

water did flow on the surface of Mars some time in the past, the water may

have originated by the melting of subsurface ice. This is supported by the

association of the apparent sources of many channels with so-called chaotic

terrain: areas of large, irregular blocks probably formed by collapse follow-

ing the removal of a subsurface material such as ground ice. Chaotic terrain

is visible in the photomosaic in Figure 1–67 at the eastern end of the Val-

lis Marineris system; a broad collapsed area extends outward from Capri

Chasma. Many of the north-trending channels in this area appear to origi-

nate in this chaotic terrain. Martian channels give the impression of having

been formed by episodic flooding of large areas, as might be expected from

the sudden release of large amounts of subsurface water. Possible terrestrial

analogs to these channels are the scablands of the Columbia plateau in the

United States and the Sandur plains in Iceland, both of which formed by

the sudden release of large quantities of glacial meltwater. The existence of

a Martian ground ice is also indicated by the unusual forms of some crater

1.18 Mars 103

Figure 1.66 Image mosaic of the shield volcanoes in the Tharsis region of Mars obtained on a single Martian afternoon by the Mars orbiter camera on board the Mars Global Surveyor. Olympus Mons is the large shield in the upper left corner. Arsia Mons, Pavonis Mons, and Ascraeus Mons lie on a line trending SW–NE near the center of the image. The great canyon system, Valles Marineris, can be seen in the lower right corner (NASA Image PIA02049).

Figure 1.67 Mars Global Surveyor image of the Valles Marineris canyon system on Mars (NASA Image PIA00422).

ejecta. Figure 1–70 shows a lobate ejecta flow surrounding an impact crater

on Mars. The unique appearance of the ejecta pattern suggests the incor-

poration of large amounts of water into the ejecta, as would occur if the

impact penetrated a ground-ice-rich subsurface. In addition to the small

amount of water currently present in the thin CO2 atmosphere of Mars,

104 Plate Tectonics

the planet presently contains water in the form of ice in its permanent or

residual polar caps, which underlie the seasonal CO2 ice caps.

Although processes associated with liquid flow may have been active only

in the past, the present surface of Mars is being actively modified by at-

mospheric erosion and deposition. It is hardly surprising, in view of the

perennial dust storms that blanket the planet, that windblown sand effec-

tively alters the present surface of Mars. Figure 1–71 is a photograph of a

large dune field on Mars. Winds are an effective means of transporting mate-

rial over the Martian surface; there are layered deposits in the polar regions

that are believed to be accumulations of material carried by the atmosphere

from other regions of the planet.

The mean density of 3950 kg m−3 and the relatively small moment of

inertia of Mars are evidence that Mars has a metallic core. The size of the

core depends on assumptions about its composition, whether the core is

Fe–FeS, for example; model values of core radius vary between 0.4 and 0.6

of the Martian radius. Even though Mars has a metallic core, it does not

have a global intrinsic magnetic field. Early magnetic field measurements

from the Mars 2, 3, and 5 spacecraft were interpreted to imply that Mars

had a small magnetic field with a dipole moment 3× 10−4 times the Earth’s

magnetic dipole moment. Data from the Mars Global Surveyor have settled

the question of the existence of a global Martian magnetic field – there is

none, but the crust of Mars has strong concentrations of remanent mag-

netism implying that Mars had a global magnetic field in the past. Crustal

magnetization on Mars is mainly confined to the ancient highlands of the

southern hemisphere and it is largely organized into east-west-trending lin-

ear features of alternating polarity extending over distances as large as 2000

km. The magnetization features are reminiscent of the magnetic stripes on

the Earth’s seafloor, and suggest the possibility of a plate tectonic regime

with seafloor spreading early in the history of Mars. The absence of crustal

magnetism near large impact basins such as Hellas and Argyre implies that

the early Martian dynamo ceased to operate before about 4 Ga. The major

evidence for an initially hot and differentiated Mars is the acceptance of

Mars as a parent body of the SNC meteorites. This is a class of meteorites

found on Earth that apparently escaped from the Martian gravity field af-

ter one or more large impacts. The radiometric ages for SNC meteorites are

about 4.6 Ga, the U/Pb isotopic composition of SNC meteorites require core

formation at about 4.6 Ga, and the old age (≥4 Ga) of the southern high-

lands suggests early crustal differentiation. Other evidence for a hot early

Mars includes water-carved features on the Martian surface suggesting early

outgassing and an early atmosphere.

1.19 Phobos and Deimos 105

1.19 Phobos and Deimos

The two satellites of Mars, Phobos and Deimos, are very small, irregularly

shaped objects. Little was known of these bodies until the Mariner 9, Viking,

and the Mars Global Surveyor missions provided detailed photographs of

them (Figure 1–72). Roughly speaking, they are triaxial ellipsoids about 10

to 20 km across. Their surfaces are heavily cratered, but not identical in

appearance. In particular, Phobos has a system of long linear depressions

or grooves each of which is about 100 to 200 m wide and 10 to 20 m deep

(Figure 1–72). There are no grooves on Deimos. The grooves on Phobos

are probably related to fractures caused by a large impact, perhaps the one

responsible for the Stickney crater (Figure 1–72). There are no craters on

Deimos as large as Stickney; this may explain the absence of grooves on

Deimos. The low mean density of Phobos, 2000 kg m−3, and its reflectance

spectrum suggest that it is made of a dark gray material similar to Types I

or II carbonaceous chondrite meteorites.

1.20 Venus

In terms of size and density Venus is the planet that is most similar to

the Earth. An obvious working hypothesis would be that the tectonics of

Venus would be similar to the tectonics of the Earth and there would be plate

tectonics. We now know that this is not the case and that mantle convection

on Venus has a different surface expression than mantle convection on the

Earth.

The cloud cover on Venus has prevented optical observations. However,

Pioneer Venus radar, Earth-based radar observations, Venera 15–16 orbital

imaging radar, and the Magellan radar images have provided clear views of

the surface of Venus. These views, along with topography and gravity data,

indicate that Earth and Venus are in fact quite different. On Earth the global

oceanic rift system and the arcuate ocean trenches are the primary surface

manifestations of plate tectonics. The almost total absence of these features

on Venus has led to the conclusion that active plate tectonics is not occurring

on that planet at this time. Clearly, any comprehensive understanding of

tectonism and volcanism on Venus requires an understanding of how heat is

transported in the absence of plate tectonics.

There are other ways in which Venus is strikingly different. It rotates in a

retrograde sense with a period of 243 days; a Venusian day is 117 Earth days.

Venus has a massive, mostly CO2 atmosphere with a surface temperature of

about 750 K and a surface pressure of nearly 10 MPa. Sulfuric acid clouds

106 Plate Tectonics

blanket the planet and prevent us from directly viewing the surface. Because

of its earthlike size and mass, Venus most likely has a metallic core similar to

Earth’s. However, magnetic field measurements by the Pioneer Venus orbiter

during 1979 and 1980 revealed that Venus does not have an intrinsic global

magnetic field; these observations determined that if Venus had an intrinsic

magnetic field, its dipole moment would have to be much less than 1019 A

m2.

Studies of the surface of Venus during the Magellan mission have provided

a wealth of data on its tectonic and volcanic processes. The radar images

of the surface are complemented by global topography and gravity data.

The planet is remarkably smooth; 64% of the surface comprises a plains

province with elevation differences of 2 km or less; highland areas stand as

much as 10 km above the plains but they constitute only about 5% of the

surface; lowlands are 2 to 3 km below the plains and occupy the remaining

31% of the surface. Figure 1–73 shows the distribution of these topographic

provinces. Although local elevation extremes on Venus and Earth are roughly

comparable, global topographic variations are much smaller on Venus; the

planet’s surface is at a nearly uniform level.

There are tectonic features on Venus that resemble major tectonic features

on the Earth. Beta Regio has many of the features of a continental rift on

Earth. It has a domal structure with a diameter of about 2000 km and a

swell amplitude of about 2 km. It has a well-defined central rift valley with

a depth of 1–2 km and there is some evidence of a three-armed planform

(aulacogen). It is dominated by two shieldlike features, Theia Mons and

Rhea Mons, which rise about 4 km above the mean level. The U.S.S.R.

Venera 9 and 10 spacecraft, which landed just east of Beta Regio, measured

a basaltic composition and a density of 2800 kg m−3 for the surface rocks.

These observations substantiate the identification of Theia Mons and Rhea

Mons as shield volcanoes. Atla, Eistla, and Bell Regiones have rift zone

characteristics similar to Beta Regio.

Most of the highlands on Venus are concentrated into two main continental-

sized areas: Ishtar Terra, the size of Australia, in the northern hemisphere,

and Aphrodite Terra, about the size of Africa, near the equator (Figures

1–73 and 1–74). Aphrodite Terra, with a length of some 1500 km, is remi-

niscent of major continental collision zones on Earth, such as the mountain

belt that extends from the Alps to the Himalayas. Ishtar Terra is a region of

elevated topography with a horizontal scale of 2000–3000 km. A major fea-

ture is Lakshmi Planum which is an elevated plateau similar to Tibet with

a mean elevation of about 4 km. This plateau is surrounded by linear moun-

1.21 The Galilean Satellites 107

tain belts. Akna, Danu, Freyja, and Maxwell Montes, reaching elevations of

10 km, are similar in scale and elevation to the Himalayas.

The gravitational anomalies associated with topographic planetary fea-

tures further constrain their origin. Gravity anomalies obtained from track-

ing Pioneer Venus provided further major surprises. Unlike on the Earth,

gravity anomalies correlate with high topography on Venus. Large posi-

tive gravity anomalies are directly associated with Beta Regio and eastern

Aphrodite Terra.

One of the most important observational constraints on the geodynamics

of Venus comes from studies of impact cratering on the surface (Figure 1–

75). Some 840 impact craters have been identified from Magellan images

with diameters ranging from 2 to 280 km. The distribution of craters on

Venus cannot be distinguished from a random distribution. Unlike the Moon

and Mars, older and younger terrains cannot be identified. The surface of

Venus appears to be of a near-uniform age. Correlations of this impact flux

with craters on the Moon, the Earth, and Mars indicate a mean surface

age of 0.5 ± 0.3 Ga. Another important observation is that 52% of the

craters are slightly fractured and only 4.5% are embayed by lava flows. These

observations led Schaber et al. (1992) to hypothesize that a global volcanic

resurfacing event had occurred at about 500 Ma and that relatively little

surface volcanism has occurred since. Further statistical tests have shown

that a large fraction of the surface of Venus (≈80–90%) was covered by fresh

volcanic flows during a period of 10–50 Myr. It is well established that the

geologic evolution of Venus is far more catastrophic than the Earth’s.

Other major features unique to Venus are the coronae. These are quasi-

circular features, 100–2600 km in diameter, with raised interiors and elevated

rims, often with annular troughs. It has been suggested that the perimeters

of several large coronae on Venus, specifically Artemis (Figure 1–76), Latona,

and Eithinoha, resemble terrestrial subduction zones in both planform and

topography. Artemis chasma has a radius of curvature similar to that of

the South Sandwich subduction zone on the Earth. Large coronae may be

incipient circular subduction zones. The foundering lithosphere is replaced

by ascending hot mantle in a manner similar to back-arc spreading on the

Earth.

1.21 The Galilean Satellites

The innermost satellites of Jupiter, in order of distance from the planet,

are Amalthea, Io, Europa, Ganymede, and Callisto. The latter four were

discovered by Galileo in 1610 and are collectively referred to as the Galilean

108 Plate Tectonics

satellites. Amalthea was discovered by Barnard in 1892. They all have nearly

circular prograde orbits lying almost exactly in Jupiter’s equatorial plane.

Our knowledge of the Galilean satellites increased considerably as a con-

sequence of the flybys of Voyagers 1 and 2 on March 5, 1979 and July 9,

1979, respectively, and the Galileo mission (1995–2000) has yielded a further

enormous jump in our knowledge of these bodies. We now know as much

about the surfaces and interiors of the Galilean satellites as we do about

some of the terrestrial planets in our inner solar system. These satellites are

very different from one another and from the terrestrial planets; many of

the physical processes occurring in their interiors and on their surfaces are

unique to these bodies.

From Appendix 2, Section D, it can be seen that Io is only slightly larger

and more massive than the Moon. Its similarity to the Moon extends no

further, however; the Voyager and Galileo missions showed Io to be the

most volcanically active body in the solar system. During the flybys of both

the Voyager and Galileo spacecraft numerous active volcanic plumes were

observed, some extending to heights of hundreds of kilometers above the

surface. Io (Figures 1–77 and 1–78) displays a great diversity of color and

albedo; spectral reflectance data suggest that its surface is dominated by

sulfur-rich materials that account for the variety of colors – orange, red,

white, black, and brown. Io’s volcanism is predominantly silicate-based as

on the terrestrial planets though sulfur-based volcanism also occurs. The

circular features on Io (Figure 1–77) are caldera-like depressions (Figure

1–78); some have diameters as large as 200 km. There are no recognizable

impact craters on the satellite, although the flux of impacting objects in the

early Jovian system is believed to be as large as it was around the terrestrial

planets. Io’s surface is geologically very young, the silicate and sulfur lavas

having only recently resurfaced the planet. Relatively few of Io’s calderas

are associated with structures of significant positive relief. Thus they are

quite unlike the calderas of the Hawaiian volcanoes or the Tharsis volcanoes

on Mars. There are isolated mountains with considerable relief on Io (∼10

km), but their exact height as well as their mode of origin is uncertain.

The source of heat for Io’s volcanism is tidal dissipation. The gravitational

interaction of Io with Europa and Ganymede forces Io into an orbit with

higher eccentricity than it would have if it were circling Jupiter by itself.

The resulting tidal flexing of Io in Jupiter’s gravity field dissipates very

large quantities of heat; Io’s surface heat flow exceeds the global heat loss

from the Earth by a factor of 3 or more. Tidal dissipation is insignificant

as a heat source for the terrestrial planets that are heated mainly by the

decay of radioactive elements. However, the special circumstances of Io’s

1.21 The Galilean Satellites 109

orbit around a massive planet makes tidal heating an unusually effective

heat source for Io.

Io’s density and moment of inertia constrain its internal structure. The

satellite has a large metallic core whose exact size is uncertain because we

do not know the composition of the core. The core radius could be as large

as about half of Io’s radius and is surrounded by a silicate mantle. Io’s

extensive volcanism suggests that the satellite has a crust and a partially

molten asthenosphere. Much of the tidal dissipative heating probably occurs

in this asthenosphere. Io is known to be in hydrostatic equilibrium under

the action of the Jovian tidal forces and its rotation. It is not known if Io

has its own magnetic field.

Europa is only slightly smaller and less massive than the Moon (Appendix

2, Section D), but it also looks quite different from our satellite. Figure 1–

79 is a Voyager 2 picture of Europa that shows the surface to consist of

two major terrain types: a uniformly bright terrain crossed by numerous

dark linear markings and a somewhat darker mottled terrain. Relatively few

impact craters exist on Europa indicating that the surface is geologically

young. The linear markings are ridges and fractures; they have little or no

vertical relief. They extend over distances as large as thousands of kilome-

ters and vary in width from several kilometers to about 100 km. Europa’s

density and moment of inertia indicate that, although it is composed mainly

of silicates, it must contain a large fraction (about 20% by mass) of water.

The water is believed to be in a surface layer about 100 km thick surround-

ing a silicate mantle and metallic core. The water layer may be completely

frozen or it may consist of ice above liquid. Infrared spectra of Europa and

its high albedo indicate that the surface is covered with water ice or frost.

High-resolution Galileo pictures show features such as ice rafts that have

rotated and separated from each other over an underlying soft ice layer or

an internal liquid ocean. The relative absence of craters on Europa may have

resulted from the freezing of a competent ice layer only after the termination

of the early phase of severe bombardment or it may be due to geologically

recent resurfacing of the satellite; the global fracture pattern may be a conse-

quence of tidal stresses and nonsynchronous rotation of Europa’s outer shell

of ice. The surfaces of Europa and, as we shall see, Ganymede and Callisto

are shaped by processes occurring in a predominantly ice shell. Although

large ice-covered regions of the Earth give us some clues about what surface

features to expect, the icy Galilean satellites provide a unique example of

surfaces shaped by global-scale ice tectonic processes at extremely low tem-

peratures (the surface temperatures of the Galilean satellites are about 150

K). The geologist studying Io must be mainly a volcanologist; the geologist

110 Plate Tectonics

investigating Europa, Ganymede, and Callisto, on the other hand, must be

mainly a glaciologist! If there is an internal ocean on Europa, the satellite

must then be considered a possible site for extra-terrestrial life. Some tidal

heating of Europa is necessary to prevent the freezing of an internal liquid

water ocean.

Ganymede and Callisto, the icy Galilean satellites, are about the size of

Mercury (Appendix 2, Sections C and D). Their low mean densities, less

than 2000 kg m−3, indicate that they are composed of silicates with very

substantial amounts of water. The fraction of water contained in these bodies

depends on the density of the silicates; as much as 50% of the satellites

could be water. Multiple flybys of Ganymede and Callisto by the Galileo

spacecraft have provided us with knowledge of the satellites’ densities and

moments of inertia from which we can infer the internal structures of the

bodies. Ganymede is fully differentiated into a metallic core, silicate mantle,

and thick (many hundreds of kilometers) outer ice shell. In contrast, Callisto

is only partially differentiated. Most of the satellite consists of a primordial

ice–rock mixture; only in the outer few hundred kilometers can the ice and

rock have separated.

The Galileo spacecraft found that Ganymede has its own magnetic field

while Callisto does not. Ganymede is the only moon in our solar system

known to have an intrinsic global magnetic field at present. Ganymede’s

magnetic field is so large that the field must be generated by dynamo action

in a liquid metallic core just as a dynamo in the Earth’s outer core produces

our magnetic field. Observations of Ganymede by the Galileo spacecraft

provide strong support for the complete differentiation of the satellite and

for the existence of a liquid metallic core in its interior.

A major unsolved question is why Ganymede is fully differentiated while

Callisto is only slightly differentiated; both bodies are about the same size

and are made up of about the same proportions of ice and rock. One possible

explanation is that Ganymede was tidally heated in the past whereas Callisto

was not. The appearances of Ganymede and Callisto are consistent with a

differentiated interior for Ganymede and a relatively primordial interior for

Callisto.

Unlike Europa and Io, Ganymede has numerous impact craters. Two ma-

jor terrain types have been identified: relatively dark heavily cratered terrain

and lighter grooved terrain. The former has a crater density comparable with

that of the lunar highlands and other ancient cratered surfaces of the inner

planets. Although the grooved terrain contains fewer craters, it nonetheless

has a crater density comparable with the oldest lunar maria and Martian

plains units. Bright-rayed impact craters are abundant on both types of ter-

References 111

rain. Almost all the grooved terrain is a mosaic of sets of grooves; groove

systems are 10 to 100 km wide and 10 to 1000 km long. Individual grooves

are a few hundred meters deep. The craters on Ganymede display a vari-

ety of morphologic forms. Craters several hundred kilometers in diameter

are found only as subdued scars on the oldest parts of Ganymede’s surface.

These presumably degraded impact craters appear today as circular bright

patches without rims or central depressions; they have been described as

crater palimpsests. Creep in a predominantly ice surface is probably respon-

sible for the severe degradation of the large craters. Craters in the younger

grooved terrain are generally better preserved that those in the older heavily

cratered areas. There is no major relief on Ganymede; that is, there are no

large mountains or basins. Galileo observations of Ganymede confirm that

tectonism has been a major factor in shaping the satellite’s surface. Tectonic

activity on Ganymede is in accord with its differentiated interior.

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114 Collateral Reading

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Cambridge, 1997), 354 pages.

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McDonald, G. A., Volcanoes (Prentice Hall, Englewood Cliffs, NJ, 1972),

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A largely descriptive and in-depth discussion of the physical aspects of vol-

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don, 1973), 358 pages.

Seven chapters deal with geomagnetism, rock magnetism, experimental

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magnetic poles.

Merrill, R. T., M. W. McElhinny, and P. L. McFadden, The Magnetic Field

of the Earth (Academic Press, San Diego, 1996), 531 pages.

A comprehensive discussion of all aspects of the Earth’s magnetic field.

Topics include the present geomagnetic field, paleomagnetism, reversals,

and dynamo theory.

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1974), 945 pages.

An introductory textbook on Earth science. The book is divided into three

major sections dealing with the geological history of the Earth and its

surface and interior.

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(McGraw-Hill, New York, 1975), 618 pages.

An advanced textbook that combines observational data from natural petrol-

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tems at high temperature and pressure to discuss the composition and

petrology of the upper mantle–crust system. There are also chapters

discussing the lower mantle and the origin and evolution of the Earth.

Williams, H., and A. R. McBirney, Volcanology (Freeman, Cooper and Com-

pany, San Francisco, 1979), 397 pages.

An advanced textbook with chapters on the physical nature of magmas, gen-

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Collateral Reading 115

pyroclastic flows, fissure eruptions, oceanic volcanism, and hydrother-

mal phenomena.

Wyllie, P. J., The Dynamic Earth: Textbook in Geosciences (John Wiley,

New York, 1971), 416 pages.

An advanced textbook designed mainly for graduate students in geology and

geochemistry. Chapters deal with the structure, composition, mineral-

ogy, and petrology of the crust and mantle, mantle phase transitions,

magma generation, plate tectonics, and the Earth’s interior.

116 Collateral Reading

Figure 1.68 Mars Global Surveyor image showing the layered structure in the walls of a mesa in southern Melas Chasma in Valles Marineris. This image covers an area 3 km wide and 8.2 km long. Erosion by landslides has exposed tens of layers several meters in thickness and has created the dark fan-shaped deposits seen near the center of the image. The floor of the canyon is dark and is covered by many parallel ridges and grooves (lower third of the image) (NASA Image PIA02398).

Collateral Reading 117

Figure 1.69 Mars Global Surveyor image showing a branching valley “net- work” in an ancient cratered terrain. This image covers an area of 11.5 by 27.4 km. The eroded valleys are bright and taken as evidence that Mars had liquid water running across its surface about 4 billion years ago (NASA Image PIA01499).

118 Collateral Reading

Figure 1.70 This Mars Global Surveyor image shows two small impact craters with dark ejecta deposits that were blown out of the craters during the impacts. The large crater has a diameter of about 89 m and the smaller crater about 36 m. The ejecta is darker than the surrounding substrate because the impacts broke through the upper, brighter surface material and penetrated to a layer of darker material beneath (NASA Image PIA01683).

Collateral Reading 119

Figure 1.71 Mars Global Surveyor image of a sand dune field. The number of impact craters in the image indicate that the dunes are quite ancient (NASA Image PIA02359).

120 Collateral Reading

Figure 1.72 This image of Phobos, the inner and larger of the two moons of Mars, was taken by the Mars Global Surveyor on August 19, 1998. It shows a close-up of the largest crater on Phobos, Stickney, 10 km in diameter, nearly half the size of the entire body. Crossing at and near the rim of Stickney are shallow, elongated depressions which may be fractures that resulted from the impact that created Stickney (NASA Image PIA01333).

Collateral Reading 121

Figure 1.73 Composite radar images of the two hemispheres of Venus. The left hemispheric view is centered at 0◦E longitude. The light region near the north pole is Maxwell Montes, the highest region on Venus. The circular structure near the center is Heng-o Corona. The light stippled region south of this is Alpha Regio. The right hemispheric view is centered at 180◦E longitude. The bright equatorial region just south of the equator on the left is Aphrodite Terra. The large circular feature just south of this is Artemis Corona (NASA Image PIA00157).

122 Collateral Reading

Figure 1.74 Magellan image of Maxwell Montes, the highest mountain on Venus, rising almost 11 km above the mean planetary radius. The promi- nent circular structure is Cleopatra, an impact basin with a diameter of about 100 km and a depth of 2.5 km (NASA Image PIA00149).

Figure 1.75 Magellan radar image of three large impact craters in the Lavinia region of Venus. The craters range in diameter from 37 km to 50 km. The bright areas are rough (radar-bright) ejecta (NASA Image PIA00086).

Collateral Reading 123

Figure 1.76 Composite Magellan radar image of Artemis corona. The near circular trough of the Artemis chasma has a diameter of 2100 km. The concentric features outside the chasma are attributed to normal faulting associated with lithospheric flexure similar to that occurring seaward of subduction zones on the Earth. The geometry of Artemis corona is generally similar to the Aleutian island arc and has been associated with an aborted subduction zone (NASA Image PIA00101).

124 Collateral Reading

Figure 1.77 High-resolution image of Jupiter’s moon Io acquired by the Galileo spacecraft on July 3, 1999. The surface is covered by volcanic cen- ters, many of them active (NASA Image PIA02308).

Collateral Reading 125

Figure 1.78 An active volcanic eruption on Jupiter’s moon Io was captured in this image taken on February 22, 2000, by the Galileo spacecraft. This picture is about 250 km across. The eruption is occurring at Tvashtar Catena, a chain of giant volcanic calderas (NASA Image PIA02550).

126 Collateral Reading

Figure 1.79 Near hemispheric image of Jupiter’s satellite Europa taken by the Voyager 2 spacecraft on July 9, 1979. The linear crack-like features are clearly illustrated as well as the darker mottled regions (NASA Image PIA01523).

2

Stress and Strain in Solids

2.1 Introduction

Plate tectonics is a consequence of the gravitational body forces acting on

the solid mantle and crust. Gravi- tational forces result in an increase of

pressure with depth in the Earth; rocks must support the weight of the

overburden that increases with depth. A static equilibrium with pressure

increasing with depth is not possible, however, because there are horizontal

variations in the gravitational body forces in the Earth’s interior. These

are caused by horizontal variations in density associated with horizontal

differences in temperature. The horizontal thermal contrasts are in turn the

inevitable consequence of the heat release by radioactivity in the rocks of the

mantle and crust. The horizontal variations of the gravitational body force

produce the differential stresses that drive the relative motions associated

with plate tectonics.

One of the main purposes of this chapter is to introduce the fundamen-

tal concepts needed for a quantitative understanding of stresses in the solid

Earth. Stresses are forces per unit area that are transmitted through a mate-

rial by interatomic force fields. Stresses that are transmitted perpendicular

to a surface are normal stresses; those that are transmitted parallel to a

surface are shear stresses. The mean value of the normal stresses is the pres-

sure. We will describe the techniques presently used to measure the state of

stress in the Earth’s crust and discuss the results of those measurements.

Stress in an elastic solid results in strain or deformation of the solid. The

simplest example of strain is the decrease in volume accompanying an in-

crease in pressure due to the compressibility of a solid. Normal strain is

defined as the ratio of the change in length of a solid to its original length.

The shear strain is defined as one-half of the decrease in a right angle in

a solid when it is deformed. The surface of the Earth is continually being

128 Stress and Strain in Solids

strained by tectonic processes. These changes in strain can be measured di-

rectly by geodetic techniques. This chapter also discusses the basic concepts

required for a quantitative understanding of strain and changes in strain in

the solid Earth.

2.2 Body Forces and Surface Forces

The forces on an element of a solid are of two types: body forces and surface

forces. Body forces act throughout the volume of the solid. The magnitude

of the body force on an element is thus directly proportional to its volume

or mass. An example is the downward force of gravity, that is, the weight of

an element, which is the product of its mass and the acceleration of gravity

g. Since density ρ is mass per unit volume, the gravitational body force

on an element is also the product of ρg and the element’s volume. Thus

the downward gravitational body force is g per unit mass and ρg per unit

volume.

The densities of some common rocks are listed in Appendix 2, Section

E. The densities of rocks depend on the pressure; the values given are zero-

pressure densities. Under the high pressures encountered deep in the mantle,

rocks are as much as 50% denser than the zero-pressure values. The varia-

tion of density with depth in the Earth is discussed in Chapter 4. Typical

mantle rocks have zero-pressure densities of 3250 kg m−3. Basalt and gab-

bro, which are the principal constituents of the oceanic crust, have densities

near 2950 kg m−3. Continental igneous rocks such as granite and diorite

are significantly lighter with densities of 2650 to 2800 kg m−3. Sedimentary

rocks are generally the lightest and have the largest variations in density, in

large part because of variations in porosity and water content in the rocks.

Surface forces act on the surface area bounding an element of volume.

They arise from interatomic forces exerted by material on one side of the

surface onto material on the opposite side. The magnitude of the surface

force is directly proportional to the area of the surface on which it acts.

It also depends on the orientation of the surface. As an example, consider

the force that must act at the base of the column of rock at a depth y

beneath the surface to support the weight of the column, as illustrated in

Figure 2–1. The weight of the column of cross-sectional area δA, is ρgyδA.

This weight must be balanced by an upward surface force σyyδA distributed

on the horizontal surface of area δA at depth y. We are assuming that no

vertical forces are acting on the lateral surfaces of the column and that

the density ρ is constant; σyy is thus the surface force per unit area acting

perpendicular to a horizontal surface, that is, stress. Since the forces on the

2.2 Body Forces and Surface Forces 129

Figure 2.1 Body and surface forces acting on a vertical column of rock.

Figure 2.2 A continental block “floating” on the fluid mantle.

column of rock must be equal if the column is in equilibrium, we find

σyy = ρgy. (2.1)

The normal force per unit area on horizontal planes increases linearly with

depth. The normal stress due to the weight of the overlying rock or over-

burden is known as the lithostatic stress or pressure.

To find a typical value for stress in the lithosphere, let us determine

the lithostatic stress on a horizontal plane at the base of the continental

crust. Assume that the crust is 35 km thick and that its mean density is

2750 kg m−3; from Equation (2–1) we find that

σyy = 2750 kg m−3 × 10 m s−2 × 3.5 × 104 m

= 9.625 × 108 Pa = 962.5 MPa.

The SI unit for pressure or stress is the pascal (Pa). Pressures and stresses

in the Earth are normally given in megapascals (Mpa); 1 megapascal =

106 pascals.

130 Stress and Strain in Solids

Think of continents as blocks of wood floating on a sea of mantle rock,

as illustrated in Figure 2–2. The mean density of the continent, say ρc =

2750 kg m−3, is less than the mean upper mantle density, say ρm = 3300 kg m−3,

so that the continent “floats.” Archimedes’ principle applies to continents;

they are buoyed up by a force equal to the weight of mantle rock displaced.

At the base of the continent σyy = ρcgh, where ρc is the density of the

continent and h is its thickness. At this depth in the mantle, σyy is ρmgb,

where ρm is the mantle density and b is the depth in the mantle to which the

continent “sinks.” Another statement of Archimedes’ principle, also known

as hydrostatic equilibrium, is that these stresses are equal. Therefore we find

ρch = ρmb. (2.2)

The height of the continent above the surrounding mantle is

h− b = h− ρc ρm

h = h

(

1 − ρc ρm

)

. (2.3)

Using the values given earlier for the densities and the thickness of the con-

tinental crust h = 35 km, we find from Equation (2–3) that h− b = 5.8 km.

This analysis is only approximately valid for determining the depth of the

oceans relative to the continents, since we have neglected the contribution

of the seawater and the oceanic crust. The application of hydrostatic equi-

librium to the continental crust is known as isostasy; it is discussed in more

detail in Chapter 5.

Problem 2.1 An average thickness of the oceanic crust is 6 km. Its density

is 2900 kg m−3. This is overlain by 5 km of water (ρw = 1000 kg m−3) in a

typical ocean basin. Determine the normal force per unit area on a horizontal

plane at the base of the oceanic crust due to the weight of the crust and the

overlying water.

Problem 2.2 A mountain range has an elevation of 5 km. Assuming that

ρm = 3300 kg m−3, ρc = 2800 kg m−3, and that the reference or nor-

mal continental crust has a thickness of 35 km, determine the thickness of

the continental crust beneath the mountain range. Assume that hydrostatic

equilibrium is applicable.

Problem 2.3 There is observational evidence from the continents that

the sea level in the Cretaceous was 200 m higher than today. After a few

thousand years, however, the seawater is in isostatic equilibrium with the

ocean basins. What was the corresponding increase in the depth of the ocean

basins? Take ρw = 1000 kg m−3 and the density of the displaced mantle to

be ρm = 3300 kg m−3.

2.2 Body Forces and Surface Forces 131

Figure 2.3 Isostasy of the continental crust relative to an ocean basin.

A more realistic model for the depth of the ocean basins is illustrated

in Figure 2–3. The continental crust has a thickness hcc and a density ρcc;

its upper surface is at sea level. The oceanic crust is covered with water of

depth hw and density ρw. The oceanic crust has a thickness hoc and density

ρoc. The mantle density is ρm. Application of the principle of isostasy to the

base of the continental crust gives

ρcchcc = ρwhw + ρochoc + ρm(hcc − hw − hoc).

(2.4)

The depth of the ocean basin relative to the continent is given by

hw = (ρm − ρcc)

(ρm − ρw) hcc −

(ρm − ρoc)

(ρm − ρw) hoc. (2.5)

Taking hcc = 35 km, hoc = 6 km, ρm = 3300 kg m−3, ρw = 1000 kg m−3,

ρcc = 2800 kg m−3, and ρoc = 2900 kg m−3, we find hw = 6.6 km.

Subsidence of the surface of the continental crust often results in the for-

mation of a sedimentary basin. Assume that the surface of the continental

crust is initially at sea level and, as it subsides, sediments are deposited

so that the surface of the sediments remains at sea level. One cause of the

subsidence is the thinning of the continental crust. As the crust is thinned,

isostasy requires that the surface subside. A simple model for this subsi-

dence applicable to some sedimentary basins is the crustal stretching model

(McKenzie, 1978). This two-dimensional model is illustrated in Figure 2–4.

A section of continental crust with an initial width w0 is stretched to a final

width wb. The stretching factor α is defined by

α = wb w0 . (2.6)

In order to conserve the volume of the stretched continental crust we assume

a constant crustal density ρcc and require that

wbhcb = w0hcc, (2.7)

132 Stress and Strain in Solids

Figure 2.4 Illustration of the crustal stretching model for the formation of a sedimentary basin. A section of continental crust of initial width w0, illustrated in (a), is stretched by a stretching factor α=4 to a final width wb to form the sedimentary basin illustrated in (b).

Figure 2.5 Thickness of a sedimentary basin hsb as a function of the crustal stretching factor α.

where hcc is the initial thickness of the continental crust and hcb is the final

thickness of the stretched crust. The combination of Equations (2–6) and

(2–7) gives

hcb = hcc α . (2.8)

The surface of this stretched continental crust subsides and is assumed to

be covered with sediments of density ρs(ρs<ρcc) to sea level. The sediments

have a thickness hsb and the lower boundary of the sediments is referred

to as basement. Application of the principle of isostasy to the base of the

2.2 Body Forces and Surface Forces 133

reference continental crust gives

ρcchcc = ρshsb + ρcchcb + ρm(hcc − hsb − hcb).

(2.9)

The combination of Equations (2–8) and (2–9) gives the thickness of the

sedimentary basin in terms of the stretching factor as

hsb = hcc

(

ρm − ρcc ρm − ρs

)(

1 − 1

α

)

. (2.10)

The thickness of the sedimentary basin is given as a function of the stretching

factor in Figure 2–5 for hcc = 35 km, ρm = 3300 kg m−3, ρcc = 2800 kg m−3,

and ρs = 2500 kg m−3. The maximum thickness of the sedimentary basin

for an infinite stretching factor is hsb = 22 km.

Problem 2.4 A sedimentary basin has a thickness of 4 km. Assuming

that the crustal stretching model is applicable and that hcc = 35 km, ρm =

3300 kg m−3, ρcc = 2750 kg m−3, and ρs = 2550 kg m−3, determine the

stretching factor.

Problem 2.5 A sedimentary basin has a thickness of 7 km. Assuming

that the crustal stretching model is applicable and that hcc = 35 km, ρm =

3300 kg m−3, ρcc = 2700 kg m−3, and ρs = 2450 kg m3, determine the

stretching factor.

Problem 2.6 A simple model for a continental mountain belt is the crustal

compression model illustrated in Figure 2–6. A section of the continental

crust of width w0 is compressed to a width wmb. The compression factor β

is defined by

β = w0

wmb . (2.11)

Show that the height of the mountain belt h is given by

h = hcc (ρm − ρcc)

ρm (β − 1). (2.12)

Assuming β = 2, hcc = 35 km, ρm = 3300 kg m−3, and ρcc = 2800 kg m−3,

determine the height of the mountain belt h and the thickness of the crustal

root b.

Just as there are normal surface forces per unit area on horizontal planes

in the Earth, there are also normal surface forces per unit area on vertical

planes, as sketched in Figure 2–7. The horizontal normal stress components

σxx and σzz can include large-scale tectonic forces, in which case σxx 6=

134 Stress and Strain in Solids

Figure 2.6 Illustration of the crustal compression model for a mountain belt. A section of continental crust of width w0, shown in (a), is compressed by compression factor β= 2 to form a mountain belt as shown in (b).

Figure 2.7 Horizontal surface forces acting on vertical planes.

Figure 2.8 Force balance on a section of continental block.

2.2 Body Forces and Surface Forces 135

Figure 2.9 The area under the stress versus depth profile is proportional to the total horizontal force on a vertical plane.

σzz 6= σyy. On the other hand, there are many instances in which rock was

heated to sufficiently high temperatures or was sufficiently weak initially

so that the three stresses σxx, σzz, and σyy are equal to the weight of the

overburden; that is,

pL ≡ σxx = σzz = σyy = ρgy. (2.13)

When the three normal stresses are equal, they are defined to be the pressure.

The balance between pressure and the weight of the overburden is known as a

lithostatic state of stress. It is completely equivalent to the hydrostatic state

of stress in a motionless body of fluid wherein pressure forces are exerted

equally in all directions and pressure increases proportionately with depth.

We will now show that the continental block illustrated in Figure 2–2

cannot simply be in a lithostatic state of stress. The force balance on the

continental block is illustrated in Figure 2–8. A horizontal force is acting on

the edge of the block Fm. We assume that this force is due to the lithostatic

pressure in the mantle rock of density ρm. The vertical distribution of this

pressure is given in Figure 2–9. The horizontal force Fm is obtained by

integrating the lithostatic pressure:

Fm =

∫ b

0 pL dy = ρmg

∫ b

0 y dy =

1

2 ρmgb

2. (2.14)

This force is per unit width of the block so that it has dimensions of force

per unit length. The total force per unit width is proportional to the area

under the stress distribution given in Figure 2–9.

We next determine the horizontal force per unit width acting at a typical

cross section in the continental block Fc. We assume that the horizontal nor-

mal stress acting in the continent σxx is made up of two parts, the lithostatic

contribution ρcgy and a constant tectonic contribution ∆σxx,

σxx = ρcgy + ∆σxx. (2.15)

136 Stress and Strain in Solids

The tectonic contribution is also known as the deviatoric stress. The hori-

zontal force Fc is obtained by integrating the horizontal normal stress

Fc =

∫ h

0 σxx dy =

∫ h

0 (ρcgy + ∆σxx) dy

= 1 2ρcgh

2 + ∆σxxh. (2.16)

In order to maintain a static balance, the two forces Fc and Fm must be

equal. Using Equations (2–2), (2–14), and (2–16), we obtain

∆σxx = 1

2

ρmgb 2

h − 1

2 ρcgh = −1

2 ρcgh

(

1 − ρc ρm

)

.

(2.17)

A horizontal tensile stress is required to maintain the integrity of the con-

tinental block. The horizontal tensile stress is a force per unit area acting on

vertical planes and tending to pull on such planes. A compressive stress is

a normal force per unit area tending to push on a plane. We consider com-

pressive stresses positive and tensile stresses negative, a convention generally

adopted in the geological literature. This is opposite to the sign convention

used in most elasticity textbooks in which positive stress is tensional. Taking

h = 35 km, ρm = 3300 kg m−3, and ρc = 2750 kg m−3, we find from Equa-

tion (2–17) that ∆σxx = −80.2 MPa. Typical values for deviatoric stresses

in the continents are of the order of 10 to 100 MPa.

Problem 2.7 Consider a continental block to have a thickness of 70 km

corresponding to a major mountain range. If the continent has a density

of 2800 kg m−3 and the mantle a density of 3300 kg m−3, determine the

tensional stress in the continental block.

Problem 2.8 Determine the deviatoric stress in the continent for the

oceanic–continental structure in Figure 2–3 by proceeding as follows. Show

that the pressure as a function of depth in the continental crust pc is

pc = ρccgy, (2.18)

and that the pressures in the water, in the oceanic crust, and in the mantle

beneath the oceanic crust are

p0 = ρwgy 0 ≤ y ≤ hw

= ρwghw + ρocg(y − hw) hw ≤ y ≤ hw + hoc

= ρwghw + ρocghoc + ρmg(y − hw − hoc)

hw + hoc ≤ y ≤ hcc. (2.19)

2.2 Body Forces and Surface Forces 137

Figure 2.10 Normal and tangential surface forces on an area element in the fault plane of a strike–slip fault.

Find the net difference in the hydrostatic pressure force between the conti-

nental and the oceanic crusts F by integrating the pressures over a depth

equal to the thickness of the continental crust. The result is

F = g[hwhcc(ρm − ρw) + hochcc(ρm − ρoc)

−hwhoc(ρm − ρoc) − 1 2h

2 w(ρm − ρw)

− 1 2h

2 oc(ρm − ρoc) − 1

2h 2 cc(ρm − ρcc)]. (2.20)

Calculate F for hw = 5 km, ρw = 1000 kg m−3, hoc = 7 km, ρoc = 2900 kg

m−3, ρcc = 2800 kg m−3, and ρm = 3300 kg m−3. Find hcc from Equation (2–

5). If the elastic stresses required to balance this force are distributed over

a depth equal to hcc, determine the stress. If the stresses are exerted in the

continental crust, are they tensional or compressional? If they act in the

oceanic lithosphere, are they tensional or compressional?

Surface forces can act parallel as well as perpendi-cular to a surface. An

example is provided by the forces acting on the area element δA lying in

the plane of a strike–slip fault, as illustrated in Figure 2–10. The normal

compressive force σxxδA acting on the fault face is the consequence of the

weight of the overburden and the tectonic forces tending to press the two

sides of the fault together. The tangential or shear force on the element

σxzδA opposes the tectonic forces driving the left-lateral motion on the fault.

This shear force is the result of the frictional resistance to motion on the

fault. The quantity σxz is the tangential surface force per unit area or the

shear stress. The first subscript refers to the direction normal to the surface

element and the second subscript to the direction of the shear force.

Another example of the resistive force due to a shear stress is the em-

138 Stress and Strain in Solids

Figure 2.11 Normal and tangential forces acting on a rock mass displaced horizontally to the right in a low-angle overthrust fault.

placement of a thrust sheet. In zones of continental collision a thin sheet

of crystalline rock is often overthrust upon adjacent continental rocks on a

low-angle thrust fault. This process is illustrated in Figure 2–11, where the

thrust sheet has been emplaced from the left as a consequence of horizontal

tectonic forces. Neglecting the influence of gravity, which is considered in

Section 8–4, we can write the total horizontal tectonic force FT due to a

horizontal tectonic stress ∆σxx as

FT = ∆σxxh, (2.21)

where h is the thickness of the thrust sheet and FT is a force per unit width

of the sheet. This tectonic driving force is resisted by the shear stress σyx acting on the base of the thrust sheet. The total resisting shear force per

unit width FR is

FR = σyxL, (2.22)

where L is the length of the thrust sheet.

In many cases it is appropriate to relate the shear stress resisting the

sliding of one surface over another to the normal force pressing the surfaces

together. Empirically we often observe that these stresses are proportional

to one another so that

σyx = fσyy, (2.23)

where σyy is the vertical normal stress acting on the base of the thrust sheet

and f , the constant of proportionality, is known as the coefficient of friction.

Assuming that σyy has the lithostatic value

σyy = ρcgh, (2.24)

and equating the driving tectonic force FT to the resisting shear force, we

2.2 Body Forces and Surface Forces 139

Figure 2.12 Gravitational sliding of a rock mass.

find that

∆σxx = fρcgL. (2.25)

This is the tectonic stress required to emplace a thrust sheet of length L.

Taking a typical value for the tectonic stress to be ∆σxx = 100 MPa and

assuming a thrust sheet length L = 100 km and ρc = 2750 kg m−3, we find

that the required coefficient of friction is f = 0.036. The existence of long

thrust sheets implies low values for the coefficient of friction.

Problem 2.9 Assume that the friction law given in Equation (2–23) is

applicable to the strike–slip fault illustrated in Figure 2–10 with f = 0.3.

Also assume that the normal stress σxx is lithostatic with ρc = 2750 kg m−3.

If the fault is 10 km deep, what is the force (per unit length of fault) resisting

motion on the fault? What is the mean tectonic shear stress over this depth

σ̄zx required to overcome this frictional resistance?

Problem 2.10 Consider a block of rock with a height of 1 m and horizontal

dimensions of 2 m. The density of the rock is 2750 kg m−3. If the coefficient

of friction is 0.8, what force is required to push the rock on a horizontal

surface?

Problem 2.11 Consider a rock mass resting on an inclined bedding plane

as shown in Figure 2–12. By balancing the forces acting on the block parallel

to the inclined plane, show that the tangential force per unit area σx′y ′ on

the plane supporting the block is ρgh sin θ (ρ is the density and h is the

thickness of the block). Show that the sliding condition is

θ = tan−1f. (2.26)

Problem 2.12 The pressure ph of fluids (water) in the pores of rocks

reduces the effective normal stress pressing the surfaces together along a

fault. Modify Equation (2–25) to incorporate this effect.

140 Stress and Strain in Solids

Figure 2.13 Surface forces acting on a small rectangular element in a two- dimensional state of stress.

2.3 Stress in Two Dimensions

In the previous section we were concerned primarily with stresses on the

surface of a material. However, stress components can be defined at any

point in a material. In order to illustrate this point, it is appropriate to

consider a small rectangular element with dimensions δx, δy, and δz defined

in accordance with a cartesian x, y, z coordinate system, as illustrated in

Figure 2–13. In this section we will consider a two-dimensional state of

stress; the state is two-dimensional in the sense that there are no surface

forces in the z direction and none of the surface forces shown vary in the

z direction. The normal stresses are σxx and σyy, and the shear stresses

are σxy and σyx. The notation adopted in labeling the stress components

allows immediate identification of the associated surface forces. The second

subscript on σ gives the direction of the force, and the first subscript gives

the direction of the normal to the surface on which the force acts.

The tangential or shear stresses σxy and σyx have associated surface forces

that tend to rotate the element in Figure 2–13 about the z axis. The moment

exerted by the surface force σxyδyδz is the product of the force and the

moment arm δx; that is, it is σxyδxδyδz. This couple is counteracted by the

moment σyxδxδyδz exerted by the surface force σyxδxδz with a moment arm

δy. Because the element cannot rotate if it is in equilibrium,

σxy = σyx. (2.27)

2.3 Stress in Two Dimensions 141

Figure 2.14 Transformation of stresses from the x, y coordinate system to the x ′, y ′ coordinate system. (a) Illustration of the coordinate systems. (b) Triangular element (with sides in the x, y, and y ′ directions) on which a static force balance is carried out.

Thus the shear stresses are symmetric in that their value is independent of

the order of the subscripts. Three independent components of stress σxx, σyy,

and σxy must be specified in order to prescribe the two-dimensional state of

stress.

The state of stress is dependent on the orientation of the coordinate sys-

tem. We will now determine the three components of stress in a coordinate

system x′, y′ inclined at an angle θ with respect to the x, y coordinate sys-

tem as illustrated in Figure 2–14a. To determine the normal stress, we carry

out a static force balance on the triangular element OAB illustrated in Fig-

ure 2–14b. The sides of the triangle lie in the x, y, and y′ directions. We first

write a force balance in the y direction. The force in the y direction on face

AO is

σyyAO,

and the force in the y direction on face OB is

σxyOB.

142 Stress and Strain in Solids

The force in the y direction on face AB is

−σx′x′AB sin θ − σx′y′AB cos θ.

The sum of these forces must be zero for the triangular element OAB to be

in equilibrium. This gives

(σx′x′ sin θ + σx′y′ cos θ)AB = σyyAO + σxyOB.

(2.28)

However, the sides of triangle OAB are in the ratios

AO

AB = sin θ

OB

AB = cos θ, (2.29)

so that

σx′x′ sin θ + σx′y′ cos θ = σyy sin θ + σxy cos θ.

(2.30)

We next write a force balance in the x direction. The force in the x

direction on face AO is

σyxAO,

and the force in the x direction on face OB is

σxxOB.

The force in the x direction on face AB is

−σx′x′AB cos θ + σx′y′AB sin θ.

Upon equating the sum of these forces to zero, we get

(σx′x′ cos θ − σx′y′ sin θ)AB = σyxAO + σxxOB.

(2.31)

With the values of AO/AB and OB/AB as given in Equation (2–29), we

find

σx′x′ cos θ − σx′y′ sin θ = σyx sin θ + σxx cos θ.

(2.32)

We multiply Equation (2–30) by sin θ, multiply Equation (2–32) by cos θ,

and add the results to obtain

σx′x′(cos 2 θ + sin2 θ) = σxx cos2 θ + σyy sin2 θ

+σxy sin θ cos θ

+σyx sin θ cos θ. (2.33)

2.3 Stress in Two Dimensions 143

This can be further simplified by using

cos2 θ + sin2 θ = 1 (2.34)

σxy = σyx (2.35)

2 sin θ cos θ = sin 2θ. (2.36)

The result is

σx′x′ = σxx cos2 θ + σyy sin2 θ + σxy sin 2θ. (2.37)

By multiplying Equation (2–30) by cos θ and subtracting the product of

Equation (2–32) with sin θ, we find

σx′y′(sin 2 θ + cos2 θ) = σyy sin θ cos θ + σxy cos2 θ

−σxx sin θ cos θ − σyx sin2 θ.

(2.38)

By using the trigonometric relations already quoted, the symmetry of the

shear stresses σxy and σyx and

cos 2θ = cos2 θ − sin2 θ, (2.39)

we can rewrite Equation (2–38) in the simpler form:

σx′y′ = 1 2 (σyy − σxx) sin 2θ + σxy cos 2θ. (2.40)

Equations (2–37) and (2–40) provide formulas for obtaining the nor-

mal and shear stresses on arbitrarily oriented elements of area in terms of

σxx, σyy, and σxy (or σyx). Thus these three stress components completely

specify the state of two-dimensional stress in a solid. When θ = 0, the equa-

tions show that σx′x′ is σxx and σx′y′ is σxy, as required.

Problem 2.13 Show that

σy ′y ′ = σxx sin2 θ + σyy cos2 θ − σxy sin 2θ. (2.41)

Problem 2.14 The state of stress at a point on a fault plane is σyy =

150 MPa, σxx = 200 MPa, and σxy = 0 (y is depth and the x axis points

westward). What are the normal stress and the tangential stress on the fault

plane if the fault strikes N–S and dips 35◦ to the west?

For any arbitrary two-dimensional state of stress σxx, σyy, σxy, it is pos-

sible to find a surface oriented in such a manner that no shear forces are

144 Stress and Strain in Solids

exerted on the surface. We need simply set σx′y′ in Equation (2–40) to zero

and solve for θ. Stress σx′y′ is zero if

tan 2θ = 2σxy

σxx − σyy . (2.42)

The direction θ defined by Equation (2–42) is known as a principal axis of

stress. If θ is a principal axis direction, then so is θ + π/2 because tan 2θ =

tan[2(θ + π/2)]; this can be seen as follows:

tan

[

2

(

θ + π

2

)]

= tan(2θ + π) = tan 2θ + tan π

1 − tan 2θ tanπ

= tan 2θ. (2.43)

The last step is true because tan π = 0. The coordinate axes defined by the

orthogonal principal axis directions are called the principal axes. There are

no shear stresses on area elements oriented perpendicular to the principal

axes.

The normal stresses in the principal axis coordinate system are known

as principal stresses. To solve for the principal stresses σ1 and σ2, substi-

tute Equation (2–42) into the expression for σx′x′ , Equation (2–37). Before

making the substitution, rewrite Equation (2–37) using the identities

sin2 θ = 1 − cos 2θ

2 (2.44)

cos2 θ = 1 + cos 2θ

2 . (2.45)

Equation (2–37) becomes

σx′x′ = σxx + σyy

2 +

1

2 cos 2θ(σxx − σyy) + σxy sin 2θ

(2.46)

or

σx′x′ = σxx + σyy

2 +

1

2 cos 2θ(σxx − σyy + 2σxy tan 2θ).

