lots of calculations in general relativity - mono.net · 3.2.1 spherical polar coordinates ... 5.2...

Lots of Calculations in General Relativity

Susan Larsen Tuesday, February 03, 2015

http://physicssusan.mono.net/9035/General%20Relativity

Contents

1 Introduction ............................................................................................................................................... 6

2 Special relativity......................................................................................................................................... 7

2.1 How to calculate𝚫𝒔𝟐: Example ......................................................................................................... 7

2.2 The four velocity and the four force: 𝒖𝒂𝑲𝒂 = 𝟎 ............................................................................. 7

3 Metric and Vector Transformations. ......................................................................................................... 8

3.1 Flat Minkowski space: ....................................................................................................................... 8

3.2 Other realizations of the flat space: .................................................................................................. 8

3.2.1 Spherical polar coordinates ....................................................................................................... 8

3.2.2 Flat space with a singularity ...................................................................................................... 9

3.2.3 Coordinate transformations .................................................................................................... 10

3.2.4 Flat space in two dimensions .................................................................................................. 11

3.2.5 The Penrose Diagram for Flat Space ........................................................................................ 11

3.3 The line-element and metric of an ellipsoid: ................................................................................... 12

3.4 The signature of a metric................................................................................................................. 12

3.5 Three-dimensional flat space in spherical coordinates and vector transformation ....................... 12

3.6 Static Weak Field Metric ................................................................................................................. 13

3.6.1 Rates of Emission and Reception............................................................................................. 13

3.7 Local inertial frames ........................................................................................................................ 14

3.7.1 The metric of a Sphere at the North Pole. ............................................................................... 14

3.8 Length, Area, Volume and Four-Volume for Diagonal Metrics ....................................................... 16

3.8.1 Area and Volume Elements of a Sphere .................................................................................. 16

3.8.2 Distance, Area and Volume in the Curved Space of a Constant Density Spherical Star or a

Homogenous Closed Universe ................................................................................................................. 17

3.8.3 Distance, Area, Volume and four-volume of a metric ............................................................. 19

3.8.4 The dimensions of a peanut .................................................................................................... 20

3.8.5 The dimensions of an egg ........................................................................................................ 21

3.8.6 Length and volume of the Schwarzschild geometry ............................................................... 21

3.8.7 Volume in the Wormhole geometry ........................................................................................ 22

4 Tensor Calculus ........................................................................................................................................ 22

4.1 Christoffel symbols. ......................................................................................................................... 22

4.1.1 𝚪𝒂𝒃𝒂 and 𝚪𝒂𝒂𝒃 in a diagonal metric ..................................................................................... 22





4.1.2 Find the Christoffel symbols of the 2-sphere with radius 𝒂 .................................................... 23

4.1.3 Find the Christoffel symbols of the Kahn-Penrose metric (Colliding gravitational waves) ..... 23

4.2 Alternative solution: Show that 𝛁𝒄𝒈𝒂𝒃 = 𝟎 .................................................................................. 24

4.3 One-forms. ....................................................................................................................................... 24

4.3.1 One-forms: why 𝒅𝟐 = 𝟎 .......................................................................................................... 24

4.3.2 The exterior derivative of a one-form. .................................................................................... 24

4.4 The geodesic equation. ................................................................................................................... 25

4.4.1 Find the geodesic equations for cylindrical coordinates ........................................................ 25

4.4.2 Use the geodesic equations to find the Christoffel symbols for the Rindler metric. .............. 26

4.4.3 Geodesics Equations of the plane in polar coordinates .......................................................... 27

4.4.4 Equations for geodesics in a Wormhole Geometry ................................................................. 28

4.4.5 New - Geodesics and Christoffel symbols of the Schwarzschild metric with 𝜽 = 𝝅𝟐 ............ 29

4.5 Solving the geodesic equation ......................................................................................................... 30

4.5.1 The travel time through a wormhole ...................................................................................... 30

4.5.2 Geodesics in the Plane Using Polar Coordinates. .................................................................... 32

4.5.3 NEW - Geodesics Equations of the plane in Cartesian coordinates. ....................................... 33

4.5.4 New - Show that the great circle is a solution of the geodesic equation of a two-dimensional

sphere. 33

4.5.5 NEW - Find all the time-like geodesic 𝑋(𝑇) of the Flat Space metric in two dimensions 𝑑𝑆2 =

−𝑋2𝑑𝑇2 + 𝑑𝑋2: ..................................................................................................................................... 35

4.6 Killing Vectors .................................................................................................................................. 37

4.6.1 Show that if the Lie derivative of the metric tensor with respect to vector X vanishes

(𝑳𝑿𝒈𝒂𝒃 = 𝟎), the vector X satisfies the Killing equation. - Alternative version ................................... 37

4.6.2 Prove that 𝛁𝒃𝛁𝒂𝑿𝒃 = 𝑹𝒂𝒄𝑿𝒄 ............................................................................................... 37

4.6.3 Constructing a Conserved Current with Killing Vectors – Alternative version: ....................... 37

4.6.4 Given a Killing vector 𝑿 the Ricci scalar satisfies𝑿𝒄𝛁𝒄𝑹 = 𝟎: ................................................ 37

4.7 The Riemann tensor ........................................................................................................................ 38

4.7.1 Independent elements in the Riemann, Ricci and Weyl tensor .............................................. 40

4.7.2 Compute the components of the Riemann tensor for the unit 2-sphere ............................... 40

4.8 Show that the Ricci scalar 𝑹 = 𝟐 for the unit 2-sphere .................................................................. 41

4.9 Proof: if a space is conformally flat, i.e. 𝒈𝒂𝒃𝒙 = 𝒇𝒙𝜼𝒂𝒃 the Weyl tensor vanishes ..................... 42

4.10 The three dimensional flat space in spherical polar coordinates.................................................... 46

4.10.1 Calculate the Christoffel symbols of the three dimensional flat space in spherical polar

coordinates .............................................................................................................................................. 46





4.10.2 The Riemann tensor of the three dimensional flat space in spherical polar coordinates ...... 47

4.10.3 A Lie derivative in the three dimensional flat space in spherical polar coordinates ............... 47

4.11 The Ricci scalar of the Penrose Kahn metric ................................................................................... 48

4.12 A metric example 1: 𝒅𝒔𝟐 = 𝒚𝟐𝐬𝐢𝐧𝒙𝒅𝒙𝟐 + 𝒙𝟐𝐭𝐚𝐧𝒚𝒅𝒚𝟐 .............................................................. 49

4.12.1 The Christoffel symbols of a metric example .......................................................................... 49

4.12.2 The Ricci scalar of a metric example ....................................................................................... 49

4.13 Calculate the Christoffel symbols for a metric example 2: 𝒅𝒔𝟐 = 𝒅𝝍𝟐 + 𝐬𝐢𝐧𝐡𝟐𝝍 𝒅𝜽𝟐 +

𝐬𝐢𝐧𝐡𝟐𝝍𝐬𝐢𝐧𝟐𝜽𝒅𝝓𝟐 ..................................................................................................................................... 50

4.14 A metric example 3: 𝒅𝒔𝟐 = 𝒖𝟐 + 𝝂𝟐𝒅𝒖𝟐 + 𝒖𝟐 + 𝝂𝟐𝒅𝝂𝟐 + 𝒖𝟐𝝂𝟐𝒅𝜽𝟐 ...................................... 51

4.14.1 Calculate the Christoffel symbols for a metric example.......................................................... 51

4.14.2 Calculate the Riemann tensor of metric example ................................................................... 52

4.14.3 Calculate the Riemann tensor of metric example – Alternative version ................................ 53

5 Cartan’s Structure Equations ................................................................................................................... 53

5.1 Ricci rotation coefficients for the Tolman-Bondi- de Sitter metric (Spherical dust with a

cosmological constant) ................................................................................................................................ 53

5.2 The curvature two forms and the Riemann tensor ......................................................................... 56

5.3 Find the Ricci scalar using Cartan’s structure equations of the 2-sphere ....................................... 56

5.4 The three dimensional flat space in spherical polar coordinates.................................................... 57

5.4.1 Ricci rotation coefficients of the three dimensional flat space in spherical polar coordinates

57

5.4.2 Transformation of the Ricci rotation coefficients 𝚪 𝒃𝒄𝒂 into the Christoffel symbols

𝚪 𝐛𝐜𝒂 of the three dimensional flat space in spherical polar coordinates .......................................... 58

5.5 Ricci rotation coefficients of the Rindler metric .............................................................................. 59

5.6 The Einstein tensor for the Tolman-Bondi- de Sitter metric ........................................................... 59

5.7 Calculate the Ricci rotation coefficients for a metric example 3: 𝒅𝒔𝟐 = 𝒅𝝍𝟐 + 𝐬𝐢𝐧𝐡𝟐𝝍 𝒅𝜽𝟐 +

𝐬𝐢𝐧𝐡𝟐𝝍𝐬𝐢𝐧𝟐𝜽𝒅𝝓𝟐 ..................................................................................................................................... 63

6 The Einstein Field Equations .................................................................................................................... 65

6.1 The vacuum Einstein equations ....................................................................................................... 65

6.2 The vacuum Einstein equations with a cosmological constant ....................................................... 65

6.3 General remarks on the Einstein equations with a cosmological constant .................................... 66

6.4 2+1 dimensions: Gravitational collapse of an inhomogeneous spherically symmetric dust cloud. 67

6.4.1 Find the components of the curvature tensor for the metric in 2+1 dimensions using Cartan’s

structure equations ................................................................................................................................. 67





6.4.2 Find the components of the curvature tensor for the metric in 2+1 dimensions using Cartan’s

structure equations – alternative solution .............................................................................................. 69

6.4.3 Find the components of the Einstein tensor in the coordinate basis for the metric in 2+1

dimensions............................................................................................................................................... 70

6.4.4 The Einstein equations of the metric in 2+1 dimensions. ....................................................... 72

6.5 Using the contracted Bianchi identities, prove that: 𝛁𝒃𝑮𝒂𝒃 = 𝟎 .................................................. 72

6.6 Ricci rotation coefficients, Ricci scalar and Einstein equations for a general 4-dimensional metric:

𝒅𝒔𝟐 = −𝒅𝒕𝟐 + 𝑳𝟐𝒕, 𝒓𝒅𝒓𝟐 + 𝑩𝟐𝒕, 𝒓𝒅𝝓𝟐 +𝑴𝟐𝒕, 𝒓𝒅𝒛𝟐 ........................................................................... 73

7 The Energy-Momentum Tensor .............................................................................................................. 78

7.1 Perfect Fluids – Alternative derivation ............................................................................................ 78

7.2 The Gödel metric ............................................................................................................................. 79

8 Null Tetrads and the Petrov Classification............................................................................................... 82

8.1 Construct a null tetrad for the flat space Minkowski metric........................................................... 82

8.2 The Brinkmann metric (Plane gravitational waves) ........................................................................ 84

9 The Schwarzschild Solution ..................................................................................................................... 94

9.1 The Riemann and Ricci tensor of the general Schwarzschild metric ............................................... 94

9.2 The Riemann tensor of the Schwarzschild metric ........................................................................... 97

9.3 Calculation of the scalar 𝑹𝒂𝒃𝒄𝒅𝑹𝒂𝒃𝒄𝒅 in the Schwarzschild metric ............................................ 98

9.4 Geodesics in the Schwarzschild Spacetime ..................................................................................... 98

9.5 The meaning of the integration constant: The choice of 𝟐𝒎 ......................................................... 99

9.6 Time Delay ..................................................................................................................................... 100

9.7 Use the geodesic equations to find the Christoffel symbols for the general Schwarzschild metric.

102

9.8 The Ricci tensor for the general time dependent Schwarzschild metric. ...................................... 104

9.9 The Schwarzschild metric with nonzero cosmological constant. .................................................. 108

9.9.1 The Ricci rotation coefficients and Ricci tensor for the Schwarzschild metric with nonzero

cosmological constant. .......................................................................................................................... 108

9.9.2 The general Schwarzschild metric in vacuum with a cosmological constant: The Ricci scalar

109

9.9.3 The general Schwarzschild metric in vacuum with a cosmological constant: Integration

constants 110

9.9.4 The general Schwarzschild metric in vacuum with a cosmological constant: The spatial part of

the line element. ................................................................................................................................... 111

9.9.5 The effect of the cosmological constant over the scale of the solar system ........................ 112

9.10 The Petrov type of the Schwarzschild spacetime .......................................................................... 113





9.11 The deflection of a light ray in a Schwarzschild metric with two different masses ...................... 120

9.12 The non-zero Weyl scalars of the Reissner-Nordström spacetime ............................................... 120

10 Black Holes ......................................................................................................................................... 130

10.1 The Path of a Radially Infalling Particle ......................................................................................... 130

10.2 The Schwarzschild metric in Kruskal Coordinates. ........................................................................ 133

10.3 The Kerr metric .............................................................................................................................. 136

10.3.1 The Kerr-Newman geometry ................................................................................................. 136

10.3.2 The inverse metric of the Kerr Spinning Black Hole .............................................................. 138

11 Cosmology ......................................................................................................................................... 140

11.1 Light travelling in the Universe ...................................................................................................... 140

11.2 Spaces of Positive, Negative, and Zero Curvature ......................................................................... 140

11.3 New – The critical density ............................................................................................................. 141

11.4 The Robertson-Walker metric ....................................................................................................... 142

11.4.1 Find the components of the Riemann tensor of the Robertson-Walker metric (Homogenous,

isotropic and expanding universe) using Cartan’s structure equations ................................................ 142

11.4.2 The Einstein tensor and Friedmann-equations for the Robertson Walker metric ................ 145

11.4.3 The Einstein tensor for the Robertson Walker metric – Alternative version. ....................... 147

11.5 Manipulating the Friedmann equations. ....................................................................................... 148

11.6 Parameters in an flat universe with positive cosmological constant: Starting with 𝒂𝟐 = 𝑪𝒂 +

𝚲𝟑𝒂𝟐 use a change of variables 𝒖 = 𝟐𝚲𝟑𝑪𝒂𝟑 ........................................................................................ 148

12 Gravitational Waves .......................................................................................................................... 150

12.1 Gauge transformation - The Einstein Gauge ................................................................................. 150

12.2 Plane waves ................................................................................................................................... 151

12.2.1 The Riemann tensor of a plane wave .................................................................................... 151

12.2.2 The line element of a plane wave in the Einstein gauge ....................................................... 156

12.2.3 The line element of a plane wave.......................................................................................... 157

12.2.4 The Rosen line element ......................................................................................................... 157

12.3 Colliding gravity waves - coordinate transformation .................................................................... 161

12.4 The delta – 𝜹(𝒖) and heavy-side –𝚯(𝒖) functions: prove that 𝒖𝜹𝒖 = 𝟎 .................................... 162

12.5 Impulsive gravitational wave Region III ......................................................................................... 163

12.6 Two interacting waves ................................................................................................................... 169

12.7 The Nariai spacetime ..................................................................................................................... 174





12.8 Collision of a gravitational wave with an electromagnetic wave – The non-zero spin coefficients

188

12.9 The Aichelburg-Sexl Solution – The passing of a black hole .......................................................... 192

12.10 Observations: The Future Gravitational Wave detectors. ......................................................... 192

Bibliografi ....................................................................................................................................................... 194

1 Introduction Working with GR means working with differential equations at four different levels. It can be very useful - whenever one comes across a GR calculation - to keep in mind, on which level you are working. The four levels of differential equations are:

1. The metric or line-element: 𝑑𝑠2 = 𝑔𝑎𝑏𝑑𝑥

𝑎𝑑𝑥𝑏 Example: Gravitational red shift1:

𝑑𝜏 = √1 −2𝑚

𝑟𝑑𝑡

Light emitted upward in a gravitational field, from an observer located at some inner radius 𝑟1 to an ob-server positioned at some outer radius 𝑟2

𝛼 =√1 −

2𝑚𝑟2

√1 −2𝑚𝑟1

2. Killing’s equations are conservation equations: ∇𝑏𝑋𝑎 + ∇𝑎𝑋𝑏 = 0

If you move along the direction of a Killing vector, then the metric does not change. This leads to conserved quantities: A free particle moving in a direction where the metric does not change will not fell any forces.

If 𝑋 is a Killing vector,𝑢 = (𝑑𝑡

𝑑𝜏,𝑑𝑟

𝑑𝜏,𝑑𝜃

𝑑𝜏,𝑑𝜙

𝑑𝜏) is the particle four velocity and 𝑝 is the particle four impulse,

then 𝑋 ⋅ 𝑢 = 𝑔𝑎𝑏𝑋𝑎𝑢𝑏 = 𝑐𝑜𝑛𝑠𝑡 and 𝑋 ⋅ 𝑝 = 𝑔𝑎𝑏𝑋

𝑎𝑝𝑏 = 𝑐𝑜𝑛𝑠𝑡 along a geodesic2. Translational symmetry: Whenever 𝜕σ∗𝑔𝜇𝜈 = 0 for some fixed 𝜎∗ (but for all 𝜇 and 𝜈) there will be a sym-

metry under translation along 𝑥𝜎∗3. Example: Killing vectors in the Schwarzschild metric4. The Killing vector that corresponds to the independence of the metric of 𝑡 is 𝜉 = (1,0,0,0) and of 𝜙 is 𝜂 =

(0,0,0,1). The conserved energy per unit rest mass: 𝑒 = −𝜉 ⋅ 𝑢 = −𝑔𝑎𝑏𝜉𝑎𝑢𝑏 = −𝑔𝑡𝑡 ⋅ 1 ⋅

𝑑𝑡

𝑑𝜏= −(1 −

2𝑚

𝑟)𝑑𝑡

𝑑𝜏. The conserved angular momentum per unit rest mass 𝑙 = 𝜂 ⋅ 𝑢 = 𝑔𝑎𝑏𝜂

𝑎𝑢𝑏 = 𝑔𝜙𝜙 ⋅ 1 ⋅𝑑𝜙

𝑑𝜏=

−𝑟2 sin2 𝜃𝑑𝜙

𝑑𝜏= −𝑟2

𝑑𝜙

𝑑𝜏 for 𝜃 =

𝜋

2

3. The Geodesic equation leads to equations of motion:

𝐾 =1

2𝑔𝑎𝑏��

𝑎��𝑏

1 (McMahon, p. 234) 2 (McMahon, p. 168) 3 (Carroll, 2004) 4 (McMahon, p. 220)





𝜕𝐾

𝜕𝑥𝑎 =

𝑑

𝑑𝑠(𝜕𝐾

𝜕��𝑎)

𝑑2𝑥𝑎

𝑑𝑠2+ Γ 𝑏𝑐

𝑎𝑑𝑥𝑏

𝑑𝑠

𝑑𝑥𝑐

𝑑𝑠 = 0

Example: Planetary orbits5 Manipulating the geodesic equations of the Schwarzschild metric leads to the following equation

(𝑑𝑢

𝑑𝜙)2

+ 𝑢2 =𝑘2 − 1

ℎ2+2𝑚

ℎ2𝑢 + 2𝑚𝑢3

Which can be interpreted in terms of elliptic functions, 𝑢 =1

𝑟, and h and k are constants of integration.

4. The Einstein equations are equations describing the spacetime.

𝐺𝑎𝑏 = 𝑅𝑎𝑏 −1

2𝑔𝑎𝑏𝑅

8𝜋𝐺𝑇𝑎𝑏 = 𝐺𝑎𝑏 + 𝑔𝑎𝑏Λ In case of a cosmological constant6:

𝑅𝑎𝑏 = ±𝑔𝑎𝑏Λ 𝑅�� = ±𝜂��Λ

If 𝑛 = 4, 𝑅𝑎𝑏𝑐𝑑 has twenty independent component – ten of which are given by 𝑅𝑎𝑏 and the remaining ten by the Weyl tensor7. Example: The Friedmann equations A homogenous, isotropic and expanding universe described by the Robertson-Walker metric8, in this case the Einstein equations becomes the Friedmann equations:

8𝜋𝜌 =3

𝑎2(𝑘 + ��2) + Λ

−8𝜋𝑃 = 2��

𝑎+1

𝑎2(𝑘 + ��2) + Λ

2 Special relativity

2.1 9How to calculate(𝚫𝒔)𝟐: Example In flat space calculate (Δ𝑠)2 for the following pair of events: 𝐸1 = (−1,3,2,4) and 𝐸2 = (4,0,−1,1)

(Δ𝑠)2 = (Δ𝑡)2 − (Δ𝑥)2 − (Δ𝑦)2 − (Δ𝑧)2 (1.11) = (−1 − 4)2 − (3 − 0)2 − (2 − (−1))

2− (4 − 1)2

= 52 − 32 − 32 − 32 = −2

2.2 10The four velocity and the four force: 𝒖𝒂𝑲𝒂 = 𝟎

The four velocity 𝑢𝑎 =𝑑𝑥𝑎

𝑑𝜏= (

𝑑𝑡

𝑑𝜏,𝑑𝑥

𝑑𝜏,𝑑𝑦

𝑑𝜏,𝑑𝑧

𝑑𝜏), the four impulse 𝑝𝑎 = 𝑚0𝑢

𝑎, the four force 𝐾𝑎 =𝑑𝑝𝑎

𝑑𝜏=

𝑚0𝑑𝑢𝑎

𝑑𝜏. Because 𝑢𝑎𝐾

𝑎 = 𝑢𝑎𝐾𝑎 we can calculate

𝑢𝑎𝐾𝑎 =

1

2(𝑢𝑎𝐾

𝑎 + 𝑢𝑎𝐾𝑎) =1

2(𝑚0𝑢𝑎

𝑑𝑢𝑎

𝑑𝜏+𝑚0𝑢

𝑎𝑑𝑢𝑎𝑑𝜏

) =1

2𝑚0

𝑑(𝑢𝑎𝑢𝑎)

𝑑𝜏

5 (A.S.Eddington, pp. 85-86) 6 (McMahon, p. 138) 7 (d'Inverno, p. 87) 8 (McMahon, p. 161) 9 (McMahon, 2006, p. 323), final exam 1. The answer to FE-1 is (c) 10 (McMahon, 2006, p. 324), final exam 4, and the answer to FE-4 is (a)





𝑢𝑎𝑢𝑎 is an invariant and

1

2𝑚0

𝑑(𝑢𝑎𝑢𝑎)

𝑑𝜏= 𝑢𝑎𝐾

𝑎 = 0

3 Metric and Vector Transformations.

3.1 11Flat Minkowski space: Flat Minkowski spacetime is the mathematical setting in which Einstein’s special theory of relativity is most conveniently formulated. In Cartesian coordinates with 𝑐 = 1 the line element is

𝑑𝑠2 = −𝑑𝑡2 + 𝑑𝑥2 + 𝑑𝑦2 + 𝑑𝑧2 and the metric

𝑔𝜇𝜈 = {

−1 0 0 00 1 0 00 0 1 00 0 0 1

}

𝑑𝑠2 < 0, time-like, inside the light cone 𝑑𝑠 = 0: null-vector, on the light cone 𝑑𝑠2 > 0, space-like, outside the light cone.

3.2 Other realizations of the flat space:

3.2.1 12Spherical polar coordinates The spherical part of the metric can be transformed into spherical polar coordinates by

𝑥 = 𝑟 sin𝜃 cos𝜙 (7.2) 𝑦 = 𝑟 sin𝜃 sin𝜙

𝑧 = 𝑟 cos 𝜃 ⇒ 𝑑𝑥 = sin𝜃 cos𝜙 𝑑𝑟 + 𝑟 cos 𝜃 cos𝜙 𝑑𝜃 − 𝑟 sin 𝜃 sin𝜙 𝑑𝜙

𝑑𝑦 = sin𝜃 sin𝜙 𝑑𝑟 + 𝑟 cos 𝜃 sin𝜙 𝑑𝜃 + 𝑟 sin𝜃 cos𝜙 𝑑𝜙 𝑑𝑧 = cos 𝜃 𝑑𝑟 − 𝑟 sin𝜃 𝑑𝜃 (7.3)

⇒ 𝑑𝑥2 = (sin𝜃 cos𝜙 𝑑𝑟 + 𝑟 cos 𝜃 cos𝜙 𝑑𝜃 − 𝑟 sin 𝜃 sin𝜙 𝑑𝜙)2 = sin2 𝜃 cos2𝜙𝑑𝑟2 + 2𝑟 sin 𝜃 cos 𝜃 cos2𝜙𝑑𝑟𝑑𝜃 − 2𝑟 sin2 𝜃 sin𝜙 cos𝜙 𝑑𝑟𝑑𝜙

+ 𝑟2 cos2 𝜃 cos2𝜙𝑑𝜃2 − 2𝑟2 sin𝜃 cos 𝜃 sin𝜙 cos𝜙 𝑑𝜃𝑑𝜙+ 𝑟2 sin2 𝜃 sin2𝜙𝑑𝜙2

𝑑𝑦2 = (sin𝜃 sin𝜙 𝑑𝑟 + 𝑟 cos 𝜃 sin𝜙 𝑑𝜃 + 𝑟 sin𝜃 cos𝜙 𝑑𝜙)2 = sin2 𝜃 sin2𝜙𝑑𝑟2 + 2𝑟 sin𝜃 cos 𝜃 sin2𝜙𝑑𝑟𝑑𝜃 + 2𝑟 sin2 𝜃 sin𝜙 cos𝜙 𝑑𝑟𝑑𝜙

+ 𝑟2 cos2 𝜃 sin2𝜙𝑑𝜃2 + 2𝑟2 cos 𝜃 sin𝜃 cos𝜙 sin𝜙 𝑑𝜃𝑑𝜙+ 𝑟2 sin2 𝜃 cos2𝜙𝑑𝜙2

𝑑𝑧2 = (cos 𝜃 𝑑𝑟 − 𝑟 sin𝜃 𝑑𝜃)2 = cos2 𝜃 𝑑𝑟2 − 2𝑟 cos 𝜃 sin𝜃 𝑑𝑟𝑑𝜃 + 𝑟2 sin2 𝜃 𝑑𝜃2 ⇒ 𝑑𝑥2 + 𝑑𝑦2 + 𝑑𝑧2

= sin2 𝜃 cos2𝜙𝑑𝑟2 + 2𝑟 sin 𝜃 cos 𝜃 cos2𝜙𝑑𝑟𝑑𝜃 − 2𝑟 sin2 𝜃 sin𝜙 cos𝜙 𝑑𝑟𝑑𝜙+ 𝑟2 cos2 𝜃 cos2𝜙𝑑𝜃2 − 2𝑟2 sin𝜃 cos 𝜃 sin𝜙 cos𝜙 𝑑𝜃𝑑𝜙+ 𝑟2 sin2 𝜃 sin2𝜙𝑑𝜙2 + sin2 𝜃 sin2𝜙𝑑𝑟2 + 2𝑟 sin 𝜃 cos 𝜃 sin2𝜙𝑑𝑟𝑑𝜃+ 2𝑟 sin2 𝜃 sin𝜙 cos𝜙 𝑑𝑟𝑑𝜙 + 𝑟2 cos2 𝜃 sin2𝜙𝑑𝜃2

+ 2𝑟2 cos 𝜃 sin 𝜃 cos𝜙 sin𝜙 𝑑𝜃𝑑𝜙 + 𝑟2 sin2 𝜃 cos2𝜙𝑑𝜙2 + cos2 𝜃 𝑑𝑟2

− 2𝑟 cos 𝜃 sin𝜃 𝑑𝑟𝑑𝜃 + 𝑟2 sin2 𝜃 𝑑𝜃2

11 (McMahon, p. 186) 12 (Hartle, 2003, p. 135)





= sin2 𝜃 cos2𝜙𝑑𝑟2 + 2𝑟 sin 𝜃 cos 𝜃 cos2𝜙𝑑𝑟𝑑𝜃 + 𝑟2 cos2 𝜃 cos2𝜙𝑑𝜃2

+ 𝑟2 sin2 𝜃 sin2𝜙𝑑𝜙2 + sin2 𝜃 sin2𝜙𝑑𝑟2 + 2𝑟 sin 𝜃 cos 𝜃 sin2𝜙𝑑𝑟𝑑𝜃+ 𝑟2 cos2 𝜃 sin2𝜙𝑑𝜃2 + 𝑟2 sin2 𝜃 cos2𝜙𝑑𝜙2 + cos2 𝜃 𝑑𝑟2

− 2𝑟 cos 𝜃 sin𝜃 𝑑𝑟𝑑𝜃 + 𝑟2 sin2 𝜃 𝑑𝜃2 = (sin2 𝜃 cos2𝜙 + sin2 𝜃 sin2𝜙 + cos2 𝜃)𝑑𝑟2

+ (2𝑟 sin 𝜃 cos 𝜃 cos2𝜙 − 2𝑟 cos 𝜃 sin 𝜃 + 2𝑟 sin𝜃 cos𝜃 sin2𝜙)𝑑𝑟𝑑𝜃+ (𝑟2 cos2 𝜃 cos2𝜙 + 𝑟2 cos2 𝜃 sin2𝜙 + 𝑟2 sin2 𝜃)𝑑𝜃2

+ (𝑟2 sin2 𝜃 sin2𝜙 + 𝑟2 sin2 𝜃 cos2𝜙)𝑑𝜙2 = (sin2 𝜃 (cos2𝜙 + sin2𝜙) + cos2 𝜃)𝑑𝑟2

+ (2𝑟 sin 𝜃 cos 𝜃 (cos2𝜙 + sin2𝜙) − 2𝑟 cos𝜃 sin𝜃)𝑑𝑟𝑑𝜃+ 𝑟2(cos2 𝜃 (cos2𝜙 + sin2𝜙) + sin2 𝜃)𝑑𝜃2

+ 𝑟2 sin2 𝜃 (sin2𝜙 + cos2𝜙)𝑑𝜙2 = (sin2 𝜃 + cos2 𝜃)𝑑𝑟2 + (2𝑟 sin𝜃 cos𝜃 − 2𝑟 cos𝜃 sin𝜃)𝑑𝑟𝑑𝜃

+ 𝑟2(cos2 𝜃 + sin2 𝜃)𝑑𝜃2 + 𝑟2 sin2 𝜃 𝑑𝜙2 = 𝑑𝑟2 + 𝑟2𝑑𝜃2 + 𝑟2 sin2 𝜃 𝑑𝜙2 The transformed line element is

𝑑𝑠2 = −𝑑𝑡2 + 𝑑𝑟2 + 𝑟2𝑑𝜃2 + 𝑟2 sin2 𝜃 𝑑𝜙2 (7.4)

3.2.2 13Flat space with a singularity Look at the line element of the two-dimensional plane in polar coordinates (𝜃 = 0)

𝑑𝑆2 = 𝑑𝑟2 + 𝑟2𝑑𝜙2 (7.6) and make the transformation, for some constant 𝑎

𝑟 =𝑎2

𝑟′

⇒ 𝑑𝑟 = 𝑑 (𝑎2

𝑟′) = −

𝑎2

𝑟′2𝑑𝑟′

⇒ 𝑑𝑟2 =𝑎4

𝑟′4𝑑𝑟′2

⇒ 𝑑𝑆2 =𝑎4

𝑟′4𝑑𝑟′2 + (

𝑎2

𝑟′)

2

𝑑𝜙2

=𝑎4

𝑟′4(𝑑𝑟′2 + 𝑟′2𝑑𝜙2) (7.7)

This line element blows up at 𝑟′ = 0. Not because something physically interesting happens here, but

simply because the coordinate transformation 𝑟 =𝑎2

𝑟′ has mapped all the points at 𝑟 → ∞ into 𝑟′ = 0.

14We can show that that the distance between 𝑟′ = 0 and a point with any finite value of 𝑟′ is infinite, which corresponds to the distance between some finite value of 𝑟 and 𝑟 → ∞:

∫𝑑𝑆 = ∫ √𝑎4

𝑟′4(𝑑𝑟′2 + 𝑟′2𝑑𝜙2)

𝑟′

0

= 𝑎2∫ 𝑑𝑟′√1

𝑟′4(1 + 𝑟′2 (

𝑑𝜙

𝑑𝑟′)2

)𝑟′

0

= 𝑎2∫

1

𝑟′2𝑑𝑟′

𝑟′

0

𝑑𝜙

𝑑𝑟′= 0

13 (Hartle, 2003, p. 136) 14 (Hartle, 2003, p. 163), problem 7.1





= [−

𝑎2

𝑟′]0

𝑟′

→ ∞

3.2.3 15Coordinate transformations The following line element corresponds to flat spacetime

𝑑𝑠2 = −𝑑𝑡2 + 2𝑑𝑥𝑑𝑡 + 𝑑𝑦2 + 𝑑𝑧2 with the metric

𝑔𝑎𝑏 = {

−1 1 0 01 0 0 00 0 1 00 0 0 1

}

Find a coordinate transformation that puts the line element in the usual flat space form 𝑑𝑠2 = −𝑑𝑡′

2+ 𝑑𝑥′2 + 𝑑𝑦′

2+ 𝑑𝑧′

2

We want to find the matrix, Λ𝑎𝑏𝑎′𝑏′ that transforms 𝑔𝑎′𝑏′ into 𝑔𝑎𝑏

𝑔𝑎𝑏 = Λ𝑎𝑏𝑎′𝑏′𝑔𝑎′𝑏′

we have

Λ𝑎𝑏𝑎′𝑏′ = Λ𝑎𝑏

𝑎′𝑏′𝑔𝑎′𝑏′𝑔𝑎′𝑏′

= 𝑔𝑎𝑏𝑔𝑎′𝑏′

= {

−1 1 0 01 0 0 00 0 1 00 0 0 1

}{

−1 0 0 00 1 0 00 0 1 00 0 0 1

}

= {

1 1 0 0−1 0 0 00 0 1 00 0 0 1

}

⇒ 𝑡 = 𝑡′ − 𝑥′ 𝑥 = 𝑡′ 𝑦 = 𝑦′ 𝑧 = 𝑧′

⇒ 𝑑𝑡 = 𝑑𝑡′ − 𝑑𝑥′ 𝑑𝑥 = 𝑑𝑡′ 𝑑𝑦 = 𝑑𝑦′ 𝑑𝑧 = 𝑑𝑧′

⇒ 𝑑𝑠2 = −𝑑𝑡2 + 2𝑑𝑥𝑑𝑡 + 𝑑𝑦2 + 𝑑𝑧2 = −(𝑑𝑡′ − 𝑑𝑥′)2 + 2(𝑑𝑡′)(𝑑𝑡′ − 𝑑𝑥′) + 𝑑𝑦′

2+ 𝑑𝑧′

2

= −𝑑𝑡′2 − 𝑑𝑥′2 + 2𝑑𝑡′𝑑𝑥′ + 2𝑑𝑡′2 − 2𝑑𝑡′𝑑𝑥′ + 𝑑𝑦′2+ 𝑑𝑧′

2

= −𝑑𝑥′2+ 𝑑𝑡′2 + 𝑑𝑦′

2+ 𝑑𝑧′

2

We check 𝑔𝑎𝑏 = Λ𝑎𝑏

𝑎′𝑏′𝑔𝑎′𝑏′

= {

1 1 0 0−1 0 0 00 0 1 00 0 0 1

}{

−1 0 0 00 1 0 00 0 1 00 0 0 1

}

= {

1 1 0 0−1 0 0 00 0 1 00 0 0 1

}

15 (Hartle, 2003, p. 164), problem 7.2





3.2.4 16Flat space in two dimensions Yet another realization of flat space in two dimensions is the line element

𝑑𝑠2 = −𝑋2𝑑𝑇2 + 𝑑𝑋2 (7.20) This can be found from the coordinate transformation

𝑡 = 𝑋 sinh(𝑇) 𝑥 = 𝑋 cosh(𝑇)

⇒ 𝑑𝑡 = sinh(𝑇)𝑑𝑋 + 𝑋 cosh(𝑇) 𝑑𝑇 𝑑𝑥 = cosh(𝑇) 𝑑𝑋 + 𝑋 sinh(𝑇) 𝑑𝑇

⇒ 𝑑𝑠2 = −𝑑𝑡2 + 𝑑𝑥2 = −(sinh(𝑇) 𝑑𝑋 + 𝑋 cosh(𝑇) 𝑑𝑇)2 + (cosh(𝑇) 𝑑𝑋 + 𝑋 sinh(𝑇) 𝑑𝑇)2 = −sinh2(𝑇) 𝑑𝑋2 − 𝑋2 cosh2(𝑇) 𝑑𝑇2 − 2𝑋 sinh(𝑇) cosh(𝑇)𝑑𝑋 𝑑𝑇 + cosh2(𝑇) 𝑑𝑋2

+ 𝑋2 sinh2(𝑇) 𝑑𝑇2 + 2𝑋 cosh(𝑇) sinh(𝑇) 𝑑𝑋𝑑𝑇 = (cosh2(𝑇) − sinh2(𝑇))𝑑𝑋2 − 𝑋2(cosh2(𝑇) − sinh2(𝑇))𝑑𝑇2 = 𝑑𝑋2 − 𝑋2𝑑𝑇2

3.2.5 17The Penrose Diagram for Flat Space A Penrose diagram is a method to map the infinite coordinates such as 𝑡, with the range −∞ < 𝑡 < +∞, and 𝑟, with the range 0 < 𝑟 < +∞, into to coordinates with finite ranges. Begin with the flat space line element in spherical polar coordinates

𝑑𝑠2 = −𝑑𝑡2 + 𝑑𝑟2 + 𝑟2𝑑𝜃2 + 𝑟2 sin2 𝜃 𝑑𝜙2 (7.4) and replace 𝑡 and 𝑟 by the coordinates

𝑢 = 𝑡 − 𝑟 𝑣 = 𝑡 + 𝑟

⇒ 𝑡 =1

2(𝑢 + 𝑣) 𝑟 =

1

2(𝑢 − 𝑣)

⇒ 𝑑𝑡 =1

2(𝑑𝑢 + 𝑑𝑣) 𝑑𝑟 =

1

2(𝑑𝑢 − 𝑑𝑣)

⇒ 𝑑𝑠2 = −𝑑𝑡2 + 𝑑𝑟2 + 𝑟2𝑑𝜃2 + 𝑟2 sin2 𝜃 𝑑𝜙2

= −(1

2(𝑑𝑢 + 𝑑𝑣))

2

+ (1

2(𝑑𝑢 − 𝑑𝑣))

2

+ (1

2(𝑢 − 𝑣))

2

(𝑑𝜃2 + sin2 𝜃 𝑑𝜙2)

= −𝑑𝑢𝑑𝑣 +

1

4((𝑢 − 𝑣))

2(𝑑𝜃2 + sin2 𝜃 𝑑𝜙2)

The (𝑢, 𝑣) axes are rotated with respect to the (𝑡, 𝑟) axes by 45 . Radial light rays travel on lines of either constant 𝑢 or constant 𝑣. Remember radial light rays has constant 𝜃 and 𝜙 and 𝑑𝑠2 = 0, which leaves us with 𝑑𝑢𝑑𝑣 = 0. Make a further transformation of 𝑢 and 𝑣 to new coordinates 𝑢′ and 𝑣′.

𝑢′ = tan−1 𝑢 𝑣′ = tan−1 𝑣 ⇒ 𝑢 = tan𝑢′ 𝑣 = tan𝑣′ ⇒ 𝑑𝑢 = (1 + tan2 𝑢′)𝑑𝑢′ 𝑑𝑣 = (1 + tan2 𝑣′)𝑑𝑣′

⇒ 𝑑𝑠2 = −𝑑𝑢𝑑𝑣 +1

4((𝑢 − 𝑣))


= −(1 + tan2 𝑢′)(1 + tan2 𝑣′)𝑑𝑢′𝑑𝑣′ +

1

4((tan𝑢′ − tan 𝑣′))


Map these coordinates into a (𝑡′, 𝑟′) diagram, where 𝑢′ = tan−1 𝑢 = 𝑡′ − 𝑟′ 𝑣′ = tan−1 𝑣 = 𝑡′ + 𝑟′

Because tan−1 𝑥 lies between −𝜋

2 and +

𝜋

2 the ranges for (𝑢′, 𝑣′) and (𝑡′, 𝑟′) are finite. This is another

example of how the infinite coordinates (𝑡, 𝑟) is mapped into a finite region.

16 (Hartle, 2003, p. 143), example 7.3 17 (Hartle, 2003, p. 137), Box 7.1





3.3 18The line-element and metric of an ellipsoid: The line-element of an ellipsoid in Cartesian coordinates

𝑑𝑠2 = 𝑎𝑑𝑥2 + 𝑏𝑑𝑦2 + 𝑐𝑑𝑧2

We use the parameterization 𝑥 = cos𝜙 sin 𝜃 𝑦 = sin𝜙 sin𝜃 𝑧 = cos 𝜃

With 0 ≤ 𝜙 ≤ 2𝜋 in the 𝑥𝑦-plane and 0 ≤ 𝜃 ≤ 𝜋 where the 𝑧-axis corresponds to 𝜃 = 0.

𝑑𝑥 = 𝑎(cos𝜙 cos 𝜃 𝑑𝜃 − sin𝜙 sin𝜃 𝑑𝜙) 𝑑𝑦 = 𝑏(sin𝜙 cos𝜃 𝑑𝜃 + cos𝜙 sin𝜃 𝑑𝜙) 𝑑𝑧 = −𝑐 sin𝜃 𝑑𝜃 𝑑𝑥2 = 𝑎2(cos𝜙 cos 𝜃 𝑑𝜃 − sin𝜙 sin 𝜃 𝑑𝜙)2

= 𝑎2[cos2𝜙 cos2 𝜃 𝑑𝜃2 + sin2𝜙 sin2 𝜃 𝑑𝜙2 − 2cos𝜙 cos 𝜃 sin𝜙 sin 𝜃 𝑑𝜃𝑑𝜙] 𝑑𝑦2 = 𝑏[sin2𝜙 cos2 𝜃 𝑑𝜃2 + cos2𝜙 sin2 𝜃 𝑑𝜙2 + 2 sin𝜙 cos𝜃 cos𝜙 sin 𝜃 𝑑𝜃𝑑𝜙] 𝑑𝑧2 = 𝑐2 sin2 𝜃 𝑑𝜃2

Collecting the results in terms of 𝑑𝜃2, 𝑑𝜙2 and 𝑑𝜃𝑑𝜙 we get the line element 𝑑𝑠2 = [cos2 𝜃 (𝑎2 cos2𝜙 + 𝑏2 sin2𝜙) + 𝑐2 sin2 𝜃]𝑑𝜃2 + sin2 𝜃 (𝑎2 sin2𝜙 + 𝑏2 cos2𝜙)𝑑𝜙2

+ 2(𝑏2 − 𝑎2)(cos𝜙 cos 𝜃 sin𝜙 sin 𝜃)𝑑𝜃𝑑𝜙 and the metric tensor

𝑔𝑎𝑏 = {cos2 𝜃 (𝑎2 cos2𝜙 + 𝑏2 sin2𝜙) + 𝑐2 sin2 𝜃 (𝑏2 − 𝑎2)(cos𝜙 cos𝜃 sin𝜙 sin𝜃)

(𝑏2 − 𝑎2)(cos𝜙 cos 𝜃 sin𝜙 sin 𝜃) sin2 𝜃 (𝑎2 sin2𝜙 + 𝑏2 cos2𝜙)}

As a funny observation you can now calculate the line-element and metric tensor of an idealized egg. For

an idealized egg we can choose 𝑎 = 𝑏 =1

2𝑐

⇒ 𝑑𝑠2 = 𝑎2[(cos2 𝜃 + 4 sin2 𝜃)𝑑𝜃2 + sin2 𝜃 𝑑𝜙2]

𝑔𝑎𝑏 = 𝑎2 {cos2 𝜃 + 4 sin2 𝜃 0

0 sin2 𝜃}

3.4 19The signature of a metric. Here we are going to investigate what happens to various quantities when a metric,𝑔𝑎𝑏,changes signa-ture.

Christoffel symbols Γ 𝑏𝑐𝑎 =

1

2𝑔𝑎𝑑 (

𝜕𝑔𝑑𝑏𝜕𝑥𝑐

+𝜕𝑔𝑑𝑐𝜕𝑥𝑏

−𝜕𝑔𝑏𝑐𝜕𝑥𝑑

) No change

Riemann tensor 𝑅 𝑏𝑐𝑑𝑎 = 𝜕𝑐Γ 𝑏𝑑

𝑎 − 𝜕𝑑Γ 𝑏𝑐𝑎 + Γ 𝑏𝑑

𝑒 Γ 𝑒𝑐𝑎 − Γ 𝑏𝑐

𝑒 Γ 𝑒𝑑𝑎 No change

Ricci tensor 𝑅𝑎𝑏 = 𝑅 𝑎𝑐𝑏𝑐 No change

Ricci scalar 𝑅 = 𝑔𝑎𝑏𝑅𝑎𝑏 Changes sign

Einstein tensor 𝐺𝑎𝑏 = 𝑅𝑎𝑏 −1

2𝑔𝑎𝑏𝑅 No change

Energy tensor 8𝜋𝐺𝑇𝑎𝑏 = 𝐺𝑎𝑏 No change

Cosmological constant 𝐺𝑎𝑏 = 𝑅𝑎𝑏 −1

2𝑔𝑎𝑏𝑅 + 𝑔𝑎𝑏Λ Changes sign

3.5 20Three-dimensional flat space in spherical coordinates and vector trans-

formation

The line element 𝑑𝑠2 = 𝑑𝑟2 + 𝑟2𝑑𝜃2 + 𝑟2 sin2 𝜃 𝑑𝜙2

18 (Hartle, 2003, p. 29), problem 2-8. 19 (McMahon, 2006, p. 36) 20 (McMahon, 2006, p. 46), Quiz 2-5, the answer to the quiz 2-5 is (c)





The metric tensor: 𝑔𝑎𝑏 = {

1𝑟2

𝑟2 sin2 𝜃

}

Given 𝑋𝑎 = (𝑟,1

rsin𝜃 ,

1

cos2 𝜃) calculate 𝑋𝑎 .

𝑋𝑎 = 𝑔𝑎𝑏𝑋𝑎 (2.17)

𝑋𝑟 = 𝑔𝑟𝑟𝑋𝑟 = (1 )(𝑟) = 𝑟

𝑋𝜃 = 𝑔𝜃𝜃𝑋

𝜃 = (𝑟2) (1

rsin𝜃 ) =

𝑟

sin𝜃

𝑋𝜙 = 𝑔𝜙𝜙𝑋𝜙 = (𝑟2 sin2 𝜃) (

1

cos2 𝜃) = 𝑟2 tan2 𝜃

3.6 21 Static Weak Field Metric In this model the flat spacetime geometry of special relativity is modified to introduce a slight curvature that will explain geometrically the behavior of clocks. Further, the world lines of extremal proper time in this modified geometry will reproduce the predictions of Newtonian mechanics for motion in a gravita-

tional potential for nonrelativistic velocities. Φ(𝑥𝑖) is a function of position satisfying the Newtonian field

equation22 ∇2Φ(��) = 4𝜋𝐺𝜇(��)and assumed to vanish at infinity. For example outside Earth Φ(𝑟) =

−𝐺𝑀⊕

𝑟. This line element is predicted by general relativity for small curvatures produced by time-inde-

pendent weak sources, and it is a good approximation to the curved spacetime geometry produced by the Sun.

𝑑𝑠2 = −(1 +2Φ(𝑥𝑖)

𝑐2) (𝑐𝑑𝑡)2 + (1 −

2Φ(𝑥𝑖)

𝑐2) (𝑑𝑥2 + 𝑑𝑦2 + 𝑑𝑧2) (6.20)

3.6.1 23Rates of Emission and Reception

We look at a system where two light signals are emitted in a system A, described by a world line (𝑐𝑡, 𝑥𝐴), with a proper time separation Δ𝜏𝐴. We want to predict: what is the proper time separation Δ𝜏𝐵 in a system

B, described by a world line (𝑐𝑡, 𝑥𝐵) in a static weak field limit where Φ

𝑐2≪ 1. This implies 𝑑𝑥 = 0, Φ(𝑥𝑖) =

Φ(𝑥𝑖, 0,0) = Φ𝑖 and 𝑑𝜏2 = −𝑑𝑠2

𝑐2. Also notice that because the metric is independent of the coordinate 𝑡,

Δ𝑡 is the same in both systems. This leads to

𝑑𝜏𝑖 2 = (1 +

2Φ𝑖𝑐2) (𝑑𝑡)2

⇒ Δ𝜏𝐴 = √(1 +2Φ𝐴𝑐2

) Δ𝑡

Δ𝜏𝐴 ~24 (1 +Φ𝐴𝑐2) Δ𝑡 (6.21)

and Δ𝜏𝐵 ~(1 +Φ𝐵𝑐2)Δ𝑡 (6.22)

Eliminating Δ𝑡 we get

21 (Hartle, 2003) 22 (Hartle, 2003) eq. (3.18) 23 (Hartle, 2003, p. 127) 24 √1 + 𝑥~1 +

1

2𝑥 if 𝑥 ≪ 1





Δ𝜏𝐵 =(1 +

Φ𝐵𝑐2)

(1 +Φ𝐴𝑐2)Δ𝜏𝐴

~25 (1 +

Φ𝐵𝑐2) (1 −

Φ𝐴𝑐2) Δ𝜏𝐴

Δ𝜏𝐵 ~(1 +Φ𝐵 −Φ𝐴

𝑐2)Δ𝜏𝐴 (6.23)

which tells us the observed fact, that when the receiver 𝐵 is at a higher gravitational potential that the emitter 𝐴, the signals will be received more slowly than they were emitted and vice versa.

3.7 26Local inertial frames A local inertial frame is defined by the conditions

𝑔𝛼𝛽′ (𝑥′𝑃) = 𝜂𝛼𝛽 and

𝜕𝑔𝛼𝛽′

𝜕𝑥′𝛾|𝑥=𝑥𝑃

= 0 (7.13)

This means that if you have a system described by a metric 𝑔𝛼𝛽it can locally in a point 𝑃 be transformed

into the flat space metric 𝜂𝛼𝛽. Furthermore the first derivatives of the transformed metric vanish. This is

best illustrated by an example.

3.7.1 27The metric of a Sphere at the North Pole.

The line element of the geometry of a sphere with radius 𝑎 is 𝑑𝑆2 = 𝑎2(𝑑𝜃2 + sin2 𝜃 𝑑𝜙2) (7.14)

At the north pole 𝜃 = 0, and the metric doesn’t look like the metric of a flat plane at all. Can we find a coordinate transformation so that

𝑑𝑆2 = 𝑑𝑥2 + 𝑦2 or 𝑔𝑖𝑗 = {1 00 1

}

Look at the coordinates 𝑥 = 𝑎𝜃 cos𝜙 𝑦 = 𝑎𝜃 sin𝜙 (7.15)

At the north pole both 𝑥 and 𝑦 are zero. Next we calculate

𝑥2 + 𝑦2 = (𝑎𝜃 cos𝜙)2 + (𝑎𝜃 sin𝜙)2 = 𝑎2𝜃2(cos2𝜙 + sin2𝜙) = 𝑎2𝜃2

⇒ 𝜃 =1

𝑎√𝑥2 + 𝑦2

(7.16)

⇒ 𝑦

𝑥 =

𝑎𝜃 sin𝜙

𝑎𝜃 cos𝜙

= tan𝜙

⇒ 𝜙 = tan−1 (𝑦

𝑥) (7.16)

The differentials

𝑑𝜃 = 𝑑 (1

𝑎√𝑥2 + 𝑦2)

=1

𝑎

1

2√𝑥2 + 𝑦2(2𝑥𝑑𝑥 + 2𝑦𝑑𝑦)

25

1

1+𝑥~1 − 𝑥 if 𝑥 ≪ 1

26 (Hartle, 2003, p. 140) 27 (Hartle, 2003, p. 141) example 7.2





=1

𝑎

1

√𝑥2 + 𝑦2(𝑥𝑑𝑥 + 𝑦𝑑𝑦)

𝑑(tan𝜙) = 𝑑 (𝑦

𝑥)

⇒ (1 + tan2 𝜙)𝑑𝜙 =1

𝑥𝑑𝑦 −

𝑦

𝑥2𝑑𝑥

⇒ (1 + (𝑦

𝑥)2

)𝑑𝜙 =1

𝑥𝑑𝑦 −

𝑦

𝑥2𝑑𝑥

⇒ (𝑥2 + 𝑦2)𝑑𝜙 = 𝑥𝑑𝑦 − 𝑦𝑑𝑥

⇒ 𝑑𝜙 =𝑥𝑑𝑦 − 𝑦𝑑𝑥

𝑥2 + 𝑦2

The line element 𝑑𝑆2 = 𝑎2(𝑑𝜃2 + sin2 𝜃 𝑑𝜙2)

= 𝑎2((1

𝑎

1

√𝑥2 + 𝑦2(𝑥𝑑𝑥 + 𝑦𝑑𝑦))

2

+ sin2 𝜃 (𝑥𝑑𝑦 − 𝑦𝑑𝑥

𝑥2 + 𝑦2)2

)

=

1

𝑥2 + 𝑦2(𝑥2𝑑𝑥2 + 𝑦2𝑑𝑦2 + 2𝑥𝑦𝑑𝑥𝑑𝑦)

+ (𝑎 sin 𝜃

𝑥2 + 𝑦2)2

(𝑥2𝑑𝑦2 + 𝑦2𝑑𝑥2 − 2𝑥𝑦𝑑𝑥𝑑𝑦)

=

1

𝑥2 + 𝑦2((𝑥2 +

𝑎2 sin2 𝜃

𝑥2 + 𝑦2𝑦2)𝑑𝑥2 + (𝑦2 +

𝑎2 sin2 𝜃

𝑥2 + 𝑦2𝑥2)𝑑𝑦2

+ 2𝑥𝑦 (1 −𝑎2 sin2 𝜃

𝑥2 + 𝑦2)𝑑𝑥𝑑𝑦)

= (

𝑥2

𝑥2 + 𝑦2+

𝑦2

(𝑥2 + 𝑦2)2𝑎2 sin2 (

1

𝑎√𝑥2 + 𝑦2))𝑑𝑥2

+ (𝑦2

𝑥2 + 𝑦2+

𝑥2

(𝑥2 + 𝑦2)2𝑎2 sin2 (

1

𝑎√𝑥2 + 𝑦2))𝑑𝑦2

+ 2(𝑥𝑦

𝑥2 + 𝑦2−

𝑥𝑦

(𝑥2 + 𝑦2)2𝑎2 sin2 (

1

𝑎√𝑥2 + 𝑦2))𝑑𝑥𝑑𝑦

The metric

𝑔𝑥𝑥 =𝑥2

𝑥2 + 𝑦2+

𝑦2

(𝑥2 + 𝑦2)2𝑎2 sin2 (

1

𝑎√𝑥2 + 𝑦2)

𝑔𝑦𝑦 =𝑦2

𝑥2 + 𝑦2+

𝑥2

(𝑥2 + 𝑦2)2𝑎2 sin2 (

1

𝑎√𝑥2 + 𝑦2)

𝑔𝑥𝑦 = 𝑔𝑦𝑥 =𝑥𝑦

𝑥2 + 𝑦2−

𝑥𝑦

(𝑥2 + 𝑦2)2𝑎2 sin2 (

1

𝑎√𝑥2 + 𝑦2)

If we evaluate these around the north pole where 𝑥 and 𝑦 are small28

28 Use wxMaxima http://maxima.sourceforge.net/to evaluate the Taylor polynomials to second order: g_xx(x,y) := x^2/(x^2+y^2)+a^2*sin(1/a*sqrt(x^2+y^2))*sin(1/a*sqrt(x^2+y^2))*y^2/(x^2+y^2)^2; g_yy(x,y) := y^2/(x^2+y^2)+a^2*sin(1/a*sqrt(x^2+y^2))*sin(1/a*sqrt(x^2+y^2))*x^2/(x^2+y^2)^2; g_xy(x,y) :=x*y/(x^2+y^2)-a^2*sin(1/a*sqrt(x^2+y^2))*sin(1/a*sqrt(x^2+y^2))*x*y/(x^2+y^2)^2; taylor(g_xx(x,y),[x,y],[0,0],[2,2]); taylor(g_yy(x,y),[x,y],[0,0],[2,2]); taylor(g_xy(x,y),[x,y],[0,0],[2,2]); g_xx(x,y)=1-y^2/(3*a^2)+... g_yy(x,y)= 1-x^2/(3*a^2)+...


http://maxima.sourceforge.net/




𝑔𝑥𝑥 = 1 −𝑦2

3𝑎2

𝑔𝑦𝑦 = 1 −𝑥2

3𝑎2

𝑔𝑥𝑦 = 𝑔𝑦𝑥 =𝑥𝑦

3𝑎2

At the north pole i.e. (𝑥, 𝑦) = (0,0)

𝑔𝛼𝛽((𝑥, 𝑦) = (0,0)) = 𝜂𝛼𝛽

𝜕𝑔𝛼𝛽

𝜕𝑥𝛾|(𝑥,𝑦)=(0,0)

= 0

We have proved that the sphere is locally flat at the North Pole. Of course, this is true for any set of coor-dinates on the sphere, because it is simply a matter of rotation.

