zotero-db/storage/BM5D5Y8J/.zotero-ft-cache

Relativity Demystiﬁed

Demystiﬁed Series

Advanced Statistics Demystiﬁed Algebra Demystiﬁed Anatomy Demystiﬁed asp.net Demystiﬁed Astronomy Demystiﬁed Biology Demystiﬁed Business Calculus Demystiﬁed Business Statistics Demystiﬁed C++ Demystiﬁed Calculus Demystiﬁed Chemistry Demystiﬁed College Algebra Demystiﬁed Databases Demystiﬁed Data Structures Demystiﬁed Differential Equations Demystiﬁed Digital Electronics Demystiﬁed Earth Science Demystiﬁed Electricity Demystiﬁed Electronics Demystiﬁed Environmental Science Demystiﬁed Everyday Math Demystiﬁed Genetics Demystiﬁed Geometry Demystiﬁed Home Networking Demystiﬁed Investing Demystiﬁed Java Demystiﬁed JavaScript Demystiﬁed Linear Algebra Demystiﬁed Macroeconomics Demystiﬁed

Math Proofs Demystiﬁed Math Word Problems Demystiﬁed Medical Terminology Demystiﬁed Meteorology Demystiﬁed Microbiology Demystiﬁed OOP Demystiﬁed Options Demystiﬁed Organic Chemistry Demystiﬁed Personal Computing Demystiﬁed Pharmacology Demystiﬁed Physics Demystiﬁed Physiology Demystiﬁed Pre-Algebra Demystiﬁed Precalculus Demystiﬁed Probability Demystiﬁed Project Management Demystiﬁed Quality Management Demystiﬁed Quantum Mechanics Demystiﬁed Relativity Demystiﬁed Robotics Demystiﬁed Six Sigma Demystiﬁed sql Demystiﬁed Statistics Demystiﬁed Trigonometry Demystiﬁed uml Demystiﬁed Visual Basic 2005 Demystiﬁed Visual C # 2005 Demystiﬁed xml Demystiﬁed

Relativity Demystiﬁed

DAVID McMAHON

New York

McGRAW-HILL
Chicago San Francisco Lisbon London Madrid Mexico City Milan New Delhi San Juan Seoul
Singapore Sydney Toronto

Copyright © 2006 by The McGraw-Hill Companies, Inc. All rights reserved. Manufactured in the United States of America. Except as permitted under the United States Copyright Act of 1976, no part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior written permission of the publisher.
0-07-148673-9
The material in this eBook also appears in the print version of this title: 0-07-145545-0.
All trademarks are trademarks of their respective owners. Rather than put a trademark symbol after every occurrence of a trademarked name, we use names in an editorial fashion only, and to the benefit of the trademark owner, with no intention of infringement of the trademark. Where such designations appear in this book, they have been printed with initial caps.
McGraw-Hill eBooks are available at special quantity discounts to use as premiums and sales promotions, or for use incorporate training programs. For more information, please contact George Hoare, Special Sales, at george_hoare@mcgraw-hill.com or (212) 904-4069.
TERMS OF USE
This is a copyrighted work and The McGraw-Hill Companies, Inc. (“McGraw-Hill”) and its licensors reserve all rights in and to the work. Use of this work is subject to these terms. Except as permitted under the Copyright Act of 1976 and the right to store and retrieve one copy of the work, you may not decompile, disassemble, reverse engineer, reproduce, modify, create derivative works based upon, transmit, distribute, disseminate, sell, publish or sublicense the work or any part of it without McGraw-Hill’s prior consent. You may use the work for your own noncommercial and personal use; any other use of the work is strictly prohibited. Your right to use the work may be terminated if you fail to comply with these terms.
THE WORK IS PROVIDED “AS IS.” McGRAW-HILL AND ITS LICENSORS MAKE NO GUARANTEES OR WARRANTIES AS TO THE ACCURACY, ADEQUACY OR COMPLETENESS OF OR RESULTS TO BE OBTAINED FROM USING THE WORK, INCLUDING ANY INFORMATION THAT CAN BE ACCESSED THROUGH THE WORK VIA HYPERLINK OR OTHERWISE, AND EXPRESSLY DISCLAIM ANY WARRANTY, EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO IMPLIED WARRANTIES OF MERCHANTABILITY OR FITNESS FOR A PARTICULAR PURPOSE. McGraw-Hill and its licensors do not warrant or guarantee that the functions contained in the work will meet your requirements or that its operation will be uninterrupted or error free. Neither McGraw-Hill nor its licensors shall be liable to you or anyone else for any inaccuracy, error or omission, regardless of cause, in the work or for any damages resulting therefrom. McGraw-Hill has no responsibility for the content of any information accessed through the work. Under no circumstances shall McGraw-Hill and/or its licensors be liable for any indirect, incidental, special, punitive, consequential or similar damages that result from the use of or inability to use the work, even if any of them has been advised of the possibility of such damages. This limitation of liability shall apply to any claim or cause whatsoever whether such claim or cause arises in contract, tort or otherwise.
DOI: 10.1036/0071455450

For more information about this title, click here

CONTENTS

CHAPTER 1 CHAPTER 2

Preface
A Quick Review of Special Relativity Frame of Reference Clock Synchronization Inertial Frames Galilean Transformations Events The Interval Postulates of Special Relativity Three Basic Physical Implications Light Cones and Spacetime Diagrams Four Vectors Relativistic Mass and Energy Quiz
Vectors, One Forms, and the Metric Vectors New Notation Four Vectors The Einstein Summation Convention Tangent Vectors, One Forms, and the
Coordinate Basis Coordinate Transformations

xi
1 5 5 6 7 7 8 9 13 17 19 20 21
23 23 25 27 28
29 31
v

vi
CHAPTER 3 CHAPTER 4

CONTENTS

The Metric

32

The Signature of a Metric

36

The Flat Space Metric

37

The Metric as a Tensor

37

Index Raising and Lowering

38

Index Gymnastics

41

The Dot Product

42

Passing Arguments to the Metric

43

Null Vectors

45

The Metric Determinant

45

Quiz

45

More on Tensors

47

Manifolds

47

Parameterized Curves

49

Tangent Vectors and One Forms, Again

50

Tensors as Functions

53

Tensor Operations

54

The Levi-Cevita Tensor

59

Quiz

59

Tensor Calculus

60

Testing Tensor Character

60

The Importance of Tensor Equations

61

The Covariant Derivative

62

The Torsion Tensor

72

The Metric and Christoffel Symbols

72

The Exterior Derivative

79

The Lie Derivative

81

The Absolute Derivative and Geodesics

82

The Riemann Tensor

85

The Ricci Tensor and Ricci Scalar

88

The Weyl Tensor and Conformal Metrics

90

Quiz

91

CONTENTS
CHAPTER 5
CHAPTER 6
CHAPTER 7

vii

Cartan’s Structure Equations

93

Introduction

93

Holonomic (Coordinate) Bases

94

Nonholonomic Bases

95

Commutation Coefﬁcients

96

Commutation Coefﬁcients and Basis

One Forms

98

Transforming between Bases

100

A Note on Notation

103

Cartan’s First Structure Equation and the

Ricci Rotation Coefﬁcients

104

Computing Curvature

112

Quiz

120

The Einstein Field Equations

122

Equivalence of Mass in Newtonian Theory 123

Test Particles

126

The Einstein Lift Experiments

126

The Weak Equivalence Principle

130

The Strong Equivalence Principle

130

The Principle of General Covariance

131

Geodesic Deviation

131

The Einstein Equations

136

The Einstein Equations with Cosmological

Constant

138

An Example Solving Einstein’s Equations

in 2 + 1 Dimensions

139

Energy Conditions

152

Quiz

152

The Energy-Momentum Tensor

155

Energy Density

156

Momentum Density and Energy Flux

156

Stress

156

Conservation Equations

157

viii
CHAPTER 8 CHAPTER 9 CHAPTER 10

CONTENTS

Dust

158

Perfect Fluids

160

Relativistic Effects on Number Density

163

More Complicated Fluids

164

Quiz

165

Killing Vectors

167

Introduction

167

Derivatives of Killing Vectors

177

Constructing a Conserved Current

with Killing Vectors

178

Quiz

178

Null Tetrads and the Petrov

Classiﬁcation

180

Null Vectors

182

A Null Tetrad

184

Extending the Formalism

190

Physical Interpretation and the Petrov

Classiﬁcation

193

Quiz

201

The Schwarzschild Solution

203

The Vacuum Equations

204

A Static, Spherically Symmetric Spacetime 204

The Curvature One Forms

206

Solving for the Curvature Tensor

209

The Vacuum Equations

211

The Meaning of the Integration Constant 214

The Schwarzschild Metric

215

The Time Coordinate

215

The Schwarzschild Radius

215

Geodesics in the Schwarzschild Spacetime 216

Particle Orbits in the Schwarzschild

Spacetime

218

CONTENTS
CHAPTER 11
CHAPTER 12 CHAPTER 13

ix

The Deﬂection of Light Rays

224

Time Delay

229

Quiz

230

Black Holes

233

Redshift in a Gravitational Field

234

Coordinate Singularities

235

Eddington-Finkelstein Coordinates

236

The Path of a Radially Infalling Particle 238

Eddington-Finkelstein Coordinates

239

Kruskal Coordinates

242

The Kerr Black Hole

244

Frame Dragging

249

The Singularity

252

A Summary of the Orbital Equations

for the Kerr Metric

252

Further Reading

253

Quiz

254

Cosmology

256

The Cosmological Principle

257

A Metric Incorporating Spatial

Homogeneity and Isotropy

257

Spaces of Positive, Negative, and

Zero Curvature

262

Useful Deﬁnitions

264

The Robertson-Walker Metric and the

Friedmann Equations

267

Different Models of the Universe

271

Quiz

276

Gravitational Waves

279

The Linearized Metric

280

Traveling Wave Solutions

284

The Canonical Form and Plane Waves

287

x

CONTENTS

The Behavior of Particles as a

Gravitational Wave Passes

291

The Weyl Scalars

294

Review: Petrov Types and the

Optical Scalars

295

pp Gravity Waves

297

Plane Waves

301

The Aichelburg-Sexl Solution

303

Colliding Gravity Waves

304

The Effects of Collision

311

More General Collisions

312

Nonzero Cosmological Constant

318

Further Reading

321

Quiz

322

Final Exam

323

Quiz and Exam Solutions

329

References and Bibliography

333

Index

337

PREFACE
The theory of relativity stands out as one of the greatest achievements in science. The “special theory”, which did not include gravity, was put forward by Einstein in 1905 to explain many troubling facts that had arisen in the study of electricity and magnetism. In particular, his postulate that the speed of light in vacuum is the same constant seen by all observers forced scientists to throw away many closely held commonsense assumptions, such as the absolute nature of the passage of time. In short, the theory of relativity challenges our notions of what reality is, and this is one of the reasons why the theory is so interesting.
Einstein published the “general” theory of relativity, which is a theory about gravity, about a decade later. This theory is far more mathematically daunting, and perhaps this is why it took Einstein so long to come up with it. This theory is more fundamental than the special theory of relativity; it is a theory of space and time itself, and it not only describes, it explains gravity. Gravity is the distortion of the structure of spacetime as caused by the presence of matter and energy, while the paths followed by matter and energy (think of bending of passing light rays by the sun) in spacetime are governed by the structure of spacetime. This great feedback loop is described by Einstein’s ﬁeld equations.
This is a book about general relativity. There is no getting around the fact that general relativity is mathematically challenging, so we cannot hope to learn the theory without mastering the mathematics. Our hope with this book is to “demystify” that mathematics so that relativity is easier to learn and more accessible to a wider audience than ever before. In this book we will not skip any of the math that relativity requires, but we will present it in what we hope to be a clear fashion and illustrate how to use it with many explicitly solved examples. Our goal is to make relativity more accessible to everyone. Therefore we hope that engineers, chemists, and mathematicians or anyone who has had basic mathematical training at the college level will ﬁnd this book useful. And
xi
Copyright © 2006 by The McGraw-Hill Companies, Inc. Click here for terms of use.

xii

PREFACE

of course the book is aimed at physicists and astronomers who want to learn the theory.
The truth is that relativity looks much harder than it is. There is a lot to learn, but once you get comfortable with the new math and new notation, you will actually ﬁnd it a bit easier than many other technical areas you have studied in the past.
This book is meant to be a self-study guide or a supplement, and not a fullblown textbook. As a result we may not cover every detail and will not provide lengthly derivations or detailed physical explanations. Those can be found in any number of ﬁne textbooks on the market. Our focus here is also in “demystifying” the mathematical framework of relativity, and so we will not include lengthly descriptions of physical arguments. At the end of the book we provide a listing of references used in the development of this manuscript, and you can select books from that list to ﬁnd the details we are leaving out. In much of the material, we take the approach in this book of stating theorems and results, and then applying them in solved problems. Similar problems in the end-of chapter quiz help you try things out yourself.
So if you are taking a relativity course, you might want to use this book to help you gain better understanding of your main textbook, or help you to see how to accomplish certain tasks. If you are interested in self-study, this book will help you get started in your own mastery of the subject and make it easier for you to read more advanced books.
While this book is taking a lighter approach than the textbooks in the ﬁeld, we are not going to cut corners on using advanced mathematics. The bottom line is you are going to need some mathematical background to ﬁnd this book useful. Calculus is a must, studies of differential equations, vector analysis and linear algebra are helpful. A background in basic physics is also helpful.
Relativity can be done in different ways using a coordinate-based approach or differential forms and Cartan’s equations. We much prefer the latter approach and will use it extensively. Again, it looks intimidating at ﬁrst because there are lots of Greek characters and fancy symbols, and it is a new way of doing things. When doing calculations it does require a bit of attention to detail. But after a bit of practice, you will ﬁnd that its not really so hard. So we hope that readers will invest the effort necessary to master this nice mathematical way of solving physics problems.