(2.47)

The determination of the principal stresses from Equation (2–47) requires

an expression for cos 2θ as well as for tan 2θ. The value of cos 2θ can be

obtained from the expression for tan 2θ using

tan2 2θ = sin2 2θ

cos2 2θ =

1 − cos2 2θ

cos2 2θ , (2.48)

2.3 Stress in Two Dimensions 145

which can be rewritten as

cos 2θ = 1

(1 + tan2 2θ)1/2 . (2.49)

By substituting Equation (2–42) into Equation (2–49), one finds

cos 2θ = σxx − σyy

±(4σ2 xy + (σxx − σyy)2)

1/2 . (2.50)

Upon substituting the expressions above for tan 2θ and cos 2θ into Equa-

tion (2–47), we get

σ1,2 = σxx + σyy

2 ± {

(σxx − σyy) 2

4 + σ2

xy

}1/2

.

(2.51)

Instead of specifying σxx, σyy, and σxy, we can describe the state of stress

at a point in a solid by giving the orientation of the principal axes and the

values of the principal stresses.

In deriving these formulas for the orientation of the principal axes and the

magnitudes of the principal stresses, we have tacitly assumed σxx−σyy 6= 0.

If σxx = σyy, then σx′y′ = σxy cos 2θ, and the principal axes have angles of

±45◦, assuming σxy 6= 0. If σxy = 0, the principal stresses are σxx and σyy.

If σxx = σyy and σxy 6= 0, the principal stresses are

(σxx + σyy)

2 ± σxy = σxx ± σxy. (2.52)

It is often convenient to have formulas for the normal and shear stresses for

an arbitrarily oriented coordinate system in terms of the principal stresses

and the angle of the coordinate system with respect to the principal axes.

To derive such formulas, consider the x, y axes in Figure 2–14 to be principal

axes so that σ1 = σxx, σ2 = σyy, and σxy = 0. The stresses σx′x′ , σx′y′ , and

σy′y′ are then given as

σx′x′ = σ1 cos2 θ + σ2 sin2 θ

= σ1 + σ2

2 +

(σ1 − σ2)

2 cos 2θ (2.53)

σx′y′ = −1 2(σ1 − σ2) sin 2θ (2.54)

σy′y′ = σ1 sin2 θ + σ2 cos2 θ

= σ1 + σ2

2 − (σ1 − σ2)

2 cos 2θ. (2.55)

At this point, there is no particular reason to retain the primes on the

146 Stress and Strain in Solids

coordinate axes. We can simplify future applications of Equations (2–53)

to (2–55) by identifying the x′, y′ coordinate axes as “new” x, y coordinate

axes. Therefore, if θ is considered to be the angle between the direction of

σ1 and the x direction (direction of σxx), we can write

σxx = σ1 + σ2

2 +

(σ1 − σ2)

2 cos 2θ (2.56)

σxy = −1 2(σ1 − σ2) sin 2θ (2.57)

σyy = σ1 + σ2

2 − (σ1 − σ2)

2 cos 2θ. (2.58)

Problem 2.15 Show that the sum of the normal stres- ses on any two

orthogonal planes is a constant. Evaluate the constant.

Problem 2.16 Show that the maximum and minimum normal stresses

act on planes that are at right angles to each other.

By differentiating Equation (2–40) with respect to θ and equating the

resulting expression to zero, we can find the angle at which the shear stress

σx′y′ is a maximum; the angle is given by

tan 2θ = (σyy − σxx)

2σxy . (2.59)

A comparison of Equations (2–42) and (2–59) shows that tan 2θ for the

principal axis orientation and tan 2θ for the maximum shear stress orien-

tation are negative reciprocals. Thus the angles 2θ differ by 90◦ and the

axes that maximize the shear stress lie at 45◦ to the principal axes. The

maximum value of the shear stress can thus be found by letting θ = π/4 in

Equation (2–57). One gets

(σxy)max = 1 2(σ1 − σ2). (2.60)

The maximum shear stress is half the difference of the principal stresses. It

is also obvious from Equation (2–57) that (σxy)max is exerted on a surface

whose normal is at 45◦ to the principal axes.

2.4 Stress in Three Dimensions

In three dimensions we require additional stress components to specify the

surface forces per unit area on surfaces of arbitrary orientation. Figure 2–15

shows the surface forces per unit area, that is, the stresses, on the faces

of a small rectangular parallelepiped. There are nine components of stress

required to describe the surface forces per unit area on the faces of the

2.4 Stress in Three Dimensions 147

Figure 2.15 Stress components on the faces of a small rectangular paral- lelepiped.

element. σxx, σyy, and σzz are the normal stresses, and σxy, σyx, σxz, σzx, σyz,

and σzy are shear stresses. If the parallelepiped is not to rotate about any

of its axes, then σxy = σyx, σxz = σzx, and σyz = σzy. Only six of the stress

components are independent.

The transformation of coordinates to principal axes can also be carried

out in three dimensions. Three orthogonal axes can always be chosen such

that all shear stress components are zero. The normal stresses on planes

perpendicular to these axes are the principal stresses, usually denoted as

σ1, σ2, and σ3. By convention these are chosen such that σ1 ≥ σ2 ≥ σ3.

Therefore, σ1 is the maximum principal stress, σ3 is the minimum principal

stress, and σ2 is the intermediate principal stress. The state of stress at a

point in a solid is completely specified by giving σxx, σyy, σzz, σxy, σyz, and

σxz or the orientation of the principal axes and the values of the principal

stresses.

Clearly two or even three of the principal stresses may be equal. When all

three are equal, the state of stress is isotropic and the principal stresses can

be identified as the pressure p = σ1 = σ2 = σ3. In any coordinate system the

normal stresses are equal to the pressure, and there are no shear stresses.

Any set of orthogonal axes qualifies as a principal axis coordinate system.

148 Stress and Strain in Solids

This is referred to as a hydrostatic state of stress. The lithostatic state of

stress is a hydrostatic state in which the stress increases proportionately

with depth at a rate controlled by the density of the rock. When the three

principal stresses are not equal, the pressure is defined to be their mean:

p= 1 3 (σ1 + σ2 + σ3). (2.61)

The pressure is invariant to the choice of coordinate system, that is, to the

orientation of the coordinate axes, so that it is equal to the mean of the

normal stresses in any coordinate system:

p= 1 3(σxx + σyy + σzz). (2.62)

Recall that we have taken normal stress to be positive for compression so

that it has the same sign as the pressure.

In studying stress in the Earth, it is often convenient to subtract the mean

stress, that is, the pressure, from the normal stress components. Accordingly,

we define deviatoric normal stresses by

σ′xx = σxx − p σ′yy = σyy − p σ′zz = σzz − p

σ′xy = σxy σ′xz = σxz σ′yz = σyz,

(2.63)

where primes refer to the deviatoric stresses. By definition, the average of

the normal deviatoric stresses is zero. Similarly deviatoric principal stresses

can be defined as

σ′1 = σ1 − p σ′2 = σ2 − p σ′3 = σ3 − p,

(2.64)

and their average is zero.

We can determine the orientation of the plane on which the shear stress

is a maximum, just as we did in the case of two-dimensional stress. The

direction of the normal to this plane bisects the angle between the directions

of the maximum and minimum principal stresses. The largest possible value

of the shear stress is (σ1 − σ3)/2.

2.5 Pressures in the Deep Interiors of Planets

Because rocks can readily deform on geologic time scales at the high temper-

atures encountered deep in planetary interiors, it is a good approximation

for many purposes to consider the planets to be in a hydrostatic state of

stress completely described by the dependence of pressure p on radius r.

2.5 Pressures in the Deep Interiors of Planets 149

Figure 2.16 Spherically symmetric model of a planet for the purpose of calculating p(r).

Pressure must increase with depth because the weight of the material above

any radius r increases as r decreases. The situation is completely analogous

to the lithostatic state of stress al- ready discussed. By differentiating Equa-

tion (2–13) with respect to y, we find that the rate of increase of pressure, or

lithostatic stress, with depth is ρg. In spherical coordinates, with spherical

symmetry, the rate of decrease of pressure with radius is given by

dp

dr = −ρg. (2.65)

In calculating the lithostatic stress near the surface of a planet, it is ade-

quate to consider g to be constant. However, deep in a planet g is a function

of radius, as shown in Figure 2–16. In addition, ρ is also generally a function

of r. The gravitational acceleration g(r) for a spherically symmetric body is

given by

g(r) = GM(r)

r2 , (2.66)

where G is the universal gravitational constant and M(r) is the mass inside

radius r.

M(r) =

∫ r

0 4πr′ 2ρ(r′) dr′. (2.67)

A further discussion of planetary gravity is given in Chapter 5. Given a

model of the density inside a planet, that is, given the form of ρ(r), one can

integrate Equation (2–67) to obtain M(r); g(r) follows from Equation (2–

66). Equation (2–65) can then be integrated to solve for p(r). In general,

however, ρ(r) is a function of p(r); so an equation of state is required.

150 Stress and Strain in Solids

For a small planet, such as the Moon, the procedure is particularly straight-

forward, since ρ can be considered a constant; M(r) is then 4 3πρr

3, and the

acceleration of gravity is

g(r) = 4 3πρGr. (2.68)

The equation for p is

dp

dr = −4

3πρ 2Gr, (2.69)

which upon integration gives

p = −2 3πρ

2Gr2 + c. (2.70)

The constant of integration c can be evaluated by equating the pressure to

zero at the surface of the body r = a. One obtains

p = 2 3πρ

2G(a2 − r2). (2.71)

Pressure is a quadratic function of radius in a small constant-density planet.

Problem 2.17 Determine the pressure at the center of the Moon. Assume

ρ = 3300 kg m−3 and a = 1738 km. What is the variation of g with radius

in the Moon?

Problem 2.18 Consider a simple two-layer model of a planet consisting

of a core of density ρc and radius b surrounded by a mantle of density ρm and thickness a − b. Show that the gravitational acceleration as a function

of radius is given by

g(r) = 4 3πρcGr 0 ≤ r ≤ b

= 4 3πG[rρm + b3(ρc − ρm)/r2] b ≤ r ≤ a.

(2.72)

and that the pressure as a function of radius is given by

p(r) = 4

3 πρmGb

3(ρc − ρm)

(

1

r − 1

a

)

+ 2

3 πGρ2

m(a2 − r2) b ≤ r ≤ a

= 2

3 πGρ2

c(b 2 − r2) +

2

3 πGρ2

m(a2 − b2)

+ 4

3 πρmGb

3(ρc − ρm)

(

1

b − 1

a

)

0 ≤ r ≤ b. (2.73)

Apply this model to the Earth. Assume ρm = 4000 kg m−3, b = 3486 km, a =

2.6 Stress Measurement 151

Figure 2.17 Schematic of overcoring technique for stress measurements. (a) A hole is drilled, and four strain gauges are installed: one on the side wall to measure σyy and three on the base to measure σxx, σxz, and σzz . It is assumed that the drilling of the hole has not affected the ambient state of stress. (b) The second annular hole is drilled. It is assumed that this annular hole completely relieves the initial stresses.

6371 km. Calculate ρc given that the total mass of the Earth is 5.97×1024 kg.

What are the pressures at the center of the Earth and at the core–mantle

boundary? What is the acceleration of gravity at r = b?

2.6 Stress Measurement

The direct measurement of stress is an important source of information on

the state of stress in the lithosphere. At shallow depths, the state of stress

is strongly affected by the presence of faults and joints, and stress measure-

ments near the surface yield little useful information on tectonic stresses in

the lithosphere. At sufficiently large depths, the lithostatic pressure closes

these zones of weakness, allowing stresses to be transmitted across them.

Stress measurements made at depth are thus directly interpretable in terms

of large-scale tectonic stresses. Stress measurements at depth are carried out

in mines and in deep boreholes. The two principal methods of making in situ

stress measurements are overcoring and hydrofracturing.

The first step in overcoring is to drill a hole in rock that is free of faults and

joints. Strain (deformation) gauges can be installed in three perpendicular

directions on the base of the hole and on the side, as illustrated in Figure 2–

17a (strain or deformation in response to stress is discussed quantitatively

in the following section). Alternatively two holes are drilled at right angles,

and strain gauges are installed on the bases of the two holes. We assume that

152 Stress and Strain in Solids

Figure 2.18 Pressure log during hydrofracturing.

the stress in the direction of the strain gauge is not affected by the drilling

of the hole. The hole is then overcored; that is, an annular hole with radius

larger than the original hole is drilled, as shown in Figure 2–17b. We assume

that the overcoring completely relieves the stresses in the isolated block of

rock to which the strain gauges have been attached. The displacements on

the strain gauges can then be used to determine the original state of stress.

An important limitation of this method is that the length of the hole used is

limited to about 1 m. To make overcoring measurements at greater depths,

it is necessary to drill the holes in mines.

The second method of direct stress measurement is hydrofracturing. In

this method a section of a borehole that is free of fractures or other porosities

is isolated using inflatable packers. The isolated section is then pressurized

by pumping fluid into it, and the pressure of the fluid is monitored. The

pressure is increased until a fracture occurs. The fluid pressure at which

the fracture occurs is referred to as the breakdown pressure pb. A typical

pressure–time history during hydrofracturing is illustrated in Figure 2–18.

If the pump is shut off immediately and the hydraulic circuit kept closed, an

instantaneous shut-in pressure (ISIP) is recorded, as illustrated in Figure 2–

18. This is the pressure that is just sufficient to hold the fracture open. If

the fluid pressure is dropped and then increased, the maximum pressure is

the ISIP when the fracture is once again opened.

Several assumptions are implicit in the interpretation of the pressure

record obtained during hydrofracturing. The first is that the resultant frac-

ture is in a vertical plane. The second is that the rock fractures in pure

tension so that the stress perpendicular to the fracture is the minimum

horizontal principal stress. With these assumptions the magnitude of the

2.6 Stress Measurement 153

Figure 2.19 Measured values of the minimum horizontal stress σmin as a function of depth y in the Cajon Pass borehole in California (Zoback and Healy, 1992). Also included in this figure are the vertical component of stress σyy shown by the solid line (assumed equal to the lithostatic pressure) and 0.6 σyy shown by the dashed line.

minimum horizontal principal stress is equal to the ISIP. Using theories for

the fracture of rock, the maximum horizontal principal stress can be deduced

from pb and ISIP, but with considerably less accuracy.

Measurements of the minimum horizontal stress σmin as a function of

depth in the Cajon Pass borehole in California are given in Figure 2–19.

This borehole is adjacent to the San Andreas fault in southern California

and was drilled to a depth of 3.5 km. A series of ISIP measurements were

carried out using both hydrofractures and preexisting fractures, and it is

assumed that these gave σmin.

In general, measurements of the vertical component of stress indicate that

it is nearly equal to the weight of the overburden, that is, the lithostatic

pressure. Using Equation (2–1) with ρg = 26.5 MPa km−1 the vertical com-

ponent of stress σyy is also given as a function of depth in Figure 2–19. The

measured stresses correlate reasonably well with 0.6 σyy.

Another technique used to determine the orientation of crustal stresses is

the observation of wellbore breakouts. Wellbore breakouts are the result of

localized failure around a borehole in response to horizontal compression.

154 Stress and Strain in Solids

Table 2.1 Stress Measurements at 200 m Depth vs. Distance from the San

Andreas Fault

Distance Maximum Minimum from Principal Principal Fault (km) Stress (MPa) Stress (MPa)

2 9 8 4 14 8

22 18 8 34 22 11

Compression produces spallation zones along the wellbore at the azimuth of

minimum principal stress where the circumferential compressive stress is a

maximum. The spallation zones can be used to infer the directions of the

horizontal principal stresses (Gough and Bell, 1981).

Observations of wellbore breakouts can be obtained from borehole tele-

viewer data. The borehole tele- viewer is an ultrasonic well-logging tool

which can image the orientation and distribution of fractures as well as the

orientation of stress-induced wellbore breakouts.

Problem 2.19 An overcoring stress measurement in a mine at a depth

of 1.5 km gives normal stresses of 62 MPa in the N–S direction, 48 MPa

in the E–W direction, and 51 MPa in the NE–SW direction. Determine the

magnitudes and directions of the principal stresses.

Problem 2.20 The measured horizontal principal stresses at a depth of

200 m are given in Table 2–1 as a function of distance from the San Andreas

fault. What are the values of maximum shear stress at each distance?

2.7 Basic Ideas about Strain

Stresses cause solids to deform; that is, the stresses produce changes in the

distances separating neighboring small elements of the solid. In the discus-

sion that follows we describe the ways in which this deformation can occur.

Implicit in our discussion is the assumption that the deformations are small.

Figure 2–20 shows a small element of the solid in the shape of a rectangular

parallelepiped. Prior to deformation it has sides δx, δy, and δz. The element

may be deformed by changing the dimensions of its sides while maintaining

its shape in the form of a rectangular parallelepiped. After deformation, the

sides of the element are δx−εxxδx, δy−εyyδy, and δz−εzzδz. The quantities

εxx, εyy, and εzz are normal components of strain; εxx is the change in length

2.7 Basic Ideas about Strain 155

Figure 2.20 A deformation that changes the dimensions of a rectangular parallelepiped but not its shape.

of the side parallel to the x axis divided by the original length of the side,

and εyy and εzz are similar fractional changes in the lengths of the sides

originally parallel to the y and z axes, respectively. The normal components

of strain εxx, εyy, and εzz are assumed, by convention, to be positive if

the deformation shortens the length of a side. This is consistent with the

convention that treats compressive stresses as positive.

If the deformation of the element in Figure 2–20 is so small that squares

and higher order products of the strain components can be neglected in

computing the change in volume of the element, the fractional change in

volume (volume change divided by original volume) is εxx + εyy + εzz. This

quantity is known as the dilatation ∆; it is positive if the volume of the

element is decreased by compression.

Problem 2.21 Uplift and subsidence of large areas are also accompanied

by horizontal or lateral strain because of the curvature of the Earth’s surface.

Show that the lateral strain ε accompanying an uplift ∆y is given by

ε = ∆y

R , (2.74)

where R is the radius of the Earth.

Problem 2.22 The porosity φ of a rock is defined as its void volume

per unit total volume. If all the pore spaces could be closed, for example,

by subjecting the rock to a sufficiently large pressure, what would be the

dilatation? For loose sand φ is about 40%, and for oil sands it is usually in

the range of 10 to 20%. Table 2–2 gives the porosities of several rocks.

The strain components of a small element of solid can be related to the

displacement of the element. In order to simplify the derivation of this re-

lationship, we consider the two-dimensional example in Figure 2–21. Prior

to deformation, the rectangular element occupies the position pqrs. After

156 Stress and Strain in Solids

Table 2.2 Rock Porosities

Rock Porosity (%)

Hasmark dolomite 3.5 Marianna limestone 13.0 Berea sandstone 18.2 Muddy shale 4.7 Repetto siltstone 5.6

Figure 2.21 Distortion of the rectangular element pqrs into the rectangular element p′q′r′s′.

deformation, the element is in the position p′q′r′s′. It is assumed to retain a

rectangular shape. The coordinates of the corner p before strain are x and

y; after strain the corner is displaced to the location denoted by p′ with

coordinates x′, y′. The displacement of the corner p as a result of the strain

or deformation is

wx(x, y) = x− x′ (2.75)

in the x direction and

wy(x, y) = y − y′ (2.76)

in the y direction. Displacements in the negative x and y directions are con-

sidered positive to agree with the sign convention in which positive strains

imply a contraction. Corner q at x + δx, y is displaced to position q′ with

coordinates x′ + δx′, y′ as a result of the deformation. Its displacement in

2.7 Basic Ideas about Strain 157

the x direction is

wx(x+ δx, y) = x+ δx− (x′ + δx′). (2.77)

Similarly, the displacement of corner s in the y direction wy(x, y + δy) is

given by the difference in the y coordinates of s′ and s

wy(x, y + δy) = y + δy − (y′ + δy′). (2.78)

In writing Equations (2–77) and (2–78), we have assumed that the strains

(δx− δx′)/δx and (δy − δy′)/δy are small.

Since δx and δy are infinitesimal, we can expand wx(x+δx, y) and wy(x, y+

δy) as

wx(x+ δx, y) = wx(x, y) + ∂wx ∂x

δx (2.79)

wy(x, y + δy) = wy(x, y) + ∂wy ∂y

δy. (2.80)

Substitution of Equation (2–79) into Equation (2–77) and subtraction of

Equation (2–75) yield

δx = δx′ + ∂wx ∂x

δx. (2.81)

Similarly, substitution of Equation (2–80) into Equation (2–78) and subtrac-

tion of Equation (2–76) yield

δy = δy′ + ∂wy ∂y

δy. (2.82)

From the definitions of the strain components and Equations (2–81) and

(2–82) we find

εxx ≡ δx− δx′

δx = ∂wx ∂x

(2.83)

εyy ≡ δy − δy′

δy = ∂wy ∂y

. (2.84)

In three-dimensional strain, the third strain component εzz is clearly given

by

εzz = δz − δz′

δz = ∂wz ∂z

. (2.85)

The components of strain in the x, y, and z directions are proportional to

the derivatives of the associated displacements in the respective directions.

The dilatation ∆ is given by

∆ = ∂wx ∂x

+ ∂wy ∂y

+ ∂wz ∂z

. (2.86)

158 Stress and Strain in Solids

Figure 2.22 Distortion of a rectangle into a parallelogram by a strain field involving shear.

We have so far considered strains or deformations that do not alter the

right angles between line elements that are mutually perpendicular in the

unstrained state. Shear strains, however, can distort the shapes of small

elements. For example, Figure 2–22 shows a rectangular element in two

dimensions that has been distorted into a parallelogram. As illustrated in

this figure, the shear strain εxy is defined to be one-half of the decrease in

the angle SPQ

εxy ≡ −1 2(φ1 + φ2), (2.87)

where φ1 and φ2 are the angles through which the sides of the original

rectangular element are rotated. The sign convention adopted here makes

εxy negative if the original right angle is altered to an acute angle. As in the

case of stress, the shear strain is symmetric so that εyx = εxy. Figure 2–22

shows that the angles φ1 and φ2 are related to the displacements by

tanφ1 = −wy(x+ δx, y)

δx = φ1 (2.88)

tanφ2 = −wx(x, y + δy)

δy = φ2. (2.89)

In Equations (2–88) and (2–89), we assume that the rotations are infinites-

imal so that the tangents of the angles are very nearly equal to the angles

themselves.

We can express wy(x + δx, y) and wx(x, y + δy) in terms of the spatial

2.7 Basic Ideas about Strain 159

derivatives of the displacements according to

wy(x+ δx, y) = ∂wy ∂x

δx (2.90)

wx(x, y + δy) = ∂wx ∂y

δy, (2.91)

Where, for simplicity, we assume wx(x, y) = 0 and wy(x, y) = 0. Substitution

of Equations (2–90) and (2–91) into Equations (2–88) and (2–89) and further

substitution of the resulting expressions for φ1 and φ2 into Equation (2–87)

yield

εxy = 1

2

(

∂wy ∂x

+ ∂wx ∂y

)

(2.92)

as the relation between shear strain and the spatial derivatives of displace-

ments. In the engineering literature, γxy = 2εxy is often used. Care should

be exercised in dealing with these quantities.

Shear strain can also lead to a solid-body rotation of the element if φ1 6= φ2. The solid-body rotation ωz is defined by the relation

ωz = −1

2 (φ1 − φ2). (2.93)

Substitution of Equations (2–88) and (2–89) into Equation (2–93) gives

ωz = 1

2

(

∂wy ∂x

− ∂wx ∂y

)

. (2.94)

The rotation of any element can be resolved in terms of the shear strain

and the solid-body rotation. From Equations (2–87) and (2–93), the angle

φ1 through which a line element parallel to the x axis is rotated is

φ1 = −(εxy + ωz), (2.95)

and the angle φ2 through which a line element in the y direction is rotated

is

φ2 = ωz − εxy. (2.96)

Thus, in the absence of solid-body rotation, εxy is the clockwise angle through

which a line element in the x direction is rotated. It is also the counterclock-

wise angle through which a line element in the y direction is rotated.

If the amount of solid-body rotation is zero, the distortion is known as

pure shear. In this case, illustrated in Figure 2–23a,

φ1 = φ2 (2.97)

∂wy ∂x

= ∂wx ∂y

(2.98)

160 Stress and Strain in Solids

Figure 2.23 Sketch of (a) pure shear strain that involves no solid-body rotation of elements and (b) simple shear strain that includes such rotation.

and the shear strain is

εxy = ∂wx ∂y

= ∂wy ∂x

. (2.99)

The case of simple shear, shown in Figure 2–23b, combines solid-body rota-

tion and shear in such a manner that

φ1 = ∂wy ∂x

= 0. (2.100)

From Equation (2–94), the amount of solid-body rotation is

ωz = −1

2

∂wx ∂y

, (2.101)

and the shear strain is

εxy = 1

2

∂wx ∂y

. (2.102)

Simple shear is often associated with strike–slip faulting.

The equations given for two-dimensional strains and solid-body rotation

about one axis can be readily generalized to three dimensions. A pure shear

2.7 Basic Ideas about Strain 161

strain in the xz plane has an associated shear strain component given by

εxz = εzx = 1

2

(

∂wz ∂x

+ ∂wx ∂z

)

(2.103)

and a pure shear strain in the yz plane corresponds to

εyz = εzy = 1

2

(

∂wz ∂y

+ ∂wy ∂z

)

. (2.104)

A solid-body rotation about the x axis ωx is related to displacement deriva-

tives by

ωx = 1

2

(

∂wz ∂y

− ∂wy ∂z

)

. (2.105)

Similarly, a solid-body rotation about the y axis is

ωy = 1

2

(

∂wx ∂z

− ∂wz ∂x

)

. (2.106)

The strain components εxx, εyy, εzz, εxy, εxz, and εyz are sufficient to de-

scribe the general infinitesimal deformation of solid elements subjected to

stresses. The solid-body rotations ωx, ωy, and ωz do not alter distances be-

tween neighboring elements of a solid and, therefore, do not involve stresses.

Accordingly, the strain components and their associated stresses are of pri-

mary concern to us in subsequent chapters.

Just as it was important to know the stresses on area elements whose

normals make arbitrary angles with respect to x, y axes, so it is essential to

know the fractional changes in length and the rotation angles of arbitrarily

inclined line elements. For simplicity we consider the two-dimensional case.

We wish to determine the strains in the x′, y′ coordinate system, which is

inclined at an angle θ with respect to the x, y coordinate system, as shown in

Figure 2–24a. As a result of the strain field εxx, εyy , εxy and the solid-body

rotation ωz, the line elements PR and PQ experience changes in length and

rotations. Line element PR is parallel to the x′ axis, and PQ is parallel to

the y′ axis. The extension in length of PR divided by the original length

δx′ is the strain component −εx′x′ ; the counterclockwise angle of rotation of

PR is the angle φ′1 = −εx′y′ − ωz′ . This is illustrated in Figure 2–24b. The

extension in length of PQ divided by the original length δy′ is the strain

component −εy ′y ′; the clockwise rotation of PQ is the angle φ′2 = ωz ′−εx′y ′ .

This is shown in Figure 2–24c.

We first determine the strain component −εx′x′ . The displacement of R

to R ′ in Figure 2–24b is the net result of the combined elongations and

rotations of δx and δy. The x component of the displacement of R ′ relative

to R arises from the elongation of δx in the x direction, −εxxδx, and the

162 Stress and Strain in Solids

rotation of δy through the clockwise angle φ2. The latter contribution to the

displacement is φ2δy, which, according to Equation (2–96), is (ωz − εxy)δy.

Thus the total x component of the displacement of R ′ with respect to R is

−εxxδx+ (ωz − εxy)δy.

The y component of the displacement of R ′ with respect to R is the sum

of the elongation of δy,−εyyδy, and the contribution from the rotation of

δx, which, with Equation (2–95), is φ1δx = −(εxy +ωz)δx. Thus the total y

component of displacement of R ′ with respect to R is

−εyyδy − (εxy + ωz)δx.

For small strains, the change in length of PR is the sum of the x component

of RR ′ projected on the line PR,

[−εxxδx+ (ωz − εxy)δy] cos θ,

and the y component of RR ′ projected on the line PR,

[−εyyδy − (εxy + ωz)δx] sin θ.

The strain component εx′x′ is thus

−εx′x′ = [−εxxδx+ (ωz − εxy)δy] cos θ

δx′

+ [−εyyδy − (εxy + ωz)δx] sin θ

δx′ . (2.107)

Since

δx

δx′ = cos θ

δy

δx′ = sin θ (2.108)

Equation (2–107) can be rewritten as

εx′x′ = εxx cos2 θ + εyy sin2 θ + 2εxy sin θ cos θ.

(2.109)

Using Equation (2–36), we can further rewrite Equation (2–109) as

εx′x′ = εxx cos2 θ + εyy sin2 θ + εxy sin 2θ. (2.110)

This has the same form as the transformation of the normal stress given in

Equation (2–37).

We next determine the strain component −εy ′y ′ . As can be seen in Fig-

ure 2–24c, the component of the displacement of Q′ with respect to Q in

2.7 Basic Ideas about Strain 163

Figure 2.24 (a) The transformation of coordinates x, y through an angle θ to x ′, y ′. (b) The transformation of the strain components onto the x ′

axis. (c) The transformation of the strain components onto the y ′ axis.

the negative x direction is the sum of the elongation of δx, −εxxδx, and the

contribution from the rotation of δy, −φ2δy = −(ωz − εxy)δy, that is,

−εxxδx− (ωz − εxy)δy.

The y component of the displacement of Q ′ with respect to Q is the sum of

the elongation of δy,−εyyδy, and the contribution due to the rotation of δx,

−φ1δx = (εxy + ωz)δx, that is,

−εyyδy + (ωz + εxy)δx.

After projection of these displacements onto the line PQ, the strain compo-

nent εy ′y′ can be written as

−εy ′ y ′ = −[εxxδx+ (ωz − εxy)δy] sin θ

δy′

164 Stress and Strain in Solids

Figure 2.25 Geometrical determination of (a) φ′1 and (b) φ′2.

+ [−εyyδy + (ωz + εxy)δx] cos θ

δy′ . (2.111)

Since

δx

δy′ = sin θ

δy

δy ′ = cos θ, (2.112)

Equation (2–111) can be put in the form

εy ′y ′ = εxx sin2 θ + εyy cos2 θ − 2εxy sin θ cos θ.

(2.113)

By substituting Equation (2–36) into Equation (2–113), we get

εy ′y ′ = εxx sin2 θ + εyy cos2 θ − εxy sin 2θ. (2.114)

Problem 2.23 Derive Equation (2–114) from Equation (2–110) by using

the substitution θ′ = θ + π/2. Why can this be done?

We now turn to the determination of the shear strain, εx′y ′ , and the solid-

body rotation ωz ′ in the new coordinate system. We first determine the angle

φ′1 = −εx′y ′ −ωz′ from the geometrical relationships shown in Figure 2–25a.

For sufficiently small strain, φ′1 is given by

φ′1 = −εx′y′ − ωz′ = R′V

δx′ . (2.115)

From Figure 2–25a we can see that

R′V = R′U − V U = R′U − TS, (2.116)

2.7 Basic Ideas about Strain 165

and

R′U = [−εyyδy − (εxy + ωz)δx] cos θ (2.117)

TS = [−εxxδx+ (ωz − εxy)δy] sin θ. (2.118)

By combining Equations (2–108) and (2–115) with (2–118), we obtain

εx′y ′ + ωz ′ = (εyy − εxx) sin θ cos θ

+ εxy(cos 2 θ − sin2 θ) + ωz. (2.119)

The angle φ′2 can be found from the geometrical relationships shown in

Figure 2–25b; it is given by

φ′2 = ωz ′ − εx′y ′ = U ′S′

δy′ . (2.120)

From Figure 2–25b it is seen that

U ′S′ = U ′T ′ + T ′S′, (2.121)

and

U ′T ′ = −[−εxxδx− (ωz − εxy)δy] cos θ (2.122)

T ′S′ = [−εyyδy + (ωz + εxy)δx] sin θ. (2.123)

By combining Equations (2–112) and (2–120) with (2–123), we obtain

ωz′ − εx′y′ = (εxx − εyy) sin θ cos θ

− εxy(cos 2 θ − sin2 θ) + ωz. (2.124)

By adding and subtracting Equations (2–119) and (2–124), we can find

separate equations for ωz ′ and εx′y ′ :

ωz ′ = ωz (2.125)

εx′y′ = (εyy − εxx) sin θ cos θ + εxy(cos 2 θ − sin2 θ).

(2.126)

The solid-body rotation is invariant to the coordinate transformation, as

expected, because it represents a rotation of an element without deforma-

tion. By introducing Equations (2–36) and (2–39) into Equation (2–126), we

obtain

εx′y ′ = 1 2(εyy − εxx) sin 2θ + εxy cos 2θ. (2.127)

This has the same form as the transformation of the shear stress given in

Equation (2–40).

Just as there are principal axes of stress in a solid, there are principal axes

166 Stress and Strain in Solids

of strain. In the principal strain axis coordinate system, shear strain com-

ponents are zero. Setting εx′y ′ = 0 in Equation (2–127) gives the direction

of one of the principal axes of strain as

tan 2θ = 2εxy

εxx − εyy . (2.128)

We have already shown, in connection with principal stress axes, that if θ is a

principal axis direction, so is θ+π/2. The fractional changes in length along

the directions of the principal strain axes are the principal strains. With θ

given by Equation (2–128), Equation (2–110) determines the principal strain

ε1 = εx′x′ . The principal strain ε2 is identified with εy ′y ′ . By a procedure

analogous to the one used in deriving Equation (2–51) we find

ε1,2 = 1 2 (εxx + εyy) ±

{

ε2xy + 1 4(εxx − εyy)

2 }1/2

.

(2.129)

It is convenient to have formulas for the normal and shear strains at an

angle θ with respect to the ε1 principal strain axis. Taking εxy = 0, εxx = ε1,

and εyy = ε2 in Equations (2–109) and (2–127), we obtain

εxx = ε1 cos2 θ + ε2 sin2 θ (2.130)

εxy = −1 2(ε1 − ε2) sin 2θ. (2.131)

Problem 2.24 Show that the principal strains are the minimum and the

maximum fractional changes in length.

Problem 2.25 Show that the maximum shear strain is given by 1 2(ε1−ε2).

What is the direction in which the shear strain is maximum?

Principal axes of strain can also be found for arbitrary three-dimensional

strain fields. With respect to these axes all shear strain components are zero.

The normal strains along these axes are the principal strains ε1, ε2, and ε3.

One can introduce the concept of deviatoric strain in analogy to deviatoric

stress by referring the strain components to a state of isotropic strain equal

to the average normal strain e. In three dimensions

e ≡ 1 3 (εxx + εyy + εzz) = 1

3∆. (2.132)

The average normal strain and the dilatation are invariant to the choice of

coordinate axes. The deviatoric strain components, denoted by primes, are

ε′xx = εxx − e ε′yy = εyy − e ε′zz = εzz − e

ε′xy = εxy ε′xz = εxz ε′yz = εyz . (2.133)

2.8 Strain Measurements 167

Figure 2.26 A fence offset by 3 m on the ranch of E. R. Strain, Marin County, California, as a result of slip along the San Andreas fault during the great 1906 earthquake (G. K. Gilbert 3028, U.S. Geological Survey.)

2.8 Strain Measurements

Strain or deformation at the Earth’s surface is often a consequence of large-

scale tectonic forces. Thus the measurement of surface strain can provide

important information on fundamental geodynamic processes. For example,

in order to understand the mechanical behavior of faults, it is essential to

determine the distribution of the coseismic surface strain as a function of

distance from the fault, a problem we discuss further in Chapter 8. Because

surface strains are generally very small, sophisticated distance-measuring

techniques are usually required to determine them. However, there are in-

stances in which surface displacements are so large that they can be easily

measured. An example is the surface offset on a fault when a great earth-

quake occurs; offsets of 10 m and more have been recorded. Tree lines, roads,

railroad tracks, pipelines, fences, and the like can be used to make such

measurements. Figure 2–26 shows a fence offset by 3 m during the 1906

earthquake on the San Andreas fault in California. Measured surface offsets

resulting from this earthquake are summarized in Figure 2–27. Although

there is considerable scatter in the data, an offset of about 4 m was ob-

served along much of the fault break. The scatter of the data illustrates one

168 Stress and Strain in Solids

of the principal problems in measuring surface strain. The Earth’s crust is

not a continuum material with uniform properties. Changes in rock type,

the presence of thick soil layers, and offsets on branching or secondary faults

all contribute to the variations in the measured offsets.

The strain field associated with the 1906 earthquake can be estimated

from the data in Figure 2–27. Since the San Andreas fault is a strike–slip

fault, we assume that the strain field is a simple shear and that it extends

40 km from the fault. The distance that the cyclic strain field extends from

the fault is considered in detail in Chapter 8. The value of 40 km is subject

to considerable uncertainty. The mean displacement of 4 m across the fault

during the earthquake is made up of 2 m displacements on opposite sides of

the fault. The shear strain εxz can thus be estimated from Equation (2–102)

as

εxz = 1

2

∂wx ∂z

≈ 1

2

2m

40,000 m = 2.5 × 10−5. (2.134)

If great earthquakes recur about every 100 years along the San Andreas

fault, the rate of shear strain accumulation on the fault ε̇xz is

ε̇xz = 2.5 × 10−5

100 yr = 0.25 × 10−6 yr−1. (2.135)

As we have already noted, surface strains of the magnitude calculated in

Equation (2–134) are difficult to measure; they require extremely accurate

determinations of distances. This has been the main concern of geodesy for

several centuries. The traditional end product of geodetic surveys is the to-

pographic map, constructed from the elevations of a network of benchmarks.

Benchmarks are spaced over much of the United States at intervals of a

few kilometers and ground surveys are used to establish accurate bench-

mark elevations in a geodetic network. Geodetic networks are systematically

resurveyed to determine the changes in elevation between benchmarks. Ver-

tical displacements of benchmarks on the order of 10 to 100 cm are often

found. In many instances, these displacements can be associated with sub-

sidence due to the removal of ground water. However, in other cases they

must be attributed to tectonic causes. Extensive geodetic measurements

have been made along the San Andreas fault, and these are used to illus-

trate the concepts discussed in this section. Because the San Andreas fault is

a strike–slip fault, the displacements associated with slip along the fault are

predominantly horizontal. Thus we restrict our attention to the horizontal

components of strain εxx, εxz , and εzz. Horizontal strains are obtained by

measuring the positions of fixed monuments.

Historically, the standard method for determining the positions of monu-

2.8 Strain Measurements 169

F ig

u re

2. 27

O bs

er ve

d su

rf a ce

o ff se

ts o n

th e

S a n

A n d re

a s

fa u lt

re su

lt in

g fr o m

th e

1 9 0 6

ea rt

h qu

a ke

(T h a tc

h er

, 1 9 7 5 ).

170 Stress and Strain in Solids

Figure 2.28 Illustration of triangulation. The x, z position of monument C can be determined from the line length AB and the angles θ1 and θ2. The positions of the other monuments can be similarly determined.

ments is triangulation, as illustrated in Figure 2–28. Assume that the abso-

lute positions of the reference points A and B have been determined. The

location of monument C can be found from the line length AB and the

two angles θ1 and θ2 as follows. Applying the law of sines to triangle ABC

produces

(AB)

sin(π − θ1 − θ2) =

(AC)

sin θ2 . (2.136)

If we assume that point A defines the origin of the x, z coordinate system

and that AB lies on the z axis, the coordinates of C(xc, zc) are given by

xc = (AC) sin θ1 zc = (AC) cos θ1. (2.137)

By solving Equation (2–136) for (AC) and substituting into Equation (2–

137), we obtain

xc = (AB) sin θ1 sin θ2 sin(π − θ1 − θ2)

(2.138)

zc = (AB) cos θ1 sin θ2 sin(π − θ1 − θ2)

. (2.139)

The locations of the other monuments in the triangulation network can be

similarly determined. The use of redundant triangles, as shown in Figure 2–

28, improves the accuracy of the results.

Problem 2.26 The coordinates xA, zA and xB , zB of monuments A and B

shown in Figure 2–29 are assumed known. Determine the coordinates xC , zC of monument C in terms of the coordinates of monuments A and B and the

angles θ1 and θ2.

The angles required for triangulation are obtained using a theodolite. The

2.8 Strain Measurements 171

Figure 2.29 Sketch for Problem 2–26.

Figure 2.30 (a) The three monuments at Mount Diablo, Mocho, and Mount Toro are part of a primary triangulation network that spans the San An- dreas fault south of San Francisco. (b) Observed changes in the angle θ between the monuments since 1882 (Savage and Burford, 1973).

accuracy to which an angle can be determined is 0.3 to 1.0 second of arc,

implying errors in distance determination of about 3 in 106. A typical max-

imum length over which a measurement is made is 50 km. The accuracy

of triangulation observations is equivalent to about 10 years of shear strain

accumulation on the San Andreas fault; see Equation (2–135). Therefore,

considerable redundancy in a network is required to obtain meaningful re-

sults.

An example of triangulation observations is given in Figure 2–30. Mount

Diablo, Mocho, and Mount Toro are three monuments in a primary trian-

gulation network that spans the San Andreas fault south of San Francisco

(Figure 2–30a). The changes in the angle θ between these monuments in

a series of surveys since 1882 are given in Figure 2–30b. Mount Toro lies

30 km southwest of the San Andreas fault, and Mocho lies 60 km north-

172 Stress and Strain in Solids

Figure 2.31 A triangulation net across the San Andreas fault (a) and the measured angle θ since 1855 (b).

east of the fault. If it is assumed that these monuments lie outside the zone

of strain accumulation and release associated with great earthquakes, the

relative motion across the San Andreas fault can be obtained from these

observations. A reasonable fit to the data is dθ/dt = −0.192 sec of arc yr−1.

The length of the line between Mocho and Mount Toro is 125 km, and

it crosses the San Andreas fault at an angle of 45◦. The calculated relative

velocity across the fault is thus

u = 125 × 106 × 0.192

3600 × 57.3 × sin 45◦ = 41 mm yr−1. (2.140)

This value is in quite good agreement with the predicted relative velocity of

46 mm yr−1 from plate tectonics (see Section 1–8).

Problem 2.27 Figure 2–31 shows three monuments on Mount Diablo,

Sonoma Mountain, and Farallon lighthouse and the change in the included

angle θ relative to the 1855 measurement. Assuming that these three mon-

uments lie outside the zone of strain accumulation and release on the San

Andreas fault, determine the relative velocity across the fault.

Problem 2.28 Triangulation measurements at monument 0 give the time

rate of change of θ1, θ̇1 and the time rate of change of θ2, θ̇2 (Figure 2–32).

Show that

ε̇xy = 1

2

(θ̇2 sec θ2 csc θ2 − θ̇1 sec θ1 csc θ1)

(tan θ2 − tan θ1) (2.141)

2.8 Strain Measurements 173

Figure 2.32 Sketch for Problem 2–28.

and

ε̇yy − ε̇xx = (θ̇2 csc2 θ2 − θ̇1 csc2 θ1)

(ctn θ1 − ctn θ2) , (2.142)

where ε̇xx = dεxx/dt, and so on.

As we have shown, the accuracy of triangulation measurements is gener-

ally insufficient to obtain useful data on strain accumulation. Fortunately

electro-optical distance-measuring instruments greatly improve the accuracy

of strain measurements. However, they also greatly increase the expense. To

make a distance measurement, a geodolite is placed on one monument and

a reflector on the second monument. The geodolite emits a modulated laser

beam that is reflected back to the instrument from the reflector. A com-

parison of the modulated phases of the emitted and returned beams deter-

mines the length of the optical path between the monuments as an unknown

number of whole modulation lengths plus a precisely determined fractional

modulation length. The unknown number of whole modulation lengths is

determined by carrying out measurements at successively lower modulation

frequencies.

Distances between measured monuments are typically 10 km, and the in-

herent accuracy of the geodimeter is about 1 mm. Therefore an accuracy of

1 part in 107 can in principle be achieved. This is about an order of magni-

tude better than triangulation measurement accuracy. In practice, however,

the accuracy of distance determinations is limited by variations in refractiv-

ity along the atmospheric path. In order to obtain accuracies approaching

1 part in 107, it is necessary to determine the temperature and humidity

along the path. This is usually done by flying a suitably equipped airplane

or helicopter along the path while the distance measurement is being car-

174 Stress and Strain in Solids

Figure 2.33 Illustration of how strain measurements between three monu- ments A, B, and C can be used to determine the strain field εxx, εzz, and εxz.

ried out. The atmospheric pressure at the two terminal monuments is also

required.

By carrying out measurements at three different wavelengths or frequen-

cies, devices with multiwavelength capabilities have eliminated the need for

meteorological observations along the optical path. This approach reduces

costs and improves accuracy to a few parts in 108.

If it is assumed that the three monuments A,B, and C in Figure 2–33 are

in a uniform strain field, measurements of the rates of change in the three line

lengths ε̇xx = −∆AB/AB, ε̇x′x′ = −∆AC/AC, and ε̇x′′x′′ = −∆BC/BC

and the angles θ1 and θ2 give the entire rate of strain field ε̇xx, ε̇zz , and ε̇xz.

From Equation (2–109) we have

ε̇x′x′ = ε̇xx cos2 θ1 + ε̇zz sin2 θ1 + 2ε̇xz sin θ1 cos θ1

(2.143)

ε̇x′′x′′ = ε̇xx cos2 θ2 + ε̇zz sin2 θ2 + 2ε̇xz sin θ2 cos θ2.

(2.144)

These equations can be solved for ε̇zz and ε̇xz; we find

ε̇zz = ε̇xx(ctn θ1 − ctn θ2) − ε̇x′x′ sec θ1 csc θ1

tan θ2 − tan θ1

+ ε̇x′′x′′ sec θ2 csc θ2 tan θ2 − tan θ1

(2.145)

ε̇xz = ε̇xx(ctn

2θ1 − ctn 2θ2) − ε̇x′x′ csc 2 θ1

2(ctn θ2 − ctn θ1)

+ ε̇x′′x′′ csc

2 θ2 2(ctn θ2 − ctn θ1)

. (2.146)

2.8 Strain Measurements 175

The results can be transformed into any other coordinate system using Equa-

tions (2–109), (2–114), and (2–127).

As an example of the direct measurement of strain accumulation, consider

the data given in Figure 2–34. We assume that the three lines measured are

in a uniform strain field. We further assume that line 7 (length 28 km) defines

the x coordinate so that ε̇xx = 8/(28×106) yr−1 = 0.29×10−6 yr−1 and that

line 10 (length 31 km) defines the z coordinate so that ε̇zz = −5/(31× 106)

yr−1 = −0.16 × 10−6 yr−1. The angle between lines 9 and 10 is 30◦. The

rate of strain on line 9 (length 25 km) is ε̇x′x′ = −1.5/(25 × 106) yr−1 =

−0.6×10−6 yr−1. The rate of shear strain ε̇xz can be determined by inverting

Equation (2–109):

ε̇xz = 1 2 (ε̇x′x′ sec θ csc θ − ε̇xxctn θ − ε̇zz tan θ).

(2.147)

With θ = 120◦ and the previously determined values of ε̇x′x′ , ε̇xx, and ε̇zz we obtain

ε̇xz = 1 2 (−0.6 × 1.15 × 2 − 0.29 × 1.73

+0.16 × 0.58)× 10−6 yr−1

= −0.90 × 10−6 yr−1. (2.148)

From Equation (2–128) the direction of one of the principal strain rate

axes relative to the x axis is θ = −38◦. Assuming that line 9 trends N–S,

the directions of the principal strain rate axes are 22◦ W of N and 22◦ N

of E. These can be compared with the principal axis directions expected for

a simple shear strain model of strain accumulation in this region; since the

San Andreas trends 45◦ W of N in this area, the expected directions are

north and east.

The values of the principal strain rates from Equation (2–129) are ε̇1.2 =

0.993 × 10−6 yr−1, −0.863 × 10−6 yr−1. Assuming ε̇xz = ε̇1 = −ε̇2 = 0.93 × 10−6 yr−1 (the average of the above two values) and that simple shear is

occurring uniformly for a distance d from the fault, d can be determined from

the shear strain rate and the relative velocity u = 46 mm yr−1 according to

d = u

4ε̇xz =

46 × 10−3

4 × 0.93 × 10−6 = 12.4 km. (2.149)

With the uniform strain assumption, the strain accumulation would be lim-

ited to a region closer to the fault than the geodetic net considered.

Problem 2.29 Given in Figure 2–35 are the line lengths between the mon-

ument at Diablo and the monuments at Hills, Skyline, and Sunol obtained

176 Stress and Strain in Solids

F igu

re 2.34

L in

e len

gth ch

a n ges

betw een

th e

m o n u m

en t a t M

t. H

a m

ilto n

a n d

th e

m o n u m

en ts

a t A

lliso n

(lin e

7 ),

L o m

a P rieta

(lin e

1 0 ),

a n d

M o rga

n (lin

e 9 )

o bta

in ed

betw een

1 9 6 1

a n d

1 9 7 1

u sin

g a

geod im

eter (S

a va

ge a n d

B u rfo

rd , 1 9 7 3 ).

2.8 Strain Measurements 177

Figure 2.35 Geodetic net and measurements for use in Problem 2–29. The line length is L.

between 1970 and 1978 using a geodimeter. Assuming a uniform strain field,

determine ε̇xx, ε̇yy, and ε̇xy. Take the Sunol–Diablo line to define the y co-

ordinate. Discuss the results in terms of strain accumulation on the San

Andreas fault, which can be assumed to trend at 45◦ with respect to the

Sunol–Diablo line (Savage and Prescott, 1978).

Advances in space geodesy have revolutionized geodetic investigations of

tectonic motions. Studies carried out in the 1980s utilized satellite laser

ranging (SLR) and very long baseline interferometry (VLBI). SLR measures

distances from a ground station to various satellites using an electro-optical

instrument similar to the geodolite previously described. Signals are reflected

from the satellite and the position of the station is determined relative to the

Earth’s center of mass. VLBI uses

interstellar emissions from quasars to obtain interferometric patterns that

determine an absolute position of a station. This technique can also be used

for studies of the motion of Earth relative to the stars. Because the signals

pass through the denser part of the atmosphere at an oblique angle, attenu-

ation problems associated with water vapor are greatly reduced. The success

of these techniques demonstrated that space-based geodetic systems could

178 Stress and Strain in Solids

provide absolute positions on the surface of the Earth with a subcentimeter

accuracy. However, both techniques have serious limi-tations due to their

use of large stationary or mobile antennas that are bulky and expensive.

These difficulties were overcome when the global positioning system (GPS)

became fully operational in the early 1990s. GPS consists of some 24 satel-

lites that interact with ground-based receivers to provide accurate surface

positions. GPS was introduced by the Department of Defense as a global

navigation system with an accuracy of meters (Hofmann-Wallenhof et al.,

1997). However, surface instruments were developed that use the carrier sig-

nals from the active GPS satellites in an interferometric mode to determine

differential positions between surface benchmarks with a subcentimeter ac-

curacy (Larson, 1996). The great advantage of GPS is the low cost and

availability of the instruments so that large numbers of surface observations

can be made.

One of the first accomplishments of space geodesy was the confirmation

that the plate tectonic velocities given in Section 1–8 are also valid on a year-

to-year basis. As a specific example, we show in Figure 2–36 the relative

displacements between the Yaragadee station (Perth, Australia) and the

Maui station (Hawaiian Islands). These SLR observations give a relative

velocity uYM = −90± 5 mm yr−1.

We next compare this value with the value predicted by the plate mo-

tions given in Table 1–6. We first obtain the motion of the Maui station

(θ′ = 90◦ − 20.7◦ = 69.3◦, ψ′ = 203.7◦) relative to the fixed Australian

plate. From Table 1–6 we find θ = 90◦ + 60.1◦ = 150.1◦, ψ = −178.3◦,

and ω = 1.07 deg Myr−1 (0.0187 rad Myr−1). Using Equation (1–18) we

find that the angle ∆PM = 82.76◦. Substitution into Equation (1–17) gives

uMP = 118 mm yr−1. This is the velocity of the Maui station relative to

the fixed Australian plate; this velocity is perpendicular to the great circle

path passing through the pole of rotation and the Maui station and is in the

counterclockwise direction.