3.8 29Length, Area, Volume and Four-Volume for Diagonal Metrics For a diagonal metric of the type

𝑑𝑠2 = 𝑔00𝑑𝑥0𝑑𝑥0 + 𝑔11𝑑𝑥

1𝑑𝑥1 + 𝑔22𝑑𝑥2𝑑𝑥2 + 𝑔33𝑑𝑥

3𝑑𝑥3 (7.27) you can define proper length elements in the various coordinates as

𝑑𝑙1 = √𝑔11𝑑𝑥1

𝑑𝑙2 = √𝑔22𝑑𝑥2

𝑑𝑙3 = √𝑔33𝑑𝑥3

The area element 𝑑𝐴 = 𝑑𝑙1𝑑𝑙2

(7.28) = √𝑔11𝑔22𝑑𝑥

1𝑑𝑥2

Notice, that it is not always the 𝑔11 and 𝑔22 that are involved in the calculation of the area. As we shall see below it can also be the 𝑔22 and 𝑔33 The three-volume element

𝑑𝒱 = 𝑑𝑙1𝑑𝑙2𝑑𝑙3 (7.29)

= √𝑔11𝑔22𝑔33𝑑𝑥1𝑑𝑥2𝑑𝑥3

The four-volume element

𝑑𝑣 = √−𝑔00𝑔11𝑔22𝑔33𝑑𝑥0𝑑𝑥1𝑑𝑥2𝑑𝑥3 (7.30)

In the case of a non-diagonal metric the four-volume element is

𝑑𝑣 = √−𝑔𝑑4𝑥

where 𝑔 is the determinant of the matrix 𝑔𝛼𝛽

3.8.1 30Area and Volume Elements of a Sphere The line element of flat space in polar coordinates

𝑑𝑠2 = −𝑑𝑡2 + 𝑑𝑟2 + 𝑟2𝑑𝜃2 + 𝑟2 sin2 𝜃 𝑑𝜙 (7.31) The proper length elements

𝑑𝑙1 = 𝑑𝑟

𝑑𝑙2 = 𝑟𝑑𝜃

𝑑𝑙3 = 𝑟 sin𝜃 𝑑𝜙 The area element

𝑑𝐴 = 𝑑𝑙2𝑑𝑙3 = 𝑟2 sin𝜃 𝑑𝜃𝑑𝜙 (7.32)

g_xy(x,y) =(y*x)/(3*a^2)+... 29 (Hartle, 2003, p. 146) 30 (Hartle, 2003, p. 147) example 7-5





The area 𝐴 = ∫ 𝑑𝐴

= ∫ ∫ 𝑟2 sin 𝜃 𝑑𝜃𝑑𝜙

= 𝑟2∫ sin𝜃 𝑑𝜃𝜋

0

∫ 𝑑𝜙2𝜋

0

= 2𝜋𝑟2[− cos 𝜃]0𝜋

= 2𝜋𝑟2(−(−1 − 1))

= 4𝜋𝑟2 The three-volume element

𝑑𝒱 = 𝑑𝑙1𝑑𝑙2𝑑𝑙3 = 𝑑𝑟𝑑𝐴 = 𝑟2 sin𝜃 𝑑𝑟𝑑𝜃𝑑𝜙 The three-volume

𝒱 = ∫ 𝑑𝒱 = ∫ ∫ ∫ 𝑟2 sin𝜃 𝑑𝑟𝑑𝜃𝑑𝜙 = ∫ ∫ 𝑟2𝑑𝑟𝑑𝐴

= 4𝜋∫ 𝑟2𝑑𝑟𝑟

0

= 4𝜋 [

1

3𝑟3]

0

𝑟

=4𝜋

3𝑟3

Collection the results 𝐴 = 4𝜋𝑟2

𝒱 =4𝜋

3𝑟3

We recognize the familiar values.

3.8.2 31Distance, Area and Volume in the Curved Space of a Constant Density Spherical Star or a Homogenous Closed Universe

The spherical line element is (𝑎 is a constant related to the density of matter)

𝑑𝑆2 =1

1 − (𝑟𝑎)

2 𝑑𝑟2 + 𝑟2𝑑𝜃2 + 𝑟2 sin2 𝜃 𝑑𝜙2

The proper length elements

𝑑𝑙1 =

1

√1 − (𝑟𝑎)

2𝑑𝑟

𝑑𝑙2 = 𝑟𝑑𝜃

𝑑𝑙3 = 𝑟 sin𝜃 𝑑𝜙

The circumference around equator where 𝑟 = 𝑅 and 𝜃 =𝜋

2

𝐶 = ∫𝑑𝑙3

= ∫ 𝑟 sin𝜃 𝑑𝜙

2𝜋

0

= 2𝜋𝑅 The distance from the center (𝑟 = 0) to surface (𝑟 = 𝑅) along a line where 𝜃 = 𝑐𝑜𝑛𝑠𝑡. and 𝜙 = 𝑐𝑜𝑛𝑠𝑡.

31 (Hartle, 2003, p. 147) example 7.6





𝑆 = ∫𝑑𝑙1

= ∫

1

√1 − (𝑟𝑎)2𝑑𝑟

𝑅

0

= ∫

𝑎

√𝑎2 − 𝑟2𝑑𝑟

𝑅

0

= 32𝑎 [sin−1

𝑟

𝑎]0

𝑅

= 𝑎 sin−1

𝑅

𝑎 (7.36)

The area of the two-surface at 𝑟 = 𝑅 𝐴 = ∫ 𝑑𝐴

= ∫ 𝑑𝑙2∫ 𝑑𝑙3

= 𝑅2∫ sin𝜃 𝑑𝜃𝜋

0

∫ 𝑑𝜙2𝜋

0

= 4𝜋𝑅2 (7.37) The volume inside 𝑟 = 𝑅 is

𝒱 = ∫ 𝑑𝒱

= ∫ ∫ ∫ 𝑑𝑙1𝑑𝑙2𝑑𝑙3

= ∫𝑟2

√1 − (𝑟𝑎)

2𝑑𝑟

𝑅

0

∫ sin𝜃 𝑑𝜃𝜋

0

∫ 𝑑𝜙2𝜋

0

= 4𝜋𝑎∫

𝑟2

√𝑎2 − 𝑟2𝑑𝑟

𝑅

0

= 334𝜋𝑎 [

𝑎2

2sin−1 (

𝑟

𝑎) −

𝑟

2√𝑎2 − 𝑟2]

0

𝑅

= 4𝜋𝑎3 [1

2sin−1 (

𝑟

𝑎) −

𝑟

2𝑎√1 − (

𝑟

𝑎)2

]

0

𝑅

= 344𝜋𝑎3 (1

2sin−1 (

𝑅

𝑎) −

𝑅

2𝑎√1 − (

𝑅

𝑎)2

) (7.38)

Collecting the results:

The circumference around equator where 𝑟 = 𝑅 and 𝜃 =𝜋

2

𝐶 = 2𝜋𝑅 The distance from the center (𝑟 = 0) to surface (𝑟 = 𝑅) along a line where 𝜃 = 𝑐𝑜𝑛𝑠𝑡. and 𝜙 = 𝑐𝑜𝑛𝑠𝑡.

𝑆 = 𝑎 sin−1𝑅

𝑎

The area of the two-surface at 𝑟 = 𝑅 𝐴 = 4𝜋𝑅2

32 (Spiegel, 1990) (14.237) 33 (Spiegel, 1990) (14.239)

34 For 𝑅/𝑎 ≪ 1: 𝒱 = 4𝜋𝑎3 (1

2sin−1 (

𝑅

𝑎) −

𝑅

2𝑎√1 − (

𝑅

𝑎)2

) = 4𝜋𝑎31

2 (𝑅

𝑎+

1

2

1

3(𝑅

𝑎)3

−𝑅

𝑎(1 −

1

2(𝑅

𝑎)2

)) =

4𝜋𝑎31

2 (1

2

1

3(𝑅

𝑎)3

+1

2(𝑅

𝑎)3

) =4𝜋

3𝑅3 (Spiegel, 1990) (20.27) (20.12)





The volume inside 𝑟 = 𝑅 is

𝒱 = 4𝜋𝑎3 (1

2sin−1 (

𝑅

𝑎) −

𝑅

2𝑎√1 − (

𝑅

𝑎)2

)

3.8.3 35Distance, Area, Volume and four-volume of a metric 𝑑𝑠2 = −(1 − 𝐴𝑟2)2𝑑𝑡2 + (1 − 𝐴𝑟2)2𝑑𝑟2 + 𝑟2𝑑𝜃2 + 𝑟2 sin2 𝜃 𝑑𝜙2 𝑑𝑙0 = (1 − 𝐴𝑟2)𝑑𝑡 𝑑𝑙1 = (1 − 𝐴𝑟2)𝑑𝑟 𝑑𝑙2 = 𝑟𝑑𝜃 𝑑𝑙3 = 𝑟 sin𝜃 𝑑𝜙

The proper distance along a radial line from the center 𝑟 = 0 to a coordinate radius 𝑟 = 𝑅 𝑙 = ∫ 𝑑𝑙1

= ∫ (1 − 𝐴𝑟2)𝑑𝑟

𝑅

0

= [𝑟 −

1

3𝐴𝑟3]

0

𝑅

= 𝑅 −

1

3𝐴𝑅3

The area of a sphere of coordinate radius 𝑟 = 𝑅 𝐴 = ∫ ∫ 𝑑𝑙2𝑑𝑙3

= 𝑅2∫ sin𝜃 𝑑𝜃

𝜋

0

∫ 𝑑𝜙2𝜋

0

= 4𝜋𝑅2 The three-volume of a sphere of coordinate radius 𝑟 = 𝑅

𝒱 = ∫ ∫ ∫ 𝑑𝑙1𝑑𝑙2𝑑𝑙3

= ∫ 𝑟2(1 − 𝐴𝑟2)𝑑𝑟𝑅

0


0

∫ 𝑑𝜙2𝜋

0

= 4𝜋 [

1

3𝑟3 −

1

5𝐴𝑟5]

0

𝑅

= 4𝜋 (

1

3𝑅3 −

1

5𝐴𝑅5)

The four-volume of a four-dimensional tube bounded by a sphere of coordinate radius 𝑅 and two 𝑡 =𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 planes separated by a time 𝑇

𝑣 = ∫ ∫ ∫ ∫ 𝑑𝑙0𝑑𝑙1𝑑𝑙2𝑑𝑙3

= ∫ 𝑑𝑡𝑇

0

∫ 𝑟2(1 − 𝐴𝑟2)2𝑑𝑟𝑅

0


0

∫ 𝑑𝜙2𝜋

0

= 4𝜋𝑇∫ 𝑟2(1 + 𝐴2𝑟4 − 2𝐴𝑟2)𝑑𝑟

𝑅

0

= 4𝜋𝑇 [

1

3𝑟3 +

𝐴2

7𝑟7 −

2𝐴

5𝑟5]

0

𝑅

=4𝜋𝑇

3(𝑅3 −

6𝐴

5𝑅5 +

3𝐴2

7𝑅7)

35 (Hartle, 2003, p. 166) problem 7-14





3.8.4 36The dimensions of a peanut The line-element of a peanut geometry is

𝑑𝑠2 = 𝑎2𝑑𝜃2 + 𝑎2𝑓2(𝜃)𝑑𝜙2

𝑓(𝜃) = sin𝜃 (1 −3

4sin2 𝜃)

𝑑𝑙1 = 𝑎𝑑𝜃 𝑑𝑙2 = 𝑎𝑓(𝜃)𝑑𝜙

The distance from pole to pole (𝜙 = 0) 𝑑 = ∫ 𝑑𝑙1

= ∫ 𝑎𝑑𝜃

𝜋

0

= 𝑎𝜋 The circumference at a constant angle 𝜃

𝐶 = ∫ 𝑑𝑙2

= ∫ 𝑎𝑓(𝜃)𝑑𝜙2π

0

= 𝑎𝑓(𝜃)∫ 𝑑𝜙

2π

0

= 2π𝑎𝑓(𝜃)

= 2π𝑎 sin 𝜃 (1 −3

4sin2 𝜃)

At the center or equator 𝜃 =𝜋

2

𝐶 = 2π𝑎 sin (𝜋

2)(1 −

3

4sin2 (

𝜋

2))

= 2π𝑎 (1 −

3

4)

=π

2𝑎

The area of a peanut 𝐴 = ∫ ∫ 𝑑𝑙1𝑑𝑙2

= 𝑎2∫ 𝑓(𝜃)𝑑𝜃

𝜋

0

∫ 𝑑𝜙2𝜋

0

= 2𝜋𝑎2∫ sin𝜃 (1 −

3

4sin2 𝜃)𝑑𝜃

𝜋

0

= 2𝜋𝑎2∫ (1 −

3

4(1 − cos2 𝜃)) sin 𝜃 𝑑𝜃

𝜋

0

= −2𝜋𝑎2∫ (1 −

3

4(1 − 𝑥2))𝑑𝑥

−1

1

= −

𝜋

2𝑎2∫ (1 + 3𝑥2)𝑑𝑥

−1

1

= −𝜋

2𝑎2[𝑥 + 𝑥3]1

−1

= −𝜋

2𝑎2(−1 − 1 − 1 − 1)

= 2𝜋𝑎2

36 (Hartle, 2003, p. 166) problem 7-15





3.8.5 37The dimensions of an egg The line-element

𝑑𝑠2 = 𝑎2[(cos2 𝜃 + 4 sin2 𝜃)𝑑𝜃2 + sin2 𝜃 𝑑𝜙2] The circumference of the egg at constant 𝜃

𝐶(𝜃) = ∫ 𝑎 sin 𝜃 𝑑𝜙2𝜋

0

= 2𝜋𝑎 sin 𝜃 The distance from pole to pole

𝑑𝑝𝑜𝑙𝑒−𝑡𝑜−𝑝𝑜𝑙𝑒 = ∫ 𝑎√cos2 𝜃 + 4 sin2 𝜃𝜋

0

𝑑𝜃

= 𝑎∫ √1 + 3 sin2 𝜃

𝜋

0

𝑑𝜃

= 4,84𝑎 = 0,77 ∗ 2𝜋𝑎 The ratio of the biggest circle around the axis to the pole-to-pole distance is

𝐶 (𝜃 =𝜋2)

𝑑𝑝𝑜𝑙𝑒−𝑡𝑜−𝑝𝑜𝑙𝑒 =

2𝜋𝑎

0,77 ∗ 2𝜋𝑎

= 1,3 The surface area of an egg

𝐴 = 𝑎2∫ sin𝜃√1 + 3 sin2 𝜃𝜋

0

𝑑𝜃∫ 𝑑𝜙2𝜋

0

= 3,41 ∗ 2𝜋 𝑎2

3.8.6 38Length and volume of the Schwarzschild geometry

The spatial part of the Schwarzschild line element is

𝑑𝑠2 = −(1 −2𝑚

𝑟)−1

𝑑𝑟2 − 𝑟2𝑑𝜃2 − 𝑟2 sin2 𝜃 𝑑𝜙2

(a) The radial distance between the sphere 𝑟 = 2𝑀 and the sphere 𝑟 = 3𝑀 𝑙 = ∫ 𝑑𝑙1

= ∫

1

√1 −2𝑀𝑟

𝑑𝑟3𝑀

2𝑀

= ∫ √

𝑟

𝑟 − 2𝑀

3𝑀

2𝑀

𝑑𝑟

= 39 [√(𝑟 − 2𝑀)𝑟]

2𝑀

3𝑀+𝑀∫

𝑑𝑟

√(𝑟 − 2𝑀)𝑟

3𝑀

2𝑀

= 40 [√(𝑟 − 2𝑀)𝑟 + 2𝑀 ln(√𝑟 + √𝑟 − 2𝑀)]

2𝑀

3𝑀

= √𝑀 ∗ 3𝑀 + 2𝑀 ln(√3𝑀 + √𝑀) − 2𝑀 ln(√2𝑀)

37 (Hartle, 2003, p. 29) problem 2-8 38 (Hartle, 2003, p. 166) problem 7-18

39 ∫√𝑝𝑥+𝑞

𝑎𝑥+𝑏𝑑𝑥 =

√(𝑎𝑥+𝑏)(𝑝𝑥+𝑞)

𝑎+

𝑎𝑞−𝑏𝑝

2𝑎∫

𝑑𝑥

√(𝑎𝑥+𝑏)(𝑝𝑥+𝑞) (Spiegel, 1990) (14.123)

40 ∫𝑑𝑥

√(𝑎𝑥+𝑏)(𝑝𝑥+𝑞)=

2

√𝑎𝑝ln(√𝑎(𝑝𝑥 + 𝑞) + √𝑝(𝑎𝑥 + 𝑏)) (Spiegel, 1990) (14.120)





= (

√3

2+ ln(

√3 + 1

√2)) ∗ 2𝑀

≈ 1,52 ∗ 2𝑀 (b) The spatial volume between the sphere 𝑟 = 2𝑀 and the sphere 𝑟 = 3𝑀

𝒱 = ∫ 𝑑𝑙1𝑑𝑙2𝑑𝑙33𝑀

2𝑀

= ∫

𝑟2

√1 −2𝑀𝑟

𝑑𝑟3𝑀

2𝑀


0

∫ 𝑑𝜙2𝜋

0

= 4𝜋 ∗ 17,15 ∗ 𝑀3

= 6,43 ∗4𝜋

3(2𝑀)3

3.8.7 41Volume in the Wormhole geometry

The three-dimensional volume on a 𝑡 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 slice of the wormhole geometry bounded by two spheres of coordinate radius 𝑅 on each side of the throat. The line-element

𝑑𝑠2 = −𝑑𝑡2 + 𝑑𝑟2 + (𝑏2 + 𝑟2)(𝑑𝜃2 + sin2 𝜃 𝑑𝜙2) (7.39) The volume

𝑉 = ∫𝑑𝑙1∫𝑑𝑙2∫𝑑𝑙3

= ∫ (𝑏2 + 𝑟2)𝑑𝑟

𝑅

−𝑅


0

∫ 𝑑𝜙2𝜋

0

= 4𝜋∫ (𝑏2 + 𝑟2)𝑑𝑟

𝑅

−𝑅

= 4𝜋 [𝑟𝑏2 +

1

3𝑟3]

−𝑅

𝑅

= 4𝜋((𝑅𝑏2 +

1

3𝑅3) − ((−𝑅)𝑏2 +

1

3(−𝑅)3))

=4𝜋

3∗ 2𝑅 ∗ (3𝑏2 + 𝑅2)

4 Tensor Calculus

4.1 Christoffel symbols.

4.1.1 42𝚪𝒂𝒃𝒂 and 𝚪𝒂𝒂𝒃 in a diagonal metric

Γ𝑎𝑏𝑐 =1

2(𝜕𝑔𝑎𝑏𝜕𝑥𝑐

+𝜕𝑔𝑎𝑐𝜕𝑥𝑏

−𝜕𝑔𝑏𝑐𝜕𝑥𝑎

) (4.15)

Γ𝑎𝑏𝑎 =1

2(𝜕𝑔𝑎𝑏𝜕𝑥𝑎

+𝜕𝑔𝑎𝑎𝜕𝑥𝑏

−𝜕𝑔𝑏𝑎𝜕𝑥𝑎

)

=1

2(𝜕𝑔𝑎𝑎𝜕𝑥𝑏

)

Γ𝑎𝑎𝑏 =1


+𝜕𝑔𝑎𝑏𝜕𝑥𝑎

−𝜕𝑔𝑎𝑏𝜕𝑥𝑎

)

41 (Hartle, 2003)problem 7-17 42 (McMahon, 2006, p. 324), final exam 6. The answer to FE-6 is (d)





=1


)

4.1.2 43Find the Christoffel symbols of the 2-sphere with radius 𝒂

The line element: 𝑑𝑠2 = 𝑎2𝑑𝜃2 + 𝑎2 sin2 𝜃 𝑑𝜙2

The metric tensor and its inverse: 𝑔𝑎𝑏 = {𝑎2

𝑎2 sin2 𝜃} 𝑔𝑎𝑏 = {

1

𝑎2

1

𝑎2 sin2 𝜃

}

We have: Γ𝑎𝑏𝑐 =1

2(𝜕𝑐𝑔𝑎𝑏 + 𝜕𝑏𝑔𝑎𝑐 − 𝜕𝑎𝑔𝑏𝑐) (4.15)

Γ 𝑏𝑐𝑎 = 𝑔𝑎𝑑Γ𝑑𝑏𝑐 (4.16)

Γ𝜃𝜙𝜙 = −1

2𝜕𝜃(𝑎

2 sin2 𝜃) = −𝑎2 sin 𝜃 cos 𝜃 ⇒ Γ 𝜙𝜙𝜃 = 𝑔𝜃𝜃Γ𝜃𝜙𝜙 = −sin𝜃 cos𝜃

Γ𝜙𝜃𝜙 = Γ𝜙𝜙𝜃 =1

2𝜕𝜃(𝑎

2 sin2 𝜃) = 𝑎2 sin𝜃 cos𝜃 ⇒ Γ 𝜃𝜙𝜙

= Γ 𝜙𝜃𝜙

= 𝑔𝜙𝜙Γ𝜙𝜃𝜙 = cot 𝜃

4.1.3 44Find the Christoffel symbols of the Kahn-Penrose metric (Colliding gravitational

waves)

The line element: 𝑑𝑠2 = 2𝑑𝑢𝑑𝑣 − (1 − 𝑢)2𝑑𝑥2 − (1 + 𝑢)2𝑑𝑦2


11

−(1 − 𝑢)2

−(1 + 𝑢)2

}

and its inverse: 𝑔𝑎𝑏 =

{

11

−1

(1 − 𝑢)2

−1

(1 + 𝑢)2}

Γ𝑎𝑏𝑐 =1

2(𝜕𝑐𝑔𝑎𝑏 + 𝜕𝑏𝑔𝑎𝑐 − 𝜕𝑎𝑔𝑏𝑐) (4.15) Γ 𝑏𝑐

𝑎 = 𝑔𝑎𝑑Γ𝑑𝑏𝑐 (4.16)

Γ𝑢𝑥𝑥 = −1

2𝜕𝑢(−(1 − 𝑢)

2) = −(1 − 𝑢) ⇒ Γ 𝑥𝑥𝑣 = 𝑔𝑣𝑢Γ𝑢𝑥𝑥 = −(1 − 𝑢)

Γ𝑥𝑢𝑥 = Γ𝑥𝑥𝑢 =1

2𝜕𝑢(−(1 − 𝑢)

2) = (1 − 𝑢) ⇒ Γ 𝑢𝑥𝑥 = Γ 𝑥𝑢

𝑥 = 𝑔𝑥𝑥Γ𝑥𝑢𝑥 =−1

1 − 𝑢

Γ𝑢𝑦𝑦 = −1

2𝜕𝑢(−(1 + 𝑢)

2) = (1 + 𝑢) ⇒ Γ 𝑦𝑦𝑣 = 𝑔𝑣𝑢Γ𝑢𝑦𝑦 = (1 + 𝑢)

Γ𝑦𝑢𝑦 = Γ𝑦𝑦𝑢 =1

2𝜕𝑢(−(1 + 𝑢)

2) = −(1 + 𝑢) ⇒ Γ 𝑢𝑦𝑦

= Γ 𝑦𝑢𝑦

= 𝑔𝑦𝑦Γ𝑦𝑢𝑦 =1

1 + 𝑢

43 (McMahon, 2006, p. 74), example 4-4 44 (McMahon, 2006, p. 75), example 4-5, quiz 4-7, the answer to quiz 4-7 is (b)





4.2 45Alternative solution: Show that 𝛁𝒄𝒈𝒂𝒃 = 𝟎 p.69: ∇𝑐𝑔𝑎𝑏 = 𝜕𝑐𝑔𝑎𝑏 − 𝑔𝑑𝑏Γ 𝑎𝑐

𝑑 − 𝑔𝑎𝑑Γ 𝑐𝑏𝑑

If 𝑔𝑑𝑏 = 𝑔𝑏𝑑: = 𝜕𝑐𝑔𝑎𝑏 − 𝑔𝑏𝑑Γ 𝑎𝑐𝑑 − 𝑔𝑎𝑑Γ 𝑐𝑏

𝑑

= 𝜕𝑐𝑔𝑎𝑏 − Γ𝑏𝑎𝑐 − Γ𝑎𝑐𝑏 since Γ𝑎𝑐𝑏 = Γ𝑎𝑏𝑐 (4.14): = 𝜕𝑐𝑔𝑎𝑏 − (Γ𝑏𝑎𝑐 + Γ𝑎𝑏𝑐)

and Γ𝑏𝑎𝑐 + Γ𝑎𝑏𝑐 = 𝜕𝑐𝑔𝑎𝑏 (p. 73): = 𝜕𝑐𝑔𝑎𝑏 − 𝜕𝑐𝑔𝑎𝑏 = 0 Q.E.D.

4.3 One-forms.

4.3.1 46One-forms: why 𝒅𝟐 = 𝟎

For general forms, let 𝛼 be a 𝑝-form and 𝛽 be a 𝑞-form, we have 𝛼 ∧ 𝛽 = (−1)𝑝𝑞𝛽 ∧ 𝛼

For one-forms this means 𝛼 ∧ 𝛽 = −𝛽 ∧ 𝛼

⇒ 𝛼 ∧ 𝛼 = −𝛼 ∧ 𝛼 = 0

This also holds for 𝑑𝑓 =𝜕𝑓

𝜕𝑥𝑎𝑑𝑥𝑎 because 𝑑𝑓 is a one-form as well.

𝑑𝑓𝑎 ∧ 𝑑𝑓𝑏 =

𝜕𝑓

𝜕𝑥𝑎𝑑𝑥𝑎 ∧

𝜕𝑓

𝜕𝑥𝑏𝑑𝑥𝑏

=

𝜕2𝑓

𝜕𝑥𝑎𝜕𝑥𝑏𝑑𝑥𝑎 ∧ 𝑑𝑥𝑏 (i)

= −

𝜕2𝑓

𝜕𝑥𝑎𝜕𝑥𝑏𝑑𝑥𝑏 ∧ 𝑑𝑥𝑎 (ii)

⇒ 𝑑𝑓𝑎 ∧ 𝑑𝑓𝑎 =𝜕2𝑓

𝜕𝑥𝑎𝜕𝑥𝑏𝑑𝑥𝑎 ∧ 𝑑𝑥𝑎 (i)

and 𝑑𝑓𝑎 ∧ 𝑑𝑓𝑎 = −𝜕2𝑓

𝜕𝑥𝑎𝜕𝑥𝑏𝑑𝑥𝑎 ∧ 𝑑𝑥𝑎 (ii)

Now, because the partial derivatives commutate, this can only be true if 𝑑𝑥𝑎 ∧ 𝑑𝑥𝑎 = 0 and 𝑑2 = 0 (4.25)

4.3.2 47The exterior derivative of a one-form.

The exterior derivative of a one-form 𝑓𝑎𝑑𝑥𝑎:

𝑑(𝑓𝑎𝑑𝑥𝑎) = 𝑑𝑓𝑎 ∧ 𝑑𝑥

𝑎 (4.26) To use this equation it is important to notice, that the right-hand side includes a summation of the partial

derivatives times the differential in the usual way: 𝑑𝑓(𝑥𝑖) =𝜕

𝜕𝑥𝑖𝑓(𝑥𝑖)𝑑𝑥𝑖.

Example 4-848 𝜎 = 𝑒𝑓(𝑟)𝑑𝑡

𝑑𝜎 = 𝑑(𝑒𝑓(𝑟)𝑑𝑡)

= 𝑑(𝑒𝑓(𝑟)) ∧ 𝑑𝑡

=𝜕

𝜕𝑟(𝑒𝑓(𝑟))𝑑𝑟 ∧ 𝑑𝑡

= 𝑓′(𝑟)𝑒𝑓(𝑟)𝑑𝑟 ∧ 𝑑𝑡

45 (McMahon, 2006, p. 78), example 4-7 46 (McMahon, 2006, p. 80) 47 (McMahon, 2006, p. 80) equation (4.26) 48 You will find a more extended use of equation (4.26) in e.g. the Brinkmann metric, chapter 9, p.195 (McMahon, 2006).





𝜌 = 𝑒𝑔(𝑟) cos 𝜃 sin𝜙 𝑑𝑟

𝑑𝜌 = 𝑑(𝑒𝑔(𝑟) cos 𝜃 sin𝜙 𝑑𝑟)

= 𝑑(𝑒𝑔(𝑟) cos 𝜃 sin𝜙) ∧ 𝑑𝑟

=𝜕

𝜕𝑟(𝑒𝑔(𝑟) cos 𝜃 sin𝜙)𝑑𝑟 ∧ 𝑑𝑟 +

𝜕

𝜕𝜃(𝑒𝑔(𝑟) cos𝜃 sin𝜙)𝑑𝜃 ∧ 𝑑𝑟

+𝜕

𝜕𝜙(𝑒𝑔(𝑟) cos 𝜃 sin𝜙)𝑑𝜙 ∧ 𝑑𝑟

=𝜕

𝜕𝜃(𝑒𝑔(𝑟) cos𝜃 sin𝜙)𝑑𝜃 ∧ 𝑑𝑟 +

𝜕

𝜕𝜙(𝑒𝑔(𝑟) cos 𝜃 sin𝜙)𝑑𝜙 ∧ 𝑑𝑟

= −𝑒𝑔(𝑟) sin 𝜃 sin𝜙 𝑑𝜃 ∧ 𝑑𝑟 + 𝑒𝑔(𝑟) cos𝜃 cos𝜙 𝑑𝜙 ∧ 𝑑𝑟

4.4 The geodesic equation.

4.4.1 49 Find the geodesic equations for cylindrical coordinates

The line element: 𝑑𝑠2 = 𝑑𝑟2 + 𝑟2𝑑𝜙2 + 𝑑𝑧2

The metric tensor and its inverse: 𝑔𝑎𝑏 = {1

𝑟2

1

} 𝑔𝑎𝑏 = {

11

𝑟2

1

}

Γ𝑎𝑏𝑐 =1



Γ𝑟𝜙𝜙 = −1

2𝜕𝑟(𝑟

2) = −𝑟 ⇒ Γ 𝜙𝜙𝑟 = 𝑔𝑟𝑟Γ𝑟𝜙𝜙 = −𝑟

Γ𝜙𝑟𝜙 = Γ𝜙𝜙𝑟 =1

2𝜕𝑟(𝑟

2) = 𝑟 ⇒ Γ 𝑟𝜙𝜙

= Γ 𝜙𝑟𝜙

= 𝑔𝜙𝜙Γ𝜙𝑟𝜙 =1

𝑟

The geodesics equation: 𝑑2𝑥𝑎


𝑎𝑑𝑥𝑏

𝑑𝑠

𝑑𝑥𝑐

𝑑𝑠 = 0 (4.33)

𝑥𝑎 = 𝑟: 𝑑2𝑟


𝑟𝑑𝑥𝑏

𝑑𝑠

𝑑𝑥𝑐

𝑑𝑠 = 0

⇒ 𝑑2𝑟

𝑑𝑠2− 𝑟 (

𝑑𝜙

𝑑𝑠)2

= 0

𝑥𝑎 = 𝜙: 𝑑2𝜙


𝜙 𝑑𝑥𝑏

𝑑𝑠

𝑑𝑥𝑐

𝑑𝑠 = 0

⇒ 𝑑2𝜙

𝑑𝑠2+ Γ 𝑟𝜙

𝜙 𝑑𝑟

𝑑𝑠

𝑑𝜙

𝑑𝑠+ Γ 𝜙𝑟

𝜙 𝑑𝜙

𝑑𝑠

𝑑𝑟

𝑑𝑠 = 0

⇒ 𝑑2𝜙

𝑑𝑠2+2

𝑟

𝑑𝑟

𝑑𝑠

𝑑𝜙

𝑑𝑠 = 0

𝑥𝑎 = 𝑧: 𝑑2𝑧


𝑧𝑑𝑥𝑏

𝑑𝑠

𝑑𝑥𝑐

𝑑𝑠 = 0

⇒ 𝑑2𝑧

𝑑𝑠2 = 0

49 (McMahon, 2006, p. 83), example 4-9





The Christoffel symbols from the geodesic equations

We have 𝐾 =1

2𝑔𝑎𝑏��

𝑎��𝑏 =1

2(��)2 +

1

2𝑟2(��)

2+1

2(��)2 (4.35)

Now we need 𝜕𝐾

𝜕𝑥𝑎 =

𝑑

𝑑𝑠(𝜕𝐾

𝜕��𝑎) (4.36)

𝑥𝑎 = 𝑟: 𝜕𝐾

𝜕𝑟 =

𝑑

𝑑𝑠(𝜕𝐾

𝜕��)

⇒ 𝑟��2 =𝑑

𝑑𝑠(��) = ��

⇔ 0 = �� − 𝑟��2

𝑥𝑎 = 𝜙: 𝜕𝐾

𝜕𝜙 =

𝑑

𝑑𝑠(𝜕𝐾

𝜕��)

0 =𝑑

𝑑𝑠(𝑟2��) = 2𝑟�� + 𝑟2��

⇔ 0 = �� +1

𝑟�� +

1

𝑟��

𝑥𝑎 = 𝑧: 𝜕𝐾

𝜕𝑧 =

𝑑

𝑑𝑠(𝜕𝐾

𝜕��)

⇒ 0 =𝑑

𝑑𝑠(��) = ��

Collecting the results

0 = �� − 𝑟��2

0 = �� +1

𝑟�� +

1

𝑟��

0 = ��

We can now find the Christoffel symbols from the geodesic equation : Γ 𝜙𝜙𝑟 = −𝑟

Γ 𝑟𝜙𝜙

=1

𝑟 Γ 𝜙𝑟

𝜙 =

1

𝑟

4.4.2 50Use the geodesic equations to find the Christoffel symbols for the Rindler metric.

The Rindler coordinate system or frame describes a uniformly accelerating frame of reference in Minkow-ski space.

The line element: 𝑑𝑠2 = 𝜉2𝑑𝜏2 − 𝑑𝜉2

The metric tensor: 𝑔𝑎𝑏 = {𝜉2

−1}

We have 𝐾 =1

2𝑔𝑎𝑏��

𝑎��𝑏 =1

2𝜉2(��)2 −

1

2��2 (4.35)






Now we need 𝜕𝐾

𝜕𝑥𝑎 =

𝑑

𝑑𝑠(𝜕𝐾

𝜕��𝑎) (4.36)

𝑥𝑎 = 𝜉: 𝜕𝐾

𝜕𝜉 =

𝑑

𝑑𝑠(𝜕𝐾

𝜕��)

⇒ 𝜉��2 =𝑑

𝑑𝑠(−��) = −��

⇔ 0 = �� + 𝜉��2

𝑥𝑎 = 𝜏: 𝜕𝐾

𝜕𝜏 =

𝑑

𝑑𝑠(𝜕𝐾

𝜕��)

0 =𝑑

𝑑𝑠(𝜉2��) = 2𝜉�� + 𝜉2��

⇔ 0 = �� +1

𝜉�� +

1

𝜉��


0 = �� + 𝜉��2 0 = 2𝜉�� + 𝜉2��

We can now find the Christoffel symbols from the geodesic equation:

Γ 𝜏𝜏𝜉 = 𝜉

Γ 𝜉𝜏𝜏 =

1

𝜉 Γ 𝜏𝜉

𝜏 =1

𝜉

4.4.3 51Geodesics Equations of the plane in polar coordinates

We use the Euler-Lagrange method.

0 =𝑑

𝑑𝑆(𝜕𝐹

𝜕��𝑎) −

𝜕𝐹

𝜕𝑥𝑎 (10.36)52

𝐹 =1

2𝑔𝑎𝑏��

𝑎��𝑏 (4.35)53

The line element 𝑑𝑆2 = 𝑑𝑟2 + 𝑟2𝑑𝜙2 (8.2)

𝐹 =1

2��2 +

1

2𝑟2��2

𝑥𝑎 = 𝑟: 𝜕𝐹

𝜕𝑟 = 𝑟��2

𝜕𝐹

𝜕�� = ��

⇒ �� = 𝑟��2

𝑥𝑎 = 𝜙: 𝜕𝐹

𝜕𝜙 = 0

𝜕𝐹

𝜕�� = 𝑟2��

51 (Hartle, 2003) Example 8-1 52 (McMahon, 2006) 53 (McMahon, 2006)





⇒ 0 =𝑑

𝑑𝑆(𝑟2��)

= 2𝑟�� + 𝑟2��

⇒ �� = −2

𝑟��

Collecting the results �� = 𝑟��2 (8.6a)

�� = −2

𝑟�� (8.6b)

54The non-zero Christoffel symbols are

Γ 𝜙𝜙𝑟 = −𝑟

(8.17) Γ 𝑟𝜙

𝜙 = Γ 𝜙𝑟

𝜙=1

𝑟

4.4.4 55Equations for geodesics in a Wormhole Geometry


0 =𝑑

𝑑𝑠(𝜕𝐹

𝜕��𝑎) −

𝜕𝐹

𝜕𝑥𝑎 (10.36)56

𝐹 =1

2𝑔𝑎𝑏��

𝑎��𝑏 (4.35)57

The line element

𝑑𝑠2 = −𝑑𝑡2 + 𝑑𝑟2 + (𝑏2 + 𝑟2)(𝑑𝜃2 + sin2 𝜃 𝑑𝜙2) (8.11)

𝐹 = −1

2��2 +

1

2��2 +

1

2(𝑏2 + 𝑟2)��2 +

1

2(𝑏2 + 𝑟2) sin2 𝜃 ��2

𝑥𝑎 = 𝑡: 𝜕𝐹

𝜕𝑡 = 0

𝜕𝐹

𝜕�� = −��

⇒ �� = 0 𝑥𝑎 = 𝑟: 𝜕𝐹

𝜕𝑟 = 𝑟(��2 + sin2 𝜃 ��2)

𝜕𝐹

𝜕�� = ��

⇒ �� = 𝑟��2 + 𝑟 sin2 𝜃 ��2 𝑥𝑎 = 𝜃:

𝜕𝐹

𝜕𝜃 = (𝑏2 + 𝑟2) sin𝜃 cos𝜃 ��2

𝜕𝐹

𝜕�� = (𝑏2 + 𝑟2)��

𝑑

𝑑𝑠(𝜕𝐹

𝜕��) = 2𝑟�� + (𝑏2 + 𝑟2)��

⇒ �� = sin𝜃 cos𝜃 ��2 −2𝑟

(𝑏2 + 𝑟2)��

𝑥𝑎 = 𝜙:

54 (Hartle, 2003) Example 8.3 55 (Hartle, 2003, p. 172), example 8.2 56 (McMahon, 2006) 57 (McMahon, 2006)





𝜕𝐹

𝜕𝜙 = 0

𝜕𝐹

𝜕�� = (𝑏2 + 𝑟2) sin2 𝜃 ��

⇒ 0 =𝑑

𝑑𝑠((𝑏2 + 𝑟2) sin2 𝜃 ��)

= 2𝑟 sin2 𝜃 �� + 2(𝑏2 + 𝑟2) sin𝜃 cos𝜃 �� + (𝑏2 + 𝑟2) sin2 𝜃 ��

⇒ �� = −2𝑟

(𝑏2 + 𝑟2)�� − 2 cot 𝜃 ��

Collecting the results �� = 0 (8.13a) �� = 𝑟��2 + 𝑟 sin2 𝜃 ��2 (8.13b)

�� = sin𝜃 cos𝜃 ��2 −2𝑟

(𝑏2 + 𝑟2)�� (8.13c)

�� = −2𝑟

(𝑏2 + 𝑟2)�� − 2 cot 𝜃 �� (8.13d)

58The non-zero Christoffel symbols are

Γ 𝜃𝜃𝑟 = −𝑟 Γ 𝜙𝜙

𝑟 = −𝑟 sin2 𝜃

(8.18) Γ 𝜙𝜙𝜃 = −sin𝜃 cos𝜃 Γ 𝑟𝜃

𝜃 = Γ 𝜃𝑟𝜃 =

𝑟

(𝑏2 + 𝑟2)

Γ 𝑟𝜙𝜙

= Γ 𝜙𝑟𝜙

=𝑟

(𝑏2 + 𝑟2) Γ 𝜃𝜙

𝜙 = Γ 𝜙𝜃

𝜙= cot 𝜃

4.4.5 59New - Geodesics and Christoffel symbols of the Schwarzschild metric with 𝜽 =𝝅

𝟐

The metric

𝑑𝑠2 = −(1 −2𝑚

𝑟)𝑑𝑡2 + (1 −

2𝑚

𝑟)−1

𝑑𝑟2 + 𝑟2𝑑𝜙2

To find the geodesic we use the Euler-Lagrange equation

0 =𝑑

𝑑𝑠(𝜕𝐹

𝜕��𝑎) −

𝜕𝐹

𝜕𝑥𝑎

where

𝐹 = −(1 −2𝑚

𝑟) ��2 + (1 −

2𝑚

𝑟)−1

��2 + 𝑟2��2

𝑥𝑎 = 𝑡: 𝜕𝐹

𝜕𝑡 = 0

𝜕𝐹

𝜕�� = −2(1 −

2𝑚

𝑟) ��

𝑑

𝑑𝑠(𝜕𝐹

𝜕��) = −

4𝑚

𝑟2�� − 2 (1 −

2𝑚

𝑟) ��

⇒ 0 =4𝑚

𝑟2�� + 2 (1 −

2𝑚

𝑟) ��

⇔ 0 = �� +2𝑚

𝑟(𝑟 − 2𝑚)��

𝑥𝑎 = 𝑟:

58 (Hartle, 2003) Example 8.3 59 (Hartle, 2003, p. 183) problem 8-3





𝜕𝐹

𝜕𝑟 = 60 −

2𝑚

𝑟2��2 −

2𝑚

𝑟2(1 −

2𝑚

𝑟)−2

��2 + 2𝑟��2

𝜕𝐹

𝜕�� = 2(1 −

2𝑚

𝑟)−1

��

𝑑

𝑑𝑠(𝜕𝐹

𝜕��) = 2(1 −

2𝑚

𝑟)−1

�� −4𝑚

𝑟2(1 −

2𝑚

𝑟)−2

��2

⇒ 0 = 2(1 −2𝑚

𝑟)−1

�� −2𝑚

𝑟2(1 −

2𝑚

𝑟)−2

��2 +2𝑚

𝑟2��2 − 2𝑟��2

⇔ 0 = �� −𝑚

𝑟(𝑟 − 2𝑚)��2 + (𝑟 − 2𝑚)

𝑚

𝑟3��2 − (𝑟 − 2𝑚)��2

𝑥𝑎 = 𝜙: 𝜕𝐹

𝜕𝜙 = 0

𝜕𝐹

𝜕�� = 2𝑟2��

𝑑

𝑑𝑠(𝜕𝐹

𝜕��) = 4𝑟�� + 2𝑟2��

⇒ 0 = 4𝑟�� + 2𝑟2��

⇔ 0 = �� +2

𝑟��


0 = �� +2𝑚

𝑟(𝑟 − 2𝑚)��

0 = �� −

𝑚

𝑟(𝑟 − 2𝑚)��2 + (𝑟 − 2𝑚)

𝑚

𝑟3��2 − (𝑟 − 2𝑚)��2

0 = �� +

2

𝑟��

We can now find the Christoffel symbols:

Γ 𝑟𝑡𝑡 =

𝑚

𝑟(𝑟 − 2𝑚) Γ 𝑡𝑟

𝑡 =𝑚

𝑟(𝑟 − 2𝑚) Γ 𝜙𝜙

𝑟 = −(𝑟 − 2𝑚)

Γ 𝑟𝑟𝑟 = −

𝑚

𝑟(𝑟 − 2𝑚) Γ 𝑡𝑡

𝑟 =𝑚(𝑟 − 2𝑚)

𝑟3

Γ 𝑟𝜙𝜙

=1

𝑟 Γ 𝜙𝑟

𝜙 =

1

𝑟

4.5 Solving the geodesic equation

4.5.1 61The travel time through a wormhole

Use the geodesic equations to calculate the proper travel time of an astronaut travelling through a worm-hole throat along the coordinate radius 𝑟 from 𝑟 = 𝑅 to 𝑟 = −𝑅. The initial radial four-velocity is 𝑢𝑟 ≡ 𝑈,

and because of spherically symmetry 𝑢𝜃 = 𝑢𝜙 = 0 The four-velocity is

𝑢𝑎 = (𝑢𝑡 , 𝑢𝑟, 𝑢𝜃, 𝑢𝜙)

= (

𝑑𝑡

𝑑𝜏,𝑑𝑟

𝑑𝜏,𝑑𝜃

𝑑𝜏,𝑑𝜙

𝑑𝜏)

60 𝜕

𝜕𝑡((1 −

2𝑚

𝑟)−1

) = −2𝑚

𝑟2(1 −

2𝑚

𝑟)−2

= −2𝑚

(𝑟−2𝑚)2

61 (Hartle, 2003) Example 8.5





= 62 (√1 + 𝑈2, 𝑈, 0, 0) (8.21)

we will only look at 𝑢𝑟. We use the geodesic equation

�� = 𝑟��2 + 𝑟 sin2 𝜃 ��2 (8.13b) which we can rewrite

𝑑2𝑟

𝑑𝜏2 = 𝑟��2 + 𝑟 sin2 𝜃 ��2

= 0

⇒ 𝑑2𝑟

𝑑𝜏2 =

𝑑

𝑑𝜏(𝑑𝑟

𝑑𝜏)

=𝑑𝑢𝑟

𝑑𝜏

= 0 (8.22) which implies that 𝑢𝑟 = 𝑈 is a constant along the astronauts world-line. So we can solve

𝑢𝑟 =𝑑𝑟

𝑑𝜏

= 𝑈

⇒ 𝑑𝜏 =1

𝑈𝑑𝑟

⇒ Δ𝜏 = ∫1

𝑈𝑑𝑟

𝑅

−𝑅

=1

𝑈[𝑟]−𝑅

𝑅

=1

𝑈(𝑅 − (−𝑅))

=2𝑅

𝑈 (8.24)

So the travel time through the wormhole Δ𝜏 =2𝑅

𝑈 is very much alike the usual time/speed calculation:

distance=time*velocity except with the velocity replaced by the four velocity.

4.5.1.1 NEW - Is the trajectory time-like or space-like? The line-element

𝑑𝜏2 = 𝑑𝑡2 − 𝑑𝑟2 − (𝑏2 + 𝑟2)(𝑑𝜃2 + sin2 𝜃 𝑑𝜙2)

= 𝑑𝑡2(1 − (𝑑𝑟

𝑑𝑡)2

− (𝑏2 + 𝑟2) ((𝑑𝜃

𝑑𝑡)2

+ sin2 𝜃 (𝑑𝜙

𝑑𝑡)2

)

= 𝑑𝑡2 (1 − (𝑑𝑟

𝑑𝜏

𝑑𝜏

𝑑𝑡)2

− (𝑏2 + 𝑟2)((0)2 + sin2 𝜃 (0)2))

= 𝑑𝑡2 (1 − (𝑈 ⋅

1

√1 + 𝑈2)2

)

= 𝑑𝑡2 (

1 + 𝑈2 − 𝑈2

1 + 𝑈2)

= 𝑑𝑡2 (

1

1 + 𝑈2) > 0

i.e. the trajectory is time-like.

62 𝑢𝑡 is found from the fact that the 𝑢𝑎𝑢𝑎 = −1 is a conserved quantity: 𝑢𝑎𝑢𝑎 = 𝑢𝑎𝜂𝑖𝑗𝑢

𝑎 = −(𝑢𝑡)2 + (𝑢𝑟)2 +

(𝑢𝜃)2+ (𝑢𝜙)

2= −(𝑢𝑡)2 + 𝑈2 = −1 ⇒ 𝑢𝑡 = √1 + 𝑈2





4.5.2 63Geodesics in the Plane Using Polar Coordinates.

We know that the geodesics in the plane are straight lines. We can show this by using the line element, the Killing vector 𝜉 and the conserved quantity 𝑙 = 𝜉 ⋅ 𝑢 = 𝑐𝑜𝑛𝑠𝑡. The line element 𝑑𝑆2 = 𝑑𝑟2 + 𝑟2𝑑𝜙2 (8.2)

⇒ 1 = (𝑑𝑟

𝑑𝑆)2

+ 𝑟2 (𝑑𝜙

𝑑𝑆)2

(8.33)

The metric is independent of 𝜙 so we have a Killing vector 𝜉 = (0,1) And a conserved quantity 𝑙 = 𝜉 ⋅ 𝑢 = 𝜉𝑎𝑢

𝑎 = 𝑔𝑎𝑏𝜉

𝑏𝑢𝑎 = 𝑔𝑟𝑟𝜉

𝑟𝑢𝑟 + 𝑔𝜙𝜙𝜉𝜙𝑢𝜙

= 1 ⋅ 0 ⋅

𝑑𝑟

𝑑𝑆+ 𝑟2 ⋅ 1 ⋅

𝑑𝜙

𝑑𝑆

= 64𝑟2

𝑑𝜙

𝑑𝑆 (8.34)

This we can insert into the line element and integrate

⇒ 1 = (𝑑𝑟

𝑑𝑆)2

+ 𝑟2 (𝑑𝜙

𝑑𝑆)2

(𝑑𝑟

𝑑𝑆)2

= 1 − 𝑟2 (𝑑𝜙

𝑑𝑆)2

= 1 − 𝑟2 (

𝑙

𝑟2)2

= 1 − (

𝑙

𝑟)2

⇒ 𝑑𝑟

𝑑𝑆 = (1 − (

𝑙

𝑟)2

)

12

(8.35)

We want to find 𝜙 as a function of 𝑟

𝑑𝜙

𝑑𝑟 =

𝑑𝜙

𝑑𝑆

𝑑𝑆

𝑑𝑟

=

𝑙

𝑟2 (1 − (𝑙𝑟)2

)

12

(8.36)

=𝑙

𝑟(𝑟2 − 𝑙2)12

⇒ 𝑑𝜙 = 𝑙∫𝑑𝑟

𝑟(𝑟2 − 𝑙2)12

⇒ 𝜙 = 65𝑙 ∗2

2√𝑙2cos−1√

𝑙2

𝑟2+ 𝜙∗

63 (Hartle, 2003) Example 8.7 64 Notice this is equivalent with the geodesic equation we found before

𝑑

𝑑𝑆(𝑟2��) = 0

65 ∫𝑑𝑥

𝑥√𝑥𝑛−𝑎𝑛=

2

𝑛√𝑎𝑛cos−1√

𝑎𝑛

𝑥𝑛 (Spiegel, 1990) (14.334)





= cos−1 (𝑙

𝑟) + 𝜙∗

⇒ 𝑙 = 𝑟 cos(𝜙 − 𝜙∗) (8.38) = 𝑟 cos𝜙 cos𝜙∗ + 𝑟 sin𝜙 sin𝜙∗ = 𝑥 cos𝜙∗ + 𝑦 sin𝜙∗

⇒ 𝑦 = −𝑥 cot𝜙∗ +𝑙

sin𝜙∗

= 𝑎𝑥 + 𝑏 And the geodesic is a straight line as expected.

4.5.3 66NEW - Geodesics Equations of the plane in Cartesian coordinates.


0 =𝑑

𝑑𝑆(𝜕𝐹

𝜕��𝑎) −

𝜕𝐹

𝜕𝑥𝑎 (10.36)67

𝐹 =1

2𝑔𝑎𝑏��

𝑎��𝑏 (4.35)68

The line element 𝑑𝑆2 = 𝑑𝑥2 + 𝑑𝑦2

𝐹 =1

2��2 +

1

2��2

𝑥𝑎 = 𝑥: 𝜕𝐹

𝜕𝑥 = 0

𝜕𝐹

𝜕�� = ��

⇒ �� = 0 𝑥𝑎 = 𝑦: 𝜕𝐹

𝜕𝑦 = 0

𝜕𝐹

𝜕�� = ��

⇒ �� = 0 Collecting the results �� = 0 �� = 0

The solution is obviously straight lines: �� = 0 �� = 0 ⇒ �� = 𝑘0 �� = 𝑐0 ⇒ 𝑥 = 𝑘0𝑆 + 𝑘1 𝑦 = 𝑐0𝑆 + 𝑐1 ⇒ 𝑦 = 𝐾0𝑥 + 𝐾1

4.5.4 69New - Show that the great circle is a solution of the geodesic equation of a two-dimen-

sional sphere.


66 (Hartle, 2003) Problem 8.1 67 (McMahon, 2006) 68 (McMahon, 2006) 69 (Hartle, 2003) Problem 8.2





0 =𝑑

𝑑𝑆(𝜕𝐹

𝜕��𝑎) −

𝜕𝐹

𝜕𝑥𝑎 (10.36)70

𝐹 =1

2𝑔𝑎𝑏��

𝑎��𝑏 (4.35)71

The line element

𝑑𝑆2 = 𝑎2𝑑𝜃2 + 𝑎2 sin2 𝜃 𝑑𝜙2

𝐹 =1

2𝑎2��2 +

1

2𝑎2 sin2 𝜃 ��2

𝑥𝑎 = 𝜃: 𝜕𝐹

𝜕𝜃 = 𝑎2 sin𝜃 cos𝜃 ��2

𝜕𝐹

𝜕�� = 𝑎2��

𝑑

𝑑𝑆(𝜕𝐹

𝜕��) = 𝑎2��

⇒ �� = sin𝜃 cos𝜃 ��2 𝑥𝑎 = 𝜙: 𝜕𝐹

𝜕𝜙 = 0

𝜕𝐹

𝜕�� = 𝑎2 sin2 𝜃 ��

𝑑

𝑑𝑆(𝜕𝐹

𝜕��) =

𝑑

𝑑𝑆(𝑎2 sin2 𝜃 ��)

= 2𝑎2 sin𝜃 cos 𝜃 �� + 𝑎2 sin2 𝜃 ��

⇒ �� = −2cot 𝜃 �� Collecting the results

�� = sin𝜃 cos𝜃 ��2 (I)

�� = −2cot 𝜃 �� (II) The non-zero Christoffel symbols are

Γ 𝜙𝜙𝜃 = −sin𝜃 cos𝜃

Γ 𝜃𝜙𝜙

= Γ 𝜙𝜃𝜙

= cot 𝜃

To show that the great circle is a solution to the geodesic equations we choose to work in the plane 𝜃 =𝜋

2, so the spherical polar coordinates are

𝑥 = 𝑎 sin 𝜃 cos𝜙 = 𝑎 cos𝜙 𝑦 = 𝑎 sin 𝜃 sin𝜙 = 𝑎 sin𝜙 𝑧 = 𝑎 cos𝜃 = 0

⇒ �� = −𝑎�� sin𝜙

�� = 𝑎�� cos𝜙

⇒ ��2 + ��2 = 𝑎2��2

The right hand side we can find from the second geodesic equation (II), which is reduced to �� = 0 ⇒ �� =𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡. Next we use the law of conservation of the four velocity in Cartesian coordinates to show that the left hand side is a constant as well.

�� ⋅ �� = 𝑔𝑎𝑏𝑑𝑥𝑎

𝑑𝑠

𝑑𝑥𝑏

𝑑𝑠

= ��2 + ��2 = 1

70 (McMahon, 2006) 71 (McMahon, 2006)





And we can conclude that the great circle is a solution.