Relativity Demystiﬁed

Demystiﬁed Series

Advanced Statistics Demystiﬁed Algebra Demystiﬁed Anatomy Demystiﬁed asp.net Demystiﬁed Astronomy Demystiﬁed Biology Demystiﬁed Business Calculus Demystiﬁed Business Statistics Demystiﬁed C++ Demystiﬁed Calculus Demystiﬁed Chemistry Demystiﬁed College Algebra Demystiﬁed Databases Demystiﬁed Data Structures Demystiﬁed Differential Equations Demystiﬁed Digital Electronics Demystiﬁed Earth Science Demystiﬁed Electricity Demystiﬁed Electronics Demystiﬁed Environmental Science Demystiﬁed Everyday Math Demystiﬁed Genetics Demystiﬁed Geometry Demystiﬁed Home Networking Demystiﬁed Investing Demystiﬁed Java Demystiﬁed JavaScript Demystiﬁed Linear Algebra Demystiﬁed Macroeconomics Demystiﬁed

Math Proofs Demystiﬁed Math Word Problems Demystiﬁed Medical Terminology Demystiﬁed Meteorology Demystiﬁed Microbiology Demystiﬁed OOP Demystiﬁed Options Demystiﬁed Organic Chemistry Demystiﬁed Personal Computing Demystiﬁed Pharmacology Demystiﬁed Physics Demystiﬁed Physiology Demystiﬁed Pre-Algebra Demystiﬁed Precalculus Demystiﬁed Probability Demystiﬁed Project Management Demystiﬁed Quality Management Demystiﬁed Quantum Mechanics Demystiﬁed Relativity Demystiﬁed Robotics Demystiﬁed Six Sigma Demystiﬁed sql Demystiﬁed Statistics Demystiﬁed Trigonometry Demystiﬁed uml Demystiﬁed Visual Basic 2005 Demystiﬁed Visual C # 2005 Demystiﬁed xml Demystiﬁed

Relativity Demystiﬁed

DAVID McMAHON

New York

McGRAW-HILL
Chicago San Francisco Lisbon London Madrid Mexico City Milan New Delhi San Juan Seoul
Singapore Sydney Toronto

Copyright © 2006 by The McGraw-Hill Companies, Inc. All rights reserved. Manufactured in the United States of America. Except as permitted under the United States Copyright Act of 1976, no part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior written permission of the publisher.
0-07-148673-9
The material in this eBook also appears in the print version of this title: 0-07-145545-0.
All trademarks are trademarks of their respective owners. Rather than put a trademark symbol after every occurrence of a trademarked name, we use names in an editorial fashion only, and to the benefit of the trademark owner, with no intention of infringement of the trademark. Where such designations appear in this book, they have been printed with initial caps.
McGraw-Hill eBooks are available at special quantity discounts to use as premiums and sales promotions, or for use incorporate training programs. For more information, please contact George Hoare, Special Sales, at george_hoare@mcgraw-hill.com or (212) 904-4069.
TERMS OF USE
This is a copyrighted work and The McGraw-Hill Companies, Inc. (“McGraw-Hill”) and its licensors reserve all rights in and to the work. Use of this work is subject to these terms. Except as permitted under the Copyright Act of 1976 and the right to store and retrieve one copy of the work, you may not decompile, disassemble, reverse engineer, reproduce, modify, create derivative works based upon, transmit, distribute, disseminate, sell, publish or sublicense the work or any part of it without McGraw-Hill’s prior consent. You may use the work for your own noncommercial and personal use; any other use of the work is strictly prohibited. Your right to use the work may be terminated if you fail to comply with these terms.
THE WORK IS PROVIDED “AS IS.” McGRAW-HILL AND ITS LICENSORS MAKE NO GUARANTEES OR WARRANTIES AS TO THE ACCURACY, ADEQUACY OR COMPLETENESS OF OR RESULTS TO BE OBTAINED FROM USING THE WORK, INCLUDING ANY INFORMATION THAT CAN BE ACCESSED THROUGH THE WORK VIA HYPERLINK OR OTHERWISE, AND EXPRESSLY DISCLAIM ANY WARRANTY, EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO IMPLIED WARRANTIES OF MERCHANTABILITY OR FITNESS FOR A PARTICULAR PURPOSE. McGraw-Hill and its licensors do not warrant or guarantee that the functions contained in the work will meet your requirements or that its operation will be uninterrupted or error free. Neither McGraw-Hill nor its licensors shall be liable to you or anyone else for any inaccuracy, error or omission, regardless of cause, in the work or for any damages resulting therefrom. McGraw-Hill has no responsibility for the content of any information accessed through the work. Under no circumstances shall McGraw-Hill and/or its licensors be liable for any indirect, incidental, special, punitive, consequential or similar damages that result from the use of or inability to use the work, even if any of them has been advised of the possibility of such damages. This limitation of liability shall apply to any claim or cause whatsoever whether such claim or cause arises in contract, tort or otherwise.
DOI: 10.1036/0071455450

For more information about this title, click here

CONTENTS

CHAPTER 1 CHAPTER 2

Preface
A Quick Review of Special Relativity Frame of Reference Clock Synchronization Inertial Frames Galilean Transformations Events The Interval Postulates of Special Relativity Three Basic Physical Implications Light Cones and Spacetime Diagrams Four Vectors Relativistic Mass and Energy Quiz
Vectors, One Forms, and the Metric Vectors New Notation Four Vectors The Einstein Summation Convention Tangent Vectors, One Forms, and the
Coordinate Basis Coordinate Transformations

xi
1 5 5 6 7 7 8 9 13 17 19 20 21
23 23 25 27 28
29 31
v

vi
CHAPTER 3 CHAPTER 4

CONTENTS

The Metric

32

The Signature of a Metric

36

The Flat Space Metric

37

The Metric as a Tensor

37

Index Raising and Lowering

38

Index Gymnastics

41

The Dot Product

42

Passing Arguments to the Metric

43

Null Vectors

45

The Metric Determinant

45

Quiz

45

More on Tensors

47

Manifolds

47

Parameterized Curves

49

Tangent Vectors and One Forms, Again

50

Tensors as Functions

53

Tensor Operations

54

The Levi-Cevita Tensor

59

Quiz

59

Tensor Calculus

60

Testing Tensor Character

60

The Importance of Tensor Equations

61

The Covariant Derivative

62

The Torsion Tensor

72

The Metric and Christoffel Symbols

72

The Exterior Derivative

79

The Lie Derivative

81

The Absolute Derivative and Geodesics

82

The Riemann Tensor

85

The Ricci Tensor and Ricci Scalar

88

The Weyl Tensor and Conformal Metrics

90

Quiz

91

CONTENTS
CHAPTER 5
CHAPTER 6
CHAPTER 7

vii

Cartan’s Structure Equations

93

Introduction

93

Holonomic (Coordinate) Bases

94

Nonholonomic Bases

95

Commutation Coefﬁcients

96

Commutation Coefﬁcients and Basis

One Forms

98

Transforming between Bases

100

A Note on Notation

103

Cartan’s First Structure Equation and the

Ricci Rotation Coefﬁcients

104

Computing Curvature

112

Quiz

120

The Einstein Field Equations

122

Equivalence of Mass in Newtonian Theory 123

Test Particles

126

The Einstein Lift Experiments

126

The Weak Equivalence Principle

130

The Strong Equivalence Principle

130

The Principle of General Covariance

131

Geodesic Deviation

131

The Einstein Equations

136

The Einstein Equations with Cosmological

Constant

138

An Example Solving Einstein’s Equations

in 2 + 1 Dimensions

139

Energy Conditions

152

Quiz

152

The Energy-Momentum Tensor

155

Energy Density

156

Momentum Density and Energy Flux

156

Stress

156

Conservation Equations

157

viii
CHAPTER 8 CHAPTER 9 CHAPTER 10

CONTENTS

Dust

158

Perfect Fluids

160

Relativistic Effects on Number Density

163

More Complicated Fluids

164

Quiz

165

Killing Vectors

167

Introduction

167

Derivatives of Killing Vectors

177

Constructing a Conserved Current

with Killing Vectors

178

Quiz

178

Null Tetrads and the Petrov

Classiﬁcation

180

Null Vectors

182

A Null Tetrad

184

Extending the Formalism

190

Physical Interpretation and the Petrov

Classiﬁcation

193

Quiz

201

The Schwarzschild Solution

203

The Vacuum Equations

204

A Static, Spherically Symmetric Spacetime 204

The Curvature One Forms

206

Solving for the Curvature Tensor

209

The Vacuum Equations

211

The Meaning of the Integration Constant 214

The Schwarzschild Metric

215

The Time Coordinate

215

The Schwarzschild Radius

215

Geodesics in the Schwarzschild Spacetime 216

Particle Orbits in the Schwarzschild

Spacetime

218

CONTENTS
CHAPTER 11
CHAPTER 12 CHAPTER 13

ix

The Deﬂection of Light Rays

224

Time Delay

229

Quiz

230

Black Holes

233

Redshift in a Gravitational Field

234

Coordinate Singularities

235

Eddington-Finkelstein Coordinates

236

The Path of a Radially Infalling Particle 238

Eddington-Finkelstein Coordinates

239

Kruskal Coordinates

242

The Kerr Black Hole

244

Frame Dragging

249

The Singularity

252

A Summary of the Orbital Equations

for the Kerr Metric

252

Further Reading

253

Quiz

254

Cosmology

256

The Cosmological Principle

257

A Metric Incorporating Spatial

Homogeneity and Isotropy

257

Spaces of Positive, Negative, and

Zero Curvature

262

Useful Deﬁnitions

264

The Robertson-Walker Metric and the

Friedmann Equations

267

Different Models of the Universe

271

Quiz

276

Gravitational Waves

279

The Linearized Metric

280

Traveling Wave Solutions

284

The Canonical Form and Plane Waves

287

x

CONTENTS

The Behavior of Particles as a

Gravitational Wave Passes

291

The Weyl Scalars

294

Review: Petrov Types and the

Optical Scalars

295

pp Gravity Waves

297

Plane Waves

301

The Aichelburg-Sexl Solution

303

Colliding Gravity Waves

304

The Effects of Collision

311

More General Collisions

312

Nonzero Cosmological Constant

318

Further Reading

321

Quiz

322

Final Exam

323

Quiz and Exam Solutions

329

References and Bibliography

333

Index

337

PREFACE
The theory of relativity stands out as one of the greatest achievements in science. The “special theory”, which did not include gravity, was put forward by Einstein in 1905 to explain many troubling facts that had arisen in the study of electricity and magnetism. In particular, his postulate that the speed of light in vacuum is the same constant seen by all observers forced scientists to throw away many closely held commonsense assumptions, such as the absolute nature of the passage of time. In short, the theory of relativity challenges our notions of what reality is, and this is one of the reasons why the theory is so interesting.
Einstein published the “general” theory of relativity, which is a theory about gravity, about a decade later. This theory is far more mathematically daunting, and perhaps this is why it took Einstein so long to come up with it. This theory is more fundamental than the special theory of relativity; it is a theory of space and time itself, and it not only describes, it explains gravity. Gravity is the distortion of the structure of spacetime as caused by the presence of matter and energy, while the paths followed by matter and energy (think of bending of passing light rays by the sun) in spacetime are governed by the structure of spacetime. This great feedback loop is described by Einstein’s ﬁeld equations.
This is a book about general relativity. There is no getting around the fact that general relativity is mathematically challenging, so we cannot hope to learn the theory without mastering the mathematics. Our hope with this book is to “demystify” that mathematics so that relativity is easier to learn and more accessible to a wider audience than ever before. In this book we will not skip any of the math that relativity requires, but we will present it in what we hope to be a clear fashion and illustrate how to use it with many explicitly solved examples. Our goal is to make relativity more accessible to everyone. Therefore we hope that engineers, chemists, and mathematicians or anyone who has had basic mathematical training at the college level will ﬁnd this book useful. And
xi
Copyright © 2006 by The McGraw-Hill Companies, Inc. Click here for terms of use.

xii

PREFACE

of course the book is aimed at physicists and astronomers who want to learn the theory.
The truth is that relativity looks much harder than it is. There is a lot to learn, but once you get comfortable with the new math and new notation, you will actually ﬁnd it a bit easier than many other technical areas you have studied in the past.
This book is meant to be a self-study guide or a supplement, and not a fullblown textbook. As a result we may not cover every detail and will not provide lengthly derivations or detailed physical explanations. Those can be found in any number of ﬁne textbooks on the market. Our focus here is also in “demystifying” the mathematical framework of relativity, and so we will not include lengthly descriptions of physical arguments. At the end of the book we provide a listing of references used in the development of this manuscript, and you can select books from that list to ﬁnd the details we are leaving out. In much of the material, we take the approach in this book of stating theorems and results, and then applying them in solved problems. Similar problems in the end-of chapter quiz help you try things out yourself.
So if you are taking a relativity course, you might want to use this book to help you gain better understanding of your main textbook, or help you to see how to accomplish certain tasks. If you are interested in self-study, this book will help you get started in your own mastery of the subject and make it easier for you to read more advanced books.
While this book is taking a lighter approach than the textbooks in the ﬁeld, we are not going to cut corners on using advanced mathematics. The bottom line is you are going to need some mathematical background to ﬁnd this book useful. Calculus is a must, studies of differential equations, vector analysis and linear algebra are helpful. A background in basic physics is also helpful.
Relativity can be done in different ways using a coordinate-based approach or differential forms and Cartan’s equations. We much prefer the latter approach and will use it extensively. Again, it looks intimidating at ﬁrst because there are lots of Greek characters and fancy symbols, and it is a new way of doing things. When doing calculations it does require a bit of attention to detail. But after a bit of practice, you will ﬁnd that its not really so hard. So we hope that readers will invest the effort necessary to master this nice mathematical way of solving physics problems.

1
CHAPTER
A Quick Review of Special Relativity
Fundamentally, our commonsense intuition about how the universe works is tied up in notions about space and time. In 1905, Einstein stunned the physics world with the special theory of relativity, a theory of space and time that challenges many of these closely held commonsense assumptions about how the world works. By accepting that the speed of light in vacuum is the same constant value for all observers, regardless of their state of motion, we are forced to throw away basic ideas about the passage of time and the lengths of rigid objects.
This book is about the general theory of relativity, Einstein’s theory of gravity. Therefore our discussion of special relativity will be a quick overview of concepts needed to understand the general theory. For a detailed discussion of special relativity, please see our list of references and suggested reading at the back of the book.
The theory of special relativity has its origins in a set of paradoxes that were discovered in the study of electromagnetic phenomena during the nineteenth
1
Copyright © 2006 by The McGraw-Hill Companies, Inc. Click here for terms of use.

2

CHAPTER 1 Special Relativity

century. In 1865, a physicist named James Clerk Maxwell published his famous set of results we now call Maxwell’s equations. Through theoretical studies alone, Maxwell discovered that there are electromagnetic waves and that they travel at one speed—the speed of light c. Let’s take a quick detour to get a glimpse into the way this idea came about. We will work in SI units.
In careful experimental studies, during the ﬁrst half of the nineteenth century,
Ampere deduced that a steady current J and the magnetic ﬁeld B were related by

∇ × B = µ0 J

(1.1)

However, this law cannot be strictly correct based on simple mathematical arguments alone. It is a fundamental result of vector calculus that the divergence of any curl vanishes; that is,

∇· ∇×A =0

(1.2)

for any vector A. So it must be true that

∇· ∇×B =0

(1.3)

However, when we apply the divergence operator to the right-hand side, we run into a problem. The problem is that the continuity equation, which is the mathematical description of the conservation of charge, tells us that

∂ρ ∂t

+

∇

·

J

=

0

(1.4)

where ρ is the current density. Therefore, when we apply the divergence operator to the right-hand side of (1.4), we obtain

∇·

µ0 J

=

µ0∇

·

J

=

−µ0

∂ρ ∂t

(1.5)

We can take this process even further. Gauss’s law tells us how to relate the charge density to the electric ﬁeld. In SI units, this law states

∇

·

E

=

1ρ ε0

(1.6)

CHAPTER 1 Special Relativity

3

This allows us to rewrite (1.5) as

∂ρ

∂

∂E

−µ0 ∂t = −µ0ε0 ∂t ∇ · E = −∇ · µ0ε0 ∂t

(1.7)

Putting our results together, we’ve found that

∇·

∇×B

= −∇ ·

µ0

ε0

∂E ∂t

(1.8)

when in fact it must be zero. Considerations like these led Maxwell to “ﬁx up” Ampere’s law. In modern form, we write it as

∇

×

B

=

µ0 J

+

µ0

ε0

∂E ∂t

(1.9)

The extra

term

µ0ε0

∂E ∂t

is

called

the

displacement

current and

its

presence

led

to one of Maxwell’s most dramatic discoveries. Using simple vector calculus,

one can show that the electric and magnetic ﬁelds satisfy the following wave

equations:

∇2E

=

µ0ε0

∂2E ∂t2

and

∇2B

=

µ0ε0

∂2B ∂t2

Now, the wave equation is

∇2 f

=

1 ∂2 f v2 ∂t2

where v is the velocity of the wave. Comparison of these equations shows that electromagnetic waves in vacuum travel at speed

v

=

1 √µ0ε0

= 3 × 108 m/s

=c

where c is nothing more than the speed of light. The key insight to gain from this derivation is that electromagnetic waves
(light) always travel at one and the same speed in vacuum. It doesn’t matter who you are or what your state of motion is, this is the speed you are going to ﬁnd. It took many years for this insight to sink in—and it was Einstein who simply accepted this result at face value.