The measured relative velocity between the Yaragadee and Maui stations,

uYM = −90± 5 mm yr−1, is in the direction of the great circle between the

two stations (the negative sign indicates a convergence). The angle β between

the two great circle paths YM and MP must be determined and the plate

motion velocity must be resolved onto the YM great circle direction. From

Figure 1–35 we see that we can determine the angle β using Equation (1–18)

with the result

cos β = cos ∆PY − cos ∆PM cos ∆YM

sin ∆PM sin ∆YM , (2.150)

2.8 Strain Measurements 179

1978 1979 1980 1981 1982 1983 1984 1985 1986 1987 1988 1989

−0.5

−0.25

0.0

0.25

0.5

Year

Figure 2.36 The geodetic time series for the change in distance along a great circle path between the Yaragadee station (Perth, Australia) and the Maui station (Hawaiian Islands). The distance changes are in meters. The data were obtained using satellite laser ranging (SLR) (Smith et al., 1990). The straight line correlation gives a velocity uym =−90±5 mm yr−1.

Figure 2.37 Observed velocity vectors for geodetic stations in southern Cal- ifornia obtained from a combined GPS and VLBI data set (Feigl et al., 1993). The velocities are given relative to a fixed Pacific plate. Error el- lipses and major faults are also illustrated.

180 Stress and Strain in Solids

where ∆PY is the angle subtended at the center of the Earth by the pole of

rotation P and the Yaragadee station Y (θ′′ = 90◦+29◦ = 119◦, ψ′′ = 115.3◦)

and ∆YM is the angle subtended by the Maui station M and the Yaragadee

station Y . From Equation (1–18) we find ∆PY = 53.6◦ and ∆YM = 98.5◦.

Substitution of these values into Equation (2–150) gives β = 51.4◦. The

relative velocity between the Yaragadee and Maui stations uYM is related

to the velocity of the Maui station relative to the Australian plate uMP by

uYM = uMP cos(90◦ − β) = uMP sinβ. (2.151)

Taking β = 51.4◦ and uMP = −118 mm yr−1, Equation (2–151) gives uYM =

−92 mm yr−1. This is in excellent agreement with the observed value uYM =

−90 ± 5 mm yr−1.

Problem 2.30 Based on SLR observations, the relative velocity between

the Greenbelt, USA (39◦N, 283.2◦E), and Weitzell, Germany (49.1◦N, 12.9◦E),

stations is 18±4 mm yr−1. Determine the expected relative velocities based

on the plate motion data given in Table 1–6.

Problem 2.31 Based on SLR observations, the relative velocity between

the Simosato, Japan (33.5◦N, 136◦E), and the Maui stations is −62±6 mm yr−1.

Determine the expected relative velocities based on the plate motion data

given in Table 1–6.

Problem 2.32 Based on SLR observations, the relative velocity between

the Easter Island (27.1◦S, 250.6◦E) and the Arequipa, Peru (16.5◦S, 288.5◦E),

stations is −62±7 mm yr−1. Determine the expected relative velocities based

on the plate motion data given in Table 1–6.

The low cost and mobility of GPS systems allow detailed determinations

of tectonic strain in active plate boundary regions. An example for central

and southern California is shown in Figure 2–37 which gives the observed

velocity vectors of geodetic stations obtained from a combined GPS and

VLBI data set (Feigl et al., 1993). The velocities are given relative to a fixed

Pacific plate. The velocity vectors of the OVRO (Owens Valley) and MOJA

(Mojave) stations are representative of the relative southwest motion of the

North American plate with respect to the Pacific plate. The virtual station-

arity of the VNDN (Vandenberg AFB) and the BLUF (San Clemente Island)

stations indicate their attachment to a rigid Pacific plate. The intermediate

motion of the JPL (Pasadena) station represents the complex displacement

field within the Los Angeles basin.

Problem 2.33 The displacement of the OVRO (Owens Valley) station

2.8 Strain Measurements 181

Figure 2.38 Interferometric pattern of the coseismic deformation associated with the magnitude 7.3 Landers, California, earthquake on June 28, 1992 (Price and Sandwell, 1998). The dark lines represent surface ruptures asso- ciated with the earthquake and the white lines represent other known faults in the region. Each interferometric fringe corresponds to a displacement of 28 mm.

is 20.1 mm yr−1 to the east and −28.0 mm yr−1 to the north. Assuming

the San Andreas fault to be pure strike slip, and that this displacement is

associated only with motion on this fault, determine the mean slip velocity

on the fault and its orientation.

Problem 2.34 The displacement of the MOJA (Mojave) station is

23.9 mm yr−1 to the east and −26.6 mm yr−1 to the north. Assuming

the San Andreas fault to be pure strike slip and that this displacement is

182 Collateral Reading

associated only with motion on this fault, determine the mean slip velocity

on the fault and its orientation.

Synthetic aperture radar interferometry (INSAR) from satellites has opened

a new era in geodetic observations. A synthetic aperture radar (SAR) im-

age is obtained using radar backscatter returns from the Earth’s surface. If

the Earth’s surface deforms between two SAR image acquisitions, a radar

interferogram can be obtained to quantify the deformation. The point-wise

product of the first image with the second image produces a fringe pattern

associated with the phase differences between the two images. Each fringe

represents a phase change of 2π radians.

An example of INSAR interferometry is given in Figure 2–38 (Price and

Sandwell, 1998). This is the pattern of images associated with the magnitude

7.3 Landers, California, earthquake which occurred on June 28, 1992, and

ruptured nearly 100 km of previously unmapped faults in the Mojave Desert,

California. The maximum measured surface displacement was 5.1 m. The

images were acquired by the ERS-1 satellite on April 24 and August 7, 1992.

The satellite was at an altitude of 785 km and the radar images were collected

along ray paths pointed west at an average angle of 23◦ from the vertical.

Each fringe corresponds to 28 mm (half the 56-mm wavelength of the ERS-1

SAR). The distribution of coseismic deformation shown in Figure 2–38 will

be considered in detail in Chapter 8.

Collateral Reading

Feigl, K. L., et al. (1993), Space geodetic measurement of crustal deformation

in central and southern California, 1984–1992, J. Geophys. Res. 98,

21,677–21,712.

Gough, D. J., and J. S. Bell (1981), Stress orientations from borehole

well fractures with examples from Colorado, east Texas, and northern

Canada, Can. J. Earth Sci. 19, 1,358–1,370.

Hofmann-Wallenhof, B., H. Lichtenegger, and J. Collins (1997), Global Po-

sitioning System, 4th Ed. (Springer, Vienna), 389p.

Larson, K. M. (1996), Geodesy, Prog. Astron. Aeronaut. 164, 539–557.

McKenzie, D. (1978), Some remarks on the development of sedimentary

basins, Earth Planet. Sci. Let. 40, 25–32.

Price, E. J., and D. T. Sandwell (1998), Small-scale deformations associated

with the 1992 Landers, California, earthquake mapped by synthetic

aperture radar interferometry phase gradients, J. Geophys. Res. 103,

27,001–27,016.

Collateral Reading 183

Savage, J. C., and R. O. Burford (1973), Geodetic determination of relative

plate motion in central California, J. Geophys. Res. 78, 832–845.

Savage, J. C., and W. H. Prescott (1978), Geodolite measurements near the

Briones Hills, California, earthquake swarm of January 8, 1977, Seis.

Soc. Am. Bull. 68, 175–180.

Smith, D. E., et al. (1990), Tectonic motion and deformation from satellite

laser ranging to LAGEOS, J. Geophys. Res. 95, 22,013–22,041.

Thatcher, W. (1975), Strain accumulation and release mechanism of the

1906 San Francisco earthquake, J. Geophys. Res. 80, 4,862–4,872.

Zoback, M. D., and J. H. Healy (1992), In situ stress measurements to 3.5 km

depth in the Cajon Pass scientific research borehole: Implications for

the mechanics of crustal faulting, J. Geophys. Res. 97, 5,039–5,057.

Collateral Reading

Bomford, G., Geodesy (Oxford University Press, London, 1962), 561 pages.

An in-depth discussion of geodetic measurement techniques, methods of

analysis, and implications of gravity observations for the figure of the

Earth, crustal structure, and the state of stress in the crust.

Heiskanen, W. A., and H. Moritz, Physical Geodesy (W. H. Freeman and

Company, San Francisco, 1967), 364 pages.

A graduate level textbook in geodesy. The contents includes chapters on

potential theory, the gravity field of the Earth, gravimetric methods,

astrogeodetic methods, and statistical and mathematical approaches in

determinations of the Earth’s figure.

Jaeger, J. C., and N. G. W. Cook, Fundamentals of Rock Mechanics (Chap-

man and Hall, London, 1976), 585 pages.

An advanced textbook presenting the mathematical and experimental foun-

dations of the mechanical behavior of rock. There are chapters on stress

and strain, friction, elasticity, rock strength, laboratory testing, duc-

tile behavior, fluid flow in rocks, fracture, state of stress underground,

measurements of underground stresses, mining and engineering appli-

cations, and geological and geophysical applications.

Jeffreys, H., The Earth, Its Origin, History and Physical Constitution (Cam-

bridge University Press, Cambridge, 1962), 438 pages.

A classic textbook on the physics of the solid Earth, which includes discus-

sions of stress, strain, elasticity, mechanical behavior of rocks, seismol-

ogy, gravity, and stress differences in the Earth.

Timoshenko, S., and J. N. Goodier, Theory of Elasticity (McGraw-Hill, New

York, 1970), 567 pages.

184 Collateral Reading

Fundamentals of the mathematical theory of elasticity with engineering ap-

plications. There are major chapters on plane stress and plane strain,

bending of beams, two-dimensional stress problems in rectangular, po-

lar, and curvilinear coordinates, solutions by the method of complex

variables, three-dimensional stress-strain problems, torsion, bending of

bars, thermal stresses, wave propagation, and finite-difference solutions.

About half the chapters include student exercises.

3

Elasticity and Flexure

3.1 Introduction

In the previous chapter we introduced the concepts of stress and strain. For

many solids it is appropriate to relate stress to strain through the laws of

elasticity. Elastic materials deform when a force is applied and return to their

original shape when the force is removed. Almost all solid materials, includ-

ing essentially all rocks at relatively low temperatures and pressures, behave

elastically when the applied forces are not too large. In addition, the elastic

strain of many rocks is linearly proportional to the applied stress. The equa-

tions of linear elasticity are greatly simplified if the material is isotropic, that

is, if its elastic properties are independent of direction. Although some meta-

morphic rocks with strong foliations are not strictly isotropic, the isotropic

approximation is usually satisfactory for the earth’s crust and mantle.

At high stress levels, or at temperatures that are a significant fraction of

the rock solidus, deviations from elastic behavior occur. At low temperatures

and confining pressures, rocks are brittle solids, and large deviatoric stresses

cause fracture. As rocks are buried more deeply in the earth, they are sub-

jected to increasingly large confining pressures due to the increasing weight

of the overburden. When the confining pressure on the rock approaches its

brittle failure strength, it deforms plastically. Plastic deformation is a contin-

uous, irreversible deformation without fracture. If the applied force causing

plastic deformation is removed, some fraction of the deformation remains.

We consider plastic deformation in Section 7–11. As discussed in Chapter

1, hot mantle rocks behave as a fluid on geological time scales; that is, they

continuously deform under an applied force.

Given that rocks behave quite differently in response to applied forces,

depending on conditions of temperature and pressure, it is important to

determine what fraction of the rocks of the crust and upper mantle behave

186 Elasticity and Flexure

Figure 3.1 (a) Structure formed immediately after rapidly pouring a very viscous fluid into a container. (b) Final shape of the fluid after a long time has elapsed.

elastically on geological time scales. One of the fundamental postulates of

plate tectonics is that the surface plates constituting the lithosphere do not

deform significantly on geological time scales. Several observations directly

confirm this postulate. We know that the transform faults connecting offset

segments of the oceanic ridge system are responsible for the major linear

fracture zones in the ocean. That these fracture zones remain linear and

at constant separation is direct evidence that the oceanic lithosphere does

not deform on a time scale of 108 years. Similar evidence comes from the

linearity of the magnetic lineaments of the seafloor (see Section 1–8).

There is yet other direct evidence of the elastic behavior of the lithosphere

on geological time scales. Although erosion destroys mountain ranges on a

time scale of 106 to 107 years, many geological structures in the continental

crust have ages greater than 109 years. The very existence of these struc-

tures is evidence of the elastic behavior of the lithosphere. If the rocks of

the crust behaved as a fluid on geological time scales, the gravitational body

force would have erased these structures. As an example, pour a very vis-

cous substance such as molasses onto the bottom of a flat pan. If the fluid is

sufficiently viscous and is poured quickly enough, a structure resembling a

mountain forms (see Figure 3–1a). However, over time, the fluid will even-

tually cover the bottom of the pan to a uniform depth (see Figure 3–1b).

The gravitational body force causes the fluid to flow so as to minimize the

gravitational potential energy.

A number of geological phenomena allow the long-term elastic behavior of

3.2 Linear Elasticity 187

the lithosphere to be studied quantitatively. In several instances the litho-

sphere bends under surface loads. Direct evidence of this bending comes

from the Hawaiian Islands and many other island chains, individual islands,

and seamounts. There is also observational evidence of the elastic bending

of the oceanic lithosphere at ocean trenches and of the continental litho-

sphere at sedimentary basins – the Michigan basin, for example. We will

make quantitative comparisons of the theoretically predicted elastic defor-

mations of these structures with the observational data in later sections of

this chapter.

One important reason for studying the elastic behavior of the lithosphere

is to determine the state of stress in the lithosphere. This stress distribu-

tion is responsible for the occurrence of earthquakes. Earthquakes are di-

rect evidence of high stress levels in the lithosphere. An earthquake relieves

accumulated strain in the lithosphere. The presence of mountains is also

evidence of high stress levels. Elastic stresses must balance the gravitational

body forces on mountains. Because of their elastic behavior, surface plates

can transmit stresses over large horizontal distances.

3.2 Linear Elasticity

A linear, isotropic, elastic solid is one in which stresses are linearly propor-

tional to strains and mechanical properties have no preferred orientations.

The principal axes of stress and strain coincide in such a medium, and the

connection between stress and strain can be conveniently written in this

coordinate system as

σ1 = (λ+ 2G)ε1 + λε2 + λε3 (3.1)

σ2 = λε1 + (λ+ 2G)ε2 + λε3 (3.2)

σ3 = λε1 + λε2 + (λ+ 2G)ε3, (3.3)

where the material properties λ and G are known as Lamé parameters; G

is also known as the modulus of rigidity. The material properties are such

that a principal strain component ε produces a stress (λ+2G)ε in the same

direction and stresses λε in mutually perpendicular directions.

Equations (3–1) to (3–3) can be written in the inverse form as

ε1 = 1

E σ1 −

ν

E σ2 −

ν

E σ3 (3.4)

ε2 = − ν

E σ1 +

1

E σ2 −

ν

E σ3 (3.5)

188 Elasticity and Flexure

Figure 3.2 Deformation under uniaxial stress.

ε3 = − ν

E σ1 −

ν

E σ2 +

1

E σ3, (3.6)

and E and ν are material properties known as Young’s modulus and Pois-

son’s ratio, respectively. A principal stress component σ produces a strain

σ/E in the same direction and strains (−νσ/E) in mutually orthogonal

directions.

The elastic behavior of a material can be characterized by specifying either

λ and G or E and ν; the sets of parameters are not independent. Analytic

formulas expressing λ and G in terms of E and ν, and vice versa, are ob-

tained in the following sections. Values of E, G, and ν for various rocks are

given in Section E of Appendix 2. Young’s modulus of rocks varies from

about 10 to 100 GPa, and Poisson’s ratio varies between 0.1 and 0.4. The

elastic properties of the earth’s mantle and core can be obtained from seis-

mic velocities and the density distribution. The elastic properties E, G, and

ν inferred from a typical seismically derived earth model are given in Section

F of Appendix 2. The absence of shear waves in the outer core (G = 0) is

taken as conclusive evidence that the outer core is a liquid. In the outer core

ν has the value 0.5, which we will see is appropriate to an incompressible

fluid.

The behavior of linear solids is more readily illustrated if we consider

idealized situations where several of the stress and strain components vanish.

These can then be applied to important geological problems.

3.3 Uniaxial Stress 189

3.3 Uniaxial Stress

In a state of uniaxial stress only one of the principal stresses, σ1 say, is

nonzero. Under this circumstance Equations (3–2) and (3–3), with σ2 =

σ3 = 0, give

ε2 = ε3 = −λ

2(λ+G) ε1. (3.7)

Not only does the stress σ1 produce a strain ε1, but it changes the linear

dimensions of elements aligned perpendicular to the axis of stress. If σ1 is a

compression, then ε1 is a decrease in length, and both ε2 and ε3 are increases

in length. The element in Figure 3–2 has been shortened in the y direction,

but its cross section in the xz plane has expanded.

Using Equations (3–4) to (3–6), we can also write

ε2 = ε3 = − ν

E σ1 = −νε1. (3.8)

By comparing Equations (3–7) and (3–8), we see that

ν = λ

2(λ+G) . (3.9)

From Equations (3–1) and (3–7) we find

σ1 = G(3λ + 2G)

(λ+G) ε1, (3.10)

which, with the help of Equation (3–8), identifies Young’s modulus as

E = G(3λ+ 2G)

(λ+G) . (3.11)

Equations (3–9) and (3–11) can be inverted to yield the following formulas

for G and λ in terms of E and ν

G = E

2(1 + ν) (3.12)

λ = Eν

(1 + ν)(1 − 2ν) . (3.13)

The relation between stress and strain in uniaxial compression or tension

from Equation (3–8),

σ1 = Eε1, (3.14)

is also known as Hooke’s law. A linear elastic solid is said to exhibit Hookean

behavior. Uniaxial compression testing in the laboratory is one of the sim-

plest methods of determining the elastic properties of rocks. Figure 3–3

190 Elasticity and Flexure

Figure 3.3 Stress–strain curves for quartzite in uniaxial compression (Bi- eniawski, 1967).

shows the data from such a test on a cylindrical sample of quartzite. The

rock deforms approximately elastically until the applied stress exceeds the

compressive strength of the rock, at which point failure occurs. Compressive

strengths of rocks are hundreds to thousands of megapascals. As we discussed

in the previous chapter, a typical tectonic stress is 10 MPa. With E = 70

GPa, this yields a typical tectonic strain in uniaxial stress of 1.4 × 10−4.

The dilatation ∆ or fractional volume change in uniaxial compression is,

according to Equation (3–8),

∆ = ε1 + ε2 + ε3 = ε1(1 − 2ν). (3.15)

The decrease in volume due to contraction in the direction of compressive

stress is offset by an increase in volume due to expansion in the orthogonal

directions. Equation (3–15) allows us to determine Poisson’s ratio for an

incompressible material, which cannot undergo a net change in volume. In

order for ∆ to equal zero in uniaxial compression, ν must equal 1/2. Under

uniaxial compression, an incompressible material contracts in the direction

of applied stress but expands exactly half as much in each of the perpendic-

ular directions.

There are some circumstances in which the formulas of uniaxial compres-

sion can be applied to calculate the strains in rocks. Consider, for example,

3.4 Uniaxial Strain 191

a rectangular column of height h that is free to expand or contract in the

horizontal; that is, it is laterally unconstrained. By this we mean that the

horizontal stresses are zero (σ2 = σ3 = 0). Then the vertical stress σ1 at a

distance y from the top of the column of rock is given by the weight of the

column,

σ1 = ρgy. (3.16)

The vertical strain as a function of the distance y from the top is

ε1 = ρgy

E . (3.17)

The slab contracts in the vertical by an amount

δh =

∫ h

0 ε1 dy =

ρg

E

∫ h

0 y dy =

ρgh2

2E . (3.18)

3.4 Uniaxial Strain

The state of uniaxial strain corresponds to only one nonzero component of

principal strain, ε1 say. With ε2 = ε3 = 0, Equations (3–1) to (3–3) give

σ1 = (λ+ 2G)ε1 (3.19)

σ2 = σ3 = λε1 = λ

(λ+ 2G) σ1. (3.20)

Equations (3–4) to (3–6) simplify to

σ2 = σ3 = ν

(1 − ν) σ1 (3.21)

σ1 = (1 − ν)Eε1

(1 + ν)(1 − 2ν) . (3.22)

By comparing Equations (3–19) to (3–22), one can also derive the relations

already given between λ,G and ν,E.

The equations of uniaxial strain can be used to determine the change in

stress due to sedimentation or erosion. We first consider sedimentation and

assume that an initial surface is covered by h km of sediments of density ρ,

as shown in Figure 3–4. We also assume that the base of the new sedimen-

tary basin is laterally confined so that the equations of uniaxial strain are

applicable. The two horizontal components of strain are zero, ε2 = ε3 = 0.

The vertical principal stress on the initial surface σ1 is given by the weight

of the overburden

σ1 = ρgh. (3.23)

192 Elasticity and Flexure

Figure 3.4 Stresses on a surface covered by sediments of thickness h.

From Equation (3–21) the horizontal normal stresses are given by

σ2 = σ3 = ν

(1 − ν) ρgh. (3.24)

The horizontal stresses are also compressive, but they are smaller than the

vertical stress.

It is of interest to determine the deviatoric stresses after sedimentation.

The pressure at depth h as defined by Equation (2–61) is

p = 1

3 (σ1 + σ2 + σ3) =

(1 + ν)

3(1 − ν) ρgh. (3.25)

The deviatoric stresses are then determined from Equations (2–63) with the

result

σ′1 = σ1 − p = 2(1 − 2ν)

3(1 − ν) ρgh (3.26)

σ′2 = σ2 − p = σ′3 = σ3 − p = −(1 − 2ν)

3(1 − ν) ρgh.

(3.27)

The horizontal deviatoric stress is tensional. For ν = 0.25 the horizontal

deviatoric stress is 2/9 of the lithostatic stress. With ρ = 3000 kg m−3 and

h = 2 km the horizontal deviatoric stress is −13.3 MPa. This stress is of the

same order as measured surface stresses.

We next consider erosion. If the initial state of stress before erosion is

that given above, erosion will result in the state of stress that existed before

sedimentation occurred. The processes of sedimentation and erosion are re-

versible. However, in many cases the initial state of stress prior to erosion is

lithostatic. Therefore at a depth h the principal stresses are

σ1 = σ2 = σ3 = ρgh. (3.28)

3.5 Plane Stress 193

After the erosion of h km of overburden the vertical stress at the surface is

σ̄1 = 0 (an overbar denotes a stress after erosion). The change in vertical

stress ∆σ1 = σ̄1 − σ1 is −ρgh. If only ε1 is nonzero, Equation (3–21) gives

∆σ2 = ∆σ3 =

(

ν

1 − ν

)

∆σ1. (3.29)

The horizontal surface stresses after erosion σ̄2 and σ̄3 are consequently given

by

σ̄2 = σ̄3 = σ2 + ∆σ2 = ρgh − ν

(1 − ν) ρgh

=

(

1 − 2ν

1 − ν

)

ρgh. (3.30)

If h = 5 km, ν = 0.25, and ρ = 3000 kg m−3, we find from Equation (3–30)

that σ̄2 = σ̄3 = 100 MPa. Erosion can result in large surface compressive

stresses due simply to the elastic behavior of the rock. This mechanism is

one explanation for the widespread occurrence of near-surface compressive

stresses in the continents.

Problem 3.1 Determine the surface stress after the erosion of 10 km of

granite. Assume that the initial state of stress is lithostatic and that ρ = 2700

kg m−3 and ν = 0.25.

Problem 3.2 An unstressed surface is covered with sediments with a den-

sity of 2500 kg m−3 to a depth of 5 km. If the surface is laterally constrained

and has a Poisson’s ratio of 0.25, what are the three components of stress

at the original surface?

Problem 3.3 A horizontal stress σ1 may be accompanied by stress in

other directions. If it is assumed that there is no displacement in the other

horizontal direction and zero stress in the vertical, find the stress σ2 in the

other horizontal direction and the strain ε3 in the vertical direction.

Problem 3.4 Assume that the earth is unconstrained in one lateral direc-

tion (σ2 = σ3) and is constrained in the other (ε1 = 0). Determine ε2 and

σ1 when y kilometers of rock of density ρ are eroded away. Assume that the

initial state of stress was lithostatic.

3.5 Plane Stress

The state of plane stress exists when there is only one zero component of

principal stress; that is, σ3 = 0, σ1 6= 0, σ2 6= 0. The situation is sketched

194 Elasticity and Flexure

Figure 3.5 Plane stress.

in Figure 3–5, which shows a thin plate loaded on its edges. The strain

components according to Equations (3–4) to (3–6) are

ε1 = 1

E (σ1 − νσ2) (3.31)

ε2 = 1

E (σ2 − νσ1) (3.32)

ε3 = −ν E

(σ1 + σ2). (3.33)

The geometry of Figure 3–5 suggests that the plane stress formulas may be

applicable to horizontal tectonic stresses in the lithosphere. Let us assume

that in addition to the lithostatic stresses there are equal horizontal com-

ponents of principal stress ∆σ1 = ∆σ2. According to Equations (3–31) to

(3–33), the horizontal tectonic stresses produce the strains

ε1 = ε2 = (1 − ν)

E ∆σ1 (3.34)

ε3 = −2ν

E ∆σ1. (3.35)

If the horizontal tectonic stresses are compressive, vertical columns of

lithosphere of initial thickness hL, horizontal area A, and density ρ will

undergo a decrease in area and an increase in thickness. The mass in a

column will remain constant, however. Therefore we can write

δ(ρAhL) = 0. (3.36)

3.5 Plane Stress 195

The weight per unit area at the base of the column ρghL will increase, as

can be seen from

δ(ρghL) = δ

(

ρghLA · 1

A

)

= 1

A δ(ρghLA) + ρghLA δ

(

1

A

)

= ρghLA

(

− 1

A2

)

δA = ρghL

(

−δA A

)

.

(3.37)

The term δ(ρghLA)/A is zero from Equation (3–36); δ(ρghL) is positive

because −δA/A is a positive quantity given by

−δA A

= ε1 + ε2 = 2(1 − ν)

E ∆σ1. (3.38)

The increase in the weight per unit area at the base of the lithospheric

column gives the increase in the vertical principal stress ∆σ3. By combining

Equations (3–37) and (3–38), we get

∆σ3 = 2(1 − ν)ρghL

E ∆σ1 (3.39)

or ∆σ3

∆σ1 =

2(1 − ν)ρghL E

. (3.40)

Taking ν = 0.25, E = 100 GPa, ρ = 3000 kg m−3, g = 10 m s−2, and hL =

100 km as typical values for the lithosphere, we find that ∆σ3/∆σ1 = 0.045.

Because the change in the vertical principal stress is small compared with

the applied horizontal principal stresses, we conclude that the plane stress

assumption is valid for the earth’s lithosphere.

Problem 3.5 Triaxial compression tests are a common laboratory tech-

nique for determining elastic properties and strengths of rocks at various

pressures p and temperatures. Figure 3–6 is a schematic of the experimen-

tal method. A cylindrical rock specimen is loaded axially by a compres-

sive stress σ1. The sample is also uniformly compressed laterally by stresses

σ2 = σ3 < σ1.

Show that

ε2 = ε3

and

σ1 − σ2 = 2G(ε1 − ε2).

196 Elasticity and Flexure

Figure 3.6 Sketch of a triaxial compression test on a cylindrical rock sam- ple.

Figure 3.7 An example of plane strain.

Thus if the measured stress difference σ1−σ2 is plotted against the measured

strain difference ε1 − ε2, the slope of the line determines 2G.

3.6 Plane Strain

In the case of plane strain, ε3 = 0, for example, and ε1 and ε2 are nonzero.

Figure 3–7 illustrates a plane strain situation. A long bar is rigidly confined

between supports so that it cannot expand or contract parallel to its length.

3.7 Pure Shear and Simple Shear 197

Figure 3.8 Principal stresses and shear stresses in the case of pure shear.

In addition, the stresses σ1 and σ2 are applied uniformly along the length

of the bar.

Equations (3–1) to (3–3) reduce to

σ1 = (λ+ 2G)ε1 + λε2 (3.41)

σ2 = λε1 + (λ+ 2G)ε2 (3.42)

σ3 = λ(ε1 + ε2). (3.43)

From Equation (3–6) it is obvious that

σ3 = ν(σ1 + σ2). (3.44)

This can be used together with Equations (3–4) and (3–5) to find

ε1 = (1 + ν)

E {σ1(1 − ν) − νσ2} (3.45)

ε2 = (1 + ν)

E {σ2(1 − ν) − νσ1}. (3.46)

3.7 Pure Shear and Simple Shear

The state of stress associated with pure shear is illustrated in Figure 3–8.

Pure shear is a special case of plane stress. One example of pure shear is

σ3 = 0 and σ1 = −σ2. From Equations (2–56) to (2–58) with θ = −45◦

(compare Figures 2–14 and 3–8), we find that σxx = σyy = 0 and σxy = σ1.

In this coordinate system only the shear stress is nonzero. From Equations

198 Elasticity and Flexure

(3–31) and (3–32) we find that

ε1 = (1 + ν)

E σ1 =

(1 + ν)

E σxy = −ε2, (3.47)

and from Equations (2–130) and (2–131) with θ = −45◦ we get εxx = εyy = 0

and εxy = ε1. Equation (3–47) then gives

σxy = E

1 + ν εxy. (3.48)

By introducing the modulus of rigidity from Equation (3–12), we can write

the shear stress as

σxy = 2Gεxy , (3.49)

which explains why the modulus of rigidity is also known as the shear mod-

ulus. (Note: In terms of γxy ≡ 2εxy, σxy = Gγxy.) These results are valid

for both pure shear and simple shear because the two states differ by a

solid-body rotation that does not affect the state of stress.

Simple shear is generally associated with displacements on a strike–slip

fault such as the San Andreas in California. In Equation (2–134) we con-

cluded that the shear strain associated with the 1906 San Francisco earth-

quake was 2.5× 10−5. With G = 30 GPa, Equation (3–49) gives the related

shear stress as 1.5 MPa. This is a very small stress drop to be associated

with a great earthquake. However, for the stress drop to have been larger,

the width of the zone of strain accumulation would have had to have been

even smaller. If the stress had been 15 MPa, the width of the zone of strain

accumulation would have had to have been 4 km on each side of the fault.

We will return to this problem in Chapter 8.

Problem 3.6 Show that Equation (3–49) can also be derived by assuming

plane strain.

3.8 Isotropic Stress

If all the principal stresses are equal σ1 = σ2 = σ3 ≡ p, then the state

of stress is isotropic, and the principal stresses are equal to the pressure.

The principal strains in a solid subjected to isotropic stresses are also equal

ε1 = ε2 = ε3 = 1 3∆; each component of strain is equal to one-third of the

dilatation. By adding Equations (3–1) to (3–3), we find

p =

(

3λ+ 2G

3

)

∆ ≡ K∆ ≡ 1

β ∆. (3.50)

3.9 Two-Dimensional Bending or Flexure of Plates 199

The quantity K is the bulk modulus, and its reciprocal is β, the compressibil-

ity. The ratio of p to the bulk modulus gives the fractional volume change

that occurs under isotropic compression.

Because the mass of a solid element with volume V and density ρ must be

conserved, any change in volume δV of the element must be accompanied by

a change in its density δρ. The fractional change in density can be related to

the fractional change in volume, the dilatation, by rearranging the equation

of mass conservation

δ(ρV ) = 0, (3.51)

which gives

ρδV + V δρ = 0 (3.52)

or −δV V

= ∆ = δρ

ρ . (3.53)

Equation (3–53) of course assumes ∆ to be small. The combination of Equa-

tions (3–50) and (3–53) gives

δρ = ρβp. (3.54)

This relationship can be used to determine the increase in density with depth

in the earth.

Using Equations (3–11) to (3–13), we can rewrite the formula for K given

in Equation (3–50) as

K = 1

β =

E

3(1 − 2ν) . (3.55)

Thus as ν tends toward 1/2, that is, as a material becomes more and more

incompressible, its bulk modulus tends to infinity.

3.9 Two-Dimensional Bending or Flexure of Plates

We have already discussed how plate tectonics implies that the near-surface

rocks are rigid and therefore behave elastically on geological time scales.

The thin elastic surface plates constitute the lithosphere, which floats on

the relatively fluid mantle beneath. The plates are subject to a variety of

loads – volcanoes, seamounts, for example – that force the lithosphere to

bend under their weights. By relating the observed flexure or bending of

the lithosphere to known surface loads, we can deduce the elastic properties

and thicknesses of the plates. In what follows, we first develop the theory of

200 Elasticity and Flexure

Figure 3.9 A thin plate of length L and thickness h pinned at its ends and bending under and applied load Va.

Figure 3.10 Forces and torques on a small section of a deflecting plate.

plate bending in response to applied forces and torques. The theory can also

be used to understand fold trains in mountain belts by modeling the folds

as deformations of elastic plates subject to horizontal compressive forces.

Other geologic applications also can be made. For example, we will apply

the theory to model the upwarping of strata overlying igneous intrusions

(Section 3–12).

A simple example of plate bending is shown in Figure 3–9. A plate of

thickness h and width L is pinned at its ends and bends under the load of a

line force Va (N m−1) applied at its center. The plate is infinitely long in the

z direction. A vertical, static force balance and the symmetry of the situation

require that equal vertical line forces Va/2 be applied at the supports. The

plate is assumed to be thin compared with its width, h≪ L, and the vertical

deflection of the plate w is taken to be small, w ≪ L. The latter assumption

is necessary to justify the use of linear elastic theory. The two-dimensional

bending of plates is also referred to as cylindrical bending because the plate

takes the form of a segment of a cylinder.

The deflection of a plate can be determined by requiring it to be in equilib-

3.9 Two-Dimensional Bending or Flexure of Plates 201

rium under the action of all the forces and torques exerted on it. The forces

and torques on a small section of the plate between horizontal locations x

and x+ dx are shown in Figure 3–10. A downward force per unit area q(x)

is exerted on the plate by whatever distributed load the plate is required

to support. Thus, the downward load, per unit length in the z direction,

between x and x + dx is q(x) dx. A net shear force V , per unit length in

the z direction, acts on the cross section of the plate normal to the plane

of the figure; it is the resultant of all the shear stresses integrated over that

cross-sectional area of the plate. A horizontal force P , per unit length in the

z direction, is applied to the plate; it is assumed that P is independent of

x. The net bending moment M , per unit length in the z direction, acting on

a cross section of the plate is the integrated effect of the moments exerted

by the normal stresses σxx, also known as the fiber stresses, on the cross

section. We relate M to the fiber stresses in the plate later in the discussion.

All quantities in Figure 3–10 are considered positive when they have the

sense shown in the figure. At location x along the plate the shear force is V ,

the bending moment is M , and the deflection is w; at x+dx, the shear force

is V + dV , the bending moment is M + dM , and the deflection is w+ dw. It

is to be emphasized that V , M , and P are per unit length in the z direction.

A force balance in the vertical direction on the element between x and

x+ dx yields

q(x) dx+ dV = 0 (3.56)

or dV

dx = −q. (3.57)

The moments M and M+dM combine to give a net counterclockwise torque

dM on the element. The forces V and V +dV are separated by a distance dx

(an infinitesimal moment arm) and exert a net torque V dx on the element

in a clockwise sense. (The change in V in going from x to x + dx can be

ignored in calculating the moment due to the shear forces.) The horizontal

forces P exert a net counterclockwise torque −P dw on the element through

their associated moment arm −dw. (Note that dw is negative in going from

x to x+ dx.) A balance of all the torques gives

dM − P dw = V dx (3.58)

or dM

dx = V + P

dw

dx . (3.59)

We can eliminate the shear force on a vertical cross section of the plate V

202 Elasticity and Flexure

Figure 3.11 The normal stresses on a cross section of a thin curved elastic plate.

from Equation (3–59) by differentiating the equation with respect to x and

substituting from Equation (3–57). One obtains

d 2M

dx2 = −q + P

d 2w

dx2 . (3.60)

Equation (3–60) can be converted into a differential equation for the deflec-

tion w if the bending moment M can be related to the deflection; we will

see that M is inversely proportional to the local radius of curvature of the

plate R and that R−1 is −d 2w/dx2.

To relate M to the curvature of the plate, we proceed as follows. If the

plate is deflected downward, as in Figure 3–11, the upper half of the plate

is contracted, and the longitudinal stress σxx is positive; the lower part of

the plate is extended, and σxx is negative. The fiber stress σxx is zero on the

midplane y = 0, which is a neutral unstrained surface. The net effect of these

stresses is to exert a counterclockwise bending moment on the cross section

of the plate. The curvature of the plate has, of course, been exaggerated in

Figure 3–11 so that x is essentially horizontal. The force on an element of

the plate’s cross section of thickness dy is σxx dy. This force exerts a torque

about the midpoint of the plate given by σxxy dy. If we integrate this torque

over the cross section of the plate, we obtain the bending moment

M =

∫ h/2

−h/2 σxxy dy, (3.61)

where h is the thickness of the plate.

The bending stress σxx is accompanied by longitudinal strain εxx that is

positive (contraction) in the upper half of the plate and negative (extension)

in the lower half. There is no strain in the direction perpendicular to the

xy plane because the plate is infinite in this direction and the bending is

two-dimensional or cylindrical; that is, εzz = 0. There is also zero stress

3.9 Two-Dimensional Bending or Flexure of Plates 203

normal to the surface of the plate; that is, σyy = 0. Because the plate is

thin, we can take σyy = 0 throughout. Thus plate bending is an example

of plane stress, and we can use Equations (3–31) and (3–32) to relate the

stresses and strains; that is,

εxx = 1

E (σxx − νσzz) (3.62)

εzz = 1

E (σzz − νσxx). (3.63)

In writing these equations, we have identified the principal strains ε1, ε2 with εxx, εzz and the principal stresses σ1, σ2 with σxx, σzz. With εzz = 0,

Equations (3–62) and (3–63) give

σxx = E

(1 − ν2) εxx. (3.64)

Equation (3–61) for the bending moment can be rewritten, using Equation

(3–64), as

M = E

(1 − ν2)

∫ h/2

−h/2 εxxy dy. (3.65)

The longitudinal strain εxx depends on the distance from the midplane

of the plate y and the local radius of curvature of the plate R. Figure 3–12

shows a bent section of the plate originally of length l (l is infinitesimal).

The length of the section measured along the midplane remains l. The small

angle φ is l/R in radians. The geometry of Figure 3–12 shows that the change

in length of the section ∆l at a distance y from the midplane is

∆l = −yφ = −y l R , (3.66)

where the minus sign is included because there is contraction when y is

positive. Thus the strain is

εxx = −∆l

l = y

R . (3.67)

Implicit in this relation is the assumption that plane sections of the plate

remain plane.

The local radius of curvature R is determined by the change in slope of the

plate midplane with horizontal distance. The geometry is shown in Figure

3–13. If w is small, −dw/dx, the slope of the midplane, is also the angular

deflection of the plate from the horizontal α. The small angle φ in Figure

3–13 is simply the change in α, that is, dα, in the small distance l or dx.

204 Elasticity and Flexure

Figure 3.12 Longitudinal extension and contraction at a distance y from the midplane of the plate.

Thus

φ = dα = dα

dx dx =

d

dx

(

−dw dx

)

dx = −d 2w

dx2 dx,

(3.68)

and we find

1

R = φ

l ≈ φ

dx = −d

2w

dx2 . (3.69)

Finally, the strain is given by

εxx = −yd 2w

dx2 , (3.70)

and the bending moment can be written

M = −E

(1 − ν2)

d 2w

dx2

∫ h/2

−h/2 y2 dy

= −E

(1 − ν2)

d 2w

dx2

(

y3

3

)h/2

−h/2

= −Eh3

12(1 − ν2)

d 2w

dx2 . (3.71)

The coefficient of −d 2w/dx2 on the right side of Equation (3–71) is called

the flexural rigidity D of the plate

D ≡ Eh3

12(1 − ν2) . (3.72)

3.10 Bending of Plates under Applied Moments and Vertical Loads 205

Figure 3.13 Sketch illustrating the geometrical relations in plate bending.

According to Equations (3–69), (3–71), and (3–72), the bending moment is

the flexural rigidity of the plate divided by its curvature

M = −Dd 2w

dx2 = D

R . (3.73)

Upon substituting the second derivative of Equation (3–73) into Equation

(3–60), we obtain the general equation for the deflection of the plate

D d4w

dx4 = q(x) − P

d 2w

dx2 . (3.74)

We next solve Equation (3–74) for plate deflection in a number of simple

cases and apply the results to the deformation of crustal strata and to the

bending of the lithosphere.

3.10 Bending of Plates under Applied Moments and Vertical

Loads

Consider a plate embedded at one end and subject to an applied torque Ma

at the other, as shown in Figure 3–14. Assume for simplicity that the plate

is weightless. With q = 0, Equation (3–57) shows that the shear stress on

a section of the plate V must be a constant. In fact, V = 0, since there is

no applied force acting on the plate. This can easily be seen by considering

206 Elasticity and Flexure

Figure 3.14 An embedded plate subject to an applied torque.

Figure 3.15 Force and torque balance on a section of the plate in Figure 3–14.

a force balance on a section of the plate, as shown in Figure 3–15. Since

P = 0 and since we have established V = 0, Equation (3–59) requires that

M = constant. The constant must be Ma, the applied torque, as shown by

a moment balance on an arbitrary section of the plate (Figure 3–15).

To determine the deflection of the plate, we could integrate Equation (3–

74) with q = P = 0. However, since we already know M ≡Ma, it is simpler

to integrate Equation (3–73), the twice integrated form of the fourth-order

differential equation. The boundary conditions are w = 0 at x = 0 and

dw/dx = 0 at x = 0. These boundary conditions at the left end of the plate

clarify what is meant by an embedded plate; the embedded end of the plate

cannot be displaced, and its slope must be zero. The integral of Equation

(3–73) subject to these boundary conditions is

w = −Max

2

2D . (3.75)

The bent plate has the shape of a parabola. w is negative according to the

convention we established if M is positive; that is, the plate is deflected

upward.

Problem 3.7 What is the displacement of a plate pinned at both ends

(w = 0 at x = 0, L) with equal and opposite bending moments applied at

the ends? The problem is illustrated in Figure 3–16.

As a second example we consider the bending of a plate embedded at

its left end and subjected to a concentrated force Va at its right end, as

illustrated in Figure 3–17. In this situation, q = 0, except at the point

x = L, and Equation (3–57) gives V = constant. The constant must be

3.10 Bending of Plates under Applied Moments and Vertical Loads 207

Figure 3.16 Bending of a plate pinned at both ends.

Figure 3.17 An embedded plate subjected to a concentrated load.

Va, as shown by the vertical force balance on the plate sketched in Figure

3–18. With P also equal to zero, Equation (3–59) for the bending moment

simplifies to

dM

dx = Va. (3.76)

This equation can be integrated to yield

M = Vax+ constant, (3.77)

and the constant can be evaluated by noting that there is no applied torque

at the end x = L; that is, M = 0 at x = L. Thus we obtain

M = Va(x− L). (3.78)

The bending moment changes linearly from −VaL at the embedded end to

zero at the free end. A simple torque balance on the section of the plate

shown in Figure 3–18 leads to Equation (3–78), since M must balance the

torque of the applied force Va acting with moment arm L− x.

The displacement can be determined by integrating Equation (3–74),

which simplifies to

d4w

dx4 = 0, (3.79)

when q = P = 0. The integral of Equation (3–79) is

d3w

dx3 = constant. (3.80)

208 Elasticity and Flexure

Figure 3.18 Forces and torques on a section of a plate loaded at its right end by a force Va.

Figure 3.19 A uniformly loaded plate embedded at one end.

The constant can be evaluated by differentiating Equation (3–73) with re-

spect to x and substituting for dM/dx from Equation (3–76). The result

is

d3w

dx3 = −Va

D . (3.81)

A second-order differential equation for w can be obtained by integrating

Equation (3–81) and evaluating the constant of integration with the bound-

ary condition d 2w/dx2 = 0 at x = L. Alternatively, the same equation can

be arrived at by substituting for M from Equation (3–78) into Equation

(3–73)

d 2w

dx2 = −Va

D (x− L). (3.82)

This equation may be integrated twice more subject to the standard bound-

ary conditions w = dw/dx = 0 at x = 0. One finds

w = Vax

2

2D

(

L− x

3

)

. (3.83)

Problem 3.8 Determine the displacement of a plate of length L pinned

at its ends with a concentrated load Va applied at its center. This problem

is illustrated in Figure 3–9.

As a third and final example, we consider the bending of a plate embedded

3.10 Bending of Plates under Applied Moments and Vertical Loads 209

at one end and subjected to a uniform loading q(x) = constant, as illustrated

in Figure 3–19. Equation (3–74), with P = 0, becomes

d4w

dx4 =

q

D . (3.84)

We need four boundary conditions to integrate Equation (3–84). Two of

them are the standard conditions w = dw/dx = 0 at the left end x = 0.

A third boundary condition is the same as the one used in the previous

example, namely, d 2w/dx2 = 0 at x = L, because there is no external

torque applied at the right end of the plate – see Equation (3–73). The fourth

boundary condition follows from Equation (3–59) with P = 0. Because there

is no applied concentrated load at x = L, V must vanish there, as must

dM/dx and from Equation (3–73), d3w/dx3. After some algebra, one finds

the solution

w = qx2

D

(

x2

24 − Lx

6 + L2

4

)

. (3.85)

The shear force at x = 0 is −D(d3w/dx3)x=0. From Equation (3–85)

this is qL, a result that also follows from a consideration of the overall

vertical equilibrium of the plate because qL is the total loading. The shear

stress on the section x = 0 is qL/h. The bending moment on the section

x = 0 is −D(d 2w/dx2)x=0 or −qL2/2. The maximum bending or fiber stress,

σmax xx = σxx at y = −h/2, is given, from Equations (3–85), (3–64), and (3–

70), by

σmax xx =

E

(1 − ν2)

h

2

d 2w

dx2 =

6

h2 D d 2w

dx2 = −6M

h2 .

(3.86)

At x = 0, σmax xx is 3qL2/h2. The ratio of the shear stress to the maximum

bending stress at x = 0 is h/3L, a rather small quantity for a thin plate. It

is implicit in the analysis of the bending of thin plates that shear stresses in

the plates are small compared with the bending stresses.

Problem 3.9 Calculate V and M by carrying out force and torque bal-

ances on the section of the uniformly loaded plate shown in Figure 3–20.

Problem 3.10 A granite plate with ρ = 2700 kg m−3 is embedded at one

end. If L = 10 m and h = 1/4 m, what is the maximum bending stress and

the shear stress at the base?

Problem 3.11 Determine the displacement of a plate that is embedded

at the end x = 0 and has a uniform loading q from x = L/2 to x = L.

210 Elasticity and Flexure

Figure 3.20 Section of a uniformly loaded plate.

Figure 3.21 Plate buckling under a horizontal force.

Problem 3.12 Determine the deflection of a plate of length L that is

embedded at x = 0 and has equal loads Va applied at x = L/2 and at

x = L.

Problem 3.13 Find the deflection of a uniformly loaded beam pinned at

the ends, x = 0, L. Where is the maximum bending moment? What is the

maximum bending stress?

Problem 3.14 A granite plate freely supported at its ends spans a gorge

20 m wide. How thick does the plate have to be if granite fails in tension at

20 MPa? Assume ρ = 2700 kg m−3.

Problem 3.15 Determine the deflection of a freely supported plate, that

is, a plate pinned at its ends, of length L and flexural rigidity D subject to

a sinusoidal load qa = q0 sinπx/L, as shown in Figure 3–21.

3.11 Buckling of a Plate under a Horizontal Load

When an elastic plate is subjected to a horizontal force P , as shown in Fig-

ure 3–22a, the plate can buckle, as illustrated in Figure 3–22b, if the applied

force is sufficiently large. Fold trains in mountain belts are believed to result

from the warping of strata under horizontal compression. We will therefore

3.11 Buckling of a Plate under a Horizontal Load 211

Figure 3.22 A freely supported plate loaded sinusoidally.

consider the simplest example of plate buckling under horizontal compres-

sion to determine the minimum force required for buckling to occur and

the form, that is, the wavelength, of the resulting deflection. In a subsequent

section we will carry out a similar calculation to determine if the lithosphere

can be expected to buckle under horizontal tectonic compression.

We consider a plate pinned at both ends and subjected to a horizontal

force P , as shown in Figure 3–22. The deflection of the plate is governed by

Equation (3–74) with q = 0:

D d4w

dx4 + P

d 2w

dx2 = 0. (3.87)

This can be integrated twice to give

D d 2w

dx2 + Pw = c1x+ c2. (3.88)

However, we require that w is zero at x = 0, L and that d 2w/dx2 = 0 at

x = 0, L, since there are no applied torques at the ends. These boundary

conditions require that c1 = c2 = 0, and Equation (3–88) reduces to

D d 2w

dx2 + Pw = 0. (3.89)

Equation (3–89) has the general solution

w = c1 sin

(

P

D

)1/2

x+ c2 cos

(

P

D

)1/2

x, (3.90)

where c1 and c2 are constants of integration. Because w is equal to zero at

x = 0, c2 must be zero, and

w = c1 sin

(

P

D

)1/2

x. (3.91)

212 Elasticity and Flexure

But w must also vanish at x = L, which implies that if c1 6= 0, then

sin

(

P

D

)1/2

L = 0. (3.92)

Thus (P/D)1/2L must be an integer multiple of π,

(

P

D

)1/2

L = nπ n = 1, 2, 3, . . . (3.93)

Solving this equation for P , we get

P = n2π2

L2 D. (3.94)

Equation (3–94) defines a series of values of P for which nonzero solutions

for w exist. The smallest such value is for n = 1 when P is given by

P = Pc = π2

L2 D. (3.95)

This is the minimum buckling load for the plate. If P is smaller than this

critical value, known as an eigenvalue, the plate will not deflect under the

applied load; that is, c1 = 0 or w = 0. When P has the value given by

Equation (3–95), the plate buckles or deflects under the horizontal load. At

the onset of deflection the plate assumes the shape of a half sine curve

w = c1 sin

(

P

D

)1/2

x

= c1 sin πx

L . (3.96)

The amplitude of the deflection cannot be determined by the linear analysis

carried out here. Nonlinear effects fix the magnitude of the deformation.

The application of plate flexure theory to fold trains in mountain belts

requires somewhat more complex models than considered here. Although a

number of effects must be incorporated to approximate reality more closely,

one of the most important is the influence of the medium surrounding a

folded stratum. The rocks above and below a folded layer exert forces on

the layer that influence the form (wavelength) of the folds and the critical

horizontal force necessary to initiate buckling.

3.12 Deformation of Strata Overlying an Igneous Intrusion

A laccolith is a sill-like igneous intrusion in the form of a round lens-shaped

body much wider than it is thick. Laccoliths are formed by magma that

3.12 Deformation of Strata Overlying an Igneous Intrusion 213

is intruded along bedding planes of flat, layered rocks at pressures so high

that the magma raises the overburden and deforms it into a domelike shape.

If the flow of magma is along a crack, a two-dimensional laccolith can be

formed. Our analysis is restricted to this case. A photograph of a laccolithic

mountain is given in Figure 3–23 along with a sketch of our model.

The overburden or elastic plate of thickness h is bent upward by the

pressure p of the magma that will form the laccolith upon solidification.

The loading of the plate q(x) is the part of the upward pressure force p in

excess of the lithostatic pressure ρgh:

q = −p+ ρgh. (3.97)

This problem is very similar to the one illustrated in Figure 3–19. In both

cases the loading is uniform so that Equation (3–84) is applicable. We take

x = 0 at the center of the laccolith. The required boundary conditions are

w = dw/dx = 0 at x = ±L/2. The solution of Equation (3–84) that satisfies

these boundary conditions is obtained after some algebra in the form

w = −(p − ρgh)

24D

(

x4 − L2x2

2 + L4

16

)

. (3.98)

Note that because of the symmetry of the problem the coefficients of x and

x3 must be zero. The maximum deflection at the center of the laccolith,

x = 0, is

w0 = −(p− ρgh)L4

384D . (3.99)

In terms of its maximum value, the deflection is given by

w = w0

(

1 − 8 x2

L2 + 16

x4

L4

)

. (3.100)

Problem 3.16 Show that the cross-sectional area of a two-dimensional

laccolith is given by (p− ρgh)L5/720D.