4.5.5 72NEW - Find all the time-like geodesic 𝑿(𝑻) of the Flat Space metric in two dimensions

𝒅𝑺𝟐 = −𝑿𝟐𝒅𝑻𝟐 + 𝒅𝑿𝟐:

To find 𝑋(𝑇) we need a few integrals which we can solve: a.The line element: 𝑑𝑆2 = −𝑋2𝑑𝑇2 + 𝑑𝑋2

⇒ 1 = (𝑑𝑋

𝑑𝑆)2

− 𝑋2 (𝑑𝑇

𝑑𝑆)2

(I)

b. Killing vectors: Because the metric is independent of 𝑇 a Killing vector is 𝝃 = (𝜉𝑇 , 𝜉𝑋) = (1,0) According to (8.32) 𝝃 ⋅ 𝒖 is a conserved quantity along a geodesic, where

𝒖 = (𝑢𝑇 , 𝑢𝑋) = (𝑑𝑇

𝑑𝑆,𝑑𝑋

𝑑𝑆)

⇒ 𝝃 ⋅ 𝒖 = 𝜉𝑎𝑢𝑎

= 𝑔𝑎𝑏𝜉𝑏𝑢𝑎

= 𝑔𝑇𝑇𝜉𝑇𝑢𝑇 + 𝑔𝑋𝑋𝜉

𝑋𝑢𝑋

= −𝑋2 ⋅ 1 ⋅𝑑𝑇

𝑑𝑆+ 1 ⋅ 0 ⋅

𝑑𝑋

𝑑𝑆

= −𝑋2

𝑑𝑇

𝑑𝑆

⇒ 𝑋2𝑑𝑇

𝑑𝑆 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 = 𝐾 (II)

Inserting (II) into (I)

1 = (𝑑𝑋

𝑑𝑆)2

− 𝑋2 (𝐾

𝑋2)2

= (

𝑑𝑋

𝑑𝑆)2

−𝐾2

𝑋2

⇒ 𝑑𝑋

𝑑𝑆 = √1 +

𝐾2

𝑋2 (III)

Dividing (III) by (II)

⇒

𝑑𝑋𝑑𝑆

𝑋2𝑑𝑇𝑑𝑆

=√1 +

𝐾2

𝑋2

𝐾

⇒ 𝑑𝑋

𝑑𝑇 =

𝑋2

𝐾√1 +

𝐾2

𝑋2

=𝑋

𝐾√𝑋2 + 𝐾2

⇒ 𝑑𝑇 =𝐾

𝑋√𝑋2 + 𝐾2𝑑𝑋

⇒ 𝑇 − 𝑇∗ = 𝐾∫𝑑𝑋

𝑋√𝑋2 + 𝐾2

= 73 − ln(

𝐾 + √𝑋2 + 𝐾2

𝑋) (

𝐾 + √𝑋2 + 𝐾2

𝑋) > 0 𝑖𝑓 𝑋 > 0

72 (Hartle, 2003, p. 184) problem 8-9

73 ∫𝑑𝑥

𝑥√𝑥2+𝑎2= −

1

𝑎ln(

𝑎+√𝑥2+𝑎2

𝑥) (Spiegel, 1990) (14.186)





Isolating 𝑋

𝐾 + √𝑋2 + 𝐾2

𝑋 = exp(−(𝑇 − 𝑇∗))

⇒ √1 + (𝐾

𝑋)2

= exp(−(𝑇 − 𝑇∗)) −𝐾

𝑋

⇒ 1 + (𝐾

𝑋)2

= (exp(−(𝑇 − 𝑇∗)) −𝐾

𝑋)2

= exp(−2(𝑇 − 𝑇∗)) + (

𝐾

𝑋)2

−2𝐾

𝑋exp(−(𝑇 − 𝑇∗))

⇒ 2𝐾

𝑋 =

exp(−2(𝑇 − 𝑇∗)) − 1

exp(−(𝑇 − 𝑇∗))

= exp(−(𝑇 − 𝑇∗)) − exp(𝑇 − 𝑇∗)

= −2sinh(𝑇 − 𝑇∗) And we find the geodesics

𝑋(𝑇) = −𝐾

sinh(𝑇 − 𝑇∗)

4.5.5.1 Are these geodesics space-like or time-like:

𝑑𝑆2 = −𝑋2𝑑𝑇2 + 𝑑𝑋2

= (−𝑋2 + (𝑑𝑋

𝑑𝑇)2

)𝑑𝑇2

=

(

−(−𝐾

sinh(𝑇 − 𝑇∗))2

+(𝑑 (−

𝐾sinh(𝑇 − 𝑇∗)

)

𝑑𝑇)

2

)

𝑑𝑇2

= (−𝐾2

sinh2(𝑇 − 𝑇∗)+ (

𝐾 cosh(𝑇 − 𝑇∗)

sinh2(𝑇 − 𝑇∗))

2

)𝑑𝑇2

=𝐾2

sinh2(𝑇 − 𝑇∗)(−1 + coth2(𝑇 − 𝑇∗))𝑑𝑇2

> 0 𝑏𝑒𝑐𝑎𝑢𝑠𝑒 coth2(𝑇 − 𝑇∗) > 1 So these geodesics are space-like, which we would expect because the metric is a flat-space metric, where the geodesics in other coordinates are straight lines.

4.5.5.2 74Is the world-line 𝑿(𝑻) = 𝑨𝒄𝒐𝒔𝒉(𝑻) time-like or space-like:

𝑑𝑆2 = −𝑋2𝑑𝑇2 + 𝑑𝑋2

= (−𝑋2 + (𝑑𝑋

𝑑𝑇)2

)𝑑𝑇2

= (−𝑋2 + (𝑑(𝐴 cosh𝑇)

𝑑𝑇)

2

)𝑑𝑇2

= 𝐴2(− cosh2 𝑇 + sinh2 𝑇)𝑑𝑇2 = −𝐴2𝑑𝑇2 < 0 i.e. the world-line is time-like.

74 (Hartle, 2003, p. 143) Example 7.3.





4.6 Killing Vectors

4.6.1 75Show that if the Lie derivative of the metric tensor with respect to vector X vanishes

(𝑳𝑿𝒈𝒂𝒃 = 𝟎), the vector X satisfies the Killing equation. - Alternative version

Expressions needed:

Killings equation: ∇b𝑋𝑎 + ∇𝑎𝑋𝑏 = 0 (8.1) The Lie derivative of a (0,2) tensor: 𝐿𝑉𝑇𝑎𝑏 = 𝑉𝑐∇𝑐𝑇𝑎𝑏 + 𝑇𝑐𝑏∇𝑎𝑉

𝑐 + 𝑇𝑎𝑐∇𝑏𝑉𝑐 (4.28)

∇𝑐𝑔𝑎𝑏 = 0 (4.18)

Now we can Calculate the Lie-derivative of 𝑔𝑎𝑏 (4.28) and using (4.18) 𝐿𝑋𝑔𝑎𝑏 = 𝑋𝑐∇𝑐𝑔𝑎𝑏 + 𝑔𝑐𝑏∇𝑎𝑋

𝑐 + 𝑔𝑎𝑐∇𝑏𝑋𝑐

= ∇𝑎(𝑔𝑐𝑏𝑋𝑐) + ∇𝑏(𝑔𝑎𝑐𝑋

𝑐) = ∇𝑎𝑋𝑏 + ∇𝑏𝑋𝑎 If 𝐿𝑋𝑔𝑎𝑏 = 0 this implies that ∇𝑎𝑋𝑏 + ∇𝑏𝑋𝑎 = 0, which is the Killing equation.

4.6.2 76Prove that 𝛁𝒃𝛁𝒂𝑿𝒃 = 𝑹𝒂𝒄𝑿

𝒄

We have ∇𝑐∇b𝑋

𝑎 = 𝑅 𝑏𝑐𝑑𝑎 𝑋𝑑 (8.6)

Contracting with 𝛿𝑎𝑐:

⇒ 𝛿𝑎𝑐(∇𝑐∇b𝑋

𝑎 ) = 𝛿𝑎𝑐(𝑅 𝑏𝑐𝑑

𝑎 𝑋𝑑)

⇒ ∇a∇b𝑋𝑎 = 𝑅 𝑏𝑎𝑑

𝑎 𝑋𝑑 ⇒ ∇a∇b𝑋

𝑎 = 𝑅𝑏𝑑𝑋𝑑 (8.7)

4.6.3 77Constructing a Conserved Current with Killing Vectors – Alternative version:

We write 𝐽 = ∇𝑎𝐽𝑎 = 𝑇𝑎𝑏(∇𝑎𝑋𝑏)

and 𝐽 = ∇b𝐽𝑏 = 𝑇𝑏𝑎(∇b𝑋𝑎)

Now 𝐽 =1

2∇𝑎𝐽

𝑎 +1

2∇b𝐽

𝑏

=1

2(𝑇𝑎𝑏(∇𝑎𝑋𝑏) + 𝑇

𝑏𝑎(∇b𝑋𝑎)) 𝑇𝑎𝑏 is symmetric so 𝑇𝑎𝑏 = 𝑇𝑏𝑎 p.155

=1

2𝑇𝑎𝑏((∇𝑎𝑋𝑏) + (∇b𝑋𝑎)) Killings equation (∇𝑎𝑋𝑏) + (∇b𝑋𝑎) = 0 (8.1)

= 0

4.6.4 78Given a Killing vector 𝑿 the Ricci scalar satisfies𝑿𝒄𝛁𝒄𝑹 = 𝟎:

We calculate the Lie derivative: 𝐿𝑋𝑅𝑎𝑏 = 𝑋𝑐∇𝑐𝑅𝑎𝑏 + 𝑅𝑐𝑏∇𝑎𝑋

𝑐 + 𝑅𝑎𝑐∇𝑏𝑋𝑐 (4.28)

Multiplikation by 𝑔𝑎𝑏: 𝑔𝑎𝑏(𝐿𝑋𝑅𝑎𝑏) = 𝑔𝑎𝑏(𝑋𝑐∇𝑐𝑅𝑎𝑏) + 𝑔

𝑎𝑏(𝑅𝑐𝑏∇𝑎𝑋𝑐) + 𝑔𝑎𝑏(𝑅𝑎𝑐∇𝑏𝑋

𝑐)

Using ∇𝑐𝑔𝑎𝑏 = 0, and 𝐿𝑋𝑔

𝑎𝑏 = 0:

⇔ 𝐿𝑋𝑔𝑎𝑏𝑅𝑎𝑏 = 𝑋𝑐∇𝑐(𝑔

𝑎𝑏𝑅𝑎𝑏) + 𝑔𝑎𝑏𝑅𝑐𝑏∇𝑎𝑋

𝑐 + 𝑔𝑎𝑏𝑅𝑎𝑐∇𝑏𝑋𝑐

⇔ 𝐿𝑋𝑅 = 𝑋𝑐∇𝑐𝑅 + 𝑔𝑎𝑏𝑅𝑐𝑏∇𝑎𝑋

𝑐 + 𝑔𝑎𝑏𝑅𝑎𝑐∇𝑏𝑋𝑐

= 𝑋𝑐∇𝑐𝑅 + 𝑅𝑐 𝑎∇𝑎𝑋

𝑐 + 𝑅 𝑐𝑏 ∇𝑏𝑋

𝑐 = 𝑋𝑐∇𝑐𝑅 + 𝑔𝑏𝑐𝑅

𝑏𝑎∇𝑎𝑋𝑐 + 𝑔𝑎𝑐𝑅

𝑏𝑎∇𝑏𝑋𝑐

= 𝑋𝑐∇𝑐𝑅 + 𝑅𝑏𝑎∇𝑎(𝑔𝑏𝑐𝑋

𝑐) + 𝑅𝑏𝑎∇𝑏(𝑔𝑎𝑐𝑋𝑐)

= 𝑋𝑐∇𝑐𝑅 + 𝑅𝑏𝑎∇𝑎𝑋𝑏 + 𝑅

𝑏𝑎∇𝑏𝑋𝑎 = 𝑋𝑐∇𝑐𝑅 + 𝑅

𝑏𝑎(∇𝑎𝑋𝑏 + ∇𝑏𝑋𝑎)

75 (McMahon, 2006, p. 168), example 8-1 76 (McMahon, 2006, p. 177) equation (8.7) 77 (McMahon, 2006, p. 178) 78 (McMahon, 2006, p. 179), quiz 8-3, the answer to Quiz 8-3 is (a)





= 𝑋𝑐∇𝑐𝑅 Now the Lie derivative of a scalar is zero 𝐿𝑋𝑅 = 0, so 𝑋𝑐∇𝑐𝑅 = 0

4.7 79The Riemann tensor Prove: 𝑅𝑎𝑏𝑐𝑑 = 𝜕𝑐Γ𝑎𝑏𝑑 − 𝜕𝑑Γ𝑎𝑏𝑐 − Γ𝑒𝑎𝑐Γ 𝑏𝑑

𝑒 + Γ𝑒𝑎𝑑Γ 𝑏𝑐𝑒 (4.42)

First we need ∇𝑐𝑔𝑎𝑏 = 𝜕𝑐𝑔𝑎𝑏 − Γ 𝑎𝑐

𝑑 𝑔𝑑𝑏 − Γ 𝑐𝑏𝑑 𝑔𝑎𝑑 = 𝜕𝑐𝑔𝑎𝑏 − Γ𝑏𝑎𝑐 − Γ𝑎𝑏𝑐 = 0 (4.18)

𝑅𝑎𝑏𝑐𝑑 = 𝑔𝑎𝑒𝑅 𝑏𝑐𝑑

𝑒 = 𝑔𝑎𝑒(𝜕𝑐Γ 𝑏𝑑

𝑒 − 𝜕𝑑Γ 𝑏𝑐𝑒 + Γ𝑓𝑏𝑑Γ 𝑓𝑐

𝑒 − Γ𝑓𝑏𝑐Γ 𝑓𝑑𝑒 ) (4.41)

= 𝑔𝑎𝑒𝜕𝑐Γ 𝑏𝑑𝑒 − 𝑔𝑎𝑒𝜕𝑑Γ 𝑏𝑐

𝑒 + 𝑔𝑎𝑒Γ𝑓𝑏𝑑Γ 𝑓𝑐

𝑒 − 𝑔𝑎𝑒Γ𝑓𝑏𝑐Γ 𝑓𝑑

𝑒

= 𝜕𝑐(𝑔𝑎𝑒Γ 𝑏𝑑𝑒 ) − (𝜕𝑐𝑔𝑎𝑒)Γ 𝑏𝑑

𝑒 − 𝜕𝑑(𝑔𝑎𝑒Γ 𝑏𝑐𝑒 ) + (𝜕𝑑𝑔𝑎𝑒)Γ 𝑏𝑐

𝑒 + Γ𝑓𝑏𝑑𝑔𝑎𝑒Γ 𝑓𝑐𝑒

− Γ𝑓𝑏𝑐𝑔𝑎𝑒Γ 𝑓𝑑𝑒

= 𝜕𝑐Γ𝑎𝑏𝑑 − (Γ𝑒𝑎𝑐 + Γ𝑎𝑒𝑐)Γ 𝑏𝑑𝑒 − 𝜕𝑑Γ𝑎𝑏𝑐 + (Γ𝑒𝑎𝑑 + Γ𝑎𝑒𝑑)Γ 𝑏𝑐

𝑒 + Γ𝑓𝑏𝑑Γ𝑎𝑓𝑐 − Γ𝑓𝑏𝑐Γ𝑎𝑓𝑑

= 𝜕𝑐Γ𝑎𝑏𝑑 − 𝜕𝑑Γ𝑎𝑏𝑐 − (Γ𝑒𝑎𝑐 + Γ𝑎𝑒𝑐)Γ 𝑏𝑑𝑒 + (Γ𝑒𝑎𝑑 + Γ𝑎𝑒𝑑)Γ 𝑏𝑐

𝑒 + Γ𝑒𝑏𝑑Γ𝑎𝑒𝑐 − Γe𝑏𝑐Γ𝑎𝑒𝑑

= 𝜕𝑐Γ𝑎𝑏𝑑 − 𝜕𝑑Γ𝑎𝑏𝑐 − Γ𝑒𝑎𝑐Γ 𝑏𝑑𝑒 + Γ𝑒𝑎𝑑Γ 𝑏𝑐

𝑒 (4.42)

Prove: 𝑅𝑎𝑏𝑐𝑑 =1

2(𝜕2𝑔𝑎𝑑

𝜕𝑥𝑐𝜕𝑥𝑏+

𝜕2𝑔𝑏𝑐

𝜕𝑥𝑑𝜕𝑥𝑎−

𝜕2𝑔𝑎𝑐

𝜕𝑥𝑑𝜕𝑥𝑏−

𝜕2𝑔𝑏𝑑

𝜕𝑥𝑐𝜕𝑥𝑎) − Γ𝑒𝑎𝑐Γ 𝑏𝑑

𝑒 + Γ𝑒𝑎𝑑Γ 𝑏𝑐𝑒 (4.43)

𝑅𝑎𝑏𝑐𝑑 = 𝜕𝑐Γ𝑎𝑏𝑑 − 𝜕𝑑Γ𝑎𝑏𝑐 − Γ𝑒𝑎𝑐Γ 𝑏𝑑𝑒 + Γ𝑒𝑎𝑑Γ 𝑏𝑐

𝑒 (4.42) =1

2𝜕𝑐 (

𝜕𝑔𝑎𝑏𝜕𝑥𝑑

+𝜕𝑔𝑎𝑑𝜕𝑥𝑏

−𝜕𝑔𝑏𝑑𝜕𝑥𝑎

) −1

2𝜕𝑑 (

𝜕𝑔𝑎𝑏𝜕𝑥𝑐

+𝜕𝑔𝑎𝑐𝜕𝑥𝑏

−𝜕𝑔𝑏𝑐𝜕𝑥𝑎

) − Γ𝑒𝑎𝑐Γ 𝑏𝑑𝑒 + Γ𝑒𝑎𝑑Γ 𝑏𝑐

𝑒

=1

2(𝜕2𝑔𝑎𝑏𝜕𝑥𝑐𝜕𝑥𝑑

+𝜕2𝑔𝑎𝑑𝜕𝑥𝑐𝜕𝑥𝑏

−𝜕2𝑔𝑏𝑑𝜕𝑥𝑐𝜕𝑥𝑎

)−1

2(𝜕2𝑔𝑎𝑏𝜕𝑥𝑑𝜕𝑥𝑐

+𝜕2𝑔𝑎𝑐𝜕𝑥𝑑𝜕𝑥𝑏

−𝜕2𝑔𝑏𝑐𝜕𝑥𝑑𝜕𝑥𝑎

) − Γ𝑒𝑎𝑐Γ 𝑏𝑑𝑒

+ Γ𝑒𝑎𝑑Γ 𝑏𝑐𝑒

=1

2(𝜕2𝑔𝑎𝑑𝜕𝑥𝑐𝜕𝑥𝑏

+𝜕2𝑔𝑏𝑐𝜕𝑥𝑑𝜕𝑥𝑎

−𝜕2𝑔𝑎𝑐𝜕𝑥𝑑𝜕𝑥𝑏



𝑒 (4.43)

Prove that (4.44):𝑅𝑎𝑏𝑐𝑑 = 𝑅𝑐𝑑𝑎𝑏 = −𝑅𝑎𝑏𝑑𝑐 = −𝑅𝑏𝑎𝑐𝑑 and 𝑅𝑎𝑏𝑐𝑑 + 𝑅𝑎𝑐𝑑𝑏 + 𝑅𝑎𝑑𝑏𝑐 = 0

𝑅𝑎𝑏𝑐𝑑 =1






𝑒 (4.43)

𝑅𝑐𝑑𝑎𝑏 =1

2(𝜕2𝑔𝑐𝑏𝜕𝑥𝑎𝜕𝑥𝑑

+𝜕2𝑔𝑑𝑎𝜕𝑥𝑏𝜕𝑥𝑐

−𝜕2𝑔𝑐𝑎𝜕𝑥𝑏𝜕𝑥𝑑

−𝜕2𝑔𝑑𝑏𝜕𝑥𝑎𝜕𝑥𝑐

) − Γ𝑒𝑐𝑎Γ 𝑑𝑏𝑒 + Γ𝑒𝑐𝑏Γ 𝑑𝑎

𝑒

=1





) − Γ𝑒𝑐𝑎Γ 𝑑𝑏𝑒 + 𝑔𝑒𝑓Γ 𝑐𝑏

𝑓𝑔𝑒𝑓Γ𝑓𝑑𝑎

=1





) − Γ𝑒𝑐𝑎Γ 𝑑𝑏𝑒 + (𝑔𝑒𝑓𝑔

𝑒𝑓)Γ 𝑐𝑏𝑓

Γ𝑓𝑑𝑎

= 801





) − Γ𝑒𝑐𝑎Γ 𝑑𝑏𝑒 + Γ 𝑐𝑏

𝑓Γ𝑓𝑑𝑎

=1





) − Γ𝑒𝑐𝑎Γ 𝑑𝑏𝑒 + Γ 𝑐𝑏

𝑒 Γ𝑒𝑑𝑎 = 𝑅𝑎𝑏𝑐𝑑 Q.E.D.

79 (McMahon, 2006, p. 86), equations (4.42), (4.43) and (4.44) 80 𝑔𝑎𝑏𝑔

𝑏𝑐 = 𝛿𝑎𝑐 if the metric is diagonal





𝑅𝑎𝑏𝑑𝑐 =1

2(𝜕2𝑔𝑎𝑐𝜕𝑥𝑑𝜕𝑥𝑏

+𝜕2𝑔𝑏𝑑𝜕𝑥𝑐𝜕𝑥𝑎

−𝜕2𝑔𝑎𝑑𝜕𝑥𝑐𝜕𝑥𝑏


) − Γ𝑒𝑎𝑑Γ 𝑏𝑐𝑒 + Γ𝑒𝑎𝑐Γ 𝑏𝑑

𝑒

= −𝑅𝑎𝑏𝑐𝑑 Q.E.D.

𝑅𝑏𝑎𝑐𝑑 =1

2(𝜕2𝑔𝑏𝑑𝜕𝑥𝑐𝜕𝑥𝑎




) − Γ𝑒𝑏𝑐Γ 𝑎𝑑𝑒 + Γ𝑒𝑏𝑑Γ 𝑎𝑐

𝑒

=1





) − 𝑔𝑒𝑓Γ 𝑏𝑐𝑓

𝑔𝑒𝑓Γ𝑓𝑎𝑑 + 𝑔𝑒𝑓Γ 𝑏𝑑𝑓

𝑔𝑒𝑓Γ𝑓𝑎𝑐

=1





) − 𝑔𝑒𝑓𝑔𝑒𝑓Γ 𝑏𝑐

𝑓Γ𝑓𝑎𝑑 + 𝑔𝑒𝑓𝑔

𝑒𝑓Γ 𝑏𝑑𝑓

Γ𝑓𝑎𝑐

=1





) − Γ 𝑏𝑐𝑓

Γ𝑓𝑎𝑑 + Γ 𝑏𝑑𝑓

Γ𝑓𝑎𝑐

=1





) − Γ 𝑏𝑐𝑒 Γ𝑒𝑎𝑑 + Γ 𝑏𝑑

𝑒 Γ𝑒𝑎𝑐


𝑅𝑎𝑐𝑑𝑏 =1


+𝜕2𝑔𝑐𝑑𝜕𝑥𝑏𝜕𝑥𝑎

−𝜕2𝑔𝑎𝑑𝜕𝑥𝑏𝜕𝑥𝑐

−𝜕2𝑔𝑐𝑏𝜕𝑥𝑑𝜕𝑥𝑎

) − Γ𝑒𝑎𝑑Γ 𝑐𝑏𝑒 + Γ𝑒𝑎𝑏Γ 𝑐𝑑

𝑒

𝑅𝑎𝑑𝑏𝑐 =1

2(𝜕2𝑔𝑎𝑐𝜕𝑥𝑏𝜕𝑥𝑑

+𝜕2𝑔𝑑𝑏𝜕𝑥𝑐𝜕𝑥𝑎

−𝜕2𝑔𝑎𝑏𝜕𝑥𝑐𝜕𝑥𝑑

−𝜕2𝑔𝑑𝑐𝜕𝑥𝑏𝜕𝑥𝑎

) − Γ𝑒𝑎𝑏Γ 𝑑𝑐𝑒 + Γ𝑒𝑎𝑐Γ 𝑑𝑏

𝑒

𝑅𝑎𝑐𝑑𝑏 + 𝑅𝑎𝑑𝑏𝑐 =1


+𝜕2𝑔𝑐𝑑𝜕𝑥𝑏𝜕𝑥𝑎



) − Γ𝑒𝑎𝑑Γ 𝑐𝑏𝑒 + Γ𝑒𝑎𝑏Γ 𝑐𝑑

𝑒

+1



−𝜕2𝑔𝑎𝑏𝜕𝑥𝑐𝜕𝑥𝑑

−𝜕2𝑔𝑑𝑐𝜕𝑥𝑏𝜕𝑥𝑎

)− Γ𝑒𝑎𝑏Γ 𝑑𝑐𝑒 + Γ𝑒𝑎𝑐Γ 𝑑𝑏

𝑒

=1





) − Γ𝑒𝑎𝑑Γ 𝑐𝑏𝑒 + Γ𝑒𝑎𝑐Γ 𝑑𝑏

𝑒


Also notice 𝑅𝑎𝑎𝑎𝑎 = 𝜕𝑐Γ𝑎𝑏𝑑 − 𝜕𝑑Γ𝑎𝑏𝑐 − Γ𝑒𝑎𝑐Γ 𝑏𝑑

𝑒 + Γ𝑒𝑎𝑑Γ 𝑏𝑐𝑒 = 𝜕𝑎Γ𝑎𝑎𝑎 − 𝜕𝑎Γ𝑎𝑎𝑎 − Γ𝑒𝑎𝑎Γ 𝑎𝑎

𝑒 + Γ𝑒𝑎𝑎Γ 𝑎𝑎𝑒 (4.42)

= 0 𝑅𝑏𝑎𝑎𝑎 = 𝜕𝑎Γ𝑏𝑎𝑎 − 𝜕𝑎Γ𝑏𝑎𝑎 − Γ𝑒𝑏𝑎Γ 𝑎𝑎

𝑒 + Γ𝑒𝑏𝑎Γ 𝑎𝑎𝑒 = 0

𝑅𝑎𝑎𝑏𝑎 =1






𝑒 (4.43)

=1

2(𝜕2𝑔𝑎𝑎𝜕𝑥𝑏𝜕𝑥𝑎

+𝜕2𝑔𝑎𝑏𝜕𝑥𝑎𝜕𝑥𝑎

−𝜕2𝑔𝑎𝑏𝜕𝑥𝑎𝜕𝑥𝑎

−𝜕2𝑔𝑎𝑎𝜕𝑥𝑏𝜕𝑥𝑎

) − Γ𝑒𝑎𝑏Γ 𝑎𝑎𝑒 + Γ𝑒𝑎𝑎Γ 𝑎𝑏

𝑒

= −𝑔𝑓𝑒Γ𝑓𝑎𝑏𝑔

𝑓𝑒Γ𝑓𝑎𝑎 + Γ𝑓𝑎𝑎Γ 𝑎𝑏𝑓

= −𝑔𝑓𝑒𝑔𝑓𝑒Γ𝑓𝑎𝑏Γ𝑓𝑎𝑎 + Γ𝑓𝑎𝑎Γ 𝑎𝑏

𝑓= 0

𝑅𝑎𝑎𝑏𝑏 =1

2(𝜕2𝑔𝑎𝑏𝜕𝑥𝑏𝜕𝑥𝑎

+𝜕2𝑔𝑎𝑏𝜕𝑥𝑏𝜕𝑥𝑎

−𝜕2𝑔𝑎𝑏𝜕𝑥𝑏𝜕𝑥𝑎

−𝜕2𝑔𝑎𝑏𝜕𝑥𝑏𝜕𝑥𝑎

) − Γ𝑒𝑎𝑏Γ 𝑎𝑏𝑒 + Γ𝑒𝑎𝑏Γ 𝑎𝑏

𝑒 = 0

𝑅𝑎𝑎𝑏𝑐 =1

2(𝜕2𝑔𝑎𝑐𝜕𝑥𝑏𝜕𝑥𝑎

+𝜕2𝑔𝑎𝑏𝜕𝑥𝑐𝜕𝑥𝑎

−𝜕2𝑔𝑎𝑏𝜕𝑥𝑐𝜕𝑥𝑎

−𝜕2𝑔𝑎𝑐𝜕𝑥𝑏𝜕𝑥𝑎

) − Γ𝑒𝑎𝑏Γ 𝑎𝑐𝑒 + Γ𝑒𝑎𝑐Γ 𝑎𝑏

𝑒 = 0

𝑅𝑎𝑏𝑐𝑐 =1

2(𝜕2𝑔𝑎𝑐𝜕𝑥𝑐𝜕𝑥𝑏

+𝜕2𝑔𝑏𝑐𝜕𝑥𝑐𝜕𝑥𝑎

−𝜕2𝑔𝑎𝑐𝜕𝑥𝑐𝜕𝑥𝑏

−𝜕2𝑔𝑏𝑐𝜕𝑥𝑐𝜕𝑥𝑎

) − Γ𝑒𝑎𝑐Γ 𝑏𝑐𝑒 + Γ𝑒𝑎𝑐Γ 𝑏𝑐

𝑒 = 0





4.7.1 81Independent elements in the Riemann, Ricci and Weyl tensor

In 𝑛 dimensions, there are 𝑁𝑅𝑖𝑒𝑚𝑎𝑛𝑛 = 𝑛2(𝑛2 − 1)/12 independent elements in the Riemann tensor. In

the Ricci tensor there are 𝑁𝑅𝑖𝑐𝑐𝑖 = 𝑛(𝑛 + 1)/2 independent elements, and in the Weyl tensor there are 𝑁𝑊𝑒𝑦𝑙 = 10 independent elements if 𝑛 = 4, if 𝑛 < 4 there are none. Summarized:

𝑁𝑅𝑖𝑒𝑚𝑎𝑛𝑛 =

𝑛2(𝑛2 − 1)

12 𝑁𝑅𝑖𝑐𝑐𝑖 =

𝑛(𝑛 + 1)

2 𝑁𝑊𝑒𝑦𝑙

𝑛 = 2 1 𝑅1212 3 𝑅11 𝑅12 𝑅22

0

𝑛 = 3 6 𝑅1212 𝑅1213 𝑅1223 𝑅1313 𝑅1323 𝑅2323

6 𝑅11 𝑅22 𝑅23 𝑅13 𝑅12 𝑅33

0

𝑛 = 4 20 𝑅𝑎𝑏𝑐𝑑 10 𝑅𝑎𝑏 = 𝑅𝑐𝑎𝑐𝑏 10 𝐶𝑎𝑏𝑐𝑑82

𝑅1212 𝑅1213 𝑅1214 𝑅1223 𝑅1224 𝑅1234 𝑅1313 𝑅1314 𝑅1323 𝑅1324 𝑅1334 𝑅1414

𝑅142383 𝑅1424 𝑅1434 𝑅2323 𝑅2324 𝑅2334 𝑅2424 𝑅2434 𝑅3434

𝑅11 𝑅22 𝑅23 𝑅24 𝑅13 𝑅14 𝑅33 𝑅12 𝑅34 𝑅44

𝐶1223 𝐶1224 𝐶1234 𝐶1323 𝐶1324 𝐶1334

𝐶142384 𝐶1424 𝐶1434 𝐶2334 𝐶2434

4.7.2 85Compute the components of the Riemann tensor for the unit 2-sphere

The number of independent elements in the Riemann tensor in a metric of dimension 𝑛 = 2 is

𝑁 =𝑛2(𝑛2−1)

12= 1 so we can choose to calculate 𝑅 𝜙𝜃𝜙

𝜃

81 (McMahon, 2006, p. 87) 82 The Weyl tensor possesses the same symmetries as the Riemann tensor: 𝐶𝑎𝑏𝑐𝑑 = −𝐶𝑎𝑏𝑑𝑐 = −𝐶𝑏𝑎𝑐𝑑 = 𝐶𝑐𝑑𝑎𝑏 and 𝐶𝑎𝑏𝑐𝑑 + 𝐶𝑎𝑑𝑏𝑐 + 𝐶𝑎𝑐𝑑𝑏 = 0. It possesses an additional symmetry: 𝐶𝑐𝑎𝑐𝑏 = 0. It follows that the Weyl tensor is trace-free, in other words, it vanishes for any pair of contracted indices. One can think of the Weyl tensor as that part of the curvature tensor for which all contractions vanish (d'Inverno, 1992, p. 88) 83 Because: 𝑅𝑎𝑏𝑐𝑑 + 𝑅𝑎𝑐𝑑𝑏 + 𝑅𝑎𝑑𝑏𝑐 = 0 84 Because: 𝐶𝑎𝑏𝑐𝑑 + 𝐶𝑎𝑐𝑑𝑏 + 𝐶𝑎𝑑𝑏𝑐 = 0 85 (McMahon, 2006, p. 87), example 4-11





The Riemann tensor 𝑅 𝑏𝑐𝑑𝑎 = 𝜕𝑐Γ 𝑏𝑑



𝑒 Γ 𝑒𝑑𝑎 (4.41)

We choose 𝑎 = 𝑐 = 𝜃, and 𝑏 = 𝑑 = 𝜙

𝑅 𝜙𝜃𝜙𝜃 = 𝜕𝜃Γ 𝜙𝜙

𝜃 − 𝜕𝜙Γ 𝜙𝜃𝜃 + Γ 𝜙𝜙

𝑒 Γ 𝑒𝜃𝜃 − Γ 𝜙𝜃

𝑒 Γ 𝑒𝜙𝜃

= 𝜕𝜃Γ 𝜙𝜙𝜃 + Γ 𝜙𝜙



Sum over 𝑒: = 𝜕𝜃Γ 𝜙𝜙𝜃 + Γ 𝜙𝜙

𝜃 Γ 𝜃𝜃𝜃 − Γ 𝜙𝜃

𝜃 Γ 𝜃𝜙𝜃 + Γ 𝜙𝜙

𝜙Γ 𝜙𝜃𝜃 − Γ 𝜙𝜃

𝜙Γ 𝜙𝜙𝜃

= 𝜕𝜃Γ 𝜙𝜙𝜃 − Γ 𝜙𝜃

𝜙Γ 𝜙𝜙𝜃

= 𝜕𝜃(− sin 𝜃 cos 𝜃) − cot 𝜃 (− sin 𝜃 cos 𝜃) = −cos2 𝜃 + sin2 𝜃 + cos2 𝜃 = sin2 𝜃

4.8 86Show that the Ricci scalar 𝑹 = 𝟐 for the unit 2-sphere

The line element: 𝑑𝑠2 = 𝑑𝜃2 + sin2 𝜃 𝑑𝜙2

The metric tensor and its inverse: 𝑔𝑎𝑏 = {1

sin2 𝜃} 𝑔𝑎𝑏 = {

11

sin2 𝜃

}

The Ricci scalar: 𝑅 = 𝑔𝑎𝑏𝑅𝑎𝑏 (4.47)

Sum over a: 𝑅 = 𝑔𝜃𝑏𝑅𝜃𝑏 + 𝑔𝜙𝑏𝑅𝜙𝑏

Sum over b: = 𝑔𝜃𝜃𝑅𝜃𝜃 + 𝑔𝜃𝜙𝑅𝜃𝜙 + 𝑔

𝜙𝜃𝑅𝜙𝜃 + 𝑔𝜙𝜙𝑅𝜙𝜙 𝑔𝜃𝜙 = 𝑔𝜙𝜃 = 0

= 𝑔𝜃𝜃𝑅𝜃𝜃 + 𝑔𝜙𝜙𝑅𝜙𝜙

= 𝑔𝜃𝜃𝑅 𝜃𝑐𝜃𝑐 + 𝑔𝜙𝜙𝑅 𝜙𝑐𝜙

𝑐 𝑅𝑎𝑏 = 𝑅 𝑎𝑐𝑏𝑐 (4.46)

Sum over c = 𝑔𝜃𝜃𝑅 𝜃𝜃𝜃𝜃 + 𝑔𝜃𝜃𝑅 𝜃𝜙𝜃

𝜙+ 𝑔𝜙𝜙𝑅 𝜙𝜃𝜙

𝜃 + 𝑔𝜙𝜙𝑅 𝜙𝜙𝜙𝜙

𝑅 𝜃𝜃𝜃𝜃 = 𝑅 𝜙𝜙𝜙

𝜙= 0

= 𝑔𝜃𝜃𝑅 𝜃𝜙𝜃𝜙

+ 𝑔𝜙𝜙𝑅 𝜙𝜃𝜙𝜃

= 𝑔𝜃𝜃𝑔𝜙𝜙𝑅𝜙𝜃𝜙𝜃 + 𝑔𝜙𝜙𝑔𝜃𝜃𝑅𝜃𝜙𝜃𝜙 𝑅𝜙𝜃𝜙𝜃 = 𝑅𝜃𝜙𝜃𝜙 (4.44)

= 2𝑔𝜃𝜃𝑔𝜙𝜙𝑅𝜃𝜙𝜃𝜙

= 2𝑔𝜙𝜙𝑅 𝜙𝜃𝜙𝜃 𝑅 𝜙𝜃𝜙

𝜃 = sin2 𝜃 ex 4-11

= 21

sin2 𝜃sin2 𝜃

= 2

Remark that

𝑅 = 2𝑔𝜙𝜙𝑅 𝜙𝜃𝜙𝜃

is a general solution for a 2-dimensional diagonal metric if we write:

𝑅 = 2𝑔22𝑅 2121 (S1)






4.9 87Proof: if a space is conformally flat, i.e. 𝒈𝒂𝒃(𝒙) = 𝒇(𝒙)𝜼𝒂𝒃 the Weyl tensor

vanishes The Weyl scalar

𝐶𝑎𝑏𝑐𝑑 = 𝑅𝑎𝑏𝑐𝑑 +1

2(𝑔𝑎𝑑𝑅𝑐𝑏 + 𝑔𝑏𝑐𝑅𝑑𝑎 − 𝑔𝑎𝑐𝑅𝑑𝑏 − 𝑔𝑏𝑑𝑅𝑐𝑎) +

1

6(𝑔𝑎𝑐𝑔𝑑𝑏 − 𝑔𝑎𝑑𝑔𝑐𝑏)𝑅 (4.49)

= 𝑅𝑎𝑏𝑐𝑑 +

1

2(𝑔𝑎𝑑𝑅

𝑒𝑐𝑒𝑏 + 𝑔𝑏𝑐𝑅

𝑒𝑑𝑒𝑎 − 𝑔𝑎𝑐𝑅

𝑒𝑑𝑒𝑏 − 𝑔𝑏𝑑𝑅

𝑒𝑐𝑒𝑎) +

1

6(𝑔𝑎𝑐𝑔𝑑𝑏 − 𝑔𝑎𝑑𝑔𝑐𝑏)𝑔

𝑒𝑓𝑅ℎ𝑒ℎ𝑓


1

2(𝑔𝑎𝑑𝑔

𝑒𝑓𝑅𝑓𝑐𝑒𝑏 + 𝑔𝑏𝑐𝑔𝑒𝑓𝑅𝑓𝑑𝑒𝑎 − 𝑔𝑎𝑐𝑔

𝑒𝑓𝑅𝑓𝑑𝑒𝑏 − 𝑔𝑏𝑑𝑔𝑒𝑓𝑅𝑓𝑐𝑒𝑎)

+1

6(𝑔𝑎𝑐𝑔𝑑𝑏 − 𝑔𝑎𝑑𝑔𝑐𝑏)𝑔

𝑒𝑓𝑔ℎ𝑗𝑅𝑗𝑒ℎ𝑓


1

2𝑓(𝑥)2(𝜂𝑎𝑑𝜂

𝑒𝑓𝑅𝑓𝑐𝑒𝑏 + 𝜂𝑏𝑐𝜂𝑒𝑓𝑅𝑓𝑑𝑒𝑎 − 𝜂𝑎𝑐𝜂

𝑒𝑓𝑅𝑓𝑑𝑒𝑏 − 𝜂𝑏𝑑𝜂𝑒𝑓𝑅𝑓𝑐𝑒𝑎)

+1

6𝑓(𝑥)4(𝜂𝑎𝑐𝜂𝑑𝑏 − 𝜂𝑎𝑑𝜂𝑐𝑏)𝜂

𝑒𝑓𝜂ℎ𝑗𝑅𝑗𝑒ℎ𝑓


1

2𝑓(𝑥)2(𝜂𝑎𝑑𝑅𝑒𝑐𝑒𝑏 + 𝜂𝑏𝑐𝑅𝑒𝑑𝑒𝑎 − 𝜂𝑎𝑐𝑅𝑒𝑑𝑒𝑏 − 𝜂𝑏𝑑𝑅𝑒𝑐𝑒𝑎)

+1

6𝑓(𝑥)4(𝜂𝑎𝑐𝜂𝑑𝑏 − 𝜂𝑎𝑑𝜂𝑐𝑏)𝑅ℎ𝑒ℎ𝑒

=1






𝑒

+1

2𝑓(𝑥)2 (𝜂𝑎𝑑 (

1

2(𝜕2𝑔𝑒𝑏𝜕𝑥𝑒𝜕𝑥𝑐

+𝜕2𝑔𝑐𝑒𝜕𝑥𝑏𝜕𝑥𝑒

−𝜕2𝑔𝑒𝑒𝜕𝑥𝑏𝜕𝑥𝑐

−𝜕2𝑔𝑐𝑏𝜕𝑥𝑒𝜕𝑥𝑒

) − Γ𝑓𝑒𝑒Γ 𝑐𝑏𝑓

+ Γ𝑓𝑒𝑏Γ 𝑐𝑒𝑓)

+ 𝜂𝑏𝑐 (1

2(𝜕2𝑔𝑒𝑎𝜕𝑥𝑒𝜕𝑥𝑑

+𝜕2𝑔𝑑𝑒𝜕𝑥𝑎𝜕𝑥𝑒

−𝜕2𝑔𝑒𝑒𝜕𝑥𝑎𝜕𝑥𝑑

−𝜕2𝑔𝑑𝑎𝜕𝑥𝑒𝜕𝑥𝑒

) − Γ𝑓𝑒𝑒Γ 𝑑𝑎𝑓

+ Γ𝑓𝑒𝑎Γ 𝑑𝑒𝑓)

− 𝜂𝑎𝑐 (1

2(𝜕2𝑔𝑒𝑏𝜕𝑥𝑒𝜕𝑥𝑑

+𝜕2𝑔𝑑𝑒𝜕𝑥𝑏𝜕𝑥𝑒

−𝜕2𝑔𝑒𝑒𝜕𝑥𝑏𝜕𝑥𝑑

−𝜕2𝑔𝑑𝑏𝜕𝑥𝑒𝜕𝑥𝑒

) − Γ𝑓𝑒𝑒Γ 𝑑𝑏𝑓

+ Γ𝑓𝑒𝑏Γ 𝑑𝑒𝑓)

− 𝜂𝑏𝑑 (1

2(𝜕2𝑔𝑒𝑎𝜕𝑥𝑒𝜕𝑥𝑐

+𝜕2𝑔𝑐𝑒𝜕𝑥𝑎𝜕𝑥𝑒

−𝜕2𝑔𝑒𝑒𝜕𝑥𝑎𝜕𝑥𝑐

−𝜕2𝑔𝑐𝑎𝜕𝑥𝑒𝜕𝑥𝑒

) − Γ𝑓𝑒𝑒Γ 𝑐𝑎𝑓

+ Γ𝑓𝑒𝑎Γ 𝑐𝑒𝑓))

+1

6𝑓(𝑥)4(𝜂𝑎𝑐𝜂𝑑𝑏 − 𝜂𝑎𝑑𝜂𝑐𝑏) (

1

2(𝜕2𝑔ℎ𝑒𝜕𝑥ℎ𝜕𝑥𝑒

+𝜕2𝑔𝑒ℎ𝜕𝑥𝑒𝜕𝑥ℎ

−𝜕2𝑔ℎℎ𝜕𝑥𝑒𝜕𝑥𝑒

−𝜕2𝑔𝑒𝑒𝜕𝑥ℎ𝜕𝑥ℎ

)

− Γ𝑓ℎℎΓ 𝑒𝑒𝑓+ Γ𝑓ℎ𝑒Γ 𝑒ℎ

𝑓)

87 (McMahon, 2006, p. 90)





= 881

2((𝜂𝑎𝑑)

2𝜕𝑐𝜕𝑏 + (𝜂𝑏𝑐)2𝜕𝑑𝜕𝑎 − (𝜂𝑎𝑐)

2𝜕𝑑𝜕𝑏 − (𝜂𝑏𝑑)2𝜕𝑐𝜕𝑎)𝑓(𝑥) − Γ𝑒𝑎𝑐Γ 𝑏𝑑

𝑒 + Γ𝑒𝑎𝑑Γ 𝑏𝑐𝑒

+1

2𝑓(𝑥)2 (𝜂𝑎𝑑 (

1

2((𝜂𝑒𝑏)

2𝜕𝑒𝜕𝑐 + (𝜂𝑐𝑒)2𝜕𝑏𝜕𝑒 − (𝜂𝑒𝑒)

2𝜕𝑏𝜕𝑐 − (𝜂𝑐𝑏)2𝜕𝑒𝜕𝑒)𝑓(𝑥) − Γ𝑓𝑒𝑒Γ 𝑐𝑏

𝑓

+ Γ𝑓𝑒𝑏Γ 𝑐𝑒𝑓)

+ 𝜂𝑏𝑐 (1

2((𝜂𝑒𝑎)

2𝜕𝑒𝜕𝑑 + (𝜂𝑑𝑒)2𝜕𝑎𝜕𝑒 − (𝜂𝑒𝑒)

2𝜕𝑎𝜕𝑑 − (𝜂𝑑𝑎)2𝜕𝑒𝜕𝑒)𝑓(𝑥) − Γ𝑓𝑒𝑒Γ 𝑑𝑎

𝑓+ Γ𝑓𝑒𝑎Γ 𝑑𝑒

𝑓)

− 𝜂𝑎𝑐 (1

2((𝜂𝑒𝑏)

2𝜕𝑒𝜕𝑑 + (𝜂𝑑𝑒)2𝜕𝑏𝜕𝑒 − (𝜂𝑒𝑒)

2𝜕𝑏𝜕𝑑 − (𝜂𝑑𝑏)2𝜕𝑒𝜕𝑒)𝑓(𝑥) − Γ𝑓𝑒𝑒Γ 𝑑𝑏

𝑓+ Γ𝑓𝑒𝑏Γ 𝑑𝑒

𝑓)

− 𝜂𝑏𝑑 (1

2((𝜂𝑒𝑎)

2𝜕𝑒𝜕𝑐 + (𝜂𝑐𝑒)2𝜕𝑎𝜕𝑒 − (𝜂𝑒𝑒)

2𝜕𝑎𝜕𝑐 − (𝜂𝑐𝑎)2𝜕𝑒𝜕𝑒)𝑓(𝑥) − Γ𝑓𝑒𝑒Γ 𝑐𝑎

𝑓+ Γ𝑓𝑒𝑎Γ 𝑐𝑒

𝑓))

+1

6𝑓(𝑥)4(𝜂𝑎𝑐𝜂𝑑𝑏 − 𝜂𝑎𝑑𝜂𝑐𝑏) (

1

2((𝜂ℎ𝑒)

2𝜕ℎ𝜕𝑒 + (𝜂𝑒ℎ)2𝜕𝑒𝜕ℎ − (𝜂ℎℎ)

2𝜕𝑒𝜕𝑒 − (𝜂𝑒𝑒)2𝜕ℎ𝜕ℎ)𝑓(𝑥)

− Γ𝑓ℎℎΓ 𝑒𝑒𝑓+ Γ𝑓ℎ𝑒Γ 𝑒ℎ

𝑓)

= −Γ𝑒𝑎𝑐Γ 𝑏𝑑𝑒 + Γ𝑒𝑎𝑑Γ 𝑏𝑐

𝑒

+1

2𝑓(𝑥)2 (𝜂𝑎𝑑(−Γ𝑓𝑒𝑒Γ 𝑐𝑏

𝑓+ Γ𝑓𝑒𝑏Γ 𝑐𝑒

𝑓) + 𝜂𝑏𝑐(−Γ𝑓𝑒𝑒Γ 𝑑𝑎

𝑓+ Γ𝑓𝑒𝑎Γ 𝑑𝑒

𝑓)

− 𝜂𝑎𝑐(−Γ𝑓𝑒𝑒Γ 𝑑𝑏𝑓

+ Γ𝑓𝑒𝑏Γ 𝑑𝑒𝑓) − 𝜂𝑏𝑑(−Γ𝑓𝑒𝑒Γ 𝑐𝑎

𝑓+ Γ𝑓𝑒𝑎Γ 𝑐𝑒

𝑓))

+1

6𝑓(𝑥)4(𝜂𝑎𝑐𝜂𝑑𝑏 − 𝜂𝑎𝑑𝜂𝑐𝑏)(−Γ𝑓ℎℎΓ 𝑒𝑒

𝑓+ Γ𝑓ℎ𝑒Γ 𝑒ℎ

𝑓)

= 𝑔𝑒𝑓(−Γ𝑒𝑎𝑐Γ𝑓𝑏𝑑 + Γ𝑒𝑎𝑑Γ𝑓𝑏𝑐)

+1

2𝑓(𝑥)2 (𝜂𝑎𝑑𝑔

𝑓ℎ(−Γ𝑓𝑒𝑒Γℎ𝑐𝑏 + Γ𝑓𝑒𝑏Γℎ𝑐𝑒) + 𝜂𝑏𝑐𝑔𝑓ℎ(−Γ𝑓𝑒𝑒Γℎ𝑑𝑎 + Γ𝑓𝑒𝑎Γℎ𝑑𝑒)

− 𝜂𝑎𝑐𝑔𝑓ℎ(−Γ𝑓𝑒𝑒Γℎ𝑑𝑏 + Γ𝑓𝑒𝑏Γℎ𝑑𝑒) − 𝜂𝑏𝑑𝑔

𝑓ℎ(−Γ𝑓𝑒𝑒Γℎ𝑐𝑎 + Γ𝑓𝑒𝑎Γℎ𝑐𝑒))

+1

6𝑓(𝑥)4𝑔𝑓𝑗(𝜂𝑎𝑐𝜂𝑑𝑏 − 𝜂𝑎𝑑𝜂𝑐𝑏)(−Γ𝑓ℎℎΓ𝑗𝑒𝑒 + Γ𝑓ℎ𝑒Γ𝑗𝑒ℎ)

= 𝑓(𝑥)𝜂𝑒𝑓(−Γ𝑒𝑎𝑐Γ𝑓𝑏𝑑 + Γ𝑒𝑎𝑑Γ𝑓𝑏𝑐)

+1

2𝑓(𝑥)3 (𝜂𝑎𝑑𝜂

𝑓ℎ(−Γ𝑓𝑒𝑒Γℎ𝑐𝑏 + Γ𝑓𝑒𝑏Γℎ𝑐𝑒) + 𝜂𝑏𝑐𝜂𝑓ℎ(−Γ𝑓𝑒𝑒Γℎ𝑑𝑎 + Γ𝑓𝑒𝑎Γℎ𝑑𝑒)

− 𝜂𝑎𝑐𝜂𝑓ℎ(−Γ𝑓𝑒𝑒Γℎ𝑑𝑏 + Γ𝑓𝑒𝑏Γℎ𝑑𝑒) − 𝜂𝑏𝑑𝜂

𝑓ℎ(−Γ𝑓𝑒𝑒Γℎ𝑐𝑎 + Γ𝑓𝑒𝑎Γℎ𝑐𝑒))

+1

6𝑓(𝑥)5𝜂𝑓𝑗(𝜂𝑎𝑐𝜂𝑑𝑏 − 𝜂𝑎𝑑𝜂𝑐𝑏)(−Γ𝑓ℎℎΓ𝑗𝑒𝑒 + Γ𝑓ℎ𝑒Γ𝑗𝑒ℎ)

88 ((𝜂𝑎𝑑)2𝜕𝑐𝜕𝑏 + (𝜂𝑏𝑐)

2𝜕𝑑𝜕𝑎 − (𝜂𝑎𝑐)2𝜕𝑑𝜕𝑏 − (𝜂𝑏𝑑)

2𝜕𝑐𝜕𝑎)𝑓(𝑥) = {1 + 1 − 1 − 1 = 0 𝑖𝑓 𝑎 = 𝑏 = 𝑐 = 𝑑 = 𝑥

0 𝑖𝑓 𝑎 𝑜𝑟 𝑏 𝑜𝑟 𝑐 𝑜𝑟 𝑑 ≠ 0





= 𝑓(𝑥)𝜂𝑒𝑓 (−(

1

2(𝜕𝑐𝑔𝑒𝑎 + 𝜕𝑎𝑔𝑒𝑐 − 𝜕𝑒𝑔𝑎𝑐))(

1

2(𝜕𝑑𝑔𝑓𝑏 + 𝜕𝑏𝑔𝑓𝑑 − 𝜕𝑓𝑔𝑏𝑑))

+ (1

2(𝜕𝑑𝑔𝑎𝑒 + 𝜕𝑎𝑔𝑒𝑑 − 𝜕𝑒𝑔𝑎𝑑))(

1

2(𝜕𝑐𝑔𝑓𝑏 + 𝜕𝑏𝑔𝑓𝑐 − 𝜕𝑓𝑔𝑏𝑐)))

+1


𝑓ℎ (−(1

2(𝜕𝑒𝑔𝑓𝑒 + 𝜕𝑒𝑔𝑓𝑒 − 𝜕𝑓𝑔𝑒𝑒)) (

1

2(𝜕𝑏𝑔ℎ𝑐 + 𝜕𝑐𝑔ℎ𝑏

− 𝜕ℎ𝑔𝑐𝑏)) + (1

2(𝜕𝑏𝑔𝑓𝑒 + 𝜕𝑒𝑔𝑓𝑏 − 𝜕𝑓𝑔𝑒𝑏)) (

1

2(𝜕𝑒𝑔ℎ𝑐 + 𝜕𝑐𝑔ℎ𝑒 − 𝜕ℎ𝑔𝑐𝑒)))

+ 𝜂𝑏𝑐𝜂𝑓ℎ (−(

1


1

2(𝜕𝑎𝑔ℎ𝑑 + 𝜕𝑑𝑔ℎ𝑎 − 𝜕ℎ𝑔𝑑𝑎))

+ (1

2(𝜕𝑎𝑔𝑓𝑒 + 𝜕𝑒𝑔𝑓𝑎 − 𝜕𝑓𝑔𝑒𝑎))(

1

2(𝜕𝑒𝑔ℎ𝑑 + 𝜕𝑑𝑔ℎ𝑒 − 𝜕ℎ𝑔𝑑𝑒)))

− 𝜂𝑎𝑐𝜂𝑓ℎ (−(

1


1

2(𝜕𝑏𝑔ℎ𝑑 + 𝜕𝑑𝑔ℎ𝑏 − 𝜕ℎ𝑔𝑑𝑏))

+ (1

2(𝜕𝑏𝑔𝑓𝑒 + 𝜕𝑒𝑔𝑓𝑏 − 𝜕𝑓𝑔𝑒𝑏)) (

1

2(𝜕𝑒𝑔ℎ𝑑 + 𝜕𝑑𝑔ℎ𝑒 − 𝜕ℎ𝑔𝑑𝑒)))

− 𝜂𝑏𝑑𝜂𝑓ℎ (−(

1


1

2(𝜕𝑎𝑔ℎ𝑐 + 𝜕𝑐𝑔ℎ𝑎 − 𝜕ℎ𝑔𝑐𝑎))

+ (1

2(𝜕𝑎𝑔𝑓𝑒 + 𝜕𝑒𝑔𝑓𝑎 − 𝜕𝑓𝑔𝑒𝑎))(

1

2(𝜕𝑒𝑔ℎ𝑐 + 𝜕𝑐𝑔ℎ𝑒 − 𝜕ℎ𝑔𝑐𝑒))))

+1

6𝑓(𝑥)5𝜂𝑓𝑗(𝜂𝑎𝑐𝜂𝑑𝑏

− 𝜂𝑎𝑑𝜂𝑐𝑏) (−(1

2(𝜕ℎ𝑔𝑓ℎ + 𝜕ℎ𝑔𝑓ℎ − 𝜕𝑓𝑔ℎℎ))(

1

2(𝜕𝑒𝑔𝑗𝑒 + 𝜕𝑒𝑔𝑗𝑒 − 𝜕𝑗𝑔𝑒𝑒))