4

CHAPTER 1 Special Relativity

We can give a quick heuristic insight as to why this result leads to the “paradoxes” of relativity. What is speed anyway (our argument here is qualitative, so we are going to be a bit sloppy here)? It is distance covered per unit time:
v= x t
The commonsense viewpoint, which is formalized mathematically in prerelativistic Newtonian physics, is that distances and times are ﬁxed—thus, how could you possibly have a constant velocity that is the same for all observers? That wouldn’t make any sense. However, the theoretical result that the speed of light in vacuum is the same for all observers is an experimental fact conﬁrmed many times over. If v is the constant speed of light seen by all observers regardless of their state of motion
c= x t
then distances and time intervals must be different for different observers. We will explore this in detail below.
In many treatments of special relativity, you will see a detailed discussion of the Michelson-Morley experiment. In a nutshell, the idea was that waves need a medium to travel through, so physicists at the time made the completely reasonable assumption that there was a medium that ﬁlled all of space, called the luminiferous ether. It was thought that the ether provided the medium necessary for the propagation of electromagnetic waves. The Michelson-Morley experiment was designed to detect the motion of the earth with respect to the ether—but it found nothing. This is one of the most famous “null” results in the history of experimental physics.
This experiment is a crucial result in the history of physics, but the record seems to indicate that Einstein based his derivations on an acceptance of Maxwell’s equations and what they tell us more so than on the MichelsonMorley experiment (in fact Einstein may not have even known very much, if anything, about the experiment at the time of his derivations). As a result, while the experiment is interesting and very important, we are going to skip it and move on to the theoretical framework of special relativity.
Interestingly, other researchers, Lorentz, Fitzgerald, and Poincare, independently derived the Lorentz transformations in order to explain the null results of the Michelson-Morley experiment. The gist of these equations is that clocks slow down and the lengths of rigid bodies contract, making it impossible to construct an experimental apparatus of any kind to detect motion with respect to the ether.

CHAPTER 1 Special Relativity

5

In addition to the arguments we have described here, Einstein used results that came out of Faraday’s law and the studies of electromagnetic induction to come up with ideas about relative motion. We won’t discuss those here, but the interested reader is encouraged to explore the references for details.
The history of the discovery of the special theory of relativity is a good lesson in the way science works—for it demonstrates the crucial interplay between theory and experiment. Careful experiments within the limits of technology available at the time led to Ampere’s law and Faraday’s law. Later, purely mathematical arguments and theoretical considerations were used to show that Ampere’s law was only approximately correct and that electromagnetic waves travel through space at the speed of light. More theoretical considerations were put forward to explain how those waves traveled through space, and then the dramatic experimental result found by Michelson and Morley forced those ideas to be thrown away. Then Einstein came along and once again used mostly theoretical arguments to derive special relativity. The bottom line is this: Physics is a science that depends on two legs—theory and experiment—and it cannot stand on either alone.
We now turn to a quick tour of the basics of special relativity, and begin with some deﬁnitions.
Frame of Reference
A frame of reference is more or less a fancy way of saying coordinate system. In our thought experiments, however, we do more than think in mathematical terms and would like to imagine a way that a frame of reference could really be constructed. This is done by physically constructing a coordinate system from measuring rods and clocks. Local clocks are positioned everywhere within the frame and can be used to read off the time of an event that occurs at that location. You might imagine that you have 1-m long measuring rods joined together to form a lattice, and that there is a clock positioned at each point where rods are joined together.
Clock Synchronization
One problem that arises in such a construction is that it is necessary to synchronize clocks that are physically separated in space. We can perform the synchronization using light rays. We can illustrate this graphically with a simple spacetime diagram (more on these below) where we represent space on the horizontal axis and time on the vertical axis. This allows us to plot the motion of

6

CHAPTER 1 Special Relativity

6

5

reflected light ray

returns to clock 1 at

4

time t2

3

2

1

reflected at clock 2, at time t'

1

2

Time of emission, t1

emitted light ray

Fig. 1-1. Clock synchronization. At time t1¯, a light beam is emitted from a clock at the origin. At time t , it reaches the position of clock 2 and is reﬂected back. At time t2¯, the reﬂected beam reaches the position of clock 1. If t is halfway between the times t1¯ and
t2¯, then the two clocks are synchronized.

objects in space and time (we are of course only considering one-dimensional motion). Imagine two clocks, clock 1 located at the origin and clock 2 located at some position we label x1 (Fig. 1-1).
To see if these clocks are synchronized, at time t1 we send a beam of light from clock 1 to clock 2. The beam of light is reﬂected back from clock 2 to clock 1 at time t2, and the reﬂected beam arrives at the location of clock 1 at time t1. If we ﬁnd that

t

=

1 2

(t1

+

t2)

then the clocks are synchronized. This process is illustrated in Fig. 1-1. As we’ll see later, light rays travel on the straight lines x = t in a spacetime diagram.

Inertial Frames
An inertial frame is a frame of reference that is moving at constant velocity. In an inertial frame, Newton’s ﬁrst law holds. In case you’ve forgotten, Newton’s ﬁrst

CHAPTER 1 Special Relativity

7

Fy

F' y'

x x'

z

z'

Fig. 1-2. Two frames in standard conﬁguration. The primed frame (F ) moves at velocity v relative to the unprimed frame F along the x-axis. In prerelativity physics,
time ﬂows at the same rate for all observers.

law states that a body at rest or in uniform motion will remain at rest or in uniform motion unless acted upon by a force. Any other frame that is moving uniformly (with constant velocity) with respect to an inertial frame is also an inertial frame.

Galilean Transformations

The study of relativity involves the study of how various physical phenomena appear to different observers. In prerelativity physics, this type of analysis is accomplished using a Galilean transformation. This is a simple mathematical approach that provides a transformation from one inertial frame to another. To study how the laws of physics look to observers in relative motion, we imagine two inertial frames, which we designate F and F . We assume that they are in the standard conﬁguration. By this we mean the frame F is moving in the x direction at constant velocity v relative to frame F. The y and z axes are the same for both observers (see Fig. 1-2). Moreover, in prerelativity physics, there is uniform passage of time throughout the universe for everyone everywhere. Therefore, we use the same time coordinate for observers in both frames.
The Galilean transformations are very simple algebraic formulas that tell us how to connect measurements in both frames. These are given by

t = t , x = x + vt, y = y , z = z

(1.10)

Events
An event is anything that can happen in spacetime. It could be two particles colliding, the emission of a ﬂash of light, a particle just passing by, or just anything else that can be imagined. We characterize each event by its spatial location and the time at which it occurrs. Idealistically, events happen at a single

8

CHAPTER 1 Special Relativity

mathematical point. That is, we assign to each event E a set of four coordinates (t, x, y, z).

The Interval
The spacetime interval gives the distance between two events in space and time. It is a generalization of the pythagorean theorem. You may recall that the distance between two points in cartesian coordinates is

d = (x1 − x2)2 + (y1 − y2)2 + (z1 − z2)2 = ( x)2 + ( y)2 + ( z)2

The interval generalizes this notion to the arena of special relativity, where we must consider distances in time together with distances in space. Consider an event that occurs at E1 = (ct1, x1, y1, z1) and a second event at E2 = (ct2, x2, y2, z2). The spacetime interval, which we denote by ( S)2, is given by
( S)2 = c2 (t1 − t2)2 − (x1 − x2)2 − (y1 − y2)2 − (z1 − z2)2

or more concisely by

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

(1.11)

An interval can be designated timelike, spacelike, or null if ( S)2 > 0, ( S)2 < 0, or ( S)2 = 0, respectively. If the distance between two events is inﬁnitesimal, i.e., x1 = x, x2 = x + dx ⇒ x = x + dx − x = dx, etc., then the interval is given by

ds2 = c2 dt2 − dx2 − dy2 − dz2

(1.12)

The proper time, which is the time measured by an observer’s own clock, is deﬁned to be

dτ 2 = −ds2 = −c2 dt2 + dx2 + dy2 + dz2

(1.13)

This is all confusing enough, but to make matters worse different physicists use different sign conventions. Some write ds2 = −c2 dt2 + dx2 + dy2 + dz2, and in that case the sign designations for timelike and spacelike are reversed. Once you get familiar with this it is not such a big deal, just keep track of what the author is using to solve a particular problem.

CHAPTER 1 Special Relativity

9

The interval is important because it is an invariant quantity. The meaning of this is as follows: While observers in motion with respect to each other will assign different values to space and time differences, they all agree on the value of the interval.

Postulates of Special Relativity

In a nutshell, special relativity is based on three simple postulates.

Postulate 1: The principle of relativity. The laws of physics are the same in all inertial reference frames.

Postulate 2: The speed of light is invariant. All observers in inertial frames will measure the same speed of light, regardless of their state of motion.

Postulate 3: Uniform motion is invariant. A particle at rest or with constant velocity in one inertial frame will be at rest or have constant velocity in all inertial frames.

We now use these postulates to seek a replacement of the Galilean transfor-

mations with the caveat that the speed of light is invariant. Again, we consider two frames Fand F in the standard conﬁguration (Fig. 1-2). The ﬁrst step is to consider Postulate 3. Uniform motion is represented by straight lines, and what this postulate tells us is that straight lines in one frame should map into straight lines in another frame that is moving uniformly with respect to it. This is another

way of saying that the transformation of coordinates must be linear. A linear transformation can be described using matrices. If we write the coordinates of frame F as a column vector

 ct 



x y



z

then the coordinates of F are related to those of F via a relationship of the form

 ct   ct 



x y



=

L



x y



z

z

(1.14)

10

CHAPTER 1 Special Relativity

where L is a 4 × 4 matrix. Given that the two frames are in standard conﬁguration, the y and z axes are coincident, which means that

y = y and z = z

To get the form of the transformation, we rely on the invariance of the speed of light as described in postulate 2. Imagine that at time t = 0 a ﬂash of light is emitted from the origin. The light moves outward from the origin as a spherical wavefront described by

c2t2 = x2 + y2 + z2

(1.15)

Subtracting the spatial components from both sides, this becomes c2t2 − x2 − y2 − z2 = 0

Invariance of the speed of light means that for an observer in a frame F moving at speed v with respect to F, the ﬂash of light is described as
c2t 2 − x 2 − y 2 − z 2 = 0

These are equal, and so c2t2 − x2 − y2 − z2 = c2t 2 − x 2 − y 2 − z 2

Since y = y and z = z, we can write c2t2 − x2 = c2t 2 − x 2

(1.16)

Now we use the fact that the transformation is linear while leaving y and z unchanged. The linearity of the transformation means it must have the form

x = Ax + Bct ct = Cx + Dct

(1.17)

We can implement this with the following matrix [see (1.14)]:

D C 0 0

L

=



B 0

A 0

0 1

0 0



0 0 01

CHAPTER 1 Special Relativity

11

Using (1.17), we rewrite the right side of (1.16) as follows:

x 2 = ( Ax + Bct)2 = A2x2 + 2 ABct x + B2c2t2 c2t 2 = (C x + Dct)2 = C2x2 + 2CDct x + D2c2t2
⇒ c2t 2 − x 2 = C2x2 + 2CDct x + D2c2t2 − A2x2 − 2 ABct x − B2c2t2 = c2 D2 − B2 t2 − A2 − C2 x2 + 2 (CD − AB) ct x

This must be equal to the left side of (1.16). Comparison leads us to conclude that

CD − AB = 0 ⇒ CD = AB
D2 − B2 = 1 A2 − C2 = 1

To obtain a solution, we recall that cosh2 φ − sinh2 φ = 1. Therefore we make the following identiﬁcation:

A = D = cosh φ

(1.18)

In some sense we would like to think of this transformation as a rotation. A rotation leads to a transformation of the form
x = x cos φ − y sin φ y = −x sin φ + y cos φ

In order that (1.17) have a similar form, we take

B = C = − sinh φ

(1.19)

With A, B, C, and D determined, the transformation matrix is

 cos hφ − sin hφ 0 0 

L

=

−

sin hφ 0

cos hφ 0

0 1

0 0



0

0 01

(1.20)

Now we solve for the parameter φ, which is called the rapidity. To ﬁnd a solution, we note that when the origins of the two frames are coincident;

12

CHAPTER 1 Special Relativity

that is, when x = 0, we have x = vt. Using this condition together with (1.17), (1.18), and (1.19), we obtain

x = 0 = x cosh φ − ct sinh φ = vt cosh φ − ct sinh φ = t (v cosh φ − c sinh φ)

and so we have v cosh φ − c sinh φ = 0, which means that

v cosh φ = c sinh φ

⇒

sinh φ cosh φ

=

tanh φ

=

v c

(1.21)

This result can be used to put the Lorentz transformations into the form shown in elementary textbooks. We have

x = cosh φx − sinh φct ct = − sinh φx + cosh φct

Looking at the transformation equation for t ﬁrst, we have

lct = − sinh φx + cosh φct = cosh φ
= cosh φ (− tanh φx + ct) = cosh φ ct − v x
c v = c cosh φ t − c2 x

− sinh φ cosh φ

x

+

ct

We also ﬁnd

v ⇒ t = cosh φ t − x
c2
x = cosh φx − sinh φct = cosh φ (x − tanh φct) = cosh φ (x − vt)

CHAPTER 1 Special Relativity

13

Now let’s do a little trick using the hyperbolic cosine function, using

cosh φ = cosh φ = co√sh φ =

cosh φ

1

1

cosh2 φ − sinh2 φ

=

(1 /

1 cosh

φ)

1 cosh2 φ − sinh2 φ

=

1

1/ cosh2 φ cosh2 φ − sinh2 φ

=

1

=

1

1 − tanh2 φ

1 − v2 /c2

This is none other than the deﬁnition used in elementary textbooks:

γ=

1

= cosh φ

1 − v2/c2

(1.22)

And so we can write the transformations in the familiar form:

t = γ t − v x/c2 , x = γ (x − vt) , y = y, z = z (1.23)

It is often common to see the notation β = v/c.

Three Basic Physical Implications
There are three physical consequences that emerge immediately from the Lorentz transformations. These are time dilation, length contraction, and a new rule for composition of velocities.

TIME DILATION
Imagine that two frames are in the standard conﬁguration so that frame F moves at uniform velocity v with respect to frame F. An interval of time t

14

CHAPTER 1 Special Relativity

as measured by an observer in F is seen by F to be

t= 1

t =γ t

1 − β2

that is, the clock of an observer whose frame is F runs slow relative to the clock of an observer whose frame is F by a factor of 1 − β2.

LENGTH CONTRACTION
We again consider two frames in the standard conﬁguration. At ﬁxed time t, measured distances along the direction of motion are related by

x= 1

x

1 − β2

that is, distances in F along the direction of motion appear to be shortened in the direction of motion by a factor of 1 − β2.

COMPOSITION OF VELOCITIES

Now imagine three frames of reference in the standard conﬁguration. Frame
F moves with velocity v1 with respect to frame F, and frame F moves with velocity v2 with respect to frame F . Newtonian physics tells us that frame F moves with velocity v3 = v1 + v2 with respect to frame F, a simple velocity addition law. However, if the velocities are signiﬁcant fraction of the speed
of light, this relation does not hold. To obtain the correct relation, we simply
compose two Lorentz transformations.

EXAMPLE 1-1 Derive the relativistic velocity composition law.