Problem 3.17 Determine the bending moment in the overburden above

the idealized two-dimensional laccolith as a function of x. Where is M a

maximum? What is the value of Mmax?

Problem 3.18 Calculate the fiber stress in the stratum overlying the two-

dimensional laccolith as a function of y (distance from the centerline of the

layer) and x. If dikes tend to form where tension is greatest in the base of

the stratum forming the roof of a laccolith, where would you expect dikes

to occur for the two-dimensional laccolith?

214 Elasticity and Flexure

(a )F igu

re 3.23

(a )

A la

cco lith

in R ed

a n d

W h ite

M o u n ta

in , C

o lo

ra d o . T

h e

o verlyin

g sed

im en

ta ry

rocks h a ve

been

erod ed

(U n iversity

o f C

o lo

ra d o , B o u ld

er). (b)

A tw

o -d

im en

sio n a l m

od el

fo r

a la

cco lith

.

3.12 Deformation of Strata Overlying an Igneous Intrusion 215

F ig

u re

3. 24

M od

el s

fo r

ca lc

u la

ti n g

th e

h yd

ro st

a ti c

re st

o ri

n g

fo rc

e o n

li th

o sp

h er

ic p la

te s

d efl

ec te

d by

a n

a p p li ed

lo a d

q a . (a

) O

ce a n ic

ca se

. (b

) C

o n ti n en

ta l ca

se .

216 Elasticity and Flexure

3.13 Application to the Earth’s Lithosphere

When applying Equation (3–74) to determine the downward deflection of

the earth’s lithosphere due to an applied load, we must be careful to include

in q(x) the hydrostatic restoring force caused by the effective replacement

of mantle rocks in a vertical column by material of smaller density. In the

case of the oceanic lithosphere, water fills in “the space vacated” by mantle

rocks moved out of the way by the deflected lithosphere. In the case of the

continental lithosphere, the rocks of the thick continental crust serve as the

fill. Figure 3–24a illustrates the oceanic case. The upper part of the figure

shows a lithospheric plate of thickness h and density ρm floating on a “fluid”

mantle also of density ρm. Water of density ρw and thickness hw overlies the

oceanic lithosphere. Suppose that an applied load deflects the lithosphere

downward a distance w and that water fills in the space above the plate, as

shown in the bottom part of Figure 3–24a. The weight per unit area of a

vertical column extending from the base of the deflected lithosphere to the

surface is

ρwg(hw + w) + ρmgh.

The pressure at a depth hw + h+w in the surrounding mantle where there

is no plate deflection is

ρwghw + ρmg(h +w).

Thus there is an upward hydrostatic force per unit area equal to

ρwghw + ρmg(h+ w) − {ρwg(hw + w) + ρmgh} = (ρm − ρw)gw (3.101)

tending to restore the deflected lithosphere to its original configuration. The

hydrostatic restoring force per unit area is equivalent to the force that re-

sults from replacing mantle rock of thickness w and density ρm by water

of thickness w and density ρw. The net force per unit area acting on the

lithospheric plate is therefore

q = qa − (ρm − ρw)gw, (3.102)

where qa is the applied load at the upper surface of the lithosphere. Equation

(3–74) for the deflection of the elastic oceanic lithosphere becomes

D d4w

dx4 + P

d 2w

dx2 + (ρm − ρw)gw = qa(x). (3.103)

Figure 3–24b illustrates the continental case. The upper part of the figure

shows the continental crust of thickness hc and density ρc separated by the

3.14 Periodic Loading 217

Moho from the rest of the lithosphere of thickness h and density ρm. The

entire continental lithosphere lies on top of a fluid mantle of density ρm. The

lower part of Figure 3–24b shows the plate deflected downward a distance

w by an applied load such as excess topography. The Moho, being a part of

the lithosphere, is also deflected downward a distance w. The space vacated

by the deflected lithosphere is filled in by crustal rocks. The crust beneath

the load is effectively thickened by the amount w by which the Moho is

depressed. The weight per unit area of a vertical column extending from the

base of the deflected plate to the surface is

ρcg(hc + w) + ρmgh.

The pressure at a depth hc + h+w in the surrounding mantle far from the

deflected plate is

ρcghc + ρmg(h + w).

The difference between these two quantities is the upward hydrostatic restor-

ing force per unit area

ρcghc + ρmg(h+ w) − {ρcg(hc + w) + ρmgh} = (ρm − ρc)gw. (3.104)

The restoring force is equivalent to the force that results from replacing

mantle rock by crustal rock in a layer of thickness w. The net force per unit

area acting on the elastic continental lithosphere is therefore

q = qa − (ρm − ρc)gw. (3.105)

Equation (3–74) for the deflection of the plate becomes

D d4w

dx4 + P

d 2w

dx2 + (ρm − ρc)gw = qa(x). (3.106)

We are now in a position to determine the elastic deflection of the lithosphere

and the accompanying internal stresses (shear and bending) for different

loading situations.

3.14 Periodic Loading

How does the positive load of a mountain or the negative load of a valley

deflect the lithosphere? To answer this question, we determine the response

of the lithosphere to a periodic load. We assume that the elevation of the

topography is given by

h = h0 sin 2π x

λ , (3.107)

218 Elasticity and Flexure

where h is the topographic height and λ is its wavelength. Positive h cor-

responds to ridges and negative h to valleys. Since the amplitude of the

topography is small compared with the thickness of the elastic lithosphere,

the influence of the topography on this thickness can be neglected. The

load on the lithosphere corresponding to the topography given by Equation

(3–107) is

qa(x) = ρcgh0 sin 2π x

λ (3.108)

where ρc is the density of the crustal rocks associated with the height vari-

ation. The equation for the deflection of the lithosphere is obtained by sub-

stituting this expression for qa(x) into Equation (3–106) and setting P = 0

to obtain

D d4w

dx4 + (ρm − ρc)gw = ρcgh0 sin 2π

x

λ . (3.109)

Because the loading is periodic in x, the response or deflection of the

lithosphere will also vary sinusoidally in x with the same wavelength as the

topography. Thus we assume a solution of the form

w = w0 sin 2π x

λ . (3.110)

By substituting Equation (3–110) into Equation (3–109), we determine the

amplitude of the deflection of the lithosphere to be

w0 = h0

ρm ρc

− 1 + D

ρcg

(

2π

λ

)4 . (3.111)

The quantity (D/ρcg) 1/4 has the dimensions of a length. It is proportional

to the natural wavelength for the flexure of the lithosphere.

If the wavelength of the topography is sufficiently short, that is, if

λ≪ 2π

(

D

ρcg

)1/4

, (3.112)

then the denominator of Equation (3–111) is much larger than unity, and

w0 ≪ h0. (3.113)

Short-wavelength topography causes virtually no deformation of the litho-

sphere. The lithosphere is infinitely rigid for loads of this scale. This case is

illustrated in Figure 3–25a. If the wavelength of the topography is sufficiently

long, that is, if

λ≫ 2π

(

D

ρcg

)1/4

, (3.114)

3.14 Periodic Loading 219

Figure 3.25 Deflection of the lithosphere under a periodic load. (a) Short-wavelength loading with no deflection of the lithosphere. (b) Long- wavelength loading with isostatic deflection of the lithosphere.

Figure 3.26 Buckling of an infinitely long plate under an applied horizontal load with a hydrostatic restoring force.

then Equation (3–111) gives

w = w0∞ = ρch0

(ρm − ρc) . (3.115)

This is the isostatic result obtained in Equation (2–3). For topography of

sufficiently long wavelength, the lithosphere has no rigidity and the topog-

raphy is fully compensated; that is, it is in hydrostatic equilibrium.

The degree of compensation C of the topographic load is the ratio of the

deflection of the lithosphere to its maximum or hydrostatic deflection

C = w0

w0∞ . (3.116)

Upon substituting Equations (3–111) and (3–115) into the equation for C,

we obtain

C = (ρm − ρc)

ρm − ρc + D

g

(

2π

λ

)4 . (3.117)

220 Elasticity and Flexure

Figure 3.27 Dependence of the degree of compensation on the nondimen- sional wavelength of periodic topography.

This dependence is illustrated in Figure 3–26. For a lithosphere with elastic

thickness 25 km, E = 70 GPa, ν = 0.25, ρm = 3300 kg m−3, and ρc =

2800 kg m−3 we find that topography is 50% compensated (C = 0.5) if

its wavelength is λ = 420 km. Topography with a shorter wavelength is

substantially supported by the rigidity of the lithosphere; topography with

a longer wavelength is only weakly supported.

3.15 Stability of the Earth’s Lithosphere under an End Load

We have already seen how a plate pinned at its ends can buckle if an applied

horizontal load exceeds the critical value given by Equation (3–95). Let us

investigate the stability of the lithosphere when it is subjected to a horizontal

force P . We will see that when P exceeds a critical value, an infinitely long

plate (L → ∞) will become unstable and deflect into the sinusoidal shape

shown in Figure 3–27.

The equation for the deflection of the plate is obtained by setting qa = 0

in Equation (3–103):

D d4w

dx4 + P

d 2w

dx2 + (ρm − ρw)gw = 0. (3.118)

This equation can be satisfied by a sinusoidal deflection of the plate as given

in Equation (3–110) if

D

(

2π

λ

)4

− P

(

2π

λ

)2

+ (ρm − ρw)g = 0, (3.119)

a result of directly substituting Equation (3–110) into Equation (3–118).

Equation (3–119) is a quadratic equation for the square of the wavelength

3.15 Stability of the Earth’s Lithosphere Under an End Load 221

of the sinusoid λ. Its solution is (

2π

λ

)2

= P ± [P 2 − 4(ρm − ρw)gD]1/2

2D . (3.120)

Because the wavelength of the deformed lithosphere must be real, there can

only be a solution if P exceeds the critical value

Pc = {4Dg(ρm − ρw)}1/2. (3.121)

Pc is the minimum value for P for which the initially horizontal lithosphere

will become unstable and acquire the sinusoidal shape. If P < Pc, the hori-

zontal lithosphere is stable and will not buckle under the end load.

The eigenvalue Pc can also be written

Pc =

(

Eh3(ρm − ρw)g

3(1 − ν2)

)1/2

= σch, (3.122)

where σc is the critical stress associated with the force Pc. Solving Equation

(3–122) for the critical stress we find

σc =

(

Eh(ρm − ρw)g

3(1 − ν2)

)1/2

. (3.123)

The wavelength of the instability that occurs when P = Pc is given by

Equation (3–120):

λc = 2π

(

2D

Pc

)1/2

= 2π

(

D

g(ρm − ρw)

)1/4

= 2π

(

Eh3

12(1 − ν2)(ρm − ρw)g

)1/4

. (3.124)

We wish to determine whether buckling of the lithosphere can lead to

the formation of a series of synclines and anticlines. We consider an elastic

lithosphere with a thickness of 50 km. Taking E = 100 GPa, ν = 0.25,

ρm = 3300 kg m−3, and ρw = 1000 kg m−3, we find from Equation (3–123)

that σc = 6.4 GPa. A 50-km-thick elastic lithosphere can support a hori-

zontal compressive stress of 6.4 GPa without buckling. Because of the very

large stress required, we conclude that such buckling does not occur. The

lithosphere fails, presumably by the development of a fault, before buckling

can take place. In general, horizontal forces have a small influence on the

bending behavior of the lithosphere. For this reason we neglect them in the

lithosphere bending studies to follow.

Horizontal forces are generally inadequate to buckle the lithosphere be-

cause of its large elastic thickness. However, the same conclusion may not

222 Elasticity and Flexure

Figure 3.28 A bathymetric profile across the Hawaiian archipelago.

apply to much thinner elastic layers, such as elastic sedimentary strata em-

bedded between strata that behave as fluids and highly thinned lithosphere

in regions of high heat flow. To evaluate the influence of horizontal forces

on the bending of such thin layers, we take h = 1 km and the other param-

eters as before and find from Equation (3–123) that σc = 900 MPa. From

Equation (3–124) we obtain λc = 28 km. We conclude that the buckling of

thin elastic layers may contribute to the formation of folded structures in

the earth’s crust.

3.16 Bending of the Elastic Lithosphere under the Loads of

Island Chains

Volcanic islands provide loads that cause the lithosphere to bend. The

Hawaiian ridge is a line of volcanic islands and seamounts that extends

thousands of kilometers across the Pacific. These volcanic rocks provide a

linear load that has a width of about 150 km and an average amplitude of

about 100 MPa. The bathymetric profile across the Hawaiian archipelago

shown in Figure 3–28 reveals a depression, the Hawaiian Deep, immediately

adjacent to the ridge and an outer peripheral bulge or upwarp.

To model the deflection of the lithosphere under linear loading, let us

consider the behavior of a plate under a line load V0 applied at x = 0, as

shown in Figure 3–29. Since the applied load is zero except at x = 0, we

take qa(x) = 0 and P = 0 in Equation (3–103) and solve

D d4w

dx4 + (ρm − ρw)gw = 0. (3.125)

The general solution of Equation (3–125) is

w = ex/α (

c1 cos x

α + c2 sin

x

α

)

+ e−x/α (

c3 cos x

α + c4 sin

x

α

)

, (3.126)

3.16 Bending of the Elastic Lithosphere under the Loads of Island Chains 223

Figure 3.29 Deflection of the elastic lithosphere under a line load.

where the constants c1, c2, c3, and c4 are determined by the boundary con-

ditions and

α =

[

4D

(ρm − ρw)g

]1/4

. (3.127)

The parameter α is known as the flexural parameter.

Because there is symmetry about x = 0, we need only determine w for

x ≥ 0. We require that w → 0 as x → ∞ and that dw/dx = 0 at x = 0.

Clearly, c1 and c2 must be zero and c3 = c4. Equation (3–126) becomes

w = c3e −x/α

(

cos x

α + sin

x

α

)

x ≥ 0. (3.128)

The constant c3 is proportional to the magnitude of the applied line load

V0. From Equation (3–81) we have

1

2 V0 = D

d3w

dx3 (x = 0) =

4Dc3 α3

. (3.129)

(Half the plate supports half the load applied at x = 0. Note also that a

downward force on the left end of the plate is negative according to the sign

convention illustrated in Figure 3–10.) Substituting for c3 from Equation

(3–129) into Equation (3–128), we obtain

w = V0α

3

8D e−x/α

(

cos x

α + sin

x

α

)

x ≥ 0.

(3.130)

The maximum amplitude of the deflection at x = 0 is given by

w0 = V0α

3

8D . (3.131)

In terms of w0, the deflection of the plate is

w = w0e −x/α

(

cos x

α + sin

x

α

)

. (3.132)

This profile is given in Figure 3–30.

The deflection of the lithosphere under a line load is characterized by a

224 Elasticity and Flexure

Figure 3.30 Half of the theoretical deflection profile for a floating elastic plate supporting a line load.

well-defined arch or forebulge. The half-width of the depression, x0, is given

by

x0 = α tan−1(−1) = 3π

4 α. (3.133)

The distance from the line load to the maximum amplitude of the forebulge,

xb, is obtained by determining where the slope of the profile is zero. Upon

differentiating Equation (3–132) and setting the result to zero

dw

dx = −2w0

α e−x/α sin

x

α = 0, (3.134)

we find

xb = α sin−1 0 = πα. (3.135)

The height of the forebulge wb is obtained by substituting this value of xb into Equation (3–132):

wb = −w0e −π = −0.0432w0. (3.136)

The amplitude of the forebulge is quite small compared with the depression

of the lithosphere under the line load.

This analysis for the line load is only approximately valid for the Hawaiian

Islands, since the island load is distributed over a width of about 150 km.

However, the distance from the center of the load to the crest of the arch can

be used to estimate the thickness of the elastic lithosphere if we assume that

it is equal to xb. A representative value of xb for the Hawaiian archipelago

is 250 km; with xb = 250 km, Equation (3–135) gives a flexural parameter

α = 80 km. For ρm−ρw = 2300 kg m−3 and g = 10 m s−2 Equation (3–127)

gives D = 2.4 × 1023 N m. Taking E = 70 GPa and ν = 0.25, we find from

Equation (3–72) that the thickness of the elastic lithosphere is h = 34 km.

3.16 Bending of the Elastic Lithosphere under the Loads of Island Chains 225

Figure 3.31 Deflection of a broken elastic lithosphere under a line load.

Problem 3.19 (a) Consider a lithospheric plate under a line load. Show

that the absolute value of the bending moment is a maximum at

xm = α cos−1 0 = π

2 α (3.137)

and that its value is

Mm = −2Dw0

α2 e−π/2 = −0.416

Dw0

α2 . (3.138)

(b) Refraction studes show that the Moho is depressed about 10 km beneath

the center of the Hawaiian Islands. Assuming that this is the value of w0

and that h = 34 km, E = 70 GPa, ν = 0.25, ρm − ρw = 2300 kg m−3, and

g = 10 m s−2, determine the maximum bending stress in the lithosphere.

Since volcanism along the Hawaiian ridge has weakened the lithosphere, it

may not be able to sustain large bending moments beneath the load. In this

case we should consider a model in which the lithosphere is fractured along

the line of the ridge. Let us accordingly determine the deflection of a semi-

infinite elastic plate floating on a fluid half-space and subjected to a line

load V0/2 at its end, as sketched in Figure 3–31. The deflection is given by

Equation (3–126), with the constants of integration yet to be determined.

Since the plate extends from x = 0 to x = ∞ and we require w → 0 as

x → ∞, c1 and c2 must again be zero. We have assumed that no external

torque is applied to the end x = 0. From Equation (3–73) we can conclude

that d 2w/dx2 = 0 at x = 0. This boundary condition requires that c4 = 0.

Finally, by equating the shear on the end x = 0 to the applied line load, we

find

1

2 V0 = D

d3w

dx3 (x = 0) =

2Dc3 α3

. (3.139)

With the value of c3 from Equation (3–139) and c1 = c2 = c4 = 0, Equation

(3–126) gives

w = V0α

3

4D e−x/α cos

x

α . (3.140)

226 Elasticity and Flexure

Figure 3.32 The deflection of the elastic lithosphere under an end load.

The maximum amplitude of the deflection at x = 0 is

w0 = V0α

3

4D . (3.141)

For the same load, the deflection amplitude of a broken lithosphere is twice

as great as it is for a lithosphere without a break. By substituting Equation

(3–141) into Equation (3–140), we can write

w = w0e −x/α cos

x

α . (3.142)

This profile is given in Figure 3–32.

The half-width of the depression and the position and amplitude of the

forebulge are given by

x0 = π

2 α (3.143)

xb = 3π

4 α (3.144)

wb = w0e −3π/4 cos

3π

4 = −0.0670w0. (3.145)

The amplitude of the forebulge for the broken lithosphere model, although

still small compared with the deflection of the lithosphere under the load, is

considerably larger than the forebulge amplitude of an unbroken lithosphere

supporting the same load.

We again evaluate the model results for the deflection of the lithosphere

caused by the Hawaiian Islands. With xb = 250 km, we find from Equation

(3–144) that α = 106 km. This result, together with ρm − ρw = 2300 kg

m−3, g = 10 m s−2, E = 70 GPa, and ν = 0.25, gives D = 7.26 × 1023 Nm

and h = 49 km. The thickness of a broken lithosphere turns out to be about

50% greater than the thickness of an unbroken lithosphere.

3.17 Bending of the Elastic Lithosphere at an Ocean Trench 227

Figure 3.33 Bending of the lithosphere at an ocean trench due to an applied vertical load and bending moment.

Problem 3.20 (a) Consider a lithospheric plate under an end load. Show

that the absolute value of the bending moment is a maximum at

xm = α tan−1 1 = π

4 α, (3.146)

and that its value is

Mm = −2Dw0

α2 e−π/4 sin

π

4 = −0.644

Dw0

α2 .

(3.147)

(b) Refraction studies show that the Moho is depressed about 10 km beneath

the center of the Hawaiian Islands. Assuming that this is the value of w0

and that h = 49 km, E = 70 GPa, ν = 0.25, ρm − ρw = 2300 kg m−3, and

g = 10 m s−2, determine the maximum bending stress in the lithosphere.

3.17 Bending of the Elastic Lithosphere at an Ocean Trench

Another example of the flexure of the oceanic elastic lithosphere is to be

found at ocean trenches. Prior to subduction, considerable bending of the

elastic lithosphere occurs. The bent lithosphere defines the oceanward side

of the trench. To model this behavior, we will consider an elastic plate acted

upon by an end load V0 and a bending moment M0, as illustrated in Figure

3–33.

The deflection of the plate is governed by Equation (3–125), and once

again the general solution is given by Equation (3–126). We require w → 0

as x→ ∞ so that c1 = c2 = 0 and

w = e−x/α (

c3 cos x

α + c4 sin

x

α

)

. (3.148)

At x = 0, the bending moment is −M0; from Equation (3–73) we obtain

c4 = −M0α

2

2D . (3.149)

228 Elasticity and Flexure

Also, at x = 0, the shear force is −V0; from Equations (3–59) and (3–73) we

find

c3 =(V0α+M0) α2

2D . (3.150)

The equation for w can now be written as

w = α2e−x/α

2D

{

−M0 sin x

α + (V0α+M0) cos

x

α

}

.

(3.151)

Equation (3–151) reduces to Equation (3–140) in the case M0 = 0. Note

that the line load here is V0; it was V0/2 in Equation (3–140).

The elastic deflection of the oceanic lithosphere in terms of the vertical

force and bending moment at the ocean trench axis is given by Equation

(3–151). The vertical force and bending moment are the result of the grav-

itational body force acting on the descending plate. Unfortunately, V0 and

M0 cannot be determined directly. Quantities that can be measured directly

are the height of the forebulge wb and the half-width of the forebulge xb –

x0, as illustrated in Figure 3–33. We therefore express the trench profile in

terms of these parameters. We can determine x0 by setting w = 0:

tan x0

α = 1 +

αV0

M0 . (3.152)

Similarly, we can determine xb by setting dw/dx = 0:

tan xb α

= −1 − 2M0

αV0 . (3.153)

The height of the forebulge is obtained by substituting this value of xb into

Equation (3–151):

wb = α2

2D e−xb/α

[

−M0 sin xb α

+ (M0 + V0α) cos xb α

]

.

(3–154)

From Equations (3–152) and (3–154) we find

tan

(

xb − x0

α

)

= sin (xb α − x0

α

)

cos (xb α − x0

α

)

= sin xb

α cos x0 α − cos xb

α sin x0 α

cos xb α cos x0

α + sin xb α sin x0

α

= tan xb

α − tan x0 α

1 + tan xb α tan x0

α

= 1 (3.155)

3.17 Bending of the Elastic Lithosphere at an Ocean Trench 229

and

xb − x0 = π

4 α. (3.156)

This half-width is a direct measure of the flexural parameter and, therefore,

of the flexural rigidity and thickness of the elastic lithosphere.

By using Equation (3–152), we can rewrite Equation (3–151) for the de-

flection of the lithosphere as

w = α2M0

2D e−x/α

(

−sin x

α + tan

x0

α cos

x

α

)

= α2M0

2D e−[(x−x0)/α]−x0/α

× {

sin x0 α cos xα− cos x0

α sin x α

cos x0 α

}

= −α 2M0

2D e−[(x−x0)/α]e−x0/α

sin (x−x0

α

)

cos (x0 α

) .

(3.157)

The height of the forebulge is thus given by

wb = −α 2M0

2D e−[(xb−x0)/α]e−x0/α

sin (

xb−x0 α

)

cos (x0 α

) .

(3.158)

Upon dividing Equation (3–157) by Equation (3–158) and eliminating α

using Equation (3–156), we obtain

w

wb =

exp [

− π 4

(

x− x0 xb − x0

)]

exp (

−π4 )

sin [

π 4

(

x− x0 xb − x0

)]

sin π 4

= √

2eπ/4exp

[

−π 4

(

x−x0

xb−x0

)]

sin

[

π

4

(

x−x0

xb−x0

)]

.

(3.159)

The plot of w/wb vs. (x − x0)/(xb/x0) shown in Figure 3–34a defines a

universal flexure profile. The profile is valid for any two-dimensional elastic

flexure of the lithosphere under end loading.

We can solve for the bending moment in terms of (x − x0)/(xb − x0) by

substituting Equation (3–159) into Equation (3–73)

M =

√ 2π2eπ/4

8

Dwb (xb − x0)2

cos

[

π(x− x0)

4(xb − x0)

]

230 Elasticity and Flexure

× exp

[

− π(x− x0)

4(xb − x0)

]

. (3.160)

The dependence of M(xb−x0) 2/Dwb on (x− x0)/(xb−x0) is shown in Fig-

ure 3–34b. The bending moment is a maximum at (x− x0)/(xb − x0) = −1.

The shear force can be determined from Equations (3–59) and (3–160) to

be

V = − √

2π3eπ/4

32

Dwb (xb − x0)3

[

cos

{

π(x− x0)

4(xb − x0)

}

+ sin

{

π(x− x0)

4(xb − x0)

}]

exp

[

− π(x− x0)

4(xb − x0)

]

.

(3.161)

The dimensionless shear force V (xb−x0) 3/Dwb is plotted vs. (x−x0)/(xb−

x0) in Figure 3–34c. The shear force is zero at (x− x0)/(xb − x0) = −1.

The universal flexure profile is compared with an observed bathymetric

profile across the Mariana trench in Figure 3–35. In making the comparison,

we take xb = 55 km and wb = 500 m (x0 = 0). From Equation (3–156) we

find that α = 70 km. With ρm − ρw = 2300 kg m−3 and g = 10 m s−2,

Equation (3–127) gives D = 1.4 × 1023 N m. From Equation (3–72) with

E = 70 GPa and ν = 0.25 we find that the thickness of the elastic lithosphere

is 28 km. This value is in quite good agreement with the thickness of the

oceanic elastic lithosphere obtained by considering island loads. The largest

bending stress is 900 MPa, and it occurs 20 km seaward of the trench axis.

This is a very large deviatoric stress, and it is doubtful that the near-surface

rocks have sufficient strength in tension. However, the yield stress of the

mantle is likely to approach this value at depth where the lithostatic pressure

is high.

Although the trench bathymetric profile given in Figure 3–35 appears

to exhibit elastic flexure, other trench profiles exhibit an excessively large

curvature near the point of the predicted maximum bending moment. This

is discussed in Chapter 7, where we associate this excess curvature with the

plastic failure of the lithosphere.

3.18 Flexure and the Structure of Sedimentary Basins

Lithospheric flexure is also associated with the structure of many sedimen-

tary basins. A sedimentary basin is a region where the earth’s surface has

been depressed and the resulting depression has been filled by sediments.

Typical sedimentary basins have depths up to 5 km, although some are as

3.18 Flexure and the Structure of Sedimentary Basins 231

F ig

u re

3. 34

U n iv

er sa

l so

lu ti o n

fo r

th e

d efl

ec ti o n

o f a n

el a st

ic li th

o sp

h er

e u n d er

a ve

rt ic

a l en

d lo a d

a n d

be n d in

g

m o m

en t.

(a )

D ep

en d en

ce o f th

e n o n d im

en si

o n a l d is

p la

ce m

en t

w /w

b o n

th e

n o n d im

en si

o n a l po

si ti o n

(x −

x 0 )/

(x b −

x 0 ).

T h e

p ro

fi le

is a ls o

sh o w n

a t a n

a m

p li fi ca

ti o n

o f 1 0

to 1

to m

o re

cl ea

rl y

sh o w

th e

st ru

ct u re

o f th

e

fo re

bu lg

e. (b

) T

h e

d im

en si

o n le

ss be

n d in

g m

o m

en t ve

rs u s

(x −

x 0 )/

(x b −

x 0 ).

(c )

T h e

d im

en si

o n le

ss ve

rt ic

a l sh

ea r

fo rc

e a s

a fu

n ct

io n

(x −

x 0 )/

(x b −

x 0 ).

232 Elasticity and Flexure

F igu

re 3.35

C o m

pa riso

n o f a

ba th

ym etric

p ro

fi le

a cro

ss th

e M

a ria

n a

tren ch

(so lid

lin e)

w ith

th e

u n iversa

l

lith o sp

h eric

d efl

ectio n

p ro

fi le

given by

E qu

a tio

n (3

– 1 5 9 )

(d a sh

ed lin

e); x b =

5 5

km a n d

w b =

0 .5

km .

3.18 Flexure and the Structure of Sedimentary Basins 233

deep as 15 km. Because sedimentary basins contain reservoirs of petroleum,

their structures have been studied in detail using seismic reflection profiling

and well logs.

Some sedimentary basins are bounded by near-vertical faults along which

the subsidence has occurred. Others, however, have a smooth basement, and

the subsidence is associated with the flexure of the elastic lithosphere. The

horizontal dimensions of these sedimentary basins, about 400 to 1000 km,

reflect the magnitude of the flexural parameter based on sediments of density

ρs replacing mantle rock of density ρm, α = [4D/(ρm − ρs)g] 1/4.

Some sedimentary basins have a nearly twodimensional structure. They

are caused by the loading of a linear mountain belt and are known as foreland

basins. Examples are the series of sedimentary basins lying east of the Andes

in South America and the Appalachian basin in the eastern United States

lying west of the Appalachian Mountains. Depth contours of the basement

beneath the Appalachian basin are given in Figure 3–36a. A basement profile

is shown in Figure 3–36b. The depth w is the depth below sea level, and the

coordinate −x is measured from the point where basement rocks are exposed

at the surface.

It is appropriate to model the structure of the Appalachian basin as a two-

dimensional lithospheric plate under a linear end load. Thus the universal

flexure profile given in Equation (3–159) is directly applicable. In order to

fit the basement profile given in Figure 3–36b we take xb = 122 km and

wb = 290 m (x0 = 0). Since the forebulge has been destroyed by erosion this

choice of parameters is not unique. They can be varied somewhat, and a

reasonable fit can still be obtained. However, these values are near the center

of the acceptable range. From Equation (3–156) we find that they correspond

to α = 155 km. As we have already noted, the flexural rigidity must be based

on the density difference between the mantle and the sediments ρm − ρs.

With ρm − ρs = 700 kg m−3 and g = 10 m s−2 we find D = 1024 N

m. From Equation (3–72) with E = 70 GPa and ν = 0.25 we find that

the thickness of the elastic continental lithosphere is h = 54 km. This is

somewhat larger than the values we obtained for the thickness of the elastic

oceanic lithosphere. Flexure studies of other sedimentary basins give similar

values of elastic thickness.

Problem 3.21 An ocean basin has a depth of 5.5 km. If it is filled to sea

level with sediments of density 2600 kg m−3, what is the maximum depth

of the resulting sedimentary basin? Assume ρm = 3300 kg m−3.

Problem 3.22 The Amazon River basin in Brazil has a width of 400 km.

Assuming that the basin is caused by a line load at its center and that the

234 Collateral Reading

Figure 3.36 (a) Contours of basement (in km) in the Appalachian basin of the eastern United States. Data are from well logs and seismic reflection studies. (b) The data points are the depths of basement below sea level as a function of the distance from the point where basement rocks are exposed at the surface along the profile given by the heavy line in (a). The solid line is the universal flexure profile from Equation (3–159) with xb = 122 km and wb = 290 m (x0 =0).

elastic lithosphere is not broken, determine the corresponding thickness of

the elastic lithosphere. Assume E = 70 GPa, ν = 0.25, and ρm − ρs = 700

kg m−3.

Collateral Reading

Bieniawski, Z. T. (1967), Mechanism of brittle fracture of rock: Part II.

Experimental studies, Int. J. Rock. Mech. Min. Sci., 4, 407–423.

Collateral Reading 235

Collateral Reading

Eringen, A. C., Mechanics of Continua (John Wiley, New York, 1967), 502

pages.

A comprehensive treatment of the mechanics of continua at a relatively

sophisticated level. The basic concepts of strain, stress, flow, thermo-

dynamics, and constitutive equations are introduced. Applications are

made to elasticity, fluid dynamics, thermoplasticity, and viscoelasticity.

Fung, Y. C., Foundations of Solid Mechanics (Prentice-Hall, Englewood

Cliffs, NJ, 1965), 525 pages.

A graduate-level textbook on the mechanics of solids. The text is mainly

concerned with the classical theory of elasticity, thermodynamics of

solids, thermoelasticity, viscoelasticity, plasticity, and finite deforma-

tion theory. The book begins with an introductory chapter on elastic

and viscoelastic behavior. Cartesian tensors are then introduced and

used in the discussions of stress, strain, and the conservation laws. Sub-

sequent chapters deal with linear elasticity, solutions of elastic prob-

lems by potentials, two-dimensional problems, energy theorems, Saint-

Venant’s principle, Hamilton’s principle, wave propagation, elasticity

and thermodynamics, thermoelasticity, viscoelasticity, and finite strain

theory. Problems for the student are included.

Jaeger, J. C., Elasticity, Fracture, and Flow, 3rd edition (Methuen, London,

1969), 268 pages.

A monograph on the mathematical foundations of elasticity, plasticity, vis-

cosity, and rheology. Chapter 1 develops the analysis of stress and strain

with emphasis on Mohr’s representations. Chapter 2 discusses stress–

strain relations for elasticity, viscosity, and plasticity, and criteria for

fracture and yield. Chapter 3 derives the equations of motion and equi-

librium. Chapters 4 and 5 deal with stresses in the earth’s crust, rock

mechanics, and applications to structural geology.

Jaeger, J. C., and N. G. W. Cook, Fundamentals of Rock Mechanics (Chap-

man and Hall, London, 1976), 585 pages.

See collateral reading list for Chapter 2.

Kraus, H., Thin Elastic Shells (John Wiley, New York, 1967).

An extensive mathematical treatment of the deformation of thin elastic

shells. It includes three chapters on the theory of thin elastic shells,

four chapters on static analysis, two chapters on dynamic analysis, and

two chapters on numerical methods.

Muskhelishvili, N. I., Some Basic Problems of the Mathematical Theory of

Elasticity, 4th edition (P. Noordhoff, Groningen, 1963), 718 pages.

236 Collateral Reading

This treatise on the mathematical theory of elasticity is divided into seven

major parts. Part 1 deals with the fundamental equations of the me-

chanics of an elastic body. It includes separate chapters on analyses

of stress and strain, relation between stress and strain, the equilibrium

equations of an elastic body, and the fundamental boundary value prob-

lems of static elasticity. Part 2 treats planar problems whose solutions

are obtained with the aid of the stress function and its complex rep-

resentation. The technique of conformal mapping is introduced. Part 3

develops the Fourier series approach to the solution of planar problems,

while Parts 4 and 5 make use of Cauchy integrals. Part 6 presents solu-

tions for special planar geometries and Part 7 deals with the extension,

torsion and bending of bars.

Novozhilov, V. V., Thin Shell Theory (P. Noordhoff, Groningen, 1964), 377

pages.

A mathematical analysis of stresses and strains in thin shells using linear

elasticity theory. There are four chapters on the general theory of thin

elastic shells, the membrane theory of shells, cylindrical shells, and shells

of revolution.

Timoshenko, S., and J. N. Goodier, Theory of Elasticity, (McGraw-Hill, New

York, 1970), 567 pages.

See collateral reading list for Chapter 2.

Timoshenko, S., and D. H. Young, Elements of Strength of Materials, 5th

edition (Van Nostrand, Princeton, NJ, 1968), 377 pages.

An undergraduate engineering textbook with an extensive treatment of the

bending of beams and elastic stability. Problems with solutions are in-

cluded.

4

Heat Transfer

4.1 Introduction

In the previous chapter we studied the elastic behavior of the outer shell of

the Earth. Our studies of the bending of the lithosphere have shown that a

near-surface region with a thickness of 25 to 50 km behaves elastically on

geological time scales. Seismic studies have shown that the entire mantle of

the Earth to a depth of 2885 km is a solid because it transmits shear waves.

In order to understand the presence of a thin elastic shell, it is necessary to

allow for variations in the rheology of the solid rock as a function of depth.

Although the behavior of the near-surface rocks is predominantly elastic, the

deeper rocks must exhibit a fluid or creep behavior on geological time scales

in order to relax the stresses. The fluid behavior of mantle rock also results

in mantle convection and the associated movement of the surface plates.

We know from both laboratory and theoretical studies that the rheology

of solids is primarily a function of temperature. Therefore, to understand the

mechanical behavior of the Earth, we must understand its thermal structure.

The rheology of mantle rocks is directly related to the temperature as a

function of depth. This, in turn, is dependent on the rate at which heat

can be lost from the interior to the surface. There are three mechanisms for

the transfer of heat: conduction, convection, and radiation. Conductive heat

transfer occurs through a medium via the net effect of molecular collisions.

It is a diffusive process wherein molecules transmit their kinetic energy to

other molecules by colliding with them. Heat is conducted through a medium

in which there is a spatial variation in the temperature. Convective heat

transport is associated with the motion of a medium. If a hot fluid flows

into a cold region, it will heat the region; similarly, if a cold fluid flows into

a hot region, it will cool it. Electromagnetic radiation can also transport

heat. An example is the radiant energy from the Sun. In the Earth, radiative

238 Heat Transfer

heat transport is only important on a small scale and its influence can be

absorbed into the definition of the thermal conductivity.

As the discussion of this chapter shows, both conduction and convection

are important heat transport mechanisms in the Earth. The temperature dis-

tribution in the continental crust and lithosphere is governed mainly by the

conductive heat loss to the surface of heat that is generated internally by the

decay of radioactive isotopes in the rocks and heat that flows upward from

the subcontinental mantle. The loss of the Earth’s internal heat through the

oceanic crust and lithosphere is controlled largely by conduction, although

convective heat transport by water circulating through the basaltic crustal

rocks is also important, especially near ridges. Intrusive igneous bodies cool

by both conduction and the convective effects of circulating groundwater.

The heating of buried sediments and the adjustment of subsurface temper-

atures to effects of surface erosion and glaciation occur via the process of

conduction. Convection plays the dominant role in the transport of heat from

the Earth’s deep mantle and in controlling the temperature of its interior.

This chapter discusses mainly heat conduction and its application to ge-

ological situations. Because convective heat transfer involves fluid motions,

we will postpone a detailed discussion of this subject to Chapter 6, where

we will develop the fundamentals of fluid mechanics. However, the conse-

quences of convective heat transport are incorporated into our discussion of

the Earth’s temperature toward the end of this chapter.

4.2 Fourier’s Law of Heat Conduction

The basic relation for conductive heat transport is Fourier’s law, which

states that the heat flux q, or the flow of heat per unit area and per unit

time, at a point in a medium is directly proportional to the temperature

gradient at the point. In one dimension, Fourier’s law takes the form

q = −kdT dy

(4.1)

where k is the coefficient of thermal conductivity and y is the coordinate

in the direction of the temperature variation. The minus sign appears in

Equation (4–1) since heat flows in the direction of decreasing temperature.

With dT/dy > 0, T increases in the positive y direction, so that heat must

flow in the negative y direction.

Figure 4–1 is a simple example of how Fourier’s law can be used to give the

heat flux through a slab of material of thickness l across which a temperature

4.2 Fourier’s Law of Heat Conduction 239

Figure 4.1 Heat transfer through a slab.

Figure 4.2 Heat flux and the local slope of the temperature profile when T(y) has nonzero curvature.

difference ∆T is maintained. In this case, the temperature gradient is

dT

dy = −∆T

l , (4.2)

and the heat flux, from Fourier’s law, is

q = k∆T

l . (4.3)

Fourier’s law applies even when the temperature distribution is not linear,

as sketched in Figure 4–2. In this case, the local slope of the temperature

profile must be used in Fourier’s law, and for constant k the heat flux is a

function of y, q = q(y). We will see that curvature in a temperature profile

implies either the occurrence of sources or sinks of heat or time dependence.

240 Heat Transfer

4.3 Measuring the Earth’s Surface Heat Flux

The average heat flux at the Earth’s surface provides important information

on the amount of heat being produced in the Earth and the temperature

distribution in its interior. In the 1800s it was recognized that the tem-

perature in caves and mines increases with depth. Typical values for this

increase are dT/dy = 20 to 30 K km−1. Since the thermal conductivities

of near-surface rocks are usually in the range 2 to 3 W m−1 K−1, the heat

flow to the surface of the Earth implied by these temperature gradients is,

according to Equation (4–1), 40 to 90 mW m−2. It is standard practice to

take the upward surface heat flow to be a positive quantity, even though

from Equation (4–1), with y measured positive downward, it has a negative

value.

Although temperature measurements in caves and mines give approximate

values for the near-surface thermal gradient, accurate measurements of the

thermal gradient in continental areas require deep drill holes. Deep mea-

surements are necessary because climatic variations in the Earth’s surface

temperature, particularly those due to ice ages, influence the temperatures

in the near-surface rocks. These effects are considered quantitatively in Sec-

tion 4–14. In order to reach the steady-state thermal structure, holes must

be drilled deeper than about 300 m.

The thermal gradient is measured by lowering a thermistor (an accu-

rate electronic thermometer) down the drill hole. Care must be exercised

to prevent the circulation of drilling fluid during drilling from affecting the

measured gradient. This can be done in either of two ways. Measurements

can be made at the bottom of the drill hole during drilling. The drilling

fluid does not have time to change the temperature at the bottom of the

hole. Alternatively, the temperature log of the hole (the measurement of

the temperature as a function of depth) can be carried out some time after

drilling has ceased. It usually takes 1 to 2 years for a drill hole to equilibrate

to the ambient geothermal gradient. Drill holes are invariably filled with

groundwater. It is the temperature of this water that is measured by the

thermistor. As long as the water is not flowing, its temperature equilibrates

with that of the surrounding rock. However, many drill holes cross aquifers

(underground channels of porous rock in which water flows), with the result

that water will flow up or down the drill hole if it is not lined and will affect

the temperature distribution.

From Fourier’s law it is clear that the determination of the heat flux re-

quires a measurement of both the temperature gradient and the thermal

conductivity of the rock. The thermal conductivity of rocks can be deter-

4.3 Measuring the Earth’s Surface Heat Flux 241

Figure 4.3 Laboratory device for measuring the thermal conductivity of a rock sample.

mined in the laboratory by subjecting samples cut from drill holes to known

heat fluxes and measuring the temperature drops across them. Figure 4–3

is a schematic of one way in which this can be done.

The rock sample of thermal conductivity kr is placed between material –

brass, for example – of thermal conductivity kb. Thermocouples measure the

temperatures of the hot and cold ends of the metal, TH and Tc, respectively,

and the temperatures at the surfaces of the metal adjoining the rock section,

T1 and T2. The contact between the rock and metal (air perhaps) involves an

unknown thermal resistance to the flow of heat. Since the same heat must be

conducted through the rock and metal in steady state, Fourier’s law (4–1)

can be used to determine kr in terms of the measurable quantities TH , Tc,

T1, T2, d, l and the conductivity kb. Thermal conductivities of a variety of

rocks are given in Section E of Appendix 2.

Problem 4.1 For the situation sketched in Figure 4–3 and discussed

above, show that the thermal conductivity of the rock sample can be deter-

mined from the equation

T1 − T2

TH − T1 = kb kr

d

l +

2δkb lkc

(4.4)

The thermal resistance of the contacts are accounted for by associating a

thermal conductivity kc and a thickness δ with each contact. By making

242 Heat Transfer

measurements on rock samples of different thicknesses and plotting (T1 − T2)/(TH − T1) vs. d, one can determine kr from the slope of the resulting

straight line without knowing either δ or kc.

We just discussed the determination of the surface heat flow in the con-

tinents. The heat flow can also be measured on the ocean floor. A large

fraction of the seafloor is covered by a layer of soft sediments. A needlelike

probe carrying a series of thermistors is dropped from a ship and penetrates

the sediments. Typically the probe has a length of 3 m. The near-surface

heat flow in the oceanic crust is almost a constant because climatic vari-

ations do not change the temperature of the seawater in the deep oceans.

This water is buffered at a temperature between 1 and 2◦C, the temperature

at which the density of the seawater is a maximum. The variation is due to

changes in salinity. In many cases, however, the near-surface heat flow in the

sediments is influenced by the hydrothermal convection of seawater through

the sediments and basaltic crustal rocks.

The thermal conductivity of the sediments can be determined using a

heater in the heat-flow probe. The record of the increase in probe tempera-

ture with time after the heater is turned on can be interpreted to give the

thermal conductivity of the sediments, as discussed later in this chapter.

Problem 4.2 Temperatures at the interfaces between sedimentary layers

of different rock types as determined from a well log are given in Table 4–1.

The measured thermal conductivity of each layer is also given. Determine

the heat flow through each layer and the mean value of the heat flow.

4.4 The Earth’s Surface Heat Flow

Tens of thousands of heat flow measurements have been made both in the

continents and the oceans. Because the oceanic crust participates in the

plate tectonic cycle and the continental crust does not, we can consider

these regions separately.

The mean heat flow for all continents is 65 ± 1.6 mW m−2. Regions of

high heat flow in the continents are generally restricted to active volcanic

areas. Examples are the lines of volcanoes associated with ocean trenches

– the Andes, for example – and regions of tensional tectonics such as the

western United States. The areas of high heat flow associated with volcanic

lines are generally quite small and do not contribute significantly to the

mean heat flow. Similarly, areas of tensional tectonics are quite small on a

global basis. Broad regions of continental tectonics, such as the collision zone

4.4 The Earth’s Surface Heat Flow 243

Table 4.1 Temperatures Between Layers of Rock Types

Depth (m) Temp. (◦C) Rock Type k (Wm--1 K--1)

380 18.362 Sandstone 3.2

402 18.871 Shale 1.7

412 19.330 Sandstone 5.3

465 20.446 Salt 6.1

475 20.580 Sandstone 3.4

510 21.331 Shale 1.9

515 21.510

extending from the Alps through the Himalayas, have near-normal surface

heat flows. Therefore, regions of active tectonics and mountain building make

a relatively small contribution to the mean continental heat flow.

In stable continental areas, the surface heat flow has a strong correlation

with the surface concentrations of the radioactive, heat-producing isotopes.

This correlation, which is discussed in detail in Section 4–8, is illustrated in

Figure 4–11. Approximately one-half of the surface heat flow in the conti-

nents can be attributed to the heat production from the radioactive isotopes

of uranium, thorium, and potassium in the continental crust. Surface heat

flow systematically decreases with the age of the surface rocks in stable con-

tinental areas. Similarly, the concentration of the radioactive isotopes in the

surface rocks also decreases with the age of the rocks. This decrease is at-

tributed to the progressive effects of erosion that remove the near-surface

rocks with the largest concentrations of the heat-producing isotopes. The

conclusion is that the decrease in surface heat flow with age in stable con-

tinental areas can be primarily attributed to the decrease in the crustal

concentrations of the heat-producing isotopes.

The mean measured heat flow for all the oceans is 101 ± 2.2 mW m−2.

The concentration of the heat-producing isotopes in the oceanic crust is

about one order of magnitude less than it is in the continental crust. Also,

the oceanic crust is about a factor of 5 thinner than the continental crust.

Therefore, the contribution of heat production by the radioactive isotopes

in the oceanic crust to the surface heat flow is negligible (∼2%).

The most striking feature of heat flow measurements in the oceans is the

244 Heat Transfer

systematic dependence of the surface heat flow on the age of the seafloor.

This can be understood as a consequence of the gradual cooling of the

oceanic lithosphere as it moves away from the mid-ocean ridge. This pro-

cess is analyzed in detail in Section 4–16, where it is shown that conductive

cooling of the initially hot oceanic mantle can explain quantitatively the

observed heat flow–age relation. The dependence of the oceanic heat flow

measurements on age is given in Figure 4–25.

The total heat flow from the interior of the Earth Q can be obtained by

multiplying the area of the continents by the mean continental heat flow and

adding the product of the oceanic area and the mean oceanic heat flow. The

continents, including the continental margins, have an area Ac = 2×108 km2.

Multiplying this by the mean observed continental heat flow, 65 mW m−2,

we get the total heat flow from the continents to be Qc = 1.30 × 1013 W.

Similarly, taking the oceans, including the marginal basins, to have an area

Ao = 3.1 × 108 km2 and a mean observed heat flow of 101 mW m−2, we

find that the total heat flow from the oceans is Qo = 3.13× 1013 W. Adding

the heat flow through the continents and the oceans, we find that the total

surface heat flow is Q = 4.43×1013 W. Dividing by the Earth’s surface area

A = 5.1 × 108 km2, we get 87 mW m−2 for the corresponding mean surface

heat flow.

4.5 Heat Generation by the Decay of Radioactive Elements

A substantial part of the heat lost through the Earth’s surface undoubtedly

originates in the interior of the Earth by the decay of the radioactive elements

uranium, thorium, and potassium. Some part of the surface heat loss must

come from the overall cooling of the Earth through geologic time. An upper

limit to the concentration of radioactive elements in the Earth can be derived

by attributing all the surface heat loss to the radioactive heat generation.

The mean heat generation per unit mass H is then given by

H = Q

M . (4.5)

If we take M = 5.97×1024 kg, the mass of the Earth, and Q = 4.43×1013 W,

we find H = 7.42 × 10−12 W kg−1. However, on the basis of geochemical

studies, we can argue that the core cannot contain a significant fraction of

the heat-producing elements. In this case, the mass in Equation (4–5) should

be the mass of the mantle, M = 4.0×1024 kg and H = 11.1×10−12 W kg−1.

A further reduction must be made in the value of H appropriate to the

mantle because a substantial fraction of the heat lost from the continents

4.5 Heat Generation by the Decay of Radioactive Elements 245

originates in the highly concentrated radioactive isotopes of the continental

crust. Although the mean continental heat flux of 65 mW m−2 is known with

some certainty, we are uncertain as to the fraction that can be attributed to

the heat-producing elements. Based on estimates of the mean concentrations

of these elements in the continental crust, we attribute 37 mW m−2 to the

heat-producing elements. The remaining 28 mW m−2 is attributed to basal

heating of the continental lithosphere by mantle convection. This heat is then

conducted through the mantle portion of the continental lithosphere to the

base of the continental crust. Radiogenic heat production in the continental

crust corresponds to a total heat flow of 7.4 × 1012 W, or 17% of the total

surface heat flow. Reduction of the mantle heat production by this amount

gives H = 9.22 × 10−12 W kg−1.

A further correction to the radiogenic heat production in the mantle must

be made because of the secular cooling of the Earth. Only a fraction of the

present-day surface heat flow can be attributed to the decay of radioactive

isotopes presently in the mantle. Because the radioactive isotopes decay

into stable isotopes, heat production due to radioactive decay is decreasing

with time. For example, we will show that the heat production 3 billion

years ago was about twice as great as it is today. Since less heat is being

generated in the Earth through time, less heat is also being convected to

the surface. Thus, the vigor of the mantle convection system is decreasing

with the age of the Earth. Because the strength of convection is dependent

on viscosity and the viscosity of the mantle is a sensitive function of its

temperature, a decrease in the heat flux with time leads to a decrease in the

mean mantle temperature. This cooling of the Earth in turn contributes to

the surface heat flow. We will consider this problem in some detail in Section

7–8 and conclude that about 80% of the present-day surface heat flow can be

attributed to the decay of radioactive isotopes presently in the Earth and

about 20% comes from the cooling of the Earth. We can thus reduce the

present-day mantle heat production accordingly so that our preferred value

is H = 7.38 × 10−12 W kg−1.

Radioactive heating of the mantle and crust is attributed to the decay

of the uranium isotopes 235U and 238U, the thorium isotope 232Th, and the

potassium isotope 40K. The rates of heat production and the half-lives τ1/2 of these isotopes are given in Table 4–2. At the present time natural uranium

is composed of 99.28% by weight 238U and 0.71% 235U. Natural thorium is

100% 232Th. Natural potassium is composed of 0.0119% 40K. The present

rates of heat production of natural uranium and potassium are also given in

Table 4–2.

The ratios of potassium to uranium and thorium to uranium are nearly

246 Heat Transfer

Table 4.2 Rates of Heat Release H and Half-Lives τ1/2 of the Important

Radioactive Isotopes in the Earth’s Interior

H τ1/2 Concentration C

Isotope (W kg--1) (yr) (kg kg--1)

238U 9.46 × 10−5 4.47 × 109 30.8 × 10−9

235U 5.69 × 10−4 7.04 × 108 0.22 × 10−9

U 9.81 × 10−5 31.0 × 10−9

232Th 2.64 × 10−5 1.40 × 1010 124 × 10−9

40K 2.92 × 10−5 1.25 × 109 36.9 × 10−9

K 3.48 × 10−9 31.0 × 10−5

Note: Heat release is based on the present mean mantle concentrations of the heat-producing elements.

constant in a wide range of terrestrial rocks. Based on these observed ra-

tios we take CK 0 /C

U 0 = 104 and CTh

0 /CU 0 = 4, where CK

0 , CTh 0 , and CU

0

are the present mass concentrations of potassium, thorium, and uranium,

respectively. The total present-day production H0 is related to the heat gen-

eration rates of the individual radioactive elements by

H0 = CU 0

(

HU + CTh

0

CU 0

HTh + CK

0

CU 0

HK )

. (4.6)

Taking H0 = 7.38× 10−12 W kg−1 and the other parameters as given above

and in Table 4–2, we find that CU 0 = 3.1×10−8 kg kg−1 or 31 ppb (parts per

billion by weight). These preferred values for the mean mantle concentrations

of heat-producing elements are also given in Table 4–2.