+ (1

2(𝜕𝑒𝑔𝑓ℎ + 𝜕ℎ𝑔𝑓𝑒 − 𝜕𝑓𝑔ℎ𝑒))(

1

2(𝜕ℎ𝑔𝑗𝑒 + 𝜕𝑒𝑔𝑗ℎ − 𝜕𝑗𝑔𝑒ℎ)))





= 𝑓(𝑥)𝜂𝑒𝑓 (−(

1

2(𝜂𝑒𝑎𝜕𝑐 + 𝜂𝑒𝑐𝜕𝑎 − 𝜂𝑎𝑐𝜕𝑒))(

1

2(𝜂𝑓𝑏𝜕𝑑 + 𝜂𝑓𝑑𝜕𝑏 − 𝜂𝑏𝑑𝜕𝑓))

+ (1

2(𝜂𝑒𝑎𝜕𝑑 + 𝜂𝑒𝑑𝜕𝑎 − 𝜂𝑎𝑑𝜕𝑒))(

1

2(𝜂𝑓𝑏𝜕𝑐 + 𝜂𝑓𝑐𝜕𝑏 − 𝜂𝑏𝑐𝜕𝑓)))𝑓(𝑥)

+1


𝑓ℎ (−(1

2(𝜂𝑓𝑒𝜕𝑒 + 𝜂𝑓𝑒𝜕𝑒 − 𝜂𝑒𝑒𝜕𝑓 )) (

1

2(𝜂ℎ𝑐𝜕𝑏 + 𝜂ℎ𝑏𝜕𝑐

− 𝜂𝑐𝑏𝜕ℎ)) + (1

2(𝜂𝑓𝑒𝜕𝑏 + 𝜂𝑓𝑏𝜕𝑒 − 𝜂𝑒𝑏𝜕𝑓)) (

1

2(𝜂ℎ𝑐𝜕𝑒 + 𝜂𝑓𝑒𝜕𝑐 − 𝜂𝑐𝑒𝜕ℎ)))


1

2(𝜂𝑓𝑒𝜕𝑒 + 𝜂𝑓𝑒𝜕𝑒 − 𝜂𝑒𝑒𝜕𝑓))(

1

2(𝜂ℎ𝑑𝜕𝑎 + 𝜂ℎ𝑎𝜕𝑑 − 𝜂𝑑𝑎𝜕ℎ))

+ (1

2(𝜂𝑓𝑒𝜕𝑎 + 𝜂𝑓𝑎𝜕𝑒 − 𝜂𝑒𝑎𝜕𝑓))(

1

2(𝜂ℎ𝑑𝜕𝑒 + 𝜂ℎ𝑒𝜕𝑑 − 𝜂𝑑𝑒𝜕ℎ)))


1

2(𝜂𝑓𝑒𝜕𝑒 + 𝜂𝑓𝑒𝜕𝑒 − 𝜂𝑒𝑒𝜕𝑓)) (

1

2(𝜂ℎ𝑑𝜕𝑏 + 𝜂ℎ𝑏𝜕𝑑 − 𝜂𝑑𝑏𝜕ℎ))

+ (1

2(𝜂𝑓𝑒𝜕𝑏 + 𝜂𝑓𝑏𝜕𝑒 − 𝜂𝑒𝑏𝜕𝑓))(

1

2(𝜂ℎ𝑑𝜕𝑒 + 𝜂ℎ𝑒𝜕𝑑 − 𝜂𝑑𝑒𝜕ℎ)))


1

2(𝜂𝑓𝑒𝜕𝑒 + 𝜂𝑓𝑒𝜕𝑒 − 𝜂𝑒𝑒𝜕𝑓)) (

1

2(𝜂ℎ𝑐𝜕𝑎 + 𝜂ℎ𝑎𝜕𝑐 − 𝜂𝑐𝑎𝜕ℎ))

+ (1

2(𝜂𝑓𝑒𝜕𝑎 + 𝜂𝑓𝑎𝜕𝑒 − 𝜂𝑒𝑎𝜕𝑓))(

1

2(𝜂ℎ𝑐𝜕𝑒 + 𝜂ℎ𝑒𝜕𝑐 − 𝜂𝑐𝑒𝜕ℎ))))𝑓(𝑥)

+1

6𝑓(𝑥)5𝜂𝑓𝑗(𝜂𝑎𝑐𝜂𝑑𝑏

− 𝜂𝑎𝑑𝜂𝑐𝑏) (−(1

2(𝜂𝑓ℎ𝜕ℎ + 𝜂𝑓ℎ𝜕ℎ − 𝜂ℎℎ𝜕𝑓))(

1

2(𝜂𝑗𝑒𝜕𝑒 + 𝜂𝑗𝑒𝜕𝑒 − 𝜂𝑒𝑒𝜕𝑗))

+ (1

2(𝜂𝑓ℎ𝜕𝑒 + 𝜂𝑓𝑒𝜕ℎ − 𝜂ℎ𝑒𝜕𝑓)) (

1

2(𝜂𝑗𝑒𝜕ℎ + 𝜂𝑗ℎ𝜕𝑒 − 𝜂𝑒ℎ𝜕𝑗)))𝑓(𝑥)





= 89𝑓(𝑥)𝜂𝑒𝑓 (−(

1

2𝜂𝑒𝑎𝜕𝑐)(

1

2𝜂𝑓𝑏𝜕𝑑) + (

1

2𝜂𝑒𝑎𝜕𝑑) (

1

2𝜂𝑓𝑏𝜕𝑐))𝑓(𝑥)

+1


𝑓ℎ (−(1

2𝜂𝑓𝑒𝜕𝑒) (

1

2𝜂ℎ𝑐𝜕𝑏) + (

1

2𝜂𝑓𝑒𝜕𝑏) (

1

2𝜂ℎ𝑐𝜕𝑒))


1


1

2𝜂ℎ𝑑𝜕𝑎) + (

1

2𝜂𝑓𝑒𝜕𝑎) (

1

2𝜂ℎ𝑑𝜕𝑒))


1


1

2𝜂ℎ𝑑𝜕𝑏) + (

1

2𝜂𝑓𝑒𝜕𝑏) (

1

2𝜂ℎ𝑑𝜕𝑒))


1


1

2𝜂ℎ𝑐𝜕𝑎) + (

1

2𝜂𝑓𝑒𝜕𝑎) (

1

2𝜂ℎ𝑐𝜕𝑒)))𝑓(𝑥)

+1

6𝑓(𝑥)5𝜂𝑓𝑗(𝜂𝑎𝑐𝜂𝑑𝑏 − 𝜂𝑎𝑑𝜂𝑐𝑏) (−(

1

2𝜂𝑓ℎ𝜕ℎ) (

1

2𝜂𝑗𝑒𝜕𝑒)

+ (1

2𝜂𝑓ℎ𝜕𝑒)(

1

2𝜂𝑗𝑒𝜕ℎ))𝑓(𝑥)

=1

4𝑓(𝑥)𝜂𝑒𝑓(−𝜂𝑒𝑎𝜂𝑓𝑏𝜕𝑐𝜕𝑑 + 𝜂𝑒𝑎𝜂𝑓𝑏𝜕𝑑𝜕𝑐)𝑓(𝑥)

+1


𝑓ℎ(−𝜂𝑓𝑒𝜂ℎ𝑐𝜕𝑒𝜕𝑏 + 𝜂𝑓𝑒𝜂ℎ𝑐𝜕𝑏𝜕𝑒)

+ 𝜂𝑏𝑐𝜂𝑓ℎ(−𝜂𝑓𝑒𝜂ℎ𝑑𝜕𝑒𝜕𝑎 + 𝜂𝑓𝑒𝜂ℎ𝑑𝜕𝑎𝜕𝑒)

− 𝜂𝑎𝑐𝜂𝑓ℎ(−𝜂𝑓𝑒𝜂ℎ𝑑𝜕𝑒𝜕𝑏 + 𝜂𝑓𝑒𝜂ℎ𝑑𝜕𝑏𝜕𝑒)

− 𝜂𝑏𝑑𝜂𝑓ℎ(−𝜂𝑓𝑒𝜂ℎ𝑐𝜕𝑒𝜕𝑎 + 𝜂𝑓𝑒𝜂ℎ𝑐𝜕𝑎𝜕𝑒))𝑓(𝑥)

+1

24𝑓(𝑥)5𝜂𝑓𝑗(𝜂𝑎𝑐𝜂𝑑𝑏 − 𝜂𝑎𝑑𝜂𝑐𝑏)(−𝜂𝑓ℎ𝜂𝑗𝑒𝜕ℎ𝜕𝑒 + 𝜂𝑓ℎ𝜂𝑗𝑒𝜕𝑒𝜕ℎ)𝑓(𝑥)

= 0

4.10 The three dimensional flat space in spherical polar coordinates.

4.10.1.1 90Calculate the Christoffel symbols of the three dimensional flat space in spherical polar

coordinates

The line element: 𝑑𝑠2 = 𝑑𝑟2 + 𝑟2𝑑𝜃2 + 𝑟2 sin2 𝜃 𝑑𝜙2

The metric tensor and its in-verse:

𝑔𝑎𝑏 = {1

𝑟2

𝑟2 sin2 𝜃

} 𝑔𝑎𝑏 =

{

1

1

𝑟2

1

𝑟2 sin2 𝜃}

Γ𝑎𝑏𝑐 =1



Γ𝑟𝜃𝜃 = −1

2𝜕𝑟(𝑟

2) = −𝑟 ⇒ Γ 𝜃𝜃𝑟 = 𝑔𝑟𝑟Γ𝑟𝜃𝜃 = −𝑟

89 (𝜂𝑒𝑐𝜕𝑎 − 𝜂𝑎𝑐𝜕𝑒)𝑓(𝑥) = {1 − 1 = 0 𝑖𝑓 𝑒 = 𝑐 = 𝑎 = 𝑥

0 𝑖𝑓 𝑒 𝑜𝑟 𝑐 𝑜𝑟 𝑎 ≠ 0

90 (McMahon, 2006, p. 91), quiz 4-2 and 4-3, the answer to quiz 4-2 is (a) −𝑟2 sin 𝜃 cos 𝜃, and the answer to quiz 4-3 is (c) cot 𝜃





Γ𝜃𝑟𝜃 = Γ𝜃𝜃𝑟 =1

2𝜕𝑟(𝑟

2) = 𝑟 ⇒ Γ 𝑟𝜃𝜃 = Γ 𝜃𝑟

𝜃 = 𝑔𝜃𝜃Γ𝜃𝑟𝜃 =1

𝑟


2𝜕𝑟(𝑟

2 sin2 𝜃) = −𝑟 sin2 𝜃 ⇒ Γ 𝜙𝜙𝑟 = 𝑔𝑟𝑟Γ𝑟𝜙𝜙 = −𝑟 sin2 𝜃


2𝜕𝑟(𝑟

2 sin2 𝜃) = 𝑟 sin2 𝜃 ⇒ Γ 𝑟𝜙𝜙

= Γ 𝜙𝑟𝜙

= 𝑔𝜙𝜙Γ𝜙𝑟𝜙 =1

𝑟


2𝜕𝜃(𝑟

2 sin2 𝜃) = −𝑟2 sin𝜃 cos 𝜃 ⇒ Γ 𝜙𝜙𝜃 = 𝑔𝜃𝜃Γ𝜃𝜙𝜙 = −sin𝜃 cos𝜃


2𝜕𝜃(𝑟

2 sin2 𝜃) = 𝑟2 sin𝜃 cos 𝜃 ⇒ Γ 𝜃𝜙𝜙

= Γ 𝜙𝜃𝜙

= 𝑔𝜙𝜙Γ𝜙𝜃𝜙 = cot 𝜃

4.10.2 91The Riemann tensor of the three dimensional flat space in spherical polar coordinates

The number of independent elements in the Riemann tensor in a metric of dimension 𝑛 = 3 is

𝑁 =𝑛2(𝑛2−1)

12= 6 and we have to calculate: 𝑅 𝜃𝑟𝜃

𝑟 ; 𝑅 𝜙𝑟𝜙𝑟 ; 𝑅 𝜙𝜃𝜙

𝜃 ; 𝑅 𝜃𝑟𝜙𝑟 ; 𝑅 𝑟𝜃𝜙

𝜃 ; 𝑅 𝑟𝜙𝜃𝜙

The Riemann tensor 𝑅 𝑏𝑐𝑑𝑎 = 𝜕𝑐Γ 𝑏𝑑



𝑒 Γ 𝑒𝑑𝑎 (4.41)

𝑎, 𝑐 = 𝑟, 𝑏, 𝑑 = 𝜃: 𝑅 𝜃𝑟𝜃𝑟 = 𝜕𝑟Γ 𝜃𝜃

𝑟 − 𝜕𝜃Γ 𝜃𝑟𝑟 + Γ 𝜃𝜃

𝑒 Γ 𝑒𝑟𝑟 − Γ 𝜃𝑟

𝑒 Γ 𝑒𝜃𝑟

𝑒 = 𝑟, 𝜃, 𝜙: = 𝜕𝑟Γ 𝜃𝜃𝑟 − Γ 𝜃𝑟

𝜃 Γ 𝜃𝜃𝑟 = −1 − (

1

𝑟) (−𝑟) = 0

𝑎, 𝑐 = 𝑟, 𝑏, 𝑑 = 𝜙: 𝑅 𝜙𝑟𝜙𝑟 = 𝜕𝑟Γ 𝜙𝜙

𝑟 − 𝜕𝜙Γ 𝜙𝑟𝑟 + Γ 𝜙𝜙

𝑒 Γ 𝑒𝑟𝑟 − Γ 𝜙𝑟

𝑒 Γ 𝑒𝜙𝑟

𝑒 = 𝑟, 𝜃, 𝜙: = 𝜕𝑟Γ 𝜙𝜙𝑟 − Γ 𝜙𝑟

𝜙Γ 𝜙𝜙𝑟 = −sin2 𝜃 − (

1

𝑟) (−𝑟 sin2 𝜃) = 0

𝑎, 𝑐 = 𝜃, 𝑏, 𝑑 = 𝜙: 𝑅 𝜙𝜃𝜙𝜃 = 𝜕𝜃Γ 𝜙𝜙

𝜃 − 𝜕𝜙Γ 𝜙𝜃𝜃 + Γ 𝜙𝜙



𝑒 = 𝑟, 𝜃, 𝜙: = 𝜕𝜃Γ 𝜙𝜙𝜃 + Γ 𝜙𝜙

𝑟 Γ 𝑟𝜃𝜃 − Γ 𝜙𝜃

𝜙Γ 𝜙𝜙𝜃

= −cos2 𝜃 + sin2 𝜃 + (−𝑟 sin2 𝜃) (1

𝑟) − (cot 𝜃)(− sin𝜃 cos 𝜃) = 0

𝑎, 𝑐 = 𝑟, 𝑏 = 𝜃, 𝑑 = 𝜙: 𝑅 𝜃𝑟𝜙𝑟 = 𝜕𝑟Γ 𝜃𝜙

𝑟 − 𝜕𝜙Γ 𝜃𝑟𝑟 + Γ 𝜃𝜙

𝑒 Γ 𝑒𝑟𝑟 − Γ 𝜃𝑟

𝑒 Γ 𝑒𝜙𝑟

𝑒 = 𝑟, 𝜃, 𝜙: = 0

𝑎, 𝑐 = 𝜃, 𝑏 = 𝑟, 𝑑 = 𝜙: 𝑅 𝑟𝜃𝜙𝜃 = 𝜕𝜃Γ 𝑟𝜙

𝜃 − 𝜕𝜙Γ 𝑟𝜃𝜃 + Γ 𝑟𝜙

𝑒 Γ 𝑒𝜃𝜃 − Γ 𝑟𝜃


𝑒 = 𝑟, 𝜃, 𝜙: = 0

𝑎, 𝑐 = 𝜙, 𝑏 = 𝑟, 𝑑 = 𝜃: 𝑅 𝑟𝜙𝜃𝜙

= 𝜕𝜙Γ 𝑟𝜃𝜙

− 𝜕𝜃Γ 𝑟𝜙𝜙

+ Γ 𝑟𝜃𝑒 Γ 𝑒𝜙

𝜙− Γ 𝑟𝜙

𝑒 Γ 𝑒𝜃𝜙

𝑒 = 𝑟, 𝜃, 𝜙: = (1

𝑟) cot 𝜃 − (

1

𝑟) cot 𝜃 = 0

𝑅𝑟𝜃𝜃𝜙 = −𝑅𝜃𝑟𝜃𝜙 = −𝑔𝜃𝜃𝑅 𝑟𝜃𝜙𝜃 = 0

4.10.3 92A Lie derivative in the three dimensional flat space in spherical polar coordinates

Let 𝑤𝑎 = (𝑟, sin𝜃 , sin𝜃 cos𝜙) and 𝑣𝑎 = (𝑟, 𝑟2 cos 𝜃 , sin𝜙)

Calculate the Lie derivative 𝑢 = 𝐿𝑣𝑤 = 𝑣𝑏∇𝑏𝑤 −𝑤𝑏∇𝑏𝑣 (4.27)

The covariant derivative: ∇𝑏𝐴𝑎 =

𝜕𝐴𝑎

𝜕𝑥𝑏+ Γ 𝑏𝑐

𝑎 𝐴𝑐 (4.6)

91 (McMahon, 2006, p. 91), quiz 4-4. We see that all the elements of the Riemann tensor equals 0, and the answer to quiz 4-4 is (d) 92 (McMahon, 2006, p. 91), quiz 4-6, the answer to quiz 4-6 is (a)





𝑢𝑎 = 𝑣𝑏∇𝑏𝑤𝑎 −𝑤𝑏∇𝑏𝑣

𝑎 = 𝑣𝑏 (𝜕𝑤𝑎


𝑎 𝑤𝑐) − 𝑤𝑏 (𝜕𝑣𝑎


𝑎 𝑣𝑐)

= 𝑣𝑏𝜕𝑤𝑎

𝜕𝑥𝑏+ 𝑣𝑏Γ 𝑏𝑐

𝑎 𝑤𝑐 −𝑤𝑏𝜕𝑣𝑎

𝜕𝑥𝑏−𝑤𝑏Γ 𝑏𝑐

𝑎 𝑣𝑐

= 𝑣𝑏𝜕𝑤𝑎

𝜕𝑥𝑏−𝑤𝑏

𝜕𝑣𝑎


𝑎 𝑤𝑐 −𝑤𝑏Γ 𝑏𝑐𝑎 𝑣𝑐

𝑎 = 𝑟: 𝑢𝑟 = 𝑣𝑏𝜕𝑤𝑟


𝜕𝑣𝑟


𝑟 𝑤𝑐 −𝑤𝑏Γ 𝑏𝑐𝑟 𝑣𝑐

𝑏 = 𝑟, 𝜃, 𝜙: = 𝑣𝜃Γ 𝜃𝜃𝑟 𝑤𝜃 −𝑤𝜃Γ 𝜃𝜃

𝑟 𝑣𝜃 + 𝑣𝜙Γ 𝜙𝜙𝑟 𝑤𝜙 −𝑤𝜙Γ 𝜙𝜙

𝑟 𝑣𝜙 = 0

𝑎 = 𝜃: 𝑢𝜃 = 𝑣𝑏𝜕𝑤𝜃


𝜕𝑣𝜃


𝜃 𝑤𝑐 −𝑤𝑏Γ 𝑏𝑐𝜃 𝑣𝑐

𝑏 = 𝑟, 𝜃, 𝜙: = 𝑣𝜃𝜕𝑤𝜃

𝜕𝜃− 𝑤𝑟

𝜕𝑣𝜃

𝜕𝑟− 𝑤𝜃

𝜕𝑣𝜃

𝜕𝜃+ 𝑣𝑟Γ 𝑟𝑐

𝜃 𝑤𝑐 −𝑤𝑟Γ 𝑟𝑐𝜃 𝑣𝑐 + 𝑣𝜃Γ 𝜃𝑐

𝜃 𝑤𝑐

−𝑤𝜃Γ 𝜃𝑐𝜃 𝑣𝑐 + 𝑣𝜙Γ 𝜙𝑐

𝜃 𝑤𝑐 −𝑤𝜙Γ 𝜙𝑐𝜃 𝑣𝑐

𝑐 = 𝑟, 𝜃, 𝜙: = 𝑣𝜃𝜕𝑤𝜃


𝜕𝑣𝜃


𝜕𝑣𝜃

𝜕𝜃+ 𝑣𝑟Γ 𝑟𝜃

𝜃 𝑤𝜃 −𝑤𝑟Γ 𝑟𝜃𝜃 𝑣𝜃 + 𝑣𝜃Γ 𝜃𝑟

𝜃 𝑤𝑟

−𝑤𝜃Γ 𝜃𝑟𝜃 𝑣𝑟 + 𝑣𝜙Γ 𝜙𝜙

𝜃 𝑤𝜙 −𝑤𝜙Γ 𝜙𝜃 𝑣𝜙

= 𝑣𝜃𝜕𝑤𝜃


𝜕𝑣𝜃


𝜕𝑣𝜃

𝜕𝜃

= (𝑟2 cos𝜃)(cos 𝜃) − 𝑟(2𝑟 cos 𝜃) − (sin 𝜃)(−𝑟2 sin 𝜃) = 𝑟2(1 − 2 cos 𝜃)

𝑎 = 𝜙: 𝑢𝜙 = 𝑣𝑏𝜕𝑤𝜙


𝜕𝑣𝜙


𝜙𝑤𝑐 −𝑤𝑏Γ 𝑏𝑐

𝜙𝑣𝑐

𝑏 = 𝑟, 𝜃, 𝜙: = 𝑣𝜃𝜕𝑤𝜙

𝜕𝑥𝜃+ 𝑣𝜙

𝜕𝑤𝜙

𝜕𝑥𝜙−𝑤𝜙

𝜕𝑣𝜙

𝜕𝑥𝜙+ 𝑣𝑟Γ 𝑟𝑐

𝜙𝑤𝑐 −𝑤𝑟Γ 𝑟𝑐

𝜙𝑣𝑐 + 𝑣𝜃Γ 𝜃𝑐

𝜙𝑤𝑐

−𝑤𝜃Γ 𝜃𝑐𝜙𝑣𝑐 + 𝑣𝜙Γ 𝜙𝑐

𝜙𝑤𝑐 −𝑤𝜙Γ 𝜙𝑐

𝜙𝑣𝑐

𝑐 = 𝑟, 𝜃, 𝜙: = 𝑣𝜃𝜕𝑤𝜙


𝜕𝑤𝜙


𝜕𝑣𝜙

𝜕𝑥𝜙+ 𝑣𝑟Γ 𝑟𝜙

𝜙𝑤𝜙 −𝑤𝑟Γ 𝑟𝜙

𝜙𝑣𝜙 + 𝑣𝜃Γ 𝜃𝜙

𝜙𝑤𝜙

−𝑤𝜃Γ 𝜃𝜙𝜙

𝑣𝜙 + 𝑣𝜙Γ 𝜙𝑟𝜙

𝑤𝑟 −𝑤𝜙Γ 𝜙𝑟𝜙

𝑣𝑟 + 𝑣𝜙Γ 𝜙𝜃𝜙

𝑤𝜃 −𝑤𝜙Γ 𝜙𝜃𝜙

𝑣𝜃

= 𝑣𝜃𝜕𝑤𝜙


𝜕𝑤𝜙


𝜕𝑣𝜙

𝜕𝑥𝜙

= (𝑟2 cos𝜃)(cos 𝜃 cos𝜙) + sin𝜙 (− sin 𝜃 sin𝜙) − sin 𝜃 cos𝜙 (cos𝜙) = 𝑟2 cos2 𝜃 cos𝜙 − sin𝜃

So we can conclude:

𝑢 = (𝑢𝑟

𝑢𝜃

𝑢𝜙) = (

0𝑟2(1 − 2 cos 𝜃)

𝑟2 cos2 𝜃 cos𝜙 − sin 𝜃)

4.11 93The Ricci scalar of the Penrose Kahn metric The Ricci scalar: 𝑅 = 𝑔𝑎𝑏𝑅𝑎𝑏 (4.47)

The Ricci tensor 𝑅𝑎𝑏 = 𝑅 𝑎𝑐𝑏𝑐 (4.46)

Sum over 𝑎 = 𝑢, 𝑣, 𝑥, 𝑦: 𝑅 = 𝑔𝑢𝑏𝑅𝑢𝑏 + 𝑔𝑣𝑏𝑅𝑣𝑏 + 𝑔

𝑥𝑏𝑅𝑥𝑏 + 𝑔𝑦𝑏𝑅𝑦𝑏

Sum over 𝑏 = 𝑢, 𝑣, 𝑥, 𝑦: = 𝑔𝑢𝑣𝑅𝑢𝑣 + 𝑔𝑣𝑢𝑅𝑣𝑢 + 𝑔

𝑥𝑥𝑅𝑥𝑥 + 𝑔𝑦𝑦𝑅𝑦𝑦

= 𝑔𝑢𝑣𝑅 𝑢𝑐𝑣𝑐 + 𝑔𝑣𝑢𝑅 𝑣𝑐𝑢

𝑐 + 𝑔𝑥𝑥𝑅 𝑥𝑐𝑥𝑐 + 𝑔𝑦𝑦𝑅 𝑦𝑐𝑦

𝑐

Sum over 𝑐 = 𝑢, 𝑣, 𝑥, 𝑦: = 𝑔𝑢𝑣𝑅 𝑢𝑥𝑣𝑥 + 𝑔𝑢𝑣𝑅 𝑢𝑦𝑣

𝑦+ 𝑔𝑣𝑢𝑅 𝑣𝑥𝑢

𝑥 + 𝑔𝑣𝑢𝑅 𝑣𝑦𝑢𝑦

+ 𝑔𝑥𝑥𝑅 𝑥𝑢𝑥𝑢 + 𝑔𝑥𝑥𝑅 𝑥𝑣𝑥

𝑣

93 (McMahon, 2006, p. 92), quiz 4-8.The answer to quiz 4-8 is (b)





+𝑔𝑥𝑥𝑅 𝑥𝑦𝑥𝑦

+ 𝑔𝑦𝑦𝑅 𝑦𝑢𝑦𝑢 + 𝑔𝑦𝑦𝑅 𝑦𝑣𝑦

𝑣 + 𝑔𝑦𝑦𝑅 𝑦𝑥𝑦𝑥

𝑔𝑢𝑣 = 𝑔𝑣𝑢: = 𝑔𝑢𝑣𝑔𝑥𝑥𝑅𝑥𝑢𝑥𝑣 + 𝑔𝑢𝑣𝑔𝑦𝑦𝑅𝑦𝑢𝑦𝑣 + 𝑔

𝑣𝑢𝑔𝑥𝑥𝑅𝑥𝑣𝑥𝑢 + 𝑔𝑣𝑢𝑔𝑦𝑦𝑅𝑦𝑣𝑦𝑢

+𝑔𝑥𝑥𝑔𝑢𝑣𝑅𝑣𝑥𝑢𝑥 + 𝑔𝑥𝑥𝑔𝑣𝑢𝑅𝑢𝑥𝑣𝑥 + 𝑔

𝑥𝑥𝑔𝑦𝑦𝑅𝑦𝑥𝑦𝑥 + 𝑔𝑦𝑦𝑔𝑢𝑣𝑅𝑣𝑦𝑢𝑦

+𝑔𝑦𝑦𝑔𝑣𝑢𝑅𝑢𝑦𝑣𝑦 + 𝑔𝑦𝑦𝑔𝑥𝑥𝑅𝑥𝑦𝑥𝑦

𝑅𝑎𝑏𝑐𝑑 = 𝑅𝑐𝑑𝑎𝑏 = −𝑅𝑏𝑎𝑐𝑑 = −𝑅𝑎𝑏𝑑𝑐:

= 4𝑔𝑢𝑣𝑔𝑥𝑥𝑅𝑥𝑢𝑥𝑣 + 4𝑔𝑢𝑣𝑔𝑦𝑦𝑅𝑦𝑢𝑦𝑣 + 2𝑔

𝑥𝑥𝑔𝑦𝑦𝑅𝑦𝑥𝑦𝑥

= 4𝑔𝑢𝑣𝑅 𝑢𝑥𝑣𝑥 + 4𝑔𝑢𝑣𝑅 𝑢𝑦𝑣

𝑦+ 2𝑔𝑥𝑥𝑅 𝑥𝑦𝑥

𝑦

Remark we can rewrite this into to a general expression for a non-diagonal metric of the type:

𝑔𝑎𝑏 =

{

𝑔12

𝑔12𝑔33

𝑔44}

We write 𝑅 = 4𝑔12𝑅 1323 + 4𝑔12𝑅 142

4 + 2𝑔33𝑅 3434 (S2)

Now we need to calculate the three elements in the Riemann tensor: 𝑅 𝑢𝑥𝑣𝑥 ; 𝑅 𝑢𝑦𝑣

𝑦; 𝑅 𝑥𝑦𝑥

𝑦

𝑅 𝑢𝑥𝑣𝑥 = 𝜕𝑥Γ 𝑢𝑣

𝑥 − 𝜕𝑣Γ 𝑢𝑥𝑥 + Γ 𝑢𝑣

𝑒 Γ 𝑒𝑥𝑥 − Γ 𝑢𝑥

𝑒 Γ 𝑒𝑣𝑥 = 0

𝑅 𝑢𝑦𝑣𝑦

= 𝜕𝑦Γ 𝑢𝑣𝑦

− 𝜕𝑣Γ 𝑢𝑦𝑦

+ Γ 𝑢𝑣𝑒 Γ 𝑒𝑦

𝑦− Γ 𝑢𝑦

𝑒 Γ 𝑒𝑣𝑦

= 0

𝑅 𝑥𝑦𝑥𝑦

= 𝜕𝑦Γ 𝑥𝑥𝑦

− 𝜕𝑥Γ 𝑥𝑦𝑦

+ Γ 𝑥𝑥𝑒 Γ 𝑒𝑦

𝑦− Γ 𝑥𝑦

𝑒 Γ 𝑒𝑥𝑦

= 0

So we can conclude 𝑅 = 0

4.12 94A metric example 1: 𝒅𝒔𝟐 = 𝒚𝟐 𝐬𝐢𝐧 𝒙𝒅𝒙𝟐 + 𝒙𝟐 𝐭𝐚𝐧 𝒚𝒅𝒚𝟐

4.12.1 The Christoffel symbols of a metric example

The line element: 𝑑𝑠2 = 𝑦2 sin 𝑥 𝑑𝑥2 + 𝑥2 tan 𝑦 𝑑𝑦2

The metric tensor and its in-verse:

𝑔𝑎𝑏 = {𝑦2 sin 𝑥

𝑥2 tan 𝑦} 𝑔𝑎𝑏 =

{

1

𝑦2 sin 𝑥1

𝑥2 tan 𝑦}

Γ𝑎𝑏𝑐 =1



Γ𝑦𝑥𝑥 =1

2(−𝜕𝑦𝑔𝑥𝑥) = −𝑦 sin 𝑥 ⇒ Γ 𝑥𝑥

𝑦 = 𝑔𝑦𝑦Γ𝑦𝑥𝑥 = −

𝑦 sin 𝑥

𝑥2 tan 𝑦

Γ𝑥𝑦𝑥 = Γ𝑥𝑥𝑦 =1

2𝑦 sin𝑥 ⇒ Γ 𝑦𝑥

𝑥 = Γ 𝑥𝑦𝑥 = 𝑔𝑥𝑥Γ𝑥𝑦𝑥 =

1

𝑦

Γ𝑥𝑥𝑥 =1

2𝑦2 cos 𝑥 ⇒ Γ 𝑥𝑥

𝑥 = 𝑔𝑥𝑥Γ𝑥𝑥𝑥 =1

2cot 𝑥

Γ𝑥𝑦𝑦 = −𝑥 tan𝑦 ⇒ Γ 𝑦𝑦𝑥 = 𝑔𝑥𝑥Γ𝑥𝑦𝑦 = −

𝑥 tan𝑦

𝑦2 sin 𝑥

Γ𝑦𝑥𝑦 = Γ𝑦𝑦𝑥 = 𝑥 tan𝑦 ⇒ Γ 𝑥𝑦𝑦

= Γ 𝑦𝑥𝑦

= 𝑔𝑦𝑦Γ𝑦𝑥𝑦 =1

𝑥

Γ𝑦𝑦𝑦 =1

2𝑥2(1 + tan2 𝑦) ⇒ Γ 𝑦𝑦

𝑦 = 𝑔𝑦𝑦Γ𝑦𝑦𝑦 =

1 + tan2 𝑦

2 tan𝑦

4.12.2 The Ricci scalar of a metric example

From example 4-12 we know that for a 2-dimensional diagonal metric: 𝑅 = 2𝑔22𝑅 2121 which

means we only have to calculate 𝑅 𝑦𝑥𝑦𝑥

94 (McMahon, 2006, p. 92), quiz 4-9, quiz 4-10. The answer to quiz 4-9 is (c), and the answer to quiz 4-10 is (b)





𝑅 𝑦𝑥𝑦𝑥 = 𝜕𝑥Γ 𝑦𝑦

𝑥 − 𝜕𝑦Γ 𝑦𝑥𝑥 + Γ 𝑦𝑦

𝑒 Γ 𝑒𝑥𝑥 − Γ 𝑦𝑥

𝑒 Γ 𝑒𝑦𝑥

𝑒 = 𝑥, 𝑦: = 𝜕𝑥Γ 𝑦𝑦𝑥 − 𝜕𝑦Γ 𝑦𝑥

𝑥 + Γ 𝑦𝑦𝑥 Γ 𝑥𝑥

𝑥 − Γ 𝑦𝑥𝑥 Γ 𝑥𝑦

𝑥 + Γ 𝑦𝑦𝑦Γ 𝑦𝑥𝑥 − Γ 𝑦𝑥

𝑦Γ 𝑦𝑦𝑥

= 𝜕𝑥 (−𝑥 tan 𝑦

𝑦2 sin 𝑥) − 𝜕𝑦 (

1

𝑦) + (−

𝑥 tan 𝑦

𝑦2 sin 𝑥) (1

2cot 𝑥) − (

1

𝑦)2

+ (1 + tan2 𝑦

2 tan 𝑦) (1

𝑦) − (

1

𝑥) (−

𝑥 tan 𝑦

𝑦2 sin 𝑥)

= −

tan𝑦

𝑦2(1

sin 𝑥−𝑥 cos 𝑥

sin2 𝑥) + (

1

𝑦)2

− (1

2

𝑥

𝑦2tan 𝑦 cos 𝑥

sin2 𝑥) − (

1

𝑦)2

+ (1 + tan2 𝑦

2𝑦 tan 𝑦)

+ (tan𝑦

𝑦2 sin 𝑥)

= (𝑥tan𝑦 cos 𝑥

𝑦2 sin2 𝑥) − (

1

2

𝑥

𝑦2tan 𝑦 cos 𝑥

sin2 𝑥) + (

1 + tan2 𝑦

2𝑦 tan 𝑦)

= (𝑥 cos 𝑥 tan2 𝑦 + 𝑦 sin2 𝑥 + 𝑦 sin2 𝑥 tan2 𝑦

2𝑦2 sin2 𝑥 tan𝑦)

⇒ 𝑅 = 2𝑔𝑦𝑦𝑅 𝑦𝑥𝑦𝑥 = 2(

1

𝑥2 tan 𝑦)(𝑥 cos 𝑥 tan2 𝑦 + 𝑦 sin2 𝑥 + 𝑦 sin2 𝑥 tan2 𝑦

2𝑦2 sin2 𝑥 tan 𝑦)

= (𝑥 cos 𝑥 tan2 𝑦 + 𝑦 sin2 𝑥 + 𝑦 sin2 𝑥 tan2 𝑦

𝑥2𝑦2 sin2 𝑥 tan2 𝑦)

4.13 95Calculate the Christoffel symbols for a metric example 2: 𝒅𝒔𝟐 = 𝒅𝝍𝟐 +

𝐬𝐢𝐧𝐡𝟐𝝍 𝒅𝜽𝟐 + 𝐬𝐢𝐧𝐡𝟐𝝍𝐬𝐢𝐧𝟐 𝜽𝒅𝝓𝟐 The line element: 𝑑𝑠2 = 𝑑𝜓2 + sinh2𝜓 𝑑𝜃2 + sinh2𝜓 sin2 𝜃 𝑑𝜙2

To find the Christoffel symbols we calculate the geodesic from the Euler-Lagrange equation

0 =𝑑

𝑑𝑠(𝜕𝐹

𝜕��𝑎) −

𝜕𝐹

𝜕𝑥𝑎 (10.36)

where

𝐹 = 𝜓2 + sinh2𝜓 ��2 + sinh2𝜓 sin2 𝜃 ��2

𝑥𝑎 = 𝜓:

𝜕𝐹

𝜕𝜓 = 2cosh𝜓 sinh𝜓 ��2 + 2cosh𝜓 sinh𝜓 sin2 𝜃 ��2

𝜕𝐹

𝜕�� = 2��

𝑑

𝑑𝑠(𝜕𝐹

𝜕��) = 2��

⇒ 0 = 2�� − 2 cosh𝜓 sinh𝜓 ��2 − 2cosh𝜓 sinh𝜓 sin2 𝜃 ��2

⇔ 0 = �� − cosh𝜓 sinh𝜓 ��2 − cosh𝜓 sinh𝜓 sin2 𝜃 ��2

𝑥𝑎 = 𝜃:

𝜕𝐹

𝜕𝜃 = −2cos 𝜃 sin𝜃 sinh2𝜓 ��2

𝜕𝐹

𝜕�� = 2 sinh2𝜓 ��

𝑑

𝑑𝑠(𝜕𝐹

𝜕��) = 4cosh𝜓 sinh𝜓 �� + 2 sinh2𝜓 ��

⇒ 0 = 2 sinh2𝜓 �� + 4 cosh𝜓 sinh𝜓 �� + 2 cos𝜃 sin𝜃 sinh2𝜓 ��2

95 (McMahon, 2006, p. 325), final exam 9. The answer to FE-9 is (a)





⇔ 0 = �� + 2 cosh𝜓 sinh𝜓 �� + cos 𝜃 sin 𝜃 ��2

𝑥𝑎 = 𝜙:

𝜕𝐹

𝜕𝜙 = 0

𝜕𝐹

𝜕�� = 2 sinh2𝜓 sin2 𝜃 ��

𝑑

𝑑𝑠(𝜕𝐹

𝜕��) = 4cosh𝜓 sinh𝜓 �� − 4 cos 𝜃 sin 𝜃 �� + 2 sinh2𝜓 sin2 𝜃 ��

⇒ 0 = 4cosh𝜓 sinh𝜓 �� − 4 cos 𝜃 sin 𝜃 �� + 2 sinh2𝜓 sin2 𝜃 ��

⇔ 0 = �� + 2coth𝜓

sin2 𝜃�� − 2

cot 𝜃

sinh2𝜓��


0 = �� − cosh𝜓 sinh𝜓 ��2 − cosh𝜓 sinh𝜓 sin2 𝜃 ��2 0 = �� + 2 cosh𝜓 sinh𝜓 �� + cos 𝜃 sin 𝜃 ��2

0 = �� + 2coth𝜓

sin2 𝜃�� − 2

cot 𝜃

sinh2𝜓��

We can now find the non-zero Christoffel symbols:

Γ 𝜃𝜃𝜓 = −cosh𝜓 sinh𝜓 Γ 𝜙𝜃

𝜃 = cosh𝜓 sinh𝜓 Γ 𝜓𝜙𝜙

=coth𝜓

sin2 𝜃

Γ 𝜙𝜙𝜓

= −cosh𝜓 sinh𝜓 sin2 𝜃 Γ 𝜙𝜙𝜃 = cos 𝜃 sin𝜃 Γ 𝜃𝜙

𝜙 = −

cot 𝜃

sinh2𝜓��

4.14 96A metric example 3: 𝒅𝒔𝟐 = (𝒖𝟐 + 𝝂𝟐)𝒅𝒖𝟐 + (𝒖𝟐 + 𝝂𝟐)𝒅𝝂𝟐 + 𝒖𝟐𝝂𝟐𝒅𝜽𝟐

4.14.1 Calculate the Christoffel symbols for a metric example

The line ele-ment:

𝑑𝑠2 = (𝑢2 + 𝜈2)𝑑𝑢2 + (𝑢2 + 𝜈2)𝑑𝜈2 + 𝑢2𝜈2𝑑𝜃2

The metric ten-sor and its in-verse:

𝑔𝑎𝑏 = {(𝑢2 + 𝜈2) 0 0

0 (𝑢2 + 𝜈2) 0

0 0 𝑢2𝜈2} 𝑔𝑎𝑏 =

{

1

(𝑢2 + 𝜈2)1

(𝑢2 + 𝜈2)1

𝑢2𝜈2}

Γ𝑎𝑏𝑐 =1



Γ𝑢𝑢𝑢 =1

2𝜕𝑢(𝑔𝑢𝑢) = 𝑢 ⇒ Γ 𝑢𝑢

𝑢 = 𝑔𝑢𝑢Γ𝑢𝑢𝑢 =𝑢

(𝑢2 + 𝜈2)

Γ𝜈𝑢𝑢 = −1

2𝜕𝜈(𝑔𝑢𝑢) = −𝜈 ⇒ Γ 𝑢𝑢

𝜈 = 𝑔𝜈𝜈Γ𝜈𝑢𝑢 = −𝜈

(𝑢2 + 𝜈2)

Γ𝑢𝜈𝑢 =1

2𝜕𝜈(𝑔𝑢𝑢) = 𝜈 ⇒ Γ 𝜈𝑢

𝑢 = 𝑔𝑢𝑢Γ𝑢𝜈𝑢 =𝜈

(𝑢2 + 𝜈2)

96 (McMahon, 2006, p. 324), final exam 7 and 8. All the independent elements of the Riemann tensor is zero and the answer to FE-8 is (d)





Γ𝑢𝜈𝜈 = −1

2𝜕𝑢(𝑔𝜈𝜈) = −𝑢 ⇒ Γ 𝜈𝜈

𝑢 = 𝑔𝑢𝑢Γ𝑢𝜈𝜈 = −𝑢

(𝑢2 + 𝜈2)

Γ𝜈𝑢𝜈 =1

2𝜕𝑢(𝑔𝜈𝜈) = 𝑢 ⇒ Γ 𝑢𝜈

𝜈 = 𝑔𝜈𝜈Γ𝑢𝜈𝜈 =𝑢

(𝑢2 + 𝜈2)

Γ𝜈𝜈𝜈 =1

2𝜕𝜈(𝑔𝜈𝜈) = 𝜈 ⇒ Γ 𝜈𝜈

𝜈 = 𝑔𝜈𝜈Γ𝜈𝜈𝜈 =𝜈

(𝑢2 + 𝜈2)

Γ𝑢𝜃𝜃 = −1

2𝜕𝑢(𝑔𝜃𝜃) = −𝑢𝜈

2 ⇒ Γ 𝜃𝜃𝑢 = 𝑔𝑢𝑢Γ𝑢𝜃𝜃 = −

𝑢𝜈2

(𝑢2 + 𝜈2)

Γ𝜃𝑢𝜃 =1

2𝜕𝑢(𝑔𝜃𝜃) = 𝑢𝜈

2 ⇒ Γ 𝑢𝜃𝜃 = 𝑔𝜃𝜃Γ𝜃𝑢𝜃 =

1

𝑢

Γ𝜈𝜃𝜃 = −1

2𝜕𝜈(𝑔𝜃𝜃) = −𝑢

2𝜈 ⇒ Γ 𝜃𝜃𝜈 = 𝑔𝜈𝜈Γ𝜈𝜃𝜃 = −

𝑢2𝜈

(𝑢2 + 𝜈2)

Γ𝜃𝜈𝜃 =1

2𝜕𝜈(𝑔𝜃𝜃) = 𝑢

2𝜈 ⇒ Γ 𝜈𝜃𝜃 = 𝑔𝜃𝜃Γ𝜃𝜈𝜃 =

1

𝜈

Collecting the results we find the non-zero Christoffel symbols

Γ 𝑢𝑢𝑢 = −Γ 𝜈𝜈

𝑢 = Γ 𝑢𝜈𝜈 =

𝑢

(𝑢2 + 𝜈2)

Γ 𝜈𝜈𝜈 = −Γ 𝑢𝑢

𝜈 = Γ 𝜈𝑢𝑢 =

𝜈

(𝑢2 + 𝜈2)

Γ 𝜃𝜃𝑢 = −

𝑢𝜈2

(𝑢2 + 𝜈2)

Γ 𝜃𝜃𝜈 = −

𝑢2𝜈

(𝑢2 + 𝜈2)

Γ 𝑢𝜃𝜃 =

1

𝑢

Γ 𝜈𝜃𝜃 =

1

𝜈

4.14.2 Calculate the Riemann tensor of metric example

In 3 dimensions the Riemann tensor has six independent elements: 𝑅1212; 𝑅1213; 𝑅1223; 𝑅1313; 𝑅1323; 𝑅2323 The Riemann tensor

𝑅𝑎𝑏𝑐𝑑 = 𝜕𝑐Γ𝑎𝑏𝑑 − 𝜕𝑑Γ𝑎𝑏𝑐 − Γ𝑒𝑎𝑐Γ 𝑏𝑑𝑒 + Γ𝑒𝑎𝑑Γ 𝑏𝑐

𝑒 (4.42) 𝑅𝑢𝜈𝑢𝜈 = 𝜕𝑢Γ𝑢𝜈𝜈 − 𝜕𝜈Γ𝑢𝜈𝑢 − Γ𝑒𝑢𝑢Γ 𝜈𝜈

𝑒 + Γ𝑒𝑢𝜈Γ 𝜈𝑢𝑒

= 𝜕𝑢(−𝑢) − 𝜕𝜈(𝜈) − Γ𝑢𝑢𝑢Γ 𝜈𝜈𝑢 + Γ𝑢𝑢𝜈Γ 𝜈𝑢

𝑢 − Γ𝜈𝑢𝑢Γ 𝜈𝜈𝜈 + Γ𝜈𝑢𝜈Γ 𝜈𝑢

𝜈

= −2 +𝑢2

(𝑢2 + 𝜈2)+

𝜈2

(𝑢2 + 𝜈2)+

𝜈2

(𝑢2 + 𝜈2)+

𝑢2

(𝑢2 + 𝜈2)

= 0 𝑅𝑢𝜈𝑢𝜃 = 𝜕𝑢Γ𝑢𝜈𝜃 − 𝜕𝜃Γ𝑢𝜈𝜃 − Γ𝑒𝑢𝑢Γ 𝜈𝜃

𝑒 + Γ𝑒𝑢𝜃Γ 𝜈𝑢𝑒 = 0

𝑅𝑢𝜈𝜈𝜃 = 𝜕𝜈Γ𝑢𝜈𝜃 − 𝜕𝜃Γ𝑢𝜈𝜈 − Γ𝑒𝑢𝜈Γ 𝜈𝜃𝑒 + Γ𝑒𝑢𝜃Γ 𝜈𝜈

𝑒 = 0 𝑅𝑢𝜃𝑢𝜃 = 𝜕𝑢Γ𝑢𝜃𝜃 − 𝜕𝜃Γ𝑢𝜃𝑢 − Γ𝑒𝑢𝑢Γ 𝜃𝜃

𝑒 + Γ𝑒𝑢𝜃Γ 𝜃𝑢𝑒

= 𝜕𝑢(−𝑢𝜈2) − Γ𝑢𝑢𝑢Γ 𝜃𝜃

𝑢 − Γ𝜈𝑢𝑢Γ 𝜃𝜃𝜈 + Γ𝜃𝑢𝜃Γ 𝜃𝑢

𝜃

= −𝜈2 +𝑢2𝜈2

(𝑢2 + 𝜈2)+

𝑢2𝜈2

(𝑢2 + 𝜈2)+ 𝜈2

= 0 𝑅𝑢𝜃𝜈𝜃 = 𝜕𝜈Γ𝑢𝜃𝜃 − 𝜕𝜃Γ𝑢𝜃𝜈 − Γ𝑒𝑢𝜈Γ 𝜃𝜃

𝑒 + Γ𝑒𝑢𝜃Γ 𝜃𝜈𝑒

= 𝜕𝜈(−𝑢𝜈2) − Γ𝑢𝑢𝜈Γ 𝜃𝜃

𝑢 − Γ𝜈𝑢𝜈Γ 𝜃𝜃𝜈 + Γ𝜃𝑢𝜃Γ 𝜃𝜈

𝜃

= −2𝑢𝜈 +

𝑢𝜈3

(𝑢2 + 𝜈2)+

𝑢3𝜈

(𝑢2 + 𝜈2)+ 𝑢𝜈





= 𝑢𝜈 (

𝜈2

(𝑢2 + 𝜈2)+

𝑢2

(𝑢2 + 𝜈2)− 1)

= 0 𝑅𝜈𝜃𝜈𝜃 = 𝜕𝜈Γ𝜈𝜃𝜃 − 𝜕𝜃Γ𝜈𝜃𝜈 − Γ𝑒𝜈𝜈Γ 𝜃𝜃

𝑒 + Γ𝑒𝜈𝜃Γ 𝜃𝜈𝑒

= 𝜕𝜈(−𝑢2𝜈) − Γ𝑢𝜈𝜈Γ 𝜃𝜃

𝑢 − Γ𝜈𝜈𝜈Γ 𝜃𝜃𝜈 + Γ𝜃𝜈𝜃Γ 𝜃𝜈

𝜃

= −𝑢2 −

𝑢2𝜈2

(𝑢2 + 𝜈2)+

𝑢2𝜈2

(𝑢2 + 𝜈2)+ 𝑢2

= 0

4.14.3 Calculate the Riemann tensor of metric example – Alternative version97

The answer to Exercise 8 in the Final Exam is given by 𝑅𝑎𝑏𝑐𝑑 = 0 for the following reason: Consider the global Minkowski spacetime 𝑑2 = 𝑑𝑢′2 + 𝑑𝑣′2 + 𝑑𝜃′2, in some coordinates (𝑢′, 𝑣′, 𝜃′) [the correspond-ing Riemannian curvature tensor identically vanishing, of course], and consider the following coordinate transformation:

𝑢′ = 𝑢𝑣 cos 𝜃 𝑣′ = 𝑢𝑣 sin 𝜃

𝜃′ =1

2(𝑢2 − 𝑣2)

The differentials are related as 𝑑𝑢′ = (𝑢𝑑𝑣 + 𝑣𝑑𝑢) cos𝜃 − 𝑢𝑣 sin 𝜃 𝑑𝜃 𝑑𝑣′ = (𝑢𝑑𝑣 + 𝑣𝑑𝑢) sin𝜃 + 𝑢𝑣 cos 𝜃 𝑑𝜃 𝑑𝜃′ = 𝑢𝑑𝑢 − 𝑣𝑑𝑣

from which it readily follows that 𝑑𝑢′2 + 𝑑𝑣′2 + 𝑑𝜃′2 = (𝑢𝑑𝑣 + 𝑣𝑑𝑢)2 + (𝑢𝑣)2𝑑𝜃2 + (𝑢𝑑𝑢 − 𝑣𝑑𝑣)2

= (𝑢2 + 𝑣2)(𝑑𝑢2 + 𝑑𝑣2) + 𝑢2𝑣2𝑑𝜃2 Thus the line element 𝑑𝑠2 = (𝑢2 + 𝑣2)(𝑑𝑢2 + 𝑑𝑣2) + 𝑢2𝑣2𝑑𝜃2 must correspond to an identically van-ishing Riemannian curvature tensor.

5 Cartan’s Structure Equations

5.1 98Ricci rotation coefficients for the Tolman-Bondi- de Sitter metric (Spheri-

cal dust with a cosmological constant) The line element: 𝑑𝑠2 = 𝑑𝑡2 − 𝑒−2𝜓(𝑡,𝑟)𝑑𝑟2 − 𝑅2(𝑡, 𝑟)𝑑𝜃2 − 𝑅2(𝑡, 𝑟) sin2 𝜃 𝑑𝜙2

The Basis one forms

𝜔�� = 𝑑𝑡

𝜂𝑖𝑗 = {

1−1

−1−1

}

𝜔�� = 𝑒−𝜓(𝑡,𝑟)𝑑𝑟 𝑑𝑟 = 𝑒𝜓(𝑡,𝑟)𝜔��

𝜔�� = 𝑅(𝑡, 𝑟)𝑑𝜃 𝑑𝜃 =1

𝑅(𝑡, 𝑟)𝜔��

𝜔�� = 𝑅(𝑡, 𝑟) sin 𝜃 𝑑𝜙 𝑑𝜙 =1

𝑅(𝑡, 𝑟) sin 𝜃𝜔��

97 Kindly provided by Mr. John Fredsted: http://johnfredsted.dk/science/publications.php 98 (McMahon, 2006, p. 106), example 5-1


http://johnfredsted.dk/science/publications.php




Cartan’s First Structure equation and the calculation of the Ricci rotation coefficients Γ ��𝑐�� :

𝑑𝜔�� = −Γ �� ∧ 𝜔�� (5.9) Γ ��

�� = Γ ��𝑐�� 𝜔𝑐 (5.10)

𝑑𝜔�� = 0

𝑑𝜔�� = 𝑑(𝑒−𝜓(𝑡,𝑟)𝑑𝑟) = −��𝑒−𝜓(𝑡,𝑟)𝑑𝑡 ∧ 𝑑𝑟 = −��𝜔�� ∧ 𝜔��

𝑑𝜔�� = 𝑑(𝑅(𝑡, 𝑟)𝑑𝜃) = ��𝑑𝑡 ∧ 𝑑𝜃 + 𝑅′𝑑𝑟 ∧ 𝑑𝜃 =��

𝑅𝜔�� ∧ 𝜔�� +

𝑅′

𝑅𝑒𝜓(𝑡,𝑟)𝜔�� ∧ 𝜔��

𝑑𝜔�� = 𝑑(𝑅(𝑡, 𝑟) sin 𝜃 𝑑𝜙)

= �� sin 𝜃 𝑑𝑡 ∧ 𝑑𝜙 + 𝑅′ sin 𝜃 𝑑𝑟 ∧ 𝑑𝜙 + 𝑅(𝑡, 𝑟) cos 𝜃 𝑑𝜃 ∧ 𝑑𝜙

=��

𝑅𝜔�� ∧ 𝜔�� +

𝑅′

𝑅𝑒𝜓(𝑡,𝑟)𝜔�� ∧ 𝜔�� +

cot 𝜃

𝑅𝜔�� ∧ 𝜔��

Summarizing the curvature one forms in a matrix:

Γ �� =

{

0 −��𝜔��

��

𝑅𝜔��

��

𝑅𝜔��

−��𝜔�� 0𝑅′

𝑅𝑒𝜓(𝑡,𝑟)𝜔��

𝑅′


��

𝑅𝜔�� −

𝑅′

𝑅𝑒𝜓(𝑡,𝑟)𝜔�� 0

cot 𝜃

𝑅𝜔��

��

𝑅𝜔�� −

𝑅′

𝑅𝑒𝜓(𝑡,𝑟)𝜔�� −

cot 𝜃

𝑅𝜔�� 0 }

Where �� refers to column and �� to row.