SOLUTION 1-1 Using β = v/c, the matrix representation of a Lorentz transformation between

F and F is

√1

L1

=



1−β12
√−β1
1−β12
0

√−β1
1−β12
√1
1−β12
0



00

0 1

0 0



(1.24)

0

0 01

CHAPTER 1 Special Relativity

15

The transformation between F and F is



√1

L2

=



1−β22
√−β2
1−β22
0

0

√−β2
1−β22
√1
1−β22
0
0



00

0 1

0 0



01

(1.25)

We can obtain the Lorentz transformation between F and F by computing L2 L1 using (1.25) and (1.24). We ﬁnd

√1



1−β12
√−β1
1−β12
0

0

√−β1
1−β12
√1
1−β12
0 0



0 0 √1

0 1

0 0





1−β22
√−β2
1−β22
0

01

0

√−β2
1−β22
√1
1−β22
0
0



00

0 1

0 0



01

 √

1+β1β2

=



(1−β12)(1−β22) √ −(β1+β2)
(1−β12)(1−β22) 0

0

√ −(β1+β2) (1−β12)(1−β22)
√ 1+β1β2 (1−β12)(1−β22) 0
0



00

0 1

0 0



01

This matrix is itself a Lorentz transformation, and so must have the form

√1

L3

=



1−β32
√−β3
1−β32
0

0

√−β3
1−β32
√1
1−β32
0
0

 00

0 1

0 0



01

We can ﬁnd β3 by equating terms. We need to consider only one term, so pick the terms in the upper left corner of each matrix and set

1 + β1β2

=

1 − β12 1 − β22

1 1 − β32

16

CHAPTER 1 Special Relativity

Let’s square both sides:

(1 + β1β2)2 1 − β12 1 − β22

=

1

1 − β32

Inverting this, we get

1 − β12 1 − β22 (1 + β1β2)2

= 1 − β32

Now we isolate the desired term β3:

β32 = 1 −

1 − β12 1 − β22 (1 + β1β2)2

On

the

right-hand

side,

we

set

1

=

, (1+β1β2)2
(1+β1β2)2

and

rewrite

the

expression

as

β32

=

(1 (1

+ +

β1β2)2 β1β2)2

−

1 − β12 1 − β22 (1 + β1β2)2

=

(1

+

β1β2)2 − (1 +

1 − β12 β1β2)2

1 − β22

Now we expand the terms on the right to get

β32

=

(1

+

β1β2)2 − (1 +

1 − β12 β1β2)2

1 − β22

=

1

+

2β1β2

+

β12β22 (1

− 1 − β12 + β1β2)2

−

β22

+

β12β22

This simpliﬁes to

β32

=

2β1β2 + β12 + β22 (1 + β1β2)2

=

(β1 + B2)2 (1 + β1β2)2

Taking square roots of both sides, we obtain

β3

=

β1 + B2 1 + β1β2

CHAPTER 1 Special Relativity

17

Now we use β = v/c to write

v3 c

=

v1/c + v2/c 1 + v1v2/c2

Multiplication of both sides by c gives the velocity composition law

v3

=

1

v1 + v2 + v1v2/c2

(1.26)

Light Cones and Spacetime Diagrams
It is often helpful to visualize spacetime by considering a ﬂash of light emitted at the origin. As we discussed earlier, such a ﬂash of light is described by a spherical wavefront. However, our minds cannot visualize four dimensions and it’s not possible to draw it on paper. So we do the next best thing and suppress one or more of the spatial dimensions. Let’s start with the simplest of all cases, where we suppress two spatial dimensions.
Doing so gives us a simple spacetime diagram (see Fig. 1- for the basic idea). In a spacetime diagram, the vertical axis represents time while one or two horizontal axes represent space. It is convenient to work in units where c = 1. The upper half plane where t > 0 represents events to the future of the origin. Past events are found in the lower half plane where t < 0. The motion of light in
t Future

x
Light moves on lines t = x
Past
Fig. 1-3. The division of spacetime into future and past regions. Light rays move on the lines t = x and t = −x. These lines deﬁne the light cone while the origin is some event E in spacetime. The inside of the light cone in the lower half plane is the past of E, where
we ﬁnd all events in the past that could affect E. The future of E, which contains all events that can be causally affected by E, is inside the light cone deﬁned in the upper
half plane. Regions outside the light cone are called spacelike.

18

CHAPTER 1 Special Relativity

World line of a

t

stationary

particle

Past (t < 0)

Future (t > 0)
x Light moves on lines t = x

Fig. 1-4. The worldline of a stationary particle is a straight line.
such a diagram is then described by lines that make a 45◦angle with the x-axis, i.e., lines that satisfy
t2 = x2
In the ﬁrst quadrant, the paths of light rays are described by the lines t = x. The motion of a particle through spacetime as depicted in a spacetime diagram
(Fig. 1-4) is called a worldline. The simplest of all particle motion is a particle just sitting somewhere. To indicate the worldline of a stationary particle on a spacetime diagram, we simply draw a straight vertical line that passes through the x-axis at the spatial location of the particle. This makes sense because the particle is located at some position x that does not change, but time keeps marching forward.

t

timelike

spacelike x

Fig. 1-5. A light cone with two spatial dimensions.

CHAPTER 1 Special Relativity

19

Special relativity tells us that a particle cannot move faster than the speed of light. On a spacetime diagram, this is indicated by the fact that particle motion is restricted to occur only inside the light cone. The region inside the light cone is called timelike. Regions outside the light cone, which are casually unrelated to the event E, are called spacelike.
More insight is gained when we draw a spacetime diagram showing two spatial dimensions. This is shown in Fig. 1-5.

Four Vectors

In special relativity we work with a uniﬁed entity called spacetime rather than viewing space as an arena with time ﬂowing in the background. As a result, a vector is going to have a time component in addition to the spatial components we are used to. This is called a four vector. There are a few four vectors that are important in relativity. The ﬁrst is the four velocity, which is denoted by u and has components

u=

dt dτ

,

dx dτ

,

dy dτ

,

dz dτ

We can differentiate this expression again with respect to proper time to obtain the four acceleration a. The norm or magnitude squared of v · v tells us if a vector
is timelike, spacelike, or null. The deﬁnition will depend on the sign convention used for the line element. If we take ds2 = c2dt2 − dx2 − dy2 − dz2, then if v · v > 0 we say that v is timelike. If v · v < 0, we say v is spacelike. When v · v = 0, we say that v is null. The four velocity is always a timelike vector.
Following this convention, we compute the dot product as

v · v = (vt )2 − (vx )2 − vy 2 − (vz)2

The dot product is an invariant, and so has the same value in all Lorentz frames.
If a particle is moving in some frame F with velocity u, we ﬁnd that energy and momentum conservation can be achieved if we take the energy to be E = γ m0c2 and deﬁne the four momentum p using

p = γ m0u

(1.27)

where m0 is the particle’s rest mass and u is the velocity four vector. In a more familiar form, the momentum four vector is given by p = E/c, px , py, pz .

20

CHAPTER 1 Special Relativity

For two frames in the standard conﬁguration, components of the momentum four vector transform as

px = γ px − βγ E/c py = py pz = pz E = γ E − βγ cpx

(1.28)

Using our sign convention for the dot product, we ﬁnd

p · p = E 2/c2 − px2 − p2y − pz2 = E 2/c2 − p2

Remember, the dot product is a Lorentz invariant. So we can ﬁnd its value
by calculating it in any frame we choose. In the rest frame of the particle, the momentum is zero (and so p2 = 0) and the energy is given by Einstein’s famous formula E = m0c2. Therefore

p · p = m20c2

Putting these results together, we obtain

E 2 − p2c2 = m20c4

(1.29)

Relativistic Mass and Energy

The rest mass of a particle is the mass of the particle as measured in the instantaneous rest frame of that particle. We designate the rest mass by m0. If a particle is moving with respect to an observer O with velocity v then O measures the mass of the particle as

m=

m0 1 − v2/c2

=

γ m0

(1.30)

Now consider the binomial expansion, which is valid for |x| < 1,

(1 + x)n ≈ 1 + nx

CHAPTER 1 Special Relativity

21

For n = −1/2,

(1

−

x )−1/2

≈

1

+

1 x

2

Setting x = v2/c2 in (1.30), we obtain

m=

m0 1 − v2/c2

≈

m0

1 v2 1+
2 c2

1 v2 = m0 + 2m0 c2

Multiplying through by c2 we obtain an expression that relates the relativistic energy to the rest mass energy plus the Newtonian kinetic energy of the particle:

mc2

=

m0c2

+

1 2

m

0v

2

1. An inertial frame is best described by (a) one that moves with constant acceleration (b) a frame that is subject to constant forces (c) a frame that moves with constant velocity (d) a frame that is subject to galilean transformations

2. The proper time dτ 2 is related to the interval via

(a) dτ 2 = −ds2

(b) dτ 2 = ds2

(c) dτ 2 = −c2 ds2

(d)

dτ 2=−ds2 c2

3. The principle of relativity can be best stated as

(a) The laws of physics differ only by a constant in all reference frames

differing by a constant acceleration.

(b) The laws of physics change from one inertial reference frames to

another.

(c) The laws of physics are the same in all inertial reference frames.

4. Rapidity is deﬁned using which of the following relationships?

(a) (b) (c)

tanh φ =

tan φ

=

v c

v c

tanh φ = −

v c

(d) v tanh φ = c

Quiz

22

CHAPTER 1 Special Relativity

5. Consider two frames in standard conﬁguration. The phenomenon of length contraction can be described by saying that distances are shortened by a factor of

(a) 1 + β2

(b) 1 − β2

(c) −

1 + β2 c2

2
CHAPTER
Vectors, One Forms, and the Metric
In this chapter we describe some of the basic objects that we will encounter in our study of relativity. While you are no doubt already familiar with vectors from studies of basic physics or calculus, we are going to be dealing with vectors in a slightly different light. We will also encounter some mysterious objects called one forms, which themselves form a vector space. Finally, we will learn how a geometry is described by the metric.
Vectors
A vector is a quantity that has both magnitude and direction. Graphically, a vector is drawn as a directed line segment with an arrow head. The length of the arrow is a graphic representation of its magnitude. (See Figure 2-1).
23
Copyright © 2006 by The McGraw-Hill Companies, Inc. Click here for terms of use.

24

CHAPTER 2 Vectors, One Forms, Metric

y

→v

x
Fig. 2-1. Your basic vector, a directed line segment drawn in the x–y plane.
The reader is no doubt familiar with the graphical methods of vector addition, scalar multiplication, and vector subtraction. We will not review these methods here because we will be looking at vectors in a more abstract kind of way. For our purposes, it is more convenient to examine vectors in terms of their components. In the plane or in ordinary three-dimensional space, the components of a vector are the projections of the vector onto the coordinate axes. In Fig. 2-2, we show a vector in the x–y plane and its projections onto the x and y axes.
The components of a vector are numbers. They can be arranged as a list. For example, in 3 dimensions, the components of a vector A can be written as A = Ax , Ay, Az . More often, one sees a vector written as an expansion in terms of a set of basis vectors. A basis vector has unit length and points along the direction of a coordinate axis. Elementary physics books typically denote the basis for cartesian coordinates by (iˆ, jˆ, kˆ), and so in ordinary three-dimensional
y

Wy

W

x Wx
Fig. 2-2. A vector W in the x−y plane, resolved into its components Wx and Wy. These are the projections of W onto the x and y axes.

CHAPTER 2 Vectors, One Forms, Metric

25

cartesian space, we can write the vector A as
A = Ax iˆ + Ay jˆ + Azkˆ
In more advanced texts a different notation is used:
A = Ax xˆ + Ay yˆ + Az zˆ
This has some advantages. First of all, it clearly indicates which basis vector points along which direction (the use of (iˆ, jˆ, kˆ) may be somewhat mysterious to some readers). Furthermore, it provides a nice notation that allows us to deﬁne a vector in a different coordinate system. After all, we could write the same vector in spherical coordinates:
A = Arrˆ + Aθ θˆ + Aφφˆ
There are two important things to note here. The ﬁrst is that the vector A is a geometric object that exists independent of coordinate system. To get its components we have to choose a coordinate system that we want to use to represent the vector. Second, the numbers that represent the vector in a given coordinate system, the components of the vector, are in general different depending on what coordinate system we use to represent the vector. So for the example we have been using so far Ax , Ay, Az = Ar , Aθ , Aφ .
New Notation
We are now going to use a different notation that will turn out to be a bit more convenient for calculation. First, we will identify the coordinates by a set of labeled indices. The letter x is going to be used to represent all coordinates, but we will write it with a superscript to indicate which particular coordinate we are referring to. For example, we will write y as x2. It is important to recognize that the “2” used here is just a label and is not an exponent. In other words,
y2 = x2 2
and so on. For the entire set of cartesian coordinates, we make the following identiﬁcation:
(x, y, z) → x1, x2, x3

26

CHAPTER 2 Vectors, One Forms, Metric

With this identiﬁcation, when we write x1 we mean x, while x2 means y and so forth. This identiﬁcation is entirely general, and we can use it to represent another coordinate system such as cylindrical coordinates. What represents what will be made clear from the context. Also, it will be convenient to move back and forth between this representation and the ones you are used to.
When using more than one coordinate system, or more importantly when considering transformations between coordinate system, we need a way to apply this representation to two different sets of coordinates. One way to do so is to put primes on the indices. As an example, suppose we are considering cartesian and spherical coordinates within the context of the same problem. If we label cartesian coordinates with (x, y, z) → x1, x2, x3 , then we add primes to the labels used for spherical coordinates and write
(r, θ, φ) → x1 , x2 , x3
We will label the components of a vector in the same way, with a raise index. In the current example, the components of a vector A in cartesian coordinates would be given by
A = A1, A2, A3
While primed coordinates would represent the same vector in spherical coordinates
A = A1 , A2 , A3
another useful notation change is to write partial derivatives in a succinct way. We write
∂f ∂x = ∂x f
or, using indices, we write
∂ ∂ xa → ∂a
There are two reasons for writing things in this apparently obscure fashion. The ﬁrst is that when you do relativity a great deal of writing is required. Hey! Anything that can cut back on that is a good thing. But we will see that the

CHAPTER 2 Vectors, One Forms, Metric

27

placement of the indices (as shown in ∂a) will prove to be more enlightening and useful. Unfortunately, at this point we’re not ready to say why, so you’ll just have to take my word for it and just keep in mind what the shorthand symbols mean.

Four Vectors

In many cases, we are not going to work with speciﬁc components of an objectlike vector, but will rather work with a general component. In this book we will label components of objectlike vectors with lowercase Latin letters. For example, we can refer to the vector A by
Aa

As we get involved with relativity, a vector will have space and time components (it will be a four vector). In that case, the time component of the vector will be labeled by the index 0, and so the components of a four vector V will be given by
V = V 0, V 1, V 2, V 3

Vector addition can be described in the following way: A + B = A0 + B0, A1 + B1, A2 + B2, A3 + B3

while scalar multiplication can be written as α A = α A0, α A1, α A2, α A3

Keep in mind that some authors prefer to use (1, 2, 3, 4) as their indices, using 4 to label time. We will stick to using 0 to label the time coordinate, however.
Many authors prefer to use the following labeling convention. When all four components (space and time) in an expression are used, Greek letters are used for indices. So if an author writes

T

µ ν

γ

the indices µ, ν, γ can range over (0, 1, 2, 3). In this context, Latin indices are reserved for spatial components only, and so in the expression

Sij

28

CHAPTER 2 Vectors, One Forms, Metric

the indices i, j will range only over (1, 2, 3). Typing a lot of Greek symbols is a bit of extra work, so we will stick to using Latin indices all the time. When possible, we will use early letters (a, b, c, . . .) to range over all possible values (0, 1, 2, 3) and use letters from the middle of the alphabet such as i, j to range only over the spatial components (1, 2, 3).
The Einstein Summation Convention
The Einstein summation convention is a way to write sums in a shorthand format. When the same index appears twice in an expression, once raised and once lowered, a sum is implied. As a speciﬁc example,
3
Ai Bi → Ai Bi = A1 B1 + A2 B2 + A3 B3
i =1
Another example is that Sa Tab is shorthand for the expression
3
Sa Tab
a=0
An index that is summed over is called a dummy index, and can be replaced by another label if it is convenient. For example,
Sa Tab = S cTcb
The index b in the previous expressions is not involved in the sum operations. Such an index is known as a free index. A free index will typically appear on both sides of an expression. For example, consider the following equation:
Aa = a b Ab
In this expression, b is once again a dummy index. The sum implied here means that
Aa = a b Ab = a 0 A0 + a 1 A1 + a 2 A2 + a 3 A3
The other index found in this expression, a , is a free index. If we elect to change the free index, it must be changed on both sides of the equation. Therefore it

CHAPTER 2 Vectors, One Forms, Metric

29

would be valid to make the change a → b , provided that we make this change on both sides; i.e.,
Ab = b b Ab
Tangent Vectors, One Forms, and the Coordinate Basis
We will often label basis vectors with the notation ea. Using the Einstein summation convention, a vector V can be written in terms of some basis as
V = V aea
In this context the notation ea makes sense, because we can use it in the summation convention (this would not be possible with the cumbersome (iˆ, jˆ, kˆ) for example).
In a given coordinate system, the basis vectors ea are tangent to the coordinate lines. (See Fig. 2-3 and Fig. 2-4) This is the reason why we can write basis vectors as partial derivatives in a particular coordinate direction (for an explanation, see Carroll, 2004). In other words, we take
∂ ea = ∂a = ∂ xa
This type of basis is called a coordinate basis. This allows us to think of a vector as an operator, one that maps a function into a new function that is related to its derivative. In particular,
V f = (V aea) = V a∂a f
A vector V can be represented with covariant components Va. This type of vector is called a one form. Basis one forms have raised indices and are often denoted by ωa. So we can write
V˜ = Vaωa
We have used a tilde to note that this is a one form and not an ordinary vector (but it is the same object, in a different representation). Later, we will see how to move between the two representations by raising and lowering indices with the

30

CHAPTER 2 Vectors, One Forms, Metric

→
V

Fig. 2-3. A tangent vector to a curve.
metric. The basis one forms form a dual vector space to ordinary vectors, in the sense that the one forms constitute a vector space in their own right and the basis one forms map basis vectors to a number, the Kronecker delta function; i.e.,

where

ωa (eb) = δba

δba =

1 0

a=b otherwise

(2.1)

In a coordinate representation, the basis one forms are given by

ωa = dxa

(2.2)

With this representation, it is easy to see why (2.1) holds. An arbitrary one form σa maps a vector V a to a number via the scalar product
σ · V = σa V a
We can think of this either way: we can visualize vectors as maps that take one forms to the real numbers via the scalar product. More generally, we can deﬁne a ( p, q) tensor as a function that takes p one forms and q vectors as input and

Tp

Fig. 2-4. An admittedly crude representation.The blob represents a manifold (basically a space of points). Tp is the tangent space at a point p. The tangent vectors live here.