The mean heat production rate of the mantle in the past can be related

to the present heat production rate using the half-lives of the radioactive

isotopes. The concentration C of a radioactive isotope at time t measured

backward from the present is related to the present concentration C0 and

the half-life of the isotope τ1/2 by

C = C0 exp

(

t ln 2

τ1/2

)

. (4.7)

Thus, the past mean mantle heat production rate is given by

H = 0.9928CU 0 H

U238 exp

t ln 2

τU238

1/2

+ 0.0071CU 0 H

U235 exp

t ln 2

τU235

1/2

4.5 Heat Generation by the Decay of Radioactive Elements 247

Figure 4.4 Mean mantle heat production rates due to the decay of the radioactive isotopes of U, Th, and K as functions of time measured back from the present.

+CTh 0 HTh exp

(

t ln 2

τTh 1/2

)

+ 1.19 × 10−4CK 0 H

K40 exp

t ln 2

τK40

1/2

.

(4.8)

The rate of mean mantle heat production based on Equation (4–8) and

parameter values in Table 4–2 is plotted as a function of time before the

present in Figure 4–4. The past contributions of the individual radioactive

elements are also shown. We see that the rate of heat production 3× 109 yr

ago was about twice the present value. Today heat is produced primarily by 238U and 232Th, but in the distant past 235U and 40K were the dominant

isotopes because of their shorter half-lives.

The concentrations of the heat-producing elements in surface rocks vary

considerably. Some typical values are given in Table 4–3. The mantle values

from Table 4–2 are included for reference. Partial melting at ocean ridges

depletes mantle rock of incompatible elements such as uranium, thorium,

and potassium. These incompatible elements are concentrated in the basaltic

partial melt fraction. As a result, the oceanic crust is enriched in these

elements by about a factor of 4 relative to the fertile mantle. Peridotites

that have been depleted in the incompatible elements are sometimes found

on the surface of the Earth. A typical example of the small concentrations

of the heat-producing elements in a “depleted” peridotite is given in Table

4–3. Processes that lead to the formation of the continental crust, such

as the volcanism associated with ocean trenches, further differentiate the

248 Heat Transfer

Table 4.3 Typical Concentrations of the Heat-Producing Elements in

Several Rock Types and the Average Concentrations in Chondritic

Meteorites

Concentration Rock Type U (ppm) Th (ppm) K (%)

Reference undepleted (fertile) mantle 0.031 0.124 0.031 “Depleted” peridotites 0.001 0.004 0.003 Tholeiitic basalt 0.07 0.19 0.088 Granite 4.7 20 4.2 Shale 3.7 12 2.7 Average continental crust 1.42 5.6 1.43 Chondritic meteorites 0.008 0.029 0.056

incompatible elements. The concentrations of the heat-producing elements

in a typical continental rock such as a granite are quite variable, but in

general they are an order of magnitude greater than in tholeiitic basalts.

Representative values of concentrations in granite are given in Table 4–3.

It is generally accepted that the chondritic class of meteorites is rep-

resentative of primitive mantle material. The average concentrations of the

heatproducing elements in chondritic meteorites are listed in Table 4–3. The

concentrations of uranium and thorium are about a factor of 4 less than our

mean mantle values, and the concentration of potassium is about a factor of

2 larger. The factor of 8 difference in the ratio CK 0 /C

U 0 is believed to repre-

sent a fundamental difference in elemental abundances between the Earth’s

mantle and chondritic meteorites.

Problem 4.3 Determine the present mean mantle concentrations of the

heat-producing elements if the present value for the mean mantle heat pro-

duction is 7.38 × 10−12 W kg−1 and CK 0 /C

U 0 = 6 × 104 and CTh

0 /CU 0 = 4.

Problem 4.4 Determine the rates of heat production for the rocks listed

in Table 4–3.

Problem 4.5 The measured concentrations of the heat-producing ele-

ments in a rock are CU = 3.2 ppb, CTh = 11.7 ppb, and CK = 2.6%.

Determine the rate of heat generation per unit mass in the rock.

4.6 One-Dimensional Steady Heat Conduction 249

Figure 4.5 Heat flow into q(y) and out of q(y+ δ y) a thin slab of thickness δy producing heat internally at the rate of H per unit mass.

4.6 One-Dimensional Steady Heat Conduction with Volumetric

Heat Production

Heat conduction theory enables us to determine the distribution of temper-

ature in a region given information about the temperatures or heat fluxes

on the boundaries of the region and the sources of heat production in the

region. In general, we can also use the theory to determine time variations

in the temperature distribution. We first develop the theory for the simple

situation in which heat is transferred in one direction only and there are no

time variations (steady state) in the temperature or heat flow. The basic

equation of conductive heat transfer theory is a mathematical statement of

conservation of energy; the equation can be derived as follows.

Consider a slab of infinitesimal thickness δy, as sketched in Figure 4–5.

The heat flux out of the slab q(y + δy) crosses the face of the slab located

at y + δy, and the heat flux into the slab q(y) crosses the face located at y.

The net heat flow out of the slab, per unit time and per unit area of the

slab’s face, is

q(y + δy) − q(y).

Since δy is infinitesimal, we can expand q(y + δy) in a Taylor series as

q(y + δy) = q(y) + δy dq

dy + · · · . (4.9)

250 Heat Transfer

Thus we find

q(y + δy) − q(y) = δy dq

dy = δy

d

dy

[

−k (

dT

dy

)]

= δy

[

−k (

d2T

dy2

)]

, (4.10)

where we have used Equation (4–1) (Fourier’s law) for q, and we have as-

sumed a constant thermal conductivity. The right side of Equation (4–10)

is the net heat flow out of a slab of thickness δy, per unit time and per unit

area. It is nonzero only when there is curvature in the temperature profile.

If there is a nonzero net heat flow per unit area out of the thin slab, as

given by Equation (4–10), this heat flow must be supplied, in steady state,

by heat generated internally in the slab. If H is the heat production rate

per unit mass, the amount of heat generated in the slab per unit time and

per unit area of the slab face is

ρHδy, (4.11)

where ρ is the density of the slab. By equating (4–10) and (4–11), one obtains

0 = k d2T

dy2 + ρH. (4.12)

This equation can be integrated to determine temperature as a function of

position y once the region of interest and appropriate boundary conditions

have been specified.

Assume that the medium is a half-space with the surface at y = 0 (see

Figure 4–6). The coordinate y increases with distance into the half-space;

thus y is a depth coordinate. One possible set of boundary conditions for

Equation (4–12) is the specification of both temperature and heat flux at the

surface. Thus we require the temperature T to be T0 at y = 0 and the heat

flux at the surface q to be −q0. The reason for the difference in sign between

q and q0 is that q is positive in the direction of positive y, that is, downward,

while q0 is assumed to be positive upward (we anticipate application to the

Earth for which the surface heat flux is indeed upward).

One integration of Equation (4–12) gives

ρHy = −kdT dy

+ c1 = q + c1, (4.13)

where c1 is a constant of integration. Since q = −q0 on y = 0, we find

c1 = q0 (4.14)

4.6 One-Dimensional Steady Heat Conduction 251

Figure 4.6 Geometry and boundary conditions for integration of Equation (4–12).

and

ρHy = −kdT dy

+ q0. (4.15)

Integration of Equation (4–15) results in

ρH y2

2 = −kT + q0y + c2, (4.16)

where c2 is another constant of integration. Since T = T0 on y = 0, we find

c2 = kT0.

and

T = T0 + q0 k y − ρH

2k y2. (4.17)

Problem 4.6 Consider a geological situation in which the subsurface is

layered, with bedding planes making an angle θ with the horizontal surface,

as shown in Figure 4–7a. Suppose that the thermal conductivity for heat

conduction parallel to BC is k1 and the conductivity for heat transport par-

allel to AB is k3. Though the bedding planes are inclined to the horizontal,

isotherms are nevertheless horizontal. Show that the upward surface heat

flow is given by

q0 = {k1 + (k3 − k1) cos2 θ}∂T ∂y

. (4.18)

HINT: Introduce coordinates ζ and η as shown in Figure 4–7b. Note that

y = η cos θ + ζ sin θ (4.19)

252 Heat Transfer

Figure 4.7 Geometry for Problem 4–6.

Figure 4.8 Temperature as a function of depth within the Earth assuming heat transport is by conduction (conduction geotherm). Also included are the solidus and liquidus of basalt and the solidus of peridotite (olivine).

and

∂T

∂η = cos θ

∂T

∂y

∂T

∂ζ = sin θ

∂T

∂y . (4.20)

Write a steady-state heat balance for the triangle ABC in Figure 4–7a. Use

Fourier’s law of heat conduction to evaluate qζ and qη. Note q0 = −qy. Assume no heat sources.

4.7 A Conduction Temperature Profile for the Mantle 253

4.7 A Conduction Temperature Profile for the Mantle

We can use Equation (4–17) to determine the temperature as a function of

depth in the Earth, that is, the geotherm, assuming heat is transported by

conduction. The depth profile of the temperature is given in Figure 4–8, as-

suming T0 = 0◦C, q0 = 70 mW m−2, ρ = 3300 kg m−3, H = 7.38× 10−12 W

kg−1, and k = 4 W m−1 K−1. Also included in Figure 4–8 are the liquidus

and solidus of basalt and the solidus of peridotite. Basalt is the low-melting-

temperature fraction of the mantle. When the temperature of the mantle

exceeds the basalt solidus, this fraction starts to melt, resulting in volcan-

ism. This is the cause of the extensive basaltic volcanism that forms the

oceanic crust. When the temperature reaches the basalt liquidus, this frac-

tion is entirely melted, leaving a high-melting-temperature residuum that is

primarily composed of the mineral olivine. When the mantle temperature

reaches the olivine solidus, the remainder of the mantle rock melts. The abil-

ity of seismic shear waves to propagate through the mantle indicates that

substantial melting does not occur. The conclusion is that this conduction

analysis does not predict the temperature in the Earth’s mantle.

In an attempt to assess the failure of the conductive mantle geotherm to

model the Earth, one may ask whether the near-surface concentration of

radioactive elements in crustal rocks can modify the analysis. (The partial

melting processes that lead to the formation of the crust concentrate the

radioactive elements.) The only way in which this could have an effect is

through a reduction in the amount of the surface heat flow q0 attributed

to mantle heat sources. Thus we must assess the contribution of crustal

radioactivity to surface heat flow. It is appropriate to do this for the oceanic

crust because the suboceanic mantle geotherm dominates the temperature

distribution of the mantle.

To determine the contribution qc to the surface heat flow of a layer of

crust of thickness hc and heat production per unit mass Hc, we proceed

as indicated in Figure 4–9. Equation (4–13) applies to this case also, with

ρ = ρc and H = Hc (subscript c refers to the crust),

ρcHcy = −kdT dy

+ c1 = q + c1. (4.21)

To evaluate c1, we note that q = −qc on y = 0 and

c1 = qc.

The heat flux in the slab satisfies

q + qc = ρcHcy. (4.22)

254 Heat Transfer

Figure 4.9 Heat flow through the top of a slab containing internal heat sources. No heat flows through the bottom of the slab.

But q = 0 at y = hc because we have assumed that no heat enters the bottom

of the slab (the appropriate boundary condition if we want to determine the

heat flowing out the top of the slab due only to radioactive isotopes contained

in it). Thus we find

qc = ρcHchc. (4.23)

The oceanic crust is primarily composed of basalts. Thus we take ρc =

2900 kg m−3, hc = 6 km, and Hc = 2.6 × 10−11 W kg−1. (The radiogenic

heat production rate per unit mass of basalts was calculated in Problem

4–4.) From Equation (4–23) the resultant contribution to the surface heat

flow is qc = 0.45 mW m−2; this is a small fraction of the mean oceanic heat

flow, which is about 100 mW m−2. The conclusion is that heat production

in the oceanic crust does not make a significant contribution to the oceanic

surface heat flow. Therefore, an alternative explanation must be found for

the failure of the simple conduction profile to model the suboceanic mantle

geotherm. In later sections we show that heat flow due to mantle convection

invalidates the conduction results.

4.8 Continental Geotherms

Whereas conductive temperature profiles fail to describe the mantle geotherm,

they successfully model the geotherm in the continental crust and litho-

sphere, where the dominant thermal processes are radiogenic heat produc-

tion and conductive heat transport to the surface. Because of the great age

4.8 Continental Geotherms 255

of the continental lithosphere, time-dependent effects can, in general, be

neglected.

The surface rocks in continental areas have considerably larger concen-

trations of radioactive elements than the rocks that make up the oceanic

crust. Although the surface rocks have a wide range of heat production, a

typical value for a granite is Hc = 9.6 × 10−10 W kg−1 (H for granite was

calculated in Problem 4–4). Taking hc = 35 km and ρc = 2700 kg m−3, one

finds that the heat flow from Equation (4–23) is qc = 91 mW m−2. Since this

value is considerably larger than the mean surface heat flow in continental

areas (65 mW m−2), we conclude that the concentration of the radioactive

elements decreases with depth in the continental crust.

For reasons that we will shortly discuss in some detail it is appropriate to

assume that the heat production due to the radioactive elements decreases

exponentially with depth,

H = H0e −y/hr . (4.24)

Thus H0 is the surface (y = 0) radiogenic heat production rate per unit

mass, and hr is a length scale for the decrease in H with depth. At the depth

y = hr, H is 1/e of its surface value. Substitution of Equation (4–24) into

the equation of energy conservation (4–12) yields the differential equation

governing the temperature distribution in the model of the continental crust:

0 = k d2T

dy2 + ρH0e

−y/hr . (4.25)

Beneath the near-surface layer of heat-producing elements we assume that

the upward heat flow at great depth is qm; that is, q → −qm as y → ∞.

This model for heat production in the continental crust is sketched in Figure

4–10.

An integration of Equation (4–25) yields

c1 = k dT

dy − ρH0hre

−y/hr = −q − ρH0hre −y/hr .

(4.26)

The constant of integration c1 can be determined from the boundary con-

dition on the heat flux at great depth, that is, from the mantle heat flux to

the base of the lithosphere

c1 = qm. (4.27)

Thus the heat flux at any depth is

q = −qm − ρH0hre −y/hr . (4.28)

256 Heat Transfer

Figure 4.10 Model of the continental crust with exponential radiogenic heat source distribution.

The surface heat flow q0 = −q(y = 0) is obtained by setting y = 0 with the

result

q0 = qm + ρhrH0. (4.29)

With an exponential depth dependence of radioactivity, the surface heat

flow is a linear function of the surface radioactive heat production rate.

In order to test the validity of the linear heat flow–heat production re-

lation (4–29), determinations of the radiogenic heat production in surface

rocks have been carried out for areas where surface heat flow measurements

have been made. Several regional correlations are given in Figure 4–11. In

each case a linear correlation appears to fit the data quite well. The cor-

responding length scale hr is the slope of the best-fit straight line and the

mantle (reduced) heat flow qm is the vertical intercept of the line. For the

Sierra Nevada data we have qm = 17 mW m−2 and hr = 10 km; for the

eastern United States data we have qm = 33 mW m−2 and hr = 7.5 km;

for the Norway and Sweden data, qm = 22 mW m−2 and hr = 7.2 km; and

for the eastern Canadian shield data, qm = 30.5 mW m−2 and hr = 7.1 km.

In all cases the length scale hr is near 10 km. The values of the mantle or

reduced heat flow qm are reasonably consistent with the mean basal heating

of the continental lithosphere qm = 28 mW m−2 given in Section 4–5.

Thus a model of the continental crust with exponentially decreasing ra-

4.8 Continental Geotherms 257

Figure 4.11 Dependence of surface heat flow q0 on the radiogenic heat production per unit volume in surface rock ρ H0 in selected geological provinces: Sierra Nevada (solid squares and very long dashed line), east- ern U.S. (solid circles and intermediate dashed line), Norway and Sweden (open circles and solid line), eastern Canadian shield (open squares and short dashed line). In each case the data are fit with the linear relationship Equation (4–29).

dioactivity can explain the linear surface heat flow–surface radioactivity

relation. The exercises to follow show that the exponential radioactivity

distribution is not unique in its ability to model the linear q0 versus ρH0

relation; other dependences of H on depth that confine radioactivity near

the surface are consistent with observations. However, the exponential dis-

tribution is the only one that preserves the linear q0 versus ρH0 relation

under differential erosion, a strong argument to support its relevance to the

continental crust. The exponential depth dependence is also consistent with

magmatic and hydrothermal differentiation processes, although a detailed

understanding of these processes in the continental crust is not available.

Problem 4.7 Table 4–4 gives a series of surface heat flow and heat produc-

tion measurements in the Sierra Nevada Mountains in California. Determine

the reduced heat flow qm and the scale depth hr.

Problem 4.8 Consider one-dimensional steady-state heat conduction in

a half-space with heat production that decreases exponentially with depth.

258 Heat Transfer

Table 4.4 Surface Heat Flow and Heat Production Data for the Sierra

Nevada Mountains

q0 ρ H0 q0 ρ H0

(mW m--2) (µWm--3) (mW m--2) (µW m--3)

18 0.3 31 1.5 25 0.8 34 2.0 25 0.9 42 2.6 29 1.3 54 3.7

The surface heat flow–heat production relation is q0 = qm + ρH0hr. What

is the heat flow–heat production relation at depth y = h∗? Let q∗ and H∗

be the upward heat flux and heat production at y = h∗.

Problem 4.9 Assume that the radioactive elements in the Earth are

uniformly distributed through a near-surface layer. The surface heat flow

is 70 mW m−2, and there is no heat flow into the base of the layer. If

k = 4 W m−1 K−1, T0 = 0◦C, and the temperature at the base of the layer

is 1200◦C, determine the thickness of the layer and the volumetric heat

production.

Problem 4.10 Consider one-dimensional steady-

state heat conduction in a half-space. The heat sources are restricted to

a surface layer of thickness b; their concentration decreases linearly with

depth so that H = H0 at the surface y = 0 and H = 0 at the depth y = b.

For y > b, H = 0 and there is a constant upward heat flux qm. What is

the q0 (upward surface heat flow)–H0 relation? Determine the temperature

profile as a function of y.

Problem 4.11 The exponential depth dependence of heat production is

preferred because it is self-

preserving upon erosion. However, many alternative models can be pre-

scribed. Consider a two-layer model with H = H1 and k = k1 for 0 ≤ y ≤ h1,

and H = H2 and k = k2 for h1 ≤ y ≤ h2. For y > h2, H = 0 and the up-

ward heat flux is qm. Determine the surface heat flow and temperature at

y = h2 for ρ1 = 2600 kg m−3, ρ2 = 3000 kg m−3, k1 = k2 = 2.4 W m−1

K−1, h1 = 8 km, h2 = 40 km, ρ1H1 = 2 µW m−3, ρ2H2 = 0.36 µW m−3,

T0 = 0 ◦C, and qm = 28 mW m−2.

A further integration of Equation (4–28) using Equation (4–1) and the

4.8 Continental Geotherms 259

Figure 4.12 A typical geotherm in the continental crust.

boundary condition T = T0 at y = 0 gives

T = T0 + qmy

k + ρH0h

2 r

k (1 − e−y/hr). (4.30)

or, alternatively, using Equation (4–29), we obtain

T = T0 + qmy

k +

(q0 − qm)hr k

(1 − e−y/hr). (4.31)

Figure 4–12 is a plot of a typical geotherm in the continental crust computed

from Equation (4–31) with T0 =10 ◦C, q0 = 56.5 mW m−2, qm = 30 mW

m−2, hr = 10 km, and k = 3.35 W m−1 K−1.

Problem 4.12 An alternative model for the continental crust is to assume

that in addition to the exponentially decreasing near-surface radioactivity

there is also a constant concentration of radioactivity H0 to the depth hc.

Show that the crustal geotherm for this model is given by

T = T0 + ρH0h

2 r

k (1 − e−y/hr) − ρHcy

2

2k

+ (qm + ρHchc)

k y for 0 ≤ y ≤ hc, (4.32)

T = T0 + ρHch

2 c

2k + ρH0h

2 r

k + qmy

k for y ≥ hc.

(4.33)

260 Heat Transfer

Figure 4.13 Heat flow into and out of a thin spherical shell with internal heat generation.

4.9 Radial Heat Conduction in a Sphere or Spherical Shell

We next consider the steady radial conduction of heat in a sphere or spher-

ical shell with volumetric heat production. The temperature distributions

in thick planetary lithospheres, such as the lithospheres of the Moon and

Mars, are properly described by solutions of the heat conduction equation in

spherical geometry. The effects of spherical geometry are not so important

for the Earth’s lithosphere, which is quite thin compared with the Earth’s

radius. However, on a small body like the Moon, the lithosphere may be a

substantial fraction of the planet’s radius. To describe heat conduction in

spherical geometry, we must derive an energy balance equation.

Consider a spherical shell of thickness δr and inner radius r, as sketched

in Figure 4–13. Assume that the conductive transport of heat occurs in a

spherically symmetric manner. The total heat flow out of the shell through

its outer surface is

4π(r + δr)2qr(r + δr),

and the total heat flow into the shell at its inner surface is

4πr2qr(r).

The subscript r on the heat flux q indicates that the flow of heat is radial.

Since δr is infinitesimal, we can expand qr(r + δr) in a Taylor series as

qr(r + δr) = qr(r) + δr dqr dr

+ · · · . (4.34)

Thus neglecting powers of δr, the net heat flow out of the spherical shell is

4.9 Radial Heat Conduction in a Sphere or Spherical Shell 261

given by

4π[(r + δr)2qr(r + δr) − r2qr(r)]

= 4πr2 (

2

r qr +

dqr dr

)

δr. (4.35)

If the net heat flow from the shell is nonzero, then, by conservation of

energy, this flow of heat must be supplied by heat generated internally in

the shell (in steady state). With the rate of heat production per unit mass

H, the total rate at which heat is produced in the spherical shell is

4πr2ρHδr,

4πr2δr being the approximate expression for the volume of the shell. By

equating the rate of heat production to the net heat flow out of the spherical

shell, Equation (4–35), we get

dqr dr

+ 2qr r

= ρH. (4.36)

The heat balance Equation (4–36) can be converted into an equation for

the temperature by relating the radial heat flux qr to the radial temperature

gradient dT/dr. Fourier’s law still applies in spherical geometry,

qr = −kdT dr . (4.37)

Upon substituting Equation (4–37) into Equation (4–36), we find

0 = k

(

d2T

dr2 +

2

r

dT

dr

)

+ ρH (4.38)

or

0 = k 1

r2 d

dr

(

r2 dT

dr

)

+ ρH. (4.39)

By twice integrating Equation (4–39), one obtains a general expression for

the temperature in a sphere or spherical shell with internal heat production

and in steady state:

T = −ρH 6k

r2 + c1 r

+ c2. (4.40)

The constants of integration c1 and c2 depend on the boundary conditions

of a particular problem. As an example, we determine the temperature dis-

tribution in a sphere of radius a that has a uniform internal rate of heat

production. The boundary condition is that the outer surface of the sphere

has a temperature T0. In order to have a finite temperature at the center

262 Heat Transfer

Figure 4.14 Steady-state temperature distribution in a sphere containing internal heat sources.

of the sphere, we must set c1 = 0. To satisfy the temperature boundary

condition at the surface of the sphere, we require

c2 = T0 + ρHa2

6k . (4.41)

The temperature in the sphere is therefore given by

T = T0 + ρH

6k (a2 − r2). (4.42)

From Equation (4–37), the surface heat flux q0 at r = a is given by

q0 = 1

3 ρHa. (4.43)

Equation (4–43) is a statement of conservation of energy that applies no

matter what the mode of internal heat transfer in the sphere is. The tem-

perature distribution in the sphere is shown in Figure 4–14.

Problem 4.13 Derive the equation q0 = ρHa/3 for a sphere with uniform

volumetric heating and density by making a simple overall steady-state heat

balance.

Problem 4.14 What would the central temperature of the Earth be if

it were modeled by a sphere with uniform volumetric heating? Take q0 =

70 mW m−2, k = 4 W m−1 K−1, and T0 = 300 K.

Problem 4.15 Derive an expression for the temperature at the center of

a planet of radius a with uniform density ρ and internal heat generation H.

4.10 Temperatures in the Moon 263

Heat transfer in the planet is by conduction only in the lithosphere, which

extends from r = b to r = a. For 0 ≤ r ≤ b heat transfer is by convection,

which maintains the temperature gradient dT/dr constant at the adiabatic

value −Γ. The surface temperature is T0. To solve for T (r), you need to

assume that T and the heat flux are continuous at r = b.

Problem 4.16 It is assumed that a constant density planetary body of

radius a has a core of radius b. There is uniform heat production in the core

but no heat production outside the core. Determine the temperature at the

center of the body in terms of a, b, k, T0 (the surface temperature), and q0 (the surface heat flow).

4.10 Temperatures in the Moon

The Moon is a relatively small planetary body so it is a good approximation

to assume that its density is constant. If we also assume that the Moon is in a

steady-state thermal balance and that the mean heat production is the same

as the value we derived for the Earth’s mantle, that is, H = 7.38× 10−12 W

kg−1, we can determine the surface heat flow on the Moon using Equation (4–

43). With ρ = 3300 kg m−3 and a = 1738 km we find that q0 = 14.1 mW

m−2. The mean of two lunar heat flow measurements on Apollos 15 and 17

is qs = 18 mW m−2. This approximate agreement suggests that the mean

lunar abundances of the radioactive isotopes are near those of the Earth.

The difference may be partially attributable to the cooling of the Moon.

Assuming that the conduction solution is applicable and that the Moon

has a uniform distribution of radioactivity, the maximum temperature at the

center of the Moon can be obtained from Equation (4–42) with the result

Tmax = 3904 K, assuming k = 3.3 W m−1 K−1 and that the surface temper-

ature is T0 = 250 K. This conduction solution indicates that a substantial

fraction of the interior of the Moon is totally melted. However, the limited

seismic results from the Apollo missions suggest that a sizable liquid core

in the Moon is unlikely. Thus, either the conductive solution is not valid or

the radioactive isotopes are not distributed uniformly throughout the Moon.

There should be some upward concentration of radioactive isotopes in the

relatively thick lunar highland crust (60 km) by analogy with the upward

concentration of radioactive isotopes in the Earth’s continental crust.

Problem 4.17 Determine the steady-state conduction temperature profile

for a spherical model of the Moon in which all the radioactivity is confined

to an outer shell whose radii are b and a (a is the lunar radius). In the outer

shell H is uniform.

264 Heat Transfer

Figure 4.15 Heat flow into and out of a rectangular element.

4.11 Steady Two- and Three-Dimensional Heat Conduction

Obviously, not all heat conduction problems of geologic interest can be solved

by assuming that heat is transported in one direction only. In this section,

we generalize the heat conduction equation to account for heat transfer

in two dimensions. The further generalization to three dimensions will be

obvious and stated without proof. The first step is to write an appropriate

energy conservation equation. If heat can be conducted in both the x and y

directions, we must consider the heat balance on a small rectangular element

with dimensions δx and δy, as illustrated in Figure 4–15.

The heat flux in the x direction is qx, and in the y direction it is qy.

The rate at which heat flows into the element in the y direction is qy(y)δxl,

where l is an arbitrary length in the third direction (in two-dimensional heat

conduction we assume that nothing varies in the third dimension). Similarly,

heat flows into the element in the x direction at the rate qx(x)δyl. The heat

flow rates out of the element are qy(y + δy)δxl and qx(x + δx)δyl. The net

heat flow rate out of the element is

{qx(x+ δx) − qx(x)}δyl + {qy(y + δy) − qy(y)}δxl

= ∂qx ∂x

δxδyl + ∂qy ∂y

δxδyl =

(

∂qx ∂x

+ ∂qy ∂y

)

δxδyl.

(4.44)

Taylor series expansions have been used for qx(x + δx) and qy(y + δy) to

simplify the expression in Equation (4–44). Partial derivative symbols ap-

4.11 Steady Two- and Three-Dimensional Heat Conduction 265

pear in Equation (4–44) because qx can depend on both x and y; similarly

qy can be a function of both x and y.

In steady state, a nonzero value of the right side of Equation (4–44) re-

quires that heat be produced internally in the rectangular element. The rate

of heat generation in the element is ρH(δxδyl); equating this to the right

side of Equation (4–44) yields

∂qx ∂x

+ ∂qy ∂y

= ρH. (4.45)

Clearly, if we had heat conduction in three dimensions, Equation (4–45)

would be replaced by

∂qx ∂x

+ ∂qy ∂y

+ ∂qz ∂z

= ρH. (4.46)

Fourier’s law of heat conduction relates the heat flow in any direction to

the temperature gradient in that direction. If we assume that the thermal

conductivity of the rock is isotropic, that is, the rock conducts heat equally

readily in any direction, Fourier’s law can be written

qx = − k∂T ∂x

(4.47)

qy = − k∂T ∂y

. (4.48)

Upon substitution of Equations (4–47) and (4–48) into Equation (4–45), we

obtain

−k (

∂2T

∂x2 + ∂2T

∂y2

)

= ρH. (4.49)

Generalizing this to three-dimensional heat conduction gives

−k (

∂2T

∂x2 + ∂2T

∂y2 + ∂2T

∂z2

)

= ρH. (4.50)

If there are no internal heat sources, the temperature satisfies

∂2T

∂x2 + ∂2T

∂y2 = 0. (4.51)

Equation (4–51) is known as Laplace’s equation. In three dimensions, Laplace’s

equation is

∂2T

∂x2 + ∂2T

∂y2 + ∂2T

∂z2 = 0. (4.52)

Laplace’s equation is encountered in many other fields, including fluid flow,

diffusion, and magnetostatics.

266 Heat Transfer

Figure 4.16 Temperature in a half-space whose surface temperature varies periodically with distance.

Problem 4.18 If the medium conducting heat is anisotropic, separate

thermal conductivities must be used for heat transfer in the x and y direc-

tions, kx and ky, respectively. What equation replaces Equation (4–49) for

determining the temperature distribution in such a medium?

4.12 Subsurface Temperature Due to Periodic Surface

Temperature and Topography

As an example of a two-dimensional heat conduction problem, we solve for

the temperatures beneath the surface in a region where there are lateral

variations in surface temperature. Such surface temperature variations can

arise as a result of topographic undulations and the altitude dependence

of temperature in the Earth’s atmosphere. Horizontal surface temperature

variations also occur at the boundaries between land and bodies of water,

such as lakes or seas. A knowledge of how surface temperature variations

affect subsurface temperature is important for the correct interpretation of

borehole temperature measurements in terms of surface heat flow.

Consider again a semi-infinite half-space in the region y ≥ 0. The surface

is defined by the plane y = 0. For simplicity, we assume that the surface

temperature Ts is a periodic function of x (horizontal distance) given by

Ts = T0 + ∆T cos 2πx

λ , (4.53)

where λ is the wavelength of the spatial temperature variation. The situa-

tion is sketched in Figure 4–16. We also assume that there are no radiogenic

heat sources in the half-space, since our interest here is in the nature of the

subsurface temperature variations caused by the periodic surface tempera-

ture. To determine the temperature distribution, we must solve Equation

(4–51) with the boundary condition given by Equation (4–53).

We assume that the method of separation of variables is appropriate; that

4.12 Subsurface Temperature 267

Figure 4.17 An undulating surface topography results in surface tempera- ture variations that extend downward.

is,

T (x, y) = T0 +X(x)Y (y). (4.54)

In order to satisfy the surface boundary condition, we must have

X(x) = cos 2πx

λ ; (4.55)

that is, the horizontal variations in temperature are the same at all depths.

When Equations (4–54) and (4–55) are substituted into Equation (4–51),

we obtain

0 = −4π2

λ2 Y +

d2Y

dy2 , (4.56)

which is an ordinary differential equation for Y . The general solution of this

equation is

Y (y) = c1e −2πy/λ + c2e

2πy/λ, (4.57)

where c1 and c2 are the constants of integration. Since the temperature

must be finite as y → ∞, we must require that c2 = 0. To satisfy the

boundary condition given in Equation (4–53), it is necessary that c1 = ∆T .

The solution for the temperature distribution in the half-space is

T (x, y) = T0 + ∆T cos 2πx

λ e−2πy/λ. (4.58)

The temperature disturbance introduced by the surface temperature vari-

ation decays exponentially with depth in a distance proportional to the

horizontal wavelength of the surface temperature variation.

The energy equation is linear in the temperature. Therefore, solutions to

the equation can be added, and the result is still a solution of the energy

equation. This is known as the principle of superposition. If the temperature

in the continental crust is given by Equation (4–30) but the surface temper-

ature has a periodic variation given by Equation (4–53), the temperature

268 Heat Transfer

distribution in the crust is obtained by adding Equations (4–30) and (4–58):

T = T0 + qmy

k + ρH0h

2 r

k (1 − e−y/hr)

+ ∆T cos 2πx

λ e−2πy/λ. (4.59)

This result satisfies the applicable energy equation (4–49) and the required

surface boundary condition (4–53).

The analysis in this section can also be used to determine the effect of

small amplitude, periodic topography on the near-surface temperature dis-

tribution. This problem is illustrated in Figure 4–17. The topography is

given by the relation

h = h0 cos 2πx

λ (4.60)

We assume that the atmosphere has a vertical temperature gradient β so

that the surface temperature Ts is given by

Ts = T0 + βy y = h. (4.61)

A typical value for β is 6.5 K km−1.

To apply these results, we must project the surface temperature values

that are known on y = h onto y = 0. This is because the temperature

given by Equation (4–59) is written in terms of ∆T , the amplitude of the

periodic temperature variation on y = 0; see Equation (4–53). Because the

topography is shallow, this can be accomplished with just the first term of

a Taylor series expansion:

T (y = 0) = T (y = h) − (

∂T

∂y

)

y=0 h. (4.62)

The temperature on y = h is given by Equation (4–61), and the temperature

gradient (∂T/∂y)y=0 is given to sufficient accuracy by the value of the gradi-

ent in the absence of topography because h is small. From Equation (4–29)

we can write (

∂T

∂y

)

y=0 = q0 k

= qm + ρhrH0

k . (4.63)

The result of substituting Equations (4–61) and (4–63) into Equation (4–62)

is

T (y = 0) = T0 + βh− (qm + ρhrH0)

k h

= T0 +

{

β − (

qm + ρhrH0

k

)}

h0 cos 2πx

λ .

4.13 One-Dimensional, Time-Dependent Heat Conduction 269

(4.64)

Comparison of Equation (4–53) and (4–64) shows that

∆T =

(

β − qm k

− ρH0hr k

)

h0. (4.65)

Finally, substitution of Equation (4–65) into Equation (4–59) gives

T = T0 + qmy

k + ρH0h

2 r

k (1 − e−y/hr )

+

(

β − qm k

− ρH0hr k

)

h0 cos 2πx

λ e−2πy/λ,

(4.66)

for the temperature distribution in the continental crust with periodic to-

pography.

Problem 4.19 If a spatially periodic surface temperature variation has a

wavelength of 1 km, at what depth is the horizontal variation 1% of that at

the surface?

Problem 4.20 A mountain range can be represented as a periodic topog-

raphy with a wavelength of 100 km and an amplitude of 1.2 km. Heat flow

in a valley is measured to be 46 mW m−2. If the atmospheric gradient is

6.5 K km−1 and k = 2.5 Wm−1 K−1, determine what the heat flow would

have been without topography; that is, make a topographic correction.

4.13 One-Dimensional, Time-Dependent Heat Conduction

Many of the important geological problems involving heat conduction are

time dependent. Examples that we consider later are the cooling of intrusive

igneous bodies, the cooling of the oceanic lithosphere, erosion or sedimenta-

tion effects on temperature, and others. Volumetric heat production usually

plays a minor role in these phenomena, and we accordingly assume H = 0.

In addition, it is adequate to consider heat conduction in one direction only.

If there are not heat sources in the medium, a net heat flow out of the

slab illustrated in Figure 4–5 must reduce its temperature. The specific heat

c of the medium is the energy required to raise the temperature of a unit

mass of material by one degree. Thus, an element of the slab of thickness δy

and unit cross-sectional area requires an energy flow per unit time given by

ρc ∂T

∂t δy

270 Heat Transfer

to maintain a temperature change at the rate ∂T/∂t(ρδy is the slab mass

per unit cross-sectional area and ρcδy is the slab’s heat capacity per unit

cross-sectional area). Thus we can equate the right side of Equation (4–10)

with −δy ρc∂T/∂t, since a net heat flow out of the slab leads to a decrease

in slab temperature

ρc ∂T

∂t = k

∂2T

∂y2 . (4.67)

Equation (4–67) is the basic equation governing the time and spatial vari-

ations of the temperature when heat is transferred in one dimension by

conduction. Partial derivatives are required because T is a function of both

time and space. We can rewrite Equation (4–67) in the form

∂T

∂t = κ

∂2T

∂y2 , (4.68)

where κ, the thermal diffusivity, is

κ = k

ρc . (4.69)

Note that κ has units of length2/time such as square meters per second.

If temperature changes occur with a characteristic time interval τ , they

will propagate a distance on the order of √ κτ . Similarly, a time l2/κ is

required for temperature changes to propagate a distance l. Such simple

considerations can be used to obtain useful estimates of thermal effects. We

now proceed to solve Equation (4–68) for a number of situations of geological

and geophysical interest.

Problem 4.21 Derive the time-dependent heat conduction equation for

a situation in which heat transport occurs radially toward or away from a

line of infinite length. The heat flux q and the temperature T depend only

on the perpendicular distance from the line r and time t (see Figure 4–18a).

HINT: Write an energy balance for a cylindrical shell of length l, inner

radius r, and outer radius r+δr. The heat flows occur over the entire lateral

surfaces of the cylindrical shell, as sketched in Figure 4–18b. Fourier’s law

of heat conduction in the form q = −k (∂T/∂r) applies. The answer is

∂T

∂t = κ

r

∂

∂r

(

r ∂T

∂r

)

. (4.70)

Problem 4.22 Derive the time-dependent heat conduction equation ap-

propriate to the situation in which heat transport is always radially toward

or away from a point. Equation (4–35) gives the net heat flow out of a thin

4.14 Periodic Heating of a Semi-Infinite Half-Space 271

Figure 4.18 Geometry for the derivation of the time-dependent heat con- duction equation in cylindrical coordinates.

spherical shell. This must be equated to the time rate of change of temper-

ature of the shell times the heat capacity of the shell. Fourier’s law in the

form of Equation (4–37) applies. The answer is

∂T

∂t = κ

r

∂2

∂r2 (rT ). (4.71)

Problem 4.23 Using the relation τ = l2/κ and taking κ = 1 mm2 s−1,

determine the characteristic times for the conductive cooling of the Earth,

Moon, Mars, Venus, and Mercury. What are the implications of these esti-

mates?

Problem 4.24 If the mean surface heat flow on the Earth (q̄0 = 87 mW

m−2) were attributed entirely to the cooling of the Earth, what would be

the mean rate of cooling? (Take c̄ = 1 kJ kg−1 K−1.)

Problem 4.25 If the mean surface heat flow on the Moon (q̄0 = 18 mW m−2)

were attributed entirely to the cooling of the Moon, what would be the mean

rate of cooling (Take c̄ = 1 kJ kg−1 K−1q0).

4.14 Periodic Heating of a Semi-Infinite Half-Space: Diurnal and

Seasonal Changes in Subsurface Temperature

The surface temperature of the Earth regularly changes with time because

of day–night variations and the changes of season. On a longer time scale,

it changes because of the quasi-periodic nature of glaciations, for example.

In this section we use the one-dimensional, time-dependent heat conduction

equation to determine how these time-periodic surface temperature changes

affect temperatures below the surface.

272 Heat Transfer

Again consider a semi-infinite half-space in the region y ≥ 0 whose surface

is defined by the plane y = 0. We assume that the surface temperature is a

periodic function of time

Ts = T0 + ∆T cosωt. (4.72)

The circular frequency ω is related to the frequency f by

ω = 2πf. (4.73)

In addition, the period of the temperature fluctuations τ is

τ = 1

f =

2π

ω . (4.74)

If this represents the daily variation of the surface temperature, then τ = 1

day, f = 1.157×10−5 s−1, and ω = 7.272×10−5 rad s−1. We also assume that

T → T0 as y → ∞; that is, very far beneath the surface, the temperature is

the average surface temperature.

To solve Equation (4–68) with this surface boundary condition, we use

the method of separation of variables:

T (y, t) = T0 + Y (y)T ′(t). (4.75)

Because the surface temperature is time-periodic, we can assume that the

subsurface temperature also varies periodically with time at the same fre-

quency. However, it is not correct to assume that T ′(t) is simply cosωt,

as that would imply that the subsurface temperatures are in exact time

phase with the surface temperature. In other words, if T ′(t) were cosωt, the

maximum and minimum temperatures would be reached at the same times,

independent of depth. In fact, we will see that the temperature changes at

different depths are not in phase; the maximum temperature at any depth

lags behind the maximum temperature at the surface, for example, because

of the finite amount of time required for the temperature maximum to dif-

fuse downward. The difference in phase between temperature variations at

the surface and those at depth can be accounted for by using both cosωt

and sinωt dependences for T ′(t). We generalize Equation (4–75) to

T (y, t) = T0 + Y1(y) cosωt+ Y2(y) sinωt. (4.76)

By substituting Equation (4–76) into Equation (4–68), we find

−ωY1 = κ d2Y2

dy2 ωY2 = κ

d2Y1

dy2 . (4.77)

These are two coupled ordinary differential equations for the unknowns Y1

4.14 Periodic Heating of a Semi-Infinite Half-Space 273

and Y2. We can solve the first of these equations for Y1 and substitute into

the second equation to obtain

d4Y2

dy4 + ω2

κ2 Y2 = 0. (4.78)

Alternatively, we could have solved the second of Equations (4–77) for Y2

and substituted into the first equation. Had we done so, we would have found

that Y1 satisfies the same fourth-order ordinary differential equation as does

Y2.

A standard technique for solving ordinary differential equations with con-

stant coefficients, of which Equation (4–78) is an example, is to assume a

solution of the form

Y2 = ceαy. (4.79)

If Equation (4–79) is to satisfy Equation (4–78), then

α4 + ω2

κ2 = 0 (4.80)

or

α = ± (

1 ± i√ 2

) √

ω

κ , (4.81)

where i is the square root of −1. Because four values of α satisfy Equation (4–

80), the general solution for Y2 (or Y1) must be written

Y2 = c1 exp

(

(1 + i)√ 2

√

ω

κ y

)

+ c2 exp

(

(1 − i)√ 2

√

ω

κ y

)

+ c3 exp

(−(1 + i)√ 2

√

ω

κ y

)

+ c4 exp

(−(1 − i)√ 2

√

ω

κ y

)

. (4.82)

Because the temperature fluctuations must decay with depth, the constants

c1 and c2 are zero, and Y2 takes the form

Y2 = exp

(

− y

√

ω

2κ

)[

c3 exp

(

− iy

√

ω

2κ

)

+ c4 exp

(

iy

√

ω

2κ

)]

. (4.83)

274 Heat Transfer

It is convenient to rewrite the solution for Y2 as

Y2 = exp

(

− y

√

ω

2κ

)(

b1 cos

√

ω

2κ y + b2 sin

√

ω

2κ y

)

,

(4.84)

where b1 and b2 are constants that can be related to c3 and c4, although it is

unnecessary to do so. The transition from Equation (4–83) to Equation (4–

84) is possible because the trigonometric functions sin x and cos x can

be written in terms of the exponentials eix and e−ix, and vice versa. The

unknown function Y1 has a similar form

Y1 = exp

(

− y

√

ω

2κ

)(

b3 cos

√

ω

2κ y + b4 sin

√

ω

2κ y

)

.

(4.85)

The remaining constants of integration can be determined as follows. If

Y1 and Y2 are to satisfy Equations (4–77), then

b2 = b3 and b1 = −b4. (4.86)

Also, the surface temperature must be of the form (4–72), which requires

b1 = 0 and b3 = ∆T. (4.87)

Thus, the temperature variation in the half-space due to a time-periodic

surface temperature is

T = T0 + ∆T exp

(

− y

√

ω

2κ

)

× (

cosωt cos y

√

ω

2κ + sinωt sin y

√

ω

2κ

)

,

(4.88)

T = T0 + ∆T exp

(

−y √

ω

2κ

)

cos

(

ωt− y

√

ω

2κ

)

.

(4.89)

Equation (4–89) shows that the amplitude of the time-dependent tem-

perature fluctuation decreases exponentially with depth. This fluctuation

decreases to 1/e of its surface value in a skin depth dω given by

dω =

(

2κ

ω

)1/2

. (4.90)

For the daily variation of temperature, the frequency is ω = 7.27 × 10−5

rad s−1. With κ = 1 mm2 s−1, the skin depth for diurnal temperature

4.14 Periodic Heating of a Semi-Infinite Half-Space 275

changes from Equation (4–90) is 0.17 m. Except for a factor of π−1/2, the

skin depth for the penetration of the surface temperature variation is just

what one would have estimated on the basis of dimensional arguments, that

is, √ κτ(τ = period). Because skin depth is inversely proportional to the

square root of frequency, it is clear that the more rapid the fluctuation in

temperature, the less it penetrates beneath the surface.

The argument of the trigonometric factor in Equation (4–89) shows that

the phase difference φ between temperature fluctuations at the surface and

those at depth y is

φ = y

√

ω

2κ . (4.91)

If the depth y is the skin depth, the fluctuations are out of phase by 1 radian

(57.3◦). Figure 4–19 illustrates how the amplitudes of the temperature vari-

ations decay with depth and how the phases of the fluctuations shift with

depth.

Problem 4.26 Assume that the yearly temperature variation is periodic.

What is the skin depth? At what depth is the temperature 180◦ out of phase

with the surface variation? Assume κ = 1 mm2 s−1.

Problem 4.27 Assume that the temperature effects of glaciations can be

represented by a periodic surface temperature with a period of 104 yr. If

it is desired to drill a hole to a depth that the temperature effect of the

glaciations is 5% of the surface value, how deep must the hole be drilled?

Assume κ = 1 mm2 s−1.

Problem 4.28 Estimate the depth to which frost penetrates in the ground

at a latitude where the annual surface temperatures vary between −5 and

25◦C. Assume that the water content of the ground is sufficiently small so

that the latent heat can be ignored on freezing and thawing. Assume κ for

the soil is 0.8 mm2 s−1.

Problem 4.29 Estimate the effects of variations in bottom water tem-

perature on measurements of oceanic heat flow by using the model of a

semi-infinite half-space subjected to periodic surface temperature fluctua-

tions. Such water temperature variations at a specific location on the ocean

floor can be due to, for example, the transport of water with variable tem-

perature past the site by deep ocean currents. Find the amplitude of water

temperature variations that cause surface heat flux variations of 40 mWm−2

above and below the mean on a time scale of 1 day. Assume that the ther-

mal conductivity of sediments is 0.8 W m−1 K−1 and the sediment thermal

diffusivity is 0.2 mm2 s−1.

276 Heat Transfer

Figure 4.19 Phase shift and amplitude decay with depth of a time-periodic surface temperature variation.

Problem 4.30 Consider a semi-infinite half-space (y≥ 0) whose surface

temperature is given by Equation (4–72). At what values of ωt is the surface

heat flow zero?

4.15 Instantaneous Heating or Cooling of a Semi-Infinite

Half-Space

A number of important geological problems can be modeled by the instan-

taneous heating or cooling of a semi-infinite half-space. In the middle of the

nineteenth century Lord Kelvin used this solution to estimate the age of

the Earth. He assumed that the surface heat flow resulted from the cool-

ing of an initially hot Earth and concluded that the age of the Earth was

about 65 million years. We now know that this estimate was in error for two

reasons – the presence of radioactive isotopes in the mantle and solid-state

thermal convection in the mantle.

4.15 Instantaneous Heating or Cooling of a Semi-Infinite Half-Space 277

Figure 4.20 Heating of a semi-infinite half-space by a sudden increase in surface temperature.

In many cases magma flows through preexisting joints or cracks. When

the flow commences, the wall rock is subjected to a sudden increase in tem-

perature. Heat flows from the hot magma into the cold country rock, thus

increasing its temperature. The temperature of the wall rock as a function of

time can be obtained by solving the one-dimensional, time-dependent heat

conduction equation for a semi-infinite half-space, initially at a uniform tem-

perature, whose surface is suddenly brought to a different temperature at

time t = 0 and maintained at this new temperature for later times.

This solution can also be used to determine the thermal structure of the

oceanic lithosphere. At the crest of an ocean ridge, hot mantle rock is sub-

jected to a cold surface temperature. As the seafloor spreads away from the

ridge crest, the near-surface rocks lose heat to the cold seawater. The cooling

near-surface rocks form the rigid oceanic lithosphere.

We now obtain the solution to Equation (4–68) in a semi-infinite half-

space defined by y > 0 whose surface is given an instantaneous change in

temperature. Initially at t = 0, the half-space has a temperature T1; for

t > 0, the surface y = 0 is maintained at a constant temperature T0. As a

result, heat is transferred into the half-space if T0 > T1, and the temperature

increases. If T1 > T0, the half-space cools, and its temperature decreases.

The situation is sketched in Figure 4–20 for the case T0 > T1.

The temperature distribution in the rock is the solution of Equation (4–

68) subject to the conditions

T = T1 at t = 0, y > 0

T = T0 at y = 0 t > 0

T → T1 as y → ∞ t > 0. (4.92)

The problem posed by Equations (4–68) and (4–92) is a familiar one in the

theory of partial differential equations. It can be solved in a rather straight-

forward way using an approach known as similarity. First, it is convenient

278 Heat Transfer

to introduce the dimensionless temperature ratio θ

θ = T − T1

T0 − T1 (4.93)

as a new unknown. The equation for θ is identical with the one for T ,

∂θ

∂t = κ

∂2θ

∂y2 , (4.94)

but the conditions on θ are simpler

θ(y, 0) = 0

θ(0, t) = 1

θ(∞, t) = 0. (4.95)

The similarity approach to determining θ is based on the idea that the only

length scale in the problem, that is, the only quantity that has the dimen-

sions of length other than y itself, is √ κt, the characteristic thermal diffusion

distance (recall that the diffusivity κ has dimensions of length2/time). It is

reasonable to suppose that, in this circumstance, θ is not a function of t and

y separately, but rather it is a function of the dimensionless ratio

η = y

2 √ κt . (4.96)

The factor of 2 is introduced to simplify the subsequent results. It is not only

reasonable that θ should depend only on η, but a theorem in dimensional

analysis shows that this must be the case.

The dimensionless parameter η is known as the similarity variable. The

solutions at different times are “similar” to each other in the sense that the

spatial dependence at one time can be obtained from the spatial dependence

at a different time by stretching the coordinate y by the square root of the

ratio of the times. We will see from the solution to this problem that the

characteristic thermal diffusion length is the distance over which the effects

of a sudden, localized change in temperature can be felt after a time t has

elapsed from the onset of the change.

The equations (4–94) and (4–95) must be rewritten in terms of η. This

requires that we determine the partial derivatives of θ with respect to t and

y in terms of derivatives with respect to η. This can be accomplished using

the chain rule for differentiation as follows:

∂θ

∂t = dθ

dη

∂η

∂t = dθ

dη

(

−1

4

y√ κt

1

t

)

= dθ

dη

(

−1

2

η

t

)

(4.97)

4.15 Instantaneous Heating or Cooling of a Semi-Infinite Half-Space 279

∂θ

∂y = dθ

dη

∂η

∂y = dθ

dη

1

2 √ κt

(4.98)

∂2θ

∂y2 =

1

2 √ κt

d2θ

dη2

∂η

∂y =

1

4

1

κt

d2θ

dη2 . (4.99)

Equation (4–94) becomes

−η dθ dη

= 1

2

d2θ

dη2 . (4.100)

The boundary conditions are easy to deal with; y = 0 maps into η = 0

and both y = ∞ and t = 0 map into η = ∞. Thus the conditions (4–95)

reduce to

θ(∞) = 0

θ(0) = 1. (4.101)

The fact that the introduction of the similarity variable reduces the par-

tial differential equation (4–94) to an ordinary differential equation (4–100)

with respect to η and reduces the separate conditions in t and y to consis-

tent conditions involving η alone, is a posteriori proof of the validity of the

approach.