Now we can read off the Ricci rotation coefficients

Γ �� = −�� Γ ��

�� = −�� Γ �� =

��

𝑅 Γ

��

�� =

��

𝑅

Γ �� =

��

𝑅 Γ ��

�� = −𝑅′

𝑅𝑒𝜓(𝑡,𝑟) Γ ��

�� =𝑅′

𝑅𝑒𝜓(𝑡,𝑟) Γ

��

�� =

𝑅′

𝑅𝑒𝜓(𝑡,𝑟)

Γ �� =

��

𝑅 Γ ��

�� = −𝑅′

𝑅𝑒𝜓(𝑡,𝑟) Γ ��

�� = −cot 𝜃

𝑅 Γ

��

�� =

cot 𝜃

𝑅

Which means the answer to quiz 5-6 is (b)

Transformation of the Ricci rotation coefficients Γ ��𝑐�� into the Christoffel symbols Γ bc

𝑎

We have the transformation Γ bc𝑎 = (Λ−1) ��

𝑎 Γ �� Λ 𝑏

�� Λ 𝑐��

(5.14)

Γ 𝑡𝑟𝑟 = (Λ−1) ��

𝑟 Γ �� Λ 𝑡

�� Λ 𝑟��

= Γ �� Λ 𝑡

�� = −�� ⋅ 1 = −��

Γ 𝑟𝑟𝑡 = (Λ−1) ��

𝑡 Γ �� Λ 𝑟

�� Λ 𝑟��

= (Λ−1) ��𝑡 Γ ��

�� (Λ 𝑟�� )

2 = 1(−��)(𝑒−𝜓(𝑡,𝑟))

2 = −��𝑒−2𝜓(𝑡,𝑟)





Γ 𝑡𝜃𝜃 = (Λ−1) ��

𝜃 Γ �� Λ 𝑡

�� Λ 𝜃��

= Γ �� Λ 𝑡

�� =��

𝑅⋅ 1 =

��

𝑅

Γ 𝜃𝜃𝑡 = (Λ−1) ��

𝑡 Γ �� Λ 𝜃

�� Λ 𝜃��

= (Λ−1) ��𝑡 Γ

�� (Λ 𝜃

�� )2 = 1 ⋅

��

𝑅𝑅2 = 𝑅��

Γ 𝑟𝜃𝜃 = (Λ−1) ��

𝜃 Γ �� Λ 𝑟

�� Λ 𝜃��

= Γ �� Λ 𝑟

�� =𝑅′

𝑅𝑒𝜓(𝑡,𝑟)𝑒−𝜓(𝑡,𝑟) =

𝑅′

𝑅

Γ 𝜃𝜃𝑟 = (Λ−1) ��

𝑟 Γ �� Λ 𝜃

�� Λ 𝜃��

= (Λ−1) ��𝑟 Γ ��

�� (Λ 𝜃�� )

2 = 𝑒𝜓

(𝑡,𝑟) (−𝑅′

𝑅𝑒𝜓(𝑡,𝑟))𝑅2 = −𝑅𝑅′𝑒2𝜓(𝑡,𝑟)

Γ 𝑡𝜙𝜙

= (Λ−1) ��

𝜙Γ �� Λ 𝑡

�� Λ 𝜙��

= Γ ��

��Λ 𝑡�� =

��

𝑅⋅ 1 =

��

𝑅

Γ 𝜙𝜙𝑡 = (Λ−1) ��

𝑡 Γ �� Λ 𝜙

�� Λ 𝜙��

= (Λ−1) ��𝑡 Γ ��

�� (Λ 𝜙��)2

= 1 ⋅��

𝑅𝑅2 sin2 𝜃 = 𝑅�� sin2 𝜃

Γ 𝑟𝜙𝜙

= (Λ−1) ��

𝜙Γ �� Λ 𝑟

�� Λ 𝜙��

= Γ ��

��Λ 𝑟�� =

𝑅′

𝑅𝑒𝜓(𝑡,𝑟)𝑒−𝜓(𝑡,𝑟) =

𝑅′

𝑅

Γ 𝜙𝜙𝑟 = (Λ−1) ��

𝑟 Γ �� Λ 𝜙

�� Λ 𝜙��

= (Λ−1) ��𝑟 Γ ��

�� (Λ 𝜙��)2

= 𝑒𝜓(𝑡,𝑟) (−

𝑅′

𝑅𝑒𝜓(𝑡,𝑟))𝑅2 sin2 𝜃 = −𝑅𝑅′𝑒2𝜓(𝑡,𝑟) sin2 𝜃

Γ 𝜃𝜙𝜙

= (Λ−1) ��

𝜙Γ �� Λ 𝜃

�� Λ 𝜙��

= Γ ��

��Λ 𝜃�� =

cot 𝜃

𝑅𝑅 = cot𝜃

Γ 𝜙𝜙𝜃 = (Λ−1) ��

𝜃 Γ �� Λ 𝜙

�� Λ 𝜙��

= (Λ−1) ��𝜃 Γ ��

�� (Λ 𝜙��)2

=1

𝑅(−

cot 𝜃

𝑅)𝑅2 sin2 𝜃 = −cos𝜃 sin 𝜃

However by this method we do not obtain Γ 𝑟𝑟𝑟 = −𝜓′

To check we calculate the Christoffel symbols directly from the metric

The line element: 𝑑𝑠2 = 𝑑𝑡2 − 𝑒−2𝜓(𝑡,𝑟)𝑑𝑟2 − 𝑅2(𝑡, 𝑟)𝑑𝜃2 − 𝑅2(𝑡, 𝑟) sin2 𝜃 𝑑𝜙2

The metric tensor 𝑔𝑎𝑏 = {

1 0 0 00 −𝑒−2𝜓(𝑡,𝑟) 0 00 0 −𝑅2(𝑡, 𝑟) 0

0 0 0 −𝑅2(𝑡, 𝑟) sin2 𝜃

}


{

1 0 0 00 −𝑒2𝜓(𝑡,𝑟) 0 0

0 0 −1

𝑅2(𝑡, 𝑟)0

0 0 0 −1

𝑅2(𝑡, 𝑟) sin2 𝜃}

Γ𝑎𝑏𝑐 =1



Γ𝑟𝑟𝑟 =1

2𝜕𝑟(−𝑒

−2𝜓(𝑡,𝑟)) = 𝜓′𝑒−2𝜓(𝑡,𝑟)

⇒ Γ 𝑟𝑟𝑟 = 𝑔𝑟𝑟Γ𝑟𝑟𝑟 = −𝑒

2𝜓(𝑡,𝑟)𝜓′𝑒−2𝜓(𝑡,𝑟) = −𝜓′

Γ𝑡𝑟𝑟 = −1

2𝜕𝑡(−𝑒

−2𝜓(𝑡,𝑟)) = −��𝑒−2𝜓(𝑡,𝑟)

⇒ Γ 𝑟𝑟𝑡 = 𝑔𝑡𝑡Γ𝑡𝑟𝑟 = 1(−��𝑒

−2𝜓(𝑡,𝑟)) = −��𝑒−2𝜓(𝑡,𝑟)

Γ𝑟𝑟𝑡 = Γ𝑟𝑡𝑟 =1

2𝜕𝑡(−𝑒

−2𝜓(𝑡,𝑟)) = ��𝑒−2𝜓(𝑡,𝑟)





⇒ Γ 𝑡𝑟𝑟 = Γ 𝑟𝑡

𝑟 = 𝑔𝑟𝑟Γ𝑟𝑡𝑟 = −𝑒2𝜓(𝑡,𝑟)��𝑒−2𝜓(𝑡,𝑟) = −��

Γ𝑡𝜃𝜃 = −1

2𝜕𝑡(−𝑅

2(𝑡, 𝑟)) = ��𝑅

⇒ Γ 𝜃𝜃𝑡 = 𝑔𝑡𝑡Γ𝑡𝜃𝜃 = 1 ⋅ ��𝑅 = ��𝑅

Γ𝜃𝑡𝜃 = Γ𝜃𝜃𝑡 =1

2𝜕𝑡(−𝑅

2(𝑡, 𝑟)) = −��𝑅

⇒ Γ 𝑡𝜃𝜃 = Γ 𝜃𝑡

𝜃 = 𝑔𝜃𝜃Γ𝜃𝑡𝜃 = −1

𝑅2(−��𝑅) =

��

𝑅

Γ𝑟𝜃𝜃 = −1

2𝜕𝑟(−𝑅

2(𝑡, 𝑟)) = 𝑅′𝑅

⇒ Γ 𝜃𝜃𝑟 = 𝑔𝑟𝑟Γ𝑟𝜃𝜃 = −𝑒2𝜓(𝑡,𝑟)𝑅′𝑅

Γ𝜃𝑟𝜃 = Γ𝜃𝜃𝑟 =1

2𝜕𝑟(−𝑅

2(𝑡, 𝑟)) = −𝑅′𝑅

⇒ Γ 𝑟𝜃𝜃 = Γ 𝜃𝑟

𝜃 = 𝑔𝜃𝜃Γ𝜃𝑟𝜃 = −1

𝑅2(−𝑅′𝑅) =

𝑅′

𝑅

Γ𝑡𝜙𝜙 = −1

2𝜕𝑡(−𝑅

2(𝑡, 𝑟) sin2 𝜃 ) = ��𝑅 sin2 𝜃

⇒ Γ 𝜙𝜙𝑡 = 𝑔𝑡𝑡Γ𝑡𝜙𝜙 = 1 ⋅ ��𝑅 sin

2 𝜃 = ��𝑅 sin2 𝜃

Γ𝜙𝑡𝜙 = Γ𝜙𝜙𝑡 =1

2𝜕𝑡(−𝑅

2(𝑡, 𝑟) sin2 𝜃 ) = −��𝑅 sin2 𝜃

⇒ Γ 𝑡𝜙𝜙

= Γ 𝜙𝑡𝜙

= 𝑔𝜙𝜙Γ𝜙𝑡𝜙 = (−1

𝑅2 sin2 𝜃) (−��𝑅 sin2 𝜃) =

��

𝑅


2𝜕𝑟(−𝑅

2(𝑡, 𝑟) sin2 𝜃 ) = 𝑅′𝑅 sin2 𝜃

⇒ Γ 𝜙𝜙𝑟 = 𝑔𝑟𝑟Γ𝑟𝜙𝜙 = −𝑒2𝜓(𝑡,𝑟)𝑅′𝑅 sin2 𝜃


2𝜕𝑟(−𝑅

2(𝑡, 𝑟) sin2 𝜃 ) = −𝑅′𝑅 sin2 𝜃

⇒ Γ 𝑟𝜙𝜙

= Γ 𝜙𝑟𝜙

= 𝑔𝜙𝜙Γ𝜙𝑟𝜙 = (−1

𝑅2 sin2 𝜃) (−𝑅′𝑅 sin2 𝜃) =

𝑅′

𝑅


2𝜕𝜃(−𝑅

2(𝑡, 𝑟) sin2 𝜃 ) = 𝑅2 sin 𝜃 cos 𝜃

⇒ Γ 𝜙𝜙𝜃 = 𝑔𝜃𝜃Γ𝜃𝜙𝜙 = −

1

𝑅2𝑅2 sin𝜃 cos𝜃 = −sin𝜃 cos𝜃


2𝜕𝜃(−𝑅

2(𝑡, 𝑟) sin2 𝜃 ) = −𝑅2 sin 𝜃 cos 𝜃

⇒ Γ 𝜃𝜙𝜙

= Γ 𝜙𝜃𝜙

= 𝑔𝜙𝜙Γ𝜙𝜃𝜙 = −1

𝑅2 sin2 𝜃(−𝑅2 sin𝜃 cos 𝜃) = cot 𝜃

5.2 99The curvature two forms and the Riemann tensor To find the Riemann tensor from the curvature two forms, it can sometimes be more convenient to use the following expression

Ω �� =

1

2𝑅 ��𝑐�� 𝜔𝑐 ∧ 𝜔�� (5.28)

= 𝑅 ��𝑐�� 𝜔𝑐 ∧ 𝜔�� no summations

5.3 100Find the Ricci scalar using Cartan’s structure equations of the 2-sphere

The line element: 𝑑𝑠2 = 𝑑𝜃2 + sin2 𝜃 𝑑𝜙2

99 (McMahon, 2006, p. 113) 100 (McMahon, 2006, p. 113), example 5-2





The Basis one forms

𝜔�� = 𝑑𝜃 𝑑𝜃 = 𝜔�� 𝜂𝑖𝑗 = { 1

1}

𝜔�� = sin𝜃 𝑑𝜙 𝑑𝜙 =1

sin𝜃𝜔��

Cartan’s First Structure equation and the calculation of the curvature one forms:

𝑑𝜔�� = −Γ �� ∧ 𝜔�� (5.9)

𝑑𝜔�� = 0

𝑑𝜔�� = 𝑑(sin𝜃 𝑑𝜙) = cos 𝜃 𝑑𝜃 ∧ 𝑑𝜙 = cot 𝜃 𝜔�� ∧ 𝜔�� ⇒ Γ ��

�� = cot 𝜃 𝜔��

Curvature two forms:

Ω �� = 𝑑Γ ��

�� + Γ 𝑐�� ∧ Γ ��

𝑐 =1

2𝑅 ��𝑐�� 𝜔𝑐 ∧ 𝜔�� (5.27), (5.28)

Ω ��

��: 𝑑Γ

��

�� = 𝑑(cot 𝜃 𝜔��) = 𝑑(cos 𝜃 𝑑𝜙 ) = − sin 𝜃 𝑑𝜃 ∧ 𝑑𝜙 = −𝜔�� ∧ 𝜔��

Γ 𝑐��∧ Γ ��

𝑐 = Γ ��

��∧ Γ ��

�� + Γ ��

��∧ Γ

��

��= 0

⇒ Ω ��

�� = −𝜔�� ∧ 𝜔��

From which we can identify the single independent element of the Riemann tensor

𝑅 �� = 1

And the Ricci scalar

𝑅 = 2𝜂��𝑅 �� = 2 (S1)

5.4 The three dimensional flat space in spherical polar coordinates

5.4.1 101Ricci rotation coefficients of the three dimensional flat space in spherical polar coor-

dinates

The line element: 𝑑𝑠2 = 𝑑𝑟2 + 𝑟2𝑑𝜃2 + 𝑟2 sin2 𝜃 𝑑𝜙2

The Basis one forms

𝜔�� = 𝑑𝑟 𝑑𝑟 = 𝜔�� 𝜂𝑖𝑗

101 (McMahon, 2006, p. 120), quiz 5-1 and 5-2, the answer to quiz 5-1 is (a) and to 5-2 is (c).





𝜔�� = 𝑟𝑑𝜃 𝑑𝜃 =1

𝑟𝜔��

= {1

11

} 𝜔�� = 𝑟 sin𝜃 𝑑𝜙 𝑑𝜙 =

1

𝑟 sin𝜃𝜔��


We have: 𝑑𝜔�� = −Γ �� ∧ 𝜔�� (5.9)

Γ �� = Γ ��𝑐

�� 𝜔𝑐 (5.10)

𝑑𝜔�� = 0

𝑑𝜔�� = 𝑑(𝑟𝑑𝜃) = 𝑑𝑟 ∧ 𝑑𝜃 =1

𝑟𝜔�� ∧ 𝜔��

𝑑𝜔�� = 𝑑(𝑟 sin𝜃 𝑑𝜙) = sin𝜃 𝑑𝑟 ∧ 𝑑𝜙 + 𝑟 cos 𝜃 𝑑𝜃 ∧ 𝑑𝜙 =1

𝑟𝜔�� ∧ 𝜔�� +

cot 𝜃

𝑟𝜔�� ∧ 𝜔��


Γ �� =

{

0

1

𝑟𝜔��

1

𝑟𝜔��

−1

𝑟𝜔�� 0

cot 𝜃

𝑟𝜔��

−1

𝑟𝜔�� −

cot 𝜃

𝑟𝜔�� 0 }



Γ �� = −

1

𝑟 Γ ��

�� =1

𝑟 Γ

��

�� =

1

𝑟

Γ �� = −

1

𝑟 Γ ��

�� = −cot 𝜃

𝑟 Γ

��

�� =

cot 𝜃

𝑟

5.4.2 102Transformation of the Ricci rotation coefficients 𝚪 �� into the Christoffel symbols

𝚪 𝐛𝐜𝒂 of the three dimensional flat space in spherical polar coordinates

We have the transformation Γ bc𝑎 = (Λ−1) ��

𝑎 Γ �� Λ 𝑏

�� Λ 𝑐��

(5.14)

Γ 𝑟𝜃𝜃 = (Λ−1) ��

𝜃 Γ �� Λ 𝑟

�� Λ 𝜃��

= Γ �� Λ 𝑟

�� =1

𝑟⋅ 1 =

1

𝑟

Γ 𝜃𝜃𝑟 = (Λ−1) ��

𝑟 Γ �� Λ 𝜃

�� Λ 𝜃��

= (Λ−1) ��𝑟 Γ ��

�� (Λ 𝜃�� )

2 = 1(−

1

𝑟) 𝑟2 = −𝑟

Γ 𝑟𝜙𝜙

= (Λ−1) ��

𝜙Γ �� Λ 𝑟

�� Λ 𝜙��

= Γ ��

��Λ 𝑟�� =

1

𝑟⋅ 1 =

1

𝑟

102 (McMahon, 2006, p. 120), quiz 5-3, the answer to quiz 5-3 is Γ 𝜙𝜙

𝑟 = −𝑟 sin2 𝜃





Γ 𝜙𝜙𝑟 = (Λ−1) ��

𝑟 Γ �� Λ 𝜙

�� Λ 𝜙��

= (Λ−1) ��𝑟 Γ ��

�� (Λ 𝜙��)2

= 1(−1

𝑟) 𝑟2 sin2 𝜃 = −𝑟 sin2 𝜃

Γ 𝜃𝜙𝜙

= (Λ−1) ��

𝜙Γ �� Λ 𝜃

�� Λ 𝜙��

= Γ ��

��Λ 𝜃�� =

cot 𝜃

𝑟⋅ 𝑟 = cot 𝜃

Γ 𝜙𝜙𝜃 = (Λ−1) ��

𝜃 Γ �� Λ 𝜙

�� Λ 𝜙��

= (Λ−1) ��𝜃 Γ ��

�� (Λ 𝜙��)2

=1

𝑟(−

cot 𝜃

𝑟) 𝑟2 sin2 𝜃 = −sin𝜃 cos𝜃

5.5 103Ricci rotation coefficients of the Rindler metric The line element: 𝑑𝑠2 = 𝑢2𝑑𝑣2 − 𝑑𝑢2

The Basis one forms

𝜔�� = 𝑑𝑢 𝑑𝑢 = 𝜔�� 𝜂𝑖𝑗 = { 1

−1}

𝜔�� = 𝑢𝑑𝑣 𝑑𝑣 =1

𝑢𝜔��


We have 𝑑𝜔�� = −Γ �� ∧ 𝜔�� (5.9)

Γ �� = Γ ��𝑐

�� 𝜔𝑐 (5.10)

𝑑𝜔�� = 0

𝑑𝜔�� = 𝑑(𝑢𝑑𝑣) = 𝑑𝑢 ∧ 𝑑𝑣 =1

𝑢𝜔�� ∧ 𝜔�� ⇒ Γ ��

�� =1

𝑢𝜔��

⇒ Γ �� =

1

𝑢

We calculate Γ �� = 𝜂��Γ�� = −𝜂��Γ�� = −𝜂��𝜂��Γ ��

�� =1

𝑢

5.6 104The Einstein tensor for the Tolman-Bondi- de Sitter metric

The curvature two forms:

Ω �� = 𝑑Γ ��

�� + Γ 𝑐�� ∧ Γ ��

𝑐 =1

2𝑅 ��𝑐�� 𝜔𝑐 ∧ 𝜔�� (5.27), (5.28)

Ω �� : 𝑑Γ ��

�� = 𝑑(−��𝜔��) = 𝑑(−��𝑒−𝜓(𝑡,𝑟)𝑑𝑟) = [−��𝑒−𝜓(𝑡,𝑟) + (��)2𝑒−𝜓(𝑡,𝑟)]𝑑𝑡 ∧ 𝑑𝑟

= [−�� + (��)2]𝜔�� ∧ 𝜔��

Γ 𝑐�� ∧ Γ ��

𝑐 = Γ �� ∧ Γ ��

�� + Γ �� ∧ Γ ��

�� + Γ �� ∧ Γ ��

�� + Γ �� ∧ Γ

�� = 0

⇒ Ω �� = [−�� + (��)

2]𝜔�� ∧ 𝜔��

103 (McMahon, 2006, p. 120), quiz 5-4, the answer to quiz 5-4 is Γ ��

�� = Γ ��𝑢 =

1

𝑢

104 (McMahon, 2006, p. 121), quiz 5-7, the answer to quiz 5-7 is (d), almost!





Ω �� : 𝑑Γ ��

�� = 𝑑 (��

𝑅𝜔��) = 𝑑(��(𝑡, 𝑟)𝑑𝜃) = ��𝑑𝑡 ∧ 𝑑𝜃 + (��)′𝑑𝑟 ∧ 𝑑𝜃

=��

𝑅𝜔�� ∧ 𝜔�� +

(��)′


Γ 𝑐�� ∧ Γ ��

𝑐 = Γ �� ∧ Γ ��

�� + Γ �� ∧ Γ ��

�� + Γ �� ∧ Γ ��

�� + Γ �� ∧ Γ

�� = Γ ��

�� ∧ Γ ��

=𝑅′

𝑅𝑒𝜓(𝑡,𝑟)𝜔�� ∧ (−��𝜔��)

⇒ Ω �� =

��

𝑅𝜔�� ∧ 𝜔�� +

(��)′+ 𝑅′��


Ω ��

𝑑Γ ��

= 𝑑(��

𝑅𝜔��) = 𝑑(��(𝑡, 𝑟) sin 𝜃 𝑑𝜙)

= �� sin 𝜃 𝑑𝑡 ∧ 𝑑𝜙 + (��)′sin𝜃 𝑑𝑟 ∧ 𝑑𝜙 + �� cos 𝜃 𝑑𝜃 ∧ 𝑑𝜙

=��

𝑅𝜔�� ∧ 𝜔�� +

(��)′

𝑅𝑒𝜓(𝑡,𝑟)𝜔�� ∧ 𝜔�� +

��

𝑅2cot 𝜃 𝜔�� ∧ 𝜔��

Γ 𝑐��∧ Γ ��

𝑐 = Γ ��∧ Γ ��

�� + Γ ��∧ Γ ��

�� + Γ ��

��∧ Γ ��

�� + Γ ��

��∧ Γ

�� = Γ ��

��∧ Γ ��

�� + Γ ��

��∧ Γ ��

��

=𝑅′

𝑅𝑒𝜓(𝑡,𝑟)𝜔�� ∧ (−��𝜔��) +

cot 𝜃

𝑅𝜔�� ∧

��

𝑅𝜔��

⇒ Ω ��

=��

𝑅𝜔�� ∧ 𝜔�� +

(��)′+ 𝑅′��


Ω �� : 𝑑Γ ��

�� = 𝑑 (𝑅′

𝑅𝑒𝜓(𝑡,𝑟)𝜔��) = 𝑑(𝑅′𝑒𝜓(𝑡,𝑟)𝑑𝜃)

= [(��)′𝑒𝜓(𝑡,𝑟) + 𝑅′��𝑒𝜓(𝑡,𝑟)] 𝑑𝑡 ∧ 𝑑𝜃 + [𝑅′′𝑒𝜓(𝑡,𝑟) + 𝑅′𝜓′𝑒𝜓(𝑡,𝑟)]𝑑𝑟 ∧ 𝑑𝜃

= [(��)′+ 𝑅′��]

𝑒𝜓(𝑡,𝑟)

𝑅 𝜔�� ∧ 𝜔�� + [𝑅′′ + 𝑅′𝜓′]

𝑒2𝜓(𝑡,𝑟)

𝑅𝜔�� ∧ 𝜔��

Γ 𝑐�� ∧ Γ ��

𝑐 = Γ ��

�� ∧ Γ �� + Γ ��

�� ∧ Γ �� + Γ ��

�� ∧ Γ �� + Γ ��

�� ∧ Γ �� = Γ ��

�� ∧ Γ �� =

��

𝑅𝜔�� ∧(−��𝜔��)

⇒ Ω �� = [(��)

′+ 𝑅′��]

𝑒𝜓(𝑡,𝑟)

𝑅 𝜔�� ∧ 𝜔�� + [(𝑅′′ + 𝑅′𝜓′)


𝑅+��

𝑅]𝜔�� ∧ 𝜔��

Ω ��: 𝑑Γ ��

�� = 𝑑 (

𝑅′

𝑅𝑒𝜓(𝑡,𝑟)𝜔��) = 𝑑(𝑅′𝑒𝜓(𝑡,𝑟) sin 𝜃 𝑑𝜙)

= [(��)′𝑒𝜓(𝑡,𝑟) sin 𝜃 + 𝑅′𝜓′𝑒𝜓(𝑡,𝑟) sin𝜃] 𝑑𝑡 ∧ 𝑑𝜙

+[𝑅′′𝑒𝜓(𝑡,𝑟) sin 𝜃 + 𝑅′𝜓′𝑒𝜓(𝑡,𝑟) sin 𝜃]𝑑𝑟 ∧ 𝑑𝜙 + 𝑅′𝑒𝜓(𝑡,𝑟) cos 𝜃 𝑑𝜃 ∧ 𝑑𝜙

= [(��)′+ 𝑅′𝜓′]

𝑒𝜓(𝑡,𝑟)

𝑅𝜔�� ∧ 𝜔�� + [𝑅′′ + 𝑅′𝜓′]


𝑅𝜔�� ∧ 𝜔�� +

𝑅′

𝑅2𝑒𝜓(𝑡,𝑟) cot 𝜃 𝜔�� ∧ 𝜔��

Γ 𝑐��∧ Γ ��

𝑐 = Γ ��∧ Γ ��

�� + Γ ��∧ Γ ��

�� + Γ ��

��∧ Γ ��

�� + Γ ��

��∧ Γ ��

��= Γ

��∧ Γ ��

�� + Γ ��

��∧ Γ ��

��

=��

𝑅𝜔�� ∧ (−��𝜔��) +

cot 𝜃

𝑅𝜔�� ∧

𝑅′


⇒ Ω ��

= [(��)′+ 𝑅′𝜓′]

𝑒𝜓(𝑡,𝑟)

𝑅𝜔�� ∧ 𝜔�� + [(𝑅′′ +𝑅′𝜓′)


𝑅+��

𝑅]𝜔�� ∧ 𝜔��

Ω ��

��: 𝑑Γ

��

�� = 𝑑 (

cot 𝜃

𝑅𝜔��) = 𝑑(cos𝜃 𝑑𝜙) = − sin𝜃 𝑑𝜃 ∧ 𝑑𝜙 = −

1

𝑅2𝜔�� ∧ 𝜔��

Γ 𝑐��∧ Γ ��

𝑐 = Γ ��∧ Γ ��

�� + Γ ��∧ Γ ��

�� + Γ ��

��∧ Γ ��

�� + Γ ��

��∧ Γ

��

��= Γ

��∧ Γ ��

�� + Γ ��∧ Γ ��

��

=��

𝑅𝜔�� ∧

��

𝑅𝜔�� +

𝑅′

𝑅𝑒𝜓(𝑡,𝑟)𝜔�� ∧ (−

𝑅′

𝑅𝑒𝜓(𝑡,𝑟)𝜔��)





⇒ Ω ��

�� = [

1

𝑅2+(��)

2

𝑅2−(𝑅′)2

𝑅2𝑒2𝜓(𝑡,𝑟)]𝜔�� ∧ 𝜔��

Summarized in a matrix:

Ω �� =

{

0 [−�� + (��)

2]𝜔�� ∧ 𝜔��

��

𝑅𝜔�� ∧ 𝜔�� +

(��)′+ 𝑅′��

𝑅𝑒𝜓𝜔�� ∧ 𝜔��

��

𝑅𝜔�� ∧ 𝜔�� +

(��)′+ 𝑅′��

𝑅𝑒𝜓𝜔�� ∧ 𝜔��

𝑆 0 [(��)′+ 𝑅′��]

𝑒𝜓

𝑅 𝜔�� ∧ 𝜔�� + [(𝑅′′ + 𝑅′𝜓′)

𝑒2𝜓

𝑅+��

𝑅]𝜔�� ∧ 𝜔�� [(��)

′+ 𝑅′𝜓′]

𝑒𝜓

𝑅𝜔�� ∧ 𝜔�� + [(𝑅′′ + 𝑅′𝜓′)

𝑒2𝜓

𝑅+��

𝑅]𝜔�� ∧ 𝜔��

𝑆 𝐴𝑆 0 [1

𝑅2+(��)

2

𝑅2−(𝑅′)2

𝑅2𝑒2𝜓(𝑡,𝑟)] 𝜔�� ∧ 𝜔��

𝑆 𝐴𝑆 𝐴𝑆 0 }

Now we can find the independent elements of the Riemann tensor in the non-coordinate basis:

R �� (𝐴)

= [��

− (��)2]

𝑅 �� (𝐵) = −

��

𝑅 𝑅

��

��(𝐵) = −

��

𝑅

𝑅 �� (𝐶) = −[(��)

′+ 𝑅′��]

𝑒𝜓(𝑡,𝑟)

𝑅 𝑅

��

�� (𝐶) = −[(��)′+ 𝑅′��]

𝑒𝜓(𝑡,𝑟)

𝑅

𝑅 �� (𝐷)

= −[(𝑅′′ + 𝑅′𝜓′)𝑒2𝜓(𝑡,𝑟)

𝑅

+��

𝑅]

𝑅 ��

��(𝐷)

= −[(𝑅′′ + 𝑅′𝜓′)𝑒2𝜓(𝑡,𝑟)

𝑅

+��

𝑅]

𝑅 ��

��(𝐸) = [

1

𝑅2+(��)

2

𝑅2−(𝑅′)2

𝑅2𝑒2𝜓(𝑡,𝑟)]

Where A,B,C,D,E will be used later, to make the calculations easier

The Ricci tensor:

𝑅�� = 𝑅 ��𝑐��𝑐 (4.46)

𝑅�� = 𝑅 ��𝑐��

𝑐 = 𝑅 �� + 𝑅 ��

�� + 𝑅 �� + 𝑅

��

��= 𝑅 ��

�� + 𝑅 �� + 𝑅

��

��= [�� − (��)

2] − 2

��

𝑅= 𝐴 + 2𝐵

𝑅�� = 𝑅 ��𝑐��𝑐 = 𝑅 ��

�� + 𝑅 �� + 𝑅 ��

�� + 𝑅 ��

��= 𝑅 ��

�� + 𝑅 ��

��

= −[(��)′+ 𝑅′��]

𝑒𝜓(𝑡,𝑟)

𝑅− [(��)

′+ 𝑅′��]

𝑒𝜓(𝑡,𝑟)

𝑅= −2 [(��)

′+ 𝑅′��]

𝑒𝜓(𝑡,𝑟)

𝑅= 2𝐶

𝑅�� = 𝑅 ��𝑐��𝑐 = 𝑅 ��

�� + 𝑅 �� + 𝑅 ��

�� + 𝑅 ��

��= 0

𝑅�� = 𝑅 ��𝑐��𝑐 = 𝑅 ��

�� + 𝑅 �� + 𝑅 ��

�� + 𝑅 ��

��= 0

𝑅�� = 𝑅 ��𝑐��𝑐 = 𝑅 ��

�� + 𝑅 �� + 𝑅 ��

�� + 𝑅 ��

��= −R ��

�� + 𝑅 �� + 𝑅

��

��

= −[�� − (��)2] − 2 [(𝑅′′ + 𝑅′𝜓′)


𝑅+��

𝑅] = −𝐴 + 2𝐷

𝑅�� = 𝑅 ��𝑐��𝑐 = 𝑅 ��

�� + 𝑅 �� + 𝑅 ��

�� + 𝑅 ��

��= −𝑅 ��

�� + 𝑅 �� + 𝑅

��

��

=��

𝑅− [(𝑅′′ + 𝑅′𝜓′)


𝑅+��

𝑅] + [

1

𝑅2+(��)

2

𝑅2−(𝑅′)2

𝑅2𝑒2𝜓(𝑡,𝑟)] = −𝐵 + 𝐷 + 𝐸





𝑅�� = 𝑅 ��𝑐��𝑐 = 𝑅 ��

�� + 𝑅 �� + 𝑅 ��

�� + 𝑅 ��

��= −𝑅

��

��+ 𝑅

��

��+ 𝑅

��

��

=��

𝑅− [(𝑅′′ + 𝑅′𝜓′)


𝑅+��

𝑅] + [

1

𝑅2+(��)

2

𝑅2−(𝑅′)2

𝑅2𝑒2𝜓(𝑡,𝑟)] = −𝐵 + 𝐷 + 𝐸 = 𝑅��


𝑅�� =

{

[�� − (��)

2] − 2

��

𝑅−2 [(��)

′+ 𝑅′��]

𝑒𝜓

𝑅0 0

𝑆 − [�� − (��)2] − 2 [(𝑅′′ + 𝑅′𝜓′)

𝑒2𝜓

𝑅+��

𝑅] 0 0

0 0��

𝑅− [(𝑅′′ + 𝑅′𝜓′)

𝑒2𝜓

𝑅+��

𝑅] + [

1

𝑅2+(��)

2

𝑅2−(𝑅′)2

𝑅2𝑒2𝜓] 0

0 0 0��

𝑅− [(𝑅′′ + 𝑅′𝜓′)

𝑒2𝜓

𝑅+��

𝑅] + [

1

𝑅2+(��)

2

𝑅2−(𝑅′)2

𝑅2𝑒2𝜓]

}

Where �� refers to column and �� to row

The Ricci scalar:

𝑅 = 𝜂��𝑅�� (4.47)

𝑅 = 𝜂��𝑅�� + 𝜂��𝑅�� + 𝜂

��𝑅�� + 𝜂��𝑅�� = 𝑅�� − 𝑅�� − 𝑅�� − 𝑅��

= 𝐴 + 2𝐵 − (−𝐴 + 2𝐷) − 2(−𝐵 + 𝐷 + 𝐸) = 2𝐴 + 4𝐵 − 4𝐷 − 2𝐸

= 2𝑅 �� + 4𝑅 ��

�� − 4𝑅 �� − 2𝑅

��

��

= 2 [�� − (��)2] − 4

��

𝑅+ 4 [(𝑅′′ + 𝑅′𝜓′)


𝑅+��

𝑅] − 2 [

1

𝑅2+(��)

2

𝑅2−(𝑅′)2


The Einstein tensor:

𝐺�� = 𝑅�� −1

2𝜂��𝑅 (4.48)

𝐺�� = 𝑅�� −

1

2𝜂��𝑅 = 𝑅�� −

1

2𝑅 = 𝐴 + 2𝐵 −

1

2(2𝐴 + 4𝐵 − 4𝐷 − 2𝐸) = 2𝐷 + 𝐸

= 2𝑅 �� + 𝑅

��

��

= −2[(𝑅′′ + 𝑅′𝜓′)𝑒2𝜓(𝑡,𝑟)

𝑅+��

𝑅] + [

1

𝑅2+(��)

2

𝑅2−(𝑅′)2


=1

𝑅2[1 − 2𝑅�� + (��)

2− (2𝑅𝑅′′ + 2𝑅𝑅′𝜓′ + (𝑅′)2)𝑒2𝜓(𝑡,𝑟)]

𝐺�� = 𝑅�� −1

2𝜂��𝑅 = 𝑅�� = 𝑅 ��

�� + 𝑅 ��

��= −2 [(��)

′+ 𝑅′��]

𝑒𝜓(𝑡,𝑟)

𝑅

𝐺�� = 𝑅�� −1

2𝜂��𝑅 = 0

𝐺�� = 𝑅�� −1

2𝜂��𝑅 = 𝑅�� +

1

2𝑅 = −𝐴 + 2𝐷 +

1

2(2𝐴 + 4𝐵 − 4𝐷 − 2𝐸) = 2𝐵 − 𝐸

= 2𝑅 �� − 𝑅

��

��= −2

��

𝑅− [

1

𝑅2+(��)

2

𝑅2−(𝑅′)2


=1

𝑅2[(𝑅′)2𝑒2𝜓(𝑡,𝑟) − 2𝑅�� − 1 − (��)

2]





𝐺�� = 𝑅�� −1

2𝜂��𝑅 = 𝑅�� +

1

2𝑅 = −𝐵 + 𝐷 + 𝐸 +

1

2(2𝐴 + 4𝐵 − 4𝐷 − 2𝐸) = 𝐴 + 𝐵 − 𝐷

= R �� + 𝑅 ��

�� − 𝑅 �� = [�� − (��)

2] −

��

𝑅+ [(𝑅′′ + 𝑅′𝜓′)


𝑅+��

𝑅]

= [�� − (��)2] +

1

𝑅[(𝑅′′ + 𝑅′𝜓′)𝑒2𝜓(𝑡,𝑟) + �� − ��]

𝐺�� = 𝐺�� = [�� − (��)2] +

1

𝑅[(𝑅′′ + 𝑅′𝜓′)𝑒2𝜓(𝑡,𝑟) + �� − ��]


𝐺�� =

{

1

𝑅2[1 − 2𝑅�� + (��)

2− (2𝑅𝑅′′ + 2𝑅𝑅′𝜓′ + (𝑅′)2)𝑒2𝜓] −2 [(��)

′+ 𝑅′��]

𝑒𝜓

𝑅0 0

𝑆1

𝑅2[(𝑅′)2𝑒2𝜓 − 2𝑅�� − 1 − (��)

2] 0 0

0 0 [�� − (��)2] +

1

𝑅[(𝑅′′ + 𝑅′𝜓′)𝑒2𝜓 + �� − ��] 0

0 0 0 [�� − (��)2] +

1

𝑅[(𝑅′′ + 𝑅′𝜓′)𝑒2𝜓 + �� − ��]}


The Einstein tensor in the coordinate basis:

The transformation: 𝐺𝑎𝑏 = Λ 𝑎𝑐 Λ 𝑏

�� 𝐺𝑐�� (6.34)

𝐺𝑡𝑡 = Λ 𝑡𝑐 Λ 𝑡

�� 𝐺𝑐�� = (Λ 𝑡�� )

2𝐺�� =

1

𝑅2[1 − 2𝑅�� + (��)

2− (2𝑅𝑅′′ + 2𝑅𝑅′𝜓′ + (𝑅′)2)𝑒2𝜓]

𝐺𝑟𝑡 = Λ 𝑟𝑐 Λ 𝑡

�� 𝐺𝑐�� = Λ 𝑟�� Λ 𝑡

�� 𝐺�� = −2[(��)

′

𝑅+𝑅′��

𝑅]

𝐺𝑟𝑟 = Λ 𝑟𝑐 Λ 𝑟

�� 𝐺𝑐�� = (Λ 𝑟�� )

2𝐺�� =

1

𝑅2[(𝑅′)2 − (2𝑅�� + 1 + (��)

2) 𝑒−2𝜓]

𝐺𝜃𝜃 = Λ 𝜃𝑐 Λ 𝜃

�� 𝐺𝑐�� = (Λ 𝜃�� )

2𝐺�� = 𝑅2 ([�� − (��)

2] +

1

𝑅[(𝑅′′ + 𝑅′𝜓′)𝑒2𝜓 + �� − ��])

𝐺𝜙𝜙 = Λ 𝜙 𝑐 Λ 𝜙

�� 𝐺𝑐�� = (Λ 𝜙 ��)2

𝐺�� = 𝑅2 sin2 𝜃 ([�� − (��)2] +

1

𝑅[(𝑅′′ + 𝑅′𝜓′)𝑒2𝜓 + �� − ��])


𝐺𝑎𝑏 =

{

1

𝑅2[1 − 2𝑅�� + (��)

2− (2𝑅𝑅′′ + 2𝑅𝑅′𝜓′ + (𝑅′)2)𝑒2𝜓] −2 [

(��)′

𝑅+𝑅′��

𝑅] 0 0

𝑆1

𝑅2[(𝑅′)2 − (2𝑅�� + 1 + (��)

2) 𝑒−2𝜓] 0 0

0 0 𝑅2 ([�� − (��)2] +

1

𝑅[(𝑅′′ +𝑅′𝜓′)𝑒2𝜓 + �� − ��]) 0

0 0 0 𝑅2 sin2 𝜃 ([�� − (��)2] +

1

𝑅[(𝑅′′ + 𝑅′𝜓′)𝑒2𝜓 + �� − ��])}

Where 𝑎 refers to column and 𝑏 to row

5.7 105Calculate the Ricci rotation coefficients for a metric example 3: 𝒅𝒔𝟐 =

𝒅𝝍𝟐 + 𝐬𝐢𝐧𝐡𝟐𝝍 𝒅𝜽𝟐 + 𝐬𝐢𝐧𝐡𝟐𝝍𝐬𝐢𝐧𝟐 𝜽𝒅𝝓𝟐 The line element: 𝑑𝑠2 = 𝑑𝜓2 + sinh2𝜓 𝑑𝜃2 + sinh2𝜓 sin2 𝜃 𝑑𝜙2

105 (McMahon, 2006, p. 325), final exam 10





The Basis one forms

𝜔�� = 𝑑𝜓 𝑑𝜓 = 𝜔��

𝜂𝑖𝑗 = {1

11

} 𝜔�� = sinh𝜓𝑑𝜃 𝑑𝜃 =1

sinh𝜓𝜔��

𝜔�� = sinh𝜓 sin𝜃 𝑑𝜙 𝑑𝜙 =1

sinh𝜓 sin 𝜃𝜔��


We have: 𝑑𝜔�� = −Γ �� ∧ 𝜔�� (5.9)

Γ �� = Γ ��𝑐

�� 𝜔𝑐 (5.10)

𝑑𝜔�� = 0

𝑑𝜔�� = 𝑑(sinh𝜓𝑑𝜃)

= cosh𝜓𝑑𝜓 ∧ 𝑑𝜃

= cosh𝜓𝜔�� ∧1

sinh𝜓𝜔��

= coth𝜓𝜔�� ∧ 𝜔��

𝑑𝜔�� = 𝑑(sinh𝜓 sin𝜃 𝑑𝜙)

= cosh𝜓𝑑𝜓 ∧ 𝑑𝜙 + cos 𝜃 𝑑𝜃 ∧ 𝑑𝜙

= cosh𝜓𝜔�� ∧1

sinh𝜓 sin𝜃𝜔�� + cos𝜃

1

sinh𝜓𝜔�� ∧

1

sinh𝜓 sin𝜃𝜔��

=coth𝜓

sin𝜃𝜔�� ∧ 𝜔�� +

cot𝜃

sinh2𝜓𝜔�� ∧ 𝜔��


Γ �� =

{

0 coth𝜓𝜔��

coth𝜓

sin 𝜃𝜔��

−coth𝜓𝜔�� 0cot 𝜃

sinh2𝜓𝜔��

−coth𝜓

sin 𝜃𝜔�� −

cot𝜃

sinh2𝜓𝜔�� 0

}


Now we can find the non-zero Ricci rotation coefficients

Γ ��

�� = −coth𝜓 Γ ��

�� = coth𝜓 Γ ��

�� =

coth𝜓

sin𝜃

Γ ��

�� = −

coth𝜓

sin𝜃 Γ ��

�� = −cot 𝜃

sinh2𝜓 Γ

��

�� =

cot 𝜃

sinh2𝜓





6 The Einstein Field Equations

6.1 106The vacuum Einstein equations Prove that the Einstein field equations 𝐺𝑎𝑏 = 𝜅𝑇𝑎𝑏 reduces to the vacuum Einstein equations

𝑅𝑎𝑏 = 0 if we set 𝑇𝑎𝑏 = 0

The Einstein tensor


2𝑔𝑎𝑏𝑅

(4.48)

Now setting 𝐺𝑎𝑏 = 𝜅𝑇𝑎𝑏 = 0 and calculating


2𝑔𝑎𝑏𝑅 = 0

⇔ 𝑅𝑎𝑏 =1

2𝑔𝑎𝑏𝑅

Multiplying by 𝑔𝑎𝑏

⇒ 𝑔𝑎𝑏𝑅𝑎𝑏 =1

2𝑔𝑎𝑏𝑔𝑎𝑏𝑅

using the definition 𝑅 = 𝑔𝑎𝑏𝑅𝑎𝑏 (4.47)

and that in 4 dimensions 𝑔𝑎𝑏𝑔𝑎𝑏 = 4

⇒ 𝑅 =1

24𝑅 = 2𝑅

Now this can only be true if 𝑅𝑎𝑏 = 0 Q.E.D.

6.2 The vacuum Einstein equations with a cosmological constant Prove that the Einstein field equations 𝐺𝑎𝑏 = 𝜅𝑇𝑎𝑏 reduces to 𝑅𝑎𝑏 = 𝑔𝑎𝑏Λ and 𝑅 = 4Λ for metrics

with positive signature and 𝑅𝑎𝑏 = −𝑔𝑎𝑏Λ and 𝑅 = −4Λ for metrics with negative signature in vacuum with a cosmological constant107.

The Einstein equation in vacuum with a cosmological constant and positive signature is

0 = 𝑅𝑎𝑏 −1

2𝑔𝑎𝑏𝑅 + 𝑔𝑎𝑏Λ (6.6)

⇒ 0 = 𝑔𝑎𝑏𝑅𝑎𝑏 −1

2𝑔𝑎𝑏𝑔𝑎𝑏𝑅 + 𝑔𝑎𝑏𝑔

𝑎𝑏Λ

= 𝑅 −

1

24𝑅 + 4Λ

⇒ 𝑅 = 4Λ Q.E.D. Next we rewrite the Einstein equation

0 = 𝑅𝑎𝑏 −1

2𝑔𝑎𝑏𝑅 + 𝑔𝑎𝑏Λ

= 𝑅𝑎𝑏 −

1

2𝑔𝑎𝑏(4Λ) + 𝑔𝑎𝑏Λ

= 𝑅𝑎𝑏 − 𝑔𝑎𝑏Λ ⇒ 𝑅𝑎𝑏 = 𝑔𝑎𝑏Λ Q.E.D.

In the non coordinate basis 𝑅�� = 𝜂��Λ

106 (McMahon, 2006, p. 138) 107 An excellent qualitative explanation of the cosmological constant, you can find in (Greene, 2004, s. 273-279)





In the case of metrics with negative signature the Einstein equation in vacuum with a cosmological con-stant

0 = 𝑅𝑎𝑏 −1

2𝑔𝑎𝑏𝑅 − 𝑔𝑎𝑏Λ (6.6)

and we can see that 𝑅 = −4Λ Q.E.D. 𝑅𝑎𝑏 = −𝑔𝑎𝑏Λ Q.E.D. In the non coordinate basis 𝑅�� = −𝜂��Λ

6.3 108General remarks on the Einstein equations with a cosmological constant If we demand that the gravitational field equations are

(1) generally covariant (2) be of second differential order in 𝑔𝑎𝑏 (3) involve the energy-momentum 𝑇𝑎𝑏 linearly it can be shown that the only equation which meets these requirements is 𝑅𝑎𝑏 + 𝜇𝑅𝑔𝑎𝑏 − Λ𝑔𝑎𝑏 = 𝜅𝑇𝑎𝑏 where 𝜇, Λ, and 𝜅 are constants. The demand that 𝑇𝑎𝑏 satisfies the conservation equation ∇𝑏𝑇

𝑎𝑏 = 0

leads to 𝜇 = −1

2

Proof: if ∇𝑏𝑇

𝑎𝑏 = 0 ⇒ ∇𝑏(𝑅

𝑎𝑏 + 𝜇𝑅𝑔𝑎𝑏 − Λ𝑔𝑎𝑏) = 0

⇒ ∇𝑏𝑅𝑎𝑏 + 𝜇∇𝑏(𝑅𝑔

𝑎𝑏) − Λ∇𝑏𝑔𝑎𝑏 = 0

⇒ 109 ∇𝑏𝑅𝑎𝑏 + 𝜇 ((∇𝑏𝑅)𝑔

𝑎𝑏 + 𝑅(∇𝑏𝑔𝑎𝑏)) = 0

⇒ ∇𝑏𝑅𝑎𝑏 + 𝜇(∇𝑏𝑅)𝑔

𝑎𝑏 = 0 Next we use the Bianchi identity: ∇𝑎𝑅𝑑𝑒𝑏𝑐 + ∇𝑏𝑅𝑑𝑒𝑐𝑎 + ∇𝑐𝑅𝑑𝑒𝑎𝑏 = 0 ⇒ 𝑔𝑑𝑏(∇𝑎𝑅𝑑𝑒𝑏𝑐 + ∇𝑏𝑅𝑑𝑒𝑐𝑎 + ∇𝑐𝑅𝑑𝑒𝑎𝑏) = 0 ⇒ ∇𝑎𝑔

𝑑𝑏𝑅𝑑𝑒𝑏𝑐 + ∇𝑏𝑔𝑑𝑏𝑅𝑑𝑒𝑐𝑎 + ∇𝑐𝑔

𝑑𝑏𝑅𝑑𝑒𝑎𝑏 = 0 ⇒ ∇𝑎𝑅

𝑏𝑒𝑏𝑐 + ∇𝑏𝑅

𝑏𝑒𝑐𝑎 + ∇𝑐𝑅

𝑏𝑒𝑎𝑏 = 0

⇒ ∇𝑎𝑅𝑒𝑐 + ∇𝑏𝑅𝑏𝑒𝑐𝑎 − ∇𝑐𝑅𝑒𝑎 = 0

⇒ 𝑔𝑎𝑒(∇𝑎𝑅𝑒𝑐 + ∇𝑏𝑅𝑏𝑒𝑐𝑎 − ∇𝑐𝑅𝑒𝑎) = 0

⇒ ∇𝑎𝑔𝑎𝑒𝑅𝑒𝑐 + ∇𝑏𝑔

𝑎𝑒𝑅𝑏𝑒𝑐𝑎 − ∇𝑐𝑔𝑎𝑒𝑅𝑒𝑎 = 0

⇒ 110 ∇𝑎𝑅𝑎𝑐 + ∇𝑏𝑅

𝑏𝑐 − ∇𝑐𝑅 = 0

⇒ 2 (∇𝑎𝑅

𝑎𝑐 −

1

2∇𝑐𝑅) = 0

⇒ 2𝑔𝑏𝑐 (∇𝑎𝑅

𝑎𝑐 −

1

2∇𝑐𝑅) = 0

⇒ 2 (∇𝑎𝑔

𝑏𝑐𝑅𝑎𝑐 −1

2(∇𝑐𝑅)𝑔

𝑏𝑐) = 0

108 (d'Inverno, 1992, p. 172) 109 ∇𝑏𝑔

𝑎𝑏 = 0 110 𝑔𝑎𝑒𝑅 𝑒𝑐𝑎

𝑏 = 𝑔𝑎𝑒𝑅𝑒 𝑎𝑐 𝑏 = 𝑅 𝑎𝑐

𝑎𝑏 = 𝑅 𝑐𝑏





⇒ 2 (∇𝑎𝑅

𝑎𝑏 −1

2∇𝑎𝑅𝑔

𝑎𝑏) = 0

Now if we compare with ∇𝑏𝑅

𝑎𝑏 + 𝜇(∇𝑏𝑅)𝑔𝑎𝑏 = 0

we see that

𝜇 = −1

2

6.4 1112+1 dimensions: Gravitational collapse of an inhomogeneous spherically

symmetric dust cloud.

6.4.1 Find the components of the curvature tensor for the metric in 2+1 dimensions using

Cartan’s structure equations

The line element: 𝑑𝑠2 = −𝑑𝑡2 + 𝑒2𝑏(𝑡,𝑟)𝑑𝑟2 + 𝑅2(𝑡, 𝑟)𝑑𝜙2

The Basis one forms

𝜔�� = 𝑑𝑡

𝜂𝑖𝑗 = {−1

11

} 𝜔�� = 𝑒𝑏(𝑡,𝑟)𝑑𝑟 𝑑𝑟 = 𝑒−𝑏(𝑡,𝑟)𝜔��

𝜔�� = 𝑅(𝑡, 𝑟)𝑑𝜙 𝑑𝜙 =1

𝑅(𝑡, 𝑟)𝜔��

Cartan’s First Structure equation and the calculation of the curvature two-forms

𝑑𝜔�� = −Γ �� ∧ 𝜔�� (5.9) Γ ��

�� = Γ ��𝑐�� 𝜔𝑐 (5.10)

𝑑𝜔�� = 0

𝑑𝜔�� = 𝑑(𝑒𝑏(𝑡,𝑟)𝑑𝑟) = ��𝑒𝑏(𝑡,𝑟)𝑑𝑡 ∧ 𝑑𝑟 = ��𝜔�� ∧ 𝜔��

𝑑𝜔�� = 𝑑(𝑅(𝑡, 𝑟)𝑑𝜙) = ��𝑑𝑡 ∧ 𝑑𝜙 + 𝑅′𝑑𝑟 ∧ 𝑑𝜙 =��

𝑅𝜔�� ∧ 𝜔�� +

𝑅′

𝑅𝑒−𝑏(𝑡,𝑟)𝜔�� ∧ 𝜔��


Γ �� =

{

0 ��𝜔��

��

𝑅𝜔��

��𝜔�� 0𝑅′

𝑅𝑒−𝑏(𝑡,𝑟)𝜔��

��

𝑅𝜔�� −

𝑅′

𝑅𝑒−𝑏(𝑡,𝑟)𝜔�� 0 }


111 (McMahon, 2006, pp. 139-150), example 6-2, example 6-3, example 6-4






Ω �� = 𝑑Γ ��

�� + Γ 𝑐�� ∧ Γ ��

𝑐 =1

2𝑅 ��𝑐�� 𝜔𝑐 ∧ 𝜔�� (5.27), (5.28)

Ω �� : 𝑑Γ ��

�� = 𝑑(��𝜔��) = 𝑑(��𝑒𝑏(𝑡,𝑟)𝑑𝑟) = [��𝑒𝑏(𝑡,𝑟) + (��)2𝑒𝑏(𝑡,𝑟)] 𝑑𝑡 ∧ 𝑑𝑟

= −[�� + (��)2]𝜔�� ∧ 𝜔��

Γ 𝑐�� ∧ Γ ��

𝑐 = Γ �� ∧ Γ ��

�� + Γ �� ∧ Γ ��

�� + Γ �� ∧ Γ

�� = 0

⇒ Ω �� = −[�� + (��)

2]𝜔�� ∧ 𝜔��

Ω ��: 𝑑Γ

��

= 𝑑 (��

𝑅𝜔��) = 𝑑(��(𝑡, 𝑟)𝑑𝜙) = ��𝑑𝑡 ∧ 𝑑𝜙 + (��)

′𝑑𝑟 ∧ 𝑑𝜙

= −��

𝑅𝜔�� ∧ 𝜔�� −

(��)′

𝑅𝑒−𝑏(𝑡,𝑟)𝜔�� ∧ 𝜔��

Γ 𝑐��∧ Γ ��

𝑐 = Γ ��∧ Γ ��

�� + Γ ��∧ Γ ��

�� + Γ ��

��∧ Γ

�� = Γ ��

��∧ Γ ��

�� =𝑅′

𝑅𝑒−𝑏(𝑡,𝑟)𝜔�� ∧ ��𝜔��

⇒ Ω ��

= −��

𝑅𝜔�� ∧ 𝜔�� − (

(��)′

𝑅−𝑅′��

𝑅)𝑒−𝑏(𝑡,𝑟)𝜔�� ∧ 𝜔��

Ω ��: 𝑑Γ ��

�� = 𝑑 (

𝑅′

𝑅𝑒−𝑏(𝑡,𝑟)𝜔��) = 𝑑(𝑅′𝑒−𝑏(𝑡,𝑟)𝑑𝜙)

= [(��)′𝑒−𝑏(𝑡,𝑟) − 𝑅′��𝑒−𝑏(𝑡,𝑟)] 𝑑𝑡 ∧ 𝑑𝜙 + [𝑅′′𝑒−𝑏(𝑡,𝑟) + 𝑅′𝑏′𝑒−𝑏(𝑡,𝑟)]𝑑𝑟 ∧ 𝑑𝜙

= −[(��)′− 𝑅′��]

𝑒−𝑏(𝑡,𝑟)

𝑅 𝜔�� ∧ 𝜔�� − [𝑅′′ + 𝑅′𝑏′]

𝑒−2𝑏(𝑡,𝑟)

𝑅𝜔�� ∧ 𝜔��

Γ 𝑐��∧ Γ ��

𝑐 = Γ ��∧ Γ ��

�� + Γ ��∧ Γ ��

�� + Γ ��

��∧ Γ ��

��= Γ

��∧ Γ ��

�� =��

𝑅𝜔�� ∧ 𝜔��

⇒ Ω ��

= −[(��)′− 𝑅′��]


𝑅 𝜔�� ∧ 𝜔�� − ([𝑅′′ + 𝑅′𝑏′]


𝑅−��

𝑅)𝜔�� ∧ 𝜔��


Ω �� =

{

0 − [�� + (��)

2] 𝜔�� ∧ 𝜔�� −

��

𝑅𝜔�� ∧ 𝜔�� − (

(��)′

𝑅−𝑅′��

𝑅) 𝑒−𝑏(𝑡,𝑟)𝜔�� ∧ 𝜔��

0 −[(��)′− 𝑅′��]


𝑅 𝜔�� ∧ 𝜔�� − ([𝑅′′ + 𝑅′𝑏′]


𝑅−��

𝑅)𝜔�� ∧ 𝜔��

0 }


R �� (𝐴) = −[�� + (��)

2]𝜔�� ∧ 𝜔�� 𝑅

��

��(𝐵) = −

��

𝑅

𝑅 ��

��(C) = −[(��)

′+ 𝑅′��]


𝑅

𝑅 ��

��(D) = −([𝑅′′ + 𝑅′𝑏′]


𝑅−��

𝑅)





Where A, B, C and D will be used later to make the calculations easier

6.4.2 Find the components of the curvature tensor for the metric in 2+1 dimensions using

Cartan’s structure equations – alternative solution

The line element: 𝑑𝑠2 = −𝑑𝑡2 + 𝑒2𝑏(𝑡,𝑟)𝑑𝑟2 + 𝑅2(𝑡, 𝑟)𝑑𝜙2

Now we can compare with the Tolman-Bondi – de Sitter line element, where the primes should not be mistaken for the derivative 𝑑/𝑑𝑟.