CHAPTER 2 Vectors, One Forms, Metric

31

maps them to the real numbers. We can write a general tensor in the following way:
T = Tabc···lmn···ωa ⊗ ωb ⊗ ωc · · · el ⊗ em ⊗ en · · ·
We will have more to say about one forms, basis vectors, and tensors later.

Coordinate Transformations

In relativity it is often necessary to change from one coordinate system to

another, or from one frame to another. A transformation of this kind is imple-

mented with a transformation matrix that we denote by

a b

.

The

placement

of the indices and where we put the prime notation will depend on the particu-

lar transformation. In a coordinate transformation, the components of

a b

are

formed by taking the partial derivative of one coordinate with respect to the

other. More speciﬁcally,

a b

=

∂xa ∂xb

(2.3)

The easiest way to get a handle on how to apply this is to simply write the formulas down and apply them in a few cases. Basis vectors transform as

ea =

b a

eb

(2.4)

We can do this because, as you know, it is possible to expand any vector in terms of some basis. What this relation gives us is an expansion of the basis vector ea in terms of the old basis eb. The components of ea in the eb basis are given by ab . Note that we are denoting the new coordinates by primes and the old coordinates are unprimed indices.
EXAMPLE 2-1 Plane polar coordinates are related to cartesian coordinates by

x = r cos θ and y = r sin θ

Describe the transformation matrix that maps cartesian coordinates to polar coordinates, and write down the polar coordinate basis vectors in terms of the basis vectors of cartesian coordinates.

32

CHAPTER 2 Vectors, One Forms, Metric

SOLUTION 2-1

Using

a b

=

∂xa ∂xb

,

the

components

of

the

transformation

matrix

are

x r

=

∂x ∂r

=

cos θ

and

x θ

=

∂x ∂θ

= −r sin θ

and

y r

=

∂y ∂r

=

sin θ

y θ

=

∂y ∂θ

=

r

cos θ

Using (2.4), we can write down the basis vectors in polar coordinates. We obtain

er = x r ex + yr ey = cos θ ex + sin θ ey eθ = x θ ex + yθ ey = −r sin θ ex + r cos θ ey

The components of a vector transform in the opposite manner to that of a basis vector (this is why, an ordinary vector is sometimes called contravariant; it transforms contrary to the basis vectors). This isn’t so surprising given the placement of the indices. In particular,

Va =

a

bV b

=

∂xa ∂xb

Vb

(2.5)

The components of a one form transform as

σa =

b a

σb

(2.6)

Basis one forms transform as

ωa = dxa = a bdxb

(2.7)

To ﬁnd the way an arbitrary tensor transforms, you just use the basic rules for vectors and one forms to transform each index (OK we aren’t transforming the indices, but you get the drift). Basically, you add an appropriate for each index. For example, the metric tensor, which we cover in the next section, transforms as

ga b =

c a

d b

gcd

The Metric
At the most fundamental level, one could say that geometry is described by the pythagorean theorem, which gives the distance between two points (see

CHAPTER 2 Vectors, One Forms, Metric

33

c b
a
Fig. 2-5. The pythagorean theorem tells us that the lengths of a, b, c are related by c = a2 + b2.
Fig. 2-5). If we call P1 = (x1, y1) and P2 = (x2, y2), then the distance d is given by
d = (x1 − x2)2 + (y1 − y2)2
Graphically, of course, the pythagorean theorem gives the length of one side of a triangle in terms of the other two sides, as shown in Fig. 2-3.
As we have seen, this notion can be readily generalized to the ﬂat spacetime of special relativity, where we must consider differences between spacetime events. If we label two events by (t1, x1, y1, z1) and (t2, x2, y2, z2), then we deﬁne the spacetime interval between the two events to be
( s)2 = (t1 − t2)2 − (x1 − x2)2 − (y1 − y2)2 − (z1 − z2)2
Now imagine that the distance between the two events is inﬁnitesimal. That is, if the ﬁrst event is simply given by the coordinates (t, x, y, z), then the second event is given by (t + dt, x + dx, y + dy, z + dz). In this case, it is clear that the differences between each term will give us (dt, dx, dy, dz). We write this inﬁnitesimal interval as
ds2 = dt2 − dx2 − dy2 − dz2
As we shall see, the form that the spacetime interval takes, which describes the geometry, is closely related to the gravitational ﬁeld. Therefore it’s going to become quite important to familiarize ourselves with the metric. In short, the interval ds2, which often goes by the name the metric, contains information about how the given space (or spacetime) deviates from a ﬂat space (or spacetime).
You are already somewhat familiar with the notion of a metric if you’ve studied calculus. In that kind of context, the quantity ds2 is often called a line element. Let’s quickly review some familiar line elements. The most familiar is

34

CHAPTER 2 Vectors, One Forms, Metric

that of ordinary cartesian coordinates. That one is given by

ds2 = dx2 + d y2 + dz2

(2.8)

For spherical coordinates, we have

ds2 = dr 2 + r 2dθ 2 + r 2 sin θ dφ2

(2.9)

Meanwhile, the line element for cylindrical coordinates is ds2 = dr 2 + r 2dφ2 + dz2

(2.10)

We can write these and other line elements in a succinct way by writing the coordinates with indices and summing. Generally, the line element is written as

ds2 = gab (x) dxadxb

(2.11)

where gab (x) are the components of a second rank tensor (note that we can write these components as a matrix) called the metric. You can remember what this

thing is by recalling that the components of the metric are given by the coefﬁ-

cient functions that multiply the differentials in the line element. For a metric

describing ordinary three-dimensional space, these components are arranged

into a matrix as follows:





g11 g12 g13 gij =  g21 g22 g23 

g31 g32 g33

For example, looking at (2.8), we see that for cartesian coordinates we can write





100

gij =  0 1 0 

001

When dealing with spacetime, we take the convention that the time coordinate is labeled by x0 and write the matrix representation of the metric as

 g00 g01 g02 g03 

gab

=



g10 g20

g11 g21

g12 g22

g13 g23



g30 g31 g32 g33

CHAPTER 2 Vectors, One Forms, Metric

35

For spherical coordinates, we make the identiﬁcation (r, θ, φ) → x1, x2, x3 and using (2.9) write





10 0

gij =  0 r 2

0

0 0 r 2 sin2 θ

(2.12)

For cylindrical coordinates, the matrix takes the form





100

gij =  0 r 2 0 

001

(2.13)

In many cases, like the ones we have considered so far, the metric has components only along the diagonal. However, be aware that this is not always the case. For example, a metric can arise in the study of gravitational radiation that is called the Bondi metric. The coordinates used are (u, r, θ, φ) and the line element can be written as

ds2 = f e2β − g2r 2e2α du2 + 2e2β dudr + 2gr 2e2αdudθ r
−r 2 e2αdθ 2 + e−2α sin2 θ dφ2

(2.14)

Here f, g, α, β are functions of the coordinates (u, r, θ, φ). With these coordinates, we can write the matrix representation of the metric as

 guu gur guθ guφ 

gab

=



gr u gθ u

gr r gθr

gr θ gθ θ

gr φ gθ φ



gφu gφr gφθ gφφ

A good piece of information to keep in the back of your mind is that the metric is symmetric; i.e., gab = gba. This information is useful when writing down components of the metric associated with the mixed terms in the line
element. For example, in this case we have

2e2β dudr = e2βdudr + e2β dr du = gur dudr + grudr du 2gr 2e2αdudθ = gr 2e2αdudθ + gr 2e2αdθ du = guθ dudθ + gθudθ du

36

CHAPTER 2 Vectors, One Forms, Metric

With this in mind, we write

gab

=

 

f r

e2β − g2r 2e2α e2β
gr 2e2α

e2β 0 0

gr 2e2α
0 −r 2e2α



0

0 0



0

0

0

−r 2e−2α sin2 θ

The metric is a coordinate-dependent function, as can be seen from the examples discussed so far. Furthermore, recall that different sign conventions are used for space and time components. As an example, consider ﬂat Minkowski space written with spherical coordinates. It is entirely appropriate to use
ds2 = dt2 − dr 2 − r 2dθ 2 − r 2 sin2 θ dφ2

and it is equally appropriate to use ds2 = −dt2 + dr 2 + r 2dθ 2 + r 2 sin2 θ dφ2

The important thing is to make a choice at the beginning and stick with it for the problem being solved. When reading textbooks or research papers, be aware of the convention that the author is using. We will use both conventions from time to time so that you can get used to seeing both conventions.

The Signature of a Metric

The sum of the diagonal elements in the metric is called the signature. If we have
ds2 = −dt2 + dx2 + dy2 + dz2

then

 −1 0 0 0 

gab

= 

0 0

1 0

0 1

0 0



0 001

The signature is found to be

−1 + 1 + 1 + 1 = 2

CHAPTER 2 Vectors, One Forms, Metric

37

The Flat Space Metric

By convention, the ﬂat metric of Minkowski spacetime is denoted by ηab. Therefore

 −1 0 0 0 

ηab

= 

0 0

1 0

0 1

0 0



0 001

provided that ds2 = −dt2 + dx2 + dy2 + dz2.

The Metric as a Tensor

So far we have casually viewed the metric as a collection of the coefﬁcients found in a given line element. But as we mentioned earlier, the metric is a symmetric second rank tensor. Let’s begin to think about it more in this light. In fact the metric g, which we sometimes loosely call the line element, is written formally as a sum over tensor products of basis one forms

g = gabdxa ⊗ dxb

First, note that the metric has an inverse, which is written with raised indices. The inverse is deﬁned via the relationship

gabgbc = δac

(2.15)

where δac is the familiar (hopefully) Kronecker delta function. When the metric is diagonal, this makes it easy to ﬁnd the inverse. For ex-
ample, looking at the metric for spherical coordinates (2.12), it is clear that all components gab = 0 when a = b. So using (2.15), we arrive at the following:

grr grr = 1 ⇒ grr = 1

gθθ gθθ = gθθ r 2 = 1

⇒

gθθ

=

1 r2

gφφ gφφ = gφφr 2 sin2 θ = 1

⇒

gφφ

=

r2

1 sin2

θ

38

CHAPTER 2 Vectors, One Forms, Metric

These components can be arranged in matrix form as





10 0

gab =  0

1 r2

0

0

0

1 r 2 sin2 θ

(2.16)

Index Raising and Lowering

In relativity it is often necessary to use the metric to manipulate expressions via index raising and lowering. That is, we can use the metric with lowered indices to lower an upper index present on another term in the expression, or use the metric with raised indices to raise a lower index present on another term. This probably sounds confusing if you have never done it before, so let’s illustrate with an example. First, consider some vector V a. We can use the metric to obtain the covariant components by writing

Va = gab V b

(2.17)

Remember, the summation convention is in effect and so this expression is shorthand for

Va = gab V b = ga0V 0 + ga1V 1 + ga2V 2 + ga3V 3

Often, but not always, the metric will be diagonal and so only one of the terms in the sum will contribute. Indices can be raised in an analogous manner:

V a = gab Vb

(2.18)

Let’s provide a simple illustration with an example.

EXAMPLE 2-2
Suppose we are working in spherical coordinates where a contravariant vector X a = (1, r, 0) and a covariant vector Ya = (0, −r 2, cos2 θ ). Find Xa and Y a.

SOLUTION 2-2

Earlier

we

showed

that

gr r

=

grr

=

1, gθθ

=

r2, gθθ

=

1 r2

,

gφφ

=

r 2 sin2 θ,

gφφ

=

r

2

1 sin2

θ

.

Now

Xa = gab X b

CHAPTER 2 Vectors, One Forms, Metric

39

and so

Xr = grr Xr = (1)(1) = 1 Xθ = gθθ X θ = r 2 (r ) = r 3 Xφ = gφφ X φ = r 2 sin2 θ (0) = 0

Therefore, we obtain Xa = 1, r 3, 0 . For the second case, we need to raise indices, so we write

Y a = gabYb

This gives

Y r = grr Yr = (1)(0) = 0

Y θ = gθθ Yθ =

1 r2

−r 2 = −1

Y φ = gφφYφ =

1 r 2 sin2 θ

(cos2 θ ) =

cos2 θ r 2 sin2 θ

=

cot2 θ r2

and so Y a =

0,

−1,

cot2 r2

θ

.

