the dynamic earth - betavak-nlt.nl · chapter 2. plate tectonics 21 2.1 the observations that led...

The Dynamic Earth

English edition

‘The Dynamic Earth’ was developed as an Advanced Integrated Science module for Dutch upper secondary students (pre-university stream, vwo), within the course known as ‘Nature Life and Technology’ (NLT). In 2008, the module received certification from the national NLT Steering Commission.

The module was developed for Junior College Utrecht (www.uu.nl/jcu ) by a team headed by Dr. M.L. Kloosterboer – van Hoeve (module coordinator) with contributions from:

Utrecht University, Faculty of Geosciences o Prof. dr. R. Wortel o Dr. P. Meijer o Dr. H. Paulssen o Dr. H. de Bresser o Dr. M. van Bergen o Dr. A. van den Berg o T.S. van der Voort

Junior College Utrecht o Dr. M.L. Kloosterboer-van Hoeve (module coordinator) o Dr. A.E. van der Valk (curriculum coordinator) o K.J. Kieviet MSc (layout)

Partner schools of JCU: o Baarnsch Lyceum: Drs. W. Theulings and Drs. I. Rijnja o Revius Lyceum Doorn: Drs. J. Hillebrand o Leidsche Rijn College: Drs. P. Duifhuis and Drs. F. Valk o Goois Lyceum Bussum: mw. M. Raaijmaakers

Translation: ms. Anne Glerum and ms. Nienke Blom, with thanks to ms. Kate Smith for editing the English text. The translation was made possible by the PRIMAS Project (www.primas-project.eu and a SPRINT grant from the Faculty of Science and the Faculty of Geosciences, Utrecht University.

© 2011 JCU/Utrecht University.

For this module a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Netherlands License applies http://creativecommons.org/licenses/by-nc-sa/3.0/nl/deed.en

The authors’ copyrights remain with lie with Utrecht University and Junior College Utrecht, P.O. box 80 000, 3508TA Utrecht, The Netherlands. Altered versions of this module may only be distributed if the fact that it is an altered version is clearly stated, along with the name of the authors who made the modifications.

In the development of this material, the authors have used materials of others. If possible, the source of the materials is mentioned and a similar or more open license is applicable. If material has been used in which the source is mentioned incorrectly, please contact Junior College Utrecht ([email protected] ).

The module has been composed with care. Utrecht University and Junior College Utrecht do not accept any responsibility for any damage that originates from this module or the use of this module.

An electronic version of this module (pdf format) can be found on http://www.uu.nl/faculty/science/EN/vwo/juniorcollege. Teacher materials such as a teacher guide as well as the module in MSWord format can be requested at [email protected].

The Dynamic Earth Table of Contents

4

Table of Contents Table of Contents 4

Introduction to the English edition 6

Introduction 8

An interdisciplinary course 8

Four kinds of exercises 8

Glossary 8

Chapter 1. Basic Theory 10

1.1 Introduction 11

1.2 How to use maps and the atlas 11

1.3 Types of rock 11

1.4 Dating of rock 15

Chapter 2. Plate Tectonics 21

2.1 The observations that led to plate tectonics theory 22

2.2 Three steps to plate tectonics. 24

2.3 Structure of the Earth 31

2.4 The engine driving plate motion 33

2.5 Types of motion at plate boundaries 33

2.6 Intraplate motion 34

2.7 Describing plate motions 35

Chapter 3. Earthquakes and tsunamis 45

3.1 Earthquake waves 46

3.2 The relationship between plate tectonics and earthquakes. 51

3.3 The development of an earthquake 53

3.4 The motion of plates during an earthquake 56

3.5 The strength (magnitude) of an earthquake 60

3.6 The relation between earthquakes and tsunamis 62

Chapter 4. Volcanoes 69

4.1 Volcano types and occurrences 70

4.2 Formation of different types of igneous rocks 72

4.3 Magma formation 74

Table of Contents The Dynamic Earth

5

4.4 Volcanoes on Iceland 77

4.5 The quantity of gas emitted during the Laki eruption 79

4.6 Acid rain in Western Europe. 83

4.7 The consequences of acid rain 85

Chapter 5. Mountain building 96

5.1 The relation between mountain formation and plate tectonics 97

5.2 Information about orogenesis: hidden in rocks 99

5.3 The relation between force, stress and deformation. 104

5.4 Fast deformation (brittle behaviour) 110

5.6 Slow deformation (ductile behaviour) 114

5.5 What different types of deformation tell us about orogenesis. 119

Chapter 6. Convection: the Earth as an heat engine 125

6.1 Introduction 126

6.2 The interior structure and chemical composition of the Earth 126

6.3 A model for the Earth’s thermal state 133

6.4 The start of convection in a viscous medium 138

6.5 The internal temperature of the Earth 140

6.6 Liquid magma within solid mantle rock 144

6.7 Recent developments 144

6.8 What have we learnt? 146

Glossary 147

The Dynamic Earth Introduction to the English edition

6

Introduction to the English edition

Junior College Utrecht, JCU, developed the original Dutch version of this module in cooperation with the Department of Earth Sciences from the Faculty of Geoscience, Utrecht University (http://www.uu.nl/faculty/geosciences/en/facultystructure/departments). JCU offers an enriched science curriculum to select groups of talented and motivated secondary school students from the Utrecht region of the Netherlands. The module was tested on JCU students as well as on students from its partner schools and, as a result of evaluations, it has been revised several times.

This module can be chosen as a part of the new integrated, advanced science course Nature Life and Technology (NLT) from the Dutch upper secondary curriculum. NLT was been introduced in 2007. It aims to expose students to modern science to encourage them to consider science studies as part of their future secondary education. This module aims to orientate students towards geology, geophysics and geochemistry, subjects that traditionally get little attention in the Dutch upper secondary curriculum.

This module and the module ‘the Molecules of Life’ have been translated into English for two reasons:

- Many Dutch upper secondary schools have bilingual programs. With these English versions, the modules can be taught in the English part of those programs

- Teachers and scientists from countries other than the Netherlands are interested in NLT. Using these two modules as examples, they can inform themselves about the pedagogy and methods used for these subjects.

On http://www.uu.nl/faculty/science/EN/vwo/juniorcollege), the modules are available in pdf format on the Internet.

For many parts of the module ‘The Dynamic Earth’, the context is Dutch (e.g. assignment 2.1 “when did the Netherlands lie on the equator?”) Someone from abroad who wants to use (parts of) this module can adapt the context so that it is suitable for to his or her own location (under the conditions of the creative commons licence). For this, a Word version of the student material is available. In addition, teacher materials are available, including a teacher guide, outcomes of assignments and PowerPoint presentations. Send a request to [email protected].

I would like to thank Anne Glerum and Nienke Blom for translating the module, Kate Smith for editing the English text, the writing team for checking the chapters they have written, Krijn Kieviet and Saskia Klaasing for the layout of the module.

Please feel free to use this lesson material in your classes. We would be very happy to hear from you and to learn about your experience of this module!

Dr. Ton van der Valk

Curriculum Coordinator Junior College Utrecht, Utrecht University

[email protected]

Introduction to the English edition The Dynamic Earth

7

The Dynamic Earth Introduction

8

Introduction The Dynamic Earth is a course intended for the last two years of upper secondary school education (grades 11 and 12). It is centred on the theory of plate tectonics (see Figure I, front page). Chapter 1 will provide you with all the tools earth scientists use to study the Earth. It is essential you read this chapter first. Chapter 2 explains the theory of plate tectonics and Chapter 3 focuses on an important consequence of the dynamic Earth: earthquakes. The first three chapters combine to form the basis of this course. You will have completed the course only when you have finished these chapters and one of the optional chapters (Chapter 4, 5 or 6). The other optional chapters can be used as additional course material or as project material during other courses.

An interdisciplinary course So how does this course relate to your other classes? Studying the Earth is an interdisciplinary science: all the natural sciences are required to understand processes within and at the surface of the Earth. This course will build on theory from your Mathematics, Physics and Chemistry classes.

The table below shows where elements of each subject are studied during the course. The red thread, plate tectonics theory, has probably already been covered in your Geography class. However, if you did not take Geography, or geology has not yet been studied within your Geography class, you can still follow this course. The glossary in the back of the reader will help you with any unfamiliar terminology.

Four kinds of exercises The reader comprises four types of exercises:

1*: An exercise with one asterisk tests the knowledge that you have acquired in other courses. The exercises assume some familiarity with the subject.

2**: An exercise with two asterisks is directly related to the text preceding it.

3***: An exercise with three asterisks requires that you to apply the theory that has just been explained to other, similar cases. You will have to use what you have learned in your other courses as well as what you have just learned from this course. You should be able to apply your new knowledge to these cases.

4****: Four asterisks denote optional exercises which are more challenging.

Every chapter ends with a final exercise. Here you will revisit the main questions of the chapter and, in answering them, test what you have learned. Also, you will write down any new questions that have come up while studying the chapter. The final exercises can be used in the examination as well.

You can use atlases, the Internet and Google Earth to help you with the exercises. Where needed, they are denoted by A (atlas), I (Internet) or G (Google Earth).

Glossary Earth scientists use many technical terms and concepts. These are underlined in the text and explained in the glossary.

Introduction The Dynamic Earth

9

This course

Previous

subject matter

Basic theory Plate tectonics Earthquakes Volcanoes Mountain building

Structure of the Earth

Classes Math, Geography

Physics, Geography

Physics, Geography

Chemistry, Geography

Math, Geography Physics

Waves and vibrations

Earthquakes General structure

Magnetism Paleomagnetism

Plate motion reconstruction

Geomagnetic field

Radioactive decay

Dating techniques

Heat source Heat source in certain layers

Forces Isostasy, plate motion

Earthquakes Isostasy,

formation of faults

Heat and phase changes

Driving plate motion

Magma generation

Liquid and viscous layers

Math and geometry

Calculating plate motion

Calculations on faults

Chemistry

Average composition

Melt diagrams

Phase diagrams

Composition of the different

layers

Chemical equilibrium

Types of magma, types of

rock

The Dynamic Earth Basic Theory

10

Chapter 1. Basic Theory The main question of this chapter is:

How do we work with the basic tools of earth scientists: maps, rocks and dating methods?

This question is addressed by answering the following section questions:

How do we use maps and the atlas? (1.2) What types of rocks are recognized? (1.3) How do we date these rocks? (1.4)

Objective: To obtain the skills necessary for this course.

Basic Theory The Dynamic Earth

11

1.1 Introduction This course discusses movements within the Earth. Some simple tools that earth scientists use when studying these movements are:

Maps and atlases (1.2): To maintain an overview of the Earth, and to see where specific processes take place, we make use of maps, atlases and the Internet. Section 1.2 explains how to use these tools.

Rocks (1.3): Rocks contain a lot of information about the Earth. Section 1.3 discusses the three main types of rocks.

Dating of rocks (1.4): Plate movements are slow. In the earth sciences we do not consider timescales of 30, 100 or even a 1000 years, but timescales of millions or billions of years. How we date rocks using such timescales is explained in section 1.4.

1.2 How to use maps and the atlas

In order for you to study the moving Earth, figures and maps are included in this textbook. We make frequent use of the Dutch Grote Bosatlas (see e.g. http://en.wikipedia.org/wiki/Bosatlas). For example, the notation GB 174A refers to map 174A of the 53rd edition of the Grote Bosatlas. Google Earth is a useful tool for most exercises too. The capital letters A, I and G indicate which tool to use for each exercise: the Atlas, the Internet, Google Earth or a combination of these. Exercise 1-1*: Where do earthquakes and volcanic eruptions occur? A, I Keep track of the occurrence of earthquakes and volcanic eruptions, while working on this module. Draw their locations in on a map. Some of the other chapters will have exercises where you can re-use this map, for example Exercise 3-3 and exercise 4-1. a. Print out the blank map of the world and mark the places where you know earthquakes and

eruptions occur. b. Check for new earthquakes and eruptions in newspapers and at www.earthweek.com and other

similar websites. On your map, mark the location, magnitude, number of casualties and, when available, the actions and precautions taken to limit the damage done by these earthquakes and eruptions. The general term for such actions and precautions is hazard management.

Exercise 1-2*: Finding the locations of earthquakes, volcanoes and mountain chains with Google Earth G a. Look up the locations of the earthquakes and volcanic eruptions found in exercise 1-1 using

Google Earth. b. Use GoogleEarth for any exercise when you are adding data to your map, such as exercise 3-3.

1.3 Types of rock Much of what we know about processes within the Earth and on its surface comes from studying rocks. There are three main types of rocks: igneous, sedimentary and metamorphic.

1.3.1 Igneous rocks Igneous rocks are formed when liquid magma or lava solidifies. When this occurs close to or at the Earth’s surface, the rock formed is called volcanic or extrusive. When this process takes place much deeper within the Earth, intrusive rock is formed.

Unlike the deeper intrusive igneous rock, volcanic rock forms at the Earth’s surface when liquid magma cools quickly. This fast cooling means there is little time for crystal growth, resulting in fine-grained rocks with small crystals. A crystal is a homogeneous solid with smooth flat surfaces called faces. This regular morphology is the result of the regular arrangement of the atoms making up the crystal.


12

An example of a rock that has cooled quickly is basalt. It is a black rock that has very few visible crystals. Basalt often forms at the bottom of the ocean, where the heat dissipates quickly in the cold seawater. When molten material cools slowly there is time for a greater number of crystals to form. If there is enough space, these crystals will be larger too. An example of a rock that has cooled slowly is granite, an intrusive rock that solidifies deep beneath the Earth’s surface. Granite cools slowly, enabling it to form large crystals, but often there is not enough space for crystals to grow into perfect crystal shapes. Several different crystals can easily be recognized within granite: quartz (transparent), feldspar (pink) and biotite (black). Biotite crystallizes first, followed by feldspar and then quartz. Quartz grows into the space left over by the other crystals. It does not obtain the crystal shape it could grow into if more space were available.

The characteristics of igneous rocks depend on the circumstances under which they formed. The composition of the source material of the magma is also important. You will learn more about this in Sections 4.1 and 4.2. Possible characteristics of igneous rocks are

A low density, like pumice, or a high density, like basalt Large crystals, especially in intrusive rocks Can contain metal ores A smooth and glassy morphology, like volcanic glass; it solidified so quickly there was no

time for crystals to form. Basalt, granite, andesite, pumice and tuff are all examples of igneous rocks.

1.3.2 Sedimentary rocks Sedimentary rock forms after the transport and deposit (sedimentation) of loose material, the result of weathering and erosion. This loose material, or sediment, can be transported and deposited by wind (Aeolian sedimentation), oceans (marine sedimentation), rivers (fluvial sedimentation) or ice (glacial sedimentation). The sediment, for example, sand, is slowly buried under new layers of loose material. The pressure of the build-up of layers compresses the material creating rock. In our example sandstone is formed. Horizontal layering and fossils are often present in sedimentary rocks. Sediment deposited on the ocean floor is usually composed of calcium carbonate particles from marine organisms and some clay; this will become limestone. Shallower waters and rivers carry more sand; therefore this is where sandstone is formed. General characteristics of sedimentary rocks are

Individual grains can be distinguished, such as sandstone A coarse surface, like sandpaper Can contain fossils A dull, matte exterior Can contain layering Rocks can contain calcium carbonate that can be detected using hydrochloric acid (HCl): if

you put some drops of HCl on limestone, the rock will react and start to fizz. Examples of sedimentary rocks are sandstone, limestone, shale and lignite.

1.3.3 Metamorphic rocks Metamorphic rock is formed when igneous and sedimentary rocks are exposed to high pressure and temperature conditions. As a result of this metamorphism, the rock is very compact and

Figure 1.1: The Rock cycle, (igneous rocks), sedimentary-and metamorphic rocks. Source: commons.wikimedia.org.


13

additional layering can form. This layering will be perpendicular to the direction of the applied pressure. This is comparable to flattening a balloon with your hands: the balloon is elongated in the vertical direction (parallel to your hands) and shortened in the horizontal direction. New minerals form during metamorphism because of the high pressure and temperature. These give the rock a shiny appearance.

Metamorphism takes place deep within the Earth’s crust where both the temperature and pressure are high. However high temperatures alone can also cause metamorphism. Rocks exposed to high temperatures from hot magma intruding into the Earth’s crust become metamorphic. This is known as contact metamorphism.

General characteristics of metamorphic rocks are

Typical shimmer or shine Characteristic layering Separate grains cannot be distinguished any more.

Examples of metamorphic rocks are marble (formed from limestone), slate (formed from clay), quartzite (formed from sandstone) and granite gneiss (formed from granite).

Exercise 1-3***: Classifying rocks: what goes where? Your teacher will supply you with a set of rock samples. Number each of them and write down as many characteristics as you can think of, e.g. colour, density, layering, smoothness or coarseness and the presence of fossils. Separate the samples into igneous, sedimentary and metamorphic rocks and try to name each sample. You should at least be able to recognize a basalt, a granite and a sandstone sample.


14

Box: Carbon

Carbon is an abundant element in the Earth. Lignite and coal are good examples of rocks containing carbon. Both form under high pressure and temperature from organic matter, the remains of plants. As a mineral, carbon is found in the form of diamond and graphite (and it can have the shape of ‘Bucky ball’, C60, but this is not considered here because they are artificial, not natural).

The crystal structures of graphite and diamond are very different (see Figure 1.2). Diamond is formed under extreme pressure in the order of 109 Pascal (1 Giga Pascal or 1 GPa). This is four orders of magnitude greater than normal air pressure of 105 Pa (1 bar or 1 atmosphere). Within the three-dimensional crystal structure of diamond, each carbon atom is connected to four other atoms. This way, diamond forms one large molecule with a continuous and stable lattice that is equally strong in every direction and contains no weak points. Atoms are connected through strong covalent bonds. The melting point of diamond is therefore very high, 3600oC. Diamond is insoluble and very strong. All the electrons are used in forming bonds between atoms, which makes diamond an excellent insulator.

In graphite, atoms form hexagonal rings. These rings are connected in the horizontal plane, forming one large molecule. The bonds between different planes are weak; they are known as vanderWaals bonds. The distance between layers is more than twice as big as the distance between atoms in the same plane. When you use a pencil or put on mascara, carbon layers slide off leaving a mark or deposit because the weak bonds between the layers have been broken. Unlike diamond, graphite has many free electrons, which makes it a good conductor. Its density is only two thirds that of diamond. At a pressure of 10,000 atmospheres, graphite can be compressed into diamond.

Diamond is formed under high pressure, while graphite forms under lower pressure. This implies that at the low pressure of the Earth’s surface graphite is stable, while diamond is not! At the Earth’s surface diamond is described as ‘metastable’. This is comparable to super-cooled water: When liquid water is cooled carefully, it can reach a temperature of e.g. –5oC without solidifying and turning into ice. With super-cooled water, only a small amount of movement is needed to initiate solidification, but for diamond, the energy needed for this phase change is much larger, the reason why it is hard to change its metastable state.

Other minerals can also be stable or metastable. At the Earth’s surface – at a standard temperature of 20oC and pressure of 1 atmosphere - sand and clay minerals such as illite and kaolinite are stable, while ruby and emerald are metastable.

Figure 1.2: Carbon bonds of a) diamond and b) graphite.


15

1.4 Dating of rock We live in the Holocene, the youngest episode, or epoch, of Earth’s history. The Holocene is part of the Cenozoic, the youngest era (see Figure 1.3). The Precambrian spans the period starting 4.6 billion years ago with the formation of the Earth, and ending 542 million years ago. Little is known of this period, which is again divided into three parts (see Figure 1-3): the Hadean, the Archean and the Proterozoic. Only a few rocks dating back to these early times have been found and little remains of the primitive life that existed at this time.

After the Precambrian, the Phanerozoic began: the eon of life. The Phanerozoic is also divided into three parts: the Paleozoic, the Mesozoic and the Cenozoic (see Figure 1.3). The Paleozoic started 542 million years ago and comprises the Cambrian, Ordovician, Silurian, Devonian, Carboniferous and Permian periods. During the Carboniferous period, coal deposits formed in many countries; salt deposits formed during the Permian period. The Mesozoic, starting 251 million years ago, consists of the Triassic, Jurassic and Cretaceous. This was the time of ammonites and dinosaurs. Mammals evolved during the Cenozoic, which started 65 million years ago.

The Cenozoic is made up of the Tertiary and Quaternary periods, of which the Quaternary period is divided into the Pleistocene and the Holocene mentioned above. Alternating glacials (cold periods) and interglacials (warm periods) characterize the Quaternary. The Holocene corresponds to the current interglacial. Modern humans evolved approximately 200,000 years ago, during the Pleistocene. If we compare the lifetime of the Earth to a 12-hour clock where 00:00:00.00 is the time the Earth came into being, man’s arrival corresponds to 11:59:58.20 a.m., or 1.8 seconds before noon (see the 12-hour clock of Figure 1.3).

How did earth scientists come to such a timescale? The age of the Earth is determined with the help of two types of dating techniques: relative dating and absolute dating. Both are discussed below.

1.4.1 Relative dating Relative dating is based on the order of geological events. Relative dating uses e.g.:

The principle of superposition: the rock layer at the bottom of a sequence is the youngest Fossils, the remains of plants and animals in rocks: fossils often characterize a certain

period of time Paleomagnetic reversals: you will learn more about reversals in Chapter 2

Figure 1.3: The geological history of the earth shown in a 12 hour clock (source:www.vob-ond.be). The Precambrian is sub divided in the Hadean, the Archean andthe Proterozoic.

Hadean (time 2.05h)Archean (time 3.24 h)Proterozoic (time 5.06h)Paleozoic (time 46 min)Mesozoic (time 29 min)Cenozoic (time 10 min)


16

The astronomical timescale: this type of dating uses small changes in the orbit of the Earth around the Sun. Mathematician Milanković calculated these changes, which can be traced back in sequential layers of rock.

Exercise 1-4**: How much time is needed to deposit 300 meters of clay? Small clay sediments are usually deposited at a rate of about 1 centimetre per 1000 years. The Dutch subsurface contains a thick layer of clay – part of the Rupel Formation – that is in parts over 300 meters thick. a. Disregarding soil settling and drainage, how long would it take for 300m of clay to be deposited? b. If you take soil settling and drainage into consideration, would this deposition take more or less time? Hint: both settling and drainage result in denser material.

1.4.2 Absolute dating The real (numerical) age of a rock can be determined using absolute dating techniques. These techniques are based on the process of radioactive decay. Certain elements have isotopes that are unstable and start to decay after a given time. As they decay, these isotopes are converted into isotopes of another element and emit radiation.

With the help of experiments, the time it takes for half of the original amount of isotope to decay have been established for every radioactive isotope. This period of time is called the half-life of an isotope. The half-life does not depend on the original amount of isotopes: a tonne of radioactive material decays to half a tonne in the same amount of time as a milligram of the same material to half a milligram. The half-life is not effected by temperature, pressure, chemical reactions or other processes. Every unstable isotope has a different half-life.

If the half-life of an isotope is known and the amount of new isotope formed can be measured, the absolute age of a rock that contains both isotopes can be determined. For dating rocks older than 50,000 years, the potassium isotope 40K is often used. Rocks containing potassium are common; the potassium fraction is made up of 7% 41K, 0.012% 40K and 93% 39K. The former and latter are stable, but 40K is radioactive. On average, 11% of 40K decays to 40Ar (argon) by electron capture: an electron from the K-orbit is pulled into the nucleus. Together with a proton, it an extra neutron is formed. The half-life of the 40K isotope decaying to 40Ar is 1.31 · 109 years. So, after 1.31 billion years, half of the 40K isotopes have decayed to 40Ar.

To use this decay reaction for dating, you must find rocks that did not contain 40Ar when they first formed. Otherwise the original 40Ar influences the age calculation, resulting in an incorrect age determination. Volcanic rock is a good candidate: argon gas is completely expelled from molten lava. It is only when lava solidifies that 40Ar forms from the decay of 40K. This 40Ar is captured in the crystal lattices of the minerals containing potassium. As the 40Ar/40K ratio can be measured, the age of the volcanic rock can be determined.

Exercise 1-5***: K/Ar dating a. Why is the K/Ar dating method of little use when dating Pleistocene deposits? b. What are the two main controls for the reliability of K/Ar dating? c. Argon is a gas and can easily escape. What will be the resulting error in your dating due to

argon escape? Explain.

Other decay reactions used in absolute dating are rubidium/strontium (87Rb to 87Sr decay, half-life of 4.9 · 1010 years) and uranium/lead (238Ur to 206Pb and 235Ur to 207Pb decay, half-life of 4.47 · 109 years and 7.04 · 108 years respectively).

Exercise 1-6**: Discrepancy between different dating techniques A geologist discovers the fossils of a fish that is characteristic for the Devonian period. The fossils are found in slightly metamorphosed rock. The age determined with Ru/Sr dating is only 70 million years. How could you explain the discrepancy between relative and absolute dating?

14C dating is used for geologically young rock, that is, rock no older than 70,000 years. This method is based on the ratio of the amount of 14C (a radioactive isotope of carbon) compared to the amount of 12C in material containing carbon. Plants and animals have a constant amount of both isotopes due to their continuous carbon exchange with the atmosphere. 14C is formed as the result of the interaction of cosmic neutrons and 14N in the atmosphere. The ratio 12C to 14C is about 1 to 1.2 · 10-12, although this is not constant through time and the ratio must be corrected for this


17

variation. When plants and animals die, the exchange of gases with the atmosphere ceases and the amount of 14C decreases due to radioactive decay. The half-life of 14C is 5730 ± 40 years. Since the atmospheric isotope ratio 12C/14C is known, the age of organic material can be determined from the ratio of carbon isotopes present in a sample. Because the half-life of 14C is relatively short, after a certain amount of time there is too little 14C to detect, hence the maximum determinable age of 70,000 years. Carbon dating can only be used on rocks and deposits containing organic material (e.g. wood, charcoal, seeds, nuts, bones and shells) and peat.

1.4.3 Radioactive decay and age calculations Radioactive dating

Scientists use radioactive carbon (14C) to date biological remains because the amount of 14C in the atmosphere has been more or less constant throughout history. As plants incorporate 14C and 12C through photosynthesis and animals through eating plants, the amount of 14C within living material is in equilibrium with the atmosphere. However, when a plant or animal dies, it no longer absorbs 14C. As 14C decays and its concentration decreases, the 12C/14C ratio within the plant or animal changes. Therefore the quantity of 14C found in biological remains depends on the length of time they have been buried.

Simple 14C-dating calculations

The factor by which the amount of 14C in organic material has decreased is used to determine the time passed since decay set in. This factor is easily established when understanding decay through time.

The half-life is symbolized by t½. Every period of time of t½ = 5730 years results in a decay factor ½. After 5730 years the original amount of 14C should thus be multiplied with ½, in which case only half of the amount remains. As one half-life has passed, only half of the original 14C isotopes are left.

So After a period of 5730 years the decay is (½)

After a period of 3 x 5730 years the decay is (1½) x (½) x (½) = (½)3

In general, after a period of n x 5730 years, decay is (½)n.

This also applies to non-integer values of n. For example, if n = ¼, after a period of ¼ x 5730 years the decay should be (½)¼ because four of those periods will give a decay of ½.

The above can be summarized in the following equation:

N(t) = N(0)*( ½)n

Here N(t) is the amount of decaying isotope left at time t, while N(0) is the original amount of that isotope. The fraction ½ indicates halving the original amount n times, and n corresponds to the n from the previous example, the number of half-lives. If n = 3, three times 5730 years have passed.

What percentage of the original amount of radioactive isotope remains after n = 3? Take N(0) = 1 and n = 3, this will give N(t) = 1*(½)3. Thus N(t) = 0.125, or 12.5%.

Example

In a peat sample, only 10% of the original amount of 14C is left. How old is the sample?

Suppose the sample is n half-lives old. As 100% = 1, 10% = 0.1. N(0) is therefore 1 and can be left out of the equation. What remains is:

1.0)2/1( n

You can find n by taking the logarithm of both sides and then using logarithmic identities.

)1.0log()2/1log( n

The age of the organic material is then n times the half-life of 5730 years, in this case 19,000 years.


18

General equation for radioactive decay (here we will use exponential functions)

N(0) is the original amount of radioactive isotope; N(t) is the amount of isotope after time t; t½ is

the half-life of the isotope; and n is the number of passed half-lives, n = 1 2

t

t . If, for example,

three times 5730 years have passed and the half-life is 5730 years, then for this t = 17,190 years and t½ = 5730 years, n = 3.

This gives:

1 2

t

t1 / 2( ) (0)N t N

You know ½ = e-ln2; you can check this on your calculator. This allows you to express the formula above as an exponential function. Usually not the half-life, but the decay constant λ is used in such a formulation:

1/ 2

ln 2

t

Therefore, the general equation for radioactive decay is written as:

( ) (0) tN t N e

Exercise 1-7***: Calculations on radioactive decay a. Derive the last equation yourself. Differentiate it to obtain an equation for the rate of change of

amount of isotope, N’(t). b. Show N’(t)/N(t) = -λ by using substitution. c. Describe the meaning of λ: λ is the ratio between…. Hint: Look at the equation in b and the

meaning of the derivative of N(t).

In addition, you can now do the optional exercise 1-1 you can find at the end of this chapter.

1.4.4 The dating of rock and the geological timescale All absolute and relative dating techniques have the same limitation: the older the tested material, the less accurate the results. Therefore as many different techniques as possible are applied to the same rock. Unfortunately only certain types of rocks are appropriate for a certain techniques: fossils are only found in sedimentary rock, 40K dating can only be done on igneous rock and 14C dating only on organic material.

Absolute and relative dating enables scientists to establish extensive and detailed geological timescales. The most commonly used timescale is from the International Commission on Stratigraphy (see www.stratigraphy.org). Simple representations of this timescale can be found in Figure 1.4 and in the atlas, map GB 192E.


19

Exercise 1-8*: Mass extinctions and the geological timescale Several times during the geological past, many (sometimes over 70%) of the species present became extinct within a short period of time. These events are called mass extinctions. The four most important mass extinctions occurred 444 million years ago, 416 million years ago, 251 million years ago and 65 million years ago. a. In what periods of the timescale of Figure 1-4 did these events occur? b. What is it about the timescale that makes these extinctions so obvious?

Final exercise Ch1. Answer the section questions and the main question a. Answer the three section questions as well as the main question from the beginning of this

chapter. b. If you find you have new questions after reading this chapter, write them down.

Optional exercise 1-9****: Determining the age of ice

(Based on a question in the Dutch final exam Physics, havo 1998-I. Question a, b and c lie outside the scope of this course, but are essential to the question as a whole.) “Greenland. A group of scientists are investigating the ice at the North Pole. They drill a hole in the ice over 3 km deep, retrieving ice cores of 2.5 m in length and 84 cm2 in cross section. By investigating the ice cores, knowledge of the past - for example of the average temperature on Earth during the formation of the ice - is gained. Past temperatures are known from the concentration of two particular oxygen isotopes. The age of the ice can be obtained by dating ash layers within the ice with C-14 dating. The concentration of C-14 in the ashes is compared to the normal concentration of C-14 in the atmosphere.”

Figure 1.4: Geological timescale. The top of the first column corresponds to the entire second columnand the top of the second column corresponds to the third column. Source:http://www.geo.ucalgary.ca/~macrae/timescale/timescale.


20

After: Mens en Wetenschap, July 1994 The temperature of an ice core is -4°C. a. Calculate the mass of the ice core. Hint: You can look up the density of ice, and calculate the

volume of the core. Oxygen isotopes are mentioned in the article. b. Name one difference and one similarity on a molecular level between the two different oxygen isotopes. C-14 dating is based on the decay of the radioactive isotope 14C. c. Write down the decay reaction of 14C. With time, the concentration of 14C in the ashes found in the ice cores decreases due to radioactive decay. Measurements show that the present concentration of 14C in the ashes is 25% of the normal concentration in the atmosphere. d. Calculate the age of the ashes.

Plate Tectonics The Dynamic Earth

21

Chapter 2. Plate Tectonics The main question of this chapter is:

Which theory lies behind the concept of plate tectonics, what drives plate tectonics and to what extent do we notice plate motion in daily life?


Which observations advanced the theory of plate tectonics? (2.1) Which three steps led to the theory of plate tectonics? (2.2) What does the interior structure of the Earth look like? (2.3) What causes plates to move? (2.4) What types of motion occur along plate boundaries? (2.5) Are movements only noted along plate boundaries? (2.6) How do we describe plate motion? (2.7)

Objective: To describe the current distribution and motion of plates by making use of the theoretical knowledge you gained from Mathematics and Physics classes.

The Dynamic Earth Plate Tectonics

22

2.1 The observations that led to plate tectonics theory Consider capturing the Earth on film from a satellite: in a day several things would clearly change. For example: different places would experience day or night, cloud coverage or perhaps tornados. But the spacing of the continents would not appear to change. However, if you were to continue recording the Earth for, say, a century, you would see even the continents move about. Continents do not move fast, they move at about the same speed as your fingernails grow.

Exercise 2-1*: When did The Netherlands lie at the equator? Your fingernails grow at about 2 to 3 mm per month, knowing this a. How long would it take for a part of a continent to move from the equator to the current

position of The Netherlands? From Section 1.3.3 ‘Metamorphic rocks’, you know that extensive coal deposits occur in The Netherlands. Since coal forms from tropical swamp vegetation, the deposits imply that The Netherlands used to lie at the equator. b. Assuming plates move with a speed of 2 to 3 mm per month, how old are the Dutch coal

deposits and during which geological period did they form? c. Research shows that coal in The Netherlands formed during the Carboniferous period (named

after the coal, which is typical for sediments of this period). Is this dating consistent with question b? If not, how do you explain the discrepancy?

How long have we known that parts of the Earth’s crust move? Before plate tectonics theory, scientists assumed only vertical crustal movements could take place. Their assumptions were based on the observation that the surface of the Earth is divided into oceans and continents. Both are relatively flat, but they are significantly offset in elevation. The bottom of the ocean is about 3 km deep, while the continents are 1 to 2 km high. Weathering and erosion should have evened out this difference in elevation by now. As the difference still exists, scientists thought that vertical movements must have preserved it. They expected the lighter continental rock to rise higher and higher with time, while the heavier oceanic rocks were supposed to sink.