𝑑𝑠2 = 𝑑𝑡′2 − 𝑒−2𝜓(𝑡′,𝑟′)𝑑𝑟′2 − 𝑅2(𝑡′, 𝑟′)𝑑𝜃′2 − 𝑅2(𝑡′, 𝑟′) sin2 𝜃′ 𝑑𝜙′2

And chose: 𝑑𝑡′ = 𝑑𝑡

𝑒−𝜓(𝑡′,𝑟′)𝑑𝑟′ = 𝑒𝑏(𝑡,𝑟)𝑑𝑟

𝑅(𝑡′, 𝑟′)𝑑𝜃′ = 0 𝑅(𝑡′, 𝑟′) sin𝜃′ 𝑑𝜙′ = 𝑅(𝑡, 𝑟)𝑑𝜙

Comparing the two metrics we see: 𝑑𝜙′ = 𝑑𝜙, 𝜃′ =𝜋

2, 𝑅(𝑡′, 𝑟′) = 𝑅(𝑡, 𝑟), 𝑑𝑡′ = 𝑑𝑡

Next we can use the former calculations of the Tolman-Bondi – de Sitter metric to find the Riemann and Einstein tensor for the 2+1 metric.

But first we need to find

�� =𝑑𝜓(𝑡′, 𝑟′)

𝑑𝑡′= 𝑒−𝜓(𝑡

′,𝑟′)𝑑

𝑑𝑡′(𝑒𝜓(𝑡

′,𝑟′)) = 𝑒𝑏(𝑡,𝑟)𝑑𝑟

𝑑𝑟′𝑑

𝑑𝑡(𝑒−𝑏(𝑡,𝑟)

𝑑𝑟′

𝑑𝑟) = −

𝑑𝑏(𝑡, 𝑟)

𝑑𝑡= −��(𝑡, 𝑟)

�� =𝑑2𝜓(𝑡′, 𝑟′)

𝑑𝑡′2=𝑑

𝑑𝑡(−��) = −��(𝑡, 𝑟)

𝜓′ =𝑑𝜓(𝑡′, 𝑟′)

𝑑𝑟′= 𝑒−𝜓(𝑡

′,𝑟′)𝑑

𝑑𝑟′(𝑒𝜓(𝑡

′,𝑟′)) = 𝑒−𝜓(𝑡′,𝑟′)

𝑑𝑟

𝑑𝑟′𝑑

𝑑𝑟(𝑒−𝑏(𝑡,𝑟)

𝑑𝑟′

𝑑𝑟)

= −𝑒−𝜓(𝑡′,𝑟′)𝑒−𝑏(𝑡,𝑟)𝑏′(𝑡, 𝑟)

�� =𝑑𝑅(𝑡′, 𝑟′)

𝑑𝑡′=𝑑𝑅(𝑡, 𝑟)

𝑑𝑡= ��(𝑡, 𝑟)

�� =𝑑2𝑅(𝑡′, 𝑟′)

𝑑𝑡′2=𝑑2𝑅(𝑡, 𝑟)

𝑑𝑡2= ��(𝑡, 𝑟)

𝑅′ =𝑑𝑅(𝑡′, 𝑟′)

𝑑𝑟′=𝑑𝑟

𝑑𝑟′𝑑𝑅(𝑡, 𝑟)

𝑑𝑟= 𝑒−𝜓(𝑡

′,𝑟′)𝑒−𝑏(𝑡,𝑟)𝑅′(𝑡, 𝑟)

��′ =𝑑2𝑅(𝑡′, 𝑟′)

𝑑𝑡′𝑑𝑟′=

𝑑

𝑑𝑟′(𝑑𝑅(𝑡′, 𝑟′)

𝑑𝑡′) =

𝑑𝑟

𝑑𝑟′𝑑

𝑑𝑟(��(𝑡, 𝑟)) = 𝑒−𝜓(𝑡

′,𝑟′)𝑒−𝑏(𝑡,𝑟)��′(𝑡, 𝑟)

The Riemann tensor

Tolman –Bondi – de Sitter 2+1

R �� = [�� − (��)

2] ⇒ R ��

�� (𝐴) = −[�� + (��)2]

𝑅 �� = −

��

𝑅

𝑅 �� = −[(��)

′+ 𝑅′��]

𝑒𝜓(𝑡,𝑟)

𝑅





𝑅 �� = −[(𝑅′′ + 𝑅′𝜓′)


𝑅+��

𝑅]

𝑅 ��

�� = −

��

𝑅 ⇒ 𝑅

��

��(𝐵) = −

��

𝑅

𝑅 ��

�� = −[(��)

′+ 𝑅′��]

𝑒𝜓(𝑡,𝑟)

𝑅 ⇒ 𝑅

��

��(𝐶) = −[(��)

′− 𝑅′��]


𝑅

𝑅 ��

�� = −[(𝑅′′ + 𝑅′𝜓′)


𝑅+��

𝑅] ⇒ 𝑅

��

�� (𝐷) = −[(𝑅′′ − 𝑅′𝑏′)𝑒−2𝑏(𝑡,𝑟)

𝑅−��

𝑅]

𝑅 ��

�� = [

1

𝑅2+(��)

2

𝑅2−(𝑅′)2


Where A, B, C and D will be used later to make the calculations easier

6.4.3 Find the components of the Einstein tensor in the coordinate basis for the metric in 2+1

dimensions.

The Ricci tensor:

𝑅�� = 𝑅 ��𝑐��𝑐 (4.46)

𝑅�� = 𝑅 ��𝑐��𝑐 = 𝑅 ��

�� + 𝑅 �� + 𝑅

��

��= 𝑅 ��

�� + 𝑅 ��

��= −[�� + (��)

2] −

��

𝑅= 𝐴 + 𝐵

𝑅�� = 𝑅 ��𝑐��𝑐 = 𝑅 ��

�� + 𝑅 �� + 𝑅

��

��= 𝑅

��

��= −[(��)

′− 𝑅′��]


𝑅= 𝐶

𝑅�� = 𝑅 ��𝑐��𝑐 = 𝑅 ��

�� + 𝑅 �� + 𝑅 ��

�� + 𝑅 ��

��= 0

𝑅�� = 𝑅 ��𝑐��𝑐 = 𝑅 ��

�� + 𝑅 �� + 𝑅

��

��= −R ��

�� + 𝑅 ��

��

= [�� + (��)2] − [(𝑅′′ − 𝑅′𝑏′)


𝑅−��

𝑅] = −𝐴 + 𝐷

𝑅�� = 𝑅 ��𝑐��𝑐 = 𝑅 ��

�� + 𝑅 �� + 𝑅

��

��= −𝑅

��

��+ 𝑅

��

��=��

𝑅− [(𝑅′′ − 𝑅′𝑏′)


𝑅−��

𝑅]

= −𝐵 + 𝐷


𝑅�� =

{

−[�� + (��)

2] −

��

𝑅− [(��)

′− 𝑅′��]

𝑒−𝑏

𝑅0

𝑆 [�� + (��)2] − [(𝑅′′ − 𝑅′𝑏′)

𝑒−2𝑏

𝑅−��

𝑅] 0

0 0��

𝑅− [(𝑅′′ −𝑅′𝑏′)

𝑒−2𝑏

𝑅−��

𝑅]}


The Ricci scalar:

𝑅 = 𝜂��𝑅�� (4.47)





𝑅 = 𝜂��𝑅�� + 𝜂��𝑅�� + 𝜂

��𝑅�� = −𝑅�� + 𝑅�� + 𝑅�� = −(𝐴 + 𝐵) + (−𝐴 + 𝐷) + (−𝐵 + 𝐷)

= −2𝐴 − 2𝐵 + 2𝐷 = −2𝑅 �� − 2𝑅

��

��+ 2𝑅

��

��

= 2 [�� + (��)2] + 2

��

𝑅− 2 [(𝑅′′ − 𝑅′𝑏′)


𝑅−��

𝑅]


𝐺�� = 𝑅�� −1

2𝜂��𝑅 (4.48)

𝐺�� = 𝑅�� −1

2𝜂��𝑅 = 𝑅�� +

1

2𝑅 = 𝐴 + 𝐵 +

1

2(−2𝐴 − 2𝐵 + 2𝐷) = 𝐷 = 𝑅

��

��

= −[(𝑅′′ − 𝑅′𝑏′)𝑒−2𝑏

𝑅−��

𝑅]

𝐺�� = 𝑅�� −1

2𝜂��𝑅 = 𝑅�� = 𝑅 ��

��= −[(��)

′− 𝑅′��]

𝑒−𝑏

𝑅

𝐺�� = 𝑅�� −1

2𝜂��𝑅 = 0

𝐺�� = 𝑅�� −1

2𝜂��𝑅 = 𝑅�� −

1

2𝑅 = −𝐴 + 𝐷 −

1

2(−2𝐴 − 2𝐵 + 2𝐷) = 𝐵 = 𝑅

��

��= −

��

𝑅

𝐺�� = 𝑅�� −1

2𝜂��𝑅 = 𝑅�� −

1

2𝑅 = −𝐵 + 𝐷 −

1

2(−2𝐴 − 2𝐵 + 2𝐷) = 𝐴 = R ��

�� = −[�� + (��)2]


𝐺�� =

{

− [(𝑅′′ − 𝑅′𝑏′)

𝑒−2𝑏

𝑅−��

𝑅] − [(��)

′− 𝑅′��]

𝑒−𝑏

𝑅0

𝑆 −��

𝑅0

0 0 − [�� + (��)2]}


The Einstein tensor in the coordinate basis:

The transformation: 𝐺𝑎𝑏 = Λ 𝑎𝑐 Λ 𝑏

�� 𝐺𝑐�� (6.34)

𝐺𝑡𝑡 = Λ 𝑡𝑐 Λ 𝑡

�� 𝐺𝑐�� = (Λ 𝑡�� )

2𝐺�� = −[(𝑅′′ − 𝑅′𝑏′)

𝑒−2𝑏

𝑅−��

𝑅]

𝐺𝑟𝑡 = Λ 𝑟𝑐 Λ 𝑡

�� 𝐺𝑐�� = Λ 𝑟�� Λ 𝑡

�� 𝐺�� = −[(��)

′− 𝑅′��]

𝑅

𝐺𝑟𝑟 = Λ 𝑟𝑐 Λ 𝑟

�� 𝐺𝑐�� = (Λ 𝑟�� )

2𝐺�� = −

��

𝑅𝑒2𝑏





𝐺𝜙𝜙 = Λ 𝜙𝑐 Λ 𝜙

�� 𝐺𝑐�� = (Λ 𝜙��)2

𝐺�� = −𝑅2 [�� + (��)2]


𝐺𝑎𝑏 =

{

− [(𝑅′′ − 𝑅′𝑏′)

𝑒−2𝑏

𝑅−��

𝑅] −

[(��)′− 𝑅′��]

𝑅0

𝑆 −��

𝑅𝑒2𝑏 0

0 0 −𝑅2 [�� + (��)2]}


6.4.4 The Einstein equations of the metric in 2+1 dimensions.

Given the Einstein equation ( if 𝑐 = 𝐺 = 1): 𝐺�� + Λ𝜂�� = 𝜅𝑇�� (6.40) with Λ = −𝜆2 you get 𝐺�� − 𝜆

2𝜂�� = 𝜅𝑇��

and the stress-energy tensor: 𝑇�� = 𝜅 {𝜌 0 00 0 00 0 0

}

You can find the Einstein – equations

{

−[(𝑅′′ − 𝑅′𝑏′)

𝑒−2𝑏

𝑅−��

𝑅] − [(��)

′− 𝑅′��]

𝑒−𝑏

𝑅0

𝑆 −��

𝑅0

0 0 − [�� + (��)2]}

− 𝜆2 {−1

1

1

}

= 𝜅 {𝜌 0 00 0 00 0 0

}

𝐺��: −[(𝑅′′ − 𝑅′𝑏′)𝑒−2𝑏

𝑅−��

𝑅] + 𝜆2 = 𝜅𝜌 p.152

𝐺��: −[(��)′− 𝑅′��]

𝑒−𝑏

𝑅 = 0

⇔ (��)′− 𝑅′�� = 0 p.152

𝐺��: −��

𝑅− 𝜆2 = 0

⇔ �� + 𝜆2𝑅 = 0 (6.41)

𝐺��: − [�� + (��)2] − 𝜆2 = 0 (6.42)

6.5 112Using the contracted Bianchi identities, prove that: 𝛁𝒃𝑮𝒂𝒃 = 𝟎

Expressions needed: Bianchi identity: 0 = ∇a𝑅𝑑𝑒𝑏𝑐 + ∇𝑏𝑅𝑑𝑒𝑐𝑎 + ∇𝑐𝑅𝑑𝑒𝑎𝑏 (4.45) Proved page 78: 0 = ∇𝑐𝑔𝑎𝑏 0 = ∇𝑐𝑔

𝑎𝑏 ⇔ 𝑔𝑑𝑎𝑔𝑒𝑏∇𝑐𝑔𝑎𝑏 = 0 ⇔ ∇𝑐𝑔𝑑𝑎𝑔𝑒𝑏𝑔

𝑎𝑏 = 0 (S3)

112 (McMahon, 2006, p. 152), quiz 6-1, the answer to quiz 6-1 is (c)





⇔ 0 = ∇𝑐𝑔𝑑𝑒 Riemann tensor: 𝑅𝑎𝑏𝑐𝑑 = −𝑅𝑎𝑏𝑑𝑐 (4.44) Ricci tensor: 𝑅 𝑎𝑐𝑏

𝑐 = 𝑅𝑎𝑏 (4.46) Ricci scalar: 𝑅 = 𝑔𝑎𝑏𝑅𝑎𝑏 = 𝑅 𝑎

𝑎 (4.47) 𝑔𝑎𝑒𝑅 𝑒𝑐𝑎

𝑏 = 𝑔𝑎𝑒𝑅𝑒 𝑎𝑐 𝑏 = 𝑅 𝑎𝑐

𝑎𝑏 = 𝑅 𝑐𝑏 (S4)

The Einstein tensor: 𝐺𝑎𝑏 = 𝑅𝑎𝑏 −1

2𝑔𝑎𝑏𝑅 (4.48)

Kronecker delta 𝑔 𝑐𝑎 = 𝑔𝑎𝑏𝑔𝑏𝑐 = 𝛿𝑐

𝑎 (2.15)

∇𝑎𝐺 𝑐𝑎 = ∇𝑎 (𝑅 𝑐

𝑎 −1

2𝑔 𝑐𝑎 𝑅) = ∇𝑎𝑅 𝑐

𝑎 −1

2∇𝑎𝑔 𝑐

𝑎 𝑅

use (2.15) = ∇𝑎𝑅 𝑐𝑎 −

1

2𝑔 𝑐𝑎 ∇𝑎𝑅 = ∇𝑎𝑅 𝑐

𝑎 −1

2𝛿𝑐𝑎∇𝑎𝑅

= ∇𝑎𝑅 𝑐𝑎 −

1

2∇𝑐𝑅 (S5)

The proof:

Multiply (4.45) by 𝑔𝑑𝑏: 0 = 𝑔𝑑𝑏(∇a𝑅𝑑𝑒𝑏𝑐 + ∇𝑏𝑅𝑑𝑒𝑐𝑎 + ∇𝑐𝑅𝑑𝑒𝑎𝑏) ⇔ (use (S3)) 0 = ∇a𝑔

𝑑𝑏𝑅𝑑𝑒𝑏𝑐 + ∇𝑏𝑔𝑑𝑏𝑅𝑑𝑒𝑐𝑎 + ∇𝑐𝑔

𝑑𝑏𝑅𝑑𝑒𝑎𝑏

⇔ 0 = ∇a𝑅 𝑒𝑏𝑐𝑏 + ∇𝑏𝑅 𝑒𝑐𝑎

𝑏 + ∇𝑐𝑅 𝑒𝑎𝑏𝑏

⇔ (use (4.44) and (4.46)) 0 = ∇𝑎𝑅𝑒𝑐 + ∇𝑏𝑅 𝑒𝑐𝑎𝑏 − ∇𝑐𝑅 𝑒𝑏𝑎

𝑏 ⇔ (use (4.46)) 0 = ∇𝑎𝑅𝑒𝑐 + ∇𝑏𝑅 𝑒𝑐𝑎

𝑏 − ∇𝑐𝑅𝑒𝑎 Multiply by 𝑔𝑎𝑒 0 = 𝑔𝑎𝑒(∇𝑎𝑅𝑒𝑐 + ∇𝑏𝑅 𝑒𝑐𝑎

𝑏 − ∇𝑐𝑅𝑒𝑎)

⇔ 0 = ∇𝑎𝑔𝑎𝑒𝑅𝑒𝑐 + ∇𝑏𝑔

𝑎𝑒𝑅 𝑒𝑐𝑎𝑏 − ∇𝑐𝑔

𝑎𝑒𝑅𝑒𝑎 ⇔ 0 = ∇𝑎𝑅 𝑐

𝑎 + ∇𝑏𝑔𝑎𝑒𝑅 𝑒𝑐𝑎

𝑏 − ∇𝑐𝑅 𝑎𝑎

⇔ (use (4.47)) 0 = ∇𝑎𝑅 𝑐𝑎 + ∇𝑏𝑔

𝑎𝑒𝑅 𝑒𝑐𝑎𝑏 − ∇𝑐𝑅

⇔(use (S4)) 0 = ∇𝑎𝑅 𝑐𝑎 + ∇𝑏𝑅 𝑐

𝑏 − ∇𝑐𝑅

⇔ 0 = 2 [∇𝑎𝑅 𝑐𝑎 −

1

2∇𝑐𝑅]

⇔ (use (S5)) 0 = ∇𝑎𝐺 𝑐𝑎

Multiply by 𝑔𝑏𝑐 0 = 𝑔𝑏𝑐∇𝑎𝐺 𝑐𝑎 = ∇𝑎𝑔

𝑏𝑐𝐺 𝑐𝑎 = ∇𝑎𝐺

𝑎𝑏

6.6 113Ricci rotation coefficients, Ricci scalar and Einstein equations for a general

4-dimensional metric: 𝒅𝒔𝟐 = −𝒅𝒕𝟐 + 𝑳𝟐(𝒕, 𝒓)𝒅𝒓𝟐 + 𝑩𝟐(𝒕, 𝒓)𝒅𝝓𝟐 +𝑴𝟐(𝒕, 𝒓)𝒅𝒛𝟐 The line element: 𝑑𝑠2 = −𝑑𝑡2 + 𝐿2(𝑡, 𝑟)𝑑𝑟2 + 𝐵2(𝑡, 𝑟)𝑑𝜙2 +𝑀2(𝑡, 𝑟)𝑑𝑧2

The Basis one forms

𝜔�� = 𝑑𝑡

𝜂𝑖𝑗 = {

−11

11

}

𝜔�� = 𝐿(𝑡, 𝑟)𝑑𝑟 𝑑𝑟 =1

𝐿(𝑡, 𝑟)𝜔��

𝜔�� = 𝐵(𝑡, 𝑟)𝑑𝜙 𝑑𝜙 =1

𝐵(𝑡, 𝑟)𝜔��

𝜔 �� = 𝑀(𝑡, 𝑟)𝑑𝑧 𝑑𝑧 =1

𝑀(𝑡, 𝑟)𝜔��

113 (McMahon, 2006, pp. 152-53), quiz 6-5, 6-6, 6-7 and 6-8, the answer to quiz 6-5 is (a) and quiz 6-6 is (c), the answer to quiz 6-7 is (a), the answer to quiz 6-8 is (a)






𝑑𝜔�� = −Γ �� ∧ 𝜔�� (5.9) Γ ��

�� = Γ ��𝑐�� 𝜔𝑐 (5.10)

𝑑𝜔�� = 0

𝑑𝜔�� = 𝑑(𝐿(𝑡, 𝑟)𝑑𝑟) = ��𝑑𝑡 ∧ 𝑑𝑟 =��

𝐿𝜔�� ∧ 𝜔��

𝑑𝜔�� = 𝑑(𝐵(𝑡, 𝑟)𝑑𝜙) = ��𝑑𝑡 ∧ 𝑑𝜙 + 𝐵′𝑑𝑟 ∧ 𝑑𝜙 =��

𝐵𝜔�� ∧ 𝜔�� +

𝐵′

𝐿𝐵𝜔�� ∧ 𝜔��

𝑑𝜔�� = 𝑑(𝑀(𝑡, 𝑟)𝑑𝑧) = ��𝑑𝑡 ∧ 𝑑𝑧 +𝑀′𝑑𝑟 ∧ 𝑑𝑧 =��

𝑀𝜔�� ∧ 𝜔�� +

𝑀′

𝐿𝑀𝜔�� ∧ 𝜔��


Γ �� =

{

0

��

𝐿𝜔��

��

𝐵𝜔��

��

𝑀𝜔��

��

𝐿𝜔�� 0

𝐵′

𝐿𝐵𝜔��

𝑀′

𝐿𝑀𝜔��

��

𝐵𝜔�� −

𝐵′

𝐿𝐵𝜔�� 0 0

��

𝑀𝜔�� −

𝑀′

𝐿𝑀𝜔�� 0 0 }



Γ �� =

��

𝐿 Γ ��

�� =��

𝐿 Γ

��

�� =

��

𝐵 Γ ��

�� =��

𝑀

Γ �� =

��

𝐵 Γ ��

�� = −𝐵′

𝐿𝐵 Γ

��

�� =

𝐵′

𝐿𝐵 Γ ��

�� =𝑀′

𝐿𝑀

Γ �� =

��

𝑀 Γ ��

�� = −𝑀′

𝐿𝑀


Ω �� = 𝑑Γ ��

�� + Γ 𝑐�� ∧ Γ ��

𝑐 =1

2𝑅 ��𝑐�� 𝜔𝑐 ∧ 𝜔�� (5.27), (5.28)

Ω �� : 𝑑Γ ��

�� = 𝑑 (��

𝐿𝜔��) = 𝑑 (��(𝑡, 𝑟)) 𝑑𝑟 = ��𝑑𝑡 ∧ 𝑑𝑟 + 𝐿′ 𝑑𝑟 ∧ 𝑑𝑟 =

��

𝐿𝜔�� ∧ 𝜔��

Γ 𝑐�� ∧ Γ ��

𝑐 = Γ �� ∧ Γ ��

�� + Γ �� ∧ Γ ��

�� + Γ �� ∧ Γ

��+ Γ ��

�� ∧ Γ �� = 0

⇒ Ω �� = −

��

𝐿𝜔�� ∧ 𝜔��





Ω ��: 𝑑Γ

��

= 𝑑 (��

𝐵𝜔��) = 𝑑(��(𝑡, 𝑟)𝑑𝜙) = ��𝑑𝑡 ∧ 𝑑𝜙 + 𝐵′ 𝑑𝑟 ∧ 𝑑𝜙

=��

𝐵𝜔�� ∧ 𝜔�� +

𝐵′

𝐵𝐿𝜔�� ∧ 𝜔��

Γ 𝑐��∧ Γ ��

𝑐 = Γ ��∧ Γ ��

�� + Γ ��∧ Γ ��

�� + Γ ��

��∧ Γ

��++Γ ��

��∧ Γ ��

�� = Γ ��∧ Γ ��

�� =𝐵´

𝐵𝐿𝜔�� ∧

��

𝐿𝜔��

⇒ Ω ��

=��

𝐵𝜔�� ∧ 𝜔�� +

𝐵′

𝐵𝐿𝜔�� ∧ 𝜔�� +

𝐵´

𝐵𝐿𝜔�� ∧

��

𝐿𝜔��

= −��

𝐵𝜔�� ∧ 𝜔�� + (

𝐵´��

𝐵𝐿2−𝐵′

𝐵𝐿)𝜔�� ∧ 𝜔��

Ω �� : 𝑑Γ ��

�� = 𝑑 (

��

𝑀𝜔��) = 𝑑(��(𝑡, 𝑟)𝑑𝑧) = ��𝑑𝑡 ∧ 𝑑𝑧 + ��′𝑑𝑟 ∧ 𝑑𝑧

=��

𝑀𝜔�� ∧ 𝜔�� +

��′

𝑀𝐿𝜔�� ∧ 𝜔��

Γ 𝑐�� ∧ Γ ��

𝑐 = Γ �� ∧ Γ ��

�� + Γ �� ∧ Γ ��

�� + Γ �� ∧ Γ

��+ Γ ��

�� ∧ Γ �� = Γ ��

�� ∧ Γ �� =

𝑀´

𝑀𝐿𝜔�� ∧

��

𝐿𝜔��

Ω �� =

��

𝑀𝜔�� ∧ 𝜔�� +

��′

𝑀𝐿𝜔�� ∧ 𝜔�� +

𝑀´

𝑀𝐿𝜔 �� ∧

��

𝐿𝜔��

= −��

𝑀𝜔 �� ∧ 𝜔�� + (

𝑀´��

𝑀𝐿2−��′

𝑀𝐿)𝜔�� ∧ 𝜔��

Ω ��: 𝑑Γ ��

�� = 𝑑 (

𝐵´

𝐵𝐿𝜔��) = 𝑑 (

𝐵´

𝐿𝑑𝜙) =

��′𝐿 − 𝐵′��

𝐿2𝑑𝑡 ∧ 𝑑𝜙 +

𝐵′′𝐿 − 𝐵′𝐿′

𝐿2𝑑𝑟 ∧ 𝑑𝜙

=𝐵′�� − ��′𝐿

𝐵𝐿2𝜔�� ∧ 𝜔�� +

𝐵′𝐿′ − 𝐵′′𝐿

𝐵𝐿3𝜔�� ∧ 𝜔��

Γ 𝑐��∧ Γ ��

𝑐 = Γ ��∧ Γ ��

�� + Γ ��∧ Γ ��

�� + Γ ��

��∧ Γ ��

��+ Γ ��

��∧ Γ ��

�� = Γ ��∧ Γ ��

�� =��

𝐵𝐿𝜔�� ∧ 𝜔��

⇒ Ω ��

=𝐵′�� − ��′𝐿

𝐵𝐿2𝜔�� ∧ 𝜔�� +

𝐵′𝐿′ − 𝐵′′𝐿

𝐵𝐿3𝜔�� ∧ 𝜔�� +

��

𝐵𝐿𝜔�� ∧ 𝜔��

=𝐵′�� − ��′𝐿

𝐵𝐿2𝜔�� ∧ 𝜔�� + (

𝐵′𝐿′ − 𝐵′′𝐿

𝐵𝐿3+��

𝐵𝐿)𝜔�� ∧ 𝜔��

Ω �� : 𝑑Γ ��

�� = 𝑑 (𝑀´

𝑀𝐿𝜔��) = 𝑑 (

𝑀´

𝐿𝑑𝑧) =

��´𝐿 − 𝑀´��

𝐿2𝑑𝑡 ∧ 𝑑𝑧 +

𝑀´´𝐿 −𝑀´𝐿´

𝐿2𝑑𝑟 ∧ 𝑑𝑧

=𝑀´�� − ��´𝐿

𝑀𝐿2𝜔�� ∧ 𝜔�� +

𝑀´𝐿´ − 𝑀´´𝐿

𝑀𝐿3𝜔�� ∧ 𝜔��

Γ 𝑐�� ∧ Γ ��

𝑐 = Γ �� ∧ Γ ��

�� + Γ �� ∧ Γ ��

�� + Γ �� ∧ Γ ��

��+ Γ ��

�� ∧ Γ �� = Γ ��

�� ∧ Γ �� =

��

𝑀𝐿𝜔�� ∧ 𝜔��

⇒ Ω �� =

𝑀´�� − ��´𝐿

𝑀𝐿2𝜔�� ∧ 𝜔�� +

𝑀´𝐿´ − 𝑀´´𝐿

𝑀𝐿3𝜔�� ∧ 𝜔�� +

��

𝑀𝐿𝜔�� ∧ 𝜔��

=𝑀´�� − ��´𝐿

𝑀𝐿2𝜔�� ∧ 𝜔�� + (

𝑀´𝐿´ − 𝑀´´𝐿

𝑀𝐿3+��

𝑀𝐿)𝜔�� ∧ 𝜔��

Ω �� : 𝑑Γ ��

�� = 0

⇒ Ω �� = Γ 𝑐

�� ∧ Γ ��𝑐 = Γ ��

�� ∧ Γ �� + Γ ��

�� ∧ Γ �� + Γ ��

�� ∧ Γ ��

��+ Γ ��

�� ∧ Γ ��

= Γ �� ∧ Γ ��

�� + Γ �� ∧ Γ ��

�� =��

𝑀𝜔�� ∧

��

𝐵𝜔�� +

𝑀′

𝐿𝑀𝜔�� ∧ (−

𝐵′

𝐿𝐵𝜔��)

= (��

𝐵𝑀−𝐵′𝑀′

𝐿2𝐵𝑀)𝜔�� ∧ 𝜔��






Ω �� =

{

0 −

��

𝐿𝜔�� ∧ 𝜔�� −

��

𝐵𝜔�� ∧ 𝜔�� + (

𝐵´��

𝐵𝐿2−𝐵′

𝐵𝐿)𝜔�� ∧ 𝜔�� −

��

𝑀𝜔�� ∧ 𝜔�� + (

𝑀´��

𝑀𝐿2−��′

𝑀𝐿)𝜔�� ∧ 𝜔��

𝑆 0 (𝐵´��

𝐵𝐿2−𝐵′

𝐵𝐿)𝜔�� ∧ 𝜔�� + (

𝐵′𝐿′ − 𝐵′′𝐿

𝐵𝐿3+��

𝐵𝐿)𝜔�� ∧ 𝜔�� (

𝑀´��

𝑀𝐿2−��′

𝑀𝐿)𝜔�� ∧ 𝜔�� + (

𝑀´𝐿´ −𝑀´´𝐿

𝑀𝐿3+��

𝑀𝐿)𝜔�� ∧ 𝜔��

𝑆 𝐴𝑆 0 (��


𝐿2𝐵𝑀)𝜔�� ∧ 𝜔��



R �� (𝐴) = −

��

𝐿 𝑅

��

��(𝐵) = −

��

𝐵 𝑅 ��𝑧��

�� (𝐶) = −��

𝑀

𝑅 ��

�� (𝐷) =𝐵´��

𝐵𝐿2−𝐵′

𝐵𝐿 𝑅 ��

�� (𝐸) =𝑀´��

𝑀𝐿2−��′

𝑀𝐿

𝑅 ��

��(𝐹) =

𝐵′𝐿′ −𝐵′′𝐿

𝐵𝐿3+��

𝐵𝐿 𝑅 ��

�� (𝐺) =𝑀´𝐿´ − 𝑀´´𝐿

𝑀𝐿3+��

𝑀𝐿

𝑅 �� (𝐻) =

��


𝐿2𝐵𝑀

Where A,B,C,D,E,F,G,H will be used later, to make the calculations easier

The Ricci tensor:

𝑅�� = 𝑅 ��𝑐��𝑐 (4.46)

𝑅�� = 𝑅 ��𝑐��𝑐 = 𝑅 ��

�� + 𝑅 �� + 𝑅

��

��+ 𝑅 ��

�� = 𝑅 �� + 𝑅

��

��+ 𝑅 ��

��

= −��

𝐿−��

𝐵−��

𝑀= 𝐴 + 𝐵 + 𝐶

𝑅�� = 𝑅 ��𝑐��𝑐 = 𝑅 ��

�� + 𝑅 �� + 𝑅

��

��+ 𝑅 ��

�� = 𝑅 ��

��+ 𝑅 ��

��

=𝐵´��

𝐵𝐿2−𝐵′

𝐵𝐿+𝑀´��

𝑀𝐿2−��′

𝑀𝐿= 𝐷 + 𝐸

𝑅�� = 𝑅 ��𝑐��𝑐 = 𝑅 ��

�� + 𝑅 �� + 𝑅

��

��+ 𝑅 ��

�� = 0

𝑅�� = 𝑅 ��𝑐��𝑐 = 𝑅 ��

�� + 𝑅 �� + 𝑅

��

��+ 𝑅 ��

�� = 0

𝑅�� = 𝑅 ��𝑐��𝑐 = 𝑅 ��

�� + 𝑅 �� + 𝑅

��

��+ 𝑅 ��

�� = −R �� + 𝑅

��

��+ 𝑅 ��

��

=��

𝐿+𝐵′𝐿′ − 𝐵′′𝐿

𝐵𝐿3+��

𝐵𝐿+𝑀´𝐿´ − 𝑀´´𝐿

𝑀𝐿3+��

𝑀𝐿= −𝐴 + 𝐹 + 𝐺

𝑅�� = 𝑅 ��𝑐��𝑐 = 𝑅 ��

�� + 𝑅 �� + 𝑅

��

��+ 𝑅 ��

�� = −𝑅 ��

��+ 𝑅

��

��+ 𝑅 ��

��

=��

𝐵+𝐵′𝐿′ − 𝐵′′𝐿

𝐵𝐿3+��

𝐵𝐿+��


𝐿2𝐵𝑀= −𝐵 + 𝐹 + 𝐻

𝑅�� = 𝑅 ��𝑐��𝑐 = 𝑅 ��

�� + 𝑅 �� + 𝑅

��

��+ 𝑅 ��

�� = −𝑅 �� + 𝑅 ��

�� + 𝑅 ��

=��

𝑀+𝑀´𝐿´ − 𝑀´´𝐿

𝑀𝐿3+��

𝑀𝐿+��


𝐿2𝐵𝑀= −𝐶 + 𝐺 + 𝐻






𝑅�� =

{

−

��

𝐿−��

𝐵−��

𝑀

𝐵´��

𝐵𝐿2−𝐵′


𝑀𝐿2−��′

𝑀𝐿0 0

𝑆��

𝐿+𝐵′𝐿′ − 𝐵′′𝐿

𝐵𝐿3+��

𝐵𝐿+𝑀´𝐿´ −𝑀´´𝐿

𝑀𝐿3+��

𝑀𝐿0 0

0 0��

𝐵+𝐵′𝐿′ −𝐵′′𝐿

𝐵𝐿3+��

𝐵𝐿+��


𝐿2𝐵𝑀0

0 0 0��

𝑀+𝑀´𝐿´ −𝑀´´𝐿

𝑀𝐿3+��

𝑀𝐿+��


𝐿2𝐵𝑀}


The Ricci scalar:

𝑅 = 𝜂��𝑅�� (4.47)

𝑅 = 𝜂��𝑅�� + 𝜂��𝑅�� + 𝜂

��𝑅�� + 𝜂��𝑅�� = −𝑅�� + 𝑅�� + 𝑅�� + 𝑅��

= −(𝐴 + 𝐵 + 𝐶) − 𝐴 + 𝐹 + 𝐺 − 𝐵 + 𝐹 + 𝐻 − 𝐶 + 𝐺 + 𝐻 = −2𝐴 − 2𝐵 − 2𝐶 + 2𝐹 + 2𝐺 + 2𝐻

= −2R �� − 2𝑅

��

��− 2𝑅 ��𝑧��

�� + 2𝑅 ��

��+ 2𝑅 ��

�� + 2𝑅 ��

= 2(��

𝐿+��

𝐵+��

𝑀+𝐵′𝐿′ − 𝐵′′𝐿

𝐵𝐿3+��


𝑀𝐿3+��

𝑀𝐿+��


𝐿2𝐵𝑀)


𝐺�� = 𝑅�� −1

2𝜂��𝑅 (4.48)

𝐺�� = 𝑅�� −1

2𝜂��𝑅 = 𝑅�� +

1

2𝑅 = 𝐴 + 𝐵 + 𝐶 +

1

2(−2𝐴 − 2𝐵 − 2𝐶 + 2𝐹 + 2𝐺 + 2𝐻) = 𝐹 + 𝐺 + 𝐻

= 𝑅 ��

��+ 𝑅 ��

�� + 𝑅 �� =

𝐵′𝐿′ − 𝐵′′𝐿

𝐵𝐿3+��


𝑀𝐿3+��

𝑀𝐿+��


𝐿2𝐵𝑀

𝐺�� = 𝑅�� −1

2𝜂��𝑅 = 𝑅�� =

𝐵´��

𝐵𝐿2−𝐵′


𝑀𝐿2−��′

𝑀𝐿

𝐺�� = 𝑅�� −1

2𝜂��𝑅 = 0

𝐺�� = 𝑅�� −1

2𝜂��𝑅 = 0

𝐺�� = 𝑅�� −

1

2𝜂��𝑅 = 𝑅�� −

1

2𝑅 = −𝐴 + 𝐹 + 𝐺 −

1

2(−2𝐴 − 2𝐵 − 2𝐶 + 2𝐹 + 2𝐺 + 2𝐻)

= 𝐵 + 𝐶 + 𝐻

= 𝑅 ��

��+ 𝑅 ��𝑧��

�� + 𝑅 �� = −

��

𝐵−��

𝑀+��


𝐿2𝐵𝑀

𝐺�� = 𝑅�� −1

2𝜂��𝑅 = 0

𝐺�� = 𝑅�� −1

2𝜂��𝑅 = 0

𝐺�� = 𝑅�� −1

2𝜂��𝑅 = 𝑅�� −

1

2𝑅 = −𝐵 + 𝐹 + 𝐻 −

1

2(−2𝐴 − 2𝐵 − 2𝐶 + 2𝐹 + 2𝐺 + 2𝐻)

= 𝐴 + 𝐶 + 𝐺 = R �� + 𝑅 ��𝑧��

�� + 𝑅 �� = −

��

𝐿−��

𝑀


𝑀𝐿3+��

𝑀𝐿





𝐺�� = 𝑅�� −1

2𝜂��𝑅 = 0

𝐺�� = 𝑅�� −

1

2𝜂��𝑅 = 𝑅�� −

1

2𝑅 = −𝐶 + 𝐺 + 𝐻 −

1

2(−2𝐴 − 2𝐵 − 2𝐶 + 2𝐹 + 2𝐺 + 2𝐻)

= 𝐴 + 𝐵 + 𝐹

= R �� + 𝑅

��

��+ 𝑅

��

��= −

��

𝐿−��

𝐵+𝐵′𝐿′ − 𝐵′′𝐿

𝐵𝐿3+��

𝐵𝐿


𝐺�� =

{

𝐵′𝐿′ − 𝐵′′𝐿

𝐵𝐿3+��


𝑀𝐿3+��

𝑀𝐿+��


𝐿2𝐵𝑀

𝐵´��

𝐵𝐿2−𝐵′


𝑀𝐿2−��′

𝑀𝐿0 0

𝑆 −��

𝐵−��

𝑀+��


𝐿2𝐵𝑀0 0

0 0 −��

𝐿−��

𝑀


𝑀𝐿3+��

𝑀𝐿0

0 0 0 −��

𝐿−��

𝐵+𝐵′𝐿′ − 𝐵′′𝐿

𝐵𝐿3+��

𝐵𝐿}


7 The Energy-Momentum Tensor

7.1 114Perfect Fluids – Alternative derivation The most general form of the stress energy tensor is

𝑇𝑎𝑏 = 𝐴𝑢𝑎𝑢𝑏 + 𝐵𝑔𝑎𝑏 (7.8) In the local frame we know that

𝑇�� = {

𝜌 0 0 00 𝑃 0 00 0 𝑃 00 0 0 𝑃

} (7.6)

and

𝑢�� = (1,0,0,0)

Then we choose the metric with negative signature

𝜂�� = {

1 0 0 00 −1 0 00 0 −1 00 0 0 −1

}

This we can use to find the constants 𝐴 and 𝐵

𝑇00 = 𝐴𝑢0𝑢0 + 𝐵𝜂00 = 𝐴 + 𝐵 = 𝜌

𝑇 �� = 𝐴𝑢��𝑢�� + 𝐵𝜂�� = −𝐵 = {𝑃0

𝑖𝑓 𝑖 = 𝑗 𝑖𝑓 𝑖 ≠ 𝑗

⇒ 𝐵 = −𝑃 and 𝐴 = 𝜌 − 𝐵 = 𝜌 + 𝑃 Which leaves us with the most general form of the stress energy tensor for a perfect fluid for a metric with negative signature

𝑇𝑎𝑏 = (𝜌 + 𝑃)𝑢𝑎𝑢𝑏 − 𝑃𝑔𝑎𝑏 (7.11) If we instead choose the metric with positive signature

𝜂�� = {

−1 0 0 00 1 0 00 0 1 00 0 0 1

}

114 (McMahon, 2006, p. 160)





𝑇00 = 𝐴𝑢0𝑢0 + 𝐵𝜂00 = 𝐴 − 𝐵 = 𝜌

𝑇 �� = 𝐴𝑢��𝑢�� + 𝐵𝜂�� = 𝐵 = {𝑃0

𝑖𝑓 𝑖 = 𝑗 𝑖𝑓 𝑖 ≠ 𝑗

⇒ 𝐵 = 𝑃 and 𝐴 = 𝜌 + 𝐵 = 𝜌 + 𝑃 Which leaves us with the most general form of the stress energy tensor for a perfect fluid for a metric with negative signature

𝑇𝑎𝑏 = (𝜌 + 𝑃)𝑢𝑎𝑢𝑏 + 𝑃𝑔𝑎𝑏 (7.12)

7.2 115The Gödel metric The Gödel metric is an exact solution of the Einstein field equations in which the stress-energy tensor contains two terms, the first representing the matter density of a homogeneous distribution of swirling dust particles, and the second associated with a nonzero cosmological constant.116

The line element: 𝑑𝑠2 =1

2𝜔2((𝑑𝑡 + 𝑒𝑥𝑑𝑧)2 − 𝑑𝑥2 − 𝑑𝑦2 −

1

2𝑒2𝑥𝑑𝑧2)

=1

2𝜔2(𝑑𝑡2 + 2𝑒𝑥𝑑𝑡𝑑𝑧 − 𝑑𝑥2 − 𝑑𝑦2 +

1

2𝑒2𝑥𝑑𝑧2)

The metric tensor 𝑔𝑎𝑏 =1

2𝜔2

{

1 0 0 𝑒𝑥

0 −1 0 00 0 −1 0

𝑒𝑥 0 01

2𝑒2𝑥

}

and its inverse 𝑔𝑎𝑏 = 2𝜔2 {

−1 0 0 2𝑒−𝑥

0 −1 0 00 0 −1 0

2𝑒−𝑥 0 0 −2𝑒−2𝑥

}

The stress energy tensor 𝑇𝑎𝑏 =𝜌

2𝜔2{

1 0 0 𝑒𝑥

0 0 0 00 0 0 0𝑒𝑥 0 0 𝑒2𝑥

}

The Einstein equation for a metric with a negative signature 8𝜋𝐺𝑇𝑎𝑏 = 𝐺𝑎𝑏 − 𝑔𝑎𝑏Λ

⇒ 117 8𝜋𝐺𝑔𝑎𝑏𝑇𝑎𝑏 = 𝑔𝑎𝑏𝐺𝑎𝑏 − 𝑔𝑎𝑏𝑔𝑎𝑏Λ

8𝜋𝐺𝜌 = 𝑔𝑎𝑏 (𝑅𝑎𝑏 −1

2𝑔𝑎𝑏𝑅) − 4Λ

= 𝑔𝑎𝑏𝑅𝑎𝑏 −1

2𝑔𝑎𝑏𝑔𝑎𝑏𝑅 − 4Λ

= 𝑅 −1

24𝑅 − 4Λ

= −𝑅 − 4Λ

⇒ Λ = −1

4(8𝜋𝐺𝜌 + 𝑅)

To find 𝑅 we work in the non-coordinate basis

The Basis one forms

115 (McMahon, 2006, p. 326), final exam 14, the answer to Final Exam quiz 14 is (a). 116 http://en.wikipedia.org/wiki/G%C3%B6del_metric 117 𝑔𝑎𝑏𝑇𝑎𝑏 =

𝜌

2𝜔22𝜔2(−1 ⋅ 1 + 2𝑒−𝑥𝑒𝑥 + 2𝑒−𝑥𝑒𝑥 − 2𝑒−2𝑥𝑒2𝑥) = 𝜌(−1 + 2 + 2 − 2) = 𝜌


http://en.wikipedia.org/wiki/Exact_solutions_in_general_relativity

http://en.wikipedia.org/wiki/Einstein_field_equations

http://en.wikipedia.org/wiki/Stress-energy_tensor

http://en.wikipedia.org/wiki/Cosmological_constant

http://en.wikipedia.org/wiki/G%C3%B6del_metric




𝜔�� =1

√2𝜔(𝑑𝑡 + 𝑒𝑥𝑑𝑧) 𝑑𝑡 = √2𝜔𝜔�� − 2𝜔𝜔��

𝜂𝑖𝑗 = {

1−1

−1−1

} 𝜔𝑥 =

1

√2𝜔𝑑𝑥 𝑑𝑥 = √2𝜔𝜔𝑥

𝜔 �� =1

√2𝜔𝑑𝑦 𝑑𝑦 = √2𝜔𝜔��

𝜔 �� =1

2𝜔𝑒𝑥𝑑𝑧 𝑑𝑧 = 2𝜔𝑒−𝑥𝜔��

Cartan’s First Structure equation:

𝑑𝜔�� = −Γ �� ∧ 𝜔�� (5.9)

𝑑𝜔�� = 𝑑1

√2𝜔(𝑑𝑡 + 𝑒𝑥𝑑𝑧)

=𝑒𝑥

√2𝜔𝑑𝑥 ∧ 𝑑𝑧

=𝑒𝑥

√2𝜔(√2𝜔𝜔𝑥) ∧ (2𝜔𝑒−𝑥𝜔��)

= 2𝜔𝜔𝑥 ∧ 𝜔�� = −𝜔𝜔�� ∧ 𝜔𝑥 +𝜔𝜔�� ∧ 𝜔��

= −Γ �� ∧ 𝜔𝑥 − Γ ��

�� ∧ 𝜔�� − Γ �� ∧ 𝜔��

𝑑𝜔�� = 0

𝑑𝜔�� = 0

𝑑𝜔�� = 𝑑 (1

2𝜔𝑒𝑥𝑑𝑧) =

1

2𝜔𝑒𝑥𝑑𝑥 ∧ 𝑑𝑧 =

1

2𝜔𝑒𝑥(√2𝜔𝜔𝑥) ∧ (2𝜔𝑒−𝑥𝜔��) = √2𝜔𝜔�� ∧ 𝜔��


Γ �� =

{

0 𝜔𝜔�� 0 −𝜔𝜔𝑥

𝜔𝜔�� 0 0 √2𝜔𝜔��

0 0 0 0

−𝜔𝜔�� −√2𝜔𝜔 �� 0 0 }


The curvature two forms and the Riemann tensor:

Ω �� = 𝑑Γ ��

�� + Γ 𝑐�� ∧ Γ ��

𝑐 =1

2𝑅 ��𝑐�� 𝜔𝑐 ∧ 𝜔�� (5.27), (5.28)

Ω �� : 𝑑Γ ��

�� = 0





Γ 𝑐�� ∧ Γ ��

𝑐 = Γ �� ∧ Γ ��

�� + Γ �� ∧ Γ ��

𝑥 + Γ �� ∧ Γ ��

��+ Γ ��

�� ∧ Γ �� = 0

⇒ Ω �� = 0

Ω ��𝑥 : 𝑑Γ ��

𝑥 = 𝑑(𝜔𝜔 ��) = 𝑑 (𝜔1

2𝜔𝑒𝑥𝑑𝑧) =

1

2𝑒𝑥𝑑𝑥 ∧ 𝑑𝑧 =

1

2𝑒𝑥(√2𝜔𝜔��) ∧ 2𝜔𝑒−𝑥𝜔��

= √2𝜔2𝜔�� ∧ 𝜔��

Γ 𝑐𝑥 ∧ Γ ��

𝑐 = Γ ��𝑥 ∧ Γ ��

�� + Γ ��𝑥 ∧ Γ ��

𝑥 + Γ �� ∧ Γ ��

��+ Γ ��

�� ∧ Γ �� = Γ ��

𝑥 ∧ Γ ��

= (−√2𝜔𝜔��) ∧ (−𝜔𝜔𝑥) = −√2𝜔2𝜔�� ∧ 𝜔��

⇒ Ω ��𝑥 = 0

Ω ��: 𝑑Γ ��

�� = 0

Γ 𝑐��∧ Γ ��

𝑐 = Γ ��∧ Γ ��

�� + Γ ��∧ Γ ��

𝑥 + Γ ��∧ Γ ��

��+ Γ ��

��∧ Γ ��

�� = 0

⇒ Ω ��

= 0

Ω �� : 𝑑Γ ��

�� = 𝑑(−𝜔𝜔𝑥) = 𝑑 (−𝜔1

√2𝜔𝑑𝑥) = 0

Γ 𝑐�� ∧ Γ ��

𝑐 = Γ �� ∧ Γ ��

�� + Γ �� ∧ Γ ��

𝑥 + Γ �� ∧ Γ ��

��+ Γ ��

�� ∧ Γ �� = 0

⇒ Ω �� = 0

Ω ��𝑥 : 𝑑Γ ��

𝑥 = 0

Γ 𝑐𝑥 ∧ Γ ��

𝑐 = Γ ��𝑥 ∧ Γ ��

�� + Γ �� ∧ Γ ��

𝑥 + Γ ��𝑥 ∧ Γ ��

��+ Γ ��

𝑥 ∧ Γ �� = 0

⇒ Ω ��𝑥 = 0

Ω ��𝑥 : 𝑑Γ ��

𝑥 = 0

Γ 𝑐𝑥 ∧ Γ ��

𝑐 = 0

⇒ Ω ��𝑥 = 0

Ω ��: 𝑑Γ ��

�� = 0

Γ 𝑐��∧ Γ ��

𝑐 = Γ ��∧ Γ ��

�� + Γ ��∧ Γ ��

𝑥 + Γ ��∧ Γ ��

��+ Γ ��

��∧ Γ ��

�� = 0

Ω ��

= 0

Ω �� : 𝑑Γ ��

�� = 𝑑(√2𝜔𝜔��) = 𝑑 (1

√2𝑒𝑥𝑑𝑥 ∧ 𝑑𝑧) =

1

√2𝑒𝑥(√2𝜔𝜔𝑥) ∧ (2𝜔𝑒−𝑥𝜔��)

= 2𝜔2𝜔�� ∧ 𝜔��

Γ 𝑐�� ∧ Γ ��

𝑐 = Γ �� ∧ Γ ��

�� + Γ �� ∧ Γ ��

𝑥 + Γ �� ∧ Γ ��

��+ Γ ��

�� ∧ Γ �� = Γ ��

�� ∧ Γ ��

= (−𝜔𝜔𝑥) ∧ (𝜔𝜔��) = −𝜔2𝜔�� ∧ 𝜔 ��

Ω �� = 𝜔2𝜔�� ∧ 𝜔��


Ω �� = {

0 0 0 00 0 0 𝜔2𝜔�� ∧ 𝜔��

0 0 0 00 −𝜔2𝜔𝑥 ∧ 𝜔�� 0 0

}


Now we can see that the nonzero elements of the Riemann tensor in the non-coordinate basis are 𝑅 ��𝑥�� =

−𝜔2

The Ricci tensor, the nonzero elements:





𝑅�� = 𝑅 ��𝑐��𝑐 (4.46)

𝑅𝑥�� = 𝑅 ��𝑐��𝑐 = 𝑅 ��𝑥

�� = 𝑅 ��𝑥��𝑥 = −𝜔2

𝑅�� = 𝑅 ��𝑐��𝑐 = 𝑅 ��𝑥��

𝑥 = −𝜔2


𝑅�� = {

0 0 0 00 −𝜔2 0 00 0 0 00 0 0 −𝜔2

}


The Ricci scalar:

𝑅 = 𝜂��𝑅�� (4.47)

𝑅 = 𝜂��𝑅�� + 𝜂��𝑅�� + 𝜂

��𝑅�� + 𝜂��𝑅�� = −𝑅�� − 𝑅�� = 2𝜔

2

Now we can find Λ

Λ = −1

4(8𝜋𝐺𝜌 + 𝑅)

= −

1

4(8𝜋𝐺𝜌 + 2𝜔2)

If we use geometrized units118 i.e. 8𝜋𝐺 = 1 we get

⇒ Λ = −1

4(𝜌 + 2𝜔2)

The first term –𝜌

4 represents the matter density of a homogeneous distribution of swirling dust particles,

and the second term Λ = −𝜔2

2 is associated with a nonzero cosmological constant.