EXAMPLE 2-3 In this example, we consider a ﬁctitious two-dimensional line element given by

ds2 = x2dx2 + 2dxdy − dy2

Write down gab, gab and then raise and lower indices on Va = (1, −1), W a = (0, 1).
SOLUTION 2-3 With lowered indices the metric can be written in matrix form as

gab =

gx x gyx

gx y gyy

We see immediately from the coefﬁcients of dx2 and dy2 in the line element that

gxx = x2 and gyy = −1

40

CHAPTER 2 Vectors, One Forms, Metric

To obtain the terms with mixed indices, we use the symmetry of the metric to write gxy = gyx and so
2dxdy = dxdy + dydx = gxydxdy + gyx dydx
Therefore, we ﬁnd that

gxy = gyx = 1

In matrix form, the metric is

gab =

x2 1

1 −1

To obtain the inverse, we use the fact that multiplying the two matrices gives the identity matrix. In other words,

gxx gxy gyx gyy

x2 1

1 −1

=

1 0

0 1

Four equations can be obtained by carrying out ordinary matrix multiplication in the above expression. These are

gxxx2 + gxy = 1 gxx − gxy = 0 ⇒ gxy = gxx gxxx2 + gyy = 0 gyy = −x2gxx gyx − gyy = 1

In addition, the symmetry of the metric provides the constraint gxy = gyx . Using gxy = gxx in the ﬁrst equation, we ﬁnd

gxx x 2 + gxx = gxx (1 + x 2) = 1

⇒

gxx

=

1

1 + x2

and so

gxy = gxx = 1 = gyx 1 + x2

CHAPTER 2 Vectors, One Forms, Metric

41

Finally, for the last term we get

gyy = −x2gyx = −x2 1 + x2

Now with this information in place, we can raise and lower indices as desired. We ﬁnd

V a = gab Vb

⇒

Vx

=

gxb Vb

=

gxx Vx

+

gxy Vy

=

1

1 (1)
+ x2

+

1

1 (−1) + x2

=

0

Vy

=

g yb Vb

=

g yx Vx

+

gyy Vy

=

1 1 + x2 (1)

−

1

x2 +x

2

(−1)

=

1+ 1+

x2 x2

=

1

⇒ V a = (0, 1)

In the other case, we have

Wa = gabW b

⇒

Wx

=

gxbW b

=

gxx W x

+

gxy W y

=

1

1 + x2 (0)

+

1

1 + x2 (1)

=

1

1 + x2

Wy

=

gybW b

=

gyx W x

+

gyyW y

=

1

1 (0)
+ x2

+

⇒ Wa =

1

1 +

x2

,

−x2 1 + x2

−x2 1 + x2

(1) = −x2 1 + x2

Index Gymnastics

Often we raise and lower indices with the metric in a more abstract fashion. The reason for doing this is to get equations in a more desirable form to derive some result, say. We will be seeing more of this later, so it will make more sense as we go along. Right now we will just provide a few examples that show how this works. We’ve already seen a bit of this with an ordinary vector
Xa = gab Xb
We can also apply this technique to a vector that is present in a more complicated expression, such as
XaY c = gab XbY c

42

CHAPTER 2 Vectors, One Forms, Metric

or we can use it with higher rank tensors. Some examples are
Sba = gac Scb T ab = gacTcb = gacgbd Tcd Rabcd = gae Rbecd
In ﬂat spacetime, the metric ηab is used to raise and lower indices. When we begin to prove results involving tensors, this technique will be used frequently.

The Dot Product

Earlier we brieﬂy mentioned the scalar or dot product. The metric also tells us how to compute the dot or scalar product in a given geometry. In particular, the dot product is written as
V · W = Va W a
Now we can use index raising and lowering to write the scalar product in a different way:
V · W = Va W a = gabV bW a = gabVa Wb

EXAMPLE 2-4 Consider the metric in plane polar coordinates with components given by

gab =

1 0

0 r2

and

gab =

1 0

0
1 r2

Let V a = (1, 1) and Wa = (0, 1). Find Va, W a, and V · W.
SOLUTION 2-4 Proceeding in the usual manner, we ﬁnd

Va = gab V b ⇒ Vr = grr V r = (1)(1) = 1 Vθ = gθθ V θ = r 2 (1) = r 2 ⇒ Va = (1, r 2)

CHAPTER 2 Vectors, One Forms, Metric

43

In an analogous manner, we obtain

W a = gabWb

⇒ W r = grr Wr = (1)(0) = 0

W θ = gθθ Wθ =

1 r2

(1)

=

1 r2

⇒ Wa =

0,

1 r2

For the dot product, we ﬁnd

V · W = gabV a W b = grr V r W r + gθθ V θ W θ = (1)(0) + r 2 1 = 0 + 1 = 1 r2

As a check, we compute

V · W = V a Wa = V r Wr + V θ Wθ = (1)(0) + (1)(1) = 0 + 1 = 1

Passing Arguments to the Metric

Thinking of a tensor as a map from vectors and one forms to the real numbers, we can think of the metric in different terms. Speciﬁcally, we can view the metric as a second rank tensor that accepts two vector arguments. The output is a real number, the dot product between the vectors

g (V, W ) = V · W

Looking at the metric tensor in this way, we see that the components of the metric tensor are found by passing the basis vectors as arguments. That is

g (ea, eb) = ea · eb = gab

(2.19)

In ﬂat space, we have ea · eb = ηab.
EXAMPLE 2-5 Given that the basis vectors in cartesian coordinates are orthnormal, i.e.,

∂x · ∂x = ∂y · ∂y = ∂z · ∂z = 1

44

CHAPTER 2 Vectors, One Forms, Metric

with all other dot products vanishing, show that the dot products of the basis vectors in spherical polar coordinates give the components of the metric.
SOLUTION 2-5 The basis vectors in spherical coordinates are written in terms of those of cartesian coordinates using the basis vector transformation law

ea =

b a

eb

where the elements of the transformation matrix are given by

b a

=

∂xb ∂xa

The coordinates are related in the familiar way:

x = r sin θ cos φ, y = r sin θ sin φ, z = r cos θ

We have

∂r

=

∂r ∂x

∂x

+

∂ ∂

r y

∂

y

+

∂ ∂

r z

∂z

= sin θ cos φ∂x + sin θ sin φ∂y + cos θ ∂z

Therefore, the dot product is

grr = ∂r · ∂r = sin2 θ cos2 φ + sin2 θ sin2 φ + cos2 θ = sin2 θ cos2 φ + sin2 φ + cos2 θ = sin2 θ + cos2 θ = 1

The basis vector ∂θ is given by ∂θ = r cos θ cos φ∂x + r cos θ sin φ∂y − r sin θ ∂z

and the dot product is

gθθ = ∂θ · ∂θ = r 2 cos2 θ cos2 φ + r 2 cos2 θ sin2 φ + r 2 sin2 θ = r 2 cos2 θ cos2 φ + sin2 φ + sin2 θ = r 2

CHAPTER 2 Vectors, One Forms, Metric

45

The last basis vector is ∂φ = −r sin θ sin φ∂x + r sin θ cos φ∂y
Therefore, the last component of the metric is gφφ = ∂φ · ∂φ = r 2 sin2 θ sin2 φ + r 2 sin2 θ cos2 φ = r 2 sin2 θ sin2 φ + cos2 φ = r 2 sin2 θ
As an exercise, you can check to verify that all the other dot products vanish.

A null vector V a is one that satisﬁes gabV a V b = 0

Null Vectors
(2.20)

The Metric Determinant

The determinant of the metric is used often. We write it as

g = det (gab)

(2.21)

Quiz

1. The following is a valid expression involving tensors:
(a) Sa Tab = ScTab (b) Sa Tab = Sa Tac (c) Sa Tab = ScTcb

2. Cylindrical coordinates are related to cartesian coordinates via x =

r cos φ, y = r sin φ, and z = z. This means that

z z

is

given

by

(a) 1

(b) −1

(c) 0

46

CHAPTER 2 Vectors, One Forms, Metric

3. If ds2 = dr 2 + r 2 dφ2 + dz2 then
(a) grr = dr, gφφ = r dφ, and gzz = dz (b) grr = 1, gφφ = r 2, and gzz = 1 (c) grr = 1, gφφ = r, and gzz = 1 (d) grr = dr 2, gφφ = r 2 dφ2, and gzz = dz2

4. The signature of

1 0 0 0 

gab

=



0 0

−1 0

0 −1

0 0



0 0 0 −1

is (a) −2 (b) 2 (c) 0 (d) 1

5. In spherical coordinates, a vector has the following components: X a =

r,

r

1 sin

θ

,

1 cos2 θ

. The component Xφ is given by

(a) 1/ cos2 θ

(b) cos2 θ

(c) r 2 tan2 θ

(d) r 2/ cos2 θ

3
CHAPTER
More on Tensors
In this chapter we continue to lay down the mathematical framework of relativity. We begin with a discussion of manifolds. We are going to loosely deﬁne only what a manifold is so that the readers will have a general idea of what this concept means in the context of relativity. Next we will review and add to our knowledge of vectors and one forms, and then learn some new tensor properties and operations.
Manifolds
To describe curved spacetime mathematically, we will use a mathematical concept known as a manifold. Basically speaking, a manifold is nothing more than a continuous space of points that may be curved (and complicated in other ways) globally, but locally it looks like plain old ﬂat space. So in a small enough neighborhood Euclidean geometry applies. Think of the surface of a sphere or the surface of the earth as an example. Globally, of course, the earth is a curved surface. Imagine drawing a triangle with sides that went from the equator to
47
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48

CHAPTER 3 More on Tensors

Fig. 3-1. The surface of a sphere is an example of a manifold. Pick a small enough patch, and space in the patch is Euclidean (ﬂat).
the North Pole. For that kind of triangle, the familiar formulas of Euclidean geometry are not going to apply. But locally it is ﬂat, and good old Euclidean geometry applies.
Other examples of manifolds include a torus (Fig. 3-2), or even a more abstract example like the set of rotations in cartesian coordinates.
A differentiable manifold is a space that is continuous and differentiable. It is intuitively obvious that we must describe spacetime by a differentiable manifold, because to do physics we need to be able to do calculus.
Generally speaking, a manifold cannot be completely covered by a single coordinate system. But we can “cover” the manifold with a set of open sets Ui called coordinate patches. Each coordinate patch can be mapped to ﬂat Euclidean space.
Fig. 3-2. A torus is an example of a manifold.

CHAPTER 3 More on Tensors

49

M
U2 U1 p
U3

Flat space

xa(p)

Image of U1

Fig. 3-3. A crude abstract illustration of a manifold. The manifold cannot be covered by
a single coordinate system, but we cover it with a set of coordinate patches that we designate by Ui . These patches may overlap. The points in each patch can be mapped to ﬂat Euclidean space. Here we illustrate a mapping for a coordinate patch we’ve called
U1¯. A point p that belongs to the manifold M (say in U1¯ ) is mapped to a coordinate that we designate x a (p). (Houghston and Todd, 1992)

Parameterized Curves

The notion of a curve as the plot of some function is a familiar one. In the context of relativity, however, it is more useful to think of curves in terms of parameterization. In this view, a point that moves through space traces out the curve.
The parameter of the curve, which we denote by λ, is a real number. We describe a curve by a set of parametric equations that give the coordinates along the curve for a given value of λ:

xa = xa (λ)

(3.1)

In n dimensions, since there are n coordinates xa, there will be n such equations.
EXAMPLE 3-1 Consider the curve traced out by the unit circle in the plane. Describe a parametric equation for the curve.
SOLUTION 3-1 We call the parameter θ. Since we are in two dimensions, we need two functions that will determine (x, y) for a given value of θ. Since the equation of a circle is given by x2 + y2 = r 2, in the case of the unit circle the equation that describes

50

CHAPTER 3 More on Tensors

Fig. 3-4. A parabola with x ≥ 0.

the curve is

x2 + y2 = 1

This is an easy equation to describe parametrically. Since we know that cos2 θ + sin2 θ = 1, we choose the parametric representation to be

x = cos θ and y = sin θ

EXAMPLE 3-2 Find a parameterization for the curve y = x2 such that x ≥ 0.
SOLUTION 3-2 The curve is shown in Fig. 3-4. To parameterize this curve with the restriction that x ≥ 0, we can take x to be some positive number λ. To ensure that the number is positive we square it; i.e.,
x = λ2

Since y = x2, it follows that y = λ4 will work.

Tangent Vectors and One Forms, Again
A curved space or spacetime is one that is going to change from place to place. As such, in a curved spacetime, one cannot speak of a vector that stretches from point to point. Instead, we must deﬁne all quantities like vectors and one forms locally. In the previous chapter, we basically tossed out the idea that a basis vector is deﬁned as a partial derivative along some coordinate direction. Now

CHAPTER 3 More on Tensors

51

let’s explore that notion a bit more carefully. We can do so with parameterized
curves. Let xa (λ) be a parameterized curve. Then the components of the tangent
vector to the curve are given by

dxa

dλ

(3.2)

EXAMPLE 3-3 Describe the tangent vectors to the curves in Examples 3-1 and 3-2.
SOLUTION 3-3 In Example 3-1, we parameterized the curve traced out by the unit circle with (cos θ, sin θ). Using (3.2), we see that the components of the tangent vector to the unit circle are given by (− sin θ, cos θ). Calling the tangent vector τ and writing the basis vectors in cartesian coordinates as ex and ey, we have

τ = − sin θ ex + cos θ ey

Now we consider Example 3-2, where the given curve was y = x2. We found that this curve could parameterize the curve with λ2, λ4 . Using (3.2), we obtain the tangent vector v = 2λ, 4λ3 . Let’s invert the relation used to describe the curve. We have
√ λ= x

Therefore, we can write v as

v

=

√ 2 xex

+

4x 3/2ey

Now that we’ve found out how to make a vector that is tangent to a parameter-

ized curve, the next step is to ﬁnd a basis. Let’s expand this idea by considering

some arbitrary continuous and differentiable function f . We can compute the

derivative of

f

in the direction of the curve by

d d

f λ

.

Therefore,

d dλ

is a vector

that

maps

f

to a real number that is given by

df dλ

.

The

chain

rule

allows

us

to

write

this as

d f = d f ∂xa = dxa ∂ f dλ dλ ∂ xa dλ ∂ xa

52

CHAPTER 3 More on Tensors

The function f is arbitrary, so we can write

d = dxa ∂ dλ dλ ∂ xa

In other words, we have expanded the vector

The components of

the vector are

dxa dλ

and

d dλ

in

terms

of

a

set

of

the basis vectors are

basis vectors.

given

by

∂ ∂x

a

.

This is the origin of the argument that the basis vectors ea are given by partial

derivatives along the coordinate directions.

EXAMPLE 3-4 The fact that the basis vectors are given in terms of partial derivatives with respect to the coordinates provides an explanation as to why we can write the transformation matrices as

a b

=

∂xa ∂xb

SOLUTION 3-4 This can be done by applying the chain rule to a basis vector. We have, in the primed coordinates,

ea

=∂ ∂xa

=∂ ∂xa

∂xb ∂xb

= ∂xb ∂ = ∂xa ∂xb

b a

eb

Similarly, going the other way, we have

∂

∂ ∂xa

eb = ∂ xb = ∂ xb ∂ xa

∂xa ∂

=

=

∂xb ∂xa

a bea

Now let’s explore some new notation that is frequently seen in books and the literature. We can write the dot product using a bracket-type notation , . In the left slot, we place a one form and in the right side we place a vector. And so we write the dot product p · v as

p · v = p˜, v = pava

(3.3)

Here p˜ is a one form. Using this notation, we can write the dot product between the basis one forms and basis vectors as

ωa, eb = δba

CHAPTER 3 More on Tensors

53

This can be seen easily writing the bases in terms of partial derivatives:

ωa, eb =

dx

a,

∂ ∂xb

=

∂xa ∂xb

=

δba

This type of notation makes it easy to ﬁnd the components of vectors and one forms. We consider the dot product between an arbitrary vector V and a basis one form:

ωa, V = ωa, V beb = V b ωa, eb = V bδba = V a
Since the components of a vector are just numbers, we are free to pull them outside of the bracket , . We can use the same method to ﬁnd the components of a one form:

σ, eb = σaωa, eb = σa ωa, eb = σaδba = σb
Now we see how we can derive the inner product between an arbitrary one form and vector:

σ, V = σaωa, V beb = σa V b ωa, eb = σa V bδba = σa V a These operations are linear. In particular,

σ, aV + bW = a σ, V + b σ, W aσ + bρ, V = a σ, V + b ρ, V

where a, b are scalars, σ, ρ are one forms, and V, W are vectors.