Equation (4–100) can be integrated by letting

φ = dθ

dη . (4.102)

Rewriting Equation (4–100), we obtain

−ηφ = 1

2

dφ

dη (4.103)

or

−η dη = 1

2

dφ

φ . (4.104)

Integration of Equation (4–104) is straightforward:

−η2 = lnφ− ln c1, (4.105)

where − ln c1 is the constant of integration. It follows that

φ = c1e −η2 =

dθ

dη . (4.106)

Upon integrating Equation (4–106), we get

θ = c1

∫ η

0 e−η

′2 dη′ + 1, (4.107)

280 Heat Transfer

where η′ is a dummy variable of integration and the condition θ(0) = 1 was

used to evaluate the second constant of integration. Since θ(∞) = 0, we

must have

0 = c1

∫ ∞

0 e−η

′2 dη′ + 1. (4.108)

The integral in Equation (4–108) is well known:

∫ ∞

0 e−η

′2 dη′ =

√ π

2 . (4.109)

Thus the constant c1 is −2/ √ π and

θ = 1 − 2√ π

∫ η

0 e−η

′2 dη′. (4.110)

The function defined by the integral in Equation (4–110) occurs so often

in solutions of physical problems that it is given a special name, the error

function erf(η)

erf(η) ≡ 2√ π

∫ η

0 e−η

′2 dη′. (4.111)

Thus we can rewrite θ as

θ = 1 − erf(η) = erfc η (4.112)

where erfc(η) is the complementary error function. Values of the error func-

tion and the complementary error function are listed in Table 4–5. The

functions are also shown in Figure 4–21.

The solution for the temperature as a function of time t and distance y is

Equation (4–112). It can be written in terms of the original variables as

T − T1

T0 − T1 = erfc

y

2 √ κt . (4.113)

At y = 0, the complementary error function is 1 and T = T0. As y → ∞ or

t = 0, erfc is 0 and T = T1. The general solution for θ or (T −T1)/(T0 −T1)

is shown as erfc η in Figure 4–21.

The near-surface region in which there is a significant temperature change

is referred to as a thermal boundary layer. The thickness of the thermal

boundary layer requires an arbitrary definition, since the temperature T

approaches the initial rock temperature T1 asymptotically. We define the

thickness of the boundary layer yT as the distance to where θ = 0.1. This

distance changes with time as the half-space heats up or cools off. The con-

dition θ = 0.1 defines a unique value of the similarity variable ηT , however.

4.15 Instantaneous Heating or Cooling of a Semi-Infinite Half-Space 281

Table 4.5 The Error Function and the Complementary Error Function

η erf η erfc η

0 0 1.0 0.02 0.022565 0.977435 0.04 0.045111 0.954889 0.06 0.067622 0.932378 0.08 0.090078 0.909922 0.10 0.112463 0.887537 0.15 0.167996 0.832004 0.20 0.222703 0.777297 0.25 0.276326 0.723674 0.30 0.328627 0.671373 0.35 0.379382 0.620618 0.40 0.428392 0.571608 0.45 0.475482 0.524518 0.50 0.520500 0.479500 0.55 0.563323 0.436677 0.60 0.603856 0.396144 0.65 0.642029 0.357971 0.70 0.677801 0.322199 0.75 0.711156 0.288844 0.80 0.742101 0.257899 0.85 0.770668 0.229332 0.90 0.796908 0.203092 0.95 0.820891 0.179109 1.0 0.842701 0.157299 1.1 0.880205 0.119795 1.2 0.910314 0.089686 1.3 0.934008 0.065992 1.4 0.952285 0.047715 1.5 0.966105 0.033895 1.6 0.976348 0.023652 1.7 0.983790 0.016210 1.8 0.989091 0.010909 1.9 0.992790 0.007210 2.0 0.995322 0.004678 2.2 0.998137 0.001863 2.4 0.999311 0.000689 2.6 0.999764 0.000236 2.8 0.999925 0.000075 3.0 0.999978 0.000022

From Equation (4–112) and Table (4–5) we obtain

ηT = erfc−10.1 = 1.16 (4.114)

282 Heat Transfer

Figure 4.21 The error function and the complementary error function.

and from Equation (4–96) we get

yT = 2ηT √ κt = 2.32

√ κt. (4.115)

The thickness of the thermal boundary layer is 2.32 times the characteristic

thermal diffusion distance √ κt.

Problem 4.31 Derive an expression for the thickness of the thermal

boundary layer if we define it to be the distance to where θ = 0.01.

Problem 4.32 If the surface temperature is increased 10 K, how long is

it before the temperature increases 2 K at a depth of 1 m (κ = 1 mm2 s−1)?

The heat flux at the surface y = 0 is given by differentiating Equation (4–

113) according to Fourier’s law and evaluating the result at y = 0 such

that

q = −k (

∂T

∂y

)

y=0

= −k(T0 − T1) ∂

∂y

(

erfc y

2 √ κt

)

y=0

= k(T0 − T1) ∂

∂y

(

erf y

2 √ κt

)

y=0

= k(T0 − T1)

2 √ κt

d

dη (erf η)η=0

= k(T0 − T1)

2 √ κt

(

2√ π e−η

2 )

η=0 = k(T0 − T1)√

πκt .

(4.116)

4.15 Instantaneous Heating or Cooling of a Semi-Infinite Half-Space 283

The surface heat flux q is infinite at t = 0 because of the sudden application

of the temperature T0 at t = 0. However, q decreases with time, and the

total heat into the semi-infinite half-space up to any time, Q, is finite; it is

given by the integral of Equation (4–116) from t = 0 to t

Q =

∫ t

0 q dt′ =

2k(T0 − T1)√ κπ

√ t. (4.117)

Except for the factor π−1/2 the heat flux into the rock is k times the tem-

perature difference (T0 − T1) divided by the thermal diffusion length √ κt.

In the mid-1800s William Thompson, later Lord Kelvin, used the theory

for the conductive cooling of a semi-infinite half-space to estimate the age

of the Earth. He hypothesized that the Earth was formed at a uniform high

temperature T1 and that its surface was subsequently maintained at the low

temperature T0. He assumed that a thin near-surface boundary layer devel-

oped as the Earth cooled. Since the boundary layer would be thin compared

with the radius of the Earth, he reasoned that the one-dimensional model

developed above could be applied. From Equation (4–116) he concluded that

the age of the Earth t0 was given by

t0 = (T1 − T0)

2

πκ(∂T/∂y)20 , (4.118)

where (∂T/∂y)0 is the present near-surface thermal gradient. With (∂T/∂y)0 =

25 K km−1, T1−T0 = 2000 K, and κ = 1 mm2 s−1, the age of the Earth from

Equation (4–118) is t0 = 65 million years. It was not until radioactivity was

discovered about 1900 that this estimate was seriously questioned.

Problem 4.33 One way of determining the effects of erosion on subsurface

temperatures is to consider the instantaneous removal of a thickness l of

ground. Prior to the removal T = T0 + βy, where y is the depth, β is

the geothermal gradient, and T0 is the surface temperature. After removal,

the new surface is maintained at temperature T0. Show that the subsurface

temperature after the removal of the surface layer is given by

T = T0 + βy + βl erf

(

y

2 √ κt

)

.

How is the surface heat flow affected by the removal of surface material?

Problem 4.34 Determine the effect of a glacial epoch on the surface

geothermal gradient as follows. At the start of the glacial epoch t = −τ , the subsurface temperature is T0 +βy. The surface is y = 0, and y increases

downward. During the period of glaciation the surface temperature drops to

284 Heat Transfer

T0 − ∆T0. At the end of the glacial period, t = 0, the surface temperature

again rises to T0. Find the subsurface temperature T (y, t) and the surface

heat flow for t > −τ . If the last glaciation began at 13,000 year BP and

ended 8000 year BP and ∆T0 = 20 K (κ = 1 mm2 s−1, k = 3.3 W m−1

K−1), determine the effect on the present surface heat flow.

HINT: Use the idea of superposition to combine the elementary solutions

to the heat conduction equation in such a way as to develop the solution of

this problem without having to solve a differential equation again.

Problem 4.35 One technique for measuring the thermal conductivity of

sediments involves the insertion of a very thin cylinder, or needle, heated by

an internal heater wire at a known and constant rate, into the sediments. A

small thermistor inside the needle measures the rise of temperature T with

time t. After the heater has been on for a short time, measurements of T

show a linear growth with ln t,

T = c1 ln t+ c2.

The sediment conductivity can be deduced from the slope of a T versus ln

t plot, c1, with the aid of a theoretical formula you can derive as follows.

Consider the temperature field due to an infinite line source that emits

Q units of heat per unit time and per unit length for times t > 0 in an

infinite medium initially at temperature T0. Determine T (r, t) by solving

Equation (4–70) subject to the appropriate initial and boundary conditions.

HINT: A similarity solution with the similarity variable η = r2/4κt works.

In fact, the solution is

T − T0 = Q

4πk

∫ ∞

η

e−η ′

η′ dη′. (4.119)

The integral ∫∞ η (1/η′)e−η

′

dη′ is known as the exponential integral E1(η).

Thus T −T0 = (Q/4πk)E1(η). The function E1(η) can be evaluated numer-

ically and tabulated, just as the error function. (Values of E1(η) are given

in Table 8–4.) Furthermore, it can be shown that for η sufficiently small (t

large enough)

E1(η) = −γ − ln η + · · · , (4.120)

where γ is Euler’s constant 0.5772156649. . . . Thus, after a sufficiently long

time,

T − T0 = Q

4πk (−γ − ln η + · · ·)

= −Qγ 4πk

− Q

4πk ln

r2

4κt + · · ·

4.16 Cooling of the Oceanic Lithosphere 285

= −Qγ 4πk

− Q

4πk ln r2

4κ +

Q

4πk ln t+ · · · .

(4.121)

The measured slope c1 is thus Q/4πk, and, with Q known, k can be deter-

mined.

Problem 4.36 Displacements along faults can bring rock masses with

different temperatures into sudden contact. Thrust sheets result in the em-

placement of buried crustal rocks above rocks that were previously at the

surface. The transform faults that offset ocean ridge segments juxtapose

oceanic lithospheres of different ages. Consider therefore how temperature

varies with time and position when two semi-infinite half-spaces initially

at temperatures T−(y < 0) and T+(y > 0) are placed adjacent to each other

along y = 0 at time t = 0. Show that T is given by

T = (T+ + T−)

2 +

(T+ − T−)

2 erf

(

y

2 √ κt

)

. (4.122)

Consider also how temperature varies with time and depth for a situation

in which the initial temperature distribution in a half-space (y > 0) is T =

T1 for 0<y<b and T = T2 for y >b. Assume that the surface y = 0 is

maintained at T =T0 for t> 0 and that T → T2 as y →∞ for t > 0. Show

that T (y, t) is given by

T = (T1 − T0)erf

(

y

2 √ κt

)

+ (T2 − T1)

2

× {

erf (y − b)

2 √ κt

+ erf (y + b)

2 √ κt

}

+ T0. (4.123)

4.16 Cooling of the Oceanic Lithosphere

As we have already noted, the solution developed in Section 4–15 is also

relevant to the cooling of the oceanic lithosphere adjacent to a mid-ocean

ridge. In Chapter 1 we discussed how the mid-ocean ridge system is asso-

ciated with ascending mantle convection. The surface plates on either side

of the ridge move horizontally with a velocity u, as illustrated in Figure 4–

22. The plates are created from the hot mantle rock that is flowing upward

beneath the ridge. This rock is cooled by the seawater and forms the rigid

plates that move away from the ridge. Since the oceanic lithosphere is the

surface plate that moves rigidly over the deeper mantle, it can be identi-

fied with the part of the upper mantle whose temperature is less than some

value below which mantle rocks do not readily deform over geologic time.

286 Heat Transfer

Figure 4.22 Schematic of the cooling oceanic lithosphere.

Figure 4.23 The solid lines are isotherms, T − T0 (K), in the oceanic lithosphere from Equation (4–125). The data points are the thicknesses of the oceanic lithosphere in the Pacific determined from studies of Rayleigh wave dispersion data (Leeds et al., 1974).

High-temperature deformation of rocks in the laboratory indicates that this

temperature is about 1600 K. Thus we can think of the lithosphere as the

region between the surface and a particular isotherm, as shown in the figure.

The depth to this isotherm increases with the age of the lithosphere; that

is, the lithosphere thickens as it moves farther from the ridge, since it has

more time to cool. We refer to the age of the lithosphere as the amount of

time t required to reach the distance x from the ridge (because of symmetry

we consider x positive); t = x/u.

The temperature of the rock at the ridge crest x = 0 and beneath the plate

is T1. The seawater cools the surface to the temperature T0. Thus, a column

of mantle is initially at temperature T1, and its surface is suddenly brought

to the temperature T0. As the column moves away from the ridge, its sur-

face temperature is maintained at T0, and it gradually cools. This problem is

identical to the sudden cooling of a half-space, treated in Section 4–15, if we

neglect horizontal heat conduction compared with vertical heat conduction.

4.16 Cooling of the Oceanic Lithosphere 287

Figure 4.23 Vertical columns of mantle and lithosphere moving horizontally away from the ridge and cooling vertically to the surface (t2 > t1 > 0).

This is a good approximation as long as the lithosphere is thin. With hori-

zontal heat conduction neglected, heat conduction is vertical in columns of

mantle and lithosphere, as it is in the half-space problem. Although a thin

column may not resemble a semi-infinite half-space, the essential feature

both have in common that makes the cooling problem identical for both

is the vertical heat conduction. Figure 4–23 illustrates columns of mantle

moving laterally away from the ridge and cooling to the surface.

To adapt the half-space sudden cooling solution to the oceanic lithosphere

cooling situation, let t = x/u, and rewrite Equation (4–113) as

T1 − T

T1 − T0 = erfc

(

y

2 √

κx/u

)

. (4.124)

This can be further rearranged as

T1 − T

T1 − T0 = 1 − T − T0

T1 − T0 = 1 − erf

(

y

2 √

κx/u

)

and T − T0

T1 − T0 = erf

(

y

2 √

κx/u

)

. (4.125)

According to Equation (4–125) the surface temperature is T0, since erf

(0) = 0 and T → T1 as the depth y → ∞, since erf(∞) = 1. Figure 4–24

shows the isotherms beneath the ocean surface as a function of the age of

the seafloor for T1 − T0 = 1300 K, and κ = 1 mm2 s−1. The isotherms

in Figure 4–24 have the shape of parabolas. The thickness of the oceanic

lithosphere yL can be obtained directly from Equation (4–115) by replacing

t with x/u:

yL = 2.32(κt)1/2 = 2.32

(

κx

u

)1/2

. (4.126)

With κ = 1 mm2 s−1 the thickness of the lithosphere at an age of 80 Myr is

288 Heat Transfer

116 km. It should be emphasized that the thickness given in Equation (4–

126) is arbitrary in that it corresponds to (T − T0)/(T1 − T0) = 0.9. Also

included in Figure 4–24 are thicknesses of the oceanic lithosphere in the

Pacific obtained from studies of Rayleigh wave dispersion.

The surface heat flux q0 as a function of age and distance from the ridge

crest is given by Equation (4–116)

q0 = k(T1 − T0)√

πκt = k(T1 − T0)

(

u

πκx

)1/2

. (4.127)

This is the surface heat flow predicted by the half-space cooling model.

Many measurements of the surface heat flow in the oceans have been

carried out and there is considerable scatter in the results. A major cause of

this scatter is hydrothermal circulations through the oceanic crust. The heat

loss due to these circulations causes observed heat flows to be systematically

low. Lister et al. (1990) considered only measurements in thick sedimentary

cover that blocked hydrothermal circulations. Their values of surface heat

flow are given in Figure 4–25 as a function of the age of the seafloor. The

results, for the half-space cooling model from Equation (4–127) are compared

with the observations taking k = 3.3 W m−1 K−1 and the other parameter

values as above. Quite good agreement is found at younger ages but the data

appear to lie above the theoretical prediction for older ages. This discrepancy

will be discussed in detail in later sections.

The cumulative area of the ocean floor A as a function of age, that is,

the area of the seafloor with ages less than a specified value, is given in

Figure 4–26. The mean age of the seafloor is 60.4 Myr. Also included in

Figure 4–26 is the cumulative area versus age for a model seafloor that has

been produced at a rate dA/dt = 0.0815 m2 s−1 and subducted at an age

τ of 120.8 Myr (dashed line). This is the average rate of seafloor accretion

over this time. It should be noted that the present rate of seafloor accretion

is about 0.090 m2 s−1; very close to the long-term average value.

For a constant rate of seafloor production and for subduction at an age

τ , the mean oceanic heat flow q̄0 is

q̄0 = 1

τ

∫ τ

0 q0 dt =

1

τ

∫ τ

0

k(T1 − T0)√ πκt

dt = 2k(T1 − T0)√

πκτ .

(4.128)

Taking τ = 120.8 Myr and the other parameters as above, we find that

the mean oceanic heat flow is q̄0 = 78.5 mW m−2. This is in reasonable

agreement with the mean value of oceanic heat flow measurements (101 mW

m−2) given in Section 4–4. This agreement is somewhat fortuitous since

4.16 Cooling of the Oceanic Lithosphere 289

Figure 4.25 Heat flow as a function of the age of the ocean floor. The data points are from sediment covered regions of the Atlantic and Pacific Oceans (Lister et al., 1990). Comparisons are made with the half-space cooling model (HSCM) from Equation (4–127) and the plate model from Equation (4–133) with yL0 =95 km (PM 95) and with yL0 =125 km (PM 125).

the data are not evenly distributed with respect to the age of the seafloor;

oceanic heat flow data are also biased toward areas of the seafloor that are

well covered by sediments. Nevertheless, we can conclude that a substantial

fraction of the heat lost from the interior of the Earth is directly attributable

to the cooling of the oceanic lithosphere. An oceanic geotherm corresponding

to the mean age of 60.4 Myr as obtained from Equation (4–125) is given in

Figure 4–27.

So far we have discussed only the oceanic lithosphere. We can also ap-

ply the one-dimensional cooling model to the continental lithosphere and

compare the surface heat flow predicted by the model with heat flow mea-

surements. The heat flow values that are relevant to this purpose are the

reduced heat flows defined in Section 4–8. Recall that the reduced heat

flux is the mantle contribution to the surface heat flow; it is deduced from

the correlation of the surface heat flow with the surface concentration of

heat-producing radioactive isotopes (see Equation (4–29) and Figure 4–11).

Reduced heat flux values for several geological provinces are plotted against

290 Heat Transfer

Figure 4.26 Cumulative area of seafloor A as a function of age t (the area of seafloor with ages younger than a given age) (solid lines). The dashed line is a cumulative area function for a model seafloor produced at a constant rate of 0.0815 m2 s−1, and subducted at an age of 120.8 Myr.

the ages of the provinces in Figure 4–28. If the mantle heat flow in conti-

nental areas were due to the conductive cooling of the lithosphere, qm would

be given by Equation (4–127) with t the age of the continental crust. The

prediction of this equation for k = 3.3 W m−1 K−1, κ = 1 mm2 s−1, and

T1−T0 = 1300 K is also shown in Figure 4–28. Clearly, the values of mantle

heat flow deduced from observations lie considerably above the conductive

cooling prediction for the older provinces. The measured values correlate

better with a constant mantle heat flow of 25 mW m−2 for ages between

500 Myr and 2.5 Gyr. This correlation is strong evidence that there is an

additional heat input to the base of the continental lithosphere caused by

mantle convection; this heat input is very close to the mean mantle heat

flow qm = 28 mW m−2 given in Section 4–5.

4.17 Plate Cooling Model of the Lithosphere

As discussed earlier, observations show that the half-space cooling model

does not predict the time evolution of the continental lithosphere. The con-

tinental lithosphere does not continue to thicken with age but instead ap-

4.17 Plate Cooling Model of the Lithosphere 291

Figure 4.27 Mean oceanic geotherm determined from Equation (4–125) with t =60.4 Myr.

proaches an equilibrium, time-independent thermal structure. This result

requires heating of the base of the continental lithosphere by mantle convec-

tion. The surface heat flow data from the ocean basins given in Figure 4–25

show that there is also basal heating of the oceanic lithosphere. To account

for this basal heating, we introduce the plate cooling model.

Consider the instantaneous heating or cooling of a finite thickness plate.

Since our application is to the lithosphere, we take the plate thickness to be

yL0, the thickness of the lithosphere at large times. The infinitely long plate

fills the region 0 ≤ y ≤ yL0. The temperature in the plate is a solution of

the one-dimensional unsteady heat conduction equation (4–68).

Initially at t = 0 the plate is at the temperature T1; for t > 0, the surface

of the plate y = 0 is maintained at the temperature T0. The base of the

plate y = yL0 is maintained at the temperature T1. These conditions can be

written as

T = T1 at t = 0, 0 ≤ y ≤ yL0

T = T0 at y = 0, t > 0

292 Heat Transfer

Figure 4.28 Dependence of the mantle heat flow on age for several conti- nental geological provinces. The solid line is the predicted heat flow for a conductively cooling continental lithosphere from Equation (4–127), and the dashed line is a constant mantle heat flow of 25 mW m−2. Data are from Sclater et al. (1980).

T = T1 at y = yL0, t > 0. (4.129)

The solution of Equation (4–68) that satisfies the initial and boundary

conditions given in Equation (4–129) can be obtained in the form of an

infinite series. A detailed derivation of the solution has been given by Carslaw

and Jaeger (1959, p. 100). The result can be written as

T = T0 + (T1 − T0)

[

y

yL0 +

2

π

∞ ∑

n=1

1

n exp

(

−κn 2π2t

y2 L0

)

sin

(

nπy

yL0

)

]

. (4.130)

At large times, t ≫ y2 L0/κ, an equilibrium linear temperature profile is es-

tablished:

T = T0 + (T1 − T0) y

yL0 . (4.131)

At small times, t≪ y2 L0/κ, the half-space cooling solution given in Section 4–

16 is recovered. The deviations from the half-space cooling solution are well

approximated if only the first two terms of the expansion given in Equation

(4–130) are retained, with the result

T = T0 + (T1 − T0)

[

y

yL0 +

2

π exp

(

−κπ 2t

y2 L0

)

sin

4.17 Plate Cooling Model of the Lithosphere 293

(

πy

yL0

)

+ 1

π exp

(

−4κπ2t

y2 L0

)

sin

(

2πy

yL0

)]

.

(4.132)

We can obtain the surface heat flow q0 as a function of age t from Equa-

tions (4–1) and (4–130) as

q0 = k(T1 − T0)

yL0

[

1 + 2 ∞ ∑

n=1

exp

(

−κn 2π2t

y2 L0

)

]

.

(4.133)

For large times, t≫ y2 L0/κ, the equilibrium value of the surface heat flow is

q0e = k(T1 − T0)

yL0 . (4.134)

We can approximate the deviations of the surface heat flow from the half-

space cooling result given in Equation (4–127) by retaining the first two

terms of the expansion in Equation (4–133) with the result

q0 = k(T1 − T0)

yL0

[

1 + 2 exp

(

−κπ 2t

y2 L0

)

+ 2exp

(

−4κπ2t

y2 L0

)]

. (4.135)

For large times t → ∞ the heat flow from Equation (4–135) approaches the

equilibrium heat flow in Equation (4–134).

We next compare the predictions of the plate model with the heat flow

compilation as a function of seafloor age given in Figure 4–25. Comparisons

are made for two values of plate thickness, yL0 = 95 and 125 km, and for

other parameter values as before. For yL0 = 95 km, the equilibrium t → ∞ heat flow from Equation (4–134) is q0e = 45 mW m−2, and for yL0 = 125 km,

we have q0e = 34 mW m−2. At ages of less than about 50 Ma, the half-space

cooling model and the plate models give the same results. At these ages,

the thickness of the thermal boundary layer is less than the thickness of the

plate, so the presence of a finite plate thickness has no effect. At older ages,

the specified plate thickness restricts the growth of the thermal boundary

layer and the heat flows predicted by the plate models are somewhat greater

than those predicted by the half-space cooling model. Further discussion

of the agreement between theories and observations will be postponed to

Section 4–23.

294 Heat Transfer

Figure 4.29 Growth of a solid layer at the surface of a cooling lava flow.

4.18 The Stefan Problem

A number of important geological problems involve the solidification of mag-

mas. We assume that the magma has a well-defined melt temperature at

which the phase change from liquid to solid occurs. Associated with this

phase change is a latent heat of fusion L. This is the amount of heat that is

liberated upon the solidification of 1 kg of magma. Heat conduction prob-

lems involving phase changes differ from problems we have already solved in

two major ways. First, we have to determine as part of the solution where

the phase change boundary, that is, the interface between solid and liquid,

is located. The position of this boundary obviously changes as solidification

proceeds. Second, we have to account for the latent heat of fusion, which

is liberated at the solid–liquid interface as solidification takes place; this

additional heat must be conducted away from the phase change boundary.

The first problem we consider is that of a horizontal layer of magma that

is solidifying from its upper surface downward as a result of being cooled

from above. We assume that the upper surface is maintained at a constant

temperature T0. An example of this would be the solidification of a lava

flow. Because of heat loss to the surface the solid layer grows thicker with

time. A lava flow also solidifies at its base. However, if we assume that the

magma is extruded at its melt temperature, then as long as there is still

a liquid region, the solidification from the top and bottom can be treated

independently. This also means that the overall flow thickness is unimportant

in describing the solidification process as long as a molten region is present.

In this section, we will consider the solidification from above; in the next

section, we will treat the solidification from below. The solidification of a

lava flow from above is essentially identical with the freezing of a lake. This

is the problem for which Stefan (1891) first obtained the solution developed

below.

4.18 The Stefan Problem 295

The problem we solve is illustrated in Figure 4–29. The flow has solidified

to the depth y = ym(t). We assume that molten material of uniform tem-

perature Tm lies everywhere below the growing surface layer. The fact that

the molten region does not extend infinitely far below the surface is of no

consequence to the solution. We must solve the heat conduction equation

(4–68) in the space 0 ≤ y ≤ ym(t) subject to the conditions T = T0 at y = 0,

T = Tm at y = ym(t), and ym = 0 at t = 0. The position of the solidification

boundary is an a priori unknown function of time. As in the case of the

sudden heating, or cooling, of a semi-infinite half-space, there is no length

scale in this problem. For this reason, we once again introduce the dimen-

sionless coordinate η = y/2 √ κt as in Equation (4–96); it is also convenient

to introduce the dimensionless temperature θ = (T − T0)/(Tm − T0) as in

Equation (4–93).

The dimensionless coordinate η is obtained by scaling the depth with the

thermal diffusion length √ κt because there is no other length scale in the

problem. Similarly, the depth of the solidification interface ym must also scale

with the thermal diffusion length in such a way that ym/ √ κt is a constant.

In other words, the depth of the solidification boundary increases with time

proportionately with the square root of time. We have used dimensional

arguments to determine the functional form of the dependence of ym on t,

a nontrivial result. Because η = y/2 √ κt and ym is proportional to

√ κt, the

solidification boundary corresponds to a constant value ηm = ym/2 √ κt of

the similarity coordinate η. We denote this constant value by ηm = λ1. Thus

we have

ym = 2λ1

√ κt. (4.136)

With our definitions of θ and η, the heat conduction equation for θ(η)

is clearly identical to Equation (4–100), whose solution we already know to

be proportional to erf(η). This form of solution automatically satisfies the

condition θ = 0(T = T0) on η = 0(y = 0). To satisfy the remaining condition

that θ = 1(T = Tm) at η = ηm(y = ym) = λ1, we need simply choose the

constant of proportionality appropriately. The solution is

θ = erf(η)

erf(λ1) . (4.137)

Equation (4–137) determines the temperature in the solidified layer 0 ≤ y ≤ ym. In the molten region y > ym, T = Tm and θ = 1.

The constant λ1 is determined by requiring that the latent heat liberated

at the solidification boundary be conducted vertically upward, away from

the interface. The situation at the solidification boundary is illustrated in

296 Heat Transfer

Figure 4.30 Latent heat released at the solidification boundary must be conducted upward through the solidified layer.

Figure 4–30. In time δt, the interface moves downward a distance (dym/dt)δt.

In so doing, a mass per unit area ρ(dym/dt)δt is solidified, thus releasing an

amount of latent heat ρL(dym/dt)δt per unit area. Conservation of energy

requires that this heat release be conducted away from the boundary at

precisely the rate at which it is liberated. The heat cannot be conducted

downward because the magma is at a constant temperature; moreover, heat

flows toward cooler temperatures that lie upward. Fourier’s law gives the

rate of upward heat conduction per unit time and per unit area at y = ym as k(∂T/∂y)y=ym . Multiplying this by δt and equating it to ρL(dym/dt)δt

gives the equation for finding λ1.

ρL dym dt

= k

(

∂T

∂y

)

y=ym

. (4.138)

From Equation (4–136) the speed of the solidification boundary is

dym dt

= λ1

√ κ√ t . (4.139)

From Equation (4–137) the temperature gradient at y = ym is (

∂T

∂y

)

y=ym

=

(

dθ

dη

)

η=ηm=λ1

(

∂η

∂y

)

(Tm − T0)

= (Tm − T0)

2 √ κT

2√ π e−λ

2 1

1

erfλ1 . (4.140)

Substituting Equations (4–139) and (4–140) into Equation (4–138), we get

L √ π

c(Tm − T0) =

e−λ 2 1

λ1 erfλ1 , (4.141)

4.18 The Stefan Problem 297

Figure 4.31 The right side of the transcendental equation for determining the growth of a solid layer at the surface of a cooling lava flow.

a transcendental equation for determining λ1. Given a numerical value for

the left side of Equation (4–141), λ1 can be found by iteratively calculating

the right side of the equation until agreement is found. Alternatively, the

right side of Equation (4–141) can be plotted as a function of λ1, as in

Figure 4–31, and the solution, for a particular value of the left side of the

equation, can be found graphically.

This theory can be compared directly with observations. The thickness

of the crusts on three lava lakes on the Hawaiian volcano Kilauea have

been measured as functions of time. Eruptions produced lava lakes in the

pit craters Kilauea Iki in 1959, Alae in 1963, and Makaopuhi in 1965. A

photograph of the initial formation of the crust on the lava lake in the Alae

pit crater is given in Figure 4–32. The thicknesses of the solidifying crusts

on the three lava lakes are given as functions of time after the eruptions in

Figure 4–33. For L = 400 kJ kg−1, c = 1 kJ kg−1 K−1, and Tm−T0 = 1050 K,

Equation (4–141) gives λ1 = 0.876. With this value of λ1 and κ = 0.7 mm2

s−1, we can determine the thickness of a solidifying crust as a function of

298 Heat Transfer

Figure 4.32 Photograph of the lava lake formed in the pit crater Alae during the 1963 eruption. A solid crust is just beginning to form on the magma (D. L. Peck 19, U.S. Geological Survey).

time from Equation (4–136). The result plotted in Figure 4–33 shows quite

good agreement between the observations and theory.

Problem 4.37 A body of water at 0◦C is subjected to a constant surface

temperature of −10◦C for 10 days. How thick is the surface layer of ice? Use

L = 320 kJ kg−1, k = 2 J m−1 s−1 K−1, c = 4 kJ kg−1 K−1, ρ = 1000 kg

m−3.

Problem 4.38 Scientists believe that early in its evolution, the Moon was

covered by a magma ocean with a depth of 50 km. Assuming that the magma

was at its melt temperature of 1500 K and that the surface of the Moon was

maintained at 500 K, how long did it take for the magma ocean to solidify

if it was cooled from the surface? Take L = 320 kJ kg−1, κ = 1 mm2 s−1,

and c = 1 kJ kg−1 K−1.

Problem 4.39 One of the estimates for the age of the Earth given by

Lord Kelvin in the 1860s assumed that Earth was initially molten at a

constant temperature Tm and that it subsequently cooled by conduction

with a constant surface temperature T0. The age of the Earth could then be

determined from the present surface thermal gradient (dT/dy)0. Reproduce

Kelvin’s result assuming Tm−T0 = 1700 K, c = 1 kJ kg−1 K−1, L = 400 kJ

kg−1, κ = 1 mm2 s−1, and (dT/dy)0 = 25 K km−1. In addition, determine

the thickness of the solidified lithosphere. Note: Since the solidified layer is

4.18 The Stefan Problem 299

Figure 4.33 The thicknesses of the solidifying crusts on the lava lakes in the three pit craters Kilauea lki (1959), Alae (1963), and Makaopuhi (1965) on the volcano Kilauea, Hawaii (Wright et al. 1976). The theoretical curve is from Equations (4–136) and (4–141).

thin compared with the Earth’s radius, the curvature of the surface may be

neglected.

Problem 4.40 Consider the solidification near the upper surface of a lava

flow. Compute the surface heat flux q0 as a function of time. Integrate q0 over time, and compare the result with the latent heat release up to that

time, ρLym.

Problem 4.41 Generalize the solution for the solidification of the top of a

lava lake to the situation where the lava is initially at a uniform temperature

Tv greater than the solidification temperature Tm. Assume that the lava

extends to great depth y → ∞ and that T → Tv as y → ∞ for all t. Also

assume that T = T0 at y = 0 for all t. Assume that the molten lava and the

solidified layer near the surface have the same thermal properties.

HINT: You have to modify the energy balance condition at the solid–liquid

interface to account for heat conduction in the liquid.

300 Heat Transfer

Figure 4.34 (a) Photograph of a large sill on Finger Mountain, Victoria, Victoria Land, Antarctica (W. B. Hamilton 384, U.S. Geological Survey). (b) Photograph of a small dike offset along a joint (J. P. Lockwood 1, U.S. Geological Survey).

Problem 4.42 The oceanic crust is believed to form from the solidification

of a large magma chamber beneath the ridge crest. Use the Stefan solution

to determine the width of the magma chamber at its base. Let L = 400 kJ

kg−1, c = 1 kJ kg−1 K−1, Tm − T0 = 1300 K, u = 50 mm yr−1, κ = 1 mm2

s−1, and the thickness of the oceanic crust is 6 km.

Problem 4.43 The mantle rocks of the asthenosphere from which the

lithosphere forms are expected to contain a small amount of magma. If the

mass fraction of magma is 0.05, determine the depth of the lithosphere–

asthenosphere boundary for oceanic lithosphere with an age of 60 Ma. As-

sume L = 400 kJ kg−1, c = 1 kJ kg−1 K−1, Tm = 1600 K, T0 = 275 K, and

κ = 1 mm2 s−1.

4.19 Solidification of a Dike or Sill

A large fraction of the magma coming from the interior of the Earth does not

reach the surface but instead solidifies as intrusive igneous bodies. Two of the

simplest types of intrusive bodies are sills and dikes. A sill is a horizontal

layer of solidified rock, and a dike is its vertical counterpart. These one-

dimensional structures are illustrated in Figure 4–34. We will now consider

the problem of the solidification of a dike or sill. The solidifying magma loses

heat by conduction to the adjacent country rock.

Our model for dike or sill solidification is illustrated in Figure 4–35. The

plane y = 0 defines the original magma–rock boundary. The dike or sill

occupies the region −2b < y < 0. Initially at t = 0 the molten rock in the

dike is at its melt temperature Tm, and the wall rock is at the temperature

4.19 Solidification of a Dike or Sill 301

Figure 4.35 Initial temperature distribution at t = 0 (T= T0 for y> 0, T= Tm for −2b< y< 0) and subsequent temperature distribution at t = t1 when the solidification boundary is at y = ym(t).

T0. At time t = 0, the dike begins solidifying at the interface y = 0. Figure 4–

35 shows the temperature distribution initially at t = 0 and at a later time

t1. The liquid part of the dike −2b < y < ym(t) is still at temperature Tm,

but the solidified part ym < y < 0 has cooled below Tm. The surrounding

rock near the dike has been warmed above T0 by the release of the latent

heat of fusion, but T → T0 far from the dike y → ∞.

We assume that the physical properties of the country rock and solid-

ified magma are the same. Therefore the temperature satisfies the one-

dimensional, time-dependent heat conduction equation (4–68) in the region

y > ym(t). The boundary conditions are that T = Tm at y = ym(t) and

T → T0 as y → ∞; the initial condition is that T = T0 for y > 0 and

ym = 0. Once again Equation (4–138) provides a balance between the heat

conducted away from the solidification interface and the heat released by

the solidification. We use the same method of solution as in the previ-

ous section and introduce the dimensionless variables η = y/(2 √ κt) and

θ = (T − T0)/(Tm − T0).

In this problem the position of the solidification boundary ym is negative

so that ηm = ym/(2 √ κt) is also negative. We denote this constant value by

ηm = −λ2. Thus we have

ym = −2λ2

√ κt. (4.142)

302 Heat Transfer

The dimensionless temperature in the region η > −λ2 satisfies Equation (4–

100). We have previously shown that a solution of this equation is propor-

tional to erfc η. Such a solution also satisfies the condition θ → 0(T → T0)

as η → ∞(y → ∞). In order to satisfy the condition that θ= 1 (T =Tm) at

η= ηm =−λ2(y= ym), we need simply to choose the constant of proportion-

ality appropriately. The solution is clearly

θ = erfc η

erfc(−λ2) . (4.143)

From the definition of erfc in Equation (4–112) and the property erf (−x) =

−erf x we obtain

erfc(−λ2) = 1 − erf(−λ2) = 1 + erf λ2. (4.144)

Substitution of Equation (4–144) into Equation (4–143) gives

θ = erfc η

1 + erf λ2 . (4.145)

The temperature of the boundary between the country rock and the solidified

magma η = 0(y = 0) is

θ(0) = 1

1 + erf λ2 . (4.146)

The temperature of this boundary is therefore constant while solidification

is occurring.

In order to determine the constant λ2, we must once again use the heat

balance at the solidification boundary given in Equation (4–138). The speed

of this boundary is obtained by differentiating Equation (4–142)

dym dt

= −λ2

(

κ

t

)1/2

. (4.147)

The temperature gradient at y = ym is obtained by differentiating Equa-

tion (4–145). (

∂T

∂y

)

y=ym

=

(

dθ

dη

)

η=−λ2

(

∂η

∂y

)

(Tm − T0)

= −(Tm − T0)

(πκt)1/2 e−λ

2 2

(1 + erf λ2) . (4.148)

Substitution of Equations (4–147) and (4–148) into Equation (4–138) gives

L √ π

c(Tm − T0) =

e−λ 2 2

λ2(1 + erf λ2) . (4.149)

4.19 Solidification of a Dike or Sill 303

Figure 4.36 The right side of the transcendental Equation (4–149) for de- termining the motion of the solidification boundary.

In terms of evaluating λ2, this equation plays the same role that Equation (4–

141) played in the Stefan problem. The right side of the equation is plotted as

a function of λ2 in Figure 4–36. Given a value for the left side of the equation,

λ2 may be determined graphically from the figure or more accurately by

iterative numerical calculations.

The time ts required to solidify a dike of width 2b can be obtained directly

from Equation (4–142). Solidification occurs symmetrically from the two

sides of the dike so that

ts = b2

4κλ2 2

. (4.150)

At t = ts all the magma has solidified.

Let us again consider a numerical example. Taking L = 320 kJ kg−1,

Tm − T0 = 1000 K, and c = 1.2 kJ kg−1 K−1, we find from Equation (4–

149) (Figure 4–36) that λ2 = 0.73. With this value of λ2, b = 1 m, and

κ = 0.5 mm2 s−1, we find from Equation (4–150) that the time required

to solidify this intrusion is 10.9 days. The temperature at the boundary

304 Heat Transfer

Figure 4.37 Temperature profiles at different times during dike solidifica- tion.

between the country rock and the solidified magma from Equation (4–139)

is T0+590 K. The temperature profiles at several times are given in Figure 4–

37.

Problem 4.44 Use the results of the sudden half-space heating problem,

Equation (4–117), to estimate the time required for dike solidification by

setting Q = ρLb. How does this time compare with the 10.9 days computed

in the example?

Problem 4.45 Consider the following model for the cooling and solidifica-

tion of an intrusive igneous body. Suppose that the region y < 0 is initially

solid with constant temperature T− and that the region y > 0 is initially

liquid with constant temperature T+. The igneous body cools and solidifies;

a solid–liquid interface at temperature Tm propagates into the region y > 0.

The melting temperature Tm is less than T+, but it is greater than T−. De-

termine ym, the position of the solidification boundary as a function of time

t. Find T in the regions y < 0, 0 < y < ym, and y >ym.

4.20 The Heat Conduction Equation in a Moving Medium:

Thermal Effects of Erosion and Sedimentation

A number of important problems in geology involve moving boundaries.

Examples include the solidification problems that we have just discussed;

4.20 The Heat Conduction Equation in a Moving Medium 305

Figure 4.38 Fixed (x, y) and moving (ξ, ζ) coordinate systems for the derivation of the heat conduction equation for a moving medium.

other examples involve erosion and sedimentation. One useful approach to

the solution of moving boundary problems is to consider the boundary fixed

and the material moving into the boundary. For this and other reasons it

is worthwhile to develop the form of the equation of heat conduction for a

moving medium. Let x, y be a fixed coordinate system, and assume that

the medium moves in the positive x direction with velocity U . Let ξ, ζ be

a coordinate system moving with the medium. The situation is sketched in

Figure 4–38. The coordinates x, y and ξ, ζ are thus related by

x = ξ + Ut y = ζ ∂2T

∂ζ2 . (4.151)

The heat conduction equation for an observer moving with the medium

is Equation (4–68) (generalized to two dimensions): (

∂T

∂t

)

ξ = κ

(

∂2T

∂ξ2 + ∂2T

∂ζ2

)

. (4.152)

To find the appropriate form of the heat conduction equation with respect to

the fixed coordinate system, we need to relate partial derivatives with respect

to ξ, ζ, t to partial derivatives with respect to x, y, t. From Equation (4–151)

we have (

∂T

∂t

)

ξ =

(

∂T

∂t

)

x + ∂T

∂x

dx

dt =

(

∂T

∂t

)

x + U

∂T

∂x ,

(4.153) (

∂T

∂ξ

)

t =

(

∂T

∂x

)

t

(

∂T

∂ζ

)

t =

(

∂T

∂y

)

t . (4.154)

Thus Equation (4–152) can be rewritten as

∂T

∂t + U

∂T

∂x = κ

(

∂2T

∂x2 + ∂2T

∂y2

)

, (4.155)

where derivatives are understood to be taken with x or t held constant as

appropriate. The term U(∂T/∂x) is the advective derivative. An observer

306 Heat Transfer

moving with the medium and measuring temperature on a recorder cannot

distinguish between temperature variations resulting from motion through a

spatially varying temperature field U(∂T/∂x) and actual temporal variations

in temperature ∂T/∂t.

As an additional example of the use of Equation (4–155), consider the

oceanic lithosphere cooling problem. With respect to an observer moving

with a column of lithosphere, the relevant heat conduction equation is (4–

152), a point of view we have already taken. Alternatively, one could take a

larger view of the situation and consider a fixed observer viewing the whole

spreading process and measuring x from the ridge. That observer sees a

steady heat transfer problem described by Equation (4–155) with ∂T/∂t = 0

(and vertical heat conduction only):

U ∂T

∂x = κ

∂2T

∂y2 . (4.156)

These alternative ways of approaching the problem are, of course, equivalent

because the age of the seafloor is x/U .

Problem 4.46 Assume that a half-space with a deep temperature T∞ is being eroded at a constant velocity U . If the erosional surface is at a

temperature T0, determine the temperature as a function of the distance

from the surface.

Problem 4.47 Assume that a half-space y > 0 with a deep temperature

T∞ is being eroded such that ym = α √ t, where ym is the depth of the

instantaneous surface measured from the location of the surface at t = 0. If

the erosional surface is at a temperature T0, determine T (y, t) and the heat

flow at the surface.

HINT: Solve the problem in the y, t system and assume an artificial (un-

known) temperature at y = 0.

Problem 4.48 Suppose that upon entering the Earth’s atmosphere, the

surface of a meteorite has been heated to the melting point and the molten

material is carried away by the flow. It is of interest to calculate the rate at

which melting removes material from the meteorite. For this purpose, con-

sider the following problem. The surface of a semi-infinite half-space moves

downward into the half-space with constant velocity V , as indicated in Fig-

ure 4–39. The surface is always at the melting temperature Tm, and melted

material above the instantaneous surface is removed from the problem. As-

sume that the surface of the half-space is melted by a constant heat flux qm into the half-space from above the surface. Assume also that far from the

melting surface the temperature is T0; that is, T → T0 as ζ → ∞. Find the

4.21 One-Dimensional, Unsteady Heat Conduction in an Infinite Region 307

Figure 4.39 Model for the ablating meteorite problem.

temperature distribution in the half-space as a function of time T (ζ, t), and

determine V in terms of qm and the thermodynamic properties of the rock.

Account for the latent heat L required to melt the material.

4.21 One-Dimensional, Unsteady Heat Conduction in an Infinite

Region

The problem solved in Section 4–19 provides the temperature distribution

up until the time the dike or sill is completely solidified. To determine the

subsequent thermal history, we must solve the problem of one-dimensional

unsteady heat conduction in an infinite region with a specified initial tem-

perature distribution.

If the temperature distribution at t = 0 is T̄ (y), the temperature distri-

bution at subsequent times is

T = 1

2 √ πκt

∫ ∞

−∞ T̄ (y′)e−(y−y′)2/4κtdy′. (4.157)

This result is known as Laplace’s solution. For the dike or sill cooling prob-

lem, the temperature at the time of total solidification would be the initial

temperature distribution T̄ (y), and Equation (4–157) could then be used to

determine the subsequent temperature distribution. To do this would require

a numerical integration.

Instead, an approximate solution is possible if the temperature far from

the dike is considered. The heat content of the dike per unit area of the

dike–country rock interface is

Q = ρ[c(Tm − T0) + L]2b (4.158)

For | y | ≫ b, one can consider the dike to be a planar heat source located

at y = 0 containing Q units of heat per unit area at t = 0. At later times

this heat diffuses away from the origin, and we are interested in determining

308 Heat Transfer

Figure 4.40 Planar heat source containing Q units of heat at t =0.

how the temperature evolves as a function of distance from the origin and

time. The situation is sketched in Figure 4–40.

The temperature T (y, t) must satisfy the one-

dimensional, time-dependent heat conduction equation (4–68) subject to

the conditions T → T0 as |y| → ∞. An essential condition on the tempera-

ture distribution is that the heat content of all space must be the original

heat content of the dike,

ρc

∫ ∞

−∞ (T − T0) dy = 2ρc

∫ ∞

0 (T − T0) dy = Q,

(4.159)

for all t. In other words, the heat pulse supplied by the dike can spread out

as it diffuses away from the origin, but no heat can be lost from the medium.

A nondimensional form of the solution must be possible. The only quantity

with dimensions of length is the thermal diffusion length √ κt so that, once

again, the solution depends on the similarity variable η = y/2 √ κt. Because

this problem has no imposed temperature drop, T must be made dimension-

less with the specified initial heat content of the dike Q. A quantity with

units of temperature obtained from Q is

Q

2ρc √ κt ,

and the appropriate dimensionless temperature for this problem is

θ ≡ T − T0

Q/(2ρc √ κt)

. (4.160)

θ as defined in this equation depends only on η. The integral constraint on

4.21 One-Dimensional, Unsteady Heat Conduction in an Infinite Region 309

temperature (4–159) can be written in terms of θ as

2ρc

∫ ∞

0

Q

2ρc √ κt θ2

√ κt dη = Q

or ∫ ∞

0 θdη =

1

2 . (4.161)

The heat conduction equation must be rewritten in terms of θ. From

Equation (4–160) we have

∂T

∂t =

Q

2ρc √ κt

dθ

dη

(−η 2t

)

+ Qθ

2ρc √ κ

(

−1

2

)

t−3/2

= −Qt−3/2

4ρc √ κ

(

η dθ

dη + θ

)

. (4.162)

Also, from Equation (4–160) one obtains

κ ∂2T

∂y2 =

κQ

2ρc √ κt

d2θ

dη2

1

4κt . (4.163)

Upon equating (4–162) and (4–163), we find

−2

(

η dθ

dη + θ

)

= d2θ

dη2 (4.164)

or

−2 d

dη (ηθ) =

d2θ

dη2 , (4.165)

which can be integrated to give

−2ηθ = dθ

dη + c1. (4.166)

The constant c1 must be zero because the temperature distribution must be

symmetric about the plane y = 0. This requires dθ/dη = 0 at η = 0. Thus

we have

−2ηdη = dθ

θ , (4.167)

which integrates to

θ = c2e −η2 . (4.168)

From the integral constraint (4–161) we can find c2 to be

∫ ∞

0 c2e

−η2dη = 1

2 = c2

√ π

2

310 Heat Transfer

or

c2 = 1√ π . (4.169)

Finally, the temperature distribution is

T − T0 = Q

2ρc √ πκt

e−y 2/4κt. (4.170)

At distances that are large compared with the width of the initial tempera-

ture distribution, the time dependence of the temperature is independent of

the initial temperature distribution and is proportional to the heat content

of the region.

The temperature at any distance y as given by Equation (4–170) increases

with time to a maximum value and then decreases. The time tmax when this

maximum occurs can be obtained by setting the time derivative of Equation

(4–170) equal to zero. The result is

tmax = y2

2κ . (4.171)

Except for a factor of 2, tmax is the thermal diffusion time corresponding

to the distance y. Substitution of Equation (4–171) into Equation (4–170)

gives the maximum temperature Tmax as a function of y,

Tmax = T0 + Q

ρcy

(

1

2πe

)1/2

. (4.172)

The maximum temperature is proportional to 1/y.

Applying these results to the example given at the end of Section 4–19,

we find Q = 8.8 × 109 J m−2 with the parameter values given for that

example and ρ = 2900 kg m−3. For the temperature distribution given in

Equation (4–170) to be valid, the time must be long compared with the

solidification time of 10.9 days. The temperature profiles from Equation (4–

170) at several times are given in Figure 4–41. The maximum temperature

as a function of distance from the center of the dike, from Equation (4–172),

is given in Figure 4–42. Calculations of this type can be used to determine

the distance from an intrusion that low-temperature metamorphic reactions

in the country rock can be expected.

4.22 Thermal Stresses

According to the laws of thermodynamics the equilibrium state of any ma-

terial is determined by any two state variables. Examples of state variables

4.22 Thermal Stresses 311

Figure 4.41 Temperatures as a function of distance from the center of a 2-m-wide intrusion at several times from Equation (4–170).

Figure 4.42 Maximum temperature as a function of distance from the cen- ter of a 2-m-wide intrusion from Equation (4–172).

include the temperature T , pressure p, and density ρ. In thermodynamics

it is often convenient to use the specific volume v (volume per unit mass)

rather than the density; the two variables are related by

v = 1

ρ . (4.173)

312 Heat Transfer

As a state variable, the specific volume can be related to the pressure and

temperature using the chain rule for partial differentiation

dv =

(

∂v

∂T

)

p dT +

(

∂v

∂p

)

T dp. (4.174)

The subscript indicates the variable that is held constant during the differ-

entiation; that is, (∂v/∂T )p is the partial derivative of volume with respect

to temperature at constant pressure.

The two partial derivatives appearing in Equation (4–174) are related

to well-known thermodynamic quantities. The isothermal compressibility β

of a material is its fractional change in volume with pressure at constant

temperature,

β = −1

v

(

∂v

∂p

)

T , (4.175)

and its volumetric coefficient of thermal expansion αv is its fractional change

in volume with temperature at constant pressure,

αv = 1

v

(

∂v

∂T

)

p . (4.176)

The coefficients β and αv are material properties that can be obtained from

laboratory experiments. We previously saw in Equation (3–55) how β is

related to the elastic properties of a material. Substitution of Equations (4–

175) and (4–176) into Equation (4–174) yields

dv = −vβ dp+ vαv dT. (4.177)

If a material is unconstrained, so that the pressure does not change (dp = 0)

when the temperature and volume change, Equation (4–177) implies

dv = vαv dT (4.178)

or

dρ = −ραv dT. (4.179)

If a material is confined, so that its volume cannot change (dv = 0), the

changes in temperature and pressure are related by

dp = αv β dT. (4.180)

Typical values of αv and β for rock are 3 × 10−5 K−1 and 10−11 Pa−1,

respectively. With these material properties and a temperature increase of

100 K, the increase in the confining pressure from Equation (4–180) is ∆p =

4.22 Thermal Stresses 313

300 MPa. The implication is that changes in temperature can lead to very

large changes in pressure or stress.

When temperature changes occur, the laws of linear elasticity must be

modified to include the thermally associated changes in volume. We have

seen that a temperature change ∆T produces a volume change ∆v = vαv∆T

in an unconstrained body. This change in volume is accompanied by the

strains

ε1 = ε2 = ε3 = −1

3 αv∆T (4.181)

if the body is isotropic. The minus sign on the right side of Equation (4–

181) arises because of the sign convention that treats contraction strains as

positive. The linear coefficient of thermal expansion αl relates the thermally

induced strains to the temperature change,

ε1 = ε2 = ε3 = −αl∆T or αl = 1

3 αv , (4.182)

so that the linear coefficient of thermal expansion is the change in the strain

in the material per degree change in temperature.

The total strain in a body is the sum of the stress-associated strains and

the temperature-associated strains. The stress-associated strains have been

written in Equations (3–4) to (3–6). If to these we add the thermal strains

of Equation (4–182), we obtain the total strain

ε1 = 1

E (σ1 − νσ2 − νσ3) − αl∆T (4.183)

ε2 = 1

E (−νσ1 + σ2 − νσ3) − αl∆T (4.184)

ε3 = 1

E (−νσ1 − νσ2 + σ3) − αl∆T. (4.185)

For a state of uniaxial stress we take σ1 = σ and σ2 = σ3 = 0. From

Equations (4–183) to (4–185) we obtain

ε1 = σ

E − αl∆T (4.186)

ε2 = ε3 = −ν E σ − αl∆T. (4.187)

For plane stress, σ3 = 0, and the equations of thermal elasticity reduce to

ε1 = 1

E (σ1 − νσ2) − αl∆T (4.188)

ε2 = 1

E (σ2 − νσ1) − αl∆T (4.189)

314 Heat Transfer

ε3 = − ν

E (σ1 + σ2) − αl∆T. (4.190)

For a state of isotropic stress σ1 = σ2 = σ3 = p, ε1 = ε2 = ε3 = ∆/3, and by

adding Equations (4–183) to (4–185) we find

∆ = 3

E (1 − 2ν)p− 3αl∆T. (4.191)

We previously identified the isothermal compressibility in Equation (3–55)

as

β = 3

E (1 − 2ν). (4.192)

This together with αl = αv/3 and ∆ = −dv/v shows that Equation (4–191)

is equivalent to Equation (4–177).

In Section 4–14, Equation (4–89), we obtained the temperature distribu-

tion in a semi-infinite half-space due to time periodic variation of the surface

temperature. Assuming that this half-space is a uniform elastic medium, we

can determine the resultant thermal stresses. Take the half-space to be con-

fined in the horizontal directions so that ε1 = ε2 = 0 and to be unconstrained

in the vertical direction so that σ3 = 0. From Equations (4–188) and (4–189)

we find

σ1 = σ2 = Eαl∆T

1 − ν . (4.193)

The temperature ∆T is measured relative to the temperature at which the

stress is zero. For the periodic heating of a semi-infinite half-space we as-

sume that at T = T0, the average temperature, the stress is zero. Therefore

substitution of Equation (4–89) into Equation (4–193) gives

σ1 = σ2 = Eαl∆T

(1 − ν) exp

(

−y √

ω

2κ

)

cos

(

ωt− y

√

ω

2κ

)

,

(4.194)

where ∆T is the actual amplitude of the periodic surface temperature vari-

ation about the average temperature. The maximum thermal stress is ob-

tained by setting y = t = 0 in Equation (4–194),

σmax = Eαl∆T

1 − ν . (4.195)

Let us take as typical values for rock E = 60 GPa, ν = 0.25, and αl =

10−5 K−1. If ∆T = 100 K, we find that σmax = 80 MPa.

In Section 3–4 we determined the elastic stresses resulting from sedimen-

tation and erosion. It was shown that the addition or removal of overburden

caused significant deviatoric stresses. However, when overburden is added or

4.22 Thermal Stresses 315

removed, the temperature at a given depth changes, and as a result thermal

stresses are generated. Because the equations of thermal elasticity are linear,

the thermal stresses can be added to those previously obtained.

We first consider near-surface rocks that have been buried to a depth h.

If sufficient time has elapsed to reestablish the normal geothermal gradient

β, the temperature increase of the surface rocks is βh. Again assuming no

horizontal strain, the thermal stress from Equation (4–193) is

σ1 = σ2 = Eαlβh

(1 − ν) . (4.196)

The elastic stress due to the addition of the overburden was given in Equa-

tion (3–24). Addition of the thermal stress to the elastic stress gives

σ1 = σ2 = h

(1 − ν) (ρgν + Eαlβ). (4.197)

To determine the deviatoric stresses after sedimentation, we determine the

pressure at depth h, noting that σ3 = ρgh; the result is

p = 1

3 (σ1 + σ2 + σ3) =

(1 + ν)

3(1 − ν) ρgh+

2

3

Ehαlβ

(1 − ν) .

(4.198)

The deviatoric stresses are obtained by subtracting this expression for p

from Equation (4–197) and from σ3 = ρgh:

σ′1 = σ′2 = −(1 − 2ν)

3(1 − ν) ρgh+

Ehαlβ

3(1 − ν) (4.199)

σ′3 = 2

3

(1 − 2ν)

(1 − ν) ρgh− 2

3

Ehαlβ

(1 − ν) . (4.200)

The differential horizontal stresses due to the elastic effect are extensional;

because of the thermal effect they are compressional. Figure 4–43 gives

σ′1/ρgh as a function of β for E = 60 GPa, ν = 0.25, αl = 10−5 K−1,

g = 10 m s−2, and ρ = 2700 kg m−3. The thermal effect is seen to be

more important than the elastic effect for thermal gradients greater than

23 K km−1. Because this is a typical thermal gradient in continental areas,

the thermal and overburden stresses are likely to be comparable under most

conditions of sedimentation.

Consider next what the surface stress is after h km of overburden have

been eroded. As in Section 3–4 we assume that the initial stresses are litho-

static so that Equation (3–30) gives the nonthermal part of the surface

stress. We also assume that a new thermal equilibrium has been established

subsequent to the removal of surface material. After the erosion of h km

316 Heat Transfer

Figure 4.43 Differential stress resulting from the addition of h km of over- burden to an initially unstressed surface.

the change in temperature of the surface rocks is −βh. Again assuming no

horizontal strain, we find the surface thermal stress from Equation (4–193)

to be

σ1 = σ2 = −Eαlβh

(1 − ν) . (4.201)

Adding the surface thermal stress to the surface elastic stress due to the

removal of overburden, Equation (3–30), we obtain

σ1 = σ2 = h

(1 − ν) [(1 − 2ν)ρg − Eαlβ]. (4.202)

As a consequence of erosion, the elastic effect causes surface compression,

and the thermal effect causes surface extension. If σ′1/ρgh given in Figure 4–

43 is multiplied by −3, one obtains σ1/ρgh due to erosion. Thus it is clear

that surface thermal stress and surface stresses resulting directly from re-

moval of surface material are comparable for typical values of the geothermal

gradient.

Problem 4.49 Determine the surface stress after 10 km of erosion. Take

E = 60 GPa, ν = 0.25, αl = 10−5 K−1, ρ = 2700 kg m−3, and β = 20 K

km−1.

Problem 4.50 If αv = 3 × 10−5 K−1 and the temperature of the Earth

increased by 100 K, what would the change in radius be?

There is an important distinction between renewable and nonrenewable

stresses. Thermal and erosional stresses are permanently relieved by irre-

versible creep and are examples of nonrenewable stresses. Examples of re-

newable stresses include bending stresses in the lithosphere, the lithosphere

stresses that drive plate tectonics, and the crustal stresses that support

mountain ranges. These stresses are not relieved by a small amount of creep.

4.23 Ocean Floor Topography 317

Figure 4.44 The principle of isostasy requires the ocean to deepen with age to offset the thermal contraction in the lithosphere.

4.23 Ocean Floor Topography

We previously solved for the thermal structure of the lithosphere adjacent

to ocean ridges in Equation (4–125). We also demonstrated in the previ-

ous section that the cooling of the oceanic lithosphere causes the density of

lithospheric rock to increase. The relationship between density and temper-

ature is given in Equation (4–179). In addition, we introduced the principle

of isostasy in Section 2–2.

The principle of isostasy states that any vertical column of material has

the same mass per unit area between the surface and some depth of com-

pensation. This is equivalent to the assumption that the lithostatic pressure

at some depth is the same over a large horizontal area. However, as shown

in Figure 4–23, columns of mantle material at different ages do not contain

the same mass per unit area. The older column contains more dense, cold

lithosphere than the younger one; the extra weight of the older lithosphere

causes it to subside. Mantle material below the lithosphere flows away to

accommodate the subsidence, and the ocean fills in the hole created at the

surface. Figure 4–44 shows the oceanic lithosphere with the overlying ocean

increasing in depth with distance from the ridge. The two columns of ages

t1 and t2 now have the same mass per unit area because the older column

contains more water to offset the added weight of dense lithospheric rock.

The ability of the mantle rock beneath the lithosphere to behave as a fluid

on geological time scales is the key to the isostatic adjustment of the oceanic

lithosphere. By calculating the masses per unit area in vertical columns

extending from the surface to the base of the lithosphere and requiring that

these be the same for columns of all ages, we can derive a formula for the

318 Heat Transfer

depth of the ocean floor w as a function of age t or distance from the ridge

x.

The mass per unit area in a column of any age is ∫ yL

0 ρ dy + wρw,

where yL is the thickness of the lithosphere and ρw is the density of water.

At the ridge crest, ρ = ρm the deep mantle density, and the mass of a column

of vertical height w + yL is ρm(w + yL). Isostasy requires that

ρm(w + yL) =

∫ yL

0 ρ dy + wρw (4.203)

or

w(ρw − ρm) +

∫ yL

0 (ρ− ρm) dy = 0. (4.204)

The first term in Equation (4–204) represents a negative mass because the

water with density ρw is less dense than the mantle rock that it has replaced

because of the subsidence of the seafloor a distance w. The second term

in the equation represents a positive mass because thermal contraction in

the cooling lithosphere causes the density ρ to be higher than the reference

hot mantle rock density ρm. Introducing the volume coefficient of thermal

expansion from Equation (4–179), we can write

ρ− ρm = ρmαv(T1 − T ). (4.205)

Upon substitution of the temperature profile from Equation (4–125) into

Equation (4–205) and that result into Equation (4–204), we obtain

w(ρm− ρw) = ρmαv(T1 −T0)

× ∫ ∞

0 erfc

[

y

2

(

u0

κx

)1/2 ]

dy. (4.206)

Because ρ → ρm and T → T1 at the base of the lithosphere, the limit on

the integral has been changed from y = yL to y = ∞. We can rewrite

Equation (4–206) by using the similarity variable η = (y/2)× (u0/κx) 1/2 so

that

w = 2ρmαv(T1 − T0)

(ρm − ρw)

(

κx

u0

)1/2 ∫ ∞

0 erfc(η) dη.

(4.207)

The definite integral has the value ∫ ∞

0 erfc(η) dη =

1√ π , (4.208)

4.23 Ocean Floor Topography 319

Figure 4.45 Seafloor depth as a function of age in the Atlantic, Pacific, and Indian Oceans. Data are from DSDP and ODP drill sites on normal ocean crust and depths have been corrected for sediment cover (Johnson and Carlson, 1992). Comparisons are made with the half-space cooling model (HSCM) from Equation (4–209) and the plate model from Equation (4–211) with yL0 =95 km (PM 95) and yL0 =125 km (PM 125).

so that

w = 2ρmαv(T1 − T0)

(ρm − ρw)

(

κx

πu0

)1/2

. (4.209)

Equation (4–209) predicts that the depth of the ocean increases with

the square root of the distance from the ridge or the square root of the

age of the ocean floor. This theoretical result is compared with seafloor

depths in Figure 4–45. The results shown are from Deep Sea Drilling Project

(DSDP) and Ocean Drilling Project (ODP) drill sites (Johnson and Carlson,

1992). Corrections have been made for sediment cover and results are given

for the Atlantic, Pacific, and Indian Oceans. Predicted seafloor depths are

included for the half-space cooling model, Equation (4–209), assuming ρm =

3300 kg m−3, ρw = 1000 kg m−3, κ = 1 mm2 s−1, T1 − T0 = 1300 K, and

αv = 3× 10−5K−1. In addition, the ridge depth is 2.5 km. For seafloor ages

greater than about 80 Myr, the predicted values are systematically deeper

than the observed values. This divergence is taken as evidence of the basal

heating of old oceanic lithosphere.

A model that approximates basal heating of the lithosphere is the plate

cooling model introduced in Section 4–17. The temperature distribution in

the oceanic lithosphere according to the plate cooling model was given in

Equation (4–130). Substitution of this temperature distribution into Equa-

320 Heat Transfer

tion (4–205) and further substitution of the resulting density distribution

into Equation (4–204) give

w = ρmαv(T1 − T0) yL0

2(ρm − ρw)

[

1 − 4

π

∫ 1

0

∞ ∑

n=1

1

n exp

(

− κn2π2t

y2 L0

)

sin(nπy′) dy′ ]

. (4.210)

Evaluation of the integral in Equation (4–210) leads to

w = ρmαv(T1 − T0) yL0

(ρm − ρw)

[

1

2 − 4

π2

∞ ∑

m=0

1

(1 + 2m)2 exp

(

−κ(1 + 2m)2π2t

y2 L0

)

]

. (4.211)

Only the terms with n = 1, 3, 5, . . . in Equation (4–210) contribute to the

result in Equation (4–211) since the terms with n = 2, 4, 6, . . . integrate to

zero. For large times, t≫ y2 L0/κ, the equilibrium depth we is given by

we = ρmαv(T1 − T0)yL0

2(ρm − ρw) . (4.212)

This is the equilibrium depth of the old ocean basins beneath the ridge crests.

This relation provides a constraint on the thickness of the plate in the plate

cooling model. In comparing the predictions of this model with observations,

we consider plate thicknesses yL0 = 95 and 125 km. With ρm = 3300 kg m−3,

ρw = 1000 kg m−3, T1 − T0 = 1300 K, and αv = 3× 10−5 K−1, we find from

Equation (4–212) that we = 2.7 km for yL0 = 95 km and we = 3.5 km

for yL0 = 125 km. With the depth of ocean ridges equal to 2.5 km, the

corresponding equilibrium depths of ocean basins are 5.2 km and 6 km,

respectively.

We can approximate the deviations of bathymetry from the half-space

cooling result given in Equation (4–209) by retaining the first term of the

expansion given in Equation (4–211), with the result

w = ρmαv(T1 − T0)yL0

ρm − ρw

[

1

2 − 4

π2 exp

(

−κπ 2t

y2 L0

)]

.

(4.213)

The n = 2 term in the temperature distribution, Equation (4–130), does not

contribute to the bathymetry. Seafloor subsidence predicted by the plate

cooling model is compared with observations and the half-space cooling

4.23 Ocean Floor Topography 321

model in Figure 4–45 for plate thicknesses of yL0 = 95 km (PM 95) and

yL0 = 125 km (PM 125). The results for a plate thickness yL0 = 125 km

are in excellent agreement with the data. While a thickness of yL0 = 95 km

is in good agreement with the heat flow data (as shown in Figure 4–25), a

thickness of yL0 = 125 km is in good agreement with the subsidence data

(Figure 4–45). Because there is generally less scatter in seafloor bathymetry

than in heat flow, we prefer the value yL0 = 125 km although we recognize

there is considerable uncertainty in the choice.

The plate model is clearly an idealization of the oceanic lithosphere. There

is no well-defined “lower plate boundary” in the mantle. The flattening of

the cooling curves can be attributed to the basal heating of the oceanic

lithosphere. For yL0 = 125 km, the required basal heating from Equation (4–

135) is qm = 34 mW m−2, which is quite close to our preferred value for

the basal heating of the continental lithosphere, qm = 28 mW m−2. The

difference between the basal heating flux qm = 34 mW m−2 and the mean

oceanic heat flux qo = 101 mW m−2 we will refer to as the plate tectonic or

subduction flux qs = 67 mW m−2.

From this value of the basal heat flux for the continental lithosphere and

with the area of the continents including continental margins Ac = 2.0 × 108 km2, we find that the total basal heating of the continental lithosphere

is Qmc = 0.56×1013 W. From the value of the basal heat flux for the oceanic

lithosphere and with the area of the oceans including marginal basins Ao =

3.1×108 km2, we find that the total basal heating of the oceanic lithosphere

is Qmo = 1.05 × 1013 W. Thus, the basal heating of the entire lithosphere

is Qm = 1.61 × 1013 W, which represents 36% of the total global heat flux

Qg = 4.43 × 1013 W. From the estimate of the fraction of the oceanic heat

flow directly associated with subduction (qs = 67 mW m−2) we find that the

total heat flux associated with subduction is Qs = 2.08× 1013 W. The total

global heat flux can be divided into three contributions: (1) radiogenic heat

production in the continental crust Qr = 0.74 × 1013 W (16.7%), (2) basal

heating of the lithosphere Qm = 1.61× 1013 W (36.3%), and (3) subduction

of the oceanic lithosphere Qs = 2.08 × 1013 W (47%).

These results show that basal heating of the lithosphere is quantitatively

large. There are two competing hypotheses for this basal heating. The first is

heat transfer from mantle plumes impinging on the base of the lithosphere,

and the second is small-scale or secondary convection in the lower litho-

sphere and underlying asthenosphere. It is generally accepted that mantle

plumes are a source of basal heating so that the only question is the magni-

tude of this heating. We will address this question in Chapter 6. Small-scale

or secondary convection is associated with an instability in the lower litho-

322 Heat Transfer

Figure 4.46 Diagram for Problem 4–52.

sphere arising from the strong temperature dependence of the lithospheric

viscosity. It is a form of delamination or foundering of the lithosphere and

contrasts with the instability of the entire lithosphere that is manifest as

lithospheric subduction at an ocean trench. The amount of heat transported

by secondary convection near the base of the lithosphere is highly uncertain.

Problem 4.51 Assume that the temperature in the subducting lithosphere

is given by Equation (4–131). Show that the plate tectonic heat flux associ-

ated with subduction Qs is given by

Qs = 1

2 ρmcyL0Fs,

where ρm is mantle density, c is the specific heat, and Fs is rate of seafloor

subduction. Determine Qs taking ρm = 3300 kg m−1, c = 1 kJ kg−1 K−1,

yL0 = 125 km, and Fs = 0.090 m2 s−1.

Problem 4.52 The ocean ridges are made up of a series of parallel seg-

ments connected by transform faults, as shown in Figure 1–12. Because of the

difference of age there is a vertical offset on the fracture zones. Assuming the

theory just derived is applicable, what is the vertical offset (a) at the ridge

crest and (b) 100 km from the ridge crest in Figure 4–46 (ρm = 3300 kg m−3,

κ = 1 mm2 s−1, αv = 3 × 10−5 K−1, T1 − T0 = 1300 K, u = 50 mm yr−1).

Problem 4.53 Because of its cooling, the seafloor subsides relative to a

continent at a passive continental margin. Determine the velocity of sub-

sidence if ρm = 3300 kg m−3, κ = 1 mm2 s−1, T1 − T0 = 1300 K, αv =

3 × 10−5 K−1, and the age is 20 Ma.

Problem 4.54 The influence of a small amount of partial melt on the

lithosphere–asthenosphere boundary has been considered in Problem 4–43.

Determine the contribution of this small degree of partial melt to the sub-

sidence of the seafloor as a function of age. The density of the magma is

ρl, its mass fraction is x, and the latent heat of fusion is L. If x = 0.05,

L = 400 kJ kg−1, c = 1kJ kg−1 K−1, T1 − T0 = 1350 K, κ = 1mm2 s−1,

ρm = 3300 kg m−3, ρl = 2800 kg m−3, and αv = 3 × 10−5 K−1, determine

4.24 Changes in Sea Level 323

Figure 4.47 Height of the sea surface in the past relative to present sea level from studies of sedimentation (Vail et al., 1978).

the ratio of the subsidence due to solidification to the subsidence due to

thermal contraction.

4.24 Changes in Sea Level

Changes in sea level are well documented from studies of sedimentation. On

short time scales (∼104 to 105 years) sea level changes can be explained by

variations in the volume of the polar ice sheets. However, on longer time

scales (∼107 to 108 years) the magnitudes of sea level changes are too large

to be understood in this way. If the polar ice sheets were completely melted,

the water added to the oceans would increase sea level by about 80 m. Yet

compared with its present level, the sea has been hundreds of meters higher

during the last 550 Ma, as shown by the record in Figure 4–47 (the value

at t = 0 of 80 m above present sea level accounts for the water in the polar

ice sheets). Sea level in the Cretaceous (80 Ma) was 300 m higher than it is

today, and water flooded about 40% of the present area of the continents.

These large, long-term changes in sea level are attributed to changes in the

average depth w̄ of the seafloor below the level of the ridge crests. If this

average depth decreases as a consequence of a decrease in the average age

τ of subduction, the volume of water contained in the deep ocean basins

decreases and the height of the sea above the ridge crests h, that is, sea

level, increases. This is supported by magnetic anomaly studies that show

324 Heat Transfer

that the Cretaceous was a time when there were more ridges and the ridges

were spreading more rapidly than at present. Since 75 Ma, an extensive

ridge system has been subducted beneath western North America. Thus, on

average, the present seafloor is older and deeper than the seafloor at 80 Ma.

The present deep ocean basins hold more water than the basins during the

Cretaceous, and the sea surface today lies well below its level in that period.

Changes in the configuration of the seafloor cannot change the total vol-

ume of water in the oceans. If we neglect the changes that occur in the area

of the oceans as sea level changes, a change δw̄ in the mean depth of the

ocean basins below the ridge crests produces an opposite change δh in the

height of the sea above the ridge crests

δh = −δw̄. (4.214)

The mean depth of the ocean floor is

w̄ = 1

τ

∫ τ

0 w dt. (4.215)

By substituting for w from Equation (4–209), we get

w̄ = 2ρmαv(T1 − T0)

(ρm − ρw)

(

κ

π

)1/2(1

τ

∫ τ

0 t1/2 dt

)

= 4

3

ρmαv(T1 − T0)

(ρm − ρw)

(

κτ

π

)1/2

. (4.216)

The mean depth of the ocean basins is directly proportional to the square

root of the mean age of subduction. Sea level changes are therefore related

to changes in the average age at which subduction occurs by

δh = −4

3

ρmαv(T1 − T0)

(ρm − ρw)

(

κ

π

)1/2

δ(τ1/2). (4.217)

Equation (4–217) enables us to estimate the mean age of subduction during

the Cretaceous. We take ρm = 3300 kg m−3, αv = 3 × 10−5 K−1, κ =

1 mm2 s−1, T1 − T0 = 1300 K, ρw = 1000 kg m−3, the present mean age of

subduction equal to 120.8 Myr, and δh = 220 m (80 m of the 300-m sea level

rise is attributed to water presently locked up in polar ice). The average age

at which seafloor subducted in the Cretaceous is found to be 100 Myr.

Sea level changes due to changes in the mean subduction age of the seafloor

can be used to infer past variations in the mean oceanic heat flow q̄0. By

combining Equations (4–128) and (4–217), we find

δ

(

1

q̄0

)

= −3π

8

(ρm − ρw)

kρmαv(T1 − T0)2 δh. (4.218)

4.25 Thermal and Subsidence History of Sedimentary Basins 325

We will see that the changes that have occurred in mean oceanic heat flow

δq̄0 are small compared with the present value q̄0ρ. The left side of Equation

(4–218) can therefore be approximated by −δq̄0/q̄2 0ρ, and we can write the

percentage variation in mean oceanic heat flow as

δq̄0

q̄0ρ =

3π

8

(ρm − ρw)q̄0ρ kρmαv(T1 − T0)2

δh. (4.219)

Higher sea levels in the past imply larger values of the mean oceanic heat

flux. This is expected from the association of higher sea levels with a younger

seafloor. With the previous parameter values, k = 3.3 W m−1K−1, and

q̄0ρ = 87 mW m−2, we find that a 26 m increase in sea level is associated

with a 1% increase in the mean oceanic heat flux.

The fractional changes in average oceanic heat flow inferred from the sea

level data of Figure 4–47 with Equation (4–219) and the above parameter

values are shown in Figure 4–48. The figure also shows the increase in average

oceanic heat flow that would be expected in the past if the heat lost through

the oceans was proportional to the increased rate of heat production from

the radioactive isotopes in the mantle – see Equation (4–8). The inferred

fractional changes that have occurred in the average oceanic heat flux during

the past 550 Ma are about 10%. These variations in oceanic heat flow are not

attributable to larger radiogenic heat production rates in the past. Instead

they are statistical variations associated with changes in the geometry and

the mean spreading rate of the oceanic ridge system.

Problem 4.55 What would be the decrease in sea level due to a 10%

reduction in the area of the continents? Assume the depth of deep ocean

basins to be 5 km.

4.25 Thermal and Subsidence History of Sedimentary Basins

Subsidence of the Earth’s surface often results in the formation of sedimen-

tary basins. We can explain the subsidence history of many sedimentary

basins by essentially the same model that we used to understand the cool-

ing, thickening, and subsidence of the oceanic lithosphere. The model is

illustrated in Figure 4–49. Consider a region of the Earth that is hot, either

because of seafloor spreading or extensive volcanism. Initially (t = 0) there

is no sediment, and the basement has a temperature T1 and a density ρm.

Surface cooling causes subsidence as the basement rocks cool and contract.

We assume that sediments fill the basin caused by the subsidence; that is,

326 Heat Transfer

Figure 4.48 Fractional variations in the mean oceanic heat flow in the past 550 Ma inferred from the sea level data in Figure 4–47 and Equation (4– 219). The dashed line is the expected increase in the mean oceanic heat flow due to the larger rate of radiogenic heat production in the past.

Figure 4.49 Sedimentary basin model.

the region 0 < y < ySB. This assumption requires an adequate supply of

sediment to prevent the formation of a deep ocean basin.

As long as the thickness of the sediment ySB is much smaller than the

lithosphere thickness yL, we can carry over the results of the cooling oceanic

lithosphere calculation. Therefore the depth of the sedimentary basin is given

by Equation (4–209) with ρs, the density of the sediments, replacing ρw, and

t, the age of the basin, replacing x/u0, the age of the oceanic lithosphere;

the result is

ySB = 2ρmαm(T1 − T0)

(ρm − ρs)

(

κmt

π

)1/2

, (4.220)

where the subscripts m on α and κ emphasize that these properties refer

4.25 Thermal and Subsidence History of Sedimentary Basins 327

Figure 4.50 Depths to sedimentary layers deposited at times ts as functions of time.

to the mantle rocks and not the sediment (αm is αv for the mantle rocks).

The depth of the sedimentary basin is proportional to the square root of

time. The subsidence of the basin is caused primarily by the cooling of the

basement or lithospheric rocks. The cooling of the sedimentary rocks is a

negligible effect when ySB/yL ≪ 1.

It is of interest to determine the subsidence history of a sedimentary layer

that was deposited at a time ts after the initiation of subsidence. At the

time ts the basement lies at a depth given by setting t = ts in Equation

(4–220). Assuming no compaction of the sediments, the layers deposited at

time ts will always be this distance above the basement. However, the depth

of the basement at time t is given directly by Equation (4–220). Therefore

the depth to the sediments deposited at time t = ts at a later time t, denoted

by ys, is given by the difference between the depth to basement at t and ts;

that is,

ys = 2ρmαm(T1 − T0)

(ρm−ρs)

(

κm π

)1/2

(t1/2− t1/2s ). (4.221)

The depth to sedimentary layers deposited at various times is given in Fig-

ure 4–50. These curves were calculated assuming that ρm = 3300 kg m−3,

κm =1 mm2s−1, T1−T0 = 1300 K, αm = 3×10−5 K−1, and ρs = 2500 kg m−3.

Because the sedimentary layer is thin, the temperature–depth profile is

essentially linear in the sediments. The sedimentary layer must transport

the heat from the cooling basement rocks. Denoting this heat flux by q0, we

have, from Equation (4–127),

q0 = km(T1 − T0)√

πκmt . (4.222)

328 Heat Transfer

From Fourier’s law of heat conduction, we know that in the sediments

q0 = ks

(

dT

dy

)

s , (4.223)

where (dT/dy)s is the constant geothermal gradient in the sediments. By

combining these last two equations, we get (

dT

dy

)

s = km ks

(T1 − T0)√ πκmt

. (4.224)

Thus the temperature distribution in the sediments is

Ts = T0 + km ks

(T1 − T0)√ πκmt

y. (4.225)

The temperature of a sedimentary layer deposited at time ts at a subsequent

time t is given by substituting Equation (4–221) into Equation (4–225)

TSL = T0 + 2

π

km ks

ρmαm(T1 − T0) 2

(ρm − ρs)

(

1 − √

ts t

)

.

(4.226)

The thermal history of a sedimentary layer can be used to determine whether

organic material in the sediments has been converted to petroleum.

The Los Angeles basin is a relatively small sedimentary basin with a

width of about 50 km and a length of about 75 km. The basin is a pull-

apart structure associated with the San Andreas fault system. It is probably

similar to the small spreading centers that offset transform faults in the

Gulf of California. During the initiation of the basin subsidence, volcanism

was occurring. Volcanic rocks from drill holes in the basin have ages of 10

to 15 Ma. This volcanism was probably similar to the volcanism presently

occurring in the Imperial Valley of southern California.

Since volcanism ceased at about 10 Ma, subsidence has continued. It is

reasonable to assume that the volcanism thinned the lithosphere and that

the subsequent subsidence is associated with the conductive cooling and

thickening of the lithosphere. The structure of the basin is clearly com-

plex, with considerable faulting. Although these faults add complexity to

the basin, they are also likely to allow free vertical subsidence of the various

fault-bounded blocks.

Let us apply our analysis of the thermal subsidence of sedimentary basins

to the southwest block, which extends roughly from Santa Monica to Long

Beach and is the site of several major oil fields. A cross section of this part of

the basin is given in Figure 4–51. The depths of various sedimentary units in

4.25 Thermal and Subsidence History of Sedimentary Basins 329

Figure 4.51 Cross section of the southwest block of the Los Angeles basin. (From California Oil and Gas Fields, Vol. 2, California Division of Oil and Gas, Report TR12, Sacramento, 1974). The sedimentary layers are Pleistocene–Holocene (PH), Upper Pliocene (UPI), Lower Pliocene (LPI), Upper Miocene (UMi), Middle Miocene (MMi), and Cretaceous or older basement (Cr). Also shown is the theoretical section from Equation (4– 221).

Figure 4.52 The crosses are the depths to the boundaries between strati- graphic units in the Wilmington oil field at the ages of the boundaries. The solid line is the subsidence predicted by Equation (4–221).

the Wilmington oil field are given as a function of their ages in Figure 4–52.

The predicted depths of these sedimentary units are given by Equation (4–

221). Taking ρm = 3300 kg m−3, ρs = 2500 kg m−3, αm = 3 × 10−5 K−1,

T1 − T0 = 1200 K, and κm = 1 mm2 s−1, we obtain the solid curve given

in Figure 4–52. The predicted theoretical section is also given in Figure 4–

52. Reasonably good agreement is obtained, although considerable tectonic

structure is clearly associated with the formation of the basin and subsequent

motion on the San Andreas fault now located to the east.

330 Heat Transfer

The present thermal gradient in the basin is predicted by Equation (4–

224). With km = 3.3 W m−1 K−1, ks = 2 W m−1 K−1, and the other

parameter values as before, we find (dT/dy)s = 59 K km−1. The measured

surface thermal gradients in the Wilmington oil field are in the range 48 to

56 K km−1. Again reasonably good agreement is obtained.

Problem 4.56 Assume that the continental lithosphere satisfies the half-

space cooling model. If a continental region has an age of 1.5 × 109 years,

how much subsidence would have been expected to occur in the last 300 Ma?

Take ρm = 3300 kg m−3, κ = 1mm2 s−1, Tm − T0 = 1300 K, and αv =

3× 10−5 K−1. Assume that the subsiding lithosphere is being covered to sea

level with sediments of density ρs = 2500 kg m−3.

Problem 4.57 If petroleum formation requires temperatures between 380

and 430 K, how deep would you drill in a sedimentary basin 20 Ma old?

Assume T0 =285 K, T1 = 1600 K, κm = 1 mm2 s−1, ks = 2 W m−1K−1, and

km = 3.3 W m−1 K−1.

In Section 2–2 we introduced the crustal stretching model to explain the

subsidence of a sedimentary basin. In this model the subsidence is caused by

the thinning of the continental crust. The model was illustrated in Figure 2–

4; a section of continental crust with an initial width w0 stretched by a

stretching factor α to a final width wb = αw0 from Equation (2–6). In order

to conserve the volume of the crust the initial thickness of the crust hcc is

reduced to hcb = hcc/α as given by Equation (2–8). The resulting depth of

the sedimentary basin hsb is given by Equation (2–10).

We now extend the crustal stretching model by assuming that the conti-

nental lithosphere within the sedimentary basin is also mechanically stretched

and thinned by the same stretching factor α as the crust. We assume that

the shape of the temperature profile in the lithosphere remains unchanged

but that its thickness is reduced by the factor 1/α.

We assume that before stretching, the temperature distribution in the

lithosphere is given by Equation (4–124). By introducing the thickness of the

unstretched lithosphere yL0 from Equation (4–126), we can rewrite Equation

(4–124) as

T1 − T

T1 − T0 = erfc(1.16y/yL0). (4.227)

In order to conserve the volume of the lithosphere we require

yLb = yL0

α , (4.228)

where yLb is the thickness of the stretched lithosphere. The temperature

4.25 Thermal and Subsidence History of Sedimentary Basins 331

distribution in the stretched and thinned lithosphere is given by

T1 − T

T1 − T0 = erfc(1.16yα/yL0). (4.229)

Whereas the thinning of the crust produces subsidence, the thinning of the

lithosphere inhibits subsidence. The thinned continental lithosphere is hotter

and less dense than the original lithosphere leading to a thermal uplift.

Application of the principle of isostasy to the base of the continental

lithosphere gives

(ρcc − ρm)hcc − ρmαν(T1 − T0)

∫ ∞

0 erfc

(

1.16y

yL0

)

dy

= (ρs − ρm)hsb + (ρcc − ρm) hcc α

− ρmαν(T1 − T0)

× ∫ ∞

0 erfc

(

1.16αy

yL0

)

dy. (4.230)

Evaluation of the integrals using Equation (4–208) gives the thickness of the

sedimentary basin hsb in terms of the stretching factor α as

hsb =

[

(ρm − ρcc)

(ρm − ρs) hcc −

1

1.16 √ π

ρmαν(T1 − T0)yL0

(ρm − ρs)

]

× (

1 − 1

α

)

. (4.231)

The dependence of the basin thickness on α is the same as that given

in Equation (2–10) for crustal thinning alone. Taking the same parame-

ter values used in Section 2–2 (hcc = 35 km, ρm = 3300 kg m−3, ρcc =

2800 kg m−3, and ρs = 2500 kg m−3) along with αν = 3 × 10−5 K−1,

T1 − T0 = 1300 K, and yL0 = 150 km, we find that hsb = 10.1 km in the

limit α → ∞. The subsidence associated with crustal thinning alone would

be 22 km, so the lithospheric thinning reduces the subsidence considerably.

In the analysis just given, we implicitly assumed that crust and lithosphere

are stretched in a time interval that is short compared with the thermal time

constant κ/y2 L of the thinned lithosphere. After the initial thinning of the

crust and lithosphere, the lithosphere will thicken because of the loss of

heat to the surface. This cooling and thickening of the lithosphere will lead

to further thermal subsidence. With the assumption that the temperature

profile in the thickening lithosphere as a function of time is given by Equation

(4–124), the thickness of the sedimentary basin as a function of time is given

332 Heat Transfer

Figure 4.53 Depth of the sedimentary basin hsb as a function of age t for several values of the stretching factor α.

by

hsb = hcc

(

ρm − ρcc ρm − ρs

)(

1 − 1

α

)

− yL0ρmαν(T1 − T0)

1.16 √ π(ρm − ρs)

× [

1 − (

1

α2 +

2.322κt

y2 L0

)1/2 ]

. (4.232)

The thickness of the sedimentary basin as a function of time is given in

Figure 4–53 for several values of the stretching factor and for the same

parameters as before with κ = 1 mm2 s−1. When the thickness of the thinned

lithosphere increases to its initial value yL0, the total subsidence will be that

given by Equation (2–10).

Problem 4.58 Assume that the continental crust and lithosphere have

been stretched by a factor α = 2. Taking hcc = 35 km, yL0 = 125 km,

ρm = 3300 kg m−3, ρcc= 2750 kg m−3, ρs = 2550 kg m−3, αv = 3× 10−5 K−1,

and T1 −T0 = 1300 K, determine the depth of the sedimentary basin. What

is the depth of the sedimentary basin when the thermal lithosphere has

thickened to its original thickness?

4.26 Heating or Cooling a Semi-Infinite Half-Space 333

Problem 4.59 Assume that the continental crust and lithosphere have

been stretched by a factor α = 4. Taking hcc = 35 km, yL0 = 150 km,

ρm = 3300 kg m−3, ρcc= 2700 kg m−3, ρs = 2450 kg m−3, αv = 3× 10−5 K−1,

and T1 −T0 = 1250 K, determine the depth of the sedimentary basin. What

is the depth of the sedimentary basin when the thermal lithosphere has

thickened to its original thickness?

Problem 4.60 The compression model for a continental mountain belt

considered in Problem 2–6 can be extended to include the compression of

the lithosphere. Assuming that the temperature in the lithosphere after com-

pression is given by

T1 − T

T1 − T0 = erfc

(

1.16y

βyL0

)

(4.233)

show that the height of the mountain belt is given by

h =

[

(ρm − ρcc)

ρm hcc −

αv(T1 − T0)yL0

1.16 √ π

]

(β − 1).

(4.234)

Assuming β = 2, hcc = 35 km, ρm = 3300 kg m−3, ρcc = 2800 kg m−3,

αv = 3 × 10−5, T1 − T0 = 1300 K, and yL0 = 150 km, determine the height

of the mountain belt and the thickness of the crustal root.

4.26 Heating or Cooling a Semi-Infinite Half-Space by a

Constant Surface Heat Flux

So far we have been primarily concerned with heat conduction problems

in which temperature boundary conditions are specified. In some geological

applications it is appropriate to specify boundary conditions on the heat flux.

If we take the partial derivative of the unsteady heat conduction equation

(4–68) with respect to y and substitute Fourier’s law (4–1), we obtain

∂q

∂t = κ

∂2q

∂y2 . (4.235)

The heat flux satisfies the same diffusion equation as does temperature.

We now consider the heating of a semi-infinite half-space by the constant

addition of heat at its surface, q = q0 at y = 0. Initially at t = 0 the

temperature in the half-space is constant T = T0, and there is no heat flow

q(0) = 0. This problem is solved by Equation (4–235) with the boundary

conditions

q = 0 at t = 0, y > 0

334 Heat Transfer

q = q0 at y = 0, t > 0

q → 0 as y→∞, t > 0. (4.236)

This problem is identical with the sudden heating or cooling of a semi-infinite

half-space. Equation (4–235) together with conditions (4–236) is equivalent

to Equation (4–94) and conditions (4–95) if we identify θ as q/q0. The solu-

tion from Equation (4–112) is

q = q0 erfc η, (4.237)

with η defined by Equation (4–96). In order to find the temperature, we

substitute Fourier’s law (4–1) into Equation (4–237) with the result

∂T

∂y = −q0

k erfc η = −q0

k erfc

(

y

2 √ κt

)

. (4.238)

We can integrate Equation (4–238) using the boundary condition T → T0

as y → ∞. We find

T = T0 − q0 k

∫ y

∞ erfc

(

y′

2 √ κt

)

dy′

= T0 + q0 k

∫ ∞

y erfc

(

y′

2 √ κt

)

dy′

= T0 + 2q0

√ κt

k

∫ ∞

η erfc η′ dη′. (4.239)

After an integration by parts we can express the temperature as

T = T0 + 2q0 k

√ κt

{

e−η 2

√ π

− η erfc η

}

= T0 + 2q0 k

{

√

κt

π e−y

2/4κt − y

2 erfc

y

2 √ κt

}

.

(4.240)

The surface temperature Ts is obtained by setting y = 0 in Equation (4–240)

Ts = T0 + 2q0 k

(

κt

π

)1/2

. (4.241)

This formula gives the increase in the surface temperature due to the uniform

addition of heat to a half-space.

Problem 4.61 The heat loss from the Earth’s surface qs due to radiation

is given by

qs = σT 4, (4.242)

4.27 Frictional Heating on Faults 335

Figure 4.54 Geometry of the descending plate. (a) Side view. (b) Vertical view.

where σ = 0.567 × 10−7 W m−2 K−4 is the Stefan-Boltzmann constant, and

T is the absolute temperature. Assuming that T = 300 K, k = 2 W m−1

K−1, and κ = 0.8 mm2 s−1, use this heat loss to determine the cooling of the

Earth’s surface during 12 hr of night. (Assume q is constant, a reasonable

approximation, and use the half-space cooling model in this section.)

4.27 Frictional Heating on Faults: Island Arc Volcanism and

Melting on the Surface of the Descending Slab

As noted in Section 1–4, ocean trenches where subduction is occurring usu-

ally have parallel chains of active volcanoes overlying the descending litho-

sphere. Since the subduction process returns cold lithospheric rocks into the

interior of the Earth, a subduction zone would be expected to have low tem-

peratures and low surface heat flows. It is quite surprising, therefore, that

extensive volcanism is associated with subduction zones.

One explanation for the high temperatures required for volcanism is fric-

tional heating on the fault zone between the descending lithosphere and the

overlying mantle. That this fault zone is the site of many large earthquakes

is indicative of a large stress on the fault. When slip occurs in the presence

of a large stress, significant frictional heating occurs. If the mean stress on

the fault is τ and the mean velocity of the descending plate is u, the mean

rate of heat production on the fault, per unit area of the fault, is

q = uτ. (4.243)

336 Heat Transfer

To assess the influence of fault heating on the descending lithosphere, let

us consider the simplified geometry illustrated in Figure 4–54. The surface

plate approaches the trench with a velocity u at an angle φ to the normal

to the trench and descends into the mantle at an angle θ to the horizontal.

The linear chain of active volcanoes lies at a distance dv above the slip zone.

An x, y coordinate system is set up in the descending plate as shown.

The solution for constant heat addition to a uniform half-space can be

used for this problem if several assumptions are made:

a. A substantial fraction of the heat produced on the fault zone is lost to the

underlying descending lithosphere. This is a good approximation because

the cold descending plate is the dominant heat sink.

b. The initial thermal structure of the lithosphere can be neglected. Because

thermal conduction problems are linear in temperature, their solutions

can be superimposed. The heat addition problem can be treated inde-

pendently of the ambient conduction problem as long as the required

boundary conditions are not violated.

c. Time t in the transient conduction problem is replaced by

t = x

u cosφ . (4–244)

Substitution of Equations (4–243) and (4–244) into Equation (4–241) gives

the temperature on the slip zone Tsz as

Tsz = T0 + 2τ

k

(

uκx

π cosφ

)1/2

. (4.245)

For surface volcanism to occur, the temperature on the slip zone beneath

the volcanoes (with dv = x sin θ) must equal the melt temperature of the

rock Tm. From Equation (4–245) we find

Tm = T0 + 2τ

k

(

κdvu

π cosφ sin θ

)1/2

. (4.246)

As a typical example of a trench system we take Tm − T0 = 1200 K, u =

100 mm yr−1, k = 4 W m−1 K−1, κ = 1 mm2 s−1, dv = 125 km, θ = 45◦, and

φ = 0◦. From Equation (4–246) we find that the mean stress level required

to produce the necessary heating is τ = 180 MPa. Although this is a high

stress, it may be a reasonable value when relatively cool rocks are carried

to depths where the lithostatic pressure is high. Stress levels on faults are

considered in some detail in Chapter 8.

Problem 4.62 Assume a constant sliding velocity uf on a fault during an

earthquake that results in a frictional heat production uf τ (τ is the stress

4.28 Mantle Geotherms and Adiabats 337

on the fault). If uf = 10 m s−1, the total displacement d = 4 m, τ = 10 MPa,

k = 4 W m−1 K−1, and κ= 1 mm2 s−1, what is the temperature increase on

the fault during the earthquake as predicted by Equation (4–245)?

Problem 4.63 The amount of heat generated by friction on a fault during

an earthquake is given by Q = bτ , where b is the slip on the fault and τ

is the mean stress on the fault. If b = 3 m and τ = 10 MPa, what is the

maximum temperature increase 1 m from the fault due to friction on the

fault (ρ = 2700 kg m−3, c = 1 kJ kg−1 K−1)?

4.28 Mantle Geotherms and Adiabats

The thermal structure of the upper mantle is dominated by the large tem-

perature gradients in the lithosphere. The thermal structure of the oceanic

lithosphere was determined in Sections 4–16 and 4–17; the temperature–

depth relation is given in Equation (4–125) or Equation (4–130). A repre-

sentative geotherm for the oceanic lithosphere was plotted in Figure 4–27

for t = 60.4 Myr, T1−T0 = 1300 K, and κ = 1 mm2 s−1. The thermal struc-

ture of stable continental crust was determined in Section 4–8 and given in

Figure 4–12.

Beneath the thermal boundary layer that defines the lithosphere, heat

transport is primarily by convection. Details of this convection and the creep

mechanisms responsible for the fluidlike behavior of hot, solid mantle rock

are discussed in later chapters. For our purposes it is sufficient to know

that in the interior of a vigorously convecting fluid the mean temperature

increases with depth approximately along an adiabat. The adiabatic tem-

perature gradient in the mantle is the rate of increase of temperature with

depth as a result of compression of the rock by the weight of the overlying

material. If an element of material is compressed and reduced in volume by

increasing pressure, it will also be heated as a result of the work done by

the pressure forces during the compression. If there is no transfer of heat

into or out of the element during this process, the compression is said to be

adiabatic, and the associated temperature rise is the adiabatic increase in

temperature.

The change in density with pressure under adiabatic conditions is given

by the adiabatic compressibility

βa = 1

ρ

(

∂ρ

∂p

)

s . (4.247)

The subscript s means that the entropy s is constant. A reversible adiabatic

process is a constant entropy or isentropic process. For a solid the adiabatic

338 Heat Transfer

Figure 4.55 Seismic velocities Vp and Vs and the density ρ are given as a function of depth.

compressibility is somewhat smaller than the isothermal compressibility de-

fined in Equation (4–175) because the temperature increases with pressure

in an adiabatic process so there is some thermal expansion. If we assume that

the adiabatic compressibility is a constant, we can integrate Equation (4–

247) with the boundary condition ρ = ρ0 at p = 0 to give

ρ = ρ0e βap . (4.248)

The increase in pressure with depth is given by

dp

dy = ρg. (4.249)

For the Earth’s mantle we can reasonably assume that the gravitational

acceleration g is a constant. By combining Equations (4–248) and (4–249)

and integrating with g constant and the boundary condition p = 0 at y = 0,

we obtain

p = −1

βa ln(1 − ρ0gβay) (4.250)

ρ = ρ0

1 − ρ0gyβa . (4.251)

These expressions for pressure and density as functions of depth are not

completely satisfactory approximations to the actual pressure and density in

the mantle. The dependence of the mantle density on depth is given in Fig-

ure 4–55. The values were deduced from the seismic velocities, which are also

4.28 Mantle Geotherms and Adiabats 339

shown in Figure 4–55 and are tabulated in Section F of Appendix 2. The den-

sity discontinuity near a depth of 410 km is attributed to a solid–solid phase

change of the mineral olivine, the dominant mineral in the mantle. Labora-

tory studies have shown that olivine transforms to a denser spinel structure

at a pressure of 13.5 GPa and a temperature of about 1700 K. The density

increase due to this phase change is ∆ρ = 200 − 300 kg m−3. Laboratory

studies have also indicated that the density discontinuity near a depth of

660 km is caused by a transformation of the spinel structure to perovskite

and magnesiowüstite. At a pressure of 23.1 GPa this transformation takes

place at a temperature of about 1875 K. The density increase due to this

transformation is ∆ρ ≈ 400 kg m−3. These density discontinuities cannot be

modeled using Equation (4–247). In addition, the adiabatic compressibility

decreases with increasing pressure in the mantle from a near-surface value of

8.7×10−12 Pa−1 to a value of 1.6×10−12 Pa−1 at the core–mantle boundary.

For these reasons Equation (4–251) is a relatively poor approximation for

the Earth’s mantle.

We now return to our discussion of the mantle geotherm. As already

noted, the temperature gradient beneath the near-surface thermal boundary

layer (the lithosphere) is very near the adiabatic gradient due to mantle

convection. The adiabatic temperature gradient can be calculated from the

thermodynamic relation between entropy per unit mass s, temperature, and

pressure

ds = cp T dT − αv

ρ dp, (4.252)

where cp is the specific heat at constant pressure and ds, dT , and dp are

infinitesimal changes in entropy, temperature, and pressure. The entropy

change in an adiabatic process is zero if the process is also reversible. Thus

the rate of increase of temperature with pressure in an adiabatic, reversible

process is obtained by putting ds = 0 in Equation (4–252), from which it

follows that (

dT

dp

)

s =

(

αvT

ρcp

)

. (4.253)

We assume that Equation (4–253) is valid in the vigorously convecting com-

pressible mantle in which heat conduction and other irreversible processes

can be neglected.

If a material is strictly incompressible, pressure forces cannot change the

volume of an element of the material. Accordingly, there can be no adiabatic

compressional heating of an incompressible material; its adiabatic tempera-

ture gradient is zero. Rocks, however, are sufficiently compressible so that

340 Heat Transfer

Figure 4.56 Representative oceanic and continental shallow upper mantle geotherms.

the large increases in pressure with depth in the mantle produce significant

adiabatic increases of temperature with depth.

The adiabatic temperature gradient in the Earth (dT/dy)s can be found

by multiplying (dT/dp)s from Equation (4–253) by dp/dy from Equation

(4–249): (

dT

dy

)

s = αvgT

cp . (4.254)

For the near-surface values αv = 3× 10−5 K−1, T = 1600 K, cp = 1 kJ kg−1

K−1, and g = 10 m s−2, Equation (4–235) yields (dT/dy)s = 0.5 K km−1.

At greater depths the volume coefficient of thermal expansion is consider-

ably smaller. To extend the temperature profile in the oceanic lithosphere

given in Figure 4–27 to greater depths in the upper mantle, we assume

that (dT/dy)s = 0.3 K km−1. Figure 4–56 shows the oceanic upper mantle

geotherm to a depth of 400 km.

The upper mantle geotherm beneath the continents is not as well under-

4.28 Mantle Geotherms and Adiabats 341

Figure 4.57 Mantle geotherms are given for whole-mantle convection “(Curve a) and layered mantle convection” (Curve b). The range of val- ues for the mantle solidus and the minimum temperatures in a subducted slab are also given.

stood as the one beneath the oceans. One way to model the temperature

distribution in the continental lithosphere would be to apply the same one-

dimensional half-space cooling solution that we applied earlier to the oceanic

lithosphere. However, as shown in Figure 4–28, the predicted mantle heat

flows are considerably lower than the observed values. Also, if old continen-

tal lithosphere continued to cool, it would also continue to subside according

to Equation (4–209) (see Problem 4–56). The result would be continental

cratons overlain by a continuously thickening sedimentary cover. This con-

dition has not been observed, so there must be a heat input into the base of

the old continental lithosphere that retards further cooling and allows old

continental lithosphere to tend toward a steady-state temperature profile.

The input of heat to the base of the lithosphere is attributed either to mantle

plumes impinging on the base of the lithosphere or to secondary convection

in the lower lithosphere or to both, as discussed in Section 4–23.

342 Heat Transfer

The influence of near-surface radioactivity on continental surface heat flow

has been considered in Section 4–8. From Equation (4–29) and the data

given in Figure 4–11 we found that the heat flow beneath the near-surface

layer of heat-producing elements is about 37 mW m−2. We assume that

heat production beneath the near-surface radioactive layer can be neglected

and that the thermal structure of the continental lithosphere has reached a

steady state. Therefore, it is appropriate to assume the heat flow through

the continental lithosphere beneath the near-surface heat-producing layer,

qm, is constant. The resulting geotherm in the continental lithosphere is

given in Figure 4–56 for qm = 37 mW m−2 and k = 3.35 W m−1 K−1. The

thickness of the continental lithosphere is about 200 km.

Our discussion so far has centered on the thermal state of the shallow up-

per mantle; the geotherms in Figure 4–56 extend only to a depth of 400 km.

If the entire mantle were homogeneous and strongly convecting, the adi-

abatic temperature gradient given by Equation (4–254) would be a good

approximation of the slope of the temperature profile throughout the man-

tle. We have noted, however, that the distribution of density with depth has

significant discontinuities near depths of 410 and 660 km (see Figure 4–55).

The density discontinuity at 410 km is associated with the transformation

of olivine to a spinel structure. The phase change from olivine to spinel is

exothermic with a heat of reaction L = 90 kJ kg−1. For adiabatic flow

downward through the phase change, the temperature of the mantle rock

increases by

∆T = L

cp . (4.255)

The heat released by the phase change increases the temperature of the

rock. For cp = 1 kJ kg−1 K−1, the increase in temperature is 90 K at a

depth of 410 km. This increase in temperature with depth for adiabatic flow

is shown in Figure 4–57, where the whole mantle geotherm is given. We will

show that the 410-km phase change enhances mantle convection; that the

associated density boundary does not block mantle convection is indicated

by the descent of the subducted lithosphere through this depth.

The density discontinuity at a depth of 660 km is attributed to the trans-

formation of the spinel structure to perovskite and magnesiowüstite. This

transformation is endothermic with a heat of reaction L = −70 kJ kg−1.

The heat absorbed by this reaction cools the rock. From Equation (4–255)

with cp = 1 kJ kg−1 K−1, the decrease in temperature is 70 K at a depth of

660 km. We will show that the 660-km transformation is expected to retard

flow through this boundary.

4.28 Mantle Geotherms and Adiabats 343

Deep-focus earthquakes provide conclusive evidence that there is active

mantle convection to depths of 660 km. Since the lower mantle is expected

to contain significant concentrations of radioactive isotopes, we expect that

mantle convection will occur in the lower mantle in order to transport the

resulting heat.

Three alternative models for mantle convection have been proposed:

a Whole mantle convection. If significant amounts of subducted lithosphere

can enter the lower mantle beneath 660 km, then there must be a comple-

mentary mantle upwelling. In this case the geotherm for the entire mantle

is likely to be adiabatic. The expected geotherm is illustrated in Figure

4–57 as curve a. The primary arguments against whole mantle convection

come from chemical geodynamic studies, which we will discuss in Chapter

10.

b Layered mantle convection. Two separate convection systems are operat-

ing in the upper and lower mantle. This would be the case if the density

discontinuity at a depth of 660 km completely blocks convection. An up-

per convective system associated with plate tectonics would be restricted

to the upper 660 km of the mantle; a lower, separate system would op-

erate between a depth of 660 km and the core–mantle boundary. In this

case a thermal boundary layer would be expected to develop at a depth

of 660 km similar to the lithosphere. However, it is very difficult to es-

timate the change in temperature associated with this boundary layer.

An expected geotherm for layered mantle convection is given as curve b

in Figure 4–57. Although deep-focus earthquakes do not occur at depths

greater than 660 km, studies using mantle tomography indicate that at

least some subducted slabs penetrate through this boundary. This is taken

as convincing evidence that there is significant material transport between

the upper and lower mantle.

c Hybrid models. Hybrid models have been proposed that involve a strong

time dependence and/or a barrier to convection within the lower mantle.

If the 660-km seismic discontinuity acts as a partial barrier to mantle

convection, then mantle “avalanches” may be triggered that would lead

to a strongly time-dependent mantle convection. Dense subducted litho-

sphere could “pile up” on the 660-km deep seismic discontinuity until a

finite-amplitude instability resulted in a mantle “overturn” or avalanche.

Episodic mantle overturns have been proposed as an explanation for ap-

parent episodicities in the geological record. It has also been proposed

that there is a compositional barrier to whole mantle convection within

the lower mantle. Studies using seismic tomography have been used to

344 Heat Transfer

argue in favor of such a barrier having considerable topography and time

dependence.

A constraint on the temperature at the base of the mantle is the seismic

evidence that the outer core is liquid. This evidence consists mainly of the

inability of shear waves to propagate through the outer core. Measured veloc-

ities of seismic compressional waves in the outer core indicate that, although

the outer core is primarily composed of iron, it must also contain significant

concentrations of one or more other constituents, the most likely of which is

sulfur. The melting temperature for the iron–sulfur eutectic mixture at the

core–mantle boundary is estimated to be 3200 K. This is an approximate

minimum value for the temperature at the core–mantle boundary. The adi-

abatic lower mantle geotherm in Figure 4–57 is in approximate agreement

with this constraint.

Just as an upper mantle thermal boundary layer, the lithosphere, inter-

venes between the surface and the interior adiabatic state of the mantle,

a lower mantle thermal boundary layer is expected to exist just above the

core–mantle boundary. Seismic studies have confirmed the existence of this

boundary layer, which is referred to as the D′′-layer. The D′′-layer has a

complex structure with a thickness of 150 to 300 km. Laboratory studies

indicate that the solidus temperature of a perovskite–magnesiowüstite as-

semblage at the core–mantle boundary would be about 4300 K. In addition

to showing the two geotherms associated with whole mantle convection a

and layered mantle convection b, Figure 4–57 gives the range of values for

the mantle solidus as well as the temperature increases associated with the

D′′-layer. Although the required heat flux through the D′′-layer can be esti-

mated,the stability of the layer which would give its thickness is difficult to

determine. There may also be compositional stratification in this layer.

Problem 4.64 How much heat is conducted along the adiabat of Figure

4–57 at depths of 1000 and 2000 km? At the core–mantle interface? Use

k = 4 W m−1 K−1, αv = 1.5 × 10−5 K−1, g = 10 m s−2, cp = 1 kJ kg−1

K−1.

Problem 4.65 If the rate at which heat flows out of the core (J s−1) is

10% of the rate at which heat is lost at the Earth’s surface, how large is

the mean temperature drop across the lower mantle thermal boundary layer

in terms of the mean temperature drop across the upper mantle thermal

boundary layer? Assume that the heat transport across a boundary layer

can be calculated from Fourier’s law of heat conduction in the simple form

4.29 Thermal Structure of the Subducted Lithosphere 345

Figure 4.58 Isotherms (◦C) in a typical descending lithosphere. The 410- km phase change is elevated in the subducted lithosphere. The position of the slip zone is also shown.

of Equation (4–3). Also assume that the upper and lower mantle boundary

layers have the same thicknesses.

4.29 Thermal Structure of the Subducted Lithosphere

The subduction of the cold oceanic lithosphere into the deep mantle is a

primary mechanism for the transport of heat from the interior of the Earth

to its surface. Hot mantle rock comes to the surface at accretional plate

boundaries (ocean ridges) and is cooled by heat loss to the seafloor. The re-

sult is a cold thermal “boundary layer,” the oceanic lithosphere. The thermal

structure of this boundary layer was determined in Sections 4–16 and 4–17.

The cold subducted lithosphere is gradually heated and eventually becomes

part of the convecting mantle. Upward convective heat transfer through the

mantle involves the sinking of cold thermal anomalies (descending litho-

sphere at ocean trenches) and the rising of hot thermal anomalies (mantle

plumes). The density differences associated with the lateral temperature

variations provide the driving force for the mantle convective circulation.

In this section we discuss the temperature distribution in the subducted

oceanic lithosphere.

Isotherms in a lithosphere descending at an angle of 45◦ into the mantle

346 Heat Transfer

Figure 4.59 The Clapeyron or equilibrium curve separating two phases of the same material.

are shown in Figure 4–58. Since the subducted lithosphere was formed on the

seafloor, its initial thermal structure upon subduction is given by Equation

(4–125). The dependence of temperature upon depth prior to subduction is

the oceanic geotherm given in Figure 4–56. As the subducted lithosphere

descends into the mantle, frictional heating occurs at its upper boundary.

The effects of frictional heating were studied in Section 4–26. As discussed

there, the temperature distribution due to frictional heating – Equation (4–

240) – can be superimposed on the initial temperature distribution to give

the isotherms in the slab. The result is shown in Figure 4–58.

The low temperatures in the descending lithosphere cause it to have a

higher density than the surrounding mantle. The higher density results in

a body force driving the descending lithosphere downward. This body force

is important in driving the plates. An additional downward body force on

the descending slab is provided by the distortion of the olivine–spinel phase

boundary in the slab, as shown in Figure 4–58.

The olivine–spinel phase boundary is elevated in the descending litho-

sphere as compared with its position in the surrounding mantle because the

pressure at which the phase change occurs depends on temperature. Fig-

ure 4–59 is a sketch of the Clapeyron curve, which gives the pressures and

temperatures at which two phases of the same material, such as olivine and

spinel, are in equilibrium. Actually, the olivine–spinel transition is not uni-

variant; it occurs over a range of temperatures and pressures. It is sufficient

for our purposes here, however, to treat the phase change as occurring at a

single temperature corresponding to a single pressure. The two phases can

coexist at any point T , p lying on the Clapeyron curve.

The slope of the Clapeyron curve γ is defined by

γ ≡ (

dp

dT

)

pc . (4.256)

4.29 Thermal Structure of the Subducted Lithosphere 347

If we divide this equation by Equation (4–249), we obtain the change dy in

the vertical location of the phase transition corresponding to a change in

temperature dT

(

dy

dT

)

pc =

γ

ρg . (4.257)

For the olivine to spinel phase change, the slope of the Clapeyron curve

is positive. Since dT is negative for the lower temperatures in the interior

of the descending lithosphere, dy is negative, and the olivine–spinel phase

change occurs at a shallower depth (lower pressure) in the slab.

With γ = 2 MPa K−1, ρ = 3600 kg m−3, and g = 10 m s−2, we find from

Equation (4–257) that (dy/dT )pc = 0.055 km K−1. If we take the maximum

temperature difference across the slab to be ∆T = 800 K, we find that the

elevation of the olivine–spinel phase boundary in the descending lithosphere

is about 44 km. This elevation is illustrated in Figure 4–58. Since spinel is

about 280 kg m−3 denser than olivine, the additional mass of the elevated

spinel in the descending lithosphere provides a significant body force for

driving the plates in addition to the downward body force provided by the

thermal contraction of the lithosphere.

This approach can also be applied to the transition of spinel to perovskite

and magnesiowüstite. In this case the slope of the Clapeyron curve is nega-

tive and the transition occurs at a deeper depth (higher pressure) in the slab.

With γ = −2.5 MPa K−1, ρ = 3700 kg m−3 and g = 10 m s−2, we find from

Equation (4–257) that (dy/dT )pc = −0.07 km K−1. If we take the maximum

temperature across the slab to be ∆T = 750 K, we find that the depression

of this phase transition is 52 km. Since perovskite–magnesiowüstite is about

400 kg m−3 denser than spinel, the buoyancy of the depressed spinel pro-

vides a significant body force that inhibits convection through the 660-km

boundary.

Problem 4.66 Estimate the downward body force on the slab per unit

length of trench due to the elevation of the olivine–spinel phase boundary

in Figure 4–58. Assume ρ(spinel)− ρ(olivine) = 300 kg m−3. Estimate the

downward body force on the slab per unit length of trench due to thermal

contraction by integrating over the temperature distribution in Figure 4–58.

Assume αv = 3 × 10−5 K−1, and consider the densification of the slab only

to depths of 660 km.

348 Heat Transfer

4.30 Culling Model for the Erosion and Deposition of Sediments

The erosion and deposition of sediments are responsible for the formation

and evolution of many landforms. A classic example is an alluvial fan caused

by the deposition of sediments on a horizontal surface. Cross sections of

alluvial fans often resemble the form of the complementary error function

given in Figure 4–21. This similarity suggests that sediment deposition may

be modeled using the heat equation.

The use of the heat equation to model sediment erosion and deposition

was first proposed by W. E. H. Culling (1960) and this approach is known

as the Culling model. The basic hypothesis is that the down slope flux of

sediments S is linearly proportional to the slope so that

S = −K∂h

∂x , (4.258)

where h is the elevation of topography above a base level, x is the horizontal

distance, and K is a constant that is called a transport coefficient. The

sediment flux S is the volume of sediment transported per unit time per

unit width. In terms of the analogy with the heat equation (4–68), the flux

equation (4–258) is directly analogous to Fourier’s law given in Equation

(4–1).

Consider an element of topography of width δx. The flux of sediment out

of this element at x + δx is S(x + δx) and the flux of sediment into this

element at x is S(x). Using Equation (4–258) and the same expansion given

in Equations (4–9) and (4–10) we have

S(x+ δx) − S(x) = δx ∂S

∂x = −δxK ∂2h

∂x2 , (4.259)

where we have assumed K to be a constant. If there is a net flow of sediment

into the element, there must be a change in elevation h given by

δx ∂h

∂t .

Since a net flux of sediment out of the element leads to a decrease in eleva-

tion, we have

∂h

∂t = K

∂2h

∂x2 , (4.260)

which is identical to the one-dimensional, time-

dependent heat conduction equation (4–68).

Let us apply the Culling model to the progradation of a river delta into

a quiet basin with a horizontal flow. Sediments are supplied to the delta by

the river forming it. Sediments are deposited near the landward edge of the

4.30 Culling Model for the Erosion and Deposition of Sediments 349

Figure 4.60 Illustration of the one-dimensional model for a prograding river delta. It is assumed that the delta progrades seaward at a constant velocity U0; its position at successive times t1 to t4 is illustrated. The height of the prograding delta above the basin floor is h.

delta and are transported down the front of the delta by creep and shallow

landslides. Our simple one-dimensional model is illustrated in Figure 4–60.

The delta front is assumed to prograde forward at a constant velocity U0

and its position at successive times t1, t2, t3, and t4 is shown.

We utilize the approach given in Section 4–19 to solve this problem. Let

ξ = x− U0t (4.261)

be a coordinate that is moving with the front of the delta. The shape of the

delta is independent of time in this coordinate system and substitution of

Equation (4–261) into Equation (4–260) gives

−U0 dh

dξ = K

d2h

dξ2 , (4.262)

with the boundary conditions h = h0 at ξ = 0 and h → 0 as ξ → ∞ where

h0 is the height of the landward edge of the delta front. The solution of

Equation (4–262) is

h = A exp

(

−U0ξ

K

)

+B, (4.263)

where A and B are constants. When the boundary conditions are satisfied

we obtain

h = h0 exp

(

−U0ξ

K

)

. (4.264)

Substitution of Equation (4–261) into Equation (4–264) gives

h = h0 exp

[

−U0

K (x− U0t)

]

. (4.265)

The height of the delta front above the floor decreases exponentially with

350 Heat Transfer

Figure 4.61 Dependence of the nondimensional height h/h0 on the nondi- mensional distance from shore U0(x−U0t)/K from Equation (4–265).

Figure 4.62 Cross-sectional profiles of the Mississippi River delta (south- west passage) at various times showing its progradation (Fisk et al., 1954).

distance from the shore. A plot of height versus distance from the shore is

given in Figure 4–61.

Comparisons with progradation data are obtained using the delta front

slope. This slope at ξ = 0 is given by (

∂h

∂x

)

ξ=0 = −U0h

K . (4.266)

Using this relation, we can obtain the transport coefficient from the progra-

dation velocity U0 and the morphology of the delta. As a specific example,

consider the southwest pass segment of the Mississippi River delta. Longitu-

dinal profiles of this delta front are shown in Figure 4–62. Taking h = 107 m,

4.30 Culling Model for the Erosion and Deposition of Sediments 351

U0 = 76 m yr−1, and (∂h/∂x)ξ=0 = −0.0096, we find from Equation (4–266)

that K = 8.5 × 105 m2 yr−1.

Problem 4.67 Consider a simplified one-dimensional model for the for-

mation of an alluvial fan. Assume that there is a uniform flux of sediment

S0 over a vertical cliff, forming a one-dimensional, time-dependent alluvial

fan. Assume that the Culling theory is applicable and use the methods of

Section 4–25 to show that

h = 2S0

K

{(

Kt

π

)1/2

exp

(

− x2

4Kt

)

− x

2 erfc

(

x

2(Kt1/2)

)}

. (4.267)

Also show that the height of the alluvial fan at the cliff (x = 0) is given by

h0 = 2S0

(

t

πK

)1/2

(4.268)

and that the slope of the alluvial fan at the cliff is given by (

∂h

∂x

)

x=0 = −S0

K (4.269)

and

h0 = −2

(

∂h

∂x

)

x=0

(

Kt

π

)1/2

. (4.270)

For the alluvial fan beneath the San Gabriel Mountains in Pasadena, Cal-

ifornia, it is appropriate to take h0 = 400 m, (∂h/∂x)x=0 = −0.075, and

t = 106 years. What is the corresponding transport coefficient K?

Problem 4.68 The Culling model can also be applied to the erosion and

deposition of a fault scarp. Assume that a vertical fault scarp of height h0

forms at t = 0 and x = 0 and subsequently erodes symmetrically. At t = 0,

h = h0 for x < 0 and h = 0 for x > 0. For t > 0, h = h0/2 at x = 0,

the region x < 0 erodes and deposition occurs in x > 0. Assume that both

erosion and deposition are governed by Eq. (4–260) withK prescribed. Show

that the height of the topography h is given by

h = h0

2 erfc

(

x

2 √ Kt

)

. (4.271)

Also show that slope at x = 0 is given by (

∂h

∂x

)

x=0 =

−h0

2(πKt)1/2 . (4.272)

352 Collateral Reading

An earthquake is known to have occurred 400 years ago; with h0 = 3 m and

(∂h/∂x)x=0 = −0.5, what is the value of the transport coefficient K?

The morphology of sedimentary landforms such as prograding river deltas,

alluvial fans, eroding fault scarps, and eroding shorelines are often in good

agreement with solutions of the heat equation. However, derived values of

the transport coefficient K have considerable variability. This is not surpris-

ing because submarine sediment transport is very different from subaerial

sediment transport. Also, both are very dependent on climate, weather, and

rock type.

Collateral Reading

Carslaw, H. S., and J. C. Jaeger (1959), Conduction of Heat in Solids, 2nd

edition, Oxford University Press, Oxford, 510 p.

Culling, W. E. H. (1960), Analytical theory of erosion, J. Geol. 68, 336–344.

Fisk, H. N., E. McFarlan, C.R. Kolb, and L. J. Wilbert (1954), Sedimentary

framework of the modern Mississippi Delta, J. Sedimen. Petrol. 24,

76–99.

Johnson, H. P., and R. L. Carlson (1992), Variation of sea floor depth with

age: A test of models based on drilling results, Geophys. Res. Lett. 19,

1971–1974.

Leeds, A. R., L. Knopoff, and E. G. Kausel (1974), Variations of upper

mantle structure under the Pacific Ocean, Science 186, 141–143.

Lister, C. R. B., J. G. Sclater, E. E. Davis, H. Villinger, and S. Nagihara

(1990), Heat flow maintained in ocean basins of great age: Investigations

in the north-equatorial west Pacific, Geophys. J. Int. 102, 603–630.

Pollack, H. N., S. J. Hurter, and J. R. Johnson (1993), Heat flow from the

Earth’s interior: Analysis of the global data set, Rev. Geophys. 31, 267–

280.

Sclater, J. G., C. Jaupart, and D. Galson (1980), The heat flow through

oceanic and continental crust and the heat loss of the Earth, Rev. Geo-

phys. Space Phys. 18, 269–311.

Stefan, J. (1891), Uber die Theorie der Eisbildung, insbesondere uber die

Eisbildung im Polarmeere, Ann. Physik Chem. 42, 269–286.

Vail, P. R., R. M. Mitchum, and S. Thompson (1978), Seismic stratigraphy

and global changes of sea level, in Seismic Stratigraphy: Applications to

Hydrocarbon Exploration, C. F. Payton, ed., American Association of

Petroleum Geologists, Memoir 26, pp. 83–97.

Collateral Reading 353

Wright, T. L., D. L. Peck, and H. R. Shaw (1976), Kilauea lava lakes: Natural

laboratories of study of cooling, crystallization and differentiation of

basaltic magma, in The Geophysics of the Pacific Ocean Basin and

its Margin, G. H. Sutton, M. H. Manghnani, and R. Moberly, eds.,

American Geophysical Union, Washington, D. C., pp. 375–390.

Collateral Reading

Burchfield, J. D., Lord Kelvin and the Age of the Earth (Science History

Publications, New York, 1975), 260 pages.

A historical account of scientific attempts to determine the age of the Earth.

The book focuses on Kelvin’s influence and the debate between physi-

cists and geologists between the mid-1800s and the early 1900s.

Carslaw, H. S., and J. C. Jaeger, Conduction of Heat in Solids, 2nd edition

(Oxford University Press, Oxford, 1959), 510 pages.

A classic textbook on the mathematical theory of heat conduction in solids.

It describes fundamental mathematical techniques for solving time-

dependent heat conduction problems in a variety of geometries. The

book contains an extensive compilation of solutions to boundary value

problems often encountered in geological and geophysical applications.

5

Gravity

5.1 Introduction

The force exerted on an element of mass at the surface of the Earth has two

principal components. One is due to the gravitational attraction of the mass

in the Earth, and the other is due to the rotation of the Earth. Gravity refers

to the combined effects of both gravitation and rotation. If the Earth were

a nonrotating spherically symmetric body, the gravitational acceleration on

its surface would be constant. However, because of the Earth’s rotation,

topography, and internal lateral density variations, the acceleration of gravity

g varies with location on the surface. The Earth’s rotation leads mainly to a

latitude dependence of the surface acceleration of gravity. Because rotation

distorts the surface by producing an equatorial bulge and a polar flattening,

gravity at the equator is about 5 parts in 1000 less than gravity at the poles.

The Earth takes the shape of an oblate spheroid. The gravitational field of

this spheroid is the reference gravitational field of the Earth. Topography

and density inhomogeneities in the Earth lead to local variations in the

surface gravity, which are referred to as gravity anomalies.

The mass of the rock associated with topography leads to surface gravity

anomalies. However, as we discussed in Chapter 2, large topographic features

have low-density crustal roots. Just as the excess mass of the topography

produces a positive gravity anomaly, the low-density root produces a nega-

tive gravity anomaly. In the mid-1800s it was observed that the gravitational

attraction of the Himalayan Mountains was considerably less than would be

expected because of the positive mass of the topography. This was the first

evidence that the crust–mantle boundary is depressed under large mountain

belts.

A dramatic example of the importance of crustal thickening is the ab-

sence of positive gravity anomalies over the continents. The positive mass

5.2 Gravitational Acceleration 355

anomaly associated with the elevation of the continents above the ocean

floor is reduced or compensated by the negative mass anomaly associated

with the thicker continental crust. We will show that compensation due to

the hydrostatic equilibrium of thick crust leads in the first approximation

to a zero value for the surface gravity anomaly. There are mechanisms for

compensation other than the simple thickening of the crust. An example

is the subsidence of the ocean floor due to the thickening of the thermal

lithosphere, as discussed in Section 4–23.

Gravity anomalies that are correlated with topography can be used to

study the flexure of the elastic lithosphere under loading. Short wavelength

loads do not depress the lithosphere, but long wavelength loads result in

flexure and a depression of the Moho. Gravity anomalies can also have im-

portant economic implications. Ore minerals are usually more dense than the

country rock in which they are found. Therefore, economic mineral deposits

are usually associated with positive gravity anomalies. Major petroleum oc-

currences are often found beneath salt domes. Since salt is less dense than

other sedimentary rocks, salt domes are usually associated with negative

gravity anomalies.

As we will see in the next chapter, mantle convection is driven by vari-

ations of density in the Earth’s mantle. These variations produce grav-

ity anomalies at the Earth’s surface. Thus, measurements of gravity at

the Earth’s surface can provide important constraints on the flow patterns

within the Earth’s interior. However, it must be emphasized that the surface

gravity does not provide a unique measure of the density distribution within

the Earth’s interior. Many different internal density distributions can give

the same surface distributions of gravity anomalies. In other words, inver-

sions of gravity data are non-unique.

5.2 Gravitational Acceleration External to the Rotationally

Distorted Earth

The gravitational force exerted on a mass m′ located at point P outside the

Earth by a small element of mass dm in the Earth is given by Newton’s law

of gravitation. As shown in Figure 5–1, the gravitational attraction dfm in

the direction from P to dm is given by

dfm = Gm′dm

b2 , (5.1)

where G is the universal gravitational constant G = 6.673 × 10−11 m3 kg−1

s−2 and b is the distance between dm and the point P . The infinitesimal

356 Gravity

Figure 5.1 Force on a mass m′ due to the gravitational attraction of an infinitesimal element of mass dm in the Earth.

gravitational acceleration at P due to the attraction of dm is the force per

unit mass exerted on m′ in the direction of P :

dgm = dfm m′ . (5.2)

By combining Equations (5–1) and (5–2) we obtain

dgm = Gdm

b2 . (5.3)

If the distribution of mass in the Earth were known exactly, the gravitational

attraction of the Earth on a unit mass outside the Earth could be obtained

by summing or integrating dgm over the entire distribution. Suppose, for

example, that the entire mass of the Earth M were concentrated at its

center. The gravitational acceleration at a distance r from the center would

then be directed radially inward and, according to Equation (5–3), it would

be given by

gm = GM

r2 . (5.4)

Following the generally accepted sign convention, we take gm to be positive,

even though it is directed in the −r direction.

We next determine the gravitational acceleration outside a spherical body

with a density distribution that is a function of radius only, ρ = ρ(r′). The

geometry is illustrated in Figure 5–2. It is clear from symmetry considera-

tions that the gravitational acceleration gm at a point P outside the mass

5.2 Gravitational Acceleration 357

Figure 5.2 Geometry for the calculation of the gravitational acceleration at a point outside a spherically symmetric mass distribution.

distribution is directed radially inward and depends only on the distance r

of point P from the center of the sphere. For convenience, we let the line

from P to O be the polar axis of a spherical coordinate system r, θ, ψ. The

gravitational acceleration at P due to an element of mass dm located in the

sphere at r′, θ′, ψ′ is directed along the line from P to dm and is given by

Equation (5–3). The component of this gravitational acceleration along the

line from P to O is

G cosαdm

b2 .

The net radially inward gravitational acceleration at P is found by integrat-

ing this expression over the entire mass distribution:

gm = G

∫

cosα dm

b2 . (5.5)

The element of mass dm is the product of the volume element dV with the

density ρ(r′) at the location of dV

dm = ρ(r′) dV. (5.6)

358 Gravity

The element of volume can be expressed in spherical coordinates as

dV = r′2 sin θ′ dθ′ dψ′ dr′. (5.7)

The integral over the spherical mass distribution in Equation (5–5) can thus

be written

gm = G

∫ a

0

∫ π

0

∫ 2π

0

ρ(r′)r′2 sin θ′ cosα dψ′ dθ′ dr′

b2 ,

(5.8)

where a is the radius of the model Earth. The integral over ψ′ is 2π, since

the quantities in the integrand of Equation (5–8) are independent of ψ′. To

carry out the integration over r′ and θ′, we need an expression for cosα.

From the law of cosines we can write

cosα = b2 + r2 − r′2

2rb . (5.9)

Because the expression for cos α involves b rather than θ′, it is more conve-

nient to rewrite Equation (5–8) so that the integration can be carried out

over b rather than over θ′. The law of cosines can be used again to find an

expression for cos θ′:

cos θ′ = r′2 + r2 − b2

2rr′ . (5.10)

By differentiating Equation (5–10) with r and r′ held constant, we find

sin θ′dθ′ = b db

rr′ . (5.11)

Upon substitution of Equations (5–9) and (5–11) into Equation (5–8), we

can write the integral expression for gm as

gm = πG

r2

∫ a

0 r′ρ(r′)

∫ r+r′

r−r′

{

r2 − r′2

b2 + 1

}

db dr′.

(5.12)

The integration over b gives 4 r′ so that Equation (5–12) becomes

gm = 4πG

r2

∫ a

0 dr′r′2ρ(r′). (5.13)

Since the total mass of the model is given by

M = 4π

∫ a

0 dr′r′2ρ(r′), (5.14)

5.2 Gravitational Acceleration 359

the gravitational acceleration is

gm = GM

r2 . (5.15)

The gravitational acceleration of a spherically symmetric mass distribu-

tion, at a point outside the mass, is identical to the acceleration obtained

by concentrating all the mass at the center of the distribution. Even though

there are lateral density variations in the Earth and the Earth’s shape is dis-

torted by rotation, the direction of the gravitational acceleration at a point

external to the Earth is very nearly radially inward toward the Earth’s cen-

ter of mass, and Equation (5–15) provides an excellent first approximation

for gm.

Problem 5.1 For a point on the surface of the Moon determine the ratio

of the acceleration of gravity due to the mass of the Earth to the acceleration

of gravity due to the mass of the Moon.

The rotational distortion of the Earth’s mass adds a small latitude-dependent

term to the gravitational acceleration. This term depends on the excess mass

in the rotational equatorial bulge of the Earth. The observed latitude depen-

dence of gm can thus be used to determine this excess mass. In addition, this

effect must be removed from observed variations in surface gravity before the

residual gravity anomalies can properly be attributed to density anomalies

in the Earth’s interior. The model we use to calculate the contribution of

rotational distortion to gravitational acceleration is sketched in Figure 5–3.

The Earth is assumed to be flattened at the poles and bulged at the equator

because of its rotation with angular velocity ω. The mass distribution is as-

sumed to be symmetrical about the rotation axis. Because of the departure

from spherical symmetry due to rotation, the gravitational acceleration at a

point P outside the Earth has both radial and tangential components. The

radial component is the sum of GM/r2 and the term g′r due to rotational

distortion of the mass distribution; the tangential component g′t is entirely

due to the rotationally induced departure from spherical symmetry. Follow-

ing our previous sign convention both GM/r2 and g′r are positive if directed

inward. Since rotation modifies the otherwise spherically symmetric model

Earth only slightly, g′r and g′t are small compared with GM/r2.

The total gravitational acceleration is {(

GM

r2 + g′r

)2

+ g′2t

}1/2

=

{(

GM

r2

)2

+ 2

(

GM

r2

)

g′r + g′2r + g′2t

}1/2

. (5.16)

360 Gravity

Figure 5.3 Geometry for calculating the contribution of rotational distor- tion to the gravitational acceleration.

It is appropriate to neglect the quadratic terms because the magnitudes of g′r and g′t are much less than GM/r2. Therefore the gravitational acceleration

is given by

{(

GM

r2

)2

+ 2

(

GM

r2

)

g′r

}1/2

=

(

GM

r2

){

1 + 2g′r

GM/r2

}1/2

=

(

GM

r2

){

1 + g′r

GM/r2

}

= GM

r2 + g′r. (5.17)

Equation (5–17) shows that the tangential component of the gravitational

acceleration is negligible; the net gravitational acceleration at a point P

external to a rotationally distorted model Earth is essentially radially inward

to the center of the mass distribution.

The radial gravitational acceleration for the rotationally distorted Earth

model can be obtained by integrating Equation (5–5) over the entire mass

distribution. We can rewrite this equation for gm by substituting expression

(5–9) for cos α with the result

gm = G

2r2

∫ {

r

b + r3

b3

(

1 − r′2

r2

)}

dm. (5.18)

5.2 Gravitational Acceleration 361

The three distances appearing in the integral of Equation (5–18) r, r′, and

b are the sides of the triangle connecting O, P , and dm in Figure 5–3. It is

helpful for carrying out the integration to eliminate b from the integrand in

terms of r, r′, and the angle β, which is opposite the side of length b in this

triangle. From the law of cosines we can write

b2 = r2 + r′2 − 2rr′ cos β, (5.19)

which can be rearranged as

r

b =

{

1 + r′2

r2 − 2r′

r cos β

}−1/2

. (5.20)

Upon substituting Equation (5–20) into Equation (5–18), we get

gm = G

2r2

∫ {

1 + r′2

r2 − 2r′

r cos β

}−1/2

× {

1 +

(

1 − r′2

r2

)(

1 + r′2

r2 − 2r′

r cos β

)−1}

dm.

(5.21)

An analytic evaluation of the integral in Equation (5–21) is not possible.

The integration is complicated because both r′ and β vary with the position

of dm. However, the integration can be made tractable by approximating

the integrand with a power series in r′/r and retaining terms only up to

(r′/r)2. For P outside the mass distribution, r′/r < 1. We will show that

the expansion in powers of (r′/r) is equivalent to an expansion of the grav-

itational field in powers of a/r. This approximation yields an expression for

gm that is sufficiently accurate for our purposes. Using the formulas

(1 + ε)−1/2 ≈ 1 − ε

2 +

3ε2

8 + · · · (5.22)

(1 + ε)−1 ≈ 1 − ε+ ε2 + · · · , (5.23)

which are approximately valid for ε < 1, we find

gm = G

r2

∫ {

1 + 2r′

r cos β +

3r′2

r2

(

1 − 3

2 sin2 β

)}

dm.

(5.24)

The integrations in Equation (5–24) can be carried out in terms of well-

known physical properties of a mass distribution. The first term is just the

integral of dm over the entire mass. The result is simply M . The integral

of r′ cos β over the mass distribution is a first moment of the distribution.

362 Gravity

It is by definition zero if the origin of the coordinate system is the center of

mass of the distribution. Thus Equation (5–24) becomes

gm = GM

r2 +

3G

r4

∫

r′2 (

1 − 3

2 sin2 β

)

dm. (5.25)

The first term on the right of Equation (5–25) is the gravitational acceler-

ation of a spherically symmetric mass distribution. The second term is the

modification due to rotationally induced oblateness of the body. If higher

order terms in Equations (5–24) and (5–23) had been retained, the expan-

sion given in Equation (5–25) would have been extended to include terms

proportional to r−5 and higher powers of r−1.

We will now express the integral appearing in Equation (5–25) in terms

of the moments of inertia of an axisymmetric body. We take C to be the

moment of inertia of the body about the rotational or z axis defined by θ = 0.

This moment of inertia is the integral over the entire mass distribution of

dm times the square of the perpendicular distance from dm to the rotational

axis. The square of this distance is x′2 + y′2 so that we can write C as

C ≡ ∫

(x′2 + y′2) dm =

∫

r′2 sin2 θ′ dm (5.26)

because

x′ = r′ sin θ′ cosψ′ (5.27)

y′ = r′ sin θ′ sinψ′. (5.28)

The moment of inertia about the x axis, which is defined by θ = π/2, ψ = 0,

is

A ≡ ∫

(y′2 + z ′2) dm

=

∫

r′2(sin2 θ′ sin2 ψ′ + cos2 θ′) dm (5.29)

because

z′ = r′ cos θ′. (5.30)

Similarly, the moment of inertia about the y axis, which is defined by θ =

π/2, ψ = π/2, is

B ≡ ∫

(x′2 + z ′2) dm

=

∫

r′2(sin2 θ′ cos2 ψ′ + cos2 θ′) dm. (5.31)

5.2 Gravitational Acceleration 363

For a body that is axisymmetric about the rotation or z axis, A = B. The

addition of Equations (5–26), (5–29), and (5–31) together with the assump-

tion of axisymmetry gives

A+B + C = 2

∫

r′2 dm = 2A+ C. (5.32)

This equation expresses the integral of r′2dm appearing in Equation (5–25)

in terms of the moments of inertia of the body.

We will next derive an expression for the integral of r′2 sin2 βdm. Because

of the axial symmetry of the body there is no loss of generality in letting

the line OP in Figure 5–3 lie in the xz plane. With the help of Equation

(5–32) we rewrite the required integral as

∫

r′2 sin2 β dm =

∫

r′2(1 − cos2 β) dm

= A+ 1

2 C −

∫

r′2 cos2 β dm.

(5.33)

The quantity r′ cos β is the projection of r′ along OP . But this is also

r′ cos β = x′ cosφ+ z′ sinφ, (5.34)

where φ is the latitude or the angle between OP and the xy plane. Note that

y′ has no projection onto OP , since OP is in the xz plane. We use Equation

(5–34) to rewrite the integral of r′2 cos2 β in the form ∫

r′2 cos2 β dm = cos2 φ

∫

x′2 dm

+ sin2 φ

∫

z ′2 dm

+2cosφ sinφ

∫

x′z′ dm. (5.35)

For an axisymmetric body, ∫

x′2 dm =

∫

y′2 dm. (5.36)

This result and Equation (5–26) give

∫

x′2 dm = 1

2

∫

(x′2 + y′2) dm = 1

2 C. (5.37)

364 Gravity

The integral of z ′2dm can be evaluated by using Equations (5–26) and (5–32) ∫

z ′2 dm =

∫

(x′2 + y′2 + z ′2) dm − ∫

(x′2 + y′2) dm

=

∫

r′2 dm− ∫

(x′2 + y′2) dm

= A− 1

2 C. (5.38)

With mass symmetry about the equatorial plane we have ∫

x′z′ dm =

∫

r′2 cos θ′ sin θ′ cosψ′ dm = 0. (5.39)

Substitution of Equations (5–37) to (5–39) into Equation (5–35) yields ∫

r′2 cos2 β dm = 1

2 C cos2 φ+

(

A− 1

2 C

)

sin2 φ.

(5.40)

When Equations (5–33) and (5–40) are combined, we find, using sin2 φ +

cos2 φ = 1, that ∫

r′2 sin2 β dm = A cos2 φ+ C sin2 φ. (5.41)

The gravitational acceleration is finally obtained by substituting Equations

(5–32) and (5–41) into Equation (5–25):

gm = GM

r2 − 3G(C −A)

2r4 (3 sin2 φ− 1). (5.42)

Equation (5–42) is a simplified form of MacCullagh’s formula for an ax-

isymmetric body. The moment of inertia about the rotational axis C is larger

than the moment of inertia about an equatorial axis A because of the ro-

tational flattening of the body. It is customary to write the difference in

moments of inertia as a fraction J2 of Ma2, that is

C −A = J2Ma2, (5.43)

where a is the Earth’s equatorial radius. In terms of J2, gm is

gm = GM

r2 − 3GMa2J2

2r4 (3 sin2 φ− 1). (5.44)

The Earth’s gravitational field can be accurately determined from the track-

ing of artificial satellites. The currently accepted values are:

a = 6378.137 km

GM = 3.98600440 × 1014 m3s−2

5.3 Centrifugal Acceleration and the Acceleration of Gravity 365

Figure 5.4 Centrifugal acceleration at a point on the Earth’s surface.

J2 = 1.0826265 × 10−3. (5.45)

Although a satellite is acted upon only by the Earth’s gravitational accel-

eration, an object on the Earth’s surface is also subjected to a centrifugal

acceleration due to the Earth’s rotation.

5.3 Centrifugal Acceleration and the Acceleration of Gravity

The force on a unit mass at the surface of the Earth due to the rotation of

the Earth with angular velocity ω is the centrifugal acceleration gω. It points

radially outward along a line perpendicular to the rotation axis and passing

through P , as shown in Figure 5–4, and is given by

gω = ω2s, (5.46)

where s is the perpendicular distance from P to the rotation axis. If r is the

radial distance from P to the center of the Earth and φ is the latitude of

point P , then

s = r cosφ (5.47)

and

gω = ω2r cosφ. (5.48)

The currently accepted value for the Earth’s angular velocity is

ω = 7.292115 × 10−5 rad s−1.

366 Gravity

Problem 5.2 Determine the ratio of the centrifugal acceleration to the

gravitational acceleration at the Earth’s equator.

The gravitational and centrifugal accelerations of a mass at the Earth’s

surface combine to yield the acceleration of gravity g. Because gω ≪ gm, it

is appropriate to add the radial component of the centrifugal acceleration

to gm to obtain g; see Equations (5–16) and (5–17). As shown in Figure 5–

4, the radial component of centrifugal acceleration points radially outward.

In agreement with our sign convention that inward radial accelerations are

positive, the radial component of the centrifugal acceleration is

g′r = −gω cosφ = −ω2r cos2 φ. (5.49)

Therefore, the acceleration of gravity g is the sum of gm in Equation (5–44)

and g′r:

g = GM

r2 − 3GMa2J2

2r4 (3 sin2 φ− 1) − ω2r cos2 φ.

(5.50)

Equation (5–50) gives the radially inward acceleration of gravity for a point

located on the surface of the model Earth at latitude φ and distance r from

the center of mass.

5.4 The Gravitational Potential and the Geoid

By virtue of its position in a gravitational field, a mass m′ has gravitational

potential energy. The energy can be regarded as the negative of the work

done on m′ by the gravitational force of attraction in bringing m′ from infin-

ity to its position in the field. The gravitational potential V is the potential

energy of m′ divided by its mass. Because the gravitational field is conser-

vative, the potential energy per unit mass V depends only on the position

in the field and not on the path through which a mass is brought to the

location. To calculate V for the rotationally distorted model Earth, we can

imagine bringing a unit mass from infinity to a distance r from the center

of the model along a radial path. The negative of the work done on the unit

mass by the gravitational field of the model is the integral of the product

of the force per unit mass gm in Equation (5–44) with the increment of dis-

tance dr (the acceleration of gravity and the increment dr are oppositely

directed):

V =

∫ r

∞

{

GM

r′2 − 3GMa2J2

2r′4 (3 sin2 φ− 1)

}

dr′

5.4 The Gravitational Potential and the Geoid 367

(5.51)

or

V = −GM r

+ GMa2J2

2r3 (3 sin2 φ− 1). (5.52)

In evaluating V , we assume that the potential energy at an infinite distance

from the Earth is zero. The gravitational potential adjacent to the Earth is

negative; Earth acts as a potential well. The first term in Equation (5–52)

is the gravitational potential of a point mass. It is also the gravitational

potential outside any spherically symmetric mass distribution. The second

term is the effect on the potential of the Earth model’s rotationally induced

oblateness. A gravitational equipotential surface is a surface on which V is a

constant. Gravitational equipotentials are spheres for spherically symmetric

mass distributions.

Problem 5.3 (a) What is the gravitational potential energy of a 1-kg mass

at the Earth’s equator? (b) If this mass fell toward the Earth from a large

distance where it had zero relative velocity, what would be the velocity at

the Earth’s surface? (c) If the available potential energy was converted into

heat that uniformly heated the mass, what would be the temperature of the

mass if its initial temperature T0 = 100 K, c = 1 kJ kg−1 K−1, Tm = 1500

K, and L = 400 kJ kg−1?

A comparison of Equations (5–44) and (5–52) shows that V is the integral

of the radial component of the gravitational acceleration gm with respect to

r. To obtain a gravity potential U which accounts for both gravitation and

the rotation of the model Earth, we can take the integral with respect to r

of the radial component of the acceleration of gravity g in Equation (5–50)

with the result that

U = − GM

r + GMa2J2

2r3 (3 sin2 φ− 1)

− 1

2 ω2r2 cos2 φ. (5.53)

A gravity equipotential is a surface on which U is a constant. Within a few

meters the sea surface defines an equipotential surface. Therefore, elevations

above or below sea level are distances above or below a reference equipoten-

tial surface.

The reference equipotential surface that defines sea level is called the geoid.

We will now obtain an expression for the geoid surface that is consistent with

our second-order expansion of the gravity potential given in Equation (5–

53). The value of the surface gravity potential at the equator is found by

368 Gravity

substituting r = a and φ = 0 in Equation (5–53) with the result

U0 = −GM a

(

1 + 1

2 J2

)

− 1

2 a2ω2. (5.54)

The value of the surface gravity potential at the poles must also be U0

because we define the surface of the model Earth to be an equipotential

surface. We substitute r = c (the Earth’s polar radius) and φ = ±π/2 into

Equation (5–53) and obtain

U0 = −GM c

[

1 − J2

(

a

c

)2]

. (5.55)

The flattening (ellipticity) of this geoid is defined by

f ≡ a− c

a . (5.56)

The flattening is very slight; that is, f ≪ 1. In order to relate the flattening

f to J2, we set Equations (5–54) and (5–55) equal and obtain

1 + 1

2 J2 +

1

2

a3ω2

GM = a

c

[

1 − J2

(

a

c

)2]

. (5.57)

Substituting c = a(1 − f) and the neglecting quadratic and higher order

terms in f and J2, because f ≪ 1 and J2 ≪ 1, we find that

f = 3

2 J2 +

1

2

a3ω2

GM . (5.58)

Taking a3ω2/GM = 3.46139 × 10−3 and J2 = 1.0826265 × 10−3 from Equa-

tion (5–45), we find from Equation (5–58) that f = 3.3546×10−3 . Retention

of higher order terms in the theory gives the more accurate value

f = 3.35281068 × 10−3 = 1

298.257222 . (5.59)

It should be emphasized that Equation (5–58) is valid only if the surface of

the planetary body is an equipotential.

The shape of the model geoid is nearly that of a spherical surface; that is,

if r0 is the distance to the geoid,

r0 ≈ a(1 − ε), (5.60)

where ε≪ 1. By setting U = U0 and r = r0 in Equation (5–53), substituting

Equation (5–54) for U0 and Equation (5–60) for r0, and neglecting quadratic

and higher order terms in f , J2, a 3ω2/GM , and ε, we obtain

ε =

(

3

2 J2 +

1

2

a3ω2

GM

)

sin2 φ. (5.61)

5.4 The Gravitational Potential and the Geoid 369

Figure 5.5 Geoid height (EGM96) above reference ellipsoid WGS84 (Lemoine et al., 1998).

The substitution of Equation (5–61) into Equation (5–60) gives the approx-

imate model equation for the geoid as

r0 = a

{

1 − (

3

2 J2 +

1

2

a3ω2

GM

)

sin2 φ

}

(5.62)

or

r0 = a(1 − f sin2 φ). (5.63)

The nondimensional quantity a3ω2/GM is a measure of the relative im-

portance of the centrifugal acceleration due to the rotation of the Earth

compared with the gravitational attraction of the mass in the Earth. The

rotational contribution is about 0.33% of the mass contribution.

In the preceding analysis we considered only terms linear in J2 and a3ω2/GM .

In order to provide a reference geoid against which geoid anomalies are mea-

sured, it is necessary to include higher order terms. By convention, the ref-

erence geoid is a spheroid (ellipsoid of revolution) defined in terms of the

equatorial and polar radii by

r20 cos2 φ

a2 + r20 sin2 φ

c2 = 1. (5.64)

370 Gravity

The eccentricity e of the spheroid is given by

e ≡ (

a2 − c2

a2

)1/2

= (2f − f2)1/2. (5.65)

It is the usual practice to express the reference geoid in terms of the equa-

torial radius and the flattening with the result

r20 cos2 φ

a2 +

r20 sin2 φ

a2(1 − f)2 = 1 (5.66)

or

r0 = a

[

1 + (2f − f2)

(1 − f)2 sin2 φ

]−1/2

. (5.67)

If Equation (5–67) is expanded in powers of f and if terms of quadratic

and higher order in f are neglected, the result agrees with Equation (5–63).

Equation (5–67) with a = 6378.137 km and f = 1/298.257222 defines the

reference geoid.

The difference in elevation between the measured geoid and the reference

geoid ∆N is referred to as a geoid anomaly. A map of geoid anomalies is

given in Figure 5–5. The maximum geoid anomalies are around 100 m; this

is about 0.5% of the 21-km difference between the equatorial and polar radii.

Clearly, the measured geoid is very close to having the spheroidal shape of

the reference geoid.

The major geoid anomalies shown in Figure 5–5 can be attributed to den-

sity inhomogeneities in the Earth. A comparison with the distribution of

surface plates given in Figure 1–1 shows that some of the major anomalies

can be directly associated with plate tectonic phenomena. Examples are the

geoid highs over New Guinea and Chile–Peru; these are clearly associated

with subduction. The excess mass of the dense subducted lithosphere causes

an elevation of the geoid. The negative geoid anomaly over China may be

associated with the continental collision between the Indian and Eurasian

plates and the geoid low over the Hudson Bay in Canada may be associated

with postglacial rebound (see Section 6–10). The largest geoid anomaly is

the negative geoid anomaly off the southern tip of India, which has an am-

plitude of 100 m. No satisfactory explanation has been given for this geoid

anomaly, which has no surface expression. A similar unexplained negative

geoid anomaly lies off the west coast of North America.

The definition of geoid anomalies relative to the reference geoid is some-

what arbitrary. The reference geoid itself includes an averaging over den-

sity anomalies within the Earth. An alternative approach is to define geoid

anomalies relative to a hydrostatic geoid. The Earth is assumed to have a

5.4 The Gravitational Potential and the Geoid 371

Figure 5.6 Relationship of measured and reference geoids and geoid anomaly ∆ N.

layered structure in terms of density, but each layer is in hydrostatic equilib-

rium relative to the rotation of the Earth. The anomaly map is significantly

different for the two approaches, but the major features remain unaffected.

One of the primary concerns in geodesy is to define topography and

bathymetry. Both are measured relative to “sea level.” Sea level is closely

approximated by an equipotential surface corresponding to a constant value

of U . As we have discussed, geoid anomalies relative to a reference spheroidal

surface can be as large as 100 m. Thus, if we define sea level by a global

spheroid we would be in error by this amount. Topography (and bathymetry)

in any local area must be measured relative to a surface that approximates

the local sea level (equipotential surface).

Problem 5.4 Assume a large geoid anomaly with a horizontal scale of

several thousand kilometers has a mantle origin and its location does not

change. Because of continental drift the passive margin of a continent passes

through the anomaly. Is there a significant change in sea level associated with

the passage of the margin through the geoid anomaly? Explain your answer.

The anomaly in the potential of the gravity field measured on the reference

geoid ∆U can be related directly to the geoid anomaly ∆N . The potential

anomaly is defined by

∆U = Um0 − U0, (5.68)

where Um0 is the measured potential at the location of the reference geoid

and U0 is the reference value of the potential defined by Equation (5–54).

The potential on the measured geoid is U0, as shown in Figure 5–6. It can

be seen from the figure that U0, Um0, and ∆N are related by

U0 = Um0 +

(

∂U

∂r

)

r= r0

∆N, (5.69)

because ∆N/a ≪ 1. Recall from the derivation of Equation (5–53) that we

obtained the potential by integrating the acceleration of gravity. Therefore,

372 Gravity

the radial derivative of the potential in Equation (5–69) is the acceleration

of gravity on the reference geoid. To the required accuracy we can write (

∂U

∂r

)

r= r0

= g0, (5.70)

where g0 is the reference acceleration of gravity on the reference geoid. Just

as the measured potential on the reference geoid differs from U0, the mea-

sured acceleration of gravity on the reference geoid differs from g0. However,

for our purposes we can use g0 in Equation (5–69) for (∂U/∂r)r= r0 because

this term is multiplied by a small quantity ∆N . Substitution of Equations

(5–69) and (5–70) into Equation (5–68) gives

∆U = −g0∆N. (5.71)

A local mass excess produces an outward warp of gravity equipotentials and

therefore a positive ∆N and a negative ∆U . Note that the measured geoid

essentially defines sea level. Deviations of sea level from the equipotential

surface are due to lunar and solar tides, winds, and ocean currents. These

effects are generally a few meters.

The reference acceleration of gravity on the reference geoid is found by

substituting the expression for r0 given by Equation (5–62) into Equation

(5–50) and simplifying the result by neglecting quadratic and higher order

terms in J2 and a3ω2/GM . One finds

g0 = GM

a2

(

1+ 3

2 J2 cos2 φ

)

+aω2(sin2 φ− cos2 φ).

(5.72)

To provide a standard reference acceleration of gravity against which gravity

anomalies are measured, we must retain higher order terms in the equation

for g0. Gravity anomalies are the differences between measured values of g on

the reference geoid and g0. By international agreement in 1980 the reference

gravity field was defined to be

g0 = 9.7803267715(1 + 0.0052790414 sin2 φ

+ 0.0000232718 sin4 φ

+ 0.0000001262 sin6 φ

+ 0.0000000007 sin8