However, some scientists suggested that horizontal movements occurred in addition to vertical motion. Explorers in the 15th and 16th century, trying to map the Earth, noticed something peculiar about the coastline of the Atlantic Ocean: the east coast of North and South America looked like a reverse cut-out of the west coast of Africa and Europe. They surmised that America and Africa were connected in the past. From 1900, more and more observations were made that supported this connection:

The same fossil species occurred on both continents (Figure 2.1) These organisms could not have crossed today’s vast oceans, so the continents must have been connected some time in the past.

There were traces of large ice sheets covering several continents. Such vast glaciations could only have occurred if all the ice-covered parts were situated at the pole at the same time.

The same rock assemblages were found on both sides of the Atlantic. Most likely, they formed in one place before the continents split up.

It was Alfred Wegener, a German meteorologist, who greatly improved our understanding of plate motion. He wrote the book The Origin of the Continents and Oceans (1925), which formed the basis of the modern view on plate motion. He suggested that all continents were connected in the past, forming the supercontinent Pangaea, but then drifted apart during the Tertiary period. Wegener called his idea the continental drift hypothesis.


23

Exercise 2-2**: Wegener’s arguments for continental drift. A The fact that the continents fit together almost perfectly, and other observations such as the distribution of fossils and the large 250-million-year-old ice sheets, fuelled Wegener’s ideas about plate motion. a. Use the atlas to name the geological period during which the large ice sheets formed. b. Figure 2.2 shows where Wegener found traces of ice covering the continents in the past. What

two conclusions can you draw based on Figure 2.2 about the previous connection between the ice-covered continents and their location?

Wegener’s hypothesis left several issues unexplained:

Which part of the crust actually moves? The continents themselves, the oceans or something deeper?

What mechanism is strong enough to cause crustal movements?

It wasn’t until 1960 that science made it possible to resolve these issues.

Exercise 2-3**: Questioning Wegener’s continental drift hypothesis a. Think of a mechanism that could explain continents drifting apart. b. Write down some questions you would ask Wegener to decide whether or not you agree with

his theory.

Figure 2.1: Because the same fossil species occurs on different continents, it is evident that the continents were juxtaposed at some time in the past. (Source: http://www.daaromevolutie.net/images)

Figure 2.2: Distribution of traces of glacial cover in 250-million-year-old sediments. (Veenvliet, 1986)


24

In Geography classes nowadays, you can learn that the continental and oceanic plates on the Earth’s surface move with respect to each other and that the interior of the Earth moves as well. These ideas form the paradigm of plate tectonics, which has been the leading paradigm in the earth sciences since the late 1960s. A paradigm is a coherent set of models and theories providing a certain mindset with which to analyse reality.

However, not everything about plate tectonics is known yet. Much uncertainty still exists about the processes active within the interior of the Earth that drive plate tectonics.

The steps needed to get from Wegener’s continental drift hypothesis to the paradigm of plate tectonics are explained in the next section.

Plate tectonics considers the movement of plates. Plates are composed of continental and/or oceanic crust and the upper part of the mantle. You will learn more about the structure of plates in Section 2.3.

2.2 Three steps to plate tectonics. 2.2.1 Determining the oceans’ bathymetry: Where does new material form? The first studies that provided insight into what causes the Earth to move came from a record of the bathymetry (relief) of the ocean floors. Mid-twentieth century equipment allowed for even the deepest parts of the oceans to be mapped. These measurements showed that in the middle of oceans there were shallower areas. These areas formed ridges over the whole length of the oceans (see Figure 2.3). Further research demonstrated these Mid-Ocean Ridges (MORs) were made of active underwater volcanoes. Researchers concluded that it is at MORs where new material rises up from the depths.

Figure 2.3: Topography and bathymetry of the Earth.


25

Exercise 2-4*: Topography of the ocean floors A Recife in Brazil and Mount Cameroon in Africa lie more or less on the same latitude at both sides of the Atlantic Ocean. a. Draw a depth profile starting from Recife to Mount Cameroon. Ascension’s highest volcano

reaches up to 859 m. Use the atlas. b. What is the difference in elevation between the highest and lowest point of this profile?

2.2.2 Paleomagnetism: How do we prove plates move? Scientists wondered if the formation of new material at MORs could have anything to do with the plates drifting apart. If so, the crust on both sides of a MOR should increase in age the further it gets from the ridge. Paleomagnetism, i.e. the rock record of the past magnetic fields of the Earth (paleo is the Greek word for past), enables the dating of the crust.

2.2.3 The Earth’s magnetic field “A wonder of such nature I experienced as a child of 4 or 5 years, when my father showed me a compass. That this needle behaved in such a determined way did not at all fit into the nature of events which could find a place in the unconscious world of concepts. I can still remember - or at least believe I can remember - that this experience made a deep and lasting impression upon me. Something deeply hidden had to be behind things ...”

Albert Einstein

The outer core of our planet consists of liquid iron and nickel (see Chapter 6), and acts as a large magnet with north and south magnetic poles. The Earth’s magnetic field resembles that of a bar magnet positioned within the Earth (see Figure 2.4: The Earth as a large bar magnet. (Source: http://stargazers.gsfc.nasa.gov)).

Physics class taught you how to work with magnetism. You will now use that knowledge to measure the magnetic field in the place where you live (Exercise 2-5). To practise working with magnetism, you can also complete Optional Exercise 2-1 (at the end of Chapter 2).

Figure 2.4: The Earth as a large bar magnet. (Source: http://stargazers.gsfc.nasa.gov)


26

Exercise 2-5**: Determining a magnetic field (including a practical) You will need: coiled wire, a magnetic needle and a variable power source. In this practical we will determine the magnitude and direction of the Earth's magnetic field strength in your hometown. This is called the magnetic induction B, being a vector i.e. having a magnitude and a direction. 1) Take a coil with loops that weave through a sheet of plastic and position it horizontally. The coil needs to be about 25 cm long with 40 loops. 2) Place a magnetic needle within the coil. Make sure the magnet can spin freely in the horizontal plane. 3) Align the middle line of the coil with the magnetic needle (see Figure 2.5) Connect the coil to the power source in series with a variable resistor and an electric current meter. Varying the strength of the current will show that at an electrical current strength of 90 mA the needle does not maintain a fixed position, but will spin freely.

a. Calculate the magnetic induction B generated by the coil. Use L

INBcoil

0 with B the

magnetic induction, μ0 the permeability of free space of 4π · 10-7, N the number of loops on the coil, I the current strength and L the length of the coil. b. Why does the needle not maintain a fixed position at 90 mA? c. Copy Figure 2.6 and mark the direction of the current in the coil. The dark part of the compass needle in Figure 2.6 denotes the north magnetic pole. Take a magnetic needle that can spin freely in the vertical plane (Figure 2.7). d. Measure the angle between the needle and the horizontal plane. e. Use this angle to calculate the direction and magnitude of the magnetic induction of the Earth's magnetic field (Btot). Hint: Use the cosine of the angle found and study Figure 2.7.

Figure 2.6

Figure 2.7 Figure 2.5

Heart line


27

The magnetic poles of the Earth’s magnetic field lie close to the geographic poles. The geographic poles, at 90° north and south of the equator, are the two locations where the Earth’s rotation axis meets the surface. The location of the magnetic poles varies slightly on a yearly basis (figure 2.8), but these variations are not big enough to influence, for example, the calculations you did to discover the magnetic field for your hometown.

Besides small yearly variations, the magnetic poles also experience variation on a larger scale: every ten to one hundred thousand years the poles reverse. During such a reversal the north geographical pole changes from south magnetic pole to north magnetic pole or vice versa (Figure 2.9). Scientists do not know why magnetic reversal occurs; most likely it is related to processes within the Earth’s core and mantle. Because we do not know why reversals occur, it is not possible to predict the next reversal.

Rocks contain a record of magnetic reversals: the magnetic minerals locked inside them are aligned with the direction of the ambient geomagnetic field at the time the rock solidified or was deposited. This is called paleomagnetism. Earth scientists can read these directions and are able to determine whether the rocks were formed during a period of normal (as it is now) or reversed (opposite to now) polarity.

Geomagnetic reversals, which occur either side of periods of normal or reversed polarity, can be dated using techniques such as radioactive dating or by referring to fossils. With the age of reversals known, a paleomagnetic timescale can be computed (see Figure 2.10).

Figure 2.9: Reversals of the Earth’s magnetic field. In the present situation (normal), the north geographic pole coincides with the south magnetic pole.

Figure 2.8: The movement of the magnetic north pole.(Source: http://www.digischool.nl/ak/)


28

Figure 2.10: Paleomagnetic reversals during the geological past. Black represents a period of normal polarity, white a period of reversed polarity. (Source: kennislink)


29

In 1950, ships were fitted with equipment especially designed for taking paleomagnetic measurements of the ocean floor. Measurements taken with this equipment showed which areas of oceanic plates had normal and which had reversed polarity; the recorded reversals showed a symmetrical pattern perpendicular to the MORs (see Figure 2.11).

To summarize: first, active volcanoes were found at MORs. Secondly, a symmetric pattern was discovered in the paleomagnetic signal of the ocean floor. By combining these observations, scientists reached the conclusion that new ocean floor is formed at a MOR (the ridge axis in Figure 2.11. New ocean floor consists of basaltic magma that rises up through the volcanoes (more information on basaltic magma can be found in Chapter 4). The moment basaltic lava solidifies; it records the magnetic polarity of that particular time. Reversal of polarities creates a symmetrical pattern around the MOR. New ocean floor is added on both sides of the MOR, while older parts migrate away from the ridge. This process is called sea-floor spreading (see Figure 2.12)

Figure 2.11: The black and white magnetic pattern surrounding theMid Atlantic Ridge, a Mid-Ocean Ridge (MOR), is symmetrical aroundthe ridge. (Source: http://www.cliffshade.com)

Figure 2.12: The formation of new oceanic crust,and sea-floor spreading; the sideways migration of newly formed and older crust.


30

Figure 2.13 is the result of paleomagnetic dating of the ocean floor.

2.2.4 Deep trenches: where oceanic crust disappears beneath continental crust We now know that new crust is constantly being formed (step a) and that the Earth’s plates move (step b). Does this mean that the total amount of crust keeps increasing? This would contradict the law of conservation of mass. Obviously, somewhere crust is being destroyed.

Ocean bathymetry does not only show ridges in the middle of the oceans, but also very deep trenches along some of the continents. These deep trenches are called subduction zones (Figure 2.14). They are areas where oceanic crust disappears into the Earth. Because of subduction, the amount of crust remains constant and hardly any oceanic crust older than 200 million years has been found (see Figure 2.13).

Combining steps a (bathymetry), b (paleomagnetism) and c (subduction zones or trenches), it can be concluded that

New crust is formed at MORs (step a and Ib) Plates move away from each other at MORs (step b) Crust disappears in subduction zones (step c)

These processes are summarized in Figure 2.14.

The paradigm, or theory, of plate tectonics combines the continental drift hypothesis and sea-floor spreading, the mechanism of formation and destruction of oceanic crust.

Figure 2.13: The age of the oceanic crust. (from: Berendsen, Fysisch geografisch onderzoek)

The age of the Ocean crust

A

B B

A


31

Exercise 2-6**: Plate movements A Study Figure 2.13, and use the atlas. The part of oceanic crust formed, say, during the Early Tertiary is larger in the Pacific Ocean than in the Atlantic Ocean. a. Find a possible cause for this. b. Where can you find the thickest sequence of oceanic sediments, at A or B of Figure 2.13?

Explain. c. Explain why the oceanic crust is no older than the Jurassic period. d. Knowing America and Africa started drifting apart during the Jurassic, calculate the average

plate motion velocity per year. Work out the velocity for each plate with respect to the ridge as well as the velocity of the plates with respect to each other.

2.3 Structure of the Earth So how do plates move? Do they float like rigid plates over a liquid interior? This used to be the generally accepted view because liquid lava from the interior of the Earth was seen to extrude during volcanic eruptions. However, the molten mass (called magma as long as it remains within the Earth and lava upon reaching the surface) exists only at a few places within the Earth. For you to understand plate movements, you need to understand the structure of the Earth first. It is discussed in this section, and more thoroughly in Chapter 6. Chapter 6 also explains how we deduced the structure of the Earth.

Research shows that the Earth is not a homogeneous sphere, but consists of several shells of different material and characteristics. From centre to surface, the Earth is divided into an inner and an outer core, a thick mantle consisting of a lower and upper mantle, and a relatively thin outer layer, the crust (see Figure 2.15). Early plate tectonics theory stated that the solid crust moved over the liquid mantle. However, it is the crust together with the upper cold part of the mantle that forms the moving plate - the lithosphere. The lithosphere is about 80 to 100 km thick. Lithospheric plates move on top of the asthenosphere, the viscous, but certainly not liquid, part of the mantle that extends to a depth of 300 km.

II III I II I

Figure 2.14: A cross section of the Earth. You can see Mid-Ocean Ridges (I) where crust is being formed, moving plates, and subduction zones (II) where crust disappears. Plate boundaries are divergent (I), convergent (II) and transform (III) (see Section 2-5). (Source: www.usgs.com)


32

Nine large and about twenty smaller tectonic plates make up the lithosphere (Figure 2.15). Some of the lithosphere forms the ocean floor and is therefore called oceanic plate. Other lithosphere forms the continents and is called continental plate. The movement of Earth’s lithospheric plates over the asthenosphere is called plate tectonics.

The composition and thickness of oceanic and continental plates are strikingly different. The crust of oceanic plates is made up of basalt, while continental crust is mainly composed of granite and sediments. More information on the formation of basalt and granite can be found in Chapter 4.

Figure 2.15: A cross section of the Earth. (Source: mediatheek.thinkquest.nl/~ll125/nl/struct_nl.htm)

Figure 2.16: Configuration of the tectonic plates. By zooming in on, for example, the Mediterranean Sea or Indonesia, even smaller plates can be seen. (Figure 4.3 of Marshak, Earth; Portrait of a Planet, 2nd edition, W.W. Norton & Co, 2005)


33

2.4 The engine driving plate motion What processes initiate plate motion? Deep within the Earth temperatures are much higher than at the surface. This temperature difference causes the viscous material of the mantle to flow very slowly. This type of flow is called convective flow. Figure 2.17 shows an example of convection. Mantle convection can be compared to water in a pot flowing because it is heated from below. However, in the mantle, viscous solid material flows instead of water, therefore mantle convection is a much slower process.

Convective flow plays an important role in the formation of new crust at a MOR, in moving tectonic plates, and during subduction, as summarized in Figure 2.17. Convection and the related processes at the surface of the Earth are investigated further in Chapter 6.

2.5 Types of motion at plate boundaries

We know that plates move and what drives this plate motion, but what do we notice of plate tectonics ourselves? The centre of a tectonic plate is relatively stable. Along plate boundaries, the movement of a plate is obvious. Three types of movements are seen: plates moving apart from each other, plates moving towards each other and plates sliding sideways past each other (see Figure 2.14).

2.5.1 Divergent plate motion: two plates moving apart from each other Examples of divergent plate motion can be found in the middle of several oceans, at the oceanic spreading ridges (MORs). When two plates move apart at these undersea mountain ridges, the newly formed empty space is filled with mantle magma rising from below. The magma cools at the ridge and new lithosphere is created. The MOR in the Atlantic Ocean is an example; this is where

Figure 2.17: Convective flow in apot of water. The flamerepresents the heat coming fromthe interior of the Earth, while thewater represents the viscousmantle.

Figure 2.18: Schematic drawing of a cross section of the Earth illustrating convective flow patterns. Convective flow can be related to processes at the Earth’s surface. (Source: http://pubs.usgs.gov/)


34

Africa and America move apart. Iceland is situated on top of a MOR (and on top of a hotspot, see Chapter 4); here, you can stand on the North American Plate with one foot, and on the Eurasian Plate with the other. Volcanoes and small earthquakes (due to the motion of MOR segments) can occur along divergent plate boundaries.

2.5.2 Convergent plate motion: two plates moving towards each other Since new lithosphere is created at MORs and the overall amount of crust does not change, old lithosphere must disappear as well: this happens when two plates move towards each other. This is known as convergent motion. Based on the kind of the plates involved, three types of convergent motion are recognized:

1. When a continental and an oceanic plate meet, the heavier oceanic plate, made up of mostly basalt, will dive below the relatively light continental plate in a process called subduction. For example, to the west of South America, the oceanic Nazca Plate subducts beneath the South American Plate.

2. When two continental plates approach each other, a mountain chain (e.g. The Himalayas) is formed.

3. When two oceanic plates move towards each other, the oldest, and therefore the coldest and heaviest, will subduct. For example, this happens in the ocean near Japan.

On or near convergent plate boundaries, earthquakes, mountain chains and volcanoes can occur.

2.5.3 Transform plate motion: two plates sliding past each other The San Andreas Fault in the state of California is a famous example of a transform fault on land. However, most transform faults are located under water, linking spreading ridge segments (the stepped geometry of MORs is used in Exercises 2-18 and 2-19). When two plates move past each other, earthquakes often take place.

It is not hard to recognize plate boundaries from topography and bathymetry when you are familiar with their morphology: Spreading ridges form long mountain chains on the ocean floor. Subduction zones are characterized by a deep-ocean trench and a mountain chain on the upper continental plate. The stepping of MORs indicates the presence of transform faults (see Figure 2.27).

Exercise 2-7**: Types of plate boundaries A, G Answer questions a-d for each of the following places: Jan Mayen; Valparaiso; Jakarta; Anchorage; and the Galapagos Islands. a. Which type of plate boundary do you expect to find here? b. Upon what data from the atlas are your expectations based? c. What geological phenomena, related to the type of plate boundary, do you expect to occur

here? d. Use Google Earth to check if the phenomena you listed at c actually occur.

2.6 Intraplate motion Plate tectonics theory states that tectonic plates move over the asthenosphere as rigid pieces, with deformational processes such as earthquakes, volcanoes and mountain building concentrated on the plate boundaries. However, when looking at the distribution of earthquakes on Earth (see atlas) you can see that some earthquakes occur far from any plate boundary, in the interior of the mostly continental plates. Examples of intraplate deformation areas are the earthquake area in Limburg, The Netherlands (see exercise 3-1), and a large area in East Africa. Nowadays we can map the behaviour of intraplate deformation zones precisely through satellite positioning (see Figure 2.19).

The new observations of intraplate deformation do not mean plate tectonics theory is no longer valid; rather plate tectonics describes a large part of the behaviour of the outer layer of the Earth, with the zones of intraplate deformation acting as exceptions to the rule.


35

Exercise 2-8**: Deformation zones A Figure 2.19 shows a large zone of intraplate deformation in Africa. a. Which African countries does this zone cover? Use the atlas. Figure 2.19 also shows that the African zone of deformation changes into a plate boundary towards the north. b. Which type of plate boundary does the deformation zone change into and what kinds of processes take place here? c. What is a possible scenario for the geological future of East and West Africa? Explain. d. Would such a scenario apply to The Netherlands and Germany as well? Use Figure 2.19 and GB 76.

2.7 Describing plate motions Since the acceptance of the theory of plate tectonics, scientists have often used it to calculate, for example, in what direction and with what velocity plates move with respect to each other or for how long certain plates (e.g. Africa and America) have been drifting apart. Besides describing motions of the past, future plate motions are calculated as well. Knowing plate velocities allows scientists to better understand the consequences of plate motion, such as earthquakes.

This section explores the movement of the African Plate with respect to the American Plate. In Exercise 2-6d you calculated the average velocity with which Africa and America diverge, about 1.08 cm per year, assuming the plates move on a flat plane. However, to be able to calculate the exact movement of Africa with respect to America, we must consider motion on a

Figure 2.19: Intraplate deformation zones mapped with the help of satellites. The colours indicate the amount of stress: blue indicates low stress, red high stress. (Source: http://gsrm.unavco.org/model/images/1.2/global_sec_invariant_wh.gif)

A

B C

D

A’

B’

C’

D’

Figure 2.20: a displaced square


36

sphere. Every displacement on a sphere can be seen as a rotation around an axis through the centre of that sphere. This will be shown in the next section as well as how to find this axis of rotation.

2.7.1 Rotation poles Plate motion can be described using rotation poles. Below, you will study the geometric concept of rotation poles. We will start off with a simple example of a square moving over a flat surface.

Exercise 2-9*: A rotated square Figure 2.20 shows a square moved from position ABCD to position A’B’C’D’. a. Cut out a square of the same size and place it on top of position ABCD. b. Putting the tip of your compass or pencil on the square allows you to rotate it easily. Try it. c. Rotating the square to position A’B’C’D’ requires putting the compass tip at exactly the right

spot. Try to find that spot and mark it P with a sharp pencil on the loose square as well as on this page. By experimenting, you have shown that pivot point P exists. P is the rotation pole of the square.

The next steps will help you find the rotation pole of a motion quickly and accurately.

Exercise 2-10**: How to find rotation pole P of a displaced square Use Figure 2.20 a. If P is the rotation pole, why should the distance between P and A equal the distance between

P and A’? b. Draw the perpendicular bisector of line segment AA’ on Figure 2.20. Why should the bisector go

through P? Hint: a perpendicular bisector cuts a line into two segments of equal length, at an angle of 90˚ to the line.

c. Also draw in the perpendicular bisector of line segment BB’. d. Now you should be able to determine the position of P exactly. Indicate its location on Figure 2-

20. e. Show that the perpendicular bisectors of lines CC’ and DD’ also go through P. Explain. To summarize: moving the square from ABCD to A’B’C’D’ equals rotating the square about rotation pole P. f. What does this say about angles APA', BPB', CPC' and DPD'?

Exercise 2-11**: Instantaneous rotation Figure 2.21 shows a moving triangle. The direction and speed of points D and E of the triangle are given by vectors d and e, respectively. Think of this motion as a displacement of the triangle to D’E’F’, with the distance travelled to D’E’F’ invisibly small. This way the perpendicular bisector of DD’ coincides with the line starting from point D, perpendicular to vector d. Also, the perpendicular bisector of line EE’ coincides with the line perpendicular to e in point E. a. Take the same steps as in Exercise 2-10 to determine the rotation pole Q. You will find Q at the

crossing point of two lines perpendicular to the vectors d and e. b. Now you can construct the direction of vector f of point F because the line QF, is perpendicular

bisector of FF’, going through Q, is orthogonal to vector f. Draw in the direction in which point F moves.

c. The lengths of the vectors d and e should have the same ratio as the lines DQ and EQ. Why? d. Are the ratios d:e and DQ:EQ equal? Calculate the length of vector f, in relation to the length

of d or e, and draw in f with its correct length.

D

E

F

d

e

Figure 2.21:


37

Exercise 2-11 shows that a starting movement of which you only know the velocity can also be seen as a rotation. When the triangle eventually starts moving, a rotation pole exists at each moment in time. The pole will not remain in the same location however, except when the movement is an ongoing rotation around that point. In general, the rotation pole only exists in a certain location for an instant of time. Thus, we speak of an instantaneous rotation and an instantaneous rotation pole. These notions are useful when describing plate motions on the Earth.

First, repeat the steps you took in exercise 2-11 and exercise 2-11 to determine a rotation pole and any unknown vector.

Step 1: Vectors

Find the vectors that represent the movement of several points of the rotating object. The vectors can either show the displacement of a point to get to its present position (exercise 2-11) or they show the velocity (direction and magnitude) with which a point is moving (exercise 2-11 and exercise 2-14a).

Step 2:

In the displacement-case, determine the perpendicular bisectors of the displacement; these are lines perpendicular to the displacement vectors. In the velocity-case, draw lines perpendicular to the velocity vectors, starting at the tail of the vectors.

Step 3: Rotation pole or pivot point

The point of intersection of the different perpendicular lines is the rotation pole.

Step 4: Unknown vector (optional)

In a case where the rotation pole is known and the travelled distance can be neglected, you can construct any unknown vector (exercise 2-11b-d). The wanted vector is perpendicular to the line running from the rotation pole to the point the vector acts on.

Exercise 2-12**: Can you always find a rotation pole? a. Think of a composition of two equally sized squares ABCD and A’B’C’D’ (not in the same

position) that does not have a rotation pole. b. Draw a triangle DEF with vectors d and e acting on points D and E that does not have a

instantaneous rotation pole.

Exercise 2-12 shows that movements on a plane sometimes do not have a rotation pole. In this case, the perpendicular bisectors of AA’ and BB’ or the perpendiculars of the vectors are parallel and do not cross each other; the movement is a parallel displacement.

2.7.2 Motion on a sphere In the examples above we used geometric figures: squares and triangles. However, we did not use any characteristics of squares and triangles to determine the rotation pole; using perpendicular bisectors or perpendiculars to vectors of velocity works on any figure. Therefore, this method can also be applied to moving lithospheric plates. Plates move on a spherical surface though, so will they always have a rotation pole?


38

Exercise 2-13***: Motion on a sphere In Figure 2.22 a rigid body moves on a sphere. Point A moves to point A’, point B to B’. a. Describe and sketch how to find the rotation pole of the body.

Finding the rotation pole is difficult as you cannot use perpendicular bisectors on a spherical surface. Instead, use the intersection of the plane perpendicularly bisecting AA’ with the sphere’s surface. This intersection is a line along the surface of the sphere; it is curved like the lines in Figure 2.22. b. Explain why the rotation pole

lies on the line of intersection.

2.7.3 Great circles The special perpendicular bisectors used for motions on a sphere can also be called cross sections. They always cut through the centre of the sphere because the distances from the centre to the two points on the surface are equal. The line of intersection is a great circle. We will discuss great circles later on.

Exercise 2-14***: Instantane-ous displacements on a sphere always have a rotation pole Comparable to the example with the triangle (exercise 2-11) in which you used lines perpendicular to a velocity vector, on a sphere you can use great circles perpendicular to the direction of rotation. There is one convenient difference with motion on a flat surface however: you can always find the rotation pole! a. Why can you always find a rotation pole for motion on a sphere, unlike for displacements on a

plane? b. Sketch how to find the rotation pole of the plate segment containing points D and E Figure 2.22

Exercise 2-15***: Other particulars of rotation poles on a sphere In fact, there are always two rotation poles; better: two intersections of a pair of great circles. The second intersection can be found exactly opposite the first one. You can draw a line from one to the other through the sphere’s centre. This line is the axis of rotation. a. Mark the two rotation poles of the plate segment with points D and E in Figure 2.22. Hint: the

poles lie on a constructed great circle, directly across from one another. b. Imagine the plate segment lies on a transparent spherical shell covering the sphere. The whole

shell then rotates about the instantaneous rotation axis. Draw in on the shell the great circle that lies exactly in between the two rotation poles; this is the equator of the spherical shell.

A

BA’

B’

D

E

Figure 2.22


39

Exercise 2-16***: Angular velocity and linear velocities An instantaneous rotation has an angular velocity (see Figure 2.23), which, when describing plate tectonics, is measured in degrees per million years. The angular velocity is denoted by ω. a. Imagine a point R on the equator of a

rotation moving with an angular velocity ω of 10 degrees per million years. How many centimetres per year does that point move?

b. Points lying off the imaginary equator have a lower actual velocity (e.g. in cm/year) than point R. Why?

c. A point bisecting the great circle segment between the rotation pole and the imaginary equator has an actual velocity of only

22

1of the velocity at the equator. Explain.

d. What point has an actual velocity half that of a point on the equator?

2.7.4 Resultant motion So far we have only considered a body rotating with respect to a sphere. Plate tectonics, however, considers the movements of plates along, over and under each other: these are the relative motions of plates. Take for example plate A and plate B of which we know the motion with respect to reference body plate C. We can use these motions to obtain the relative motion of A with respect to B and vice versa.

Let us look at motion on a plane first: In Figure 2.24 a transparent red copy of the plane rotates about point A, while a transparent green copy rotates about B. The rotations of plate A about rotation pole A are indicated by the red arrows, while the green arrows indicate the rotation of plate B about pole B. It is evident from the arrows, which indicate the actual local velocity, that the rotation velocities of both plates differ.

Figure 2.23

Figure 2.24:


40

The black arrows represent the resultant velocities in which we are interested. Somewhere along the line between A and B lies a point where the magnitude and direction of a green arrow equals those of a red one. As the rotation about B is larger than that about A, this point will be closer to B than to A. The black arrows indicate a rotation about that point, the rotation pole of plate A with respect to plate B. This pole and its angular velocity are characteristics of the relative motion of lithospheric plates.

Exercise 2-17***: Determining the rotation pole of plate C and D Figure 2.25 shows rotation poles C and D: the rotations have the same direction, but the angular velocity of D is twice that of C. Find the point where the actual velocities of plate C and D are equal. In other words, find the rotation pole of the relative motion of C and D. Hint: Consider Figure 2.24 and the distance between the rotation pole and both A and B. Also, D rotates twice as fast as C, but, at the rotation pole, velocities are equal. Using the formula velocity = distance/time for the same amount of time will give you…?

For motions on a plane as well as motions on the surface of a sphere, the following is valid: the relative motion of two rotating plates can be seen as a rotation of one plate with respect to the other. This can be shown in the plane with simple trigonometry. The rotation pole and angular velocity of a relative motion in the plane can be calculated easily as well. Because the same concepts apply to motion on a sphere, it is also possible to calculate the rotation pole and angular velocity of plate motions. However, such calculations are more difficult, we leave them to specially designed computer programs.

Rotation calculations have proved to be useful not only in plate tectonics, but in determining the higher-velocity relative motion of stars and planets as well!

2.7.5 Determining rotation poles The motion of a plate over the surface of the Earth can be represented by a rotation about an imaginary axis with a certain angular velocity. The two points where the axis of rotation cuts through the surface are called rotation poles – they are not the same as the rotation poles of the Earth itself (see left figure of Figure 2.26).

You can determine the location of a rotation pole by studying an oceanic spreading ridge: there is a clear geometric relation between the orientation of the segments of a spreading ridge and the corresponding transform faults and fracture zones, and the location of the rotation pole (see Figure 2.26). The next two exercises discuss this geometric relation.

Figure 2.26: The motion of plates on a sphere. The left globe shows the difference between the Earth’s rotation axis and the axis of rotation of aplate. The right figure shows a spreading zone including a spreading ridgeand transform faults.

Figure 2.25


41

Exercise 2-18**: Relation between spreading ridge and rotation pole a. Explain the sentence in bold from the last paragraph in your own words. Clarify with a drawing. b. Use Figure 2.26 to describe the relation between the orientation of the transform faults

(perpendicular to the rifts) and the location of the rotation pole.

Figure 2.27 is a detailed map of the bathymetry of the central part of the Atlantic Ocean. Oceanic lithosphere in this area formed through spreading at the Mid-Atlantic Ridge. The white lines are added to indicate the spreading ridge segments, otherwise they are difficult to discern.

Transform faults are obvious from bathymetric maps, as is the way the faults link ridge segments. Therefore, transform faults are used in determining the location of rotation poles.

Exercise 2-19****: Determining the rotation pole of the Atlantic Ocean Using a computer program and a polarity timescale, we will describe the motion of two plates. This exercise consists of two parts: I. Where does the rotation pole of the motion of Africa with respect to South America lie? Using the bathymetric map of the central part of the Atlantic Ocean (Figure 2.27), we will find the rotation pole of the motion of Africa with respect to South America. II. What is the rotation velocity? We will use the paleomagnetic timescale (Figure 2.28) to calculate the rotation velocity.

Part I. Where does the rotation pole of the motion of Africa with respect to South America lie? First use Figure 2.27 to construct at least 10 great circles perpendicular to transform faults. The point of intersection of the great circles determines the rotation pole. The last part of this exercise requires the use of a computer program. Make sure to follow the next steps: a. Find several obvious transform faults in Figure 2.27 Remember that at a transform fault plates

move past each other. Determine (1) the coordinates of the point where ridge and transform fault cross, and (2) the azimuth of the transform fault. The azimuth is the orientation of the fault; it is given by the angle between the fault and true north, measured in a clockwise direction. Use your protractor to measure the angle.

Figure 2.27: Bathymetric map of the central part of the Atlantic Ocean. The white lines indicate the spreading ridge segments. The transform faults are perpendicular to thesesegments.


42

b. Convert the orientation of the transform faults into that of the perpendiculars of the faults by subtracting 90˚. This step is needed because we are looking for great circles perpendicular to the transform faults. Use the coordinates and orientations you find as inputs for the computer program (www.geo.uu.nl/jcu - plate tectonics) to produce a map of great circles. The rotation pole of the motion of Africa with respect to South America lies at the intersection of the great circles. Read off the coordinates of the pole and estimate the uncertainty in the coordinates.

Part II. What is the rotation velocity? In 1968, scientists Dickson, Pitman and Heirtzler published one of the first magnetic profiles that ran perpendicular to the spreading ridge in the southern Atlantic Ocean. Their profile (Figure 2.28), called V18, crossed the Mid-Atlantic Ridge at -30.5°N en -13.5°O. The black and white sequence at the bottom of Figure 2.28 is important. It shows the interpretation of the magnetic curves. d. Use Figure 2.28 together with the paleomagnetic timescale of Figure 2.29 to calculate the

average spreading velocity of the last 3.4 million years at the latitude of profile V18. Be careful: the spreading has two directions. B (Brunhes) is the youngest and GIL (Gilbert) the oldest magnetic period of Figure 2.28.

The velocity of rotation is expressed in degrees per million years. When calculating this velocity, you assume that the angle between the rotation axis and the line from the centre of the Earth to the point where the velocity is determined is exactly 90˚. This point thus lies on the equator of the rotation.