Tensors as Functions
A tensor is a function that maps vectors and one forms to the real numbers. The components of a tensor are found by passing basis one forms and basis vectors as arguments. For example, we consider a rank 2 tensor T . If we pass two basis one forms as argument, we get
T ωa, ωb = T ab

54

CHAPTER 3 More on Tensors

A tensor with raised indices has contravariant components and is therefore expanded in terms of basis vectors; i.e.,

T = T abea ⊗ eb

We have already seen an example of a tensor with lowered indices, the metric tensor

gabωa ⊗ ωb = gab dxa ⊗ dx b

(in a coordinate basis). A tensor is not ﬁxed with raised or lowered indices; we recall from the last chapter that we can raise or lower indices using the metric. We can also have tensors with mixed indices. For each raised index we need a basis vector, and for each lowered index we need a basis one form when writing out the tensor as an expansion (the way you would write a vector expanding in a basis). For example

S = Sabc ea ⊗ eb ⊗ ωc

We get the components in the opposite way; that is, to get the upper index pass a one form, and to get the lower index pass a basis vector:

S

ab c

=

S

ωa, ωb, ec

We can pass arbitrary vectors and one forms to a tensor. Remember, the components of vectors and one forms are just numbers. So we can write

S (σ, ρ, V ) = S

σaωa, ρbωb, V cec

= σaρbV c S

ωa, ωb, ec

=

σa ρb V

c

S

ab c

The quantity σaρbV c Sabc is a number, which is consistent with the notion that a tensor maps vectors and one forms to numbers. Note that the summation convention is used.

Tensor Operations

We now summarize a few basic algebraic operations that can be carried out with tensors to produce new tensors. These operations basically mirror the types of things you can carry out with vectors. For example, we can add two tensors of the same type to get a new tensor:

R

ab c

=

Sabc

+

T

ab c

CHAPTER 3 More on Tensors

55

It follows that we can subtract two tensors of the same type to get a new tensor of the same type:

Qab

=

Sab

−

Ta a

b

We can also multiply a tensor by a scalar a to get a new tensor

Sab = aTab

Note that in these examples, the placement of indices and number of indices are
arbitrary. We are simply providing speciﬁc examples. The only requirement is
that all of the tensors in these types of operations have to be of the same type.
We can use addition, subtraction, and scalar multiplication to derive the symmetric and antisymmetric parts of a tensor. A tensor is symmetric if Bab = Bba and antisymmetric if Tab = −Tba. The symmetric part of a tensor is given by

T(ab)

=

1 2

(Tab

+

Tba )

(3.4)

and the antisymmetric part of a tensor is

1 T[ab] = 2 (Tab − Tba)

(3.5)

We can extend this to more indices, but we won’t worry about that for the time being. Often, the notation is extended to include multiple tensors. For instance,

V(a Wb)

=

1 2

(Va Wb

+

Wb Va )

Tensors of different types can be multiplied together. If we multiply a tensor of type (m, n) by a tensor of type ( p, q), the result is a tensor of type (m + p, n + q). For example

Rab Scde

=

T

abc d

e

Contraction can be used to turn an (m, n) tensor into an (m− 1, n− 1) tensor. This is done by setting a raised and lowered index equal:

Rab = Rcacb

(3.6)

Remember, repeated indices indicate a sum.

56

CHAPTER 3 More on Tensors

The Kronecker delta can be used to manipulate tensor expressions. Use the following rule: When a raised index in a tensor matches the lowered index in the Kronecker delta, change it to the value of the raised index of the Kronecker delta. This sounds confusing, so we demonstrate it with an example

δba T bcd = T acd

Now consider the opposite. When a lowered index in a tensor matches a raised index in the Kronecker delta, set that index to the value of the lowered index of the Kronkecker delta:

δdc T abc = T abd

EXAMPLE 3-5 Show that if a tensor is symmetric then it is independent of basis.

SOLUTION 3-5 We can work this out easily using the tensor transformation properties. Considering Bab = Bba, we work out the left side:

Bab =

c a

d b Bc d

=

∂xc ∂xd ∂xa ∂xb Bc d

For the other side, since we can move the derivatives around, we ﬁnd

Bba =

d b

c a Bd c

=

∂xd ∂xb

∂xc ∂ x a Bd c

=

∂xc ∂xd ∂xa ∂xb

Bd c

Equating both terms, it immediately follows that Bc d = Bd c . It is also true that if Bab = Bba, then Bcd = Bdc. Working this out,

Bcd = gca Bad = gca gdb Bab = gca gdb Bba = gca Bd a = Bdc

EXAMPLE 3-6 Let T ab be antisymmetric. Show that

S[a Tbc]

=

1 3

(Sa Tbc

−

Sb Tac

+

Sc Tab )

CHAPTER 3 More on Tensors

57

SOLUTION 3-6 Since T ab is antisymmetric, we know that Tab = −Tba. The expression A[abc]
is given by

A[abc]

=

1 6

( Aabc

+

Abca

+

Acab

−

Abac

−

Aacb

−

Acba )

Therefore, we ﬁnd

1 S[a Tbc] = 6 (Sa Tbc + SbTca + ScTab − SbTac − Sa Tcb − ScTba) Now we use the antisymmetry of T , Tab = −Tba, to write

S[a Tbc]

=

1 6

(Sa Tbc

−

Sb Tac

+

Sc Tab

−

Sb Tac

+

Sa Tbc

+

Sc Tab )

1 = 6 (2Sa Tbc − 2SbTac + 2ScTab)

=

1 3

(Sa Tbc

−

Sb Tac

+

Sc Tab )

EXAMPLE 3-7 Let Qab = Qba be a symmetric tensor and Rab = −Rba be an antisymmetric
tensor. Show that

Qab Rab = 0

SOLUTION 3-7 Since Rab = −Rba, we can write

Rab

=

1 2

( Rab

+

Rab)

=

1 2

( Rab

−

Rba )

Therefore, we have

Qab Rab

=

1 Qab 2

( Rab

−

Rba )

=

1 2

Qab Rab − Qab Rba

Note that the indices a, b are repeated in both terms. This means they are dummy indices and we are free to change them. In the second term, we make the switch

58

CHAPTER 3 More on Tensors

a ↔ b, which gives

Qab Rab

=

1 2

Qab Rab − Qab Rba

=1 2

Qab Rab − Qba Rab

Now use the symmetry of Qab to change the second term, giving the desired result:

1 2

Qab Rab − Qba Rab

=1 2

Qab Rab − Qab Rab

=

1 Qab 2

( Rab

−

Rab)

=

0

EXAMPLE 3-8 Show that if Qab = Qba is a symmetric tensor and Tab is arbitrary, then

Tab Qab

=

1 2

Qab

(Tab

+

Tba )

SOLUTION 3-8 Using the symmetry of Q, we have

Tab Qab

=

1 Tab 2

Qab + Qab

=

Tab

1 2

Qab + Qba

Now multiply it by T , which gives

1 Tab 2

Qab + Qba

=1 2

Tab Qab + Tab Qba

Again, the indices a, b are repeated in both expressions. Therefore, they are dummy indices that can be changed. We swap them in the second term a ↔ b, which gives

1 2

Tab Qab + Tab Qba

=1 2

Tab Qab + Tba Qab

=

1 Qab 2

(Tab

+

Tba )

CHAPTER 3 More on Tensors

59

The Levi-Cevita Tensor

No book on relativity can really give you a good enough headache without

mention of our friend, the Levi-Cevita tensor. This is   +1 for an even permutation of 0123

εabcd

=



−1 0

for an odd permutation of 0123 otherwise

(3.7)

We will see more of this in future chapters.

1. If T ab = −T ba then

(a)

QabTab =

1 2

Qab + Qba

Tab

(b)

QabTab =

1 2

Qab − Qba

Tab

(c)

QabTab =

1 2

Qab − Qba

Tab

(d)

QabTab =

1 2

(Qab

−

Q ba )

Tab

2. Let xa (λ) be a parameterized curve. The components of the tangent

vector to this curve are best described by

(a) dλ/dx

(b) f = x (λ)

(c) dxa/dλ

(d) x = f (λ)

3.

A

tensor

with

components

Ta

b c

has

an

expansion

in

terms

of

basis

vectors

and one forms, given by

(a)

T

=

Ta

b c

ωa

⊗ ωb ⊗ ωc

(b)

T

=

Ta

b c

ωa

⊗ eb ⊗ ωc

(c)

T

=

Ta

b c

ea

⊗

ωb

⊗

ec

4. The symmetric part of Va Wb is best written as

(a)

V(a Wb) =

1 2

(Va Wb

−

Wb Va )

(b)

V(a Wb) =

1 2

V a Wb + W bVa

(c)

V(a Wb) =

1 2

(Va Wb

+

Wb Va )

5. The Kronecker delta acts as

(a)

δba T bcd

=

−T

ac d

(b) δba T bcd = Tacd

(c)

δba T bcd

=

Ta

c d

(d)

δba T bcd

=

T

ac d

Quiz

4
CHAPTER
Tensor Calculus
In this chapter we turn to the problem of ﬁnding the derivative of a tensor. In a curved space or spacetime, this is a bit of a thorny issue. As we will see, properly ﬁnding the derivative of a tensor, which should give us back a new tensor, is going to require some additional mathematical formalism. We will show how this works and then describe the metric tensor, which plays a central role in the study of gravity. Next we will introduce some quantities that are important in Einstein’s equation.
Testing Tensor Character
As we will see below, it is sometimes necessary to determine whether a given object is a tensor or not. The most straightforward way to determine whether an object is a given type of tensor is to check how it transforms. There are, however, a few useful tips that can serve as a guide as to whether or not a given quantity is a tensor.
The ﬁrst test relies on the inner product. If the inner product σa V a = φ
60
Copyright © 2006 by The McGraw-Hill Companies, Inc. Click here for terms of use.

CHAPTER 4 Tensor Calculus

61

where φ is a scalar and is invariant for all vectors V a, then σa is a one form. We can carry this process further. If
Tab V a = Ub
such that Ub is a one form, then Tab is a (0, 2) tensor. Another test on Tab is
Tab V a W b = φ
where V a and W b are vectors and φ is an invariant scalar, then Tab is a (0, 2) tensor.

The Importance of Tensor Equations

In physics we seek invariance, i.e., we seek laws of physics written in an invariant form that is true for all observers. Tensors are a key ingredient in this recipe, because if a tensor equation is true in one coordinate system then it is true in all coordinate systems. This can greatly simplify analysis because we can often transform to a coordinate system where the mathematics will be easier. There we can ﬁnd the form of a result we need and then express it in another coordinate system if desired.
A simple example of this is provided by the vacuum ﬁeld equations. We will see that these can be expressed in the form of a (0, 2) tensor called the Ricci tensor, where

Rab = 0

It is immediately obvious that this equation is true in any coordinate system. Let’s transform to a different coordinate system using the [Lambda] matrix discussed

Rab =

c a

d b Rc d

=

∂xc ∂xa

∂xd ∂xb

Rcd

On the right-hand side, the transformation of 0 is of course just 0, so we have

∂xc ∂xd ∂xa ∂xb

Rcd

=0

62

CHAPTER 4 Tensor Calculus

We

can

divide

both

sides

by

∂xc ∂xd

∂xd ∂xb

,

which

gives

Rc d = 0

The Covariant Derivative

Consider the problem of taking the derivative of a vector. In ordinary cartesian coordinates, this is not really that complicated of an issue because the basis vectors are constant. So we can get the derivative of a vector by differentiating its components. However, in general, things are not so straightforward. In curved spaces the basis vectors themselves may vary from point to point. This means that when we take a derivative of a vector, we will have to differentiate the basis vectors as well.
Consider some vector A. To compute the derivative, we expand the vector in a basis and then apply the Leibniz rule ( fg) = f g + g f to obtain

∂A = ∂ ∂xa ∂xa

Abeb

=

∂ Ab ∂xa eb

+

Ab ∂ ∂xa

(eb)

In cartesian coordinates, we could just throw away the second term. But this isn’t true in general. To see how this works in practice, we turn to a speciﬁc example.
We start with something familiar—the spherical polar coordinates. Cartesian coordinates are related to spherical coordinates in the following way:

x = r sin θ cos φ

y = r sin θ sin φ

(4.1)

z = r cos θ

The ﬁrst step in this exercise will be to work out the basis vectors in spherical coordinates in terms of the cartesian basis vectors. We’ll do this using the procedures outlined in Chapter 2. This means we will need the transformation matrices that allow us to move back and forth between the two coordinate systems. Again, we denote the transformation matrix by the symbol ab , where an element of this matrix is given by
∂xa ∂xb

CHAPTER 4 Tensor Calculus

63

We will consider the unprimed coordinates to be (x, y, z) and the primed coordinates to be (r, θ, φ). In that case, the transformation matrix assumes the
form

 ∂x ∂y ∂z 

a b

=



∂r
∂x ∂θ

∂r ∂y ∂θ

∂r
∂z ∂θ



∂x ∂y ∂z

∂φ ∂φ ∂φ

(4.2)

Using the relationship among the coordinates described by (4.1), we obtain

 sin θ cos φ sin θ sin φ cos θ 

a b

=  r cos θ cos φ

r cos θ sin φ

−r sin θ 

(4.3)

−r sin θ sin φ r sin θ cos φ 0

Now we have the machinery we need to work out the form of the basis

vectors. We have already seen many times that the basis vectors transform as

eb =

a b

ea .

So

let’s

write

down

the

basis

vectors

for

spherical

coordinates

as

expansions in terms of the cartesian basis using (4.3). We get

er =

b r

eb

=

x r

ex

+

y r

e

y

+

z r

ez

= sin θ cos φ ex + sin θ sin φ ey + cos θ ez

eθ =

b θ

eb

=

x θ

ex

+

yθ ey +

zθ ez

= r cos θ cos φ ex + r cos θ sin φ ey − r sin θ ez

eφ = bφeb = x φex + yφey + zφez

= −r sin θ sin φ ex + r sin θ cos φ ey

(4.4)

Now let’s see what happens when we differentiate these basis vectors. As an
example, we compute the derivatives of er with respect to each of the spherical coordinates. Remember, the cartesian basis vectors ex , ey, ez are constant, so we don’t have to worry about them. Proceeding, we have

∂er = ∂ ∂r ∂r

sin θ cos φ ex + sin θ sin φ ey + cos θ ez

=0

64

CHAPTER 4 Tensor Calculus

Our ﬁrst attempt hasn’t yielded anything suspicious. But let’s compute the θ derivative. This one gives

∂er = ∂ ∂θ ∂θ

sin θ cos φ ex + sin θ sin φ ey + cos θ ez

= cos θ cos φ ex + cos θ sin φ ey − sin θ ez

1 = r eθ

Now that’s a bit more interesting. Instead of computing the derivative and getting zero, we ﬁnd another basis vector, scaled by 1/r. Let’s go further and proceed by computing the derivative with respect to φ. In this case, we get

∂ er ∂φ

=

∂ ∂φ

sin θ cos φ ex + sin θ sin φ ey + cos θ ez

= − sin θ sin φ ex + sin θ cos φ ey

1 = r eφ

Again, we’ve arrived at another basis vector. It turns out that when differentiating basis vectors, there is a general relation-
ship that gives the derivative of a basis vector in terms of a weighted sum. The sum is just an expansion in terms of the basis vectors with weighting coefﬁcients denoted by abc.

∂ ea ∂xb

=

c ab

ec

(4.5)

The

a bc

are

functions

of

the

coordinates,

as

we

saw

in

the

examples

we’ve

calculated so far. Looking at the results we got above, we can identify the results

in the following way:

∂ er ∂θ

=

1 r eθ

⇒

θ rθ

=

1 r

∂ er ∂φ

=

1 r eφ

⇒

φ rφ

=

1 r

CHAPTER 4 Tensor Calculus

65

Consider another derivative

∂ eφ ∂θ

=

∂ ∂θ

−r sin θ sin φ ex + r sin θ cos φ ey

= −r cos θ sin φ ex + r cos θ cos φ ey

= cos θ −r sin φ ex + r cos φ ey

=

cos θ sin θ

−r sin θ sin φ ex + r sin θ cos φ ey

= cot θ eφ

Comparison with (4.5) leads us to conclude that

φ φθ

=

cot θ

The coefﬁcient functions we have derived here are known as Christoffel symbols

or an afﬁne connection.