8 Null Tetrads and the Petrov Classification

8.1 119Construct a null tetrad for the flat space Minkowski metric The line element: 𝑑𝑠2 = 𝑑𝑡2 − 𝑑𝑟2 − 𝑟2𝑑𝜃2 − 𝑟2 sin2 𝜃 𝑑𝜙2


1−1

−𝑟2

−𝑟2 sin2 𝜃

}


{

1

−1

−1

𝑟2

−1

𝑟2 sin2 𝜃}

118 http://en.wikipedia.org/wiki/Geometrized_unit_system 119 (McMahon, 2006, p. 186), example 9-3 and 9-4


http://en.wikipedia.org/wiki/Cosmological_constant

http://en.wikipedia.org/wiki/Geometrized_unit_system




The basis one forms

𝜔�� = 𝑑𝑡

𝜔�� = 𝑑𝑟

𝜔�� = 𝑟𝑑𝜃

𝜔�� = 𝑟 sin𝜃 𝑑𝜙

The null tetrad

Now we can use the basis one-forms to construct a null tetrad

(

𝑙𝑛𝑚��

) =1

√2(

1 1 0 01 −1 0 00 0 1 𝑖0 0 1 −𝑖

)

(

𝜔��

𝜔��

𝜔��

𝜔��)

=1

√2(

𝑑𝑡 + 𝑑𝑟𝑑𝑡 − 𝑑𝑟

𝑟𝑑𝜃 + 𝑖𝑟 sin 𝜃 𝑑𝜙𝑟𝑑𝜃 − 𝑖𝑟 sin 𝜃 𝑑𝜙

) (9.10)

Written in terms of the coordinate basis

𝑙𝑎 =1

√2(1, 1, 0, 0) 𝑛𝑎 =

1

√2(1, −1, 0, 0)

𝑚𝑎 =1

√2(0, 0, 𝑟, 𝑖𝑟 sin𝜃) 𝑚𝑎 =

1

√2(0, 0, 𝑟, −𝑖𝑟 sin𝜃)

Next we use the metric to rise the indices

𝑙𝑡 = 𝑔𝑎𝑡𝑙𝑎 = 𝑔𝑡𝑡𝑙𝑡 = 1 ⋅

1

√2=1

√2

𝑙𝑟 = 𝑔𝑎𝑟𝑙𝑎 = 𝑔𝑟𝑟𝑙𝑟 = (−1) ⋅

1

√2= −

1

√2

𝑙𝜃 = 𝑙𝜙 = 0

𝑛𝑡 = 𝑔𝑎𝑡𝑛𝑎 = 𝑔𝑡𝑡𝑛𝑡 = 1 ⋅

1

√2=1

√2

𝑛𝑟 = 𝑔𝑎𝑟𝑛𝑎 = 𝑔𝑟𝑟𝑛𝑟 = (−1) ⋅ (−

1

√2) =

1

√2

𝑛𝜃 = 𝑛𝜙 = 0

𝑚𝑡 = 𝑚𝑟 = 0

𝑚𝜃 = 𝑔𝑎𝜃𝑚𝑎 = 𝑔𝜃𝜃𝑚𝜃 = (−

1

𝑟2) ⋅

𝑟

√2= −

1

𝑟√2

𝑚𝜙 = 𝑔𝑎𝜙𝑚𝑎 = 𝑔𝜙𝜙𝑚𝜙 = (−

1

𝑟2 sin2 𝜃) ⋅𝑖𝑟 sin 𝜃

√2= −𝑖

1

𝑟 sin𝜃 √2


𝑙𝑎 =1

√2(1, 1, 0, 0) 𝑙𝑎 =

1

√2(1, −1, 0, 0)

𝑛𝑎 =1

√2(1, −1, 0, 0) 𝑛𝑎 =

1

√2(1, 1, 0, 0)

𝑚𝑎 =1

√2(0, 0, 𝑟, 𝑖𝑟 sin𝜃) 𝑚𝑎 =

1

√2(0, 0, −

1

𝑟, −

𝑖

𝑟 sin𝜃)





𝑚𝑎 =1

√2(0, 0, 𝑟, −𝑖𝑟 sin𝜃) 𝑚𝑎 =

1

√2(0, 0, −

1

𝑟,

𝑖

𝑟 sin𝜃)

8.2 120The Brinkmann metric (Plane gravitational waves) The line element: 𝑑𝑠2 = 𝐻(𝑢, 𝑥, 𝑦)𝑑𝑢2 + 2𝑑𝑢𝑑𝑣 − 𝑑𝑥2 − 𝑑𝑦2


𝐻 11

−1−1

}

and its inverse: 𝑔𝑎𝑏 = {

11 −𝐻

−1−1

}

The basis one forms

Finding the basis one forms is not so obvious, we write:

𝑑𝑠2 = 𝐻(𝑢, 𝑥, 𝑦)𝑑𝑢2 + 2𝑑𝑢𝑑𝑣 − 𝑑𝑥2 − 𝑑𝑦2 = (𝜔��)2− (𝜔��)

2− (𝜔��)

2− (𝜔 ��)

2

⇒ 𝑑𝑢[𝐻(𝑢, 𝑥, 𝑦)𝑑𝑢 + 2𝑑𝑣] − 𝑑𝑥2 − 𝑑𝑦2 = (𝜔�� +𝜔��)(𝜔�� −𝜔��) − (𝜔��)2− (𝜔��)

2

⇒ 𝜔�� +𝜔�� = 𝑑𝑢 𝜔�� −𝜔�� = 𝐻𝑑𝑢 + 2𝑑𝑣 𝜔𝑥 = 𝑑𝑥 𝜔�� = 𝑑𝑦

⇒ 𝜔�� =1

2(𝐻 + 1)𝑑𝑢 + 𝑑𝑣 𝑑𝑢 = 𝜔�� +𝜔��

𝜂𝑖𝑗 = {

1−1

−1−1

} 𝜔�� =1

2(1 − 𝐻)𝑑𝑢 − 𝑑𝑣 𝑑𝑣 =

1

2(1 − 𝐻)𝜔�� −

1

2(1 + 𝐻)𝜔��

𝜔�� = 𝑑𝑥 𝑑𝑥 = 𝜔𝑥 𝜔�� = 𝑑𝑦 𝑑𝑦 = 𝜔��

The orthonormal null tetrad

Now we can use the basis one-forms to construct a orthonormal null tetrad

(

𝑙𝑛𝑚��

) =1

√2(

1 1 0 01 −1 0 00 0 1 𝑖0 0 1 −𝑖

)(

𝜔��

𝜔��

𝜔��

𝜔��

) =1

√2(

𝜔�� +𝜔��

𝜔�� −𝜔��

𝜔�� + 𝑖𝜔��

𝜔�� − 𝑖𝜔��

) =1

√2(

𝑑𝑢𝐻𝑑𝑢 + 2𝑑𝑣𝑑𝑥 + 𝑖𝑑𝑦𝑑𝑥 − 𝑖𝑑𝑦

) (9.10)


𝑙𝑎 =1

√2(1, 0, 0, 0) 𝑛𝑎 =

1

√2(𝐻, 2, 0, 0)

𝑚𝑎 =1

√2(0, 0, 1, 𝑖) 𝑚𝑎 =

1

√2(0, 0, 1, −𝑖)

120 (McMahon, 2006, p. 195), example 9-5. The answer to quiz 9-3 is (a) and to quiz 9-4 is (d)





Next we use the metric to rise the indices 𝑙𝑢 = 𝑔𝑎𝑢𝑙𝑎 = 𝑔

𝑣𝑢𝑙𝑣 = 1 ⋅ 0 = 0

𝑙𝑣 = 𝑔𝑎𝑣𝑙𝑎 = 𝑔𝑢𝑣𝑙𝑢 + 𝑔

𝑣𝑣𝑙𝑣 = 1 ⋅ (1

√2) + (−𝐻) ⋅ 0 =

1

√2

𝑙𝑥 = 𝑙𝑦 = 0

𝑛𝑢 = 𝑔𝑎𝑢𝑛𝑎 = 𝑔𝑣𝑢𝑛𝑣 = 1 ⋅ (

2

√2) = √2

𝑛𝑣 = 𝑔𝑎𝑣𝑛𝑎 = 𝑔𝑢𝑣𝑛𝑢 + 𝑔

𝑣𝑣𝑛𝑣 = 1 ⋅ (1

√2𝐻) + (−𝐻) ⋅ (

2

√2) = −

1

√2𝐻

𝑛𝑥 = 𝑛𝑦 = 0

𝑚𝑢 = 𝑚𝑣 = 0

𝑚𝑥 = 𝑔𝑎𝑥𝑚𝑎 = 𝑔𝑥𝑥𝑚𝑥 = (−1) ⋅

1

√2= −

1

√2

𝑚𝑦 = 𝑔𝑎𝑦𝑚𝑎 = 𝑔𝑦𝑦𝑚𝑦 = (−1) ⋅ 𝑖

1

√2= −𝑖

1

√2


𝑙𝑎 =1

√2(1, 0, 0, 0) 𝑙𝑎 =

1

√2(0, 1, 0, 0)

𝑛𝑎 =1

√2(𝐻, 2, 0, 0) 𝑛𝑎 =

1

√2(2, −𝐻, 0, 0)

𝑚𝑎 =1

√2(0, 0, 1, 𝑖) 𝑚𝑎 =

1

√2(0, 0, −1, −𝑖)

𝑚𝑎 =1

√2(0, 0, 1, −𝑖) ��𝑎 =

1

√2(0, 0, −1, 𝑖)

The non-zero Christoffel symbols

Γ𝑎𝑏𝑐 =1



Γ𝑢𝑢𝑢 =1

2

𝜕𝐻

𝜕𝑢 ⇒ Γ 𝑢𝑢

𝑣 = 𝑔𝑣𝑢Γ𝑢𝑢𝑢 =1

2

𝜕𝐻

𝜕𝑢

Γ𝑢𝑢𝑥 = Γ𝑢𝑥𝑢 =1

2

𝜕𝐻

𝜕𝑥 ⇒ Γ 𝑢𝑥

𝑣 = Γ 𝑥𝑢𝑣 = 𝑔𝑣𝑢Γ𝑢𝑢𝑥 =

1

2

𝜕𝐻

𝜕𝑥

Γ𝑥𝑢𝑢 = −1

2

𝜕𝐻

𝜕𝑥 ⇒ Γ 𝑢𝑢

𝑥 = 𝑔𝑥𝑥Γ𝑥𝑢𝑢 =1

2

𝜕𝐻

𝜕𝑥

Γ𝑢𝑢𝑦 = Γ𝑢𝑦𝑢 =1

2

𝜕𝐻

𝜕𝑦 ⇒ Γ 𝑢𝑦

𝑣 = Γ 𝑦𝑢𝑣 = 𝑔𝑣𝑢Γ𝑢𝑢𝑦 =

1

2

𝜕𝐻

𝜕𝑦

Γ𝑦𝑢𝑢 = −1

2

𝜕𝐻

𝜕𝑦 ⇒ Γ 𝑢𝑢

𝑦 = 𝑔𝑦𝑦Γ𝑦𝑢𝑢 =

1

2

𝜕𝐻

𝜕𝑦

The spin coefficients calculated from the null tetrad

𝜋 = −∇𝑏𝑛𝑎��𝑎𝑙𝑏 𝜅 = ∇𝑏𝑙𝑎𝑚

𝑎𝑙𝑏 휀 =1

2(∇𝑏𝑙𝑎𝑛

𝑎𝑙𝑏 − ∇𝑏𝑚𝑎��𝑎𝑙𝑏) (9.15)





𝜈 = −∇𝑏𝑛𝑎��𝑎𝑛𝑏 𝜏 = ∇𝑏𝑙𝑎𝑚

𝑎𝑛𝑏 𝛾 =1


𝑎𝑛𝑏 − ∇𝑏𝑚𝑎��𝑎𝑛𝑏)

𝜆 = −∇𝑏𝑛𝑎��𝑎��𝑏 𝜌 = ∇𝑏𝑙𝑎𝑚

𝑎��𝑏 𝛼 =1


𝑎��𝑏 − ∇𝑏𝑚𝑎��𝑎��𝑏)

𝜇 = −∇𝑏𝑛𝑎��𝑎𝑚𝑏 𝜎 = ∇𝑏𝑙𝑎𝑚

𝑎𝑚𝑏 𝛽 =1


𝑎𝑚𝑏 − ∇𝑏𝑚𝑎��𝑎𝑚𝑏)

Calculating the spin-coefficients

𝜋 = −∇𝑏𝑛𝑎��𝑎𝑙𝑏 = −∇𝑣𝑛𝑎��

𝑎𝑙𝑣 = −∇𝑣𝑛𝑥��𝑥𝑙𝑣 − ∇𝑣𝑛𝑦��

𝑦𝑙𝑣

= −(𝜕𝑣𝑛𝑥 − Γ 𝑥𝑣𝑐 𝑛𝑐)��

𝑥𝑙𝑣 − (𝜕𝑣𝑛𝑦 − Γ 𝑦𝑣𝑐 𝑛𝑐)��

𝑦𝑙𝑣 = 0

𝜈 = −∇𝑏𝑛𝑎��𝑎𝑛𝑏 = −∇𝑢𝑛𝑎��

𝑎𝑛𝑢 − ∇𝑣𝑛𝑎��𝑎𝑛𝑣

= −∇𝑢𝑛𝑥��𝑥𝑛𝑢 − ∇𝑢𝑛𝑦��

𝑦𝑛𝑢 − ∇𝑣𝑛𝑥��𝑥𝑛𝑣 − ∇𝑣𝑛𝑦��

𝑦𝑛𝑣

= −(𝜕𝑢𝑛𝑥 − Γ 𝑥𝑢𝑐 𝑛𝑐)��

𝑥𝑛𝑢 − (𝜕𝑢𝑛𝑦 − Γ 𝑦𝑢𝑐 𝑛𝑐)��

𝑦𝑛𝑢 − (𝜕𝑣𝑛𝑥 − Γ 𝑥𝑣𝑐 𝑛𝑐)��

𝑥𝑛𝑣 − (𝜕𝑣𝑛𝑦 − Γ 𝑦𝑣𝑐 𝑛𝑐)��

𝑦𝑛𝑣

= Γ 𝑥𝑢𝑣 𝑛𝑣��

𝑥𝑛𝑢 + Γ 𝑦𝑢𝑣 𝑛𝑣��

𝑦𝑛𝑢

= (Γ 𝑥𝑢

𝑣 ��𝑥 + Γ 𝑦𝑢𝑣 ��𝑦)𝑛𝑣𝑛

𝑢 = (1

2

𝜕𝐻

𝜕𝑥(−1

√2) +

1

2

𝜕𝐻

𝜕𝑦𝑖1

√2)√2 ⋅ √2 =

1

√2(−

𝜕𝐻

𝜕𝑥+ 𝑖

𝜕𝐻

𝜕𝑦)

𝜆 = −∇𝑏𝑛𝑎��𝑎��𝑏 = −∇𝑥𝑛𝑎��

𝑎��𝑥 − ∇𝑦𝑛𝑎��𝑎��𝑦

= −∇𝑥𝑛𝑥��𝑥��𝑥 − ∇𝑥𝑛𝑦��

𝑦��𝑥 − ∇𝑦𝑛𝑥��𝑥��𝑦 − ∇𝑦𝑛𝑦��

𝑦��𝑦

= −(𝜕𝑥𝑛𝑥 − Γ 𝑥𝑥𝑐 𝑛𝑐)��

𝑥��𝑥 − (𝜕𝑥𝑛𝑦 − Γ 𝑥𝑦𝑐 𝑛𝑐)��

𝑦��𝑥 − (𝜕𝑦𝑛𝑥 − Γ 𝑦𝑥𝑐 𝑛𝑐)��

𝑥��𝑦

− (𝜕𝑦𝑛𝑦 − Γ 𝑦𝑦𝑐 𝑛𝑐)��

𝑦��𝑦

= 0 𝜇 = −∇𝑏𝑛𝑎��

𝑎𝑚𝑏 = 0

𝜅 = ∇𝑏𝑙𝑎𝑚𝑎𝑙𝑏 = ∇𝑣𝑙𝑎𝑚

𝑎𝑙𝑣 = ∇𝑣𝑙𝑥𝑚𝑥𝑙𝑣 + ∇𝑣𝑙𝑦𝑚

𝑦𝑙𝑣 = (∂𝑣𝑙𝑥 − Γ 𝑣𝑥𝑐 𝑙𝑐)𝑚

𝑥𝑙𝑣 + (∂𝑣𝑙𝑦 − Γ 𝑣𝑦𝑐 𝑙𝑐)𝑚

𝑦𝑙𝑣

= 0

𝜏 = ∇𝑏𝑙𝑎𝑚𝑎𝑛𝑏 = ∇𝑢𝑙𝑎𝑚

𝑎𝑛𝑢 + ∇𝑣𝑙𝑎𝑚𝑎𝑛𝑣 = ∇𝑢𝑙𝑥𝑚

𝑥𝑛𝑢 + ∇𝑣𝑙𝑥𝑚𝑥𝑛𝑣 + ∇𝑢𝑙𝑦𝑚

𝑦𝑛𝑢 + ∇𝑣𝑙𝑦𝑚𝑦𝑛𝑣

= (𝜕𝑢𝑙𝑥 − Γ 𝑢𝑥𝑐 𝑙𝑐)𝑚

𝑥𝑛𝑢 + (𝜕𝑣𝑙𝑥 − Γ 𝑣𝑥𝑐 𝑙𝑐)𝑚

𝑥𝑛𝑣 + (𝜕𝑢𝑙𝑦 − Γ 𝑢𝑦𝑐 𝑙𝑐)𝑚

𝑦𝑛𝑢

+ (𝜕𝑣𝑙𝑦 − Γ 𝑣𝑦𝑐 𝑙𝑐)∇𝑣𝑙𝑦𝑚

𝑦𝑛𝑣

= 0

𝜌 = ∇𝑏𝑙𝑎𝑚𝑎��𝑏 = ∇𝑥𝑙𝑎𝑚

𝑎��𝑥 + ∇𝑦𝑙𝑎𝑚𝑎��𝑦 = ∇𝑥𝑙𝑥𝑚

𝑥��𝑥 + ∇𝑦𝑙𝑥𝑚𝑥��𝑦 + ∇𝑥𝑙𝑦𝑚

𝑦��𝑥 + ∇𝑦𝑙𝑦𝑚𝑦��𝑦

= (𝜕𝑥𝑙𝑥 − Γ 𝑥𝑥𝑐 𝑙𝑐)𝑚

𝑥��𝑥 + (𝜕𝑦𝑙𝑥 − Γ 𝑦𝑥𝑐 𝑙𝑐)𝑚

𝑥��𝑦 + (𝜕𝑥𝑙𝑦 − Γ 𝑥𝑦𝑐 𝑙𝑐)𝑚

𝑦��𝑥 + (𝜕𝑦𝑙𝑦 − Γ 𝑦𝑦𝑐 𝑙𝑐)𝑚

𝑦��𝑦

= 0 𝜎 = ∇𝑏𝑙𝑎𝑚

𝑎𝑚𝑏 = 0

휀 =1


𝑎𝑙𝑏 − ∇𝑏𝑚𝑎��𝑎𝑙𝑏) =

1

2(∇𝑣𝑙𝑎𝑛

𝑎𝑙𝑣 − ∇𝑣𝑚𝑎��𝑎𝑙𝑣)

=1

2(∇𝑣𝑙𝑢𝑛

𝑢𝑙𝑣 − ∇𝑣𝑚𝑥��𝑥𝑙𝑣) +

1

2(∇𝑣𝑙𝑣𝑛

𝑣𝑙𝑣 − ∇𝑣𝑚𝑦��𝑦𝑙𝑣)

=1

2((𝜕𝑣𝑙𝑢 − Γ 𝑣𝑢

𝑐 𝑙𝑐)𝑛𝑢𝑙𝑣 − (𝜕𝑣𝑚𝑥 − Γ 𝑣𝑥

𝑐 𝑚𝑐)��𝑥𝑙𝑣)

+1

2((𝜕𝑣𝑙𝑣 − Γ 𝑣𝑣

𝑐 𝑙𝑐)𝑛𝑣𝑙𝑣 − (𝜕𝑣𝑚𝑦 − Γ 𝑣𝑦

𝑐 𝑚𝑐)��𝑦𝑙𝑣)

= 0

𝛾 =1


𝑎𝑛𝑏 − ∇𝑏𝑚𝑎��𝑎𝑛𝑏) =

1

2(∇𝑢𝑙𝑎𝑛

𝑎𝑛𝑢 − ∇𝑢𝑚𝑎��𝑎𝑛𝑢) +

1

2(∇𝑣𝑙𝑎𝑛

𝑎𝑛𝑣 − ∇𝑣𝑚𝑎��𝑎𝑛𝑣)

=1

2(∇𝑢𝑙𝑢𝑛

𝑢𝑛𝑢 − ∇𝑢𝑚𝑥��𝑥𝑛𝑢) +

1

2(∇𝑣𝑙𝑢𝑛

𝑢𝑛𝑣 − ∇𝑣𝑚𝑥��𝑥𝑛𝑣) +

1

2(∇𝑢𝑙𝑣𝑛

𝑣𝑛𝑢 − ∇𝑢𝑚𝑦��𝑦𝑛𝑢)

+1

2(∇𝑣𝑙𝑣𝑛

𝑣𝑛𝑣 − ∇𝑣𝑚𝑦��𝑦𝑛𝑣)

=1

2((𝜕𝑢𝑙𝑢 − Γ 𝑢𝑢

𝑐 𝑙𝑐)𝑛𝑢𝑛𝑢 − (𝜕𝑢𝑚𝑥 − Γ 𝑢𝑥

𝑐 𝑚𝑐)��𝑥𝑛𝑢)





+1

2((𝜕𝑣𝑙𝑢 − Γ 𝑣𝑢

𝑐 𝑙𝑐)𝑛𝑢𝑛𝑣 − (𝜕𝑣𝑚𝑥 − Γ 𝑣𝑥

𝑐 𝑚𝑐)��𝑥𝑛𝑣)

+1

2((𝜕𝑢𝑙𝑣 − Γ 𝑢𝑣

𝑐 𝑙𝑐)𝑛𝑣𝑛𝑢 − (𝜕𝑢𝑚𝑦 − Γ 𝑢𝑦

𝑐 𝑚𝑐)��𝑦𝑛𝑢)

+1

2((𝜕𝑣𝑙𝑣 − Γ 𝑣𝑣

𝑐 𝑙𝑐)𝑛𝑣𝑛𝑣 − (𝜕𝑣𝑚𝑦 − Γ 𝑣𝑦

𝑐 𝑚𝑐)��𝑦𝑛𝑣)

= 0

𝛼 =1


𝑎��𝑏 − ∇𝑏𝑚𝑎��𝑎��𝑏) =

1

2(∇𝑥𝑙𝑎𝑛

𝑎��𝑥 − ∇𝑥𝑚𝑎��𝑎��𝑥) +

1

2(∇𝑦𝑙𝑎𝑛

𝑎��𝑦 − ∇𝑦𝑚𝑎��𝑎��𝑦)

=1

2(∇𝑥𝑙𝑢𝑛

𝑢��𝑥 − ∇𝑥𝑚𝑥��𝑥��𝑥) +

1

2(∇𝑦𝑙𝑢𝑛

𝑢��𝑦 − ∇𝑦𝑚𝑥��𝑥��𝑦) +

1

2(∇𝑥𝑙𝑣𝑛

𝑣��𝑥 − ∇𝑥𝑚𝑦��𝑦��𝑥)

+1

2(∇𝑦𝑙𝑣𝑛

𝑣��𝑦 − ∇𝑦𝑚𝑦��𝑦��𝑦)

=1

2((𝜕𝑥𝑙𝑢 − Γ 𝑥𝑢

𝑐 𝑙𝑐)𝑛𝑢��𝑥 − (𝜕𝑥𝑚𝑥 − Γ 𝑥𝑥

𝑐 𝑚𝑐)��𝑥��𝑥)

+1

2((𝜕𝑦𝑙𝑢 − Γ 𝑦𝑢

𝑐 𝑙𝑐)𝑛𝑢��𝑦 − (𝜕𝑦𝑚𝑥 − Γ 𝑦𝑥

𝑐 𝑚𝑐)��𝑥��𝑦)

+1

2((𝜕𝑥𝑙𝑣 − Γ 𝑥𝑣

𝑐 𝑙𝑐)𝑛𝑣��𝑥 − (𝜕𝑥𝑚𝑦 − Γ 𝑥𝑦

𝑐 𝑚𝑐)��𝑦��𝑥)

+1

2((𝜕𝑦𝑙𝑣 − Γ 𝑦𝑣

𝑐 𝑙𝑐)𝑛𝑣��𝑦 − (𝜕𝑦𝑚𝑦 − Γ 𝑦𝑦

𝑐 𝑚𝑐)��𝑦��𝑦)

= 0

𝛽 =1


𝑎𝑚𝑏 − ∇𝑏𝑚𝑎��𝑎𝑚𝑏) = 0

Collecting the results 𝜋 = 0 𝜅 = 0 휀 = 0

𝜈 =1

√2(−

𝜕𝐻

𝜕𝑥+ 𝑖

𝜕𝐻

𝜕𝑦) 𝜏 = 0 𝛾 = 0

𝜆 = 0 𝜌 = 0 𝛼 = 0 𝜇 = 0 𝜎 = 0 𝛽 = 0

The Weyl Scalars and Petrov classification

Ψ0 = 𝐷𝜎 − 𝛿𝜅 − 𝜎(𝜌 + ��) − 𝜎(3휀 − 휀) + 𝜅(𝜋 − �� + �� + 3𝛽) (13.22) Ψ1 = 𝐷𝛽 − 𝛿휀 − 𝜎(𝛼 + 𝜋) − 𝛽(�� − 휀) + 𝜅(𝜇 + 𝛾) + 휀(�� − ��) (13.23) Ψ2 = 𝛿𝜏 − Δ𝜌 − 𝜌�� − 𝜎𝜆 + 𝜏(�� − 𝛼 − ��) + 𝜌(𝛾 + ��) + 𝜅𝜈 − 2Λ (13.24)

Ψ3 = 𝛿𝛾 − Δα + 𝜈(𝜌 + 휀) − 𝜆(𝜏 + 𝛽) + 𝛼(�� − ��) + 𝛾(�� − ��) (13.25)

Ψ4 = 𝛿𝜈 − Δλ + 𝜆(𝜇 + ��) − 𝜆(3𝛾 − ��) + 𝜈(3𝛼 + �� + 𝜋 − ��) (13.26)

Where 𝐷 = 𝑙𝑎∇𝑎 Δ = 𝑛𝑎∇𝑎 𝛿 = 𝑚𝑎∇𝑎 𝛿 = ��𝑎∇𝑎 (9.13)

We see that Ψ0 = Ψ1 = Ψ2 = Ψ3 = 0 and

Ψ4 = 𝛿𝜈 = ��𝑎∇𝑎𝜈 = ��𝑎 ∂𝑎 (

1

√2(−

𝜕𝐻

𝜕𝑥+ 𝑖

𝜕𝐻

𝜕𝑦))

=1

√2[��𝑥 ∂x (−

𝜕𝐻

𝜕𝑥+ 𝑖

𝜕𝐻

𝜕𝑦) + ��𝑦𝜕y (−

𝜕𝐻

𝜕𝑥+ 𝑖

𝜕𝐻

𝜕𝑦)]





=1

√2[(−

1

√2) (−

𝜕2𝐻

𝜕𝑥2+ 𝑖

𝜕2𝐻

𝜕𝑥𝜕𝑦) + (𝑖

1

√2)(−

𝜕2𝐻

𝜕𝑥𝜕𝑦+ 𝑖

𝜕2𝐻

𝜕𝑦2)] =

1

2[𝜕2𝐻

𝜕𝑥2−𝜕2𝐻

𝜕𝑦2− 2𝑖

𝜕2𝐻

𝜕𝑥𝜕𝑦]

Ψ4 ≠ 0: This is a Petrov type N, which means there is a single principal null direction of multiplicity 4. This corresponds to transverse gravity waves.

The Ricci tensor

Φ22 = 𝛿𝜈 − Δ𝜇 − 𝜇2 − 𝜆�� − 𝜇(𝛾 + ��) + ��𝜋 − 𝜈(𝜏 − 3𝛽 − ��) = 𝛿𝜈 = 𝑚𝑎∇𝑎𝜈 (9.24)

= 𝑚𝑎𝜕𝑎 (1

√2(−

𝜕𝐻

𝜕𝑥+ 𝑖

𝜕𝐻

𝜕𝑦))

= 𝑚𝑥𝜕𝑥 (1

√2(−

𝜕𝐻

𝜕𝑥+ 𝑖

𝜕𝐻

𝜕𝑦)) +𝑚𝑦𝜕𝑦 (

1

√2(−

𝜕𝐻

𝜕𝑥+ 𝑖

𝜕𝐻

𝜕𝑦))

= (−1

√2) (

1

√2)(−

𝜕2𝐻

𝜕𝑥2+ 𝑖

𝜕2𝐻

𝜕𝑥𝜕𝑦) + (−𝑖

1

√2) (

1

√2) (−

𝜕2𝐻


𝜕2𝐻

𝜕𝑦2)

=1

2[𝜕2𝐻

𝜕𝑥2+𝜕2𝐻

𝜕𝑦2]

Φ22 = −1

2𝑅𝑎𝑏𝑛

𝑎𝑛𝑏 = −1

2𝑅𝑢𝑏𝑛

𝑢𝑛𝑏 −1

2𝑅𝑣𝑏𝑛

𝑣𝑛𝑏 (9.22)

= −1

2𝑅𝑢𝑢𝑛

𝑢𝑛𝑢 −1

2𝑅𝑣𝑢𝑛

𝑣𝑛𝑢 −1

2𝑅𝑢𝑣𝑛

𝑢𝑛𝑣 −1

2𝑅𝑣𝑣𝑛

𝑣𝑛𝑣

= −1

2𝑅𝑢𝑢√2 ⋅ √2 − 𝑅𝑢𝑣√2 ⋅ (−

1

√2𝐻) −

1

2𝑅𝑣𝑣 (−

1

√2𝐻) ⋅ (−

1

√2𝐻)

= −𝑅𝑢𝑢 +𝐻𝑅𝑢𝑣 −𝐻2

4𝑅𝑣𝑣 =

1

2[𝜕2𝐻

𝜕𝑥2+𝜕2𝐻

𝜕𝑦2]

⇒ 𝑅121𝑢𝑢= −

1

2[𝜕2𝐻

𝜕𝑥2+𝜕2𝐻

𝜕𝑦2]

⇒ 𝑅𝑢𝑣 = 𝑅𝑣𝑣 = 0

The Weyl tensor calculated from the null tetrad found in example 9-5. This calculation show that the spin coefficients and the Weyl scalars depend on the chosen null tetrad, and the Ricci tensor does not (of course).

The null tetrad (9.26), (9.27), (9.28) 𝑙𝑎 = (1, 0, 0, 0) 𝑙𝑎 = (0, 1, 0, 0)

𝑛𝑎 = (1

2𝐻, 1, 0, 0) 𝑛𝑎 = (1, −

1

2𝐻, 0, 0)

𝑚𝑎 =1

√2(0, 0, 1, −𝑖) 𝑚𝑎 =

1

√2(0, 0, −1, 𝑖)

𝑚𝑎 =1

√2(0, 0, 1, 𝑖) ��𝑎 =

1

√2(0, 0, −1, −𝑖)

121 According to the Cartan calculation further below the sign is wrong






𝑎𝑛𝑢 − ∇𝑣𝑛𝑎��𝑎𝑛𝑣

= −∇𝑢𝑛𝑥��𝑥𝑛𝑢 − ∇𝑢𝑛𝑦��

𝑦𝑛𝑢 − ∇𝑣𝑛𝑥��𝑥𝑛𝑣 − ∇𝑣𝑛𝑦��

𝑦𝑛𝑣 = −(𝜕𝑢𝑛𝑥 − Γ 𝑥𝑢

𝑐 𝑛𝑐)��𝑥𝑛𝑢 − (𝜕𝑢𝑛𝑦 − Γ 𝑦𝑢

𝑐 𝑛𝑐)��𝑦𝑛𝑢 − (𝜕𝑣𝑛𝑥 − Γ 𝑥𝑣

𝑐 𝑛𝑐)��𝑥𝑛𝑣

− (𝜕𝑣𝑛𝑦 − Γ 𝑦𝑣𝑐 𝑛𝑐)��

𝑦𝑛𝑣

= Γ 𝑥𝑢𝑣 𝑛𝑣��

𝑥𝑛𝑢 + Γ 𝑦𝑢𝑣 𝑛𝑣��

𝑦𝑛𝑢 = (Γ 𝑥𝑢𝑣 ��𝑥 + Γ 𝑦𝑢

𝑣 ��𝑦)𝑛𝑣𝑛𝑢

= (

1

2

𝜕𝐻

𝜕𝑥(−1

√2) +

1

2

𝜕𝐻

𝜕𝑦(−𝑖

1

√2))1 ⋅ 1 = −

1

2√2(𝜕𝐻

𝜕𝑥+ 𝑖

𝜕𝐻

𝜕𝑦) (9.30)

Ψ4 = 𝛿𝜈 = ��𝑎∇𝑎𝜈 = ��𝑎 ∂𝑎 (−

1

2√2(𝜕𝐻

𝜕𝑥+ 𝑖

𝜕𝐻

𝜕𝑦))

= (−1

2√2) [��𝑥 ∂x (

𝜕𝐻

𝜕𝑥+ 𝑖

𝜕𝐻

𝜕𝑦) + ��𝑦𝜕y (

𝜕𝐻

𝜕𝑥+ 𝑖

𝜕𝐻

𝜕𝑦)]

= (−

1

2√2) [(−

1

√2)(𝜕2𝐻

𝜕𝑥2+ 𝑖

𝜕2𝐻

𝜕𝑥𝜕𝑦) + (−𝑖

1

√2) (

𝜕2𝐻


𝜕2𝐻

𝜕𝑦2)]

=1

4[𝜕2𝐻

𝜕𝑥2−𝜕2𝐻

𝜕𝑦2+ 2𝑖

𝜕2𝐻

𝜕𝑥𝜕𝑦] (9.31)

Φ22 = 𝛿𝜈 = 𝑚𝑎∇𝑎𝜈 = 𝑚𝑎𝜕𝑎 (−

1

2√2(𝜕𝐻

𝜕𝑥+ 𝑖

𝜕𝐻

𝜕𝑦)) (9.24)

= 𝑚𝑥𝜕𝑥 (−1

2√2(𝜕𝐻

𝜕𝑥+ 𝑖

𝜕𝐻

𝜕𝑦)) +𝑚𝑦𝜕𝑦 (−

1

2√2(𝜕𝐻

𝜕𝑥+ 𝑖

𝜕𝐻

𝜕𝑦))

= (−1

2√2) [(

−1

√2)(𝜕2𝐻

𝜕𝑥2+ 𝑖

𝜕2𝐻

𝜕𝑥𝜕𝑦) + (𝑖

1

√2) (

𝜕2𝐻


𝜕2𝐻

𝜕𝑦2)] =

1

4[𝜕2𝐻

𝜕𝑥2+𝜕2𝐻

𝜕𝑦2] (9.32)

Φ22 = −1

2𝑅𝑎𝑏𝑛

𝑎𝑛𝑏 (9.22) 𝑛𝑎 = (1, −1

2𝐻, 0, 0)

= −1

2𝑅𝑢𝑢𝑛

𝑢𝑛𝑢 −1

2𝑅𝑣𝑢𝑛

𝑣𝑛𝑢 −1

2𝑅𝑢𝑣𝑛

𝑢𝑛𝑣 −1

2𝑅𝑣𝑣𝑛

𝑣𝑛𝑣

= −1

2𝑅𝑢𝑢1 ⋅ 1 − 𝑅𝑣𝑢1(−

1

2𝐻) −

1

2𝑅𝑣𝑣 (−

1

2𝐻)

2

= −1

2𝑅𝑢𝑢 +

𝐻

2𝑅𝑣𝑢 −

𝐻2

8𝑅𝑣𝑣 =

1

4[𝜕2𝐻

𝜕𝑥2+𝜕2𝐻

𝜕𝑦2]

⇒ 𝑅122𝑢𝑢

= −1

2[𝜕2𝐻

𝜕𝑥2+𝜕2𝐻

𝜕𝑦2]

⇒ 𝑅𝑢𝑣 = 𝑅𝑣𝑣 = 0

Finding the Ricci tensor of the Brinkmann metric using Cartan’s structure equation

Cartan’s First Structure equation and the calculation of the curvature one-forms

122 According to the Cartan calculation further below the sign is wrong





𝑑𝜔�� = −Γ �� ∧ 𝜔�� (5.9) Γ ��

�� = Γ ��𝑐�� 𝜔𝑐 (5.10)

𝑑𝜔�� = 𝑑 (1

2(𝐻(𝑢, 𝑥, 𝑦) + 1)𝑑𝑢 + 𝑑𝑣) =

1

2(𝜕𝐻

𝜕𝑥𝑑𝑥 ∧ 𝑑𝑢 +

𝜕𝐻

𝜕𝑦𝑑𝑦 ∧ 𝑑𝑢)

=1

2(𝜕𝐻

𝜕𝑥𝜔�� ∧ (𝜔�� + 𝜔��) +

𝜕𝐻

𝜕𝑦𝜔�� ∧ (𝜔�� +𝜔��))

𝑑𝜔�� = 𝑑 (1

2(1 − 𝐻)𝑑𝑢 − 𝑑𝑣) = −

1

2(𝜕𝐻

𝜕𝑥𝑑𝑥 ∧ 𝑑𝑢 +

𝜕𝐻

𝜕𝑦𝑑𝑦 ∧ 𝑑𝑢)

= −1

2(𝜕𝐻

𝜕𝑥𝜔𝑥 ∧ (𝜔�� +𝜔��) +

𝜕𝐻

𝜕𝑦𝜔�� ∧ (𝜔�� +𝜔��))

𝑑𝜔𝑥 = 0

𝑑𝜔�� = 0

The curvature one-forms summarized in a matrix:

Γ �� =

{

0 0

1

2

𝜕𝐻

𝜕𝑥(𝜔�� +𝜔��)(𝐴)

1

2

𝜕𝐻

𝜕𝑦(𝜔�� +𝜔��)(𝐵)

0 01

2

𝜕𝐻

𝜕𝑥(𝜔�� +𝜔��)(𝐴)

1

2

𝜕𝐻

𝜕𝑦(𝜔�� +𝜔��)(𝐵)

1

2

𝜕𝐻

𝜕𝑥(𝜔�� +𝜔��)(𝐴) −

1

2

𝜕𝐻

𝜕𝑥(𝜔�� +𝜔��)(−𝐴) 0 0

1

2

𝜕𝐻

𝜕𝑦(𝜔�� +𝜔��)(𝐵) −

1

2

𝜕𝐻

𝜕𝑦(𝜔�� +𝜔��)(−𝐵) 0 0

}

Where �� refers to column and �� to row and A and B will be used later, to make the calculations easier


Ω �� = 𝑑Γ ��

�� + Γ 𝑐�� ∧ Γ ��

𝑐 =1

2𝑅 ��𝑐�� 𝜔𝑐 ∧ 𝜔�� (5.27), (5.28)

First we will calculate

𝐴 ∧ 𝐴 =1

2

𝜕𝐻

𝜕𝑥(𝜔�� +𝜔��) ∧

1

2

𝜕𝐻

𝜕𝑥(𝜔�� +𝜔��)

= (1

2

𝜕𝐻

𝜕𝑥)2

(𝜔�� ∧ 𝜔�� +𝜔�� ∧ 𝜔�� +𝜔�� ∧ 𝜔�� +𝜔�� ∧ 𝜔��)

= (1

2

𝜕𝐻

𝜕𝑥)2

(𝜔�� ∧ 𝜔�� +𝜔�� ∧ 𝜔��) = (1

2

𝜕𝐻

𝜕𝑥)2

(𝜔�� ∧ 𝜔�� −𝜔�� ∧ 𝜔��) = 0

𝐵 ∧ 𝐵 =1

2

𝜕𝐻

𝜕𝑦(𝜔�� +𝜔��) ∧

1

2

𝜕𝐻

𝜕𝑦(𝜔�� +𝜔��) = 0

𝑑𝐴 = 𝑑 (1

2

𝜕𝐻

𝜕𝑥(𝜔�� +𝜔��)) = 𝑑 (

1

2

𝜕𝐻(𝑢, 𝑥, 𝑦)

𝜕𝑥𝑑𝑢)





=1

2

𝜕2𝐻(𝑢, 𝑥, 𝑦)

𝜕𝑥2𝑑𝑥 ∧ 𝑑𝑢 +

1

2

𝜕2𝐻(𝑢, 𝑥, 𝑦)

𝜕𝑥𝜕𝑦𝑑𝑦 ∧ 𝑑𝑢

=1

2

𝜕2𝐻(𝑢, 𝑥, 𝑦)

𝜕𝑥2𝜔�� ∧ (𝜔�� +𝜔��) +

1

2

𝜕2𝐻(𝑢, 𝑥, 𝑦)

𝜕𝑥𝜕𝑦𝜔�� ∧ (𝜔�� + 𝜔��)

𝑑𝐵 = 𝑑 (1

2

𝜕𝐻

𝜕𝑦(𝜔�� +𝜔��)) = 𝑑 (

1

2

𝜕𝐻(𝑢, 𝑥, 𝑦)

𝜕𝑦𝑑𝑢) =

1

2

𝜕2𝐻

𝜕𝑥𝜕𝑦𝑑𝑥 ∧ 𝑑𝑢 +

1

2

𝜕2𝐻

𝜕𝑦2𝑑𝑦 ∧ 𝑑𝑢

=1

2

𝜕2𝐻

𝜕𝑥𝜕𝑦𝜔𝑥 ∧ (𝜔�� +𝜔��) +

1

2

𝜕2𝐻

𝜕𝑦2𝜔�� ∧ (𝜔�� +𝜔��)

Now we are ready to calculate the curvature two-forms

Ω �� : 𝑑Γ ��

�� = 0

Γ 𝑐�� ∧ Γ ��

𝑐 = Γ �� ∧ Γ ��

�� + Γ �� ∧ Γ ��

�� + Γ �� ∧ Γ ��

𝑥 + Γ �� ∧ Γ ��

��= Γ ��

�� ∧ Γ �� 𝑥 + Γ ��

�� ∧ Γ ��

= 𝐴 ∧ 𝐴 + 𝐵 ∧ 𝐵 = 0

Ω �� = 𝑑Γ ��

�� + Γ 𝑐�� ∧ Γ ��

𝑐 = 0

Ω �� : 𝑑Γ ��

�� = 0

Γ 𝑐�� ∧ Γ ��

𝑐 = Γ �� ∧ Γ ��

�� + Γ �� ∧ Γ ��

�� + Γ �� ∧ Γ ��

𝑥 + Γ �� ∧ Γ ��

��= Γ ��

�� ∧ Γ �� 𝑥 + Γ ��

�� ∧ Γ ��

= (−𝐴) ∧ (𝐴) + (−𝐵) ∧ (𝐵) = 0

⇒ Ω �� = 0

Ω ��𝑥 : 𝑑Γ ��

𝑥 = 𝑑𝐴 =1

2

𝜕2𝐻

𝜕𝑥2𝜔�� ∧ (𝜔�� +𝜔��) +

1

2

𝜕2𝐻

𝜕𝑥𝜕𝑦𝜔�� ∧ (𝜔�� +𝜔��)

Γ 𝑐𝑥 ∧ Γ ��

𝑐 = Γ ��𝑥 ∧ Γ ��

�� + Γ ��𝑥 ∧ Γ ��

�� + Γ �� ∧ Γ ��

𝑥 + Γ ��𝑥 ∧ Γ ��

��= 0

⇒ Ω ��𝑥 =

1

2

𝜕2𝐻

𝜕𝑥2𝜔𝑥 ∧ (𝜔�� +𝜔��) +

1

2

𝜕2𝐻

𝜕𝑥𝜕𝑦𝜔�� ∧ (𝜔�� +𝜔��)

Ω ��: 𝑑Γ ��

�� = 𝑑𝐵 =

1

2

𝜕2𝐻

𝜕𝑥𝜕𝑦𝜔𝑥 ∧ (𝜔�� +𝜔��) +

1

2

𝜕2𝐻

𝜕𝑦2𝜔�� ∧ (𝜔�� +𝜔��)

Γ 𝑐��∧ Γ ��

𝑐 = Γ ��∧ Γ ��

�� + Γ ��∧ Γ ��

�� + Γ ��∧ Γ ��

𝑥 + Γ ��∧ Γ ��

��= 0

⇒ Ω ��

=1

2

𝜕2𝐻

𝜕𝑥𝜕𝑦𝜔𝑥 ∧ (𝜔�� +𝜔��) +

1

2

𝜕2𝐻

𝜕𝑦2𝜔�� ∧ (𝜔�� +𝜔��)

Ω �� 𝑑Γ ��

�� = 0

Γ 𝑐�� ∧ Γ ��

𝑐 = Γ �� ∧ Γ ��

�� + Γ �� ∧ Γ ��

�� + Γ �� ∧ Γ ��

𝑥 + Γ �� ∧ Γ ��

��= Γ ��

�� ∧ Γ �� 𝑥 + Γ ��

�� ∧ Γ ��

= −(𝐴) ∧ (𝐴) − (𝐵) ∧ (𝐵) = 0

⇒ Ω �� = 0

Ω ��𝑥 : 𝑑Γ ��

�� = 𝑑𝐴 =1

2

𝜕2𝐻

𝜕𝑥2𝜔�� ∧ (𝜔�� +𝜔��) +

1

2

𝜕2𝐻

𝜕𝑥𝜕𝑦𝜔�� ∧ (𝜔�� +𝜔��)

Γ 𝑐𝑥 ∧ Γ ��

𝑐 = Γ ��𝑥 ∧ Γ ��

�� + Γ ��𝑥 ∧ Γ ��

�� + Γ �� ∧ Γ ��

𝑥 + Γ ��𝑥 ∧ Γ ��

��= 0

⇒ Ω ��𝑥 =

1

2

𝜕2𝐻

𝜕𝑥2𝜔𝑥 ∧ (𝜔�� +𝜔��) +

1

2

𝜕2𝐻

𝜕𝑥𝜕𝑦𝜔�� ∧ (𝜔�� +𝜔��)

Ω ��: 𝑑Γ ��

�� = 𝑑𝐵 =

1

2

𝜕2𝐻

𝜕𝑥𝜕𝑦𝜔𝑥 ∧ (𝜔�� +𝜔��) +

1

2

𝜕2𝐻

𝜕𝑦2𝜔�� ∧ (𝜔�� +𝜔��)

Γ 𝑐��∧ Γ ��

𝑐 = Γ ��∧ Γ ��

�� + Γ ��∧ Γ ��

�� + Γ ��∧ Γ ��

𝑥 + Γ ��∧ Γ ��

��= 0

⇒ Ω ��

=1

2

𝜕2𝐻

𝜕𝑥𝜕𝑦𝜔𝑥 ∧ (𝜔�� +𝜔��) +

1

2

𝜕2𝐻

𝜕𝑦2𝜔�� ∧ (𝜔�� +𝜔��)

Ω ��𝑥 : 𝑑Γ ��

�� = 0

Γ 𝑐𝑥 ∧ Γ ��

𝑐 = Γ ��𝑥 ∧ Γ ��

�� + Γ ��𝑥 ∧ Γ ��

�� + Γ ��𝑥 ∧ Γ ��

𝑥 + Γ ��𝑥 ∧ Γ ��

��= (𝐴) ∧ (𝐴) − (𝐴) ∧ (𝐴) = 0

⇒ Ω ��𝑥 = 0





Ω ��: 𝑑Γ ��

�� = 0

Γ 𝑐��∧ Γ ��

𝑐 = Γ ��∧ Γ ��

�� + Γ ��∧ Γ ��

�� + Γ ��∧ Γ ��

𝑥 + Γ ��∧ Γ ��

��= (𝐵) ∧ (𝐴) − (𝐵) ∧ (𝐴) = 0

⇒ Ω ��

= 0

⇒ Ω ��

= 0


Ω �� =

{

0 0

1

2

𝜕2𝐻

𝜕𝑥2𝜔𝑥 ∧ (𝜔𝑢 + 𝜔��) +

1

2

𝜕2𝐻

𝜕𝑥𝜕𝑦𝜔�� ∧ (𝜔𝑢 + 𝜔��)

1

2

𝜕2𝐻

𝜕𝑥𝜕𝑦𝜔𝑥 ∧ (𝜔𝑢 + 𝜔��) +

1

2

𝜕2𝐻

𝜕𝑦2𝜔�� ∧ (𝜔𝑢 + 𝜔��)

0 01

2

𝜕2𝐻

𝜕𝑥2𝜔𝑥 ∧ (𝜔𝑢 + 𝜔��) +

1

2

𝜕2𝐻

𝜕𝑥𝜕𝑦𝜔�� ∧ (𝜔𝑢 + 𝜔��)

1

2

𝜕2𝐻

𝜕𝑥𝜕𝑦𝜔𝑥 ∧ (𝜔𝑢 + 𝜔��) +

1

2

𝜕2𝐻

𝜕𝑦2𝜔�� ∧ (𝜔𝑢 + 𝜔��)

𝑆 𝐴𝑆 0 0𝑆 𝐴𝑆 0 0 }

Now we can write down the independent elements of the Riemann tensor in the non-coordinate basis:

R ��𝑥 (𝐶) =

1

2

𝜕2𝐻

𝜕𝑥2 R ��

�� (𝐸) =1

2

𝜕2𝐻

𝜕𝑦2

R ��𝑥 (𝐶) =

1

2

𝜕2𝐻

𝜕𝑥2 R ��

�� (𝐸) =1

2

𝜕2𝐻

𝜕𝑦2

R ��𝑥 (𝐷) =

1

2

𝜕2𝐻

𝜕𝑥𝜕𝑦 R ��

�� (𝐸) =1

2

𝜕2𝐻

𝜕𝑦2

R ��𝑥 (𝐷) =

1

2

𝜕2𝐻

𝜕𝑥𝜕𝑦

R ��𝑥 (𝐶) =

1

2

𝜕2𝐻

𝜕𝑥2

R ��𝑥 (𝐷) =

1

2

𝜕2𝐻

𝜕𝑥𝜕𝑦

R ��𝑥 (𝐷) =

1

2

𝜕2𝐻

𝜕𝑥𝜕𝑦

Where C,D and E will be used later, to make the calculations easier

The Ricci tensor:

𝑅�� = 𝑅 ��𝑐��𝑐 (4.46)

𝑅�� = 𝑅 ��𝑐��𝑐 = 𝑅 ��

�� + 𝑅 �� + 𝑅 ��

�� + 𝑅 ��

= 𝐶 + 𝐸 =1

2

𝜕2𝐻

𝜕𝑥2+1

2

𝜕2𝐻

𝜕𝑦2

𝑅�� = 𝑅 ��𝑐��𝑐 = 𝑅 ��

�� + 𝑅 �� + 𝑅 ��

�� + 𝑅 ��

= 𝐶 + 𝐸 =1

2

𝜕2𝐻

𝜕𝑥2+1

2

𝜕2𝐻

𝜕𝑦2

𝑅�� = 0 𝑅�� = 0

𝑅�� = 𝑅 ��𝑐��𝑐 = 𝑅 ��

�� + 𝑅 �� + 𝑅 ��

�� + 𝑅 ��

= 𝐶 + 𝐸 =1

2

𝜕2𝐻

𝜕𝑥2+1

2

𝜕2𝐻

𝜕𝑦2

𝑅𝑥�� = 0 𝑅�� = 0





𝑅𝑥�� = 𝑅 ��𝑐��𝑐 = 𝑅 ��

�� + 𝑅 �� + 𝑅 ��

�� + 𝑅 ��

= −𝐶 + 𝐶 = 0

𝑅��𝑥 = 𝑅 ��𝑐��𝑐 = 𝑅 ��

�� + 𝑅 �� + 𝑅 ��

�� + 𝑅 ��

= −𝐷 +𝐷 = 0

𝑅�� = 𝑅 ��𝑐��𝑐 = 𝑅 ��

�� + 𝑅 �� + 𝑅 ��

�� + 𝑅 ��

= −𝐸 + 𝐸 = 0


𝑅�� =

{

1

2

𝜕2𝐻

𝜕𝑥2+1

2

𝜕2𝐻

𝜕𝑦21

2

𝜕2𝐻

𝜕𝑥2+1

2

𝜕2𝐻

𝜕𝑦20 0

1

2

𝜕2𝐻

𝜕𝑥2+1

2

𝜕2𝐻

𝜕𝑦21

2

𝜕2𝐻

𝜕𝑥2+1

2

𝜕2𝐻

𝜕𝑦20 0

0 0 0 00 0 0 0}


The Ricci tensor in the coordinate basis:

The transformation: 𝑅𝑎𝑏 = Λ 𝑎𝑐 Λ 𝑏

�� 𝑅𝑐��

𝑅𝑢𝑢 = Λ 𝑢𝑐 Λ 𝑢

�� 𝑅𝑐�� = Λ 𝑢

�� Λ 𝑢�� 𝑅�� + Λ 𝑢

�� Λ 𝑢�� 𝑅��

= Λ 𝑢�� Λ 𝑢

�� 𝑅�� + Λ 𝑢�� Λ 𝑢

�� 𝑅�� + Λ 𝑢�� Λ 𝑢

�� 𝑅�� + Λ 𝑢�� Λ 𝑢

�� 𝑅��

= (1

2(𝐻 + 1))

2

(𝐶 + 𝐸) +1

2(𝐻 + 1)

1

2(1 − 𝐻)(𝐶 + 𝐸) +

1

2(1 − 𝐻)

1

2(𝐻 + 1)(𝐶 + 𝐸) + (

1

2(1 − 𝐻))

2

(𝐶 + 𝐸)

= (𝐶 + 𝐸)

=1

2

𝜕2𝐻

𝜕𝑥2+1

2

𝜕2𝐻

𝜕𝑦2

𝑅𝑢𝑣 = Λ 𝑢𝑐 Λ 𝑣

�� 𝑅𝑐�� = Λ 𝑢

�� Λ 𝑣�� 𝑅�� + Λ 𝑢

�� Λ 𝑣�� 𝑅��

= Λ 𝑢�� Λ 𝑣

�� 𝑅�� + Λ 𝑢�� Λ 𝑣

�� 𝑅�� + Λ 𝑢�� Λ 𝑣

�� 𝑅�� + Λ 𝑢�� Λ 𝑣

�� 𝑅��

=1

2(𝐻 + 1) ⋅ 1(𝐶 + 𝐸) +

1

2(𝐻 + 1)(−1)(𝐶 + 𝐸) +

1

2(1 − 𝐻) ⋅ 1(𝐶 + 𝐸)

+1

2(1 − 𝐻)(−1)(𝐶 + 𝐸)

= 0 𝑅𝑣𝑣 = Λ 𝑣

𝑐 Λ 𝑣�� 𝑅𝑐��

= Λ 𝑣�� Λ 𝑣

�� 𝑅�� + Λ 𝑣�� Λ 𝑣

�� 𝑅�� = Λ 𝑣

�� Λ 𝑣�� 𝑅�� + Λ 𝑣

�� Λ 𝑣�� 𝑅�� + Λ 𝑣

�� Λ 𝑣�� 𝑅�� + Λ 𝑣

�� Λ 𝑣�� 𝑅��

= 1 ⋅ 1(𝐶 + 𝐸) + 1(−1)(𝐶 + 𝐸) + (−1)1(𝐶 + 𝐸) + (−1)(−1)(𝐶 + 𝐸) = 0 𝑅𝑥𝑥 = 𝑅𝑦𝑦 = 𝑅𝑥𝑦 = 0

⇒ 𝑅 = 𝑔𝑎𝑏𝑅𝑎𝑏 = 0 ⋅ 𝑅𝑢𝑢 = 0 Summarized in a matrix:





𝑅𝑎𝑏 =

{

1

2

𝜕2𝐻

𝜕𝑥2+1

2

𝜕2𝐻

𝜕𝑦20 0 0

0 0 0 00 0 0 00 0 0 0}


⇒ 𝐺𝑢𝑢 =1

2

𝜕2𝐻

𝜕𝑥2+1

2

𝜕2𝐻

𝜕𝑦2

9 The Schwarzschild Solution

9.1 123The Riemann and Ricci tensor of the general Schwarzschild metric

The line element: 𝑑𝑠2 = 𝑒2𝜈(𝑟)𝑑𝑡2 − 𝑒2𝜆(𝑟)𝑑𝑟2 − 𝑟2𝑑𝜃2 − 𝑟2 sin2 𝜃 𝑑𝜙2

The Basis one forms

𝜔�� = 𝑒𝜈(𝑟)𝑑𝑡 𝑑𝑡 = 𝑒−𝜈(𝑟)𝜔��

𝜂𝑖𝑗 = {

1−1

−1−1

}

𝜔�� = 𝑒𝜆(𝑟)𝑑𝑟 𝑑𝑟 = 𝑒−𝜆(𝑟)𝜔��


𝑟𝜔��

𝜔�� = 𝑟 sin𝜃 𝑑𝜙 𝑑𝜙 =1



𝑑𝜔�� = −Γ �� ∧ 𝜔�� (5.9) Γ ��

�� = Γ ��𝑐�� 𝜔𝑐 (5.10)

𝑑𝜔�� = 𝑑(𝑒𝜈(𝑟)𝑑𝑡) = 𝜈′𝑒𝜈(𝑟)𝑑𝑟 ∧ 𝑑𝑡 = 𝜈′𝑒−𝜆(𝑟)𝜔�� ∧ 𝜔��

𝑑𝜔�� = 𝑑(𝑒𝜆(𝑟)𝑑𝑟) = 0

𝑑𝜔�� = 𝑑(𝑟𝑑𝜃) = 𝑑𝑟 ∧ 𝑑𝜃 =1

𝑟𝑒−𝜆(𝑟)𝜔�� ∧ 𝜔��


𝑟𝑒−𝜆(𝑟)𝜔�� ∧ 𝜔�� +

cot 𝜃

𝑟𝜔�� ∧ 𝜔��


123 (McMahon, 2006, p. 204)





Γ �� =

{

0 𝜈′𝑒−𝜆(𝑟)𝜔�� 0 0

𝜈′𝑒−𝜆(𝑟)𝜔�� 01

𝑟𝑒−𝜆(𝑟)𝜔��

1


0 −1

𝑟𝑒−𝜆(𝑟)𝜔�� 0

cot 𝜃

𝑟𝜔��

0 −1

𝑟𝑒−𝜆(𝑟)𝜔�� −

cot 𝜃

𝑟𝜔�� 0 }



Γ �� = 𝜈′𝑒−𝜆(𝑟) Γ ��

�� = 𝜈′𝑒−𝜆(𝑟) Γ �� =

1

𝑟𝑒−𝜆(𝑟) Γ

��

�� =

1

𝑟𝑒−𝜆(𝑟)

Γ �� = −

1

𝑟𝑒−𝜆(𝑟) Γ ��

�� = −cot 𝜃

𝑟 Γ

��

�� =

cot 𝜃

𝑟

Γ �� = −

1

𝑟𝑒−𝜆(𝑟)


Ω �� = 𝑑Γ ��

�� + Γ 𝑐�� ∧ Γ ��

𝑐 =1

2𝑅 ��𝑐�� 𝜔𝑐 ∧ 𝜔�� (5.27), (5.28)

Ω �� : 𝑑Γ ��

�� = 𝑑(𝜈′𝑒−𝜆(𝑟)𝜔��) = 𝑑(𝜈′𝑒𝜈(𝑟)−𝜆(𝑟)𝑑𝑡) = (𝜈′′ + 𝜈′(𝜈′ − 𝜆′))𝑒𝜈(𝑟)−𝜆(𝑟) 𝑑𝑟 ∧ 𝑑𝑡

= (𝜈′′ + 𝜈′(𝜈′ − 𝜆′))𝑒−2𝜆(𝑟)𝜔�� ∧ 𝜔��

Γ 𝑐�� ∧ Γ ��

𝑐 = Γ �� ∧ Γ ��

�� + Γ �� ∧ Γ ��

�� + Γ �� ∧ Γ ��

�� + Γ �� ∧ Γ

�� = 0

⇒ Ω �� = (𝜈′′ + 𝜈′(𝜈′ − 𝜆′))𝑒−2𝜆(𝑟)𝜔�� ∧ 𝜔��

Ω �� : 𝑑Γ ��

�� = 0

Γ 𝑐�� ∧ Γ ��

𝑐 = Γ �� ∧ Γ ��

�� + Γ �� ∧ Γ ��

�� + Γ �� ∧ Γ ��

�� + Γ �� ∧ Γ

�� = Γ ��

�� ∧ Γ ��

=1

𝑟𝑒−𝜆(𝑟)𝜔�� ∧ 𝜈′𝑒−𝜆(𝑟)𝜔�� =

𝜈′

𝑟𝑒−2𝜆(𝑟)𝜔�� ∧ 𝜔��

⇒ Ω �� =

𝜈′

𝑟𝑒−2𝜆(𝑟)𝜔�� ∧ 𝜔��

Ω ��: 𝑑Γ

��

= 0

Γ 𝑐��∧ Γ ��

𝑐 = Γ ��∧ Γ ��

�� + Γ ��∧ Γ ��

�� + Γ ��

��∧ Γ ��

�� + Γ ��

��∧ Γ

�� = Γ ��

��∧ Γ ��

��

=𝜈′

𝑟𝑒−2𝜆(𝑟)𝜔�� ∧ 𝜔��

⇒ Ω ��

=𝜈′

𝑟𝑒−2𝜆(𝑟)𝜔�� ∧ 𝜔��

Ω �� : 𝑑Γ ��

�� = 𝑑 (1

𝑟𝑒−𝜆(𝑟)𝜔��) = 𝑑(𝑒−𝜆(𝑟)𝑑𝜃) = −𝜆′𝑒−𝜆(𝑟)𝑑𝑟 ∧ 𝑑𝜃 =

𝜆′

𝑟𝑒−2𝜆(𝑟)𝜔�� ∧ 𝜔��

Γ 𝑐�� ∧ Γ ��

𝑐 = Γ �� ∧ Γ ��

�� + Γ �� ∧ Γ ��

�� + Γ �� ∧ Γ ��

�� + Γ �� ∧ Γ ��

��= 0

⇒ Ω �� =

𝜆′

𝑟𝑒−2𝜆(𝑟)𝜔�� ∧ 𝜔��





Ω ��: 𝑑Γ ��

�� = 𝑑 (

1

𝑟𝑒−𝜆(𝑟)𝜔��) = 𝑑(𝑒−𝜆(𝑟) sin𝜃 𝑑𝜙)

= −𝜆′𝑒−𝜆(𝑟) sin𝜃 𝑑𝑟 ∧ 𝑑𝜙 + 𝑒−𝜆(𝑟) cos𝜃 𝑑𝜃 ∧ 𝑑𝜙

= −𝜆′

𝑟𝑒−2𝜆(𝑟)𝜔𝑟 ∧ 𝜔�� +

1

𝑟2𝑒−𝜆(𝑟) cot 𝜃𝜔�� ∧ 𝜔��

Γ 𝑐��∧ Γ ��

𝑐 = Γ ��∧ Γ ��

�� + Γ ��∧ Γ ��

�� + Γ ��

��∧ Γ ��

�� + Γ ��

��∧ Γ ��

��= Γ

��

��∧ Γ ��

��

=cot 𝜃

𝑟𝜔�� ∧

1


⇒ Ω ��

=𝜆′

𝑟𝑒−2𝜆(𝑟)𝜔�� ∧ 𝜔𝑟

Ω ��

��: 𝑑Γ

��

�� = 𝑑 (

cot 𝜃

𝑟𝜔��) = 𝑑(cos𝜃 𝑑𝜙) = − sin𝜃 𝑑𝜃 ∧ 𝑑𝜙 = −

1

𝑟2𝜔�� ∧ 𝜔��

Γ 𝑐��∧ Γ ��

𝑐 = Γ ��∧ Γ ��

�� + Γ ��∧ Γ ��

�� + Γ ��

��∧ Γ ��

�� + Γ ��

��∧ Γ

��

��= Γ ��

��∧ Γ ��

��

= −1

𝑟2𝑒−2𝜆(𝑟)𝜔�� ∧ 𝜔��

⇒ Ω ��

�� =

(1 − 𝑒−2𝜆(𝑟))

𝑟2𝜔�� ∧ 𝜔��


Ω �� =

{

0 (𝜈′′ + 𝜈′(𝜈′ − 𝜆′))𝑒−2𝜆(𝑟)𝜔�� ∧ 𝜔��

𝜈′

𝑟𝑒−2𝜆(𝑟)𝜔�� ∧ 𝜔��

𝜈′

𝑟𝑒−2𝜆(𝑟)𝜔�� ∧ 𝜔��

𝑆 0𝜆′

𝑟𝑒−2𝜆(𝑟)𝜔�� ∧ 𝜔��

𝜆′

𝑟𝑒−2𝜆(𝑟)𝜔�� ∧ 𝜔𝑟

𝑆 𝐴𝑆 0(1 − 𝑒−2𝜆(𝑟))

𝑟2𝜔�� ∧ 𝜔��



R �� = (𝜈′′ + 𝜈′(𝜈′ − 𝜆′))𝑒−2𝜆(𝑟) 𝑅 ��

�� =𝜈′

𝑟𝑒−2𝜆(𝑟) 𝑅

��

�� =

𝜈′

𝑟𝑒−2𝜆(𝑟)

𝑅 �� =

𝜆′

𝑟𝑒−2𝜆(𝑟) 𝑅

��

�� =

𝜆′

𝑟𝑒−2𝜆(𝑟)

𝑅 ��

�� =

(1 − 𝑒−2𝜆(𝑟))

𝑟2

The Ricci tensor:

𝑅�� = 𝑅 ��𝑐��𝑐 (4.46)

𝑅�� = 𝑅 ��𝑐��𝑐 = 𝑅 ��

�� + 𝑅 �� + 𝑅 ��

�� + 𝑅 ��

��= 𝑅 ��

�� + 𝑅 �� + 𝑅

��

��

= (𝜈′′ + 𝜈′(𝜈′ − 𝜆′) + 2𝜈′

𝑟) 𝑒−2𝜆(𝑟)

𝑅�� = 𝑅 ��𝑐��𝑐 = 𝑅 ��

�� + 𝑅 �� + 𝑅 ��

�� + 𝑅 ��

��= −R ��

�� + 𝑅 �� + 𝑅

��

��

= −(𝜈′′ + 𝜈′(𝜈′ − 𝜆′) − 2𝜆′

𝑟) 𝑒−2𝜆(𝑟)





𝑅�� = 𝑅 ��𝑐��𝑐 = 𝑅 ��

�� + 𝑅 �� + 𝑅 ��

�� + 𝑅 ��

��= −𝑅 ��

�� + 𝑅 �� + 𝑅

��

��

= −𝜈′

𝑟𝑒−2𝜆(𝑟) +

𝜆′


(1 − 𝑒−2𝜆(𝑟))

𝑟2

𝑅�� = 𝑅 ��𝑐��𝑐 = 𝑅 ��

�� + 𝑅 �� + 𝑅 ��

�� + 𝑅 ��

��= −𝑅

��

��+ 𝑅

��

��+ 𝑅

��

��

= −𝜈′


𝜆′


(1 − 𝑒−2𝜆(𝑟))

𝑟2


𝑅�� =

{

(𝜈′′ + 𝜈′(𝜈′ − 𝜆′) + 2

𝜈′

𝑟)𝑒−2𝜆 0 0 0

0 −(𝜈′′ + 𝜈′(𝜈′ − 𝜆′) − 2𝜆′

𝑟)𝑒−2𝜆 0 0

0 0 −𝜈′

𝑟𝑒−2𝜆 +

𝜆′

𝑟𝑒−2𝜆 +

(1 − 𝑒−2𝜆)

𝑟20

0 0 0 −𝜈′

𝑟𝑒−2𝜆 +

𝜆′

𝑟𝑒−2𝜆 +

(1 − 𝑒−2𝜆)

𝑟2 }


9.2 124The Riemann tensor of the Schwarzschild metric

Solving the vacuum equations we find 𝜈 = −𝜆 and 𝑒2𝜈 = 1 −2𝑚

𝑟, 𝑒2𝜆 = (1 −

2𝑚

𝑟)−1

The line element: 𝑑𝑠2 = (1 −2𝑚

𝑟)𝑑𝑡2 − (1 −

2𝑚

𝑟)−1


Now we can find the independent elements of the Riemann tensor in the non-coordinate basis: First we calculate

𝜈′ =1

2𝑒−2𝜈

𝑑

𝑑𝑟(𝑒2𝜈) =

1

2𝑒−2𝜈

𝑑

𝑑𝑟(1 −

2𝑚

𝑟) =

1

2𝑒−2𝜈 (

2𝑚

𝑟2) =

𝑚

𝑟2 − 𝑟2𝑚

𝜈′′ =𝑑

𝑑𝑟(1

2𝑒−2𝜈

𝑑

𝑑𝑟(𝑒2𝜈)) = −𝜈′𝑒−2𝜈

𝑑

𝑑𝑟(𝑒2𝜈) +

1

2𝑒−2𝜈

𝑑2

𝑑𝑟2(𝑒2𝜈)

= −2𝜈′𝜈′ +1

2𝑒−2𝜈

𝑑

𝑑𝑟(2𝑚

𝑟2) = −2𝜈′𝜈′ −

2𝑚

𝑟3𝑒−2𝜈

R �� = (𝜈′′ + 𝜈′(𝜈′ − 𝜆′))𝑒−2𝜆(𝑟) = (𝜈′′ + 2𝜈′𝜈′)𝑒2𝜈(𝑟) = (−2𝜈′𝜈′ −

2𝑚

𝑟3𝑒−2𝜈 + 2𝜈′𝜈′) 𝑒2𝜈(𝑟)

= −2𝑚

𝑟3

𝑅 �� =

𝜈′

𝑟𝑒−2𝜆(𝑟) =

𝜈′

𝑟𝑒2𝜈(𝑟) =

1

𝑟

1

2𝑒−2𝜈 (

2𝑚

𝑟2)𝑒2𝜈(𝑟) =

𝑚

𝑟3

𝑅 �� =

𝜆′

𝑟𝑒−2𝜆(𝑟) = −

𝑚

𝑟3

𝑅 ��

�� =

𝜈′

𝑟𝑒−2𝜆(𝑟) =

𝑚

𝑟3

𝑅 ��

�� =

𝜆′

𝑟𝑒−2𝜆(𝑟) = −

𝑚

𝑟3

124 (McMahon, 2006, p. 215)





𝑅 ��

�� =(1 − 𝑒−2𝜆(𝑟))

𝑟2=(1 − 𝑒2𝜈(𝑟))

𝑟2=

(1 − (1 −2𝑚𝑟 ))

𝑟2=2𝑚

𝑟3


R �� = −

2𝑚

𝑟3 𝑅 ��

�� =𝑚

𝑟3 𝑅

��

�� =

𝑚

𝑟3

𝑅 �� = −

𝑚

𝑟3 𝑅

��

�� = −

𝑚

𝑟3

𝑅 ��

�� =

2𝑚

𝑟3

9.3 125Calculation of the scalar 𝑹𝒂𝒃𝒄𝒅𝑹𝒂𝒃𝒄𝒅 in the Schwarzschild metric

𝑅𝑎𝑏𝑐𝑑𝑅𝑎𝑏𝑐𝑑 = 𝑅��𝑅

�� + 𝑅��𝑅�� + 𝑅��𝑅

�� + 𝑅��𝑅��

+𝑅��𝑅�� + 𝑅��𝑅

�� + 𝑅��𝑅�� + 𝑅��𝑅

��

+𝑅��𝑅�� + 𝑅��𝑅

�� + 𝑅��𝑅�� + 𝑅��𝑅

��

+𝑅��𝑅�� + 𝑅��𝑅

�� + 𝑅��𝑅�� + 𝑅��𝑅

��

+𝑅��𝑅�� + 𝑅��𝑅

�� + 𝑅��𝑅�� + 𝑅��𝑅

��

+𝑅��𝑅�� + 𝑅��𝑅

�� + 𝑅��𝑅�� + 𝑅��𝑅

��

= (

2𝑚

𝑟3) (2𝑚

𝑟3) + (−

2𝑚

𝑟3) (−

2𝑚

𝑟3) + (

2𝑚

𝑟3) (2𝑚

𝑟3) + (−

2𝑚

𝑟3)(−

2𝑚

𝑟3)

+(−𝑚

𝑟3) (−

𝑚

𝑟3) + (

𝑚

𝑟3) (𝑚

𝑟3) + (

𝑚

𝑟3) (𝑚

𝑟3) + (−

𝑚

𝑟3) (−

𝑚

𝑟3)

+(𝑚

𝑟3) (𝑚

𝑟3) + (−

𝑚

𝑟3) (−

𝑚

𝑟3) + (

𝑚

𝑟3) (𝑚

𝑟3) + (−

𝑚

𝑟3) (−

𝑚

𝑟3)

+(−𝑚

𝑟3) (−

𝑚

𝑟3) + (

𝑚

𝑟3) (𝑚

𝑟3) + (−

𝑚

𝑟3) (−

𝑚

𝑟3) + (

𝑚

𝑟3) (𝑚

𝑟3)

+(𝑚

𝑟3) (𝑚

𝑟3) + (−

𝑚

𝑟3) (−

𝑚

𝑟3) + (

𝑚

𝑟3) (𝑚

𝑟3) + (−

𝑚

𝑟3) (−

𝑚

𝑟3)

+(−

2𝑚

𝑟3) (−

2𝑚

𝑟3) + (

2𝑚

𝑟3)(2𝑚

𝑟3) + (−

2𝑚

𝑟3) (−

2𝑚

𝑟3) + (

2𝑚

𝑟3) (2𝑚

𝑟3)

=48𝑚2

𝑟6 (10.35)

9.4 126Geodesics in the Schwarzschild Spacetime To find the geodesic we use the Euler-Lagrange equation

0 =𝑑

𝑑𝑠(𝜕𝐹

𝜕��𝑎) −

𝜕𝐹

𝜕𝑥𝑎 (10.36)

where

𝐹 = (1 −2𝑚

𝑟) ��2 − (1 −

2𝑚

𝑟)−1

��2 − 𝑟2��2 − 𝑟2 sin2 𝜃 ��2

𝑥𝑎 = 𝑡:

𝜕𝐹

𝜕𝑡 = 0

𝜕𝐹

𝜕�� = 2(1 −

2𝑚

𝑟) ��

125 (McMahon, 2006, p. 216), equation (10.35) 126 (McMahon, 2006, p. 216)





𝑑

𝑑𝑠(𝜕𝐹

𝜕��) =

4𝑚

𝑟2�� + 2 (1 −

2𝑚

𝑟) ��

⇒ 0 =4𝑚

𝑟2�� + 2 (1 −

2𝑚

𝑟) ��

⇔ 0 = �� +2𝑚

𝑟(𝑟 − 2𝑚)�� (10.37)

𝑥𝑎 = 𝑟:

𝜕𝐹

𝜕𝑟 =

2𝑚

𝑟2��2 +

2𝑚

(𝑟 − 2𝑚)2��2 − 2𝑟��2 − 2𝑟 sin2 𝜃 ��2

𝜕𝐹

𝜕�� = −2(1 −

2𝑚

𝑟)−1

��

𝑑

𝑑𝑠(𝜕𝐹

𝜕��) = −2(1 −

2𝑚

𝑟)−1

�� + 2 (1 −2𝑚

𝑟)−2 2𝑚

𝑟2��2

= −2(1 −2𝑚

𝑟)−1

�� +4𝑚

(𝑟 − 2𝑚)2��2

⇒ 0 = −2(1 −2𝑚

𝑟)−1

�� +2𝑚

(𝑟 − 2𝑚)2��2 −

2𝑚

𝑟2��2 + 2𝑟��2 + 2𝑟 sin2 𝜃 ��2

⇔ 0 = �� −𝑚

𝑟(𝑟 − 2𝑚)��2 +

𝑚(𝑟 − 2𝑚)

𝑟3��2 − (𝑟 − 2𝑚)��2 − (𝑟 − 2𝑚) sin2 𝜃 ��2 (10.38)

𝑥𝑎 = 𝜃:

𝜕𝐹

𝜕𝜃 = −2𝑟2 cos 𝜃 sin𝜃 ��2

𝜕𝐹

𝜕�� = −2𝑟2��

𝑑

𝑑𝑠(𝜕𝐹

𝜕��) = −4𝑟�� − 2𝑟2��

⇒ 0 = −4𝑟�� − 2𝑟2�� + 2𝑟2 cos 𝜃 sin 𝜃 ��2

⇔ 0 = �� +2

𝑟�� − cos𝜃 sin𝜃 ��2 (10.39)

𝑥𝑎 = 𝜙:

𝜕𝐹

𝜕𝜙 = 0

𝜕𝐹

𝜕�� = −2𝑟2 sin2 𝜃 ��

𝑑

𝑑𝑠(𝜕𝐹

𝜕��) = −4𝑟 sin2 𝜃 �� − 4𝑟2 cos 𝜃 sin 𝜃 �� − 2𝑟2 sin2 𝜃 ��

⇒ 0 = −4𝑟 sin2 𝜃 �� − 4𝑟2 cos 𝜃 sin 𝜃 �� − 2𝑟2 sin2 𝜃 ��

⇔ 0 = �� +2

𝑟�� + 2 cot 𝜃 �� (10.40)

9.5 127The meaning of the integration constant: The choice of 𝟐𝒎 We can use the geodesic equations to justify the choice of 2𝑚 by investigating the geodesic equations in

the classical limit i.e. 𝑟 ≫ 2𝑚, 𝑑𝑡

𝑑𝜏→ 1 and 𝑣 =

𝑑𝑟

𝑑𝜏≪ 𝑐, where 𝑣 is the velocity and 𝜏 is the proper time.

We want to investigate the case of a radially infalling particle i.e. 𝑑𝜃 = 0 and 𝑑𝜙 = 0. We also want to

work in SI-units so we have to substitute 𝑚 by 𝐺𝑚

𝑐2. Also remember that �� =

𝑑𝑡

𝑑𝑠=

𝑑𝑡

𝑐𝑑𝜏→

1

𝑐, �� =

𝑑𝑟

𝑑𝑠=

𝑑𝑟

𝑐𝑑𝜏=

𝑣

𝑐 and �� =

𝑑2𝑟

𝑑𝑠2=

𝑑2𝑟

𝑐2𝑑𝜏2=

𝑎

𝑐2, where 𝑎 is the particle acceleration. We use equation (10.38):

127 (McMahon, 2006, p. 218)





0 = �� −

𝑚

𝑟(𝑟 − 2𝑚)��2 +

𝑚(𝑟 − 2𝑚)

𝑟3��2 − (𝑟 − 2𝑚)��2 − (𝑟 − 2𝑚) sin2 𝜃 ��2

Now before we carry on with the physics we also have to be sure that each term in this equation has the same dimension. It turns out that they don’t and therefore the third term has to be multiplied by 𝑐2, in

which case each term gets the dimension128 1

𝑙𝑒𝑛𝑔𝑡ℎ.

⇒ 0 = �� −𝑚

𝑟(𝑟 − 2𝑚)��2 +

𝑚(𝑟 − 2𝑚)

𝑟3𝑐2��2 − (𝑟 − 2𝑚)��2 − (𝑟 − 2𝑚) sin2 𝜃 ��2

⇒ 0 = �� −𝑚

𝑟(𝑟 − 2𝑚)��2 +

𝑚(𝑟 − 2𝑚)

𝑟3𝑐2��2 𝑑𝜃 = 0, 𝑑𝜙 = 0

⇒ 0 = �� −𝑚

𝑟2��2 +

𝑚

𝑟2𝑐2��2 𝑟 ≫ 2𝑚

⇒ 0 =𝑎

𝑐2−𝑚

𝑟2𝑣2

𝑐2 +𝑚

𝑟2=1

𝑐2(𝑎 −

𝑚

𝑟2𝑣2 +

𝑚𝑐2

𝑟2) �� =

1

𝑐, �� =

𝑣

𝑐, �� =

𝑎

𝑐2

⇒ 0 = 𝑎 −𝐺𝑚

𝑟2(𝑣

𝑐)2

+𝐺𝑚

𝑟2 𝑚 →

𝐺𝑚

𝑐2

⇒ 𝑎 = −𝐺𝑚

𝑟2 𝑣 ≪ 𝑐

Multiplying with 𝑀 on both sides we get precisely the Newtonian gravitational law

⇒ 𝐹 = 𝑀𝑎 = −𝐺𝑀𝑚

𝑟2

9.6 129Time Delay To describe the time delay of a light ray outside a massive body, like the Sun we can use the Schwarzschild

metric. We can choose to work in the plane with 𝜃 =𝜋

2 , and together with 𝑑𝑠2 = 0 (light rays are placed

on the cone) we can rewrite the Schwarzschild line element:

0 = (1 −2𝑚

𝑟)𝑑𝑡2 − (1 −

2𝑚

𝑟)−1

𝑑𝑟2 − 𝑟2𝑑𝜙2

Now we would like to describe the time delay of the light ray as a function of the distance, 𝑟 from the massive body, and therefore we have to get rid of the 𝜙’s. It’s not that difficult. If we use polar coordinates the least distance, 𝑟0, the light ray passes the massive body is:

𝑟0 = 𝑟 sin𝜙

⇒ 𝜙 = sin−1𝑟0𝑟

⇒

𝑑𝜙

𝑑𝑟 =

𝑑

𝑑𝑟(sin−1

𝑟0𝑟) = 130 −

𝑟0𝑟2

1

√1 − (𝑟0𝑟 )

2

⇒ (𝑑𝜙

𝑑𝑟)2

=𝑟02

𝑟41

1 − (𝑟0𝑟 )

2

Now we can begin rewriting the Schwarzschild line element and solve the differential equation

0 = (1 −2𝑚

𝑟)𝑑𝑡2 − (1 −

2𝑚

𝑟)−1

𝑑𝑟2 − 𝑟2𝑑𝜙2

128 This actually originates from the line element of the Schwarzschild metric itself, because in order to get the same

dimension of each term, the first term has to be multiplied by 𝑐2: 𝑑𝑠2 = (1 −2𝑚

𝑟) 𝑐2𝑑𝑡2 − (1 −

2𝑚

𝑟)−1

𝑑𝑟2 −

𝑟2𝑑𝜃2 − 𝑟2 sin2 𝜃 𝑑𝜙2 129 (McMahon, 2006, p. 229) 130

𝑑

𝑑𝑥sin−1 𝑢 =

1

√1−𝑢2

𝑑𝑢

𝑑𝑥 (13.20) (Spiegel, 1990)





⇒ (1 −2𝑚

𝑟)2

(𝑑𝑡

𝑑𝑟)2

= 1 + 𝑟2 (1 −

2𝑚

𝑟) (𝑑𝜙

𝑑𝑟)2

= 1 + 𝑟2 (1 −2𝑚

𝑟)𝑟02

𝑟41

1 − (𝑟0𝑟)2

= 1 +(1 −

2𝑚𝑟)(𝑟0𝑟)2

1 − (𝑟0𝑟)2 =

1 − (𝑟0𝑟)2+ (1 −

2𝑚𝑟)(𝑟0𝑟)2

1 − (𝑟0𝑟)2 =

(1 −2𝑚𝑟0

2

𝑟3)

1 − (𝑟0𝑟)2

⇒ (𝑑𝑡

𝑑𝑟)2

=(1 −

2𝑚𝑟02

𝑟3)

(1 − (𝑟0𝑟 )

2) (1 −

2𝑚𝑟 )

2

⇒ 𝑑𝑡 =𝑑𝑟√1 −

2𝑚𝑟02

𝑟3

(1 −2𝑚𝑟 )

√1 − (𝑟0𝑟 )

2→ 131

𝑑𝑟√1 −2𝑚𝑟0

2

𝑟3

√1 − (𝑟0𝑟 )

2(1 +

2𝑚

𝑟)

→ 132

𝑑𝑟

√1 − (𝑟0𝑟)2(1 +

2𝑚

𝑟)(1 −

𝑚𝑟02

𝑟3)

→

𝑑𝑟

√1 − (𝑟0𝑟)2(1 +

2𝑚

𝑟−𝑚𝑟0

2

𝑟3)

To get the total time delay between points (𝑟1, 𝑟2) we have to integrate from 𝑟0 to 𝑟1 and from 𝑟0 to 𝑟2

⇒ ∫ 𝑑𝑡𝑡2

𝑡1

= ∫(1 +

2𝑚𝑟 −

𝑚𝑟02

𝑟3)𝑑𝑟

√1 − (𝑟0𝑟)2

𝑟1

𝑟0

+∫(1 +

2𝑚𝑟 −

𝑚𝑟02

𝑟3)𝑑𝑟

√1 − (𝑟0𝑟)2

𝑟2

𝑟0

= ∫

1

√1 − (𝑟0𝑟)2𝑑𝑟

𝑟1

𝑟0

+∫2𝑚

𝑟√1 − (𝑟0𝑟)2𝑑𝑟

𝑟1

𝑟0

−∫𝑚𝑟0

2

𝑟3√1− (𝑟0𝑟)2𝑑𝑟

𝑟1

𝑟0

+∫

1

√1 − (𝑟0𝑟 )

2𝑑𝑟

𝑟2

𝑟0

+∫2𝑚

𝑟√1 − (𝑟0𝑟 )

2𝑑𝑟

𝑟2

𝑟0

−∫𝑚𝑟0

2

𝑟3√1− (𝑟0𝑟 )

2𝑑𝑟

𝑟2

𝑟0

= ∫𝑟

√𝑟2 − 𝑟02𝑑𝑟

𝑟1

𝑟0

+ 2𝑚∫1

√𝑟2 − 𝑟02𝑑𝑟

𝑟1

𝑟0

−𝑚𝑟02∫

1

𝑟2√𝑟2 − 𝑟02𝑑𝑟

𝑟1

𝑟0

+∫𝑟

√𝑟2 − 𝑟02𝑑𝑟

𝑟2

𝑟0

+ 2𝑚∫1

√𝑟2 − 𝑟02𝑑𝑟

𝑟2

𝑟0

−𝑚𝑟02∫

1

𝑟2√𝑟2 − 𝑟02𝑑𝑟

𝑟2

𝑟0

⇒ 𝑡2 − 𝑡1 = 133√𝑟12 − 𝑟0

2 −√𝑟02 − 𝑟0

2

+1342𝑚 [ln (𝑟1 +√𝑟12 − 𝑟0

2) − ln (𝑟0 +√𝑟02 − 𝑟0

2)]

131

1

1−𝜖→ 1 + 𝜖 𝑓𝑜𝑟 𝜖 → 0 (20.8) (Spiegel, 1990)

132 √1 − 𝜖 → 1 −𝜖

2 𝑓𝑜𝑟 𝜖 → 0 (20.12) (Spiegel, 1990)

133∫𝑥𝑑𝑥

√𝑥2−𝑎2= √𝑥2 − 𝑎2 (14.210) (Spiegel, 1990)

134 ∫𝑑𝑥

√𝑥2−𝑎2= ln(𝑥 + √𝑥2 − 𝑎2) (14.210) (Spiegel, 1990)





−135𝑚𝑟02 [√𝑟1

2 − 𝑟02

𝑟02𝑟1

−√𝑟0

2 − 𝑟02

𝑟02𝑟0

] + √𝑟22 − 𝑟0

2 −√𝑟02 − 𝑟0

2

+2𝑚 [ln (𝑟2 +√𝑟22 − 𝑟0

2) − ln (𝑟0 +√𝑟02 − 𝑟0

2)]

−𝑚𝑟02 [√𝑟2

2 − 𝑟02

𝑟02𝑟2

−√𝑟0

2 − 𝑟02

𝑟02𝑟0

]

= √𝑟12 − 𝑟0

2 +√𝑟22 − 𝑟0

2

+𝑚[2 ln((𝑟1 +√𝑟1

2 − 𝑟02)(𝑟2 +√𝑟2

2 − 𝑟02)

𝑟02 ) −

√𝑟12 − 𝑟0

2

𝑟1−√𝑟2

2 − 𝑟02

𝑟2]

We can use this formula to calculate the time delay from e.g. Venus to Earth. The first term is the ordi-nary flat space distance, and the delay is characterized by the remaining terms. We have 𝑟2 = 𝑟𝑒, 𝑟1 = 𝑟𝑣.

Remember to get the right unit we must multiply by 𝐺

𝑐3.

⇒ 𝑡𝑑𝑒𝑙𝑎𝑦 =𝑚𝐺

𝑐3[2 ln(

(𝑟𝑣 +√𝑟𝑣2 − 𝑟02)(𝑟𝑒 +√𝑟𝑒2 − 𝑟02)

𝑟02 ) −

√𝑟𝑣2 − 𝑟02

𝑟𝑣−√𝑟𝑒2 − 𝑟02

𝑟𝑒]

9.7 136Use the geodesic equations to find the Christoffel symbols for the general

Schwarzschild metric. To find the geodesic we use

0 =𝑑

𝑑𝑠(𝜕𝐾

𝜕��𝑎) −

𝜕𝐾

𝜕𝑥𝑎 (4.36)

where

𝐾 =1

2𝑔𝑎𝑏��

𝑎��𝑏 =1

2𝑒2𝜈(𝑟)��2 −

1

2𝑒2𝜆(𝑟)��2 −

1

2𝑟2��2 −

1

2𝑟2 sin2 𝜃 ��2 (4.35)

𝑥𝑎 = 𝑡:

𝜕𝐾

𝜕𝑡 = 0

𝜕𝐾

𝜕�� = 𝑒2𝜈(𝑟)��

𝑑

𝑑𝑠(𝜕𝐾

𝜕��) = 2𝑒2𝜈(𝑟)

𝑑𝜈

𝑑𝑟�� + 𝑒2𝜈(𝑟)��

⇒ 0 = 2𝑒2𝜈(𝑟)𝑑𝜈

𝑑𝑟�� + 𝑒2𝜈(𝑟)��

⇔ 0 = �� + 2𝑑𝜈

𝑑𝑟��

𝑥𝑎 = 𝑟:

𝜕𝐾

𝜕𝑟 = 𝑒2𝜈(𝑟)

𝑑𝜈

𝑑𝑟��2 − 𝑒2𝜆(𝑟)

𝑑𝜆

𝑑𝑟��2 − 𝑟��2 − 𝑟 sin2 𝜃 ��2

𝜕𝐾

𝜕�� = −𝑒2𝜆(𝑟)��

𝑑

𝑑𝑠(𝜕𝐾

𝜕��) = −2𝑒2𝜆(𝑟)

𝑑𝜆

𝑑𝑟��2 − 𝑒2𝜆(𝑟)��

⇒ 0 = −𝑒2𝜆(𝑟)𝑑𝜆

𝑑𝑟��2 − 𝑒2𝜆(𝑟)�� − 𝑒2𝜈(𝑟)

𝑑𝜈

𝑑𝑟��2 + 𝑟��2 + 𝑟 sin2 𝜃 ��2

135 ∫𝑑𝑥

𝑥2√𝑥2−𝑎2=

√𝑥2−𝑎2

𝑎2𝑥 (14.214) (Spiegel, 1990)

136 (McMahon, 2006, p. 231) quiz 10-1. Answer: 10-1 is (a)





⇔ 0 = �� +𝑑𝜆

𝑑𝑟��2 + 𝑒2(𝜈(𝑟)−𝜆(𝑟))

𝑑𝜈

𝑑𝑟��2 − 𝑟𝑒−2𝜆(𝑟)��2 − 𝑟 sin2 𝜃 𝑒−2𝜆(𝑟)��2

𝑥𝑎 = 𝜃:

𝜕𝐾

𝜕𝜃 = −𝑟2 cos 𝜃 sin 𝜃 ��2

𝜕𝐾

𝜕�� = −𝑟2��

𝑑

𝑑𝑠(𝜕𝐾

𝜕��) = −2𝑟�� − 𝑟2��

⇒ 0 = −2𝑟�� − 𝑟2�� + 𝑟2 cos𝜃 sin𝜃 ��2

⇔ 0 = �� +2

𝑟�� − cos𝜃 sin𝜃 ��2

𝑥𝑎 = 𝜙:

𝜕𝐾

𝜕𝜙 = 0

𝜕𝐾

𝜕�� = −𝑟2 sin2 𝜃 ��

𝑑

𝑑𝑠(𝜕𝐾


⇒ 0 = −2𝑟 sin2 𝜃 �� − 2𝑟2 cos 𝜃 sin 𝜃 �� − 𝑟2 sin2 𝜃 ��

⇔ 0 = �� +2

𝑟�� + 2 cot 𝜃 ��


0 = �� + 2𝑑𝜈

𝑑𝑟��

0 = �� +𝑑𝜆

𝑑𝑟��2 + 𝑒2(𝜈(𝑟)−𝜆(𝑟))

𝑑𝜈

𝑑𝑟��2 − 𝑟𝑒−2𝜆(𝑟)��2 − 𝑟 sin2 𝜃 𝑒−2𝜆(𝑟)��2

0 = �� +2


0 = �� +2

𝑟�� + 2 cot 𝜃 ��

Now we can find the Christoffel symbols from the equation

𝑑2𝑥𝑎


𝑎𝑑𝑥𝑏

𝑑𝑠

𝑑𝑥𝑐

𝑑𝑠 = 0 (4.33)

Γ 𝑟𝑡𝑡 = Γ 𝑡𝑟

𝑡 =𝑑𝜈

𝑑𝑟 Γ 𝑟𝑟

𝑟 =𝑑𝜆

𝑑𝑟 Γ 𝑟𝜃

𝜃 = Γ 𝜃𝑟𝜃 =

1

𝑟 Γ 𝑟𝜙

𝜙 = Γ 𝜙𝑟

𝜙=1

𝑟

Γ 𝑡𝑡𝑟 = 𝑒2(𝜈(𝑟)−𝜆(𝑟))

𝑑𝜈

𝑑𝑟 Γ 𝜙𝜙

𝜃 = −cos 𝜃 sin𝜃 Γ 𝜃𝜙𝜙

= Γ 𝜙𝜃𝜙

= cot 𝜃

Γ 𝜃𝜃𝑟 = −𝑟𝑒−2𝜆(𝑟)

Γ 𝜙𝜙𝑟 = −𝑟 sin2 𝜃 𝑒−2𝜆(𝑟)





9.8 137The Ricci tensor for the general time dependent Schwarzschild metric.

The line element: 𝑑𝑠2 = 𝑒2𝜈(𝑡,𝑟)𝑑𝑡2 − 𝑒2𝜆(𝑡,𝑟)𝑑𝑟2 − 𝑟2𝑑𝜃2 − 𝑟2 sin2 𝜃 𝑑𝜙2

The Basis one forms

𝜔�� = 𝑒𝜈(𝑡,𝑟)𝑑𝑡 𝑑𝑡 = 𝑒−𝜈(𝑡,𝑟)𝜔��

𝜂𝑖𝑗 = {

1−1

−1−1

}

𝜔�� = 𝑒𝜆(𝑡,𝑟)𝑑𝑟 𝑑𝑟 = 𝑒−𝜆(𝑡,𝑟)𝜔��


𝑟𝜔��



Cartan’s First Structure equation and the calculation of the Cartan structure coefficients Γ �� :

𝑑𝜔�� = −Γ �� ∧ 𝜔�� (5.9) Γ ��

�� = Γ ��𝑐�� 𝜔𝑐 (5.10)

𝑑𝜔�� = 𝑑(𝑒𝜈(𝑡,𝑟)𝑑𝑡) = 𝜈′𝑒𝜈(𝑡,𝑟)𝑑𝑟 ∧ 𝑑𝑡 = 𝜈′𝑒−𝜆(𝑡,𝑟)𝜔�� ∧ 𝜔��

𝑑𝜔�� = 𝑑(𝑒𝜆(𝑡,𝑟)𝑑𝑟) = ��𝑒𝜆(𝑡,𝑟)𝑑𝑡 ∧ 𝑑𝑟 = ��𝑒−𝜈(𝑡,𝑟)𝜔�� ∧ 𝜔��

𝑑𝜔�� = 𝑑(𝑟𝑑𝜃) = 𝑑𝑟 ∧ 𝑑𝜃 =1

𝑟𝑒−𝜆(𝑟,𝑡)𝜔�� ∧ 𝜔��


𝑟𝑒−𝜆(𝑟,𝑡)𝜔�� ∧ 𝜔�� +

cot 𝜃

𝑟𝜔�� ∧ 𝜔��

In this case we have to be particular careful in reading off the curvature one forms. The curvature one-

forms are antisymmetric in the sense that: Γ�� = −Γ�� (5.11). This means that Γ �� = 𝜂��Γ�� =

−𝜂��Γ�� = −𝜂��𝜂��Γ ��

�� = Γ �� . But in the former calculation we found that Γ ��

�� = 𝜈′𝑒−𝜆(𝑡,𝑟)𝜔�� and Γ �� =

��𝑒−𝜈(𝑡,𝑟)𝜔��, which means that we in order to fulfill the antisymmetric properties need to require that

Γ �� = Γ ��

�� = ��𝑒−𝜈(𝑡,𝑟)𝜔�� + 𝜈′𝑒−𝜆(𝑡,𝑟)𝜔��, because Γ �� = 𝜈′𝑒−𝜆(𝑡,𝑟)𝜔�� + (something that makes Γ�� anti-

symmetric), and Γ �� = ��𝑒−𝜈(𝑡,𝑟)𝜔�� + (something that makes Γ�� antisymmetric).


Γ �� =

{

0 ��𝑒−𝜈(𝑡,𝑟)𝜔�� + 𝜈′𝑒−𝜆(𝑡,𝑟)𝜔�� 0 0

��𝑒−𝜈(𝑡,𝑟)𝜔�� + 𝜈′𝑒−𝜆(𝑡,𝑟)𝜔�� 01

𝑟𝑒−𝜆(𝑡,𝑟)𝜔��

1

𝑟𝑒−𝜆(𝑡,𝑟)𝜔��

0 −1

𝑟𝑒−𝜆(𝑡,𝑟)𝜔�� 0

cot 𝜃

𝑟𝜔��

0 −1

𝑟𝑒−𝜆(𝑡,𝑟)𝜔�� −

cot 𝜃

𝑟𝜔�� 0 }


137 (McMahon, 2006, p. 231), quiz 10-2. Answer to quiz 10-2: 𝑅𝑟𝑡 =

2

𝑟(𝑑𝜆

𝑑𝑡)






Ω �� = 𝑑Γ ��

�� + Γ 𝑐�� ∧ Γ ��

𝑐 =1

2𝑅 ��𝑐�� 𝜔𝑐 ∧ 𝜔�� (5.27), (5.28)

Ω �� : 𝑑Γ ��

�� = 𝑑(��𝑒−𝜈(𝑡,𝑟)𝜔�� + 𝜈′𝑒−𝜆(𝑡,𝑟)𝜔��) = 𝑑(��𝑒𝜆(𝑡,𝑟)−𝜈(𝑡,𝑟)𝑑𝑟 + 𝜈′𝑒𝜈(𝑡,𝑟)−𝜆(𝑡,𝑟)𝑑𝑡)

= (�� + ��(�� − ��)𝑒𝜆(𝑡,𝑟)−𝜈(𝑡,𝑟)𝑑𝑡 ∧ 𝑑𝑟 + (𝜈′′ + 𝜈′(𝜈′ − 𝜆′))𝑒𝜈(𝑡,𝑟)−𝜆(𝑡,𝑟) 𝑑𝑟 ∧ 𝑑𝑡

= [ −(�� + ��(�� − ��)) 𝑒−2𝜈(𝑡,𝑟) + (𝜈′′ + 𝜈′(𝜈′ − 𝜆′))𝑒−2𝜆(𝑡,𝑟)]𝜔�� ∧ 𝜔��

Γ 𝑐�� ∧ Γ ��

𝑐 = Γ �� ∧ Γ ��

�� + Γ �� ∧ Γ ��

�� + Γ �� ∧ Γ ��

�� + Γ �� ∧ Γ

�� = 0

⇒ Ω �� = [ −(�� + ��(�� − ��)) 𝑒−2𝜈(𝑡,𝑟) + (𝜈′′ + 𝜈′(𝜈′ − 𝜆′))𝑒−2𝜆(𝑡,𝑟)]𝜔�� ∧ 𝜔��

Ω �� : 𝑑Γ ��

�� = 0

Γ 𝑐�� ∧ Γ ��

𝑐 = Γ �� ∧ Γ ��

�� + Γ �� ∧ Γ ��

�� + Γ �� ∧ Γ ��

�� + Γ �� ∧ Γ

�� = Γ ��

�� ∧ Γ ��

=1

𝑟𝑒−𝜆(𝑡,𝑟)𝜔�� ∧ (��𝑒−𝜈(𝑡,𝑟)𝜔�� + 𝜈′𝑒−𝜆(𝑡,𝑟)𝜔��)

⇒ Ω �� =

��

𝑟𝑒−𝜆(𝑡,𝑟)−𝜈(𝑡,𝑟)𝜔�� ∧ 𝜔�� +

𝜈′

𝑟𝑒−2𝜆(𝑡,𝑟)𝜔�� ∧ 𝜔��

Ω ��: 𝑑Γ

��

= 0

Γ 𝑐��∧ Γ ��

𝑐 = Γ ��∧ Γ ��

�� + Γ ��∧ Γ ��

�� + Γ ��

��∧ Γ ��

�� + Γ ��

��∧ Γ

�� = Γ ��

��∧ Γ ��

��

=1

𝑟𝑒−𝜆(𝑡,𝑟)𝜔�� ∧ (��𝑒−𝜈(𝑡,𝑟)𝜔�� + 𝜈′𝑒−𝜆(𝑡,𝑟)𝜔��)

⇒ Ω ��

=��

𝑟𝑒−𝜆(𝑡,𝑟)−𝜈(𝑡,𝑟)𝜔�� ∧ 𝜔�� +

𝜈′

𝑟𝑒−2𝜆(𝑡,𝑟)𝜔�� ∧ 𝜔��

Ω �� : 𝑑Γ ��

�� = 𝑑 (1

𝑟𝑒−𝜆(𝑡,𝑟)𝜔��) = 𝑑(𝑒−𝜆(𝑡,𝑟)𝑑𝜃) = −��𝑒−𝜆(𝑡,𝑟)𝑑𝑡 ∧ 𝑑𝜃 − 𝜆′𝑒−𝜆(𝑡,𝑟)𝑑𝑟 ∧ 𝑑𝜃

=��

𝑟𝑒−𝜈(𝑡,𝑟)−𝜆(𝑡,𝑟)𝜔�� ∧ 𝜔�� +

𝜆′

𝑟𝑒−2𝜆(𝑡,𝑟)𝜔�� ∧ 𝜔��

Γ 𝑐�� ∧ Γ ��

𝑐 = Γ �� ∧ Γ ��

�� + Γ �� ∧ Γ ��

�� + Γ �� ∧ Γ ��

�� + Γ �� ∧ Γ ��

��= 0

⇒ Ω �� =

��

𝑟𝑒−𝜈(𝑡,𝑟)−𝜆(𝑡,𝑟)𝜔�� ∧ 𝜔�� +

𝜆′

𝑟𝑒−2𝜆(𝑡,𝑟)𝜔�� ∧ 𝜔��

Ω ��: 𝑑Γ ��

�� = 𝑑 (

1

𝑟𝑒−𝜆(𝑡,𝑟)𝜔��) = 𝑑(𝑒−𝜆(𝑡,𝑟) sin𝜃 𝑑𝜙)

= −��𝑒−𝜆(𝑡,𝑟) sin𝜃 𝑑𝑡 ∧ 𝑑𝜙 − 𝜆′𝑒−𝜆(𝑡,𝑟) sin𝜃 𝑑𝑟 ∧ 𝑑𝜙 + 𝑒−𝜆(𝑡,𝑟) cos 𝜃 𝑑𝜃 ∧ 𝑑𝜙

=��

𝑟𝑒−𝜈(𝑡,𝑟)−𝜆(𝑡,𝑟)𝜔�� ∧ 𝜔�� +

𝜆′

𝑟𝑒−2𝜆(𝑡,𝑟)𝜔�� ∧ 𝜔𝑟 +

1

𝑟2𝑒−𝜆(𝑡,𝑟) cot 𝜃𝜔�� ∧ 𝜔��

Γ 𝑐��∧ Γ ��

𝑐 = Γ ��∧ Γ ��

�� + Γ ��∧ Γ ��

�� + Γ ��

��∧ Γ ��

�� + Γ ��

��∧ Γ ��

��= Γ

��

��∧ Γ ��

��

=cot 𝜃

𝑟𝜔�� ∧

1


⇒ Ω ��

=��

𝑟𝑒−𝜈(𝑡,𝑟)−𝜆(𝑡,𝑟)𝜔�� ∧ 𝜔�� +

𝜆′

𝑟𝑒−2𝜆(𝑡,𝑟)𝜔�� ∧ 𝜔𝑟

Ω ��

��: 𝑑Γ

��

�� = 𝑑 (

cot 𝜃

𝑟𝜔��) = 𝑑(cos𝜃 𝑑𝜙) = − sin𝜃 𝑑𝜃 ∧ 𝑑𝜙 = −

1

𝑟2𝜔�� ∧ 𝜔��

Γ 𝑐��∧ Γ ��

𝑐 = Γ ��∧ Γ ��

�� + Γ ��∧ Γ ��

�� + Γ ��

��∧ Γ ��

�� + Γ ��

��∧ Γ

��

��= Γ ��

��∧ Γ ��

��





= −1

𝑟2𝑒−2𝜆(𝑡,𝑟)𝜔�� ∧ 𝜔��

⇒ Ω ��

�� =

(1 − 𝑒−2𝜆(𝑡,𝑟))

𝑟2𝜔�� ∧ 𝜔��


Ω �� =

{

0 [ − (�� + ��(�� − ��)) 𝑒−2𝜈(𝑡,𝑟) + (𝜈′′ + 𝜈′(𝜈′ − 𝜆′))𝑒−2𝜆(𝑡,𝑟)] 𝜔�� ∧ 𝜔��

��

𝑟𝑒−𝜆(𝑡,𝑟)−𝜈(𝑡,𝑟)𝜔�� ∧ 𝜔�� +

𝜈′

𝑟𝑒−2𝜆(𝑡,𝑟)𝜔�� ∧ 𝜔��

��

𝑟𝑒−𝜆(𝑡,𝑟)−𝜈(𝑡,𝑟)𝜔�� ∧ 𝜔�� +

𝜈′

𝑟𝑒−2𝜆(𝑡,𝑟)𝜔�� ∧ 𝜔��

𝑆 0��

𝑟𝑒−𝜈(𝑡,𝑟)−𝜆(𝑡,𝑟)𝜔�� ∧ 𝜔�� +

𝜆′

𝑟𝑒−2𝜆(𝑡,𝑟)𝜔�� ∧ 𝜔��

��

𝑟𝑒−𝜈(𝑡,𝑟)−𝜆(𝑡,𝑟)𝜔�� ∧ 𝜔�� +

𝜆′

𝑟𝑒−2𝜆(𝑡,𝑟)𝜔�� ∧ 𝜔��

𝑆 𝐴𝑆 0(1 − 𝑒−2𝜆(𝑡,𝑟))

𝑟2𝜔�� ∧ 𝜔��


Now we can find the independent elements of the Riemann tensor in the coordinate basis:

R ��

= − (�� + ��(�� − ��)) 𝑒−2𝜈(𝑡,𝑟)

+ (𝜈′′ + 𝜈′(𝜈′ − 𝜆′))𝑒−2𝜆(𝑡,𝑟) 𝑅 �� =

𝜈′

𝑟𝑒−2𝜆(𝑡,𝑟) 𝑅

��

�� =

𝜈′

𝑟𝑒−2𝜆(𝑡,𝑟)

𝑅 �� =

��

𝑟𝑒−𝜆(𝑡,𝑟)−𝜈(𝑡,𝑟) 𝑅

��

�� =

��

𝑟𝑒−𝜆(𝑡,𝑟)−𝜈(𝑡,𝑟)

𝑅 �� =

𝜆′

𝑟𝑒−2𝜆(𝑡,𝑟) 𝑅

��

�� =

𝜆′

𝑟𝑒−2𝜆(𝑡,𝑟)

𝑅 ��

�� =

(1 − 𝑒−2𝜆(𝑡,𝑟))

𝑟2

The Ricci tensor:

𝑅�� = 𝑅 ��𝑐��𝑐 (4.46)

𝑅�� = 𝑅 ��𝑐��𝑐 = 𝑅 ��

�� + 𝑅 �� + 𝑅 ��

�� + 𝑅 ��

��= 𝑅 ��

�� + 𝑅 �� + 𝑅

��

��

= −(�� + ��(�� − ��)) 𝑒−2𝜈(𝑡,𝑟) + (𝜈′′ + 𝜈′(𝜈′ − 𝜆′) + 2𝜈′

𝑟) 𝑒−2𝜆(𝑡,𝑟)

𝑅�� = 𝑅 ��𝑐��𝑐 = 𝑅 ��

�� + 𝑅 �� + 𝑅 ��

�� + 𝑅 ��

��= 𝑅 ��

�� + 𝑅 ��

��= 2

��

𝑟𝑒−𝜆(𝑡,𝑟)−𝜈(𝑡,𝑟)

𝑅�� = 𝑅 ��𝑐��𝑐 = 𝑅 ��

�� + 𝑅 �� + 𝑅 ��

�� + 𝑅 ��

��= −R ��

�� + 𝑅 �� + 𝑅

��

��

= (�� + ��(�� − ��)) 𝑒−2𝜈(𝑡,𝑟) − (𝜈′′ + 𝜈′(𝜈′ − 𝜆′) − 2𝜆′

𝑟) 𝑒−2𝜆(𝑡,𝑟)

𝑅�� = 𝑅 ��𝑐��𝑐 = 𝑅 ��

�� + 𝑅 �� + 𝑅 ��

�� + 𝑅 ��

��= −𝑅 ��

�� + 𝑅 �� + 𝑅

��

��

= (−𝜈′

𝑟+𝜆′

𝑟 ) 𝑒−2𝜆(𝑡,𝑟) +

(1 − 𝑒−2𝜆(𝑡,𝑟))

𝑟2

𝑅�� = 𝑅 ��𝑐��𝑐 = 𝑅 ��

�� + 𝑅 �� + 𝑅 ��

�� + 𝑅 ��

��= −𝑅

��

��+ 𝑅

��

��+ 𝑅

��

��

= (−𝜈′

𝑟+𝜆′

𝑟 ) 𝑒−2𝜆(𝑡,𝑟) +

(1 − 𝑒−2𝜆(𝑡,𝑟))

𝑟2






𝑅�� =

{

−(�� + ��(�� − ��)) 𝑒−2𝜈 + (𝜈′′ + 𝜈′(𝜈′ − 𝜆′) + 2

𝜈′

𝑟) 𝑒−2𝜆 2

��

𝑟𝑒−𝜆−𝜈 0 0

2��

𝑟𝑒−𝜆−𝜈 (�� + ��(�� − ��)) 𝑒−2𝜈 − (𝜈′′ + 𝜈′(𝜈′ − 𝜆′) − 2

𝜆′

𝑟) 𝑒−2𝜆 0 0

0 0 (−𝜈′

𝑟+𝜆′

𝑟 ) 𝑒−2𝜆 +

(1 − 𝑒−2𝜆)

𝑟20

0 0 0 (−𝜈′

𝑟+𝜆′

𝑟 ) 𝑒−2𝜆 +

(1 − 𝑒−2𝜆)

𝑟2 }


The Ricci tensor in the coordinate basis:

The transformation: 𝑅𝑎𝑏 = Λ 𝑎𝑐 Λ 𝑏

�� 𝑅𝑐��

𝑅𝑡𝑡 = Λ 𝑡𝑐 Λ 𝑡

�� 𝑅𝑐�� = (Λ 𝑡�� )

2𝑅��

= 𝑒2𝜈(𝑡,𝑟) (−(�� + ��(�� − ��)) 𝑒−2𝜈(𝑡,𝑟) + (𝜈′′ + 𝜈′(𝜈′ − 𝜆′) + 2𝜈′

𝑟) 𝑒−2𝜆(𝑡,𝑟))

= −(�� + ��(�� − ��)) + (𝜈′′ + 𝜈′(𝜈′ − 𝜆′) + 2𝜈′

𝑟) 𝑒2𝜈(𝑡,𝑟)−2𝜆(𝑡,𝑟)

𝑅𝑟𝑡 = Λ 𝑟𝑐 Λ 𝑡

�� 𝑅𝑐�� = Λ 𝑟�� Λ 𝑡

�� 𝑅�� = 𝑒𝜆(𝑡,𝑟)𝑒𝜈(𝑡,𝑟)2

��

𝑟𝑒−𝜆(𝑡,𝑟)−𝜈(𝑡,𝑟) = 2

��

𝑟

𝑅𝑟𝑟 = Λ 𝑟𝑐 Λ 𝑟

�� 𝑅𝑐�� = (Λ 𝑟�� )

2𝑅��

= 𝑒2𝜆(𝑡,𝑟) ((�� + ��(�� − ��)) 𝑒−2𝜈(𝑡,𝑟) − (𝜈′′ + 𝜈′(𝜈′ − 𝜆′) − 2𝜆′

𝑟)𝑒−2𝜆(𝑡,𝑟))

= (�� + ��(�� − ��)) 𝑒−2𝜈(𝑡,𝑟)+2𝜆(𝑡,𝑟) − (𝜈′′ + 𝜈′(𝜈′ − 𝜆′) − 2𝜆′

𝑟)

𝑅𝜃𝜃 = Λ 𝜃𝑐 Λ 𝜃

�� 𝑅𝑐�� = (Λ 𝜃�� )

2𝑅�� = 𝑟

2 ((−𝜈′

𝑟+𝜆′

𝑟 ) 𝑒−2𝜆(𝑡,𝑟) +

(1 − 𝑒−2𝜆(𝑡,𝑟))

𝑟2)

= ((−𝜈′ + 𝜆′ )𝑟 − 1)𝑒−2𝜆(𝑡,𝑟) + 1

𝑅𝜙𝜙 = Λ 𝜙𝑐 Λ 𝜙

�� 𝑅𝑐�� = (Λ 𝜙��)2

𝑅�� = 𝑟2 sin2 𝜃 ((−

𝜈′

𝑟+𝜆′

𝑟 ) 𝑒−2𝜆(𝑡,𝑟) +

(1 − 𝑒−2𝜆(𝑡,𝑟))

𝑟2)

= (((−𝜈′ + 𝜆′ )𝑟 − 1)𝑒−2𝜆(𝑡,𝑟) + 1) sin2 𝜃


𝑅𝑎𝑏 =

{

−(�� + ��(�� − ��)) + (𝜈′′ + 𝜈′(𝜈′