Figure 2.28: The magnetic profile perpendicular to the spreading ridge. The zero on the upper axis represents the spreading ridge, from there the distance to the ridge is given in kilometres. The middle curve shows the observed magnetic signal. The upper curve represents this signal mirrored with respect to the spreading ridge and the lower curve was calculated from a model, approximating the observations well. The black and white sequence at the bottom shows the interpretation of the magnetic curves. Black blocks indicate a normal polarization of the oceanic crust, while white blocks represent a reversed polarity state. The letters below the sequence refer to the names of the geomagnetic periods, for example, B = Brunhes Chron. See Figure 2.29 for the names and ages of the other geomagnetic periods. (Figure 7 of Dickson, Pitman and Heirtzler, Journal of Geophysical Research, 73, 2087-2100, 1968)


43

e. What is the velocity of rotation about the pole determined in Exercise 2-19c? Hint: Use velocity [kilometres/million years] = rotation velocity [radians/million years] · radius turning-circle [kilometres]. Take the radius of the Earth, about 6400 km, as the radius of the turning-circle. Since rotation velocity is always expressed in degrees per million years, you will have to convert the velocity from radials to degrees per million years with PI radials = 180 degrees.

f. Why is the assumption of a 90-degree angle so important in the previous question? By how many degrees per million years would the rotation velocity differ for an angle smaller than 90˚?

Final exercise Ch2. Answer the section questions and the main question g. Answer the seven section questions as well as the main question from the beginning of this

chapter. h. If you find you have new questions after reading this chapter, write them down.

Figure 2.29: Geomagnetic polarity timescale showing the ages of several geomagnetic periods. Black blocks represent a normal polarity state, white blocks a reversed polarity state.


44

Optional exercise 2-1: The magnetic field of the Earth (From a Dutch Physics final exam question, 1998-I, VWO) To be able to complete this exercise on the Earth’s magnetic field, you are expected to be familiar

with oscillation, amplitude, equilibrium position, harmonic oscillation, max

2 Av

T

, magnetic

induction B expressed in tesla (T), magnetic flux (Φ) and the relation between (a change in) flux and voltage. A horizontally positioned compass needle points in the direction of the horizontal component Bh of the geomagnetic field strength B. The needle is displaced from its equilibrium position; the tip of the needle starts to oscillate harmonically with an amplitude of 3.0 mm and an oscillation period of 1.8 s. a. Calculate the velocity of the tip of the compass needle upon passing the equilibrium position.

A coil is placed parallel to Bh and connected to a variable regulated power source. The compass needle is positioned in the middle of the coil (see Figure 2.30). An electric current runs through the coil such that the direction of the coil’s magnetic field is opposite to the direction of Bh. At an electrical current strength of 2.2 mA in the coil, the magnetic field strength of the coil is equal to Bh. Therefore, the resultant magnetic field strength within the coil is zero. If the needle is set off from its equilibrium position now, it will not oscillate, but spin. The coil has 1600 loops and is 25 cm long. The magnitude of the magnetic field strength within the coil can be calculated with

l

INBcoil

0

with 0 the permeability of free space of 4π · 10-7, N the number of loops on the coil, I the strength of the current in the coil and l the coil’s length.

b. Calculate the magnitude of h . The power source and compass needle are now removed. The ends P and Q of the coil are instead connected to a device that records the voltage as a function of time. Starting from the position drawn in Figure 2-31, the coil is then rotated several times in the vertical plane with constant angular velocity. The sense of rotation is also shown in Figure 2.31. The vertical

component v of the geomagnetic field is directed downwards.

c. Use Figure 2.31 to explain which end of the coil (P or Q) has the highest potential the moment the coil passes the position drawn in Figure 2-31.

The voltage produced by rotating the coil is plotted against time t in Figure 2.32 One of the moments where the coil passes the position of Figure 2.31 is set as t = 0.

d. Use Figure 2.32 to show that the direction of the geomagnetic field strength is at a 68-degree angle with the horizontal plane.

e. Calculate the magnitude of the magnetic field strength B of the Earth’s magnetic field.

Figure 2.30

Figure 2.31

Ucoil

Figure 2.32

Earthquakes and tsunamis The Dynamic Earth

45

Chapter 3. Earthquakes and tsunamis The main question of this chapter is:

How does plate motion cause earthquakes and tsunamis and how do we record the effects of plate motion?


How can we use the waves generated during an earthquake to assess the location and strength of the earthquake? (3.1)

How are plate tectonics and earthquakes related? (3.2) What causes an earthquake? (3.3) How do plates move during an earthquake? (3.4) How do we quantify the strength, or magnitude, of an earthquake? (3.5) How do earthquakes cause tsunamis? (3.6)

Objective: To describe earthquakes and tsunamis, to understand what causes them, how they develop, and, if possible, predict future occurrences. For this, you can use the knowledge you have gained from Mathematics and Physics classes.

The Dynamic Earth Earthquakes and tsunamis

46

3.1 Earthquake waves This section describes how the location of an earthquake can be determined by examining recordings of the waves generated during the earthquake.

From Chapter 1 and 2, we know that tectonic plates move. Plate motion is a continuous process that results in the build up of stress on the plate boundaries and on faults within the plates. This stress is released through sudden, jerky, displacements. These sudden movements along the fault plane generate vibrations. It is these vibrations that we call an earthquake. The point within the Earth where an earthquake originates is called the focus or hypocentre; the location on the Earth’s surface, directly above the hypocentre, is the epicentre (see Figure 3.1) Figure 3.2 shows where earthquakes, recorded during a period of only 3 weeks in 2009, occur on Earth.

Solid rock vibrates during an earthquake. The vibrations, or waves, travel either through the Earth (body waves) or along the surface of the Earth (surface waves). Both types of waves are recorded by seismographs (see Figure 3.3).

Figure 3.3 shows the wave pattern recorded by a seismograph. Two waves can easily be distinguished: the P-wave and the S-wave. Both waves are body waves, travelling in three dimensions through the interior of the Earth. P-waves and S-waves contain a lot of information about the earthquake that generated them.

(See http://www.geo.mtu.edu/UPSeis/waves.html for more information on and applets of the motions caused by P- and S-waves.)

Figure 3.2: The location of the earthquakes registered by the US Geological Survey between June 6 and June 28, 2009. (Source: http://neic.usgs.gov/neis/qed/)

Figure 3.1: The hypocentre and epicentre of an earthquake. (Source: www.knmi.nl)


47

The P-wave, the primary wave, arrives at a seismic station first; it is faster than all other waves generated by an earthquake. P-waves are longitudinal waves, this means they vibrate in the same direction as the direction in which they travel (see Figure 3.4). In other words, particles of the material through which a P-wave passes move parallel to the direction in which the wave itself moves. Sound waves move in the same way through air. Within the Earth’s crust, P-waves have a velocity of 5-8 km/s, depending on the density of the rock through which they pass.

The S-wave, the secondary wave, is a transverse wave. Transverse waves vibrate perpendicular to their direction of motion (see Figure 3.5). S-wave velocity is almost half the velocity of P-waves. Therefore, it takes longer for S-waves to arrive at a seismic station.

There are numerous seismograph stations around the world. The data they record is used to determine the exact location of an earthquake, its focus, and to gain insight into the nature and extent of an earthquake. This data helps us to better understand earthquakes and enables us, for example, to predict future earthquake locations. Unfortunately, because the exact timing of earthquakes is still impossible to predict, it is not possible to issue precise earthquake warnings.

To practise working with seismic waves, you can complete Optional Exercise 3-1 at the end of this chapter.

We can use seismograms to determine the exact location of an earthquake. Because P- and S-waves have different velocities, they arrive at a seismic station at different times. The difference in

Figure 3.3: A schematic representation of a seismograph recording, called a seismogram. This seismogram is analogue: it is made by tracing a pen over a sheet of paper. On the website of the Seismology group of Utrecht University (www.geo.uu.nl/Research/Seismology/) you can see the digital seismic signal recorded in Utrecht.

Figure 3.4: A longitudinal wave (P-wave).


48

arrival times is a measure of the distance between the earthquake’s epicentre and the seismograph.

In Exercise 3-1, five seismograms of the same earthquake, recorded by seismographs at different locations on Earth, are used to locate the earthquake’s epicentre.

Exercise 3-1**: Locating an earthquake in The Netherlands On April 13, 1992, at about 3:20 a.m. local time, an earthquake occurred in The Netherlands. Afterwards, interpretation of the seismic data afterwards showed that this earthquake was the largest ever recorded in The Netherlands. In this exercise you will answer the following questions: a. Where was the epicentre of the earthquake? b. When exactly did the earthquake initiate? To answer these questions, use the information below: - 3:20 a.m. local time corresponds to 01:20 Universal Time (UT), or Greenwich Mean Time (GMT) (24-hour clock notation). - Figure 3.6 shows five seismograms of the April 13 earthquake. The seismograms were recorded by seismographs in Germany, Belgium and The Netherlands. Figure 3.6 is only a selection of the many recordings done worldwide. For example, the earthquake was recorded in California and Australia as well. To accurately determine the earthquake’s epicentre, we use the recordings of the seismic stations closest to the earthquake. - The seismograms in Figure 3.6 starts at reference time 01:20:15 GMT. Note that this time is arbitrary; it is not the actual origin time of the earthquake, because the actual moment it began is still unknown. - Figure 3.8 shows the locations of the seismographs that produced the five seismograms. - Figure 3.9 shows a graph of P- and S-wave travel times for places within a 500 km radius from the epicentre. This graph was made under the assumption that the earthquake took place at a depth of about 18 km. You can see that the time difference between P- and S-wave arrivals increases for seismometers farther from the epicentre. Thus, using the difference between the P- and S-wave arrival times, we can calculate the distance between seismograph and epicentre. This is called the epicentral distance. Answer questions a. and b. If needed, you can use the hints below.

Hints to answer question a. Where was the epicentre of the earthquake? i Mark the time of arrival of the P-wave in each of the seismograms and calculate the

difference between P- and S-wave arrival times. Identification of S-wave arrival times is difficult so they have already been indicated.

ii Use Figure 3.9 to determine the epicentral distance for each seismic station. iii Figure 3.8: Draw a circle around each station on the map with a radius equal to each

station’s epicentral distance. iv The epicentre of the earthquake lies where the different circles intersect.

Figure 3.5: A transverse wave (S-wave)


49

Hints to answer question b. What was the origin time of the earthquake? i Determine the arrival time of the P-wave at one of the seismometers. ii Calculate the travel time of the P-wave for that seismometer’s epicentral distance (Figure

3.9). iii Subtract the travel time from the time of arrival. This will give the exact origin time of the

earthquake.

Final questions a. Why don’t the circles around the seismic stations intersect at one exact point? b. Why don’t the lines in the travel time graph go through the origin?

Figure 3.6: The seismograms belonging to Exercise 3-1.

Time from 01:20:15 GMT (s)

Time from 01:20:15 GMT (s) Time from 01:20:15 GMT (s)


50

A larger version of Figures 3-6 to 3-9 is available as a PDF-file. Print these figures out and use them to draw on.

Figure 3.7: The locations of the seismographs of Exercise 3-1.

Figure 3.8: The location of two large Sumatra earthquakes. The earthquake that caused a tsunami in December 2004 is denoted by a yellow star and a second large earthquake that occurred shortly after the first by a red star. The part of the fault plane that was active during each earthquake, called the rupture zone, is also indicated. (Source: US Geological Survey (USGS))


51

3.2 The relationship between plate tectonics and earthquakes.

Exercise 3-2*: Where do earthquakes occur? A, I Read the newspaper article fragment below:

Central America, October 9, 2005

The earthquake in El Salvador and Guatemala had a magnitude of at least 5.8 on the Richter scale. The epicentre of the tremor lies at a depth of 28 kilometres in the Pacific Ocean, about 51 kilometres from the Salvadorian city Barra Salada.

At least 610 people were killed.

a. Mark the location of the El Salvador and Guatemala earthquake on the map you used in Exercise 1-1. Mark the earthquakes studied in Exercise 3-1 and Optional Exercise 3-1 as well.

b. Mark the location of more recent earthquakes. Use, for example, the website http://www.earthweek.com.

c. Take a look at GB 192B (GB 174B). Compare the distribution of earthquakes with the location of the plate boundaries. Can you find a correlation between the two? Are there any exceptions? If so, how do you explain these exceptions?

d. What mistake is made in the newspaper article fragment above?

Earthquakes are most common along convergent plate boundaries (two plates moving towards each other) and transform plate boundaries (two plates sliding past each other). Little seismic activity is found along the third type of plate boundary, the spreading ridge, where plates move away from each other.

Exercise 3-3**: Plate motion near Japan A From a Dutch final exam question in Geography (2007-II) Many more earthquakes have occurred along the north-western boundary of the Philippine Plate than along the eastern boundary. a. Use the data from map GB 157 (GB 140) and GB 192B (GB 174B) to explain the difference in

the numbers of earthquakes between the Philippine Plate boundaries.

Figure 3.9: The fault planes to the southwest of Sumatra that were active during different earthquakes. For each fault plane, the date and the magnitude (inseismic moment, see Section 3.5) is given. (Source: K. Sieh, Caltech, USA)


52

It is evident from maps GB 157A and D (GB 140A and D) that the depth of the hypocentre of an earthquake in the vicinity of Japan and the distance of the earthquake from the Japan Trench are related. b. What is the connection between the hypocentre depth of an earthquake and the distance to the

Japan Trench? Do not consider earthquakes with a hypocentre depth less than 50 km.

Earthquakes along subduction zones are often very powerful, which makes them easier to study. We will investigate subduction zone earthquakes further by zooming in on earthquakes near Sumatra, Indonesia.

The large Sumatran earthquake of December 26, 2004, and the tsunami generated by it were the result of the active subduction zone along which Sumatra lies. The magnitude of the earthquake attracted a lot of attention from the scientific community; as did the enormous amount of data that modern technology was able to capture. The data record of the Sumatra earthquake helps us to learn more about earthquakes in general.

The Sumatran earthquake from December 2004 is one of the most powerful earthquakes that has occurred since seismographic recording began in around 1900. Like all other earthquakes, it was caused by the relative motion of parts of the Earth along a pre-existing fault plane. The Sumatran earthquake fault plane has been active for a very long time (millions of years). However, during the December 2004 earthquake over 1000 km of the plane was active.

The area where the Sumatran earthquake happened has a known pattern of plate motion: The Australian-Indian Plate subducts underneath the Eurasian Plate on which Indonesia lies. The Sumatra earthquake thus took place on a

convergent plate boundary setting, in a subduction zone

The subduction zone runs along the south-western shore of Sumatra and the southern shore of Java (Figure 3.9 and Figure 3.10) south of the red line lies the deep-sea trench. At the latitude of northwest Sumatra, the Australian-Indian Plate subducts in a north-northeast direction underneath the Eurasian Plate with a velocity of 6 cm/year.

Figure 3.10: The motion of the Australian-Indian Plate with respect to the Eurasian Plate. The orange arrows indicate the direction of motion of the tectonic plates, the red arrows the direction of the December 2004 earthquake at Sumatra. (Source: Tsunami Laboratory, Novosibirsk, Russia)

Eurasian Plate

Australian –Indian Plate


53

Figure 3.11: The three phases of earthquake generation. Stress builds up (upper and middle figure) and is then released resulting in an earthquake (lower figure). (Source: www.usgs.gov)

After an earthquake, scientists try to characterize the motion that has taken place as quickly as possible. They want to find out how the plates involved in the earthquake moved with respect to each other. Motion along a fault is an expression of the relative motion between the two blocks that meet at the fault plane. When the blocks are part of lithospheric plates, the direction of motion along the fault plane, called earthquake slip vector, or simply slip, is an expression of the relative plate motion. The slip can be determined within an hour after an earthquake. To do this, scientists first analyse the seismograms of different seismic stations to produce a focal mechanism diagram. This characterizes the orientation of the fault and the direction of the motion along the fault. Additional later analysis can improve on the first estimation.

3.3 The development of an earthquake The Sumatran earthquake of December 2004 measured magnitude 9. The data collected from it is very useful when studying how earthquakes are generated (see Section 3.5 for more information on the term magnitude). The magnitude of an earthquake depends on the extent of the fault plane along which slip took place and on the amount of slip. During the Sumatran earthquake, 2 to 20 metres of slip occurred along approximately 1200 km of the fault plane (an extremely large segment of the fault).

The earthquake started in the south-eastern part of the activated fault plane and extended in a northwesterly direction. Displacement along the fault progressed for 12 minutes, i.e. the slip propagated in a northwesterly direction along the fault plane over a distance of 1200 km in only 12 minutes (this amounts to 1.7 km/s). In some parts of the active fault plane the vertical displacements were up to 15-20 m, while in other parts displacements reached only 2 m.

How can we explain the large 20 m displacements as well as the smaller displacements? Why is the displacement not the same everywhere? On average, the velocity of relative plate motion is several centimetres per year. So how can plates move 20 m in 12 minutes so suddenly? During the years that precede an earthquake, stress builds up because the subducting plate sinks underneath the upper plate several centimetres per year. This sinking, however, is hindered by friction. It is only when the built up stress overcomes the friction that the fault slips in a sudden, fast movement. Think of an earthquake as being like trying to move a large, heavy cabinet. First, you push very hard and stress builds up, then, suddenly, the cabinet moves.

How do we explain the difference in displacements along the fault plane? In some parts of the fault the relative plate motion has been blocked for longer periods than in other parts because of


54

variations in the material properties. This results in differences in the amount of friction.

Thus, in some parts more stress builds up than in other parts. Relaxation in an area where there has been a small build up of stress results in a small displacement (2 m), while in areas where there has been a large build up of stress, the result is large displacements (20 m). The generation of an earthquake can be divided into several phases, which are depicted in Figure 3.11. Stress builds up at depth along a fault during the first phase. The plates deform along their boundaries (phase 2) until the stress is relaxed by means of an earthquake (phase three): in a sudden movement, the plates move past each other along the fault. The results of this motion can be seen at the surface of the continents or ocean floors Involved. The building up of stress and the relaxation of that stress can be modelled with simple friction experiments using elastic strain.

Practical/demonstration: stick-slip behaviour

A simple experiment can show how a continuous motion (plate motion) can result in jerky, shock like movements (earthquakes) along the plane of contact (the fault plane). The pictures Figure 3.12 show how a glider that is rough on the bottom moves over a rough surface. An elastic band is used to build up stress on the glider. The friction between the glider and the rough surface initially prevents the glider from moving. However, when the stress on the elastic band is increased further, this resistance is overcome and the glider moves with sudden jerks, in a start/stop manner. This process is called stick-slip behaviour. By placing weights on the glider, the pressure on the plane of contact and, thus, the friction are increased. More friction reduces the number of shocks, but increases the magnitude of the shocks at the same time.

The above experiment demonstrates how the properties of the fault and the amount of pressure on the fault can influence the magnitude of an earthquake. A rough fault surface with many asperities increases friction and therefore blocks slip longer. Increased pressure on the fault has the same effect. The greater the friction, the stronger the earthquake is.

Figure 3.12: Three photographs of a stick-slip behaviour demonstration.


55

Exercise 3-4**: Stick-slip behaviour When pulling a snow sledge, it is hard to begin pulling it. Once you get the sledge moving (with constant velocity), pulling it is relatively easy. The biggest pull is needed to get the sledge started; you have to accelerate, but, more importantly, you have to overcome the friction between the sledge and the snow. Friction between solid surfaces even exists when the surfaces seem as smooth as a mirror because, on a microscopic scale, even smooth surfaces have protrusions and indentations. These are known as asperities. You may be aware of several different types of friction: rolling friction, sliding friction and air friction. Within these groups, however, we further distinguish static and kinetic friction. Static friction exists between the surfaces of two solid objects that are in contact with each other but are not moving with respect to each other (the objects are static). Imagine a book lying on a table: Unless you exert a horizontal force on the book, it will remain still. When you exert a force (pull or push) on the book and it still does not move, it is static friction that prevents the movement. As long as the book does not move (its velocity being constant and zero), there is no resultant force, the static friction is equal to your pull or push: Ftotal = 0. To move the book, the force you apply must overcome the maximal static friction. The direction of static friction is always opposite to the applied push or pull. The magnitude of the friction can vary between zero and maximal static friction, according to the relation: Ff ≤ μs · Fn. Static friction Ff is given in Newtons (N), μs is the dimensionless coefficient of static friction, and Fn is the normal force, also in Newtons (N). The subscript f denotes friction, the s static friction and the n the normal force. The ≤-sign implies that the static friction is either smaller or greater than the coefficient times the normal force. The first is valid when an object is not yet moving, the latter when an object is about to move. In other words, static friction increases in proportion to applied force, until it reaches its maximum. At that point, the static friction switches to kinetic friction. When the book in the example above moves over the table, kinetic friction has taken over. Overcoming kinetic friction requires less force than static friction; moving an object is easier when the object is already moving. The direction of kinetic friction is opposite to the direction of motion because it works against that motion. The magnitude of friction is determined by the roughness of the surface of the objects in contact and is proportional to the normal force (in the example, the table exerts a normal force on the book). This proportionality relation is reliable, but it is not a law in physics. The equation Ff = μk · Fn governs kinetic friction, with Ff the kinetic friction (in N), μk the dimensionless coefficient of kinetic friction and Fn the normal force (in N). The subscript f denotes friction, k stands for kinetic friction and n for the normal direction of the force. Ff is constant, so here an = sign is used. The table below lists values of coefficients of friction for various materials. It is evident that μk < μs. Note that the coefficient depends on the humidity of the material and varies with lubrication of the surface. a. The coefficient of static friction is always larger or equal to that of kinetic friction, it cannot be

smaller. Explain.

Surfaces in contact μs μk

Wood-wood 0.4 0.2

Ice-ice 0.1 0.03

Metal-metal, lubricated 0.15 0.07

Metal-metal, non-lubricated 0.7 0.6

Rubber-dry concrete 1.0 0.8

Rubber-wet concrete 0.7 0.5

Rubber-other surfaces 1-4 1

Ball-bearings < 0.01 <0.01

Human joints 0.01 0.01


56

Figure 3.13:

b. Study the sketch above. What should be expressed at the letters A through E?

3.4 The motion of plates during an earthquake The direction of the motion of the Sumatran earthquake was not the same as the relative motion between the tectonic plates, which was north-northeast. The displacement along the fault, the slip, was directed towards the east, perpendicular to the trench along the southwest coast of Sumatra. This difference in direction is possible through strain partitioning. Strain partitioning allows the shortening in one direction to be decomposed into shortening in two perpendicular directions. Thus, the relative plate motion is not only facilitated by the subduction contact (see Figure 3.14, left side), but also by a second transform-type fault that cuts through the leading edge of the upper plate (Figure 3.14, right side). Thus, there are two strain components: one component is perpendicular to the plate contact (subduction) and one is parallel to the contact (transverse motion along transform fault).

Near Sumatra, the two components of motion did not take place at the same time. The 2004 Sumatra earthquake was the expression of the first component, the subduction motion (red rectangle on right-hand side of Figure 3.14). An earthquake on March 6, 2007, (see newspaper article below) represented the second component (the yellow rectangle moves with respect to the


57

area on the far right). The second component took place along a separate, vertical fault, called the Great Sumatra Fault, which runs parallel to the southwest coast of Sumatra. This transverse fault was not yet active during the 2004 earthquake.

In summary, the decomposition of motion into two components, along the subduction contact and along the transform contact, is called strain partitioning.

Sumatra earthquakes cause tens of deaths (Source: www.nrc.nl, March 6, 2007)

Padang, March 6. This morning, two earthquakes on the Indonesian island Sumatra caused at least 69 deaths. The number of casualties is expected to rise, as hundreds of buildings have collapsed. Also, casualties and damage have not yet been reported in the more remote areas.

The epicentre of the earthquake is located on land, near Solok, a city north of the West-Sumatran capital Padang. The first earthquake, at 11:00 a.m. local time, had a magnitude of 6.3 on the Richter scale, the second, at 1:00 p.m. local time, a magnitude of 6.0. Both fall in the category ‘strong earthquakes’ and were felt as far away as Malaysia and Singapore.

In areas with frequent earthquakes, subsequent earthquakes are often related. This can be in the form of the strain partitioning as explained above, but strain partitioning can also occur for earthquakes with slip in the same direction, but at different locations.

The latter type of relationship also took place on Sumatra. The December 26, 2004, earthquake was followed by an earthquake of the same type (subduction – see red rectangle in Figure 3.14) of magnitude 8.7 on March 28, 2005. This earthquake occurred to the southeast of the 2004 fault plane. Could the second earthquake have been caused by the first, because the first earthquake added more stress to an area already under stress? Generally speaking, stress is released during an earthquake. In some cases, however, earthquakes add stress locally, usually near the end of the active fault. The stress levels near the end of an active fault are often already close to the critical value, i.e. the boundary between no motion and motion, no earthquake and earthquake. In a situation such as this, adding stress can easily cause another earthquake; the 2004 earthquake brought about the 2005 earthquake in this way.

In exercise 3-6 we will break down relative plate motion vectors. To do this, we will continue to study the motion on a sphere as we did in Section 2.6.

Exercise 2-17: a discussion of the rotation pole of two plates moving with respect to each other.

Both in two dimensions and on a sphere, the following is valid:

Figure 3.14: A schematic view of (the development of) an earthquake. The left figure shows the plate motion and the fault plane near Sumatra. The redrectangle in the right figure shows the motion during the 2004 Sumatraearthquake, while the yellow rectangle shows the motion of the Sumatraearthquake of March 6, 2007.


58

The relative motion of two rotating plates can be seen as a rotation of one plate with respect to the other.

The rotation pole of the relative motion lies on a line (or great circle in case of motion on a sphere) that connects both original rotation poles.

The ratio of the two rotation velocities determines where on the line (or great circle) this rotation pole lies.

For a plane, the above concepts can be proven with trigonometry; you can easily calculate the rotation pole and angular velocity of the relative motion. It is more difficult on a sphere but the same concepts apply.

Additionally, on a sphere you can use the following:

You can use vector addition when you use vectors in the direction of the rotation axis with a length proportional to the angular velocity to represent rotations,. After breaking down the vectors into x- and y-components, sum up the components separately to find the resultant vector.

It is better to leave calculations on a sphere to specially designed computer programs. You can find such programs on www.geo.uu.nl/jcu under the heading Sumatra earthquake. Be aware that it is usually not easy to see that the new rotation pole lies on the great circle through the two poles just by looking at the coordinates calculated by the program.

Exercise 3-5**: Determining the rotation pole of Plate C and D, using the computer program on www.geo.uu.nl/jcu a. Investigate the results of the computer program in the case of Exercise 2-17, with C the point

(0°, 0°) and D the point (1°, -1°) on the Earth. With this example, you should be able to see whether the result lies on the great circle through C and D.

b. Now use the same coordinates for C, together with D = (90°, 0°). Take the relative velocities equal or opposite and check the four concepts of relative motion stated above.

Now you can complete Optional Exercise 3-2: Resultant rotations as resultant vectors, see the end of this chapter.

The following also applies to motion on a sphere (see also Exercise 2-17, 2-18, 2-19 and 3-5):

A rotation is determined by three numbers: 1) the longitude of the rotation pole, 2) the latitude of the rotation pole, and 3) the corresponding angular velocity in [°/My]. Together, these numbers define the rotation vector with its head in the direction of the rotation axis and its length equal to the angular velocity in [rad/My] (radians per million years).

Rotations can be combined. Think, for example, of the motion of Africa with respect to Eurasia. The motions of Africa and of Eurasia with respect to North America are known, and both are used to calculate the rotation pole of Africa with respect to Eurasia: ωAfr(Eur) = ωAfr(NAm) + ωNAm(Eur) = ωAfr(NAm) + -1 · ωEur(NAm)

The last term of the equation above shows that changing the order of the plates (e.g.

ωNAm(Eur) to ωEur(NAm)) equals changing the sign of the angular velocity.

Exercise 3-6***: Breaking down relative plate motion The goal of this exercise is to find the rotation pole of the convergent motion near Indonesia. In Indonesia, the Australian-Indian Plate subducts underneath the Eurasian Plate. There, strain partitioning in different directions occurs. Figure 3.15 shows a map of Southeast Asia and the boundaries of the tectonic plates. The central area, which contains most of Indonesia, can be seen as part of the Eurasian Plate. Table 3-1 gives the coordinates of the rotation pole and the (angular) velocity according to the global velocity model NUVEL-1A, of each of the more important tectonic plates. The velocity given is a relative velocity: the poles describe the motion of the plates with respect to the Pacific Plate. a. Calculate the rotation pole describing the subduction underneath Indonesia. Use the

information in Table 3-1 and the computer program at www.geo.uu.nl/jcu (called Sumatra Earthquake). Note: the program requires velocities in degrees per millions of years.


59

b. Now use the program to determine the direction and magnitude of the velocity at the plate boundary. Do this for (at least) three locations including the points on the plate boundary with longitude 100°E, 105°E and 110°E. Draw in the velocity vectors on Figure 3.15 using the scale 1 cm = 10 mm/year. Are there any major differences in velocity between the three locations? Why?

c. Determine the expected trench-parallel component for each of the three locations with the help of the geometric construction on the map on Figure 3.15. Where on the boundary of the upper plate do you expect trench-parallel transform faulting? In what direction will the small strip of lithospheric plate between the subduction zone and the transform fault move?

d. At 100°E longitude, the subduction direction determined from earthquake focal mechanisms is 31° east of north. Determine the magnitude of the velocity of subduction and trench-parallel transform faulting.

Plate Degrees north Degrees east Velocity [˚/My] African Plate 59.160 -73.174 0.9270 Antarctic Plate 64.315 -83.984 0.8695 Arabian Plate 59.658 -33.193 1.1107 Australian Plate 60.080 1.742 1.0744 Caribbean Plate 54.195 -80.802 0.8160 Cocos Plate 36.823 251.371 1.9975 Eurasian Plate 61.066 -85.819 0.8591 Indian Plate 60.494 -30.403 1.1034 North American Plate 48.709 -78.167 0.7486 Nazca Plate 55.578 -90.096 1.3599 South American Plate 54.999 -85.752 0.6365 Table 3-1 The rotation pole and velocity of several plates with respect to the Pacific Plate, according to the NUVEL-1A velocity model. (De Mets et al, 1994)


60

3.5 The strength (magnitude) of an earthquake There are several ways to measure the strength of an earthquake. Here, we will discuss the three most common scales: the Mercalli scale, the Richter scale and the seismic moment. The Richter scale is the most well known magnitude scale. Before Richter established it, the Mercalli scale was used.

MERCALLI SCALE

This scale quantifies the effects of an earthquake by describing the intensity of earthquake vibrations. Most of the immediate damage caused by an earthquake is due to vibration.

Italian Giuseppe Mercalli (1850-1914) designed this scale that gives a number to the different levels of intensity of vibrations. The scale indicates the observed effects (damage) of an earthquake on humans, objects, buildings and the landscape. The intensity varies with distance from the epicentre and depends on the type of rock and soil in the subsurface. The greater the epicentral distance, the less the ground will move as a result of an earthquake and the smaller the damage. Thus the intensity should also be smaller. However, when the local subsurface intensifies the seismic vibrations, intensity can increase far from the epicentre. This happened during the powerful earthquake in Mexico in 1985.

The Mercalli scale is divided into 12 intensities, each denoted with Roman numerals starting from I (no tremors felt, earthquake only recorded by seismic instruments) to XII (cataclysmic). Scientists often use the Mercalli scale to estimate the strength of earthquakes that occurred prior to 1900 before seismographs were invented. Written reports on the earthquakes provide the observations needed to apply the Mercalli scale.

More information on the Mercalli scale can be found on the website http://earthquake.usgs.gov/learn/topics/mercalli.php.

Figure 3.15: A map of Southeast Asia showing the boundaries of the different tectonic plates.


61

In 1935 the American seismologist Charles Richter proposed a scale based on the strength of earthquake vibrations as measured by a seismograph.

RICHTER SCALE

The magnitude of an earthquake (its strength expressed in Richter scale units) is calculated using the amplitude of the deflections on seismograph recordings. Seismologists must first correct the amplitudes to account for the distance between epicentre and seismic station. This is because with distance, seismic waves lose amplitude through geometric spreading and absorption.

In other words, with greater distance from the epicentre, vibrations become weaker for two reasons. Firstly, the waves and their energy are spread out over a larger area, so, on average, a wave has less energy. This process is called geometric spreading. Secondly, the ground the waves travel through absorbs the energy of the waves due to friction of the ground particles.

To visualize geometric spreading and absorption, think of what happens when you drop a rock in a pool of water. The ripples spread out in circles. Where the rock hits the water, the ripples are higher than those further away. The spreading of the waves and the friction of the water particles causes the wave height to decrease.

The Richter scale is logarithmic: a ten-fold increase in deflection on the seismogram corresponds to an increase of one unit of magnitude. Thus an earthquake of magnitude 8 on the Richter scale has amplitudes that are ten times greater than those of an earthquake of magnitude 7.

Large earthquakes occur less often than smaller ones:

Magnitude on the Richter scale Occurrence >8 once a year 7-8 18 times a year 6-7 108 times a year 5-6 800 times a year 4-5 6200 times a year 3-4 49000 times a year 2-3 300000 times a year

There is an obvious difference between the intensity (Mercalli scale) and the magnitude (Richter scale) of an earthquake. The intensity depends on the place of measurement, whereas the magnitude is independent of the location of the measurement and, therefore, characteristic of the strength of an earthquake itself.

Exercise 3-7**: The Richter scale The equation 10logE = 5.24 + 1.44M shows the relationship between the energy of an earthquake and its Richter scale magnitude as established by experiments. E stands for the energy of an earthquake in Joules and M for the corresponding Richter scale magnitude. a. How much energy is released during an earthquake of magnitude 5 on the Richter scale? Hint:

If 10log(E) = a, then E = 10a. This follows from standard logarithmic identities. Every day, the Sun provides the Earth with 1022 J of energy. b. What magnitude does an earthquake have that releases as much energy as the Earth receives

in sunlight each day? It is sometimes stated that small earthquakes act much like safety valves, releasing a little bit of energy at a time and preventing the build up of stress to catastrophic levels. Assuming we indeed need to release a fixed amount of energy, c. How many magnitude 5 earthquakes are needed to prevent a magnitude 8 earthquake?

Compare the amounts of energy released in Joules.

Scientists often use the seismic moment to quantify the strength of an earthquake. This is because this expression provides a physical measure of the earthquake itself.

SEISMIC MOMENT AND MOMENT MAGNITUDE

More recently, a third measure of the strength of an earthquake was developed: the seismic moment. From this, the moment magnitude followed.


62

The seismic moment (M0) is calculated by multiplying the shear modulus G of the rock bordering the fault plane with the average displacement, or slip, along the fault dav and the total fault area A: M0 = G x dav x A. M0 is given in Newton meters [N·m].

The most recent magnitude scale (1977) makes use of the seismic moment and is called the moment magnitude. It is calculated as follows: Mw = log M0 /1.5 – 6.1. The moment magnitude of a certain earthquake is similar to the Richter magnitude of that earthquake.

The seismic moment depends on the shear modulus and the slip along fault plane with its size. It thus depends only on characteristics of the source area of the earthquake, not on effects observed away from the fault. Therefore, the seismic moment is drastically different from the Mercalli and Richter scale.

The seismic moment can either be determined from seismograms or from field observations at the fault.

The moment magnitude is especially important for very strong (> 7.5) earthquakes, because the Richter scale underestimates the strength of such earthquakes. The moment magnitude of the Sumatra earthquake is 9.3, which means that it would not be the fifth strongest earthquake ever (based on the Richter scale), but the second strongest. However, the strength of the 1960 Chilli earthquake and the 1964 Alaska earthquake will have to be recalculated before we can say so for certain. Unfortunately, recalculations are not easy since instruments in the sixties were not as sensitive as nowadays.

3.6 The relation between earthquakes and tsunamis Tsunamis are potentially devastating waves of water, caused by sub-marine earthquakes in which fault slip leads to vertical displacement of the seafloor. Sumatra’s 2004 tsunami is probably the best known example of a tsunami.

The uplift of the ocean floor due to fault slip is accompanied by an uplift of the entire column of ocean water above it. This causes the tsunami (Figure 3.11). The uplift increases the potential energy of the water mass and generates waves that travel in all directions (see Figure 3.16). The waves have very long wavelengths that dissipate slowly. The first motion of a tsunami can be either upwards or downwards. In the latter case, a drawdown of coastal waters occurs exposing the ocean floor beneath.

Figure 3.16: The travel times (in hours) of the Sumatra tsunami. The tsunami was generated by a submarine earthquake, and it soon hit the coasts of Sri Lanka, India, Malaysia, Thailand and Indonesia and later those of the Middle East, Madagascar and Africa. (Source: Kenji Satake, National Institute of Science & Technology, Japan.


63

Away from the coast the velocity of a tsunami is a simple function of the depth of the ocean:

hgv

with v the velocity of the wave, g the gravity acceleration (9.81 m/s2) and h the depth of the ocean floor. The average velocity of the 2004 Sumatra tsunami was 750 km/h.

It is obvious from the velocity equation that when water depth decreases (when the ocean becomes shallower) the tsunami wave slows down. In the open ocean, the amplitude (height) of the Sumatran tsunami was approximately half a meter; the damage that would be caused by such a wave would be relatively minor. However, upon approaching the coast, the amplitude of the wave increased significantly due to the decreasing water depth. The resulting wave was several meters high and had catastrophic effects.

Research (Margaritondo, G., 2005) shows that, in general, wave height A is inversely proportional

to water depth H to the power ¼, or 1/ 4A cH . From this equation you can see that when water depth H decreases upon arrival at a coastline, wave height A increases.

The Japanese word ‘tsunami’ translates into ‘harbour wave’. In the past a tsunami could not be seen in the open ocean because of its minor amplitude and large wavelength. It was only noticed when it reached a harbour and had obtained an amplitude of several meters. Today tsunami wave heights can be measured in the open ocean using satellites.

Exercise 3-8a****: Tsunami propagation calculations In this exercise you will calculate the propagation of the Sumatran tsunami through time by answering questions such as `Where was the tsunami wave after 6 hours?’ and `How long did it take for the wave to reach the coastline of Madagascar?’ For this exercise use the simple tsunami velocity equation given above. Figure 3.18 is a bathymetric map (a topographic map of the ocean floor) of the Indian Ocean and part of the Atlantic Ocean. Its shows three paths A, B and C, along which different parts of the expanding wave front of the 2004 tsunami travelled. The tsunami paths start at the earthquake’s epicentre and are simplified for this exercise. Along each path in Figure 3-16, the white dots denote distances of 1000 km. Figure 3.17 shows the bathymetric profile along path A, as measured every 100 km. The horizontal axis represents the distance from the epicentre. a. Divide the first 5000 km of the bathymetric profile along path A into 3 to 5 segments and

determine the average water depth of each segment. Hint 1: Take a relatively linear part of the ocean floor to determine the average water depth. Hint 2: Neglect the large fluctuations in depth in the first 1500 km and take this part as one segment.

b. Now calculate the average tsunami velocity for each segment. Hint: Use the equation from the previous page.

c. Calculate the time it takes for the tsunami to bridge each segment. Plot the distance travelled by the tsunami along path A against time. Hint: Use x = v · t, or, in words, distance is velocity multiplied with time.

d. Repeat steps a through c for the rest of path A. e. The bathymetry along path B is similar to that of path A for the first 5000 km (figure 3-20).

Why do you think the bathymetry differs after 5000 km? What effect does the different bathymetry have on the propagation of the tsunami wave? Repeat steps a through c for path B, starting from 5000 km.

f. Now use the profile along path C (Figure 3.20) to determine the time between the earthquake and the arrival of the tsunami wave at the coast of Madagascar.


64

Figure 3.18: A bathymetric map of the Indian Ocean and part of the Atlantic Ocean.Three paths A, B and C are indicated along which different segments of theDecember 26, 2004, Sumatra tsunami travelled.

Figure 3.17: The bathymetric profile along path A.


65

Exercise 3-8b****: Tsunami propagation calculations with Excel The tsunami propagation through time can also be determined precisely by using the bathymetry data with 100 km step size and EXCEL. Directions for the calculations with the original bathymetry data can be found below.

A. The data files For each of the three tsunami paths there is a data file with bathymetry measurements: profileA.dat, profileB.dat and profileC.dat. The first few lines of profileA.dat are: 95.85 3.32 0 -756.407 95.2708 2.63133 100 -1781.25 The first column gives the longitude in degrees east, the second column the latitude in degrees north, the third column the distance from the epicentre in kilometres and the fourth column the water depth D as a negative number in meters.

B: Reading data into EXCEL (for EXCEL 2003) You can read in the data per column as follows: i Open a new EXCEL file. Put your cursor on cell A4; this way you leave some lines open above

the data. ii Choose Data >> Import external Data >> Import Data. Browse for the right data file and

choose All Files for the option File Types. Click Open. iii Every time you click Next, new options will appear. The Data Format screen is especially

important. Select (with Shift-Click) all columns, choose Advanced and switch the decimal separator to . (point) and the thousands separator to ‘ (quotation mark).

iv Click Finish and check to see that the starting cell is A4. [This is, of course, not obligatory, but the example below assumes your data starts at A4.]

C: The bathymetric profile You can image the bathymetric profile right away: Click on the depth D column and choose Insert >> Chart >> Line. You do not need the graph for any calculations, but you might want to use graphs for step E.

D: The calculations Methods in short: Determine the velocity in each 100-km segment; determine the time needed for the tsunami to cross each segment; sum the segment travel times. You can give these calculations in the first data line and then copy them up to the last line of data. EXCEL will then automatically do the other calculations for you. (Be careful to start calculations with an = sign.) Methods in detail:

– Use cell F4 for the tsunami velocity ( hgv ). Type: = SQRT(9.81 · - D4) [the reference to

cell D4 can be given by clicking on D4]. – Use cell G4 for the travel time per segment. Each segment is 100 km long and the velocity in F4 is in meter per second. Compose a proper equation for the travel time in seconds with a reference to cell F4. – Sum up travel times in column H. In the cells of column H, you sum up the value of the neighbouring G-cell and the H-cell above. So, for the first data line, you sum up G4 and H3. H3 is empty, but EXCEL will attribute it value 0, or you can assign it value 0 yourself. - Now you have your first complete line of calculations. Copy these calculations by selecting cells F4-G4-H4, putting your cursor on the right lower corner of the selection and dragging it to the last line of data. Column H will now tell you when the wave passed each distance in column C. – Column H gives the time in seconds; make an extra column that gives the time in hours. E: Improvements and further investigations a. Do you think your calculations would be more accurate if you were to use the average of the

velocities at the beginning and end of each segment as segment velocity? You can try this method with EXCEL or predict its effects yourself.

b. Plot the bathymetry of each path on the spreadsheet and compare the three profiles.


66

c. You can also construct other graphs. For example, a plot of distance versus time. Do this by copying column C to the right of your last column, selecting both columns and making a Chart of the scatter-type.

Figure 3.20: The bathymetric profile along path C.

Figure 3.20: The bathymetric profile along path B.


67

Final exercise Ch3. Answer the section questions and the main question a. Answer the six section questions and the main question from the beginning of this chapter. b. If you find you have new questions after reading this chapter, write them down.

Optional exercise 3-1: Earthquakes and seismic waves (From a final exam in Physics, Havo 1999-I, question 6) During an earthquake, longitudinal and transverse waves (the P- and S-waves of Section 3.1) travel through the Earth. a. What is the difference between longitudinal and transverse waves? In a certain type of rock, transverse waves have a velocity of 3.4 km/s and a frequency of 1.2 Hz. b. Calculate the wavelength of the transverse waves in this rock. Seismographs register earthquake vibrations. Figure 3.21 shows a simple type of seismograph: A heavy block hangs suspended on a spring and can move freely, but only in the vertical plane (because of hinge A). During an earthquake, the spring-block system is not allowed to resonate with the earthquake vibrations. To prevent resonance, the frequency of the spring and block is small (only 0.37 Hz) compared to the frequency of the earthquake’s vibrations. The mass of the block is 4.2 kg. c. Calculate the spring constant.

The velocity of longitudinal waves differs from that of transverse waves. Due to this difference, the waves do not arrive at a seismic station at the same time. Figure 3.22 shows a recording of the seismograph of an earthquake in Greece, measured by the KNMI in De Bilt, The Netherlands. The L denotes the arrival of the longitudinal waves, the T that of the transverse waves. You can see that the longitudinal waves arrived first. It is assumed that both types of waves followed the same path. The earthquake took place 2300 km away from the seismograph. Take the average velocity of transverse waves as 3.4 km/s. d. Determine the average velocity of the longitudinal waves in two significant digits.

Figure 3.22: Registration of an earthquake in Greece.

Figure 3.21: A seismographs.

Spring

Hinge


68

Optional exercise 3-2: Resultant rotations as resultant vectors In this chapter we have investigated the resultant motion of two plates, the motion of one plate with respect to the other. Focus on the equators especially on the equators belonging to those motions, to prove the fourth point stated just above Exercise 3-5: when you use vectors in the direction of the rotation axis with a length proportional to the angular velocity to represent rotations, you can use vector addition.

Study Figure 3.23. Rotation poles A and B are indicated together with their angular velocities ωA and ωB. (In this case, ωA : ωB = 2 : 1, but the following deductions are valid for every velocity ratio.) Equators cA and cB belong to poles A and B, respectively, and cross in point P. The actual velocity of P as a point on the A-shell is vA and that of P as part of the B-shell is vB. a. Explain why |vA| : |vB| = ωA : ωB. Hint:

Use v = ω·R, with R the radius of the Earth (see BINAS).

b. Explain why the angle between the axis of rotation of A and that of B is equal to the angle between vA and vB.

Using the above definitions, we can draw two vectors, starting in the centre M, along the rotation axes of A and B with their length ratio equal to the ratio ωA : ωB. Call the vectors mA and mB; mA is already drawn in. c. Also draw in mB. Vector couples vA-vB and mA-mB are similar (they have equal angles and proportional lengths). The resultant velocity vC, the difference between the two velocity vectors vA and vB, also works on point P. We know the resultant motion has a rotation pole C as well. C should lie on the great circle through A and B, because in C the motion around A should equal the motion around B. d. We are going to try to draw C and the arc CP. CP is perpendicular to vC. Why? e. Draw the equator of C and then construct arc CP and rotation pole C. The angular velocity ωC is determined by the magnitude of vC. The following relationship holds, which is also valid if you replace B by A: |vC| : |vB| = ωC : ωB. f. Explain why the above relationship holds. Starting at M you can draw resultant vector mC. g. Why does mC lie along the rotation axis of C? Using mC, you can determine C and angular velocity ωC directly, without using P. h. Draw in vectors mA and mB along the axes of A and B with the same magnitude ratio as ωA and

ωB. i. Determine resultant vector mC. The direction of vector mC gives the position of C and its magnitude gives ωC. To summarize: The arrows in the direction of the axes of the rotation poles and with magnitudes proportional to the angular velocities act like vectors in case of resultant motions. You can determine the resultant vector through vector addition.

Figure 3.23

Volcanoes The Dynamic Earth

69

Chapter 4. Volcanoes The main questions for this optional chapter are:

What types of volcanoes exist, what do they emit during an eruption, and what do volcanic products look like at the Earth's surface?

These issues will be addressed by answering the following section questions:

What types of volcanoes exist, and where on Earth do we find them? (4.1) How are different kinds of igneous rocks formed? (4.2) How is magma formed? (4.3) How do volcanoes affect Iceland? (4.4) How much gas was emitted during the Laki eruption of 1783? (4.5) How did this eruption lead to an increase of acidic components in the atmosphere? (4.6) What are possible effects of volcanic acid rain? (4.7)

Objective:

In this chapter you will learn about the processes that occur in and around volcanoes. You will use theoretical knowledge you have acquired during your chemistry lessons.

The Dynamic Earth Volcanoes

70

4.1 Volcano types and occurrences Exercise 4-1*: Volcanoes and plate tectonics A, I Eruption: Indonesian volcano Karangetan, October 29th, 2007 “Karangetan volcano on the Indonesian island Siao erupted this morning. Hours before, hundreds of villagers were evacuated from the slopes of the 1.700 m high volcano. Villages, farms and trees were thickly covered in ash, but no substantial damage or casualties have been reported. Karangetan is one of the most active volcanoes on the Indonesian archipelago.” (From www.trouw.nl) a. Mark the volcano on your empty world map. b. Mark on your map any recent volcanic eruptions. Use www.earthweek.com for more

information. c. Now take a look at GB 192B (GB 174B). When you consider plate boundaries, does anything

strike you when looking at the locations of volcanoes that you have found? Are there any exceptions? If yes, can you explain these?

The epicentres of earthquakes are concentrated around plate boundaries. The majority of volcanoes can also be found in these zones (see Figure 4.1 and GB 192B and D (174B and D). In this chapter we will study the relationship between plate tectonics and volcanism. First we will look at the different types and occurrences of volcanoes.

4.1.1 Volcano types There are many types of volcanoes but the two main types are shield volcanoes and composite or stratovolcanoes. They differ in their form: a shield volcano is flat and wide – hence its name; it looks like a shield that is lying flat. A stratovolcano, however, has a cone shape. Whether a shield or a stratovolcano will form is mainly determined by the composition of the material that is emitted during an eruption.

Shield volcanoes (Figure 4.2): shield volcanoes are formed where eruptions produce mainly lava flows with a basalt composition. Basalt magmas crystallize at a relatively high temperature, contain much iron and magnesium but little sodium and potassium, have a relatively high density, and are relatively fluid (not viscous). This latter characteristic is especially important in determining

Figure 4.1: Active volcanoes in the world: The Smithsonian Global Volcanism Program. Source: http://volcano.si.edu/world/find_regions.cfm


71

the shape of the volcano. A low viscosity means that basaltic lava flows easily, and will therefore spread over a wide area. This results in a volcano with a relatively flat and wide shape. Basaltic magma is formed when the upper mantle partially melts. This happens especially at divergent plate boundaries and at so-called ‘hotspots’ (more about hotspots in the next section). Examples of basaltic volcanism are found in Iceland (a divergent plate boundary and a hotspot) and Hawaii (a hotspot).

Stratovolcanoes (Figure 4.3): This type of volcano is formed where the magma is composed primarily of andesite. Andesitic magmas contain more silicon, sodium and potassium and less iron and magnesium. They are more viscous, resulting in steeper slopes on the volcano, and have a lower density than basaltic magmas. Another difference is that a stratovolcano is formed from alternating layers of ash and lava (this is why they are also called ‘composite volcanoes’). Because of the high viscosity and the resulting low flowing velocity, andesitic lavas do not travel far but stay relatively close to their source.

Stratovolcanoes are known for their explosive eruptions. Different factors

explain this explosivity. Low viscosity causes rising magma to slow down or stagnate, so that outflow can be blocked for extended periods of time. Pressure builds up until it is finally released in an eruption. The magma also contains a lot of gas that cannot easily escape due to its high viscosity. If the gas remains trapped, extra high pressure develops. When it finally erupts, it can be extremely explosive. Some stratovolcanoes are characterised by a caldera. Here the explosive eruption was so voluminous and fast that the whole roof of the underlying magma chamber collapsed, leaving a cauldron-shaped feature in the landscape.

Stratovolcanoes are found mainly in subduction zones (places where an oceanic plate is pushed beneath a continental or another oceanic plate), for example in a string of locations around the Pacific. Here, the magma is also formed in the upper part mantle, but its composition changes from basaltic to (usually) andesitic as the magma travels to the surface. This is a process called magmatic differentiation, and will be explained later. The differences in magmatic composition are elaborated in section 4.2.

Figure 4.2: A shield volcano. Source: Grotzinger et al,Understanding Earth (2005)

Eruption Crater

Room

Figure 4.3: A stratovolcano. Source: Grotzinger et al,Understanding Earth (2005)

Ash layers

Passages

Crater


72

Exercise 4-2**: The Pinatubo eruption A Pinatubo erupts (source: www.nieuwsdossier.nl) June 15, 1991 – The eruption of Pinatubo volcano in the Philippines was one of the most violent eruptions of the 20th century. The first shocks were felt on March 15th. Seismologists placed instruments around the volcano and were able to warn the population. The first magma erupted on June 7th. A few days later another eruption followed, whereby ash was spewed up to great heights (24 kilometres). Eruptions kept coming until the biggest one in June, which lasted for three hours. The ash covered most of the Philippines and Central-Luzon island was completely obscured. The ash travelled as far as Vietnam and Cambodia. 300 people died as a result of the eruption, and afterward the summit of the mountain was 260 m lower than before. Because of the particles in the stratosphere, the average global temperature decreased by 0.4 degrees Centigrade. a. Is Pinatubo a shield volcano or a stratovolcano? b. Was the Pinatubo eruption explosive? Illustrate your answer using data from the above source. c. Why was the pressure in the magma chamber so large? d. Can we expect further eruptions from Pinatubo? Explain your answer.

Hotspots: In exercise 4-1 you learned that the presence of some volcanoes is not related to plate boundaries but to hotspots instead. These are isolated areas where slowly rising material from the mantle (a mantle plume) reaches the Earth’s crust. Enormous amounts of basalt can be extruded when magma from such a mantle plume pierces the crust.

A whole island can be constructed from erupted basalt, which could essentially be the upper part of a giant shield volcano if a hotspot in the ocean has remained active for a long time. If the plate on which the island lies moves, and because the hotspot beneath it usually stays at a fixed position, a chain of volcanic islands will often be formed. A well-known example is the Hawaiian island chain.

When sited under land, a hotspot can push up a whole region. If the pressure caused by the hotspot continues for a long time, a whole landmass can start breaking apart. Fault-lines will form, and basaltic mantle-derived magma will reach the surface. Silicic magmas can also be extruded if deeper parts of the continental crust are melting as well. Examples of such continental hotspots are Yellowstone in the United States and Mount Cameroon in Cameroon (Africa).

Exercise 4-3**: Volcanoes and plate motion reconstructions A, G Look up the Midway Islands in your atlas (if you are using Google Earth or Google Maps, search for Sand Island) and the island of Hawaii. a. What types of volcanoes are located on these islands, and how were they formed? The Midway Islands were formed 27.2 million years ago, and Hawaii 0 to 0.4 million years ago. b. Use this information and your atlas to calculate the average velocity of the Pacific Plate (in

whole centimetres per year), and the direction of movement of the Pacific plate during the last 27.2 million years. Show your calculation. (You may assume the Earth to be flat for this calculation, as it involves only a small part of the Earth’s surface).

4.2 Formation of different types of igneous rocks The previous section described how a shield volcano consists mainly of basalt and a stratovolcano mainly of andesite. In reality, many more types of igneous rocks can be distinguished (as you saw in chapter 1). We will now address these differences in more detail by looking at the composition of magma. This composition gives us important information about the relationship between volcanism and plate tectonics.


73

Igneous rocks are formed when liquid magma solidifies (see Chapter 1). These rocks are marked by differences in texture. For instance, basalt is much finer grained than granite. If it is not possible to identify any crystals at all, the rock consists of volcanic glass. The texture of the rock tells us how fast the magma cooled. Liquid magma needs time to form crystals. When the liquid starts cooling, small crystals of one or more minerals will form and will grow gradually as ions are attached to the surfaces. The longer the cooling takes, the more of these crystals will grow. If magma cools very quickly, no crystals (volcanic glass) or only very small crystals (e.g. in basalt) will form. Magma cools quickly when it reaches the Earth’s surface; the result is an

extrusive rock (like basalt). Slower cooling rates result in larger crystals; this happens when the magma stays deep in the Earth’s crust where it remains warm over a much longer period of time. Such rocks are called ‘intrusive rocks’ (like granite).

You can obtain more information by looking at the chemical composition and mineral content of a rock. Igneous rocks can also be classified according to their chemical composition and the type of silicates they contain. Examples of such minerals are quartz, feldspar, muscovite, biotite, amphibole, pyroxene, and olivine. Two types of igneous rocks can be distinguished, based on their mineral content: mafic rocks and felsic rocks (see Figure 4.4). Basaltic magmas (section 4.1) form mafic rocks like basalt (extrusive rock) or gabbro (intrusive rock). Examples of felsic rocks are rhyolite (extrusive rock) and granite (intrusive rock).

Mafic rocks (ma = magnesium (Mg), f = iron (Fe)) have a relatively low silicon content but contain relatively large amounts of magnesium and iron. These are important building blocks for pyroxenes and olivines. These minerals have a dark colour, which is why mafic rocks are usually dark. Basalt is the best known and most abundant mafic rock. The entire ocean floor consists of basalt. On some continents, thick layers of basalt can be found. For example, the Deccan Traps in India and the Siberian Traps in Russia.

Felsic rocks (fel = feldspar, si = silica) have a high silicon content and contain little iron and magnesium. These rocks are much lighter in colour because they contain light-coloured minerals like feldspar and quartz. Granite is a felsic rock. It is one of the most common igneous rocks found on Earth.

Figure 4.4: Different types of igneous rocks. Source: Grotzinger et al, Understanding Earth (2007)


74

Looking at texture, chemistry and mineral content, it is possible to make the following rough classification (compare to Figure 4.4):

Mafic Felsic

Fine-grained – extrusive Basalt Rhyolite

Coarse grained – intrusive Gabbro Granite

Andesite, the extrusive rock found in many stratovolcanoes, is not mentioned in the table. Its composition lies somewhere between basalt and rhyolite.

There are also rocks that we call ultramafic. These contain relatively little silicon but a lot of magnesium and iron. The most common ultramafic rock is peridotite. This coarse grained, dark green rock is the dominant rock in the mantle, and is the source of basaltic magmas (see chapter 6). It consists mainly of olivine and pyroxene.

4.2.1 Do felsic and mafic magmas have different properties? You may wonder why a distinction between felsic and mafic rocks is useful. There is a relationship between the composition of a rock and its melting temperature. Mafic rocks melt at higher temperatures, felsic rocks at lower temperatures. Mafic magmas are usually hotter and start crystallising at higher temperatures than felsic magmas. The more silicon is added, or the lower the temperature gets, the higher the viscosity of the magma will become. Viscosity is a measure for how difficult it is for a liquid to flow. So, the more felsic or colder a magma is, the more inhibited is its capacity to flow, especially when it carries abundant crystals. In contrast, a hot mafic magma, which is low in silicon, has a low viscosity and will flow easily.

4.3 Magma formation We know how magma is formed, but much remains still to be discovered. The processes take place so deep inside the Earth that direct observation is impossible. Instead, they must be simulated in lab experiments. These experiments show that the melting point of a rock not only depends on its composition, but also on the pressure applied to it.

Rocks consist of various minerals, each with its own chemical composition, and do not usually melt completely when heated. The minerals contribute to the melt in certain proportions that may change when melting proceeds; thus, magma that is created by partial melting of a rock will have a chemical composition other than the rock itself. This is the reason why basaltic magma (mafic) can be formed from a mantle rock that was originally a peridotite (ultramafic). Depending on the amount of melting, magmas will have different compositions, even if the source rock from which they are formed is the same. (Of course, magma compositions can also differ simply because their sources are different; for example, if not the mantle but the lower crust is melting).

Theoretically, there are three ways to melt a rock in the mantle:

1. By increasing the temperature. Surprisingly, this is not an important cause of magma formation and volcanism.

2. By lowering the pressure. This is the cause of volcanism at mid-oceanic ridges and at hotspots. In both cases hot mantle rock rises very slowly; in the case of mid-oceanic ridges because of the upward motion in mantle convection, and in the case of hotspots because of the rise of an isolated mantle plume. The rock remains hot (thermal conductivity in rocks is very low), while the pressure decreases, causing the rock to melt.

3. By lowering the melting temperature. This happens when water comes into contact with the mantle rocks (just like when salt is sprinkled on roads during winter to lower the melting point of water so that it will not freeze). It is the main explanation for the origin of subduction-zone volcanism (discussed below).

4.3.1 The composition of first-formed magma and final igneous rock Magmas of different composition are formed, depending on source rock and melting conditions. The next question is: will the same magma always yield the same igneous rock? The answer is no. Lab research shows us that different factors play a role:


75

- Firstly, the crystallisation history, which will determine the texture of the rock, in particular the sizes of the visible grains. Rising magma can accumulate in a magma chamber, for example somewhere within the crust. Because this environment is usually colder than where the magma came from, it will start cooling and crystals will form. Ultimately, it could crystallise completely inside the chamber and form a coarse-grained intrusive rock such as granite or gabbro. But (part of) the magma could also find its way further up during this process; it could rise to the surface and solidify there after a volcanic eruption. It would then form a fine grained or even glassy extrusive rock (for example basalt or rhyolite) that will often carry larger grains of minerals that crystallised previously at depth.

- Secondly, magma could change chemically in the course of its ascent to the surface. This is known as magmatic differentiation. While it resides in a magma chamber, different minerals will crystallise if the temperature drops. If these crystals do not stay floating in the liquid but sink to the bottom, the remaining magma will change in composition. It is “differentiated” because these removed minerals are different in composition than the original melt in which they crystallised. Therefore, a rock resulting from such a differentiated magma has not the same chemical composition as the first magma that formed during melting of the deep source.

In a basaltic magma, olivine will start to crystallise first, followed by pyroxene and other minerals. The composition of the remaining magma changes, depending upon the type of minerals formed (provided that these do indeed sink to the bottom). Bowen’s reaction series (Figure 4.5: The varying composition of igneous rocks is explained by fractional crystallisation. Source: Grotzinger et al, Understanding Earth (2007)) describes the order in which minerals crystallise. It usually holds quite well but it is no more than a general guideline. Several factors (such as the depth of crystallisation) can affect the order of crystallisation.

Basalt and granite are the two most common igneous rocks. Is there a relationship between the formation of these two rock types? We have seen that their compositions differ greatly: basalt is mafic and granite is felsic. Long ago it was thought that all granite originates from basalt by magmatic differentiation. As a basaltic magma crystallises, olivine and pyroxene (dark minerals, rich in magnesium and iron) are removed from it. What remains is a magma that is rich in silicon and poor in magnesium and iron, yielding a light coloured rock (like granite or rhyolite). The

Figure 4.5: The varying composition of igneous rocks is explained by fractional crystallisation. Source: Grotzinger et al, Understanding Earth (2007)


76

problem is that you would need an enormous amount of basaltic magma to account for all the granitic rocks on Earth.

The melting of entirely different source rocks could explain the difference between basalt and granite: when upper-mantle rocks melts partially, a basaltic magma will form. Granite can be formed through differentiation of basaltic magma, but also directly when the lower part of the continental crust melts. This part of the crust can consist of all kinds of sedimentary, igneous and metamorphic rocks, and has thus a composition that is very different from an ultramafic mantle rock.

4.3.2 Magma formation and plate tectonics So how does magma formation fit into the framework of plate tectonics? In section 4.1 we saw that magma is formed at two types of plate boundaries: MORs and subduction zones

At MORs (Mid-Oceanic Ridges), the source material is the upper mantle, which is mostly peridotite. The pressure is reduced as the hot rock rises slowly, causing it to melt partially. The difference in density between the newly formed basaltic melt and the surrounding rocks will cause the melt to escape and rise even faster independently, until it accumulates in a magma chamber in the crust. There, it will partially crystallise as a gabbro, but some of it will reach the ocean floor and form so-called pillow lavas. These pillow-shaped blobs of basalt with a glassy crust are created as outpouring lava cools quickly in the seawater. Gabbros and pillow lavas are present in MORs below all oceans. If a basaltic magma is extruded above sea level it may form a shield volcano.

At subduction zones, oceanic (basaltic) crust is subducted beneath a continental or another oceanic plate. Not only the basaltic part of oceanic crust is subducted, but a thin layer of deep-sea sediments on top of it may go down as well. All the subducting material is gradually brought to depths where the temperature and pressure become higher and higher. The rocks will be metamorphosed because minerals of which they consist are transformed into other minerals that are better “adapted” to the conditions at greater depths. During these reactions, minerals that carry fluids in their crystal structure break down so that the fluids (mostly water) are released. The water is driven out of the subducting slab, moves upward and enters the mantle of the overlying plate. The addition of water lowers the melting point of the mantle material so much that it will melt, which usually occurs at a depth of about 100-150 km. Again, the source rock is upper-mantle peridotite so that the initial magma has a basaltic composition also here. This usually changes when it rises and concentrates in magma chambers, as we discussed before. The concentration of silicon in the magma increases, so that andesitic or even felsic magmas are formed. Compositions can also become felsic if portions of the continental crust melt because of the heat of basaltic magma that came up from the mantle. Granite forms if such a felsic magma crystallises at depth. If it reaches the surface, violent eruptions may occur. The increased silica content makes the magma very viscous and gases remain trapped inside. Enormous pressures can build up, which are released in violent eruptions.

Figure 4.6 shows a schematic overview of the relationship between plate tectonics and the different types of volcanism. Note that hot-spot volcanism is not related to any plate boundary.

Figure 4.6: Relation between plate tectonics and volcanism. Source: Grotzinger, et al, Understanding Earth, 2007.


77

Exercise 4-4**: Igneous rocks a. Imagine you could drill a hole through the Earth’s crust at a MOR; describe the rocks that you would find: i What rock would you find at the sea floor? Is it an intrusive or an extrusive rock? ii What rock would you find in the lower crust? Is it intrusive or extrusive? iii What rock would you find in the upper mantle? Most of the oceanic crust and almost the entire upper mantle consists of mafic and ultramafic rocks, respectively. b. How do you explain that much felsic rock is found in the Earth’s continents? c. Where are the building blocks of felsic rocks? A significant amount of water is stored in seafloor sediments and in the oceanic crust near subduction zones. d. Explain how the role of water influences the melting process in subduction zones.

4.4 Volcanoes on Iceland We will now take a closer look at Iceland. This special piece of land is one of the few volcanic areas on a MOR that lie above sea level. This situation is exceptional because both a divergent plate boundary and a hotspot are present. Icelandic volcanism also has a great influence over a wide area. A large eruption in 1783 had a big impact in Western Europe.

In geological terms, Iceland is very young. The oldest exposed rocks are ‘only’ about 16 million years old. These are found at the extreme east and west sides of the island (see Figure 4.7.)

Many different types of volcanoes occur on Iceland, both shield and stratovolcanoes. The shield volcanoes have mild slopes, as we would expect. The biggest shield volcanoes, Skjaldbreidur and Ok, lie in the western part of the country. An example of a stratovolcano (typical for its steep slopes) is Hvannadalshnúkur, the highest point of the island (2100 m) on the south side of the Vatnajökull glacier. A third type of volcanic feature is the fissure or linear volcano. This is a linear

Figure 4.7: Geological map of Iceland. Source: Landmaelingar Islands.


78

fracture in the landscape where lava erupts along the entire length or from several vents, usually without much explosive activity. The fissures are generally no more than several meters wide but can be kilometres long. The black lines in Figure 4.8 show the location and orientation of some of them.

Exercise 4-5*: Volcanoes on Iceland a. Explain why the oldest rocks are found in the extreme south-east and the extreme north-west

of the country.

b. Think of an explanation why there are fissures, and explain their orientation.

Iceland lies on a mid-oceanic ridge, the Mid Atlantic Ridge. This ridge of the seafloor stretches from north to south in the central part of the Atlantic Ocean where the Eurasian and American plates diverge. Here, new oceanic crust is gradually formed from the basaltic magmas that rise from the mantle below the ridge. Iceland is a part of the ridge that lies above sea level. It grows from its centre outward to the east and west.

Why does Iceland lie above sea level? Because a hotspot immediately below the island generates more volcanic activity than elsewhere along the Mid Atlantic Ridge. This explains the extreme rate of magma production on the island.

Usually, a mantle plume beneath a moving oceanic plate will create a row of islands (like in the case of Hawaii, see exercise 4.3). However, the magma producing hotspot below Iceland lies exactly beneath the location where the plates diverge, so that the island grows in both directions. The location of the hotspot itself is stable. It fuels the magma chambers of many volcanic systems on Iceland.

Exercise 4-6*: Islands on a MOR A, I Iceland is not the only island on a MOR above sea level. a. Look in your atlas for five more islands that lie on a MOR. b. What is the difference between these islands and Iceland? Can you explain this difference?

Figure 4.8: Iceland with some of its most important volcanoes. The Skjaldbreidur and Ok volcanoes lie northwest of Geysir. Hvannadalshnúkur lies mostly hidden beneath the ice east of the Laki (or Lakigigar) volcano (see Öræfajökull in Figure 4.7)


79

Now that we know more about Icelandic volcanism we will take a look at one specific eruption that had a devastating effect.

In 1783 and 1784 the Laki volcano (see Figure 4.8) erupted for eight consecutive months. This volcano produced one of the largest lava flows in human history. At least 15 km3 of magma erupted from a fissure in the crust. When the magma travelled to the surface, gases that were initially dissolved in it were able to escape, because their solubility in molten rock decreases with decreasing pressure. These gases contained highly toxic components, causing severe health problems in the population, poisoning trees and vegetation, and killing more than 60% of the cattle on Iceland. A widespread famine that lasted many years killed about 20% of the population.

The effects were felt not only on Iceland. A dry fog hung over the north and west of Europe for months. The summer of 1783 was very hot, while the following winter was extremely cold in Europe and North America. Winters continued to be unusually cold in the following years and many harvests failed. Famine was widespread, and more people died than under normal conditions.

A large amount of volatile components were brought into the atmosphere during the eruption. Among these, sulphur dioxide (SO2) and hydrogen chloride (HCl) are potentially harmful. If their concentrations in the air are high, rainwater will become acid. We will discuss this further in the next sections. First, we will introduce a simplified model of the eruption and estimate the amounts of SO2 and HCl that were released. In §4.6 we will explain how and why volcanic acid rain is formed, and calculate the pH of rainwater affected by such an eruption. Furthermore, we will discuss examples of possible effects of acid rain.

4.5 The quantity of gas emitted during the Laki eruption In order to assess the effects of the Laki eruption, we need to know how much of each gas component was emitted. We will focus on SO2 and HCl, as these components were most harmful. Because the quantities of released volatile compounds were not directly measured when the eruption took place, we will have to reconstruct the data by analyzing the erupted materials that are still there. For this Thordarson and co-workers (1996) used a simplified model, which we will follow here. (see Figure 4.9)

Figure 4.9 shows a hypothetical magma chamber of a certain volume. The magma inside contains the volatile elements sulphur and chlorine (from here called ‘volatiles’ in short) in certain concentrations (vi). Here, the volatiles are completely dissolved in the liquid. When magma reaches the surface, it cools and solidifies in two different forms: a pile of tephra (ash and rock fragments) and a lava flow. Part of each volatile remains inside these solidified products, but most is expelled into the atmosphere. Some of this released portion stays in the vicinity of the volcano as a local fog. The rest is spewed high into the atmosphere, and is especially relevant in our calculations because this gas is transported away and has a damaging impact at great distances from the volcano.

Schematically, we assume that the sequence of events was as follows:

Magma rises from the magma chamber with all the dissolved volatiles present. The beginning of the eruption is quite explosive; during this time the tephra is emplaced. It

represents part of the original amount of magma. Much gas is released in the eruption plume. Therefore, the concentrations of volatiles in the rest of the magma that remains underground decrease.

This ‘degassed’ magma then rises to the surface as well, but, as it contains less gas, it flows out as a lava in a relatively calm way.

As the lava flows out, it will lose more of these still present volatiles. Low concentrations of volatiles stay behind as a residue, both in the solidified tephra and in the

lava, because ost of the gas escaped to the atmosphere.


80

Concentrations of a volatile (S or Cl, in ppm):

vi = in melt inclusions (representing original magma in the magma chamber) vt = in tephra vl = in lava (during its outflow) vc = in the solidified lava (after its emplacement) vs = total concentration in the solidified products Mass of a volatile, magma, lava or tephra: mr = original mass of a volatile in the magma chamber mv =mass of a volatile released into the atmosphere at the crater ml = mass of a volatile released into the atmosphere during the lava outflow mc = mass of a volatile released into the atmosphere after the lava emplacement ms = mass of a volatile in solid eruption products mtot(r) = total mass of magma degassed at the crater mtot(l) = total mass of the lava mtot(t) = total mass of the tephra

All information on the amount of gas release during the eruption can only come from the solid eruption products that are still there: the lava and the tephra. The residual concentrations of volatiles in these rocks can be determined in the lab, by analysing samples taken at the site of the eruption. If we then also know the initial concentrations of volatiles in the magma chamber it is easy to calculate how much gas was emitted in total.

‘Melt inclusions’ are small quantities of the original magma that are locked inside a crystal as a solidified microscopic ‘droplet’. This sometimes happens when crystals grow in a magma chamber. Because the droplets are shut off from the surrounding magma, their composition (including the concentrations of volatile elements) will not change anymore. The crystals reach the surface with the erupting magma, and will be present in the solid eruption products. Geologists use melt inclusions to determine the concentrations of volatiles present in a magma before it is (completely) degassed. Despite the small size of the ‘droplets’ (sometimes no more than a few micrometres)

magma chamber

vi

vt

mv = mtot(r) (vi - vt)

ml+mc = mtot(l) (vt –vc)

lava

crust

mr = mtot(r) vi

Eruption column and distant fog

Local fog

Figure 4.9: Schematic view of the emission of a volatile gas during the Laki eruption. After Thordarson et al. (1996).


81

advanced analytical techniques are capable to measure the concentrations of sulphur, chlorine and other volatiles.

Exercise 4-7*: Melt inclusions Basaltic magma consists of many different chemical components. When a basaltic liquid of say 1300 °C cools, it will start crystallising at some point (as in section 4.2). Crystals of different minerals will form at different temperatures. Olivine, (Mg, Fe)2SiO4, is usually the first mineral to crystallise in a basaltic melt. Melt inclusions representing original basaltic magma are locked into early formed crystals (like olivine). Explain why this will not be the case for minerals that crystallise at a lower temperature (like quartz).

You can find out more about melt inclusions in table 4.1 and 4.2.

We can calculate the amount of volatiles that entered the atmosphere from the difference between their concentrations in the magma chamber and their concentrations in the erupted products, and by taking the total amount of erupted magma into account. In the following exercises we will do this for sulphur (S) and chlorine (Cl), because these two volcanic volatiles had a bad impact on the atmosphere.

Exercise 4-8**: How would you theoretically calculate how much of a volatile element went into the air during an eruption in the past? Formulate an equation for the total mass of a volatile that was emitted during an eruption in the past (mb). Only use variables that we are able to determine now, after the eruption took place. Use Figure 4.9 as a guide. Hint: quantity before (inside the magma chamber) – quantity after eruption (left in the products) = quantity escaped to the atmosphere.

Exercise 4-9***: How much were the losses of volatile concentrations in the magma during the Laki eruption? In the above section we described how degassing proceeded as a stepwise process. Here, we will determine the concentration losses of sulphur and chlorine for each different step. Use the measured concentrations of these volatiles as given in table 4-1. Complete table 4-2 by filling in the decrease in concentration of both volatiles.

S Cl Melt inclusions, vi 1675 310 Tephra, vt 490 225 Lava during transport, vl 350 185 Solidified lava after emplacement, vc 195 150 All solid eruption products (lava + tephra averaged), vs 205 150 Table 4-1: Average concentrations (mass ppm = mg/kg) of sulphur and chlorine in eruption products and melt inclusions.

ΔS ΔCl Decrease due to total degassing (vi – vs) Decrease due to degassing at crater (vi - vt) Decrease in the lava during transport (vt – vl) Decrease in the lava after transport (vl – vc) Table 4-2: Decrease in the concentrations of sulphur and chlorine (in mass ppm) during each different phase of the eruption.

Exercise 4-10***: How much SO2 and HCl gas was emitted? In tables 1 and 2 the concentrations of the volatiles were given in terms of the ‘pure’ elements sulphur and chlorine, as measured by advanced methods for chemical analysis. However, they do not enter the atmosphere as pure elements, but usually as molecules such as SO2 and HCl. We will now calculate the amounts of SO2 and HCl gas emitted during each phase of the eruption. Use your answers, the data in table 4-1, table 4-2 and Figure 4.9 and the example below to calculate the variables in table 4-3 (use Excel for this exercise). You can calculate the mass of each gas component that was emitted only at the crater as follows:


82

rvvVvvmm txixrtirtotv ,,)(

with Vr = total volume of erupted magma = 15.1 km3

ρ = density of the magma = 2750 kg/m3

vx,i = mass fraction of element “x” in the melt inclusion (thus in the original magma) vx,t = mass fraction of element “x” in the tephra r = a constant to convert the mass of the pure element into the mass of the gas molecule: molecule mass / element mass

You can find the molar masses of the elements in BINAS (book with tables) or on the Internet (Wikipedia).

Exercise 4-11**: Difference between the percentages of S and Cl that escaped What are the percentages of S and Cl that escaped during the eruption? Use the data given in the previous exercise. Try to think of a reason why there is a difference.

SO2 (kg) HCl (kg)

Original total mass inside the magma chamber

Total mass escaped during the eruption

Mass of the remaining portion in the solid eruption products (lava + tephra)

Mass emitted during the eruption

Table 4-3: Mass of SO2 and HCl (in kg).

0

20

40

60

80

100

120

140

Door de Laki activiteit (8 jun 1783 - 7 feb 1784)

Door alle vulkanen perjaar

Door de mens wereldwijdper jaar

Meg

ato

n (

= 1

09

kg)

SO2 HCl

Figure 4.10: Gas emissions into the atmosphere: the Laki emissions compared to the annual emissions of all volcanoes put together and emissions caused by human activity (in the year 2000).

Through Laki activity Through all vulcano’s per year

Through humans worldwide

per year


83

Exercise 4-12**: The Laki eruption in perspective In Figure 4.10 the Laki eruption is compared, in terms of emitted gases, to all volcanoes on Earth and to human activities. It is estimated that volcanoes produce about 18 x 109 kg HCl and 6 x 109 kg SO2 annually. Human activity clearly contributes more: 68 x 109 kg SO2 and 13 x 109 kg HCl. The burning of coal, oil and other industrial sources are the main sources of SO2, and the burning of biomass and fossil fuels for HCl. a. Use your answers for exercise 4-10 to calculate how much bigger or smaller the SO2 and HCl

emissions from the Laki eruption were than the average total annual volcano emissions. b. Do the same, but now compare your answers to the total emissions due to human activity.

4.6 Acid rain in Western Europe. The eruption plume of the Laki eruption reached a height of 9-13 km. The gases that escaped from the crater could have reached the tropopause, the boundary between the stratosphere and the troposphere. This is where a permanent strong wind blows: the polar jet stream. This ‘jet stream’ transported much of the gas eastward.

Examine the weather maps above (Figure 4.12 and Figure 4.11 to see how the particles travelled towards Western Europe. As they travelled in the atmosphere, the gas molecules could have reacted with water. However, the two gas molecules behave quite differently. HCl reacts with water immediately so that it usually rains out quickly. SO2 is converted much more slowly to H2SO4 and can therefore travel over much greater distances. This is why it could have reached Western Europe. From now on we will consider only SO2.

Within 50 hours of the eruption the jet stream had transported the SO2 molecules all the way to the Netherlands. However, it would take another 1-3 weeks to see the effects: a dense, dry fog lying over the country and a red sun. This delay can be explained by the slow transformation of SO2 into H2SO4. The tiny droplets that were formed are called ‘aerosols’. To understand the process by which they are formed we will examine acid rain chemistry.

Rain is part of the water cycle. Sunlight heats the oceans and causes the water to evaporate. Wind blows this vapour around, and water is precipitated as rain or snow. In a ‘clean’ atmosphere, fresh water would fill lakes and rivers, and supply whole ecosystems. You would expect this water to

Figure 4.11: Cross section of point A to point B in figure 1-11. This figure shows the eruption column of the Laki volcano and the resulting transportation of the column to the European mainland. Source: Thordarson & Self, 2003

Figure 4.12: Meteorological map ofEurope on 23 June 1783, includingthe atmospheric pressure (thin lines) and the jet stream directions (boldlines). Source: Thordarson & Self,2003


84

have a pH of about 7. However, the pH is lower than 7 because CO2 is present in the atmosphere. Water reacts with CO2 in the following way:

H2O + CO2 H2CO3 (pKCO2 (25°C) = 1.47)

H2CO3 + H2O H3O+ + HCO3

- (aq) (pKz1 (25°C) = 6.35)

Even though H2CO3 is a weak acid, enough H3O+ is formed to lower the pH of fresh water to 5,66.

Exercise 4-13***: A pH of 5,66 The concentration of CO2 in the atmosphere is denoted with the partial pressure PCO2. We will take PCO2 = 10-3.5 atm. Use this information to show how the pH of fresh water drops to 5.66: a. Give the reaction equations for all equilibria and use PCO2 = [H2O][CO2]. b. Then use substitution to solve the equation. Note that pH = -log[H3O

+]

Since the beginning of the Industrial Revolution (about 1850), man-made emissions of compounds such as SOx and NOx have caused a strong increase in the acidification of rain. Rainwater that mixes with these compounds will be lowered in pH of so that it will become acid.

But even without the Industrial Revolution, volcanoes are already a natural source of acidification, because volcanic SO2 could create acid rain. How does this happen?

We have already seen that SO2 is emitted from magma as a gas. This SO2 reacts with various oxidants (like ozone, O3), forming SO3. This compound then dissolves in water, producing H2SO4 (sulphuric acid):

SO2(g) + O3(g) O2(g) + SO3(g)

SO3(g) + H2O(l) SO3(aq)

SO3(aq) + H2O H2SO4

Sulphuric acid can then react with water to form H3O+:

H2SO4 + H2O H3O+ + HSO4

-

HSO4- + H2O H3O

+ + SO42-

SO2 can also react directly with water droplets, forming sulphurous acid:

SO2(g) + H2O(l) SO2(aq)

SO2(aq) + H2O H2SO3

Exercise 4-14*: Dissociation of H2SO4

H2SO4 will dissociate in water. Write down the chemical equations describing the dissociation of H2SO4. Hint: in every reaction, the molecule loses a proton: H+. i H2SO4 + H2O ...

In the next sections we will focus on H2SO4. Being the strongest acid involved, it has the strongest influence on the pH. Assume that the Laki volcano emitted 122 megatonnes of SO2. Also assume that only 5,0% of the total emission reached the Netherlands, and that this was contained by 32.6 x 1015 kg of air.

Exercise 4-15**: The concentration of SO2 in air. What would have been the final concentration of SO2 in the air, using these data? Express your answer in mass ppm (mass ppm = mg/kg)


85

Exercise 4-16**: Does the SO2 solubility in water limit the pH of rain? SO2 dissolves in water, but there is a limitation to its solubility, which is controlled by the temperature. Will this limitation determine the ultimate acidity of the raindrops in which the SO2 dissolves? Figure 4.13 to answer this question.

Exercise 4-17***: The pH of rainwater affected by the eruption Assume that all the SO2 will dissolve in the water that is present in the air (9 gram H2O per kg air). a. Calculate the concentration of SO2 in mole/L. Use the molecular mass (64 mole/g) and the SO2

concentration you found in question 14. b. Calculate the H3O

+ concentration and thus the pH of the rain water. Use the molar mass of SO2 and the density of water (look these up on Wikipedia). Hint: use the chemical equation of the reaction of SO2 to SO4

2- (aq) and assume that this reaction proceeds completely.

4.7 The consequences of acid rain Acid rain has many negative effects:

1. A lower pH of surface water is harmful to fish populations and other water animals. At a pH < 5 fish eggs will not hatch and fish die. Biodiversity will decrease.

2. Trees weaken, which makes them more vulnerable to diseases and storms. Entire forests might die in this way.

3. Acid rain depletes the soil in important components. H3O+ ions cause soil minerals to

dissolve in water, releasing cations (e.g. Ca2+, Al3+) that are transported away. Important nutrients become depleted, or plants could be poisoned by overdoses of cations.

4. Buildings and monuments are damaged by the acid.

We will now take a closer look at how the damage to buildings and monuments happens. Limestone and marble are examples of rocks that have been used as construction materials since

Oplosbaarheid SO2

0

5

10

15

20

25

0 20 40 60 80 100 120

T[graden C]

op

losb

aarh

eid

[g/1

00m

l]

Figure 4.13: Solubility of SO2 and water at different temperature

SOLUBILITY

Solubility

T (centigrades °)


86

ages. Both rocks mainly consist of the minerals calcium carbonate (CaCO3) or magnesium carbonate (MgCO3).

Limestone is a sedimentary rock, which transforms into the metamorphic rock “marble” when put under high pressure and temperature. The crystalline structure of these rocks is therefore different. Limestone consists of smaller crystals and is more porous, making it lighter and easier to work with. It is used as a standard building material. Marble, on the other hand, has larger crystals and a low porosity, and can be polished to a great lustre. It is often preferred for monuments and statues for obvious reasons.

Both rocks are considered to be strong and durable. Yet, in recent times they can be strongly affected by acid rain. What happens?

Exercise 4-18*: Dissolution of (Ca,Mg)CO3 Limestone is affected by acid rain. This is due to the reaction between calcium (or magnesium) carbonate and a solution of sulphuric acid. Describe the reactions.

Exercise 4-19*: Climate effects of acid rain?

“Volcanic acid rain contributes to climate change” What do you think about this statement? Explain.

The exterior of buildings and monuments are primarily affected by acid rain. Acid rain can easily destroy the details on a relief (see Figure 4.15 for example). The structural cohesion of buildings is usually not influenced.

The extent of the damage is determined not only by the pH of the acid rain, but also by the amount of water that reaches a certain surface. Areas that are protected by roofing will not be damaged nearly as much as decorations that are out in the open. But protected areas can still be affected in other ways. When the water evaporates, it leaves all the ions behind that were dissolved in it. Residues containing calcium and sulphate ions will cause the formation of gypsum, CaSO4▪2H2O(s). This mineral contains water molecules in its crystal structure. Gypsum is very soft (hardness 2 on the Mohs scale) and forms sheet-like or needle-like crystals. It has an open structure so that it forms porous layers. Because the mineral is soluble in water, it will wash away in places where a lot of rain falls. In sheltered locations, however, it will accumulate. This attracts dust, carbon particles, dry ash and other pollutants, promoted by its open crystal structure (see Figure 4.18). Surfaces on which gypsum accumulates will turn black.

Figure 4.15: This photo of a statuedecorating Lincoln cathedral in theUnited Kingdom, was taken in 1910Source: Humphreys, 2003

Figure 4.15: In 1984, only 74 years later, acid rain affected the statue so severely that it is now barely recognisable. Source: Humphreys, 2003


87

A serious problem arises when water containing dissolved sulphate ions enters deep into the limestone pores. All combined, these pores have a surface area that is many times that of the outer surface area of a rock, so that any reactions will be speeded up. When the acid water enters the pores, gypsum crystals will form inside. Because the crystals have a slightly larger volume than the pores, they put the rock under pressure, which may crack and break. If the gypsum is later dissolved and washed away in fresh water, a very unstable rock remains. Porosity is therefore an

important factor affecting the durability and strength of a building material.

Exercise 4-20**: Negative effects of acid rain on buildings Consider the scenario that a large volcanic eruption produces acid rain that falls on the Dam Monument in Amsterdam and on the Dom tower in Utrecht (both in The Netherlands). Which of the two will be affected most by such an event? Hint: use the following websites(in Dutch): http://www.nitg.tno.nl/ned/products/stenen%20rond%20de%20dom%208%20pag.pdf http://www.monumentenonderhoud.nl/projecten/dam2.html

Exercise 4-21*: How does the pH change during the reaction? You have seen that calcium carbonate reacts with H2SO4. What happens to the pH, and why?

Exercise 4-22***: How much CO2 is emitted? Assume that the pH of the acid rain formed during the eruption is 3.19 and that all of the H2SO4 is consumed during a reaction with calcium carbonate. Use -log[H3O

+] = pH, and use the chemical equation. a. How much CaCO3 is dissolved per litre of rainwater? Assume that the molar mass of CaCO3 is 100.09 g/mol. We have seen that CO2 is released during the reaction between acid rain and limestone. b. How much CO2 is released during the reaction (per litre of rain water)? Assume the molar mass

of CO2= 44.01 g/mole

Exercise 4-23***: Dissolving the Dam Monument The Dam Monument in Amsterdam is 22m high. Imagine the simplified form of a cylinder with a diameter of 4.5m. Now calculate how much acid rain would be needed to completely dissolve the Dam Monument: a. What is the total mass of the monument? Take the density or the rock as 2.93 g/cm3 is. Hint:

What is the volume of the cylinder?

Figure 4.18: Black tarnish on a Chicago building made oflimestone Source: USGS, 1997

Figure 4.18: Gypsum crust on a balustrade in Washington D.C. Source: USGS, 1997

Figure 4.18: Electron microscopy photograph of gypsum crystals with visible dirt particles ‘trapped’ in the open structure. Source: USGS, 1997


88

b. What is the volume of the rain water that would be needed to completely dissolve the monument? Use your answer from exercise 21a.

Final assignment CH4. Main and section questions a. Answer each of the seven section questions as well as the main questions asked at the

beginning of this chapter. b. Did this chapter inspire you to thing about new questions? Write these down.


89

Attachment 4-1 Melt inclusions Many people believe that the solid crust of the Earth floats on a ‘sea’ of burning magma. This is incorrect. Magma is formed only in specific locations; the rest of the mantle is solid rock. Rock starts melting when conditions change, for example when temperature increases or pressure decreases. Generally this happens in the upper part of the mantle.

The molten rock rises because its density is lower than the surrounding solid rock. Usually the melt will accumulate in a magma chamber before it rises further to the surface during an eruption. If the magma starts cooling in a magma chamber, crystals of one or more mineral types will grow. In the 1920s, Norman L. Bowen discovered that these minerals crystallise in a certain order. In a basaltic magma, olivine will usually crystallise first (see section 4.2).

If a crystal grows fast enough, tiny droplets of the melt can be locked inside. A small volume like this is called a ‘melt inclusion’ (see Figure 4.19). The moment the inclusions are ‘trapped’, they are protected from contact with the surrounding magma that usually will change its composition. In

this way, a melt inclusion records the original composition of a magma.

Melt inclusions are incredibly small, with a diameter of only 1-300 µm (µ = 10-6), and therefore difficult to study. Advanced methods to analyze them have been developed (such as the electron microprobe or ion microprobe). Using these, much information can be gathered from the melt inclusions. For instance:

The original composition of the magma can be determined because the inclusions are locked in during the growth of crystals in the magma chamber. For example, it will tell us in what concentrations volatile elements were present in the magma before it arrived at the surface. This is important information. As you have seen, geologists use these data to calculate the amounts of gas that were emitted during historic eruptions.

Melt inclusions also provide information about the pressure and temperature conditions under which magmas crystallise. They tell us more about the history of a magma.

If the inclusions themselves are cooled slowly, they may crystallise as well. Figure 4.20 shows what happens, depending on the cooling rate.

Figure 4.19: Photo B: Different elongate melt inclusions, most without gas bubbles. The photo is about 700µm wide. Photo C: crystallised melt inclusion in quartz. Photo E: three glassy meltinclusions of various sizes, each without bubbles. Source: Lowenstern, 2003.


90

A melt inclusion consisting of glass with a gas bubble gives us information about the depth at which the crystal started forming, while crystallised melt inclusions tell us more about the speed of eruption and/or about the cooling rate.

Sources:

Lowenstern, J.B. USGS Melt Inclusion Page. August 2003. 1 December 2006.

http://volcanoes.usgs.gov/staff/jlowenstern/Melt%20Inc%20Page/melt_inclusion_page.html

Marshak, S. Earth: Portrait of a Planet. New York: W.W. Norton & Company, 2001. pp. 137-147.

Figure 4.20: schematic illustration of four melt inclusions that cooled at different rates. A: At high cooling rates, no crystalsand no gas bubbles will form and the inclusions solidifies to aglass. B: gas bubbles may form when the cooling rate is slightly lower. C: when the inclusion cools slowly, gas bubbles can grow through diffusion and crystals can start forming. D: theinclusion may crystallise completely if it is allowed to cool very slowly. Source: Lowenstern (2003)


91

Attachment 4-2 The Earth’s atmosphere The Earth is covered with a layer of air which we call the atmosphere. It stretches up to 700 km above the Earth’s surface. In the past it was mainly studied for relating it to the weather and weather patterns. Today, the use of modern space instruments enables us to gain a much more profound understanding of how the atmosphere functions.

The atmosphere is of vital importance to life on Earth. It contains oxygen, necessary for organisms to breathe. It protects us against the Sun’s ultraviolet radiation and against (small) meteorites. It maintains the Earth’s surface temperature, keeping it warm enough for life. A constant water cycle (water to vapour to rain, and back to surface water) is kept in place by the displacement of air in the troposphere. This hydrological cycle is a crucial component to sustain life.

The atmosphere can be divided into various layers (see Figure 4.21). They are identified according to their thermal properties (differences in temperature), chemical composition and density (see BINAS or other book with tables).

Troposphere

The troposphere is the layer closest to the surface and the one in which we live. It is between 8 and 17 km thick. It has the highest density and contains 75% of the gases present in the atmosphere. It is warmest near the ground and cools quickly towards the tropopause, its upper boundary. The temperature drops from 17 oC at the Earth’s surface to about -60 oC at an altitude of 15 km. This causes great instability inside the layer. Hot air is relatively light and will rise, whereas cold air is dense and will sink. Consequently, our weather is mostly created within the troposphere.

Stratosphere

The stratosphere is bounded by the tropopause beneath it and the stratopause, which lies at 50 km above the Earth’s surface, above it. Travelling upward through the stratosphere, the temperature increases from -60 to about 10 oC near the stratopause.

The stratosphere contains 19% of all the atmosphere’s gases and very little water vapour. Movements within the stratosphere are very calm compared to those in the troposphere, and mostly occur in a horizontal direction. This explains the layered structure of high stratus clouds. The Ozone Layer is also situated in the stratosphere. Ozone blocks UV radiation by absorbing it.

Mesosphere

Next is the mesosphere, 50 to 80 km above the Earth’s surface. The air here is too thin to absorb any solar radiation, so the temperature drops here to -120 oC, the lowest value in the atmosphere. Nevertheless, it is thick enough to slow or even burn meteorites that enter the atmosphere. A glowing meteorite is the “falling star” that we see in the sky at night.

Figure 4.21: Schematic picture of the subdivision of the atmosphere into various layers. The red line depicts the temperature distribution.


92

Thermosphere

The temperature starts to rise again in the next layer, called the thermosphere. Heat transport through conduction is important here. The air is even thinner than in the mesosphere, but it still absorbs UV radiation from the sun. Temperatures in the upper part of this layer (700 km) can be as high as 2000 oC.

Ionosphere: The Ionosphere is part of the thermosphere but is also distinct layer in itself. It consists of ionised gas particles produced by the UV light. This layer is important because it reflects radio signals that are emitted from the Earth back to the surface. This is why radio signals can be received in all parts of the world.

Exosphere: This is the outermost layer of the atmosphere and lies between 700 and 800 km above the Earth’s surface. The air is so thin that it gradually disappears into space as you go upward. Most of this part of the atmosphere is made up of hydrogen and helium (but concentrations are extremely low).

The atmosphere’s composition

The atmosphere consists mainly of nitrogen (N2, 78%), oxygen (O2, 21%), and argon (Ar, 1%). Other important constituents are water (H2O, 0-7%), ozone (O3, 0-0.01%), and carbon dioxide (CO2, 0.01-0.1%).

Sources:

ThinkQuest. SpaceShip Earth. , http://mediatheek.thinkquest.nl/~ll125/nl/home_nl.htm>

NASA. Liftoff to Space Exploration. http://liftoff.msfc.nasa.gov/

Encyclopædia Britannica 2005 Ultimate Reference Suite DVD. Atmosphere. Copyright © 1994-2003. Encyclopædia Britannica, Inc. November 26, 2006.


93

Attachment 4-3 Aerosols Aerosols are tiny particles floating in the air. Some have a natural origin, coming from volcanic eruptions, dust storms (e.g. the Sahara sand that sometimes reaches Northern Europe), forest fires, vegetation or sea salt. Human activity, such as the burning of fossil fuels and changing natural landscapes, is another important source (Figure 4.22).

Aerosols can be divided into five categories according to their particle size and composition: dust, soot, sulphate, sea salt (also called marine aerosol) and organic aerosol. Dust and sea salt are generally larger than 1 µm; soot, sulphate and organic aerosols are usually smaller than 1 µm.

Ten percent of the total amount of aerosols in the atmosphere is produced by human activity. Most of this is concentrated on the Northern Hemisphere, especially near industrial zones, in areas where farm land is created by forest burning, and in overgrazed pasture land.

We do not know exactly how aerosols influence the climate, nor do we have a good estimate of the

relative contributions of natural and human aerosols. Specifically, little is understood about the general effect of aerosols: do they actually cool or heat the planet up?

So why are we concerned about aerosols?

Measuring their concentrations is important for two reasons. Firstly, aerosols cause smog (= smoke and fog) that can lead to respiration problems (see Figure 4.22). This is evident in big cities where the large use of fossil fuels produces many aerosols.

Figure 4.22: Aerosols larger than 1 micrometer are produced by sea salt and dust which are transported by the wind. Smaller aerosols are formed by processes of condensation like the reaction of SO2 to sulphate particles, and also when soot and smoke are formed in a fire. After they are formed, aerosols are mixed and transported through the atmosphere. The most important process that removes aerosols is precipitation: rain, snow or hail. Source: NASA


94

Secondly, aerosols play a role in the Greenhouse effect. Depending in the type, they can have a cooling or a warming effect on the thermal balance of the Earth. Just like greenhouse gases, they can absorb the infrared radiation that the Earth produces.

Cooling can be a direct or an indirect effect.

Directly: aerosols reflect solar radiation and consequently decrease the amount of sunlight that

Figure 4.25b: In the case of a high aerosol concentration, there are many nucleation sites for rain drops. Up to 90% of the visible spectrum can be reflected. Source: NASA

Figure 4.25a: Clouds with a low aerosolconcentration and few droplets reflect thelight only poorly: most of the sunlight willreach the Earth’s surface. Source: NASA

Figure 4.23: Smog above Mexico City. Source: KNMI


95

reaches the Earth’s surface (see Figure 4.25 Figure 4.25b). The magnitude of this effect depends on the size and reflectivity of the aerosols.

Indirectly: aerosols change the properties of clouds. Without aerosols, we would not have any clouds at all. It is extremely difficult to form clouds without small aerosols that act as nuclei for the formation of droplets. More aerosols mean more clouds. Because the total amount of condensed water inside a cloud does not change very much, the droplets will be smaller on average. This has two effects: clouds with smaller droplets reflect more sunlight (see Figure 4.25b) and these clouds remain longer in the atmosphere because it takes more time for the droplets to grow large enough to fall as rain. Both effects increase the reflection of sunlight back into space.

Under normal circumstances, the majority of aerosols form a thin fog in the low atmosphere. They will rain out within a week. But aerosols are also found in the stratosphere (see Figure 4.26). For example, a volcano can spew large amounts of aerosols into the atmosphere. Because it does not rain in the stratosphere, aerosols can remain there for years. The effects are not only beautiful sunsets, but also cooler temperatures, especially in the summer.

In the past thirty years much has been discovered about aerosols. Scientists can distinguish different types, and can estimate the amounts that are present for each season and locality. But much remains to be explored. Essential details about quantities and properties are missing, so that it is not possible yet to accurately determine the effect of aerosols on temperature at the Earth’s surface on a global scale.

Based on:

NASA Earth Observatory. Aerosols and Climate Change.

<http://earthobservatory.nasa.gov/Library/Aerosols/>

KNMI. Aërosolen. <http://www.knmi.nl/globe/informatie/aerosol.html>

Figure 4.26: A volcano spewing aerosols. Source: NASA.

The Dynamic Earth Mountain building

96

Chapter 5. Mountain building The main question for this optional chapter is:

How are mountains formed and how do rocks behave in this process?

We will answer this main question by addressing the following section questions:

What is the relation between the process of mountain building and plate tectonics? (5.1) What do rocks tell us about the way mountains are formed? (5.2) What is the relation between force, stress and deformation? (5.3) What happens when deformation takes place quickly (brittle behaviour)? (5.4) What happens when deformation takes place slowly? (ductile behaviour) (5.5) What does this tell us about the way mountains are formed and about plate tectonics? (5.6)

Objective: To be able to understand and describe the (small scale) processes that are important for the formation of mountains, and to discover how these processes work. For this you need some knowledge from mathematics, physics and chemistry.

Mountain building The Dynamic Earth

97

5.1 The relation between mountain formation and plate tectonics Mountains are found only in certain places. Well-known examples of mountain ranges are the Alps, the Himalayas and the Rockies. But mountains play an important role for the geography everywhere in the world. Soils all over the world originate as rocks and particles eroded from mountain slopes, and without this erosion and transport, many areas in the world would lie below sea level. To complete the cycle, current basins could form the source material for future mountain ranges which are yet to form. To determine how this exactly works, we will first take a look at how mountains are formed and how this relates to plate tectonics.

In this section, we will discuss the relation between mountains and plate tectonics. Next, we will see what kind of information we can obtain from rocks and what they tell us about the processes that are important for mountain formation.

Exercise 5-1*: Mountains and plate tectonics A, I When plates of the same type collide, a contact is formed between two plates of the same weight. The clearest example is India, which was once connected to the South Pole. After migrating to the north, it connected to Asia. Everything that once lay between the two continents is now heavily deformed and folded, and currently makes up the Himalayas, the highest mountain range in the world. Proof for this theory is, amongst others, the presence of fossil sea shells that are found in the rocks. a. Why will you find no news in the papers about the event described in the above text? b. Draw the mountain range which is described in the text on your empty world map. Also draw

the Alps, the Ardennes, the Andes, the Rocky Mountains and the Ethiopian Highlands, and any other mountain range you might have been to.

c. Surf to www.earthweek.com to see whether any earthquakes or volcano eruptions have occurred anywhere near these mountain ranges.

d. What do you conclude? Is there a relationship between mountain ranges, volcanoes and earthquakes?

e. Now take a look at GB 192B. What do you notice about the locations of the Himalayas and the Andes compared to the type of plate boundaries that are indicated on the map?

To form a mountain range some source material must be present. Delta areas such as The Netherlands or the Po basin could be the start of a new mountain range, because a lot of sediment is present (and is still being deposited) in these regions. This deposition or accumulation of large amounts of sediments generally takes place in coastal areas where rivers carry a lot of new material to the sea. The basin (in the ideal case one which subsides slowly) supplies the rivers with space for deposition. As you can see, basically two things are necessary: firstly, the supply of a lot of sediment and secondly, room for the sediment to be deposited: usually some basin. This is step one in the process of building a mountain range.

Earth Scientists call the process of mountain building orogenesis. It takes place at convergent plate boundaries, and a distinction between two types of orogenesis can be made:

(1) Orogenesis as a result of convergence of an oceanic and a continental plate: in this case, orogenesis takes place inside the continental plate - the oceanic plate only subducts. An example of such an orogenic belt is the Andes. Sedimentary, volcanic and other (magmatic) rocks are pressed together as a result of the compressional force of the subducting plate pushing against the continental plate. This results in the formation of faults and folds, and part of the rock is pushed up. In this case, the processes of accumulation and orogenesis take place simultaneously: the material is pushed together and up because of the compressional forces which are caused by the orogenesis.

(2) Orogenesis as a result of convergence of two continental plates: in this case, the rocks of both plates are deformed. Both plates have the same weight and neither subducts, so that there is no way out but up. An example of such an orogenic belt is the Himalayas, formed when the continent India collided against the Eurasian plate. In fact, this process is still happening today.


98

(Convergence of two oceanic plates will always result in subduction: it never happens that both plates have exactly the same weight. This is because one of the plates is always older and therefore heavier than the other. The older an oceanic plate gets, the more it cools and the higher its density becomes. The older plate of the two will therefore always subduct beneath the younger one.

Orogenic belts are formed through uplift. However, as soon as relief is formed, the process of erosion begins. In the case of strong uplift (like in the Himalayas), the mountains will become higher and higher because the uplift outweighs the erosion. As soon as the erosion starts compensating for the uplift, the mountains stop growing.

So mountains are formed when plates converge. However, if the convergent motion stops, this doesn’t mean that vertical motion stops as well. This vertical motion continues until isostasy is reached: the gravitational equilibrium between the Earth’s hard crust and the underlying mantle material which deforms much more easily.

You can visualise isostasy by thinking of the crust ‘floating’ on the mantle (see chapter 6 for a more elaborate discussion). When the crust becomes heavier, as is the case when mountains are built, it must sink in order to compensate for this extra weight. Think of a boat lying deeper in the water when it is filled with cargo.

This is how it works: equilibrium is achieved when the mass of the crust lying on the mantle is equal to the mass of the mantle material which is displaced. In other words, if an amount of crust material of mass A is added to the crust, an amount of mantle material of the same mass A must be displaced.

Think of the following simplified situation:

A 2 cm thick piece of wood floats in a bucket of water. About 1 cm of the wood is submerged in the water. Think of the water as the mantle material and of the wood as the crust. Now suppose you take a piece of the same type of wood, but this one is 4 cm thick. As you may have guessed, it sinks deeper in the water, so that 2 cm will be beneath the surface: half of its height just like for the other piece of wood.

Exactly the same is happening inside the Earth, only on a bigger scale.

The crust will move vertically until it is in isostatic equilibrium with the mantle. If the crust becomes heavier, it will sink deeper into the mantle to achieve equilibrium. As a result, mountain ranges have a so-called mountain belt root (see Figure 5.1).

Figure 5-1: Mountain building by convergence of the Oceanic and continentalplate. On the continental plate, both vulcanism and mountain building can benoticed.http://plaattektoniek.htmlplanet.com/plaattektoniek/botsingzones.htm

Oceanic plate Continental Plate

Crust

lithosphere

asthenosphere


99

As long as the plate motion is quick enough, as is the case in the Himalayas, no isostatic equilibrium will be reached. This is because the mantle may be plastic, but only on a very long timescale. It cannot move rapidly to compensate for the new extra mass, so the Himalayas are so high only because of the rapid convergence of the Indian plate towards the Eurasian plate. As long as India keeps pushing, no isostatic equilibrium will be reached. When the orogenesis stops, however, the system is bound to reach equilibrium at some point.

But even when the root has sunk deep enough to compensate for the overload, vertical motion won’t stop. Erosion at the Earth’s surface causes the overload to decrease, so that the isostatic equilibrium is disrupted and the root will move up again. New relief is formed in this way, causing erosion, and more upward motion. In this way, rocks that originally lay kilometres beneath the Earth’s crust can reach the surface. Mantle material will flow back to where the root lay before. Vertical motion will continue to take place until all of the mountain range has been eroded.

Exercise 5-2**: Motion around the Netherlands A a. Look up in your atlas where the Ardennes are located and when they have been formed. b. What convergent plate motion caused the Ardennes to form? During the Permian geological period, all continents were connected and formed one supercontinent called Pangaea (see GB 193). In the past, various cycles have taken place in which continents were formed and drifted apart consecutively. At least two other supercontinents have been discovered dating from before Pangaea. Such a cycle is called a Wilson Cycle. c. Since the Archaean (2500 million years ago), at least four Wilson Cycles have been

distinguished. You can calculate the average duration of such a cycle. Given this information, when would the next supercontinent be formed?

d. Take a look at GB 76. Explain how you think Western Europe will be positioned in the future relative to Eastern Europe. Do you think mountains will be formed in this process?

5.2 Information about orogenesis: hidden in rocks To understand how orogenesis takes place we can’t just wait until it happens somewhere: it takes place over timescales which are far too long for human life. We have to retrieve as much information as possible from rocks that we find in mountains which are already there. These tell us about how they were formed and how the mechanisms work that take – or took – place at great depths. Deep inside the Earth, where temperature and pressure are much higher than at the surface, materials behave in a completely different way. Rocks can be deformed in many different ways: fast or slow, under high pressure and temperature or not so much, in the presence of water or not. All these factors influence the rocks, and many different structures can arise because of them. This is why geologists go into the field to study the rocks and take samples. In places where rocks surface that once lay deep beneath the Earth’s surface, much can be learned about their behaviour deep inside the Earth’s crust. This way we can learn about how mountains are formed and deformed.

Apart from just studying rocks from and in the field, experiments are carried out to investigate how a material behaves under different circumstances. This knowledge not only helps us to understand the processes resulting in the deformed rocks we see in the field, but also has important industrial and social applications where materials play a role.

Figure 5.1: Scheme of a mountain root. Source: Earth Science, Tarbuck and Lutgens, p. 158.


100

Opdracht 5-3*: Looking at rocks We will start looking at rocks. Look at the photos on the next few pages and try to answer the following questions for each of them. Fill in the form you find after Figure 5.5. a. What kind of structure do you see? b. What kind of process caused this structure? Answer the following questions for all the photos together: c. Can you group the different examples of structures and processes? Explain your classification. d. When you have divided them in groups, you can see that there are (roughly) two types of

deformational behaviour. What types of behaviour?

Figure 5.2: Rocks


101

Figure 5.4: In the city. Left: Facade of a building at the Oude Gracht in Utrecht, the Netherlands. Right: Iglesia de San Jeronimo, a church in Masaya, Nicaragua

Figure 5.3: In the field left: Playa Marsella, Nicaragua, right; Lubrín, South Spain. Picture’s: Geerke Floor


102

Figure 5.5: Upper picture: Cilindric samples of marble in an experiment: the two samples on the right are pressed in vertical position within a laboratory. Picture from ‘Structural Geology’, Twiss & Moores, 1992 Lower picture: Imitation of mountain building by Henry Moubry Cadell, 1887 http://earth.leeds.ac.uk/assyntgeology/cadell/mountains-gallery/image00.htm


103

Observation sheet opdracht 5-3.

Visible structures What process could have caused this structure to form?

Figure 5.2 Stones

Top photo:

Bottom photo:

Figure 5.3. In the field

left photo:

right photo:

Figure 5.4 In the city

Left photo:

Right photo:

Figure 5.5 In an experiment

Top photo:

Bottom photo:

In sections 5.4 and 5.5 we will take a closer look at the processes causing the most important types of deformational behaviour (the answer to opdracht 5-3). First we will take a closer look at how deformation is influenced by force and stress.


104

5.3 The relation between force, stress and deformation. Force and stress are important quantities to describe deformation. Force is a quantity that is used for many problems in physics. It describes how an object is pushed or pulled – its unit is the Newton (N). Apart from that, we have pressure and stress. In the Earth Sciences, the term pressure is only used in the case of all-sided pressure: a pressure which has the same magnitude in all directions. Think of swimming under water: you feel the same pressure from all sides. This pressure can be calculated with pw = ρgh. Here, pw is the pressure in the water (called hydrostatic pressure), h is the height of the water column above you, ρ is the density of the water and g is the gravity acceleration. (see Figure 5.6).

Stress is a term used in the Earth and Material Sciences to denote the force per unit surface (N/m2 or Pa, pascal) – just like pressure, however, there may be differences in stress in, e.g., the horizontal and vertical directions. Two examples can illustrate the significance of stress:

1. You can imagine that a person on high heels has a higher risk of damaging the parquet floor on which she’s walking than the same person on sneakers. This is because the stress on the floor is higher in the first case: the force of her weight is concentrated on a smaller surface.

2. See Figure 5.7

An object (in our case: a piece of rock) on which different stresses are exerted in the different directions will tend to deform. Stresses inside the Earth are much greater than 1 Pa, so we usually use Megapascal (MPa: 1*106 Pa) or even Gigapascal (GPa: 1*109 Pa). As a reference: even the air pressure at the Earth’s surface is already 1*105 Pa, or 0.1 MPa.

Figure 5.7: The power is the same, only the force (force vs surface) is different. A: a dinosaur can balance on a pilar with a cross cut of surface A1. B: The same dinosaur falls down if it stands on the pilar, where the foot has a smaller surface. A2. Source: ‘Structural geology’, Twiss & Moores, 1992

pw =

Figure 5.6: The pressure on a subject in the water


105

5.3.1 Stress and Mohr diagrams We have seen that stress is defined as the force on a surface. For a given force, it depends on the orientation of the plane what the exact stress on the surface is.

So if you take a small block of rock you can imagine that the force which is exerted on it causes a different stress on each of its sides. When rotating the block, you will find that there is always one direction in which the stress is highest – this is called the maximum

principal stress, denoted with 1. The direction in which the stress is lowest is called the minimum principal stress, or

3. These directions can be proven to be perpendicular to each other (try the mathematical proof yourself if you like!). As the world is three-dimensional, you can imagine there to be a third direction: the intermediate principal stress. This

stress, denoted with 2, is perpendicular

to both 1 and 3 (see Figure 5.8). We

always define 3 2 1. In Figure 5.8 the principal stresses are given on the axes of the coordinate system.

5.3.2 Stress on any plane

If you know the three principal stresses and their directions, you can calculate the stress on any

other plane you want. Triangle a1a2a3. with area A, is marked gray in Figure 5.8. The orientation of the triangle is denoted by its normal vector: a vector perpendicular to the surface starting in the

origin O. The normal touches the surface of the triangle in point h and makes an angle 3 with the

surface of σ1 and σ2. The angles 1 and 2 are defined in the same manner – they have been omitted in the figure for clarity. We don’t need to know the magnitude of all three angles: only two

are necessary to calculate the third because it is known that (sin 1)2 + (sin 2)

2 + (sin 3)2 = 1.

(You can prove this by expressing the distance between h and the surface of σ1 and σ2 in Oh and

sin3. If you do that for all three directions and if you think of Pythagoras’ theorem in three dimensions, you’re almost there.)

We want to determine the stress on triangle a1a2a3. The stress on a plane can be expressed as a vector with components in the three principal directions. The value of each component can be determined in the following way: we need to know what the ‘effect’ of triangle a1a2a3 is in that direction. If you look at the triangle from the direction of σ3, for example, the triangle you see has the size of triangle Oa1a2. Look at the figure and try to visualise this.

Exercise 5-4**: Stress on triangle a1a2a3

a. Angle 3 is also present at point a3. Draw this in the figure. Hint: draw the large triangle O e3 a3 and draw the small one (Oe3h) in it.

b. Now show that Oe3 = sin 3 a3e3 and therefore that surface(a1a2O) = sin 3 A.

Now that we know how to calculate the component of the stress on the surface in the direction of σ3, we can fill in the whole stress vector. Stress is defined as force per unit area, so the component in the stress vector is 3 sin 3 (to see this: you just found that the area in the direction of 3 was A sin 3 so you multiply this by the stress in this direction - 3 – and divide it by the area of the whole triangle – A). You can calculate the components in the other two directions in the same

Figure 5.8: Three components of stress on a surface

(area of ∆ a1a2a3)


106

way. The whole stress vector, written in standard vector notation (x component, y component, z component), in the direction of triangle a1a2a3 becomes:

( 1 sin 1, 2 sin 2, 3 sin 3)

5.3.3 Normal stress and shear stress

The stress on a plane depends on the principal stresses (1, 2, 3) and on the orientation of the surface (given by its normal). It is seldom the case that the stress is exactly perpendicular to the plane. For each stress vector, we can however determine to what extent it is perpendicular to the plane, and to what extent it is parallel. For a given plane, we can decompose the stress vector in exactly those components: the normal stress perpendicular to the plane and the shear stress parallel to it (see Figure 5.9).

Exercise 5-5**: Shear stress

Prove the following: If triangle a1a2a3 is parallel to one of the principal stresses, it follows that the

shear stress on a1a2a3 is zero.

Using some heavier mathematical techniques (a branch of mathematics called Linear Algebra) we can express the stress components in terms of σ and θ. This is rather cumbersome, so we will just discuss the two-dimensional case, where only σ1 and σ2 are involved. In reality, such a simplification proves to be very useful.

5.3.4 The 2D case

Consider the following two-dimensional figure (Figure 5.10).

In Figure 5.10 line A is the 2D section of a 3D surface like we used in Figure 5.10: Two-dimensional view of . Because this is a 2D case, we only need one angle θ. Here, it is the angle between the normal vector on line A and the direction of σ1. The stress on A is indicated by the vector σA, decomposed into τ and σn.

Figure 5.9: Deformation as a result of different types of stress.

Figure 5.10: Two-dimensional view of Figure 5.8

(Lenght of a1a2)


107

Exercise 5-6**: Stress vectors

a. Show for the 2D case that A = ( 1 cos , 2 sin ). Use vector notation: (x component, y

component). In other words, prove that the x component is equal to 1 cos , and the y

component equal to 2 sin This means that you have to decompose A in an x and a y component. Hint: remember that stress vector = force vector / area. And the force vector is

force: length. After that you can substitute using the trigonometric functions: cos = ...

b. How can you be sure that the 1 component of vector σA is smaller than the σ2 component? (the quantities of both components are not given numerically, but you can see that θ in the

figure is clearly smaller than 45

In the optional exercise 5.1 (see end of this chapter) you can derive yourself how τ and σn are derived from σ1, σ2 and θ. This derivation results in the following equations:

1 2 1 2-cos(2 )

2 2n

1 2 sin(2 )2

where Θ is the angle between the normal stress (perpendicular to the plane) and the direction of σ1.

5.3.5 Mohr’s circle

These equations look very complicated and difficult to work with. That is why the German civil engineer Christian Otto Mohr devised a method to visualise stress situations in a material.

To see how this works, think of a circle of radius r. You may know that the coordinates in every point on a circle are defined by x = r cos (α) and y = r sin (α). (Check this!)

If we move the whole circle in the positive x direction by a distance p, the y value for every point in the circle stays the same, but every x coordinate changes by an amount p, resulting in:

x = p + r cos (α) and y = r sin (α), see Figure 5.11 (right hand).

x

y (p+r cos α, r sin α)

α

p r r

y

α

(r cos α, r sin α )

X

Figure 5.11: Coordinates in a circle.


108

Now compare the equations for a circle to the equations we have for the normal and shear stress:

circle stresses

X coordinate p + r cos (α)

Y coordinate r sin (α)

If p is equal to 2

21 , r to

221

and α = 2θ then the equations are actually the same! This

means that we have to take σn for the x coordinate and τ for the y coordinate, and if we fill everything in, we end up with the following Figure 5.12: The Mohr circle cirkel.) – we call this Mohr’s Circle:

If you know the principal stresses σ1 and σ2 in a situation, you can draw the Mohr diagram by

putting them on the σn-axis and drawing a circle through both of them, with its centre exactly in between the maximum and minimum principal stress. Using this diagram, you can very easily determine the normal and shear stress for any plane. You just need to known the angle between the plane and the principal stresses. So the only thing you have to do is determine the angle and just read the stress values from the circle.

You have to be aware of a few pitfalls here: the angle that you measure in the diagram is twice as large as θ, so be certain to always fill in 2θ! Also remember that this diagram is only valid for the two-dimensional case!

The meaning of σ1 , σ2 and σ3.

In most of the exercises we will just consider the maximum and minimum principal stresses in two dimensions. In exercise 5.13, however, we will consider a three dimensional situation using σ1, σ2, and σ3: the maximum, intermediate and minimum principal stress. Think of a very small block/cube which is oriented in exactly such a way that no shear stress is exerted on any of its sides. (see Figure 5.13).

In the Earth, nearly always one of the principal stresses is vertical. The vertical principal stress on a piece of rock can be determined easily from the weight of the ‘pile’ of rocks lying on top of it. As the block cannot just ‘escape’ to the sides, the stress in the horizontal won’t differ a lot to the vertical principal stress. This is called the “confining pressure”, which you can compare to the pressure you feel from all sides when swimming under water. In the case that σ1 = σ2 = σ3, a rock will only change in volume, not in shape.

2θ

p r

Figure 5.12: The Mohr


109

In many cases however, the principal stresses are not completely the same. The Earth’s crust is subjected to all kinds of forces (related to plate tectonics) and as a result, the principal stresses will always differ slightly in magnitude. The difference in magnitude between the maximum and minimum principal stresses, σ1 – σ3 , determines whether a rock will deform or not.

The vertical principal stress (in this case σ1) can be calculated using σ1 = ρ g h (see section 5.3)

where ρ = density of the overlying rock h = depth of the rock g = gravity acceleration

You are now ready to do Optional Exercise 5-2 (CO2 storage and earthquakes) at the end of this chapter.

In this section you have learned about stress and how to work with it. We’ve discussed the principal stresses, normal and shear stress, the Mohr diagram and stress in the Earth. To conclude this section we give you a step-by-step guide how to draw a Mohr diagram:

4. Make a graph with two axes: σn as the horizontal and as the vertical axis. 1. Find the principal stresses σ1 and σ2. Indicate both on the horizontal axis. 2. Determine the centre of the circle – this is exactly halfway between the maximum and

minimum principal stress. The radius of the circle is the distance from the centre to either of the principal stresses. You can now draw the circle.

3. To determine the stresses on a certain plane, find θ – the angle between the plane and σ1. Remember to use 2θ in your Mohr diagram! Measure the angle from σ1, anticlockwise, the way you see in Figure 5.11 and Figure 5.12

4. Now you can simply read the normal and shear stress from the horizontal and vertical axes, respectively.

Figure 5.13: Pressure on a cube of rocks (source: The Earth – An Introduction to Physical Geology, E.J. Tarbuck & F.K. Lutgens, Macmillan Publishing)


110

5.4 Fast deformation (brittle behaviour) The Mohr diagram is a useful means to find out more about brittle behaviour of rocks. So what does ‘brittle behaviour’ mean exactly and what kind of processes are involved?

Exercise 5-7***: Practicing with the Mohr diagram In Figure 5.14 you see a rock with the two principal stresses indicated. The rock broke along the indicated line: this is the fracture plane. We want to know what the normal and shear stresses on the surface were at the moment the rock broke – this can tell us more about the strength of the material. a. Draw Mohr’s circle on the graph paper below, using the

provided information and the step-by-step guide provided at the end of the previous section.

b. Now determine what the normal and shear stresses on the indicated plane were the moment the rock broke.

c. Now consider another rock under the same circumstances. For this rock we happen to know that it broke at = 15 MPa and σn = 35 MPa. Calculate the orientation of the fracture plane.

σ2 = 20 Mpa

25º

σ1 = 50 Mpa

Figure 5.14: Principal stresses on a rock


111

5.4.1 Experiments In a laboratory, we can simulate situations to study how processes work. Experiments are conducted to see how rocks behave when a force is applied to them. We want to do that in a way which is as close as possible to the ‘natural’ conditions under which rocks in the Earth deform. For example, experiments can be conducted at high temperature (~200 up to 1200°C). But the speed of deformation can be varied as well to see how this influences the processes taking place.

Special instruments (like the one you see in Figure 5.15) are used to achieve the enormous forces necessary for deformation of rocks.

Figure 5.15: Deformation device in the High Pressure and Temperature (HPT) rock deformation laboratory of Utrecht University. The right picture shows a close-up. The rock sample (the white cylinder) is contained in between two massive iron cylinders that can move towards each other in a vertical direction. The speed and the force needed can be controlled and measured. A small oven that is placed around the sample makes it possible to reach high temperatures. In this small device, the oven consists of two parts that can be unfolded (right picture).


112

Exercise 5-8***: Experiment with sandstone An experiment has been conducted with a cylinder of sandstone (see Figure 5.16). In table 5-1 you see the normal stresses (σn) and shear

stresses (), both in MPa, that were measured at the moment the rock fractured. In some experiments, the angle of the normal to the direction of σ1 was measured as well.

a. Make a graph of the shear stress (vertical) plotted against the normal stress (horizontal). Use the information from Table 1. You can use the graph paper provided below. Make sure to use the same scale on the horizontal and vertical axes! Take σn as x coordinate and τ as the y coordinate. Try to draw a smooth line through all the points.

b. Now draw Mohr’s circle for experiments 2 and 4 in the graph. Use a compass to draw the

circles and make use of the data in the table (don’t forget the third column!). Also determine the values for the principal stresses σ1 and σ2 in these experiments (remember the convention: σ1 > σ2 ). Hint: make use of the step-by-step guide in section 5.3 and take a look at exercise 5-7.

σn angle

1 10.0 7.5 58º

2 20.0 12.5 58º

3 30.0 17.5

4 40.0 22.5 58º

5 50.0 27.5

6 60.0 32.0

7 70.0 36.0

8 80.0 39.0

9 90.0 41.0

10 100.0 42.5

Table 5-1:

Normal and shear stress on the fracture plane at the moment of fracturing.

Figure 5.16: Left: a shear fracture in deformed sand stone, Right: Rock fractures out of an marble sample. http://www.emeraldinsight.com/fig/0830120103002.png,


113

The line you drew through the σn- points (question a and b) defines the so-called yield criterion. As soon as a point on Mohr’s circle for a certain stress regime touches this line, the rock will ‘yield’: it will fracture. This happens in the plane which corresponds to the place where the line and the circle touch – in other words, where the normal and shear stress are equal to the stresses denoted by the line. We can now determine for every combination of σ1 and σ2 whether a rock will break or not, what the orientation of the fracture plane will be and what the stresses in that plane will be. (to determine all these things, you just need to draw the Mohr diagram for that particular situation). c. What happens to this rock when σ1 = 40 MPa and σ2 = 20 MPa? And when σ1 = 50 MPa and σ2

= 17 MPa? Hint; draw the corresponding Mohr diagram. And remember, if they intersect... The slope of the line determines another material property called the ‘coefficient of internal friction’. d. Now predict what the orientation of the fracture plane in experiment 9 will be (σn = 90, =

41). Sketch the broken sample. Use Figure 5.16 as an example. We’ve now only discussed ‘dry’ rocks. It turns out that the presence of water makes a big difference. Many rocks have small pores in between the grains which are usually interconnected. When these pores are filled with water, an internal pressure (Pf) starts working against σ1 and σ2. The stress which is ‘felt’ by the rock is then reduced to the ‘effective stress’. The effective stress is smaller by Pf) in all directions compared to the ‘dry’ rock. e. Will the effect of Pf differ for the directions σ1 and σ2? (In other words, is Pf different in different

directions?) f. What happens to the Mohr diagram when a rock has an internal fluid pressure Pf (and all

stresses become smaller by an amount Pf)? g. Determine the fluid pressure Pf necessary inside the rock to make it break at σ1 = 90 MPa and

σ2 = 50 MPa. h. What kind of place is more likely to develop earthquakes? A dry spot or a place where the

Earth’s crust is ‘wet’? Explain.


114

5.6 Slow deformation (ductile behaviour) In the previous section we discussed fast deformation, or brittle behaviour. Here we will study ductile deformation: what happens when rocks are deformed at a very slow rate. We shall see that rocks do not fracture in this case.

Exercise 5-9*: Introduction to ductile behavious What factors do you think would influence ductile behaviour in rocks?

Here you will see how you can recognise slow deformation in rocks. Apart from that, we will also discuss the underlying mechanisms that cause this kind of behaviour.

Exercise 5-10***: Maizena – liquid or solid? (including practical) What you need (see Figure 5.17) : 200 grams of Maizena – about 150 mL water – a round bowl – a spatula and/or a fork. Preparation: Pour the maizena in the bowl. Add the water little by little and stir continuously but slowly! Stop when the mixture feels as a thick syrup and can move slowly. Tip: add some lemonade or pigment to the mixture! In this exercise we will discuss the term viscosity. This is a measure for the resistance of a fluid to flow (to deformation), also called ‘thickness’ for true liquids like water or honey. The official definition for viscosity is the ratio of shear stress τ and the speed of deformation. Play with the Maizena and see what happens when you vary the speed at which you stir it. You will see that when you stir at a higher speed (with more force), the viscosity of the mixture increases (it feels more like a solid). If you stir more slowly (with a smaller force) you will see that the viscosity of the mixture decreases (it feels more like a liquid). Note: a maizena solution is a so-called ‘non-Newtonian fluid’ – a fluid whose viscosity varies. This is unlike the behaviour of normal fluids.

Figure 5.17: Demonstration with Maizena


115

Back to rocks: a. What do you think will happen to the viscosity of rocks when the temperature is raised? Why? b. Would rocks need to be melted before they can flow, or can they flow as a solid as well?

Exercise 5-11***: Flowing glass? Glass in old cathedrals (like in Figure 5.18 from the 12th century) is often wider in section at the bottom than at the top. Many people think this is due to plastic deformation (deformation without fracture), where the glass ‘flows’ down due to gravity. a. Do you think this is a plausible explanation? Motivate your answer. We can easily test this hypothesis by doing some calculations. The ‘characteristic time’ t necessary

for deformation is given by: tG

where η = the viscosity (dependent upon temperature); G =

the shear modulus (for glass with a value of 30 GPa). The shear modulus is a property describing the elasticity of a material. It indicates how easily (and fast) a material can deform as a result of a stress. The important thing for elastic deformation is that it’s not permanent: the material will move back to its original shape as soon as the applied stress is lifted.

Viscosity is dependent upon temperature, for which the following law holds: 0

expA

T

with η0 = a constant property of a material (you could call it ‘initial viscosity’. For glass this is η0 = 9 x 10-6 Pa s); A = another material property (not relevant here), with a value of A=3.2 x 104 K; T = the temperature in K. Note: exp [A/T] means: e to the power [A/T] or e[A/T]. b. Calculate t. Take the average temperature to be 20 degrees Centigrade c. What does this answer tell you about the probability of the explanation for the glass thickness

as discussed in the introduction?

So how does plastic deformation work? To understand this, we must zoom in on a process that takes place on a very small scale: intracrystalline slip. Atoms in a crystalline material are arranged in an extremely regular way. One of the mechanisms of plastic deformation is that part of the atoms move to a new position. This must be an equilibrium position: a position in which the overall

Figure 5.18: Church windows in Utrecht (Dom and Jans Church Picture: Geerke Floor.


116

system has a low energy. This happens through translational slip: part of the crystal slips along a plane. In Figure 5.19 the top row of atoms (spacing b) slips over the bottom row due to a stress which is applied to it.

You can calculate the shear stress necessary to displace a row of atoms in a perfect crystal using the following formula:

2

G

where G is the shear modulus

of the material.

A material which is important in geology and is easy to deform is rock salt. For this material, the shear modulus G is ~10 GPa, which would result in a necessary shear stress of about 1.6 GPa. However, lab experiments have shown that a shear stress of only 10 MPa (0.001 GPa) is necessary to deform rock salt (at 200 C)! This clearly shows that the model of intracrystalline slip – where whole rows of atoms move together to a new position – is not valid. There must be another mechanism explaining the deformation of rock salt. More research has shown that instead, deformations happens in a stepwise manner, using so-called ‘dislocations’. A dislocation is a defect in the crystal lattice (see Figure 5.20 and Figure 5.21). A much smaller stress is necessary to deform the crystal as a result of these defects.

Figure 5.19: Translational slip

Figure 5.20: Dislocations in the crystal lattice. The row with the black colored atoms stops halfway the lattice; so, there the lattice has a defect along the line.


117

Exercise 5-12**: The influence of temperature a. What do you think the influence of temperature is on ductile behaviour? Hint: how would

vibrations in the atoms influence deformation? And what is the effect of temperature on vibrations?

b. Would temperature influence brittle behaviour in the same way?

The homologous temperature is used to compare materials. This temperature is defined as the ratio ‘temperature of the material (in Kelvin) / melting point of the material (in Kelvin)’. See Table 5-).

Material Melting point in °C

Basalt (rock) 1000-1200

Granite (rock) 650-800

Olivine (mineral in the Earth’s mantle)

1867

Quartz (mineral in the Earth’s crust) 1650

Aluminium 660

Iron 1538

Ice 0

Table 5-2: The melting point for various substances.

What deforms easier? Granite at 300°C or ice in a glacier? Why? It turns out that the material deforms easier when the ratio mentioned above has a higher value. In general, a material will only deform in a ductile manner if the homologous temperature (the ratio T/Tmelt) is larger than 0.4. Note: always take T in Kelvin! If you do the calculations you will see that ice flows more easily than granite at 300°C.

The speed of deformation is a measure for how fast deformation takes place. ´Deformation´ is defined as a change in size and/or shape. The speed of deformation is the parameter describing the speed at which this happens and it has unit s-1. To see why, consider a bar of 100 mm (L0). It is deformed so that after 1 second, its length is only 99 mm (L1).

The deformation is defined as: e = ( L0 - L1 ) / Lo

Figure 5.21: Steps at the transport of a dislocation. In the first vertical row, the connection between two atoms breaks. A new connection is made: connection with anatom of next row. This process repeats, so eventually the upper half of the crystallattice has slipped over one atomic distance. If the process continues, the crystal will eventually change shape visibly.


118

The speed of deformation is then e per unit time. In this example the deformation e is 1 mm / 100 mm = 1% = 0.01. The speed of deformation then becomes 0.01 s-1.

Inside the Earth, deformation takes place at a very low rate. Geological processes usually take place at about 0.000 000 000 000 01 s-1, or ~1 x 10-14 s-1.

Exercise 5-13***: Experiments with salt Rock salt is a material which deforms relatively easily. In a lab experiment, rock salt has been deformed at different temperatures and at different rates. The results are indicated in tabel 5-.

a. Make a graph of the speed of deformation (horizontal axis) and the stress (vertical axis). The stress is the differential stress σ1-σ3. As you can see, the results vary up to about a factor 100, so it’s best to use a logarithmic scale. Do this for both of your axes.

b. Can you see any trends in your graph? Describe them. The standard law for intracrystalline slip is a so-called ‘flow law’ (there are other flow laws):

Q

RTnA e

With the speed of deformation (in s-1), σ the differential stress σ1-σ3 (in MPa), T the temperature

(in K) and Q the energy that is necessary to initiate the process (the activation energy). R is the

gas constant (8.314 J/mol K) and A and n are constants in flow laws. c. Rewrite this flow law in such a way that you get a straight line for every temperature on a

graph with two logarithmic axes. Hint: the kind of formula that gives a straight line is Y a bX

d. What is the big difference between a lab experiment and the geological reality?(see Figure 5.22)? Think of a way to avoid this problem.

temperature (ºC) σ1 - σ3 (MPa)

Speed of deformation (s-1)

1 100 13.6 5.3 x 10-7

2 100 17.9 2.8 x 10-6

3 100 22.1 7.5 x 10-6

4 100 32.1 3.3 x 10-5

5 150 7.2 5.3 x 10-7

6 150 13.3 4.5 x 10-6

7 150 20.0 3.6 x 10-5

8 200 5.0 5.0 x 10-7

9 200 7.5 2.4 x 10-6

10 200 13.0 2.4 x 10-5

Tabel 5-3: Deformation of rock salt at various pressures and temperatures.


119

5.5 What different types of deformation tell us about orogenesis. So how does this all relate to the way mountains are built? You have learned a lot about the way materials can deform, and now you will see how Earth Scientists use this knowledge to better understand the processes of orogenesis.

An example: we can tell what the stress state in a rock was from the directions of the faults we see. Using this information, we can reconstruct how tectonic plates moved in the past. If we see evidence of ductile deformation, we know that the rocks we see must have been subject to high temperatures. Using deformation structures, we can even calculate the time it took to form a certain structure, and under which circumstances this took place. We are also able to say more about the height of mountain range and the depth of its root. If you forgot what this was all about, reread section 1 from this chapter. We will now do some calculations about the root of mountains (Exercise 5-14).

Exercise 5-14***: Mountains and their roots You’ve seen in Chapter 2 that the Earth is made of roughly three concentric shells: the Core on the inside, the Mantle, and the Crust on the outside. The mantle is not liquid, but we know that the outer, upper parts are more or less fluid, ductile. Therefore you could say that the rigid crust ‘floats’ on this ductile mantle. The crust consists of plates. If for some reason or another a continent becomes heavier, it just sinks a little deeper into the mantle. In the North Sea, sediments like clay and sand are deposited, adding mass to the crust. Here the crust becomes heavier and sinks. On the other hand, the continent becomes lighter when erosion breaks down mountains. It will start to rise. This is what happens with the Alps. Generally, the Earth ‘tries’ to obtain a state of equilibrium. Plates rise and sink with respect to the mantle until equilibrium has been reached. Compare this to the growth and melt of glaciers. And, again just like glaciers, you can only see the tip of the iceberg. Mountains have deep roots. You also know that a fluid will have the same level everywhere in the reservoir (or in a set of connected reservoirs). Same level, same pressure. In this exercise you will calculate the depth of a continental root. a. What would be the best depth to calculate the pressure for? In other words, what is the best

reference point? b. The pressure at that depth must be the same beneath a continental plate and beneath an

oceanic plate. Make an equation that compares the pressure beneath both types of plate. Calculate the depth of the root using this equation. Use figure 5.23 and figure 5.24.

Figure 5.22: Deformation of salt rocks (Cardona, Spain)


120

Figure 5.23: The depth of the mountain root can be calculated using these data.

+ 8000 m

mountain

0 m

continent - 2500 m

~~~~~~~~~~~~~~~~ water ~~~~~~~ 1030 kg/m3

2700 kg/m3 - 9500 m

oceanic crust 2900 kg/m3

root

~ ~ ~

? m

~ ~ ~

~ ~ mantle rock

3300 kg/m3

~

Figure 5.24: Data and formulas to calculate the depth of a mountain root

ρcrust = 2.7 g/cm3 ρmantle = 3.3 g/cm3

H = height of mountain

W = depth of the root

crustmantle

crusthw


121

Geologists go on field work (Figure 5.25) to study rocks in the field. They also take home samples to study using advanced instruments such as light and electron microscopes. They try to make a reconstruction of the way the mountains have formed using these observations. As you see, observations on many different scales are used: from defects on the scale of the crystal lattice (micrometre and smaller!), up to faults and folds in a mountain side (kms). The larger the number of observations, the more accurately you can reconstruct the process of orogenesis. However, you must make a very careful distinction between small scale movements and the overall motions of tectonic plates.

Experimental lab work on the behaviour of materials complements field and rock sample data. Experimental work can give Earth Scientists a better understanding of the way rocks deform. It gives insight into the processes underlying deformation and into the circumstances necessary for deformation to take place.

Final exercise chapter 5. Main questions in this chapter. a. Answer each of the section questions separately. b. Did you form new questions after studying this chapter? Write them down.

Figure 5.25: Fieldwork in the Alps, Utrecht University, summer 2007. Photograph by Hans de Bresser


122

Optional Exercise 5-1: Derivation of the equations for shear stress and normal stress in the 2D case. The Mohr diagram.

Look at Figure 5.26. n , , the principal stresses 1 and 2 and A are indicated.

We know the vector components of A: A = (1 cos , 2 sin )

To find the relationship between n , and the principal stresses 1, 2 and the angle , we use a

new figure in which only the principal axes, the normal stress and the angle are given. All vectors are drawn with their tail in the origin. All vectors point in the opposite direction with respect to figure 5.24, but this doesn’t matter for their relationship

a. The tips of n, A, and the point O make a rectangle in the figure. Why? Draw the rectangle in your figure (but not too visibly).

b. The angle between and the axis of 2 is also Why is that? Indicate this angle with the correct letter.

Figure 5.27: Same situation as in figure 5.26, but in a different coordinate system.

Figure 5.26: Twodimensional representation ofFigure 5.8

(Lenght of a1a2)


123

The essence of the derivation is the fact that you end up with two different expressions for A. One

is A = ( 1 cos , 2 sin ) , the other is A = n + Note: this latter notation is a vector addition! The thin lines in the figure help you to see the different components of each of the

vectors. The σ1-components of n and must add up to the σ1-component of A; they must

therefore be equal to 1 cos . A similar argumentation holds for the σ2-components.

Next we will use the letters n and for the lengths of the vectors. Using the figure you can see that the following holds:

sin n cos 1 cos – cos n sin 2 sin c. Check this by marking the six parts of the above formulas in the figure

d. Using these two equations, we can obtain the formulas for n and τ. We do this by first ‘extracting’ Mohr’s circle from the equations you found. We write both equations as formulas

for a line, but instead of using x and y, we use n and. Show that the result is the following

We will now draw these lines in a graph with n and on the axes. The intersection of the graphs is

the values of (n , that we are looking for. e. The second equation seems the easiest one. What is the slope of this graph? And therefore,

what is the angle it makes to the n axis? Where does this line intersect with the n axis? f. The lines are perpendicular to each other! Why is that? Hint: multiply the slopes of both

graphs.

g. The line corresponding to the first equation intersects with the n axis at 1. Why?

Figure 5.28 shows another graph with n and on the horizontal and vertical axes; the points 1

and 2 are indicated. h. Also draw the line corresponding to the second equation. i. Indicate the intersection of both lines with P and show they are perpendicular.

j. The point P is on the circle with the middle of line 21 as its centre. What geometric rule says so?

k. Express the coordinates of the centre of the circle M and of the radius of the circle in 1 and 2.

l. The line MP makes an angle of 2the n. axis! Use geometry to prove this. m. Show how the formulas between exercises 5.9 and 5.10 can be read directly from this figure.

Figure 5.28: Coordinate system with n and. The line corresponding to the equation indicated is drawn.

1tan

-----------– n 1– =

tan n 2– =


124

Optional Exercise 5-2: CO2 storage and earthquakes It is generally accepted that carbon dioxide (CO2) is a greenhouse gas which contributes to global warming. Many countries actively try to decrease their CO2 emissions. One way to achieve this is to store CO2 in the subsurface, and a lot of research is carried out to investigate the feasibility and the best way of doing this.

One option is to store the CO2 in old, depleted gas reservoirs (see Figure 5.29. These are often sandstones with a lot of pore space, which could potentially hold a lot of carbon dioxide. Such reservoirs must however be examined carefully in order to check whether the gas can’t escape in some way and to see whether or not earthquakes would take place as a result of the storage. You work for a consultancy agency and your current project is to offer advice concerning CO2 storage in an exhausted reservoir rock in the north of the Netherlands. You will examine the risk of earthquakes as a result of the storage. A too large risk would make the reservoir unsuitable for storage. You have the following information: the reservoir lies at 1850 m below the surface and the temperature at depth is 60 °C. The experiments in exercise 5.8 were conducted to obtain information about the material properties of the reservoir rock. The density of the rock is 2280 kg/m3. Apart from these material properties you know the following: σ1 is the pressure as a result of the overlying rocks.

σ3 = 23.3 MPa Pf = 0 Use the yield criterion (graph) for sandstone which you constructed in exercise 5.8. a. Is the reservoir stable, or is there a chance the rock may fracture under its load? Hint;

pressure as a result of the overlying rock: σ1 = ρ g h, and σ3 are given. Use a Mohr diagram like the ones you used in exercise 5.8.

b. CO2 is injected into the reservoir until it reaches an internal gas pressure of 30 MPa. How stable is the reservoir now? How would you explain this?

c. Now imagine that the internal friction of the reservoir rock was measured incorrectly. In reality its value turns out to be 1,5 times as high as the experiments seemed to show. How does this influence the stability of the reservoir (with and without CO2)?

d. And what would happen if the temperature in the reservoir turned out to be 90 °C instead of 60 °C?

Figure 5.29: Map of the Netherlands with oil and gas reservoirs Empty reservoirs may be suitable for CO2 storage.Source: TNO.

Convection: the Earth as an heat engine The Dynamic Earth

125

Chapter 6. Convection: the Earth as an heat engine Within the interior of the Earth, heat is produced by the decay of radioactive isotopes. This heat must be transported to the surface of the Earth, because only there heat can be lost to space. The transport of heat results in material flow (convection) within the Earth, which forms the basis of all plate tectonic processes.

The main question of this optional chapter is:

How can we use the Earth’s heat transport mechanisms to explain plate tectonics and other large-scale processes?


What is the interior structure of the Earth? What sources provide us with information about the Earth’s chemical and mineralogical composition and internal temperature?

How can internal heat production (by radioactive decay) and heat transport (by conduction and convection) explain the internal temperature of the Earth and the heat flow through its surface?

What role does convective flow play in the mantle of the Earth and how is convection related to plate tectonics?

The Dynamic Earth Convection: the Earth as an heat engine

126

6.1 Introduction Plate tectonics (Chapter 2) and geological phenomena like mountain building (Chapter 5) suggest that large-scale processes are at play within the Earth. Scientists in the Geodynamics field of study try to explain these processes with the help of physical principles. Internal heat production, heat transport (conduction and convection) and the resulting temperature distribution, all directly linked to the Earth’s thermal state, are key to explaining geodynamical processes, which shape the Dynamic Earth on a geological timescale.

This chapter discusses the internal dynamics of the Earth. We construct several physical models to explain plate tectonics and consequent mountain building and volcanism and test them against observations (mostly from seismology).

To construct a model of the Earth, it is crucial to be familiar with the Earth’s interior structure. Therefore, Section 6.2 considers what we can learn about the structure and chemical composition of the Earth from Astronomy, Seismology, Mineralogy and other fields of study.

Section 6.3 assumes that the natural decay of radioactive isotopes acts as an internal heat source of the Earth. Based on this assumption, two models of the Earth’s thermal state are tested. The first model only considers radioactive decay and heat conduction. Calculations for this model show that the only (radioactive) heat source of this model can provide about two thirds of the observed heat flow through the Earth’s surface. The model also predicts that the rock making up the interior of the Earth is molten. This contradicts seismological and petrological observations that state the Earth mantle is solid.

In addition to conduction, the second model incorporates convective flow in the mantle. Model predictions show a lower internal temperature than for the first model; the Earth's mantle remains solid because convection can transport heat upwards efficiently enough to prevent the temperature from rising above the melting point. However, is flow possible in a solid mantle? Yes, it is: seismological observations and lab experiments with mantle rock demonstrate that rock at the pressure and temperature of the mantle moves very slowly (this is called creep behaviour). You can compare this movement with the flow of ice in a glacier.

With the help of seismology and laboratory data, we determine the temperature at different depths within the Earth in Section 6.4. The results of the second model fit in well with these temperatures.

Additionally, in Section 6.5, the convection model is tested against geophysical observations. To do so, we study images of the mantle’s three-dimensional structure obtained through seismic tomography. We also check if the heat flow through the Earth surface corresponds to the large-scale convective flow in the mantle. Finally, the temperature distribution of the convection models and the ‘dynamo theory’ that states the Earth magnetic field is generated in the liquid iron outer core are compared.

6.2 The interior structure and chemical composition of the Earth 6.2.1 The large-scale structure of the Earth Throughout the last century, the large-scale structure of the Earth (crust, mantle and core) has been defined and subsequently refined more precisely. Most of what is known about the Earth’s structure comes from seismological research. Figure 6.1 shows the structure of the Earth schematically. The curved lines represent the paths along which wave energy travels from earthquake focus to seismic station. The lines are called seismic rays; they are comparable to light rays that also refract or reflect at the interface between two different media. At sharp transitions (for example, from air to glass), refraction shows as an angle in the path of the light ray. The angle of refraction results from the difference in the velocity of light between both media and, thus, from the difference in refractive index. Seismic waves in the Earth experience both sharp and more gradual transitions. At a gradual transition, a ray path is curved instead of broken at an angle, because the wave’s velocity changes only slowly.


127

Exercise 6:1**: The refractive index and wave velocities a. What is the refractive index of a material in the case of light travelling through it? How can you

calculate the refractive index from Snell’s law? b. How are the refractive index and the velocity of light related? c. Indicate at which depth within the Earth sharp transitions occur in the refractive index

according to Figure 6.1. Why would such transitions occur at those depths? d. Where can you find more gradual transitions? What could cause this type of transition there?

Figure 3.13 shows that seismic rays refract at a greater angle to the normal (a line perpendicular to the media’s interface) deeper within the mantle. This means that wave velocity increases with depth. However, at the boundary between mantle and core, waves refract at a lesser angle: the velocity in the core is lower than in the mantle. Within the core itself, rays again refract away from the normal with increasing depth, because velocity increases.

Between 104o and 140o (starting from the epicentre, which, in this case, is the North Pole, in a clockwise direction along the surface), no seismic waves reach the Earth’s surface (Figure 6.1). This core shadow is caused by the refraction of waves entering and leaving the core.

With the starting time of an earthquake known, the travel time of a seismic wave can be determined from a seismogram (see Chapter 3). Figure 6.2 shows a seismogram recorded in Utrecht, The Netherlands. You can find seismograms of more recent earthquakes at the website of the Seismology group of Utrecht University: www.geo.uu.nl/Research/Seismology.

Figure 6.1: The structure of the Earth (starting from the centre): the inner core, theouter core, the mantle and the crust. The seismic rays of an earthquake at the NorthPole are shown together with the seismograms recorded for this earthquake at different seismic stations.


128

From the travel times of a wave recorded at different seismic stations, the change of velocity of that wave with depth can be determined. Knowing the velocity helps study the earthquake that caused the wave, but also provides information about the deep Earth. Seismic studies have shown that the Earth has a core with a 3480-kilometer radius. The core can be divided into a solid inner core with a radius of 1221 km and a liquid outer core. Transverse waves cannot travel through a fluid, so, because they do not cross the outer core, we know the outer core is liquid. The liquid outer core also explains the Earth tide, the motion of the Earth in response to the gravitation of the Moon and Sun. The Earth is not a perfect sphere, but an ellipsoid (a flattened sphere) that, through the Earth tide, changes shape (several tens of centimetres) periodically. The amplitude of this motion can only be explained by a (partially) liquid core.

Figure 6.3: Left panel: The Preliminary Reference Earth Model (PREM) with density ρ, primary and secondary seismic wave velocities vp and vs, gravity acceleration g and pressure P (Anderson & Dziewonski 1981). Right panel: The simplified mineral structure of the Earth (see Section 6.4). The transitions at a depth of 410 and 660 km correspond to sudden changes in the curves of material properties ρ, vp and vs (Yuen et al. 2007).

Figure 6.2: A seismogram showing P- and S-signal phases corresponding to longitudinal and transverse body waves, respectively. Body waves travel through the Earth. The later, larger amplitude signal phases arise from surface waves travelling along the surface between earthquake and seismic station.


129

The structure of the Earth is described in the standard Preliminary Reference Earth Model (PREM) (see Figure 6.3). The PREM model is mostly based on seismological observations. It shows the distribution of important physical quantities, such as density ρ, pressure P and seismic wave velocities vp and vs, with depth within the Earth. In Figure 6.3 the core-mantle boundary (CMB) is clearly visible at 2900 km, because here both wave velocities change, with the secondary wave velocity vs even becoming zero.

Exercise 6-2: Increasing seismic wave velocities Figure 6.3 shows that seismic wave velocities vp and vs increase more or less gradually towards the centre of the Earth, but suddenly decrease at the mantle-to-core transition. a. Explain why the curvature of the wave rays in Figure 6.1 corresponds to a velocity increase with depth according to Snell’s law. b. At a depth of 2900 km, vs becomes zero. Why?

P- and S-waves and elastic deformation Waves deform the medium they travel through. The elastic parameters of a medium describe its resistance to this deformation. Together with the density of the medium, the elastic properties also determine the velocities of longitudinal and transverse seismic waves, vp and vs. Two elastic parameters are important: shear modulus G and incompressibility K. The shear modulus gives the resistance of a medium to changing shape, while the incompressibility describes the resistance to a change in volume. When we consider a change in shape, we imply that the volume of the medium remains the same; for example, changing the shape of a cube results in a parallelepiped with the same volume as the original cube. Consequently, considering a change in volume means that the cube remains a cube, but does become bigger or smaller. A change in shape is caused by shear strain, a change in volume by (hydrostatic) pressure.

Because both longitudinal and transverse waves deform the shape of a medium, elastic parameter G occurs in the velocity expression of both types of wave. However, only longitudinal waves result in a change in volume of the medium, because they also stretch and compress rocks. Remember, particles of the medium a longitudinal wave travels through vibrate in the direction of the wave, while for transverse waves particle motion is perpendicular to the direction of the wave. Such vibrations cannot cause a change in volume, so the equation for transverse wave velocity does not contain incompressibility K. So, the velocity expressions are:

s

Gv

Because a liquid cannot resist a change in shape, its stiffness is zero. The shear modulus G in a liquid is therefore zero. Since vs depends directly on G, transverse waves have no velocity in a liquid and cannot travel through a fluid medium.

Wave type Property

Longitudinal Transverse

Arrival Primary Secondary

Change in shape? Yes, cube becomes a rectangular box

Yes, cube becomes a parallelepiped

Change in volume? Yes, through compression No

Propagation through a solid medium?

Yes Yes

Propagation through a liquid medium?

Yes No

Velocity in water 1500 m/s 0

Velocity in granite 5500 m/s 3000 m/s

Table 6-1: Several properties of longitudinal (P-) and transverse (S-) waves.


130

Early in the 20th century, analyses of seismic wave travel times already showed that wave velocities increase with depth within the Earth. In Fig. 6.1 this is evident from the wave rays: the rays refract with increasing angle to the normal with depth. Could the velocity increase depend on density ρ? The density of rocks increases with depth, so if incompressibility K and shear modulus G are constant, velocity should decrease with depth according to the two expressions above (check this). However, because velocity increases with depth, K and G should increase with depth even more than the rocks’ density does. They do so because pressure also increases with depth: and the resistance to deformation becomes larger.

Exercise 6-3**: The velocity of longitudinal and transverse waves a. What is the vp/vs ratio at a depth of about 2000 km? Use the PREM model (Figure 6.3) to

answer this question.

b. Explain why vp and vs decrease for increasing density ρ when the elastic parameters are held constant.

c. Deeper within the Earth, density is higher. How can the wave velocities increase as well?

d. Write down two observations that suggest a liquid outer core.

Years of seismic observations have enabled scientists to establish a detailed depth profile of the seismic velocities. The profile shows large increases in velocity and density, especially in the 400-700 km depth range. This zone is called the transition zone and can be seen in the PREM model. The right panel in Figure 6.3 shows the transition zones; terms like ‘ringwoodite’ indicate differences in the chemical and mineralogical composition of the mantle material.

The chemical composition of the Earth How can we gain insight in the chemical composition of the deep Earth? Not through drilling: the deepest drilling so far only reached 12 km, which does not even cross the continental crust (continental crust has an average thickness of 35 km). Surface rock does not give any direct information about the composition of the mantle either, because it was brought up from the crust or shallow mantle. Volcanic rock is also derived from crustal or shallow mantle material. Moreover, the magma composition changes upon rising slowly through the crust. Minerals in the magma with a high melting point solidify at a certain depth in the crust or shallow mantle and, thus, do not reach the surface together with the remaining liquid magma. Only a few volcanic deposits and several locations of past continent-continent collision show rocks from several hundreds of kilometres of depth.

Figure 6.4: A chondrite from Mexico. Thespherical chondrules are clearly visible.


131

Thus, we are left to indirect data to determine the composition of the deep Earth: the composition of meteorites.

There are two major classes of meteorites: iron meteorites and stony meteorites. Iron meteorites are composed of at least 95% iron and nickel. Stony meteorites have a more variable composition and consist of over 90% of O, Fe, Si and Mg (oxygen, iron, silicon and magnesium). Some meteorites contain small spherical particles called chondrules; these meteorites are therefore called chondrites (see Figure 6.4).

Chondrites are seen as the oldest material in our solar system. The chondrules are thought to represent frozen droplets from the cooling dust and gas cloud the solar system was formed from. Scientists assume that the terrestrial planets (Mercury, Venus, Earth and Mars) are mainly composed of chondritic material.

Spectral analyses of the sunlight show that the atomic composition of the non-volatile particles of the photosphere, the outer layer of the Sun, strongly resembles that of chondritic meteorites (see left panel of Figure 6.5). Also, the average composition of the Earth’s crust and mantle rock is clearly similar to the chondritic composition. This chondritic composition consists for over 90% of only a few elements (Fe, O, Si and Mg) as shown in the diagrams of Figure 6.5 (right panel).

Figure 6.5:

Left panel: The atomic com-position of the non-volatile elements of the Sun’s photosphere (‘solarphoto-sphere abundance’) compared to the average composition of chondritic meteo-rites (‘meteorite abundance’). Composit-ions are normalized to 106 for silicon.

Right panel: The composition of the Earth according to the chondritic hypothesis (Brown & Musset 1993).

(a) percentage of mass

(b) percentage of atoms


132

Exercise 6-4*: Comparing the composition of the solar photosphere and meteorites How does Figure 6.5, left panel, show that iron is more abundant in the photosphere than

magnesium?

Which non-volatile element is most abundant (atomically) in meteorites?

What percentage of the atoms of a meteorite is oxygen?

Explain why oxygen is not included in the left panel of Figure 6.5. If you would include it, where would you expect it?

Scientists think that during the formation of the solar system the Earth formed through accretion of chondritic material. This theory is supported by the age determination of chondrites and terrestrial rocks: radioactive dating shows that chondrites are about 4.5 billion years (4.5 Gy) old, and the oldest dated rocks from Earth are 4.4 Gy old zircon grains.

Crust and mantle rocks have a lower concentration of iron than chondritic meteorites; this deficit is called iron depletion. During the accretion of chondritic material, the impact energy of meteorites was transformed into heat, leading to an increase in temperature. Most of the iron and nickel in the chondritic material melted and, due to a high density and low viscosity, sank to the centre of the forming Earth in a process of core-mantle differentiation.

The ‘sinking’ of liquid metal was a self-sustaining process: sinking released more potential energy and, thus, increased temperatures in the mantle and led to more melting of metals. This process most likely resulted in a complete melting of the Earth’s mantle and the formation of a magma ocean (which, obviously, solidified again later). The settling of iron from this ocean into the core explains the iron depletion of crust and mantle rock.

Exercise 6-5*: Similarities between the composition of meteorites and the Earth Compare the composition of the different layers of the Earth with the composition of the two meteorite classes (iron and stony meteorites). Which type of meteorite corresponds to which layer? Explain.

The mineralogical composition of the Earth Meteorite studies and lab experiments with silicon rocks at high pressures and temperatures (HPT experiments) demonstrate that the upper mantle (to a depth of 660 km) consists of peridotite. The term peridotite includes all mantle rocks made up of a certain mixture of minerals, mainly olivine (Mg;Fe)2SiO4, pyroxene (Mg;Fe)SiO3 and garnet (Mg;Fe)3Al2Si3O12. The iron content of peridotite is about 10%, but this as well as the concentration of other elements can vary.

The material of the Earth’s mantle consists of about 95% of magnesium, iron, silicon and oxygen. This average composition is well explained by the chondritic hypothesis that the Earth formed through the accretion of chondrites.

The chemical composition of a rock is often expressed in the different oxides from which the rock can be synthesized. The average composition of mantle rock peridotite is given in Table 6-2.

Synthesis of rock from the oxide mixture is accomplished by heating the oxide mixture above the melting points of the individual oxides at a pressure of 1 bar. From the melt, we can crystallize the minerals olivine (Mg;Fe)2SiO4, orthopyroxene (Mg;Fe)SiO3, clinopyroxene (Ca;Mg;Fe)2Si2O6 and garnet (Mg;Fe)3Al2Si3O12. Iron, aluminium and calcium are present in these minerals as a solid solution. For olivine, Fe / (Mg+Fe) ~ 0.1. This means that one out of ten magnesium atoms in the crystal lattice of olivine is replaced by an iron atom. The weight percentages of olivine, pyroxene and garnet in representative mantle material are about 60, 30 and 10 percent, respectively.

Oxide Weight percentage [wt%]

SiO2 46.1

MgO 37.6

FeO 8.2

Al2O3 4.3

CaO 3.1

Na2O 0.4

TiO2 0.2

K2O 0.03

Table 6-2: The composition of the theoretical mantle material peridotite according to Ringwood (1975).


133

So, olivine is a silicate with in total two atoms Mg and/or Fe per SiO44- ion. Pyroxene is a silicate

with in total two atoms Mg and/or Fe per Si2O64- ion. Because the Mg and Fe content of olivine and

pyroxene vary, we refer to them as solid solutions series rather than pure materials. The crystal shape of both minerals changes with increasing pressure (thus, with increasing depth). Because this variation occurs in a step-like manner, the Earth is built up of shells with a different mineralogical composition (see PREM model, Figure 6.3b). This we know from experimental HPT research and seismological data.

With each phase change, the density of olivine increases. At 1 bar, α-olivine forms. α-olivine changes into β-olivine (wadsleyite) at about 15 GPa (1 GPa = 109 Pa) at a depth of 410 km. At 17 GPa (or a depth of 520 km), γ-olivine (ringwoodite) forms. Ringwoodite dissociates into the minerals periclase (MgO) and perovskite (MgSiO3) at 24 GPa or about 660 km.

Exercise 6-6*: Seismological determination of the layered mantle structure How do you think the contrasts in seismic wave velocities between the layers of different mineral phases (Figure 6.3, right panel) are expressed in seismic recordings?

6.3 A model for the Earth’s thermal state To find out `how the Earth works’, we have to study the thermal state of our planet. In the 19th century, the English physicist Kelvin proposed a mathematical model for the thermal evolution of the Earth. Kelvin assumed that the Earth had cooled gradually since the time it solidified from the magma ocean. His calculations resulted in an age of the Earth of maximally 40 million years (My) (since solidification). This age was much smaller than that estimated from geological data. For example, the thickness of sediment sequences combined with observed erosion and sedimentation rates suggested an age of several hundreds of millions of years. Even calculations using the burning of fossil fuels as an internal heat source (even though this was thought to be an improbable process) did not result in a longer lifespan for the Earth.

Internal heat production A solution to this problem was found when at the beginning of the 20th century natural radioactivity was discovered. Radioactivity is a source of internal heat production for the Earth much greater than external heat sources such as sunlight, meteorite impacts, tidal flows in seas and oceans or the tidal deformation of the solid Earth. Internal heat production through radioactivity greatly slowed down the cooling of the Earth, so, as expected, cooling lasted much longer than predicted by the Kelvin model. Nowadays we presume the Earth formed 4.6 billion years ago (4.6 Gy) from chondritic material and, since then, cooled very slowly.

In Chapter 1 you used radioactivity and half-lives to determine the age of rocks. Here we will use information about the radioactivity within the Earth’s interior to estimate the internal heat production. Radioactive isotopes naturally decay to stable isotopes in several steps.

Figure 6.6: Evolution of the heat production of a radioactive fuel mix of chondritic composition with time. Time starts at t = 0 with the formation of the Earth and runs till the present day 4.5 Gy later. The upper black curve shows the heat production of the combined isotopes.


134

This exponential decay releases a substantial amount of heat in the mantle (and crust). From Chapter 1 you know:

1/ 210

2

t

tN t N

with N(t) the number of radioactive particles at time t, N(0) the number of particles at t = 0, t the time passed and t½ the half-life. The half-life is the time it takes for half of the radioactive isotopes to decay to a daughter isotope.

Radioactivity always releases energy; within the Earth, this energy is almost completely transformed into heat. Therefore, the heat production density H (in W/kg), the power produced by decay (in watt) per mass (in kg), is proportional to the number of particles that decay per unit of time: H ~ dN/dt. So:

1/ 210

2

t

tH t H

The half-life t½ differs for every isotope (see Table 25 of BINAS). Because one element decays faster than the other, the composition of the mix of radioactive elements changes continuously.

From the chondritic hypothesis for the Earth’s composition, we can learn much about the radioactive isotopes in the mantle. When the Earth was young, about 4.6 to 3 Gy ago, decay of isotopes with a relatively short half-life dominated the radioactive heat production. Long half-life isotopes only became more important later on, when the short-lived isotopes had mostly decayed. This is illustrated in Figure 6.6, where the heat production of the different isotopes in a chondritic composition is plotted against the Earth’s lifespan. You can see that the relatively short-lived uranium and potassium isotopes 235U and 40K were the dominant heat source in the early Earth. In the present Earth, however, the thorium isotope 232Th is the principal heat source, and 238U is the second most important.

Exercise 6-7**: The nuclear activity of the Earth Study Figure 6.6 and BINAS. a. Estimate (without using your calculator) how long it will take for the current nuclear fuel of the

Earth to fall below a heat production density of 1 . 10-12 W/kg. Only consider the isotope that is most important now and in the future and take the current heat production density at 5 . 10-12 W/kg.

b. Now use your calculator to calculate this period accurately. c. Calculate the current heat production of the whole Earth, assuming a chondritic composition

with H = 5 . 10-12 W/kg. The mass of the Earth is 6 . 1024 kg. Models for the internal heating of the Earth usually assume an average isotope composition with an average half-life. These models suggest a current heat production density of about 5 . 10-12 W/kg and a half-life of 2.4 Gy. d. ***Calculate the amount of energy produced through radioactive decay by material of this

average composition in the 4.5 Gy that have passed since the formation of the Earth. First consider 1 kg of material, then the total mass of the Earth.

A first step to modelling the Earth’s heat regulation: a conductive model The processes that shape the Earth take place on very long geological and cosmological time scales of millions and billions of years. It is practically impossible to reach the interior of the Earth for direct observations of the composition, temperature or heat production at depth. Thus, we rely on indirect observations through seismology and geological field studies. More so, we use numerical modelling to investigate the Earth’s interior.

Model studies are used to explore possible scenarios for the evolution of the Earth on geological time scales as well as for the current situation of the Earth’s interior. The results (model predictions) are tested against geological and geophysical observations.

We will start with a simple model of the Earth’s thermal state before considering a more complex one. The production and transport of heat result in high temperatures deep in the Earth and low


135

temperatures at the surface. Vice versa temperature differences lead to the transport of heat from high temperature regions to low temperature zones. This heat transport within the Earth drives plate tectonics, mountain building, volcanism and the generation of the Earth magnetic field.

There are three ways of transporting heat: conduction (vibrations are passed on by atoms in the crystal lattice), thermal radiation (through electromagnetic wave energy) and convection (flow). When material convects, energy is transported by material transport. In stagnant solids, heat transport takes place through a combination of radiation and conduction, however. The combined process can be described effectively as conduction.

The first, simple type of model of the Earth’s thermal state, that was especially considered in the first half of the last century, assumes a homogeneous, spherically symmetric Earth that experiences only conductive heat transport. A homogeneous Earth has the same composition throughout, so it is not divided into a core, mantle and crust. Spherical symmetry implies that at a given radius, temperature is the same everywhere. In the special case of such a model that we investigate here, the temperature at each point within the Earth is also constant through time; it is a stationary situation, called steady-state conduction. At each moment in time, the internal heat production is in equilibrium with the heat loss to oceans and air at the Earth’s surface.

The conduction model is based on Fourier’s heat conduction law that describes the diffusion of heat. Diffusion considers heat flux density J (in W/m2), the energy per unit time (in watts) that flows through an area of 1 m2. The heat flux is generated by temperature differences between different regions:

J = - k dT/dr

T is the (absolute) temperature at a certain depth and r is the distance from the centre of the Earth (the radius). J is proportional to the temperature gradient dT/dr, the derivative of temperature T to distance r from the centre. Thermal conductivity k [Wm-1K-1] is a material property: it has a characteristic value for every different medium (see BINAS, Tables 9 and 10).

The model assumes a uniform distribution of the heat production density H [W/kg] throughout the sphere that corresponds to the chondritic hypothesis for the Earth’s composition. Within the whole sphere of radius R, we first consider the heat produced within a smaller sphere with radius r (see Figure 6.7). The heat production per unit of time in the sphere of radius r is called Qin. In the case of thermal equilibrium, it should be equal to the total heat flux Qout through the surface of the sphere with radius r.

Exercise 6-8*: Thermal equilibrium What happens to the temperature if the heat production in the radius-r sphere does not equal the total heat flux through the sphere’s surface?

Heat production Qin can be calculated with: Qin = H . m = H . . 4/3 r3

Here, H is the heat production density, m the mass of a sphere with radius r and density :

m = . V = . 4/3 r3

The heat loss through the surface, Qout, is: Qout = J . 4 r2 = - k dT/dr . 4 r2

With Fourier’s law of heat conduction: J = - k dT/dr

In a steady-state model: Qin = Qout

Or: H . . 4/3 r3 = - k dT/dr . 4 r2

Thus: dT / dr = - H r / 3k

Figure 6.7: A uniformly heated spherewith radius R and a smaller spherewith radius r within.


136

Exercise 6-9***: Temperature distribution and heat production According to the equation deduced above, the Earth’s internal temperature decreases from centre to surface. a. How can you see from this equation that the absolute temperature gradient increases when going from centre to surface? b. What should you measure in chondrites to determine the magnitude of the gradient at the Earth’s surface when assuming a chondritic composition for the Earth?

An equation that contains the derivative of a function (here the function is T(r) and dT/dr is the derivative) is called a differential equation. In our case, the derivative of temperature T with respect to radius r is a linear function of r. We want to find the radial temperature distribution T(r), or geotherm, to be able to calculate the temperature at any given distance from the centre. Because of the spherical symmetry of our model, the geotherm is constant along a concentric spherical surface. Therefore, temperature is the same for all points with the same given radius. How can we determine T(r) from differential equation dT/dr = - H r / (3k)?

We need a function T(r) with a derivative dT/dr that is a linear function of r. T(r) is the primitive function of dT/dr. Because the derivative of r2 is 2r, the geotherm T(r) must be proportional to ½ r2. An at first arbitrary constant must be added. This integration constant is determined when the temperature in one point is specified. For this purpose we will use the temperature at the surface of the Earth of 10oC: T(R) = TR = 283 K.

Exercise 6-10***: Solving for the geotherm The solution of the differential equation for the geotherm is T(r) = T(R) + ρH(R2 - r2) / 6k. Check the solution (the primitive of dT/dr) by solving the integral

' ' ''

( ) ( ) ( )3

R R

r r

dT Hdr T R T r r dr

dr k

with boundaries R, the distance from the centre to the Earth’s surface, and r, the distance from the centre to some point r within the Earth.

Exercise 6-11***: The temperature distribution in the conductive spherical model Use the following data for this exercise: the radius of the Earth is 6378 km, the temperature at the Earth’s surface is 283 K, k is 4 WK-1m-1, the average density of the Earth is 5500 kgm-3 and H is 5 . 10-12 Wkg-1. Also use the equation for the geotherm given in Exercise 6-10. a. Sketch the temperature with distance from the centre of the Earth if the geotherm were to be correct. b. Determine the temperature at the Earth’s centre. c. Determine the temperature at the point midway between the centre and the surface. This point approximates the depth of the core-mantle boundary.

Exercise 6-11 gives temperatures far above the melting points of rocks and metals. On the one hand, this shows that assuming heat production through radioactive decay can explain high temperatures within the Earth for long periods of time, which is an important improvement of the Kelvin model. On the other hand, the high temperatures found are not realistic, because they require a significant part of the Earth’s interior to be liquid or even to be in the gas phase. This contradicts seismological observations that show that the outer core is liquid and the mantle solid.

In conclusion: the conductive equilibrium model is a good step forward in modelling the Earth’s thermal state, but does not yet suffice. The uniform heat production density results in a temperature contrast between centre and surface that is unrealistically high. This high contrast is the result of the required heat flux through the surface and the low thermal conductivity (high thermal resistance) of mantle rock on which conductive heat transport depends.

Exercise 6-12**: Heat production within the Earth Measurements show that thermal power that 'leaks' through the surface of the Earth is approximately 44 · 1012 W. This observation can be related to the conductive model under consideration. Use the parameters given in Exercise 6-11 to complete this exercise. a. Calculate from this the heat flux through the Earth’s surface predicted by the model b. Calculate the total amount of heat produced in the model Earth by radioactive decay.


137

c. Compare the result of b. with the total actual amount of heat lost through the Earth’s surface determined by the measurements. Did you expect this? d. Assume the difference of c. is related related to the cooling of the Earth. Show that the current cooling rate of the Earth is about 72 K per billion years.

An extended conductive model?

We could modify the conductive model by assuming that the heat producing radioactive isotopes are not uniformly distributed, but concentrated in a spherical shell with R - d < r < R. Results of such a model show lower interior temperatures, because heat is mainly produced at shallow depths. But, why would radioactive elements be concentrated in the outer layers of the Earth? Scientists have tried to explain this layering with the solidification of mantle rock in the young Earth. From Petrology Studies it is known that radioactive elements in a magma concentrate in the remaining melt upon crystallization of the magma. When the melt is less dense than the solidified material, the decreasing remainder of the melt that becomes enriched in radioactive isotopes over time, will collect in the shallow mantle during the solidification of an early magma ocean.

A model with layered internal heating requires that the layering remains intact. However, plate tectonics, subduction and mountain building represent vertical motions in the Earth that destroy such layering. Therefore, the layered conductive model is not plausible.

Modelling the Earth’s thermal state: a convection model Conduction alone cannot explain the Earth’s thermal state; therefore, we will now discuss a model that also includes convective flow. We assume the Earth model is divided into a core and a mantle. Convection takes places in the mantle. Could this model lead to a geotherm lower than the melting temperatures of mantle rock, like seismological observations suggest?

Geological (mountain building, Chapter 5) and geophysical (plate tectonics, Chapter 2) phenomena show that the Earth’s mantle is not static. Take, for example, the process of postglacial rebound. During the last glacial period, Northern Europe and North America were covered with a thick layer of ice. Melting of the ice layer disturbed the equilibrium situation. Since then, Scandinavia (among others) is slowly pushed upwards to re-establish this equilibrium. Another example of mantle flow is the slow, large-scale cycle of lithosphere production (at MORs) and destruction (in subduction zones). The flow of mantle material takes place even though the mantle is solid; this is possible because solids like ice and rocks creep. For example, consider the flow of ice in a glacier, or the deformation of rocks (Chapter 5).

Obviously, flow of a solid is much slower than flow of a liquid, under conditions prevailing in the Earth's mantle, namely only several centimetres per year. Solid materials that display creep behaviour are very viscous. Viscosity is denoted by the Greek letter η. Already in 1935, the geologist Haskel estimated the viscosity of the mantle (from postglacial rebound data from Scandinavia) to be in the order of 1021 Pa s. This is much larger than the viscosity of water, which is in the order of 10-3 Pa s.


138

6.4 The start of convection in a viscous medium Imagine a viscous medium, like syrup, that is heated from below. Heating of the bottom results in a local temperature increase. The medium expands in the heated region, so, locally, density decreases. The density difference with the surroundings results in an upward force (Archimedes’ law) that, when big enough, causes upward flow. The rising hot material is cooled at the surface. Therefore, it will sink again: a convection cell is formed.

The bottom of the Earth’s mantle is formed by the hot boundary with the outer core, while the boundary with the crust represents the mantle’s cold top.

Figure 6.8 shows a possible convection cell in the mantle: flow is driven by the temperature difference between the lower mantle and the Earth’s surface.

It is not hard to imagine a convection cell in a liquid or in the air (for example, in a thundercloud on a warm day), but in the solid mantle? Scientists have shown that flow in rocks is possible using a specific equation. To give you an idea where that equation came from and what it represents, we will take another look at the viscous syrup layer mentioned above. In a laboratory experiment, we heat the bottom of the layer just a little. The small temperature contrast between top and bottom of the layer is not enough to cause the fluid to flow: convection only starts when the temperature contrast is larger than a critical value. Rayleigh and Benard have studied this phenomenon extensively.

Before the critical value for the temperature difference is reached, the viscous layer only conducts heat. In case the critical value is exceeded, however, the liquid starts to move: hot liquid flows up, while cool liquid sinks. Convection increases the upward heat flux. If the temperature difference increases further, convective heat transport will dominate over conduction.

Exercise 6-13: Heat transport within a viscous fluid Consider a viscous fluid heated from below. The liquid is at rest. a. Which type of heat transport dominates in the current state of the liquid? b. How can you cause the fluid to start moving? c. Which type of heat transport then dominates?

Rayleigh showed that the convective behaviour of a viscous fluid heated from below can be described with only one dimensionless number: the Rayleigh number. This number depends on the type of material (with its characteristic density ρ, thermal expansion coefficient α, diffusion coefficient κ [kappa] and viscosity η [èta]), the gravity acceleration g, the applied temperature difference ΔT and the thickness h of the fluid layer:

3Thg

Ra

Below the critical value of Ra ~ 1000, no material flow occurs. So, there will be no heat transfer through convection, but only through conduction. Above the critical value, the heat transport capacity of the viscous material increases greatly because of the convective flow.

Within the mantle, the Rayleigh number must exceed its critical value for convection to occur. Using the mantle viscosity known from postglacial rebound and the estimated temperature contrast between the top and bottom of the mantle, you find that the Rayleigh number is indeed large

Figure 6.8: A schematic drawing of the large-scale convective flow in the Earth’s mantle. Convection is driven by the temperature difference between the coldlithosphere and the warmer core-mantle boundary. Oceanic crust is formed at a mid-ocean ridge and is recycled at the trench of a subduction zone.


139

enough (Ra = 106-107). Thus, convection cells can form in the Earth’s mantle. We can now use the Rayleigh number to describe the cells in detail.

Computer calculations of a Rayleigh-Benard convection cell

Figure 6.9 shows the results of computer simulations of a Rayleigh-Benard cell at three different values of the Rayleigh number: 104, 105 and 106. The bottom of the model domain is kept warm, the top cool: the bottom temperature is set to 1, the top temperature to 0 (in arbitrary units). Depth is 0 at the top and 1 at the bottom.

Figure 6.9a gives the temperature distribution in the cell. For each Rayleigh number, a hot plume is seen to rise up from the warm lower boundary (left red region). On the right side of the cell, you can also see a cold downwelling (blue region). The warm plumes are often thought to represent hot spots in the mantle, such as those underneath Hawaii and Iceland. The cold downwellings represent the subduction of an oceanic plate. With increasing Rayleigh number, the hot and cold structures become thinner, and the well-mixed interior of the cell becomes larger (white region). The interior has a temperature close to the average temperature of top and bottom.

Figure 6.9: From left to right: Figure a, b and c. a) The temperature distribution in a Rayleigh-Benard convection cell. Red represents high temperatures, blue low ones. b) The horizontally averaged temperature at different depths within the cell. c) The ratio of the heat flux Q and the conductive part of the heat flux Qconduct.

Figure 6.9b shows the horizontally averaged temperature with depth for the corresponding convection cells. The graphs resemble the geotherm that gives the (horizontally) averaged temperature distribution within the Earth. They show that the temperature in the upper and lower boundary layer changes strongly with depth. In the thick layer in between the boundary layers, where convection occurs, the average temperature hardly changes though (it remains close to 0.5). Thus, the largest temperature variation occurs in the boundary layers. The boundary layers become thinner for increasing Rayleigh numbers. Within these layers, vertical heat transport can only occur through conduction, because there is no vertical flow.

The decreasing thickness of the boundary layers with increasing Rayleigh number results in a higher heat flux (with the interior temperature constant at about 0.5). This can be explained with Fourier’s law: The temperature increase over the boundary layers remains the same, but the layers become thinner. Because heat flux is proportional to ~1/Δl, with Δl the thickness of the boundary layer, the flux increases for higher Rayleigh numbers.

Rayleigh number Temperature


140

Figure 6.9c gives the ratio between the total heat flux Q and the heat flux Qcond that represents the heat flux through the layer if only conduction takes place. The graph shows that convection is a critical phenomenon. When the Rayleigh number is below the critical value of approximately 780, Q / Qcond = 1 because only conduction takes place. Above the critical value, the heat flux quickly increases with the Rayleigh number due to the contribution of convection. For example, for Ra = 106, the heat flux through the cell is about 23 times larger than in a cell without convection.

The Rayleigh-Benard convection simulations and the estimates of the mantle’s Rayleigh number (Ra > 106) indicate that convective heat transport is much more effective than heat conduction alone. Therefore, this process could play an important part in controlling the thermal state of the Earth. In the next section, we will investigate whether Rayleigh-Benard convection calculations can predict realistic values for the Earth’s interior temperature. Can this model explain why the outer core is liquid and the inner core is solid? And what about the layering of seismic velocities and density found in the PREM model?

6.5 The internal temperature of the Earth 6.5.1 Composition and temperature of the Earth’s interior Information about the deep Earth is obtained from seismology, gravimetry (the study of the Earth’s gravitational field) and laboratory experiments. When the composition of a part of the mantle is known, HPT experiments can be used to assess the influence of high temperatures and pressures on the phase of the material. Phase transitions within the Earth (solid-solid and solid-liquid) can be recognized from sharp changes in the material parameters density and seismic wave velocity in the PREM model. At a solid-solid phase transition, one crystal shape changes into another with the same chemical composition. For example, olivine, the most important mantle mineral (see Section 6.2.4), experiences several solid-solid phase transitions with increasing depth, such as the spinel-post-spinel transition (see Figure 6.10).

When HPT experiments have demonstrated how the pressure at which a solid-solid phase transition occurs depends on temperature, a line can be constructed that represents the phase boundary in a pressure-temperature phase diagram for a given material (see Figure 6.10 for an example).

We found sharp transitions in mantle material properties with increasing depth in the mantle (see PREM model, Figure 6.10). We can connect these transitions to experimentally determined phase changes in olivine. The temperature of the phase change can then be established as follows:

1) Determine the location/depth of the phase transition (from seismological data and the PREM model).

2) Use the PREM model to determine the pressure at this depth.

3) Apply the pressure found in 2) to look up the temperature in the, experimental phase diagram as indicated by the arrows in Figure 6.10.

We will use this method to determine several fixed points for the temperature distribution within the Earth, the geotherm. This way we can use phase transitions as a thermometer for mantle temperatures.

6.5.2 Clues to the Earth’s temperature distribution Assuming a chondritic composition for the mantle and core, the mantle mainly consists of peridotite (with minerals such as olivine and garnet, see Section 6.2.4) and the core of iron and nickel (see Section 6.2.3). We can use HPT experiments to study the phase changes of these materials. Many rocks have been investigated at the temperatures and pressures of the mantle and core. For most rocks, the pressure at which a phase transition occurs can be approximated as a linear function of temperature (see Figure 6.10, for example).

Figure 6.10: Schematic pressure-temperature phase diagram for the phase change at about 660 km depth that forms the boundary between upper and lower mantle.

Temperature T

Phase boundary


141

Determining the temperature at the boundary between upper and lower mantle

We will use the method described above to determine the temperature at the transition from upper to lower mantle at a depth of about 660 km. Seismological data shows that primary and secondary wave velocities as well as density suddenly increase at this depth; indicating that olivine undergoes a phase transition here. HPT experiments show that the high pressure phase of olivine, called ringwoodite ((Mg;Fe)2SiO4 with spinel structure), changes into post-spinel, which consists of the minerals perovskite (Mg;Fe)SiO3 and magnesiowüstite (Mg;Fe)O.

The spinel-post-spinel phase transition takes place at a pressure of about 24 · 109 Pa (you can read off the pressure from the PREM-model, left panel of Figure 6.3, the right panel gives the corresponding mineralogical interpretation of the phase change). From the phase diagram for the spinel-post-spinel phase change, we find the phase change temperature: at P660 = 23.9 GPa, T660 = 1900 ± 100 K.

Determining the temperature at the inner and outer core boundary

In a similar way, the temperature of the solid-liquid phase transition at the solid inner core and liquid outer core boundary is determined with HPT experiments and the PREM model. The PREM model shows a pressure at the inner-outer core boundary of 330 GPa. From the experimentally determined melting temperature of the core material (mainly Fe, Ni and S) and the pressure value of 330 GPa pressure, it is estimated that the temperature at the boundary is about 4850 K.

Figure 6.11 shows the three fixed points of the geotherm:

1. The temperature at the Earth’s surface, about 283 K.

2. The temperature of the phase transition at a depth of 660 km (upper-lower mantle boundary), about 1900 K (the left dot).

3. The temperature of the phase transition at a depth of 5150 km (inner-outer core boundary), about 4850 K (the right dot).

The estimated temperature curve is interpolated between the three fixed points. Melting temperatures of the mantle and core material are also indicated (dashed lines, denoted by ‘melting’). These lines show that temperatures in the mantle are below melting temperatures, while

temperatures in the outer core are higher than the melting curve.

Figure 6.11: a: A graph of the averagedtemperature with depth, called the geotherm(solid line). The pressure is also indicated atimportant boundaries. The dashed lines givethe melting temperature with depth. UM =upper mantle, LM = lower mantle, OC = outercore and IC = inner core.

b: The temperature, denoted by core adiabatin the liquid outer core against pressure P.The boundary between the outer and inner core is determined from the intersection of the melting temperature (Fe-O-S melting- and the geotherm. CMB = core-mantle boundary and ICB = inner core boundary. (R. Boehler 1996)


142

Exercise 6-14**: The inner-outer core boundary a. The caption of Figure 6.11b says: “The boundary between the outer and inner core is determined from the intersection of the melting temperature (‘Fe-O-S melting’) and the geotherm.” Explain. Hint: How do the inner and outer core differ? b. Explain why the solid inner core has grown through time at the expense of the outer core.

Determining the temperature at the core-mantle boundary

We can also determine the temperature jump at the core-mantle boundary at a depth of 2900 km (see

Figure 6.11b). This jump controls the heat flux out of the Earth’s core; therefore, we first have to discuss the core-mantle boundary and the type of heat transport at this boundary. From now on, we assume that heat transport in both the mantle and core only takes places through convection, so we ignore conductive heat transport. We will follow the heat flux from the core to the Earth’s surface and pay special attention to what happens to the heat flux at the core-mantle boundary. At this boundary, no vertical convective flow can take place.

Convective heat transport is very efficient in the outer core because of the strong upwellings of liquid core material (the viscosity of liquid iron is low, so there is little resistance to flow). When a convective material flow arrives at the bottom of the core-mantle boundary layer, it collides with the solid mantle material that is only half as dense. Therefore, the flow is deflected horizontally: no upward convective flow is possible here. The solid mantle material above the boundary has a very high viscosity. Here, convection is still possible (as creep), but at a speed of only several centimetres per year.

In the core-mantle boundary layer, heat flow is only possible through conduction. The layer around the core acts as an insulator, because conductive heat flow is much less efficient than convective flow. At both the bottom and the top of the boundary layer, there is a large change in temperature. At the core side of the layer, temperatures are about 4000 K, while at the mantle side they are about 2500 K. This means that the temperature decreases about 1500 K in only 200 km (7.5 K/km)! For comparison, in the outer core (over 2000 km thick), temperature decreases only 500 K (0.25 K/km). That is 30 times less! Because convective heat flow is 30 times faster than conductive heat flow, the assumption that we can neglect conduction in case of convection is allowed.

Temperatures also decrease strongly in the lithosphere, from the upper boundary of the mantle to the surface of the crust. Within the mantle and core, however, temperatures increase only slowly with depth; this corresponds to the convection cells of Figure 6.9.

Exercise 6-15**: Explaining Figure 6.11 Use the questions below to find out if you understand Figure 6.11 better now you have read the text above. a. What do the black dots on the solid line represent? b. How are the fixed temperatures interpolated and what assumptions have been made in doing

so? c. Which types of heat transport were neglected? d. At what depth does the core-mantle boundary lie? What happens to the geotherm at this

boundary? e. What causes the geotherm behaviour at the CMB?

Exercise 6-16**: The temperature in mines Miners know from experience that temperature increases with depth. In the shallow marl mines in southern Limburg, The Netherlands, temperature is constant at 14oC. But in the old coal mines hundreds of metres deep, temperatures are much higher. a. Calculate the temperature increase per kilometre of depth with the help of Fourier’s law.

Assume that temperature increases linearly with depth. Use typical crust and shallow mantle properties: J = 70 · 10-3 W m-2 and k = 3.0 W m-1 K-1.

b. What is the temperature in a mine shaft at 1.5 km depth?


143

Exercise 6-17**: The temperature distribution in the outer core Use the data in Figure 6.11 to estimate the temperature increase per kilometre in the outer core.

6.5.3 Heat flux and the Earth’s magnetic field The liquid outer core mainly consists of the ferromagnetic metals iron and nickel that enable the generation of the Earth’s magnetic field, an important property of the outer core. Does the convective model for the Earth's thermal state that we have developed so far allow for this ‘magneto-hydrodynamic' process, or geodynamo?

The geodynamo is generated by convective flows in the liquid iron-nickel outer core that are accompanied by electric currents in the electrically conducting material. These electric currents generate a magnetic field. Subsequently, interaction of the fluid flows and the magnetic field generates an electrical potential field and corresponding electric current through electro-magnetic induction. To maintain this geodynamo process, the heat flux out of the core cannot become too high or too low.

Because of the insulating effect of the core-mantle boundary layer, the core loses heat slowly. This way, the outer core can remain liquid and convection is sustained. But, heat loss cannot be too low either. Research shows that the core should lose at least 1012 W to the mantle in order to keep the dynamo process working. This heat must then be brought up to the crust by convective flow in the mantle to be lost to space. According to the convection model, most heat is brought up where there is a convective flow upwards. From plate tectonics, we know where to find such a flow: underneath a mid-ocean ridge. Figure 6.12 gives the heat flux distribution at the Earth’s surface. You can see that most heat is indeed lost at mid-ocean ridges, where material rises from depth.

Figure 6.12: The heat flux at the Earth surface. The white lines show the continent contours, the dashed black lines the mid-ocean ridges. The colouring represents the magnitude of the heat flux q0 in MW.m-2 given to the earth surface. The total heat flux can be calculated by: Q = 44.1012

W. There can be seen that the mid-ocean ridges function as a heat flux for the convectering mantel system; there, the heat flux is the largest and within the deep sea trough the smallest.


144

Exercise 6-18**: The convection model and the Earth’s magnetic field The convection model states that there is an insulating boundary layer between the core and the mantle. Temperature decreases strongly through this layer (7.5 K/km). a. Use Fourier’s law to calculate whether enough heat (at least 1012 W) passes the core-mantle boundary layer to sustain the geodynamo. Use a thermal conductivity of 3 WK-1m-1 to calculate the heat flux density J [Wm-2]. The core has a radius of 3300 km. b. Why is upwelling of mantle material at mid-ocean ridges needed to maintain the Earth’s magnetic field? c. Does the convection model contradict the generation of the Earth’s magnetic field?

The removal of heat from the core through the core-mantle boundary and from the mantle above is essential to the survival of the geodynamo. Mars, a neighbouring planet, lost its magnetic field about 4 Ga. However, there are reasons to believe that Mars still has a liquid iron core. A plausible explanation for the disappearance of the dynamo on Mars lies in the decreased heat transport when early plate tectonics ceased and Mars became a one-plate planet.

Computer models of the geodynamo are nowadays able to reproduce the important characteristics of the geomagnetic field, such as its dipole nature and varying strength and the irregular reversals of the field with an average period of several hundred thousands years (for example, see http://www.es.ucsc.edu/~glatz/geodynamo.html).

6.6 Liquid magma within solid mantle rock According to seismological data, the mantle and crust are predominantly made up of solid rock. Still there are regions where rock is molten, mainly underneath spreading ridges. This is known from volcanoes from, for example, Iceland.

At mid-ocean ridges, the molten rock forms new basaltic oceanic crust. How can solid rock that cools while it rises become liquid? Chapter 5 has touched upon this subject already, but you will understand the process better with the knowledge you have gained from this chapter.

We can explain the melting of rising mantle rock with the help of a (melting) phase diagram of mantle material (peridotite). When the pressure on a sample of peridotite decreases because the rock reaches shallower depths, its melting temperature decreases as well. Figure 6.13 shows the formation of melt underneath an oceanic spreading ridge schematically. On the left side, you see a symmetric upward flow of mantle rock underneath the ridge. The vertical dashed line indicates the model’s symmetry axis. The right figure, with a corresponding depth/pressure axis, shows the melting phase diagram of peridotite. The thick solid line shows that the melting temperature increases with pressure, so, also with depth. The geotherm indicates the temperature of the rising mantle rock. This line and the thick phase line intersect at a depth of about 70 km. Starting from this depth, the rock will gradually melt while it rises farther. The heat needed for melting is extracted from the rock itself. Therefore, melting rock, rising to the surface, cools a little faster than before the onset of melting, hence the bend in the geotherm at 70 km depth. Melting rock has a porosity of about 1 volume percent; therefore, melt rises faster than solid rock. The formed magma concentrates in magma chambers underneath the spreading ridge. The outflow of basaltic lava from the chambers feeds the formation of new oceanic crust.

Exercise 6-19*: Is the interior of the Earth liquid? Many people argue that because molten rock flows out of volcanoes, the interior of the Earth is molten as well. Would you argue differently? Explain.

6.7 Recent developments 6.7.1 Observing subduction in the mantle The convective flows depicted in Figure 6.8 correspond to a mantle circulation pattern of upward flow underneath MORs and downward flow at subduction zones, where lithosphere sinks into the deep mantle. We can actually see that cool oceanic lithosphere can sink deep into the mantle from images of the mantle obtained with seismic travel time tomography. Such an image is shown in Figure 6.14. Two subduction zones are visible in this cross section running from the Aegean Sea to Japan. You can recognize subducting lithospheric slabs as thin blue structures. The blue colour


145

indicates a higher seismic wave velocity, which corresponds to lower temperatures. The slabs enter the mantle at an angle and are visible up to great depths.

Figure 6-1: A tomografic division of seismic speed deviation in a mantle cross cut of the Aegean Sea towards the Japanese islands. The dotted lines of approximately 400 to 660 km depth point out the location of two solid phase transitions of the olivin mantle material. The 660 km phase transition line marks the boundary between the upper and lower mantle. Subduction zones come to an expression as sharply bordered areas with high wave speed. The white symbols near Japan point out earthquake locations in the subduction zone.

Figure 6.13: Left panel: Schematic representation of the formation of basaltic oceanic crust through partial melting of rising mantle rock underneath a spreading ridge. Right panel: The schematic phase diagram of peridotite mantle rock. The intersection of the geotherm with the phase line determines the depth above which melting occurs.

Upwards mantle stream

Depth first melting

Spreading Mid Ocean Ridges

Temperature


146

6.7.2 An additional fixed point for the temperature distribution in the mantle In Section 6.4.2 we learned that the temperature difference ΔT over the core-mantle boundary layer is about 1500 K. The discovery of another phase transition in the mantle provides an additional fixed point for the mantle’s temperature distribution. This phase transition is indicated in Figure 6.3 with ‘new phase’. During this phase transition, the mineral perovskite changes into the denser mineral post-perovskite. Experiments have demonstrated that this phase transition takes place at 120 GPa and about 2500 K, and seismological observations have confirmed it's presence in the deep mantle.

With more detailed information about the temperature structure directly above the CMB, it is possible to determine the heat flux out of the core. This is important for our understanding of the magneto-hydrodynamic convection process in the liquid outer core that maintains the Earth’s magnetic field. Rough first estimates (see Exercise 6-18 as well) point to a global CMB heat flux of 1013 W. This estimate is still quite uncertain, because the thermal conductivity close to the core is not very well constrained.

6.8 What have we learnt? In this chapter we have tried to find the engine driving the motion within the Earth and at its surface. We turned to the Earth’s thermal state to uncover how the internal heat produced by natural radioactive decay can be lost at the Earth’s surface without the interior temperature rising above the melting temperature of mantle rock. Slow convective flow in the mantle solves the Earth’s heat problem and the resulting mantle circulation explains the observed motion of oceanic and continental plates and, thus, plate tectonics. We have also seen that the mantle convection model coincides with the measured distribution of the heat flux through the surface of the Earth and that the thermal boundary at the CMB, predicted by the model, is confirmed by data from seismology and mineral physics.

So, did we find all the answers? No, as in most natural sciences, in geodynamics answering one question brings up another. For example, these questions remain unanswered (for now):

Has large-scale plate tectonics always been active on Earth? Or did plate tectonics only start when the Earth had cooled sufficiently? If so, was a smaller scale chaotic type of mantle circulation responsible for the heat loss of the young Earth?

Why do the other terrestrial planets Mercury, Venus and Mars and the Moon no longer show any signs of plate tectonics? This is one of the main questions of comparative planetology, a branch of science that has expanded over the last few years due to the enormous amount of data available from planetary research missions. Comparative planetology has recently been extended to planets outside our solar system i(see http://www.exoplanet.eu and http://kepler.nasa.gov).

Glossary The Dynamic Earth

147

Glossary Accumulation of sediment: the build up of loose deposits.

Anomaly: a deviation from the average or reference value (for example for gravity).

Asthenosphere: the viscous part of the mantle that lies directly underneath the lithosphere, found between a depth of approximately 80 and 300 km . The tectonic plates move over the asthenosphere.

Azimuth: the horizontal angle measured clockwise from the north of the fault orientation at the surface.

Basalt: dark extrusive rock formed during volcanic eruptions. Basalt is rock that has cooled fast, so few or no visible crystals are formed. During cooling, basalt usually fractures into polygonal columns. In The Netherlands, these columns are used to reinforce dikes.

Bathymetry: depth measurements of the ocean floor, or a map showing ocean depths.

Caldera: a circular depression of the landscape formed by an explosive volcanic eruption. A caldera is formed when the roof of a volcano’s large, shallow magma reservoir collapses due to the extrusion of large amounts of magma over a short period of time.

Chondrites: the largest group of meteorites. They contain chondrules, small spherical particles, after which they are named. The chemical composition of the Earth resembles that of the chondrites.

Conduction: the transport of heat through a substance; the passing on of vibrations by atoms in a lattice.

Convection: The transfer of heat by flow of material caused by hotter, less dense material rising and colder, denser material sinking under the influence of gravity.

Convergent plate boundary: the margin between two tectonic plates where the plates move toward each other, one of which subducts underneath the other.

Core: the central part of the Earth, starting at a depth of 3450 km. The core is divided into a liquid outer core and a solid inner core.

Crust: the outer layer of the Earth. The crust is relatively thin: oceanic crust is about 7 km thick and continental crust 30 to 50 km thick.

Crustal root: the thickened crust underneath a mountain chain that allows the mountains to maintain isostatic equilibrium.

Crystal: a homogeneous solid with well-developed surfaces which express the regular internal arrangement of the crystal’s atoms. The more time and space available for the cooling of a melt, the larger a crystal will grow.

Deformation: the change in the shape, volume, position or orientation of a rock.

Divergent plate boundary: the margin between two plates that move away from each other.

Dome: a swell resulting from the slow extrusion of viscous lava from a volcano the build-up of viscous lava in the crater of a volcano. Sometimes this build-up hampers the extrusion of magma, causing it to slow down.

Epicentre: the location on the Earth’s surface directly above the hypocentre of an earthquake. The effect of an earthquake is felt most strongly at its epicentre.

Erosion: the removal of rock through physical processes such as water flowing, ice or wind.

Focus of an earthquake (= hypocentre): the point of origin of the earthquake within the Earth .

The Dynamic Earth Glossary

148

Force: this describes the push or pull exerted on an object. The unit of force is Newton [N].

Fracture zone: a region of cracks in the ocean floor continuing from a transform fault between two mid-ocean ridge segments.

Granite: coarse, crystalline igneous rock. Granite cooled slowly beneath the Earth’s surface and had enough time to grow large crystals.

Great circle: a special circle on the surface of a sphere. It is the largest circle on a sphere and dividing it into two equal halves. A great circle therefore has the same centre as its sphere.

Half-life: the time it takes for a certain amount of radioactive isotope to decay to half its original value. Every radioactive isotope has a unique half-life.

Hazard management: the measures taken to control or reduce the effects of natural hazards - such as risk assessments, taking precautions and writing disaster scenarios.

Hotspot: a stationary plume in the mantle made up of hot mantle rock rising from large depths of the Earth. Volcanism occurs in the region above the plume, where material starts to melt at shallower levels. Hawaii, for example, was formed in this way.

HPT experiments: High Pressure and Temperature experiments to model processes taking place in the interior of the Earth.

Hypocentre: the location within the Earth’s crust where earthquake vibrations originate.

Insolation: solar radiation that reaches the surface of the Earth.

Isotopes: atoms of the same element that have a different atomic mass. Isotopes have the same number of protons in their nucleus, but a different number of neutrons.

Isostasy: the concept that where the crust is in gravitational equilibrium with the mantle. Because the Earth’s crust has a lower density than the mantle, it ‘floats’ on top of the mantle (like wood floating on water).

Lava: liquid rock that extrudes from a volcano, flows over the Earth’s surface and eventually solidifies.

Limestone: sedimentary rock mainly consisting of calcite (CaCO3).

Lithosphere: the crust and the upper part of the mantle that form the tectonic plate. Beneath the ocean the lithosphere is about 70 km thick, whereas a continental lithosphere can be well over 125 km thick.

Longitudinal wave (primary wave, P-wave): a seismic wave that oscillates in the same direction as its propagation direction. The P-wave is the first wave to arrive at a seismic station.

Magma: liquid rock below the Earth’s surface. When magma extrudes slowly, not explosively, and flows over the Earth’s surface, it is called lava.

Magma chamber: part of the crust where magma is contained in a closed reservoir. This chamber is the source of volcanism.

Mantle: the 2900-kilometre thick layer between the Earth’s crust and core. The solid mantle is divided into an upper mantle and a lower mantle.

Marble: metamorphic limestone consisting of carbonate crystals (mostly calcite, CaCO3). Marble comes in several different colours and is often used for tiles and statues.

Metamorphic rock: the rock formed due to transition of the mineral content of the original rock (for example, sedimentary or igneous rock) as a result of high pressures and/or temperatures.

Mid-Atlantic Ridge and Mid-Oceanic Ridge (MOR, spreading ridge): the plate boundary underneath an ocean where two plates diverge and new oceanic crust is formed.

Milankovitch: The Serbian mathematician who described the variation in the Earth’s orbit around the Sun with the help of mathematical equations. The Earth’s orbit varies in three ways: its eccentricity varies with a period of 100,000 years, its obliquity has a period of 40,000 years and its precession a period of 20,000 years. These kinds of orbital variations change the insolation of the Earth, which in turn affects the Earth’s climate. Thus Milankovitch linked the Earth’s orbital variations to the ice ages.

Orogenesis: mountain building.

Paleomagnetic timescale: the timescale based on paleomagnetic datings that use the Earth’s magnetic field and its reversals to determine the age of rocks.

Glossary The Dynamic Earth

149

Phase change: the transition of a material from the liquid phase to a solid or gas phase or vice versa. Solid material can also exist in different phases. For example, carbon can be found in a diamond form and a graphite form.

Plates: the Earth’s surface is divided into several tectonic plates. These plates are made up of the lithosphere, which consists of oceanic and/or continental crust, and the underlying upper part of the mantle.

Plate tectonics: the theory that states that the Earth’s surface is divided into several large, rigid plates that move slowly with respect to each other. Along the plate boundaries, geological phenomena such as volcanism, earthquakes and mountain building take place.

Polar jet stream: the strong, concentrated winds in the troposphere that drive weather systems.

Pressure: force per unit area exerted on an object. Pressure is stress equal in all directions.

Sedimentary rock: rock formed through the transport and deposition (sedimentation) of loose material.

Seismograph (seismometer): a device that records seismic waves.

Shield volcano: a shield-shaped volcano made of mostly basaltic lava flows. Shield volcanoes are characterized by slow extrusions that result in gently sloping flanks and are commonly found at MORs and hotspots.

Shear modulus: the material property of elasticity with respect to shear deformation.

Slip: the size and direction of the relative motion along a fault plane.

Stress: the force exerted per unit area, in Pascal (1 Pa = 1 N/m2).

Igneous rock (extrusive or intrusive rock): the rock formed through the crystallization of magma.

Strain partitioning: plate motion is not accommodated perpendicular to the plate contact, but the relative motion is partitioned into two components, one perpendicular and one parallel to the contact.

Stratovolcano: a conical volcano built up of alternating layers of lava and tephra (ashes, pyroclastic flow deposits and more). The volcano forms through explosive eruptions and is most commonly found along subduction zones.

Subduction: the process by which the ocean floor dives underneath a continent or island arc.

Tephra (pyroclastic deposit): the collective term for deposits of fragmented volcanic material that are produced in an explosive volcanic eruption, such as ash, volcanic bombs and blocks and mud flows.

Tectonics: motion and deformation of (parts of) the Earth’s crust or lithosphere.

Thermal boundary layer: the lower mantle layer above the core-mantle boundary that is characterized by a large contrast in temperature.

Tomographic map: map or image based on the spatial variation of seismic wave speeds in the mantle.

Transform fault: a fault formed when two tectonic plates slide past each other.

Transverse wave (secondary wave, S-wave): a wave that oscillates in the direction perpendicular to its propagation direction. Transverse waves arrive at seismic stations after the longitudinal body waves.

Tsunami: extremely long waves of water generated by vertical movement of the ocean floor.

Volatile: the constituent part of magma that can easily escape when it is in its gas phase (when the magma rises or cools). For example, water (H2O), CO2, SO2, HCl and H2S.

Weathering: the disintegration of rock due to vegetation and the weather.

The Dynamic Earth Glossary

150

the dynamic earth - betavak-nlt.nl · chapter 2. plate tectonics 21 2.1 the observations that led...

Documents