Basically, these quantities represent correction terms. A derivative operator

needs to differentiate a tensor and give a result that is another tensor. In partic-

ular, ﬁeld

the derivative of

of valence

m n+1

a .

tensor ﬁeld that has valence We have seen one reason why

m n

should give a tensor

we need the correction

terms: outside of ordinary cartesian coordinates, the derivative of a vector is

going to involve derivatives of the basis vectors as well. Another reason for

introducing the Christoffel symbols is that the partial derivative of a tensor is

not a tensor. First let’s remind ourselves how the components of a vector trans-

form:

Xa

=

∂xa ∂xb

Xb

Keeping this in mind, we have

∂c X a

=

∂ ∂xc

∂xa Xb ∂xb

= ∂ ∂xd ∂xc ∂xd

∂xa ∂xb

Xb

66

CHAPTER 4 Tensor Calculus

∂xd ∂ =

∂xa Xb

∂xc ∂xd ∂xb

=

∂xd ∂xc

∂2xa ∂xd∂xb

Xb

+

∂xd ∂xc

∂xa ∂xb

∂Xb ∂xd

Now how does a (1, 1) tensor transform? It does so like this:

Ta b

=

∂xa ∂xc

∂xd ∂xb

T

c d

That’s the kind of transformation we got from the partial derivative above, in the second term on the last line:

∂c X a

=

∂xd ∂xc

∂2xa ∂xd∂x

b

X

b

+

∂xd ∂xc

∂xa ∂xb

∂Xb ∂xd

But the ﬁrst term leaves us out of luck as far as getting another tensor—thus the need for a correction term. Let’s go back and look at the formula we had for the derivative of a vector A:

∂A = ∂ ∂xa ∂xa

Ab

eb

+

Ab ∂ ∂xa

(eb)

Now let’s use (4.5) to rewrite the second term. This gives

Ab ∂ ∂xa

(eb)

=

Ab

cba ec

Putting this into the formula that gives the derivative of a vector, we have

∂A ∂ =
∂xa ∂xa

Ab eb +

cba Abec

We rearranged the order of quantities in the second term for later convenience.

Now, recall that any repeated indices that are both up and down in an expression

are dummy indices, and so can be relabeled. In the second term, we swap b ↔ c

to change

c ba

Ab

ec

→

b ca

Ac

eb,

and

the

expression

for

the

derivative

of

a

CHAPTER 4 Tensor Calculus

67

vector becomes

∂A ∂ =
∂xa ∂xa

Ab eb +

b ca

Aceb

=

∂ Ab +
∂xa

b ca

Ac

eb

The expression in parentheses is the covariant derivative of a vector A. We denote the covariant derivative by ∇b Aa:

∇b Aa

=

∂ Aa ∂xb

+

bca Ac

(4.6)

It can be veriﬁed that this object is a (1, 1) tensor by checking how it transforms. To see how the Christoffel symbols transform, we look at (4.5). Writing this in primed coordinates, we have

c ab

ec

=

∂ ea ∂xb

(4.7)

Now, the basis vectors transform according to ec =

d c

ed

=

∂xd ∂xc

ed .

The

left-

hand side then becomes

c ab

ec

=

c ab

d c

ed

=

c

a

b

∂xd ∂xc

ed

(4.8)

Now let’s tackle the right-hand side of (4.7). We obtain

∂ ea ∂xb

=

∂ ∂xb

(

m a

em )

=

∂ ∂xb

∂xm ∂ xa em

= ∂xn ∂ ∂xb ∂xn

∂xm ∂ xa em

∂xn =
∂xb

∂2xm ∂xn∂xa

em

+

∂xm ∂xa

∂ em ∂xn

Now

we

know

that

∂ em ∂xn

gives

another

Christoffel

symbol,

so

we

write

this

as

∂ ea ∂xb

=

∂xn ∂xb

∂2xm ∂xn∂xa

em

+

∂xm ∂xa

l mnel

68

CHAPTER 4 Tensor Calculus

In the ﬁrst term inside parentheses, we are going to swap the dummy indices. We change m → d and now write the expression as

∂ea = ∂ xn ∂xb ∂xb

∂2xd ∂xn∂xa

ed

+

∂xm ∂xa

l mnel

In the second term, there is a dummy index l that is associated with the basis vector. We would like to factor out the basis vector and so we change l → d, which gives

∂ ea ∂xb

=

∂xn ∂xb

∂2xd ∂xn∂xa

ed

+

∂xm ∂xa

d mned

=

∂xn ∂xb

∂2xd ∂xn∂xa

+

∂xm ∂xa

d mn

ed

(4.9)

=

∂xn ∂xb

∂2xd ∂xn∂xa

+

∂xn ∂xb

∂xm ∂xa

d mn

ed

We started out with (4.7). Setting the result for the left side of (4.7) given in (4.8) equal to the result found in (4.9) gives us the following:

c

ab

∂xd ∂xc

ed

=

∂xn ∂2xd ∂xb ∂xn∂xa

+

∂xn ∂xb

∂xm ∂xa

d mn

ed

The basis vector ed appears on both sides, so we drop it and write

c ∂xd ab ∂xc

=

∂xn ∂2xd ∂xb ∂xn∂xa

+

∂xn ∂xb

∂xm ∂xa

d mn

Finally, we obtain the transformation law for the Christoffel symbols by

dividing

both

sides

through

by

∂xd ∂xc

.

This

gives

c ab

∂xc ∂xn ∂2xd =
∂xd ∂xb ∂xn∂xa

∂xc ∂xn ∂xm +
∂xd ∂xb ∂xa

d mn

(4.10)

Now we see why the Christoffel symbols act as correction terms. The ﬁrst piece of this can be used to cancel out the extra piece found in the transformation of a partial derivative of a tensor. That way, the covariant derivative transforms as it should, as a (1, 1) tensor.

CHAPTER 4 Tensor Calculus

69

Of course we are going to differentiate other objects besides vectors. The covariant derivative of a one form is given by

∇bσa = ∂bσa −

c ab

σc

(4.11)

This suggests the procedure to be used for the covariant derivative of an arbitrary

tensor: First take the partial derivative of the tensor, and then add a

c ab

term

for each contravariant index and subtract a

c ab

term

for

each

covariant

index.

For example,

∇cT ab = ∂cT ab +

acd T

d b

−

d bcT ad

∇cTab = ∂cTab − d acT db − d cbT ad

∇cT ab = ∂cT ab + acd T db + bdcT ad

The covariant derivative of a scalar function is just the partial derivative:

∇aφ = ∂aφ

(4.12)

EXAMPLE 4-1 Consider polar coordinates. Find the covariant derivative ∇a V a of V = r 2 cos θ er − sin θ eθ .
SOLUTION 4-1 The summation convention is in effect. Writing it out explicitly, we have

∇a V a = ∇r V r + ∇θ V θ

Using (4.6), we have

∇r V r

=

∂Vr ∂r

+

r cr

V

c

=

∂Vr ∂r

+

r rr

Vr

+

rθr V θ

∇θ V θ

=

∂Vθ ∂θ

+

θ cθ V c

=

∂Vθ ∂θ

+

θrθVr +

θθθV θ

One can show that the Christoffel symbols for polar coordinates are (Exercise)

θ rθ

=

θ θr

=

1 r

and

r θθ

=

−r

70

CHAPTER 4 Tensor Calculus

while all other components are zero, and so we obtain

∇r V r

=

∂Vr ∂r

+

rrr V r +

rθr V θ

=

∂Vr ∂r

∇θ V θ

=

∂Vθ ∂θ

+

θrθVr +

θθθV θ

=

∂Vθ ∂θ

+

1Vr r

The sum then becomes

∇a V a

=

∇r V r

+

∇θ V θ

=

∂Vr ∂r

+

1Vr r

+

∂Vθ ∂θ

For V = r 2 cos θ er − sin θ eθ , this results in

∇a V a

=

∂Vr ∂r

+

1Vr+∂Vθ

r

∂θ

= 2r cos θ

+ r cos θ

− cos θ

= 3r cos θ

− cos θ

= cos θ (3r − 1)

EXAMPLE 4-2 Suppose that for some vector ﬁeld U a, U aUa is a constant and ∇aUb − ∇bUa = 0. Show that U a∇aU b = 0.
SOLUTION 4-2 Since U aUa is a constant, the derivative must vanish. We compute the derivative of this product as
∇b (U aUa) = U a∇bUa + Ua∇bU a
Now we use ∇aUb − ∇bUa = 0 to rewrite the ﬁrst term on the right-hand side, and then use index raising and lowering with the metric to write Ub = gbcU c, which gives
U a∇bUa + Ua∇bU a = U a∇aUb + Ua∇bU a = U a∇a gbcU c + Ua∇bU a = U aU c∇a gbc + gbcU a∇aU c + Ua∇bU a = gbcU a∇aU c + Ua∇bU a

CHAPTER 4 Tensor Calculus

71

To obtain the last line, we used ∇a gbc = 0. We now concentrate on the second term. To use ∇aUb − ∇bUa = 0 to change the indices, ﬁrst we need to lower the index on U a inside the derivative. Remember, we are free to pull gab outside the derivative since ∇a gbc = 0. So we get
gbcU a∇aU c + Ua∇bU a = gbcU a∇aU c + Ua∇bgacUc = gbcU a∇aU c + gacUa∇bUc = gbcU a∇aU c + gacUa∇cUb
The ﬁrst term is already in the form we need to prove the result. To get the second term in that form, we are going to have to raise the indices on the U ’s. We do this and recall that gabgbc = δca, which gives
gbcU a∇aU c + gacUa∇cUb = gbcU a∇aU c + gacgaeU e∇cgbdU d = gbcU a∇aU c + gacgaegbdU e∇cU d = gbcU a∇aU c + δecgbdU e∇cU d = gbcU a∇aU c + gbdU c∇cU d
To obtain the last line, we used the Kronecker delta to set e → c. Now note that c is a dummy index. We change it to a to match the ﬁrst term
gbcU a∇aU c + gbdU c∇cU d = gbcU a∇aU c + gbdU a∇aU d
Now we focus on the other dummy index d. We’re free to change it so we set it equal to c and have
gbcU a∇aU c + gbdU a∇aU d = gbcU a∇aU c + gbcU a∇aU c = gbc (U a∇aU c + U a∇aU c) = 2gbcU a∇aU c
We started by taking the derivative of a constant, which is zero, ∇b (U aUa) = 0, and so this result must vanish. We divide the above expression by 2 and get
gbcU a∇aU c = 0

72

CHAPTER 4 Tensor Calculus

There are two possibilities: If gbc = 0, then the equation is trivially zero. If it is not zero, we can divide both sides by gbc and get the desired result
U a∇aU c = 0

The Torsion Tensor

The torsion tensor is deﬁned in terms of the connection as

T abc =

a bc

−

a cb

=

2

a [bc]

(4.13)

It is easy to show that even though the connection is not a tensor, the difference between two connections is a tensor. In general relativity, the torsion tensor is taken to vanish, which means that the connection is symmetric; i.e.,

a bc

=

a cb

(4.14)

In some theories of gravity (known as Einstein-Cartan theories), the torsion tensor does not vanish. We will not study such cases in this book.

The Metric and Christoffel Symbols

In n dimensions there are n3 functions called Christoffel symbols of the ﬁrst kind, abc. In a coordinate basis (see Chapter 5) these functions can be derived from the metric tensor using the relationship

1 abc = 2

∂gbc + ∂gca − ∂gab ∂xa ∂xb ∂xc

(4.15)

More generally,

abc

=

1 2

∂ gbc ∂xa

+

∂ gca ∂xb

−

∂ gab ∂xc

+ Cabc

+ Cacb

− Cbca

where the Cabc are called commutation coefﬁcients (see Chapter 5).

CHAPTER 4 Tensor Calculus

73

EXAMPLE 4-3 Show that

∂gab = ∂xc

abc +

bca

SOLUTION 4-3 This is a simple problem to solve. We simply use (4.15) and permute indices, which gives

abc

=

1 2

∂ gbc ∂xa

+

∂ gca ∂xb

−

∂ gab ∂xc

bca

=

1 2

∂gca + ∂gab − ∂gbc ∂xb ∂xc ∂xa

Adding, we ﬁnd

abc +

bca

=

1 2

∂ gbc ∂xa

+

∂ gca ∂xb

−

∂ gab ∂xc

+1 2

∂ gca ∂xb

+

∂ gab ∂xc

−

∂ gbc ∂xa

= 1 ∂gbc − ∂gbc + ∂gca + ∂gca − ∂gab + ∂gab 2 ∂xa ∂xa ∂xb ∂xb ∂xc ∂xc

=1 2

2

∂ gca ∂xb

=

∂ gca ∂xb

We can obtain the Christoffel symbols of the second kind (which are usually referred to simply as the Christoffel symbols) by raising an index with the metric

a bc

=

gad

dbc

Using this we can write (4.15) in a more popular form:

a bc

=

1 gad 2

∂ gbd¯ ∂xa

+

∂ gcd¯ ∂xb

−

∂ gab ∂ x d¯

(4.16)

74

CHAPTER 4 Tensor Calculus

EXAMPLE 4-4 Find the Christoffel symbols for the 2-sphere of radius a
ds2 = a2 dθ 2 + a2 sin2 θ dφ2

SOLUTION 4-4 The metric is given by

a2

0

gab = 0 a2 sin2 θ

and so gθθ = a2 and gφφ = a2 sin2 θ . The inverse metric is

1
gab = a2 0

0
1 a2 sin2 θ

Therefore,

gθθ

=

1 a2

and

gφφ

=

a2

1 sin2

θ

.

The

only

nonzero

derivative

is

∂gφφ = ∂ a2 sin2 θ = 2a2 sin θ cos θ ∂θ ∂θ

(4.17)

We ﬁnd the Christoffel symbols using (4.16). Considering the ﬁrst nonzero term of gab, we set a = d = θ in (4.16) and obtain

θ bc

=

1 gθθ 2

∂gbθ + ∂gcθ − ∂gbc ∂xc ∂xb ∂θ

= −1gθθ 2

∂ gbc ∂θ

We

set

∂ gcθ ∂xb

=

∂ gθ b ∂xc

= 0 because gθφ

=

gφθ

= 0,

gθ θ

= a2, where a is a constant,

and so all derivatives of these terms vanish. The only possibility is

θ φφ

=

−1 gθθ 2

∂ gφφ ∂θ

= −1 2

1 a2

2a2 sin θ cos θ = − sin θ cos θ

The only other nonzero possibility for this metric is the term involving gφφ, and so we set a = d = φ in (4.16). This gives

φ bc

=

1 gφφ 2

∂gbφ + ∂gcφ − ∂gbc ∂xc ∂xb ∂φ

CHAPTER 4 Tensor Calculus

75

Only a derivative with respect to θ is going to be nonzero. Therefore we drop the ﬁrst term and consider

φ bc

=

1 gφφ 2

∂gcφ − ∂gbc ∂xb ∂φ

First we take b = θ, c = φ, which gives

φ φθ

=

1 gφφ 2

∂φ ∂ gθ φ

−

∂θ ∂ gφφ

=

1 2

gφφ

∂ gφφ ∂θ

11

= 2

a2 sin2 θ

2a2 sin θ cos θ = cos θ = cot θ sin θ

A similar procedure with b = θ, c = φ gives

φ θφ

=

cot

θ.

All

other

Christof-

fel symbols are zero.

EXAMPLE 4-5
A metric that is used in the study of colliding gravitational waves is the KahnPenrose metric (see Fig. 4-1). The coordinates used are (u, v, x, y). With u ≥ 0 and v < 0, this metric assumes the form

ds2 = 2 du dv − (1 − u)2 dx2 − (1 + u)2 dy2

Find the Christoffel symbols of the ﬁrst kind (using (4.15) for this metric.

v u Region 4

Region 2

Region 3

Time

u = 0

Region 1

v = 0

Fig. 4-1. The division of spacetime for the Kahn-Penrose solution used in the study of gravitational waves. We consider the metric in region 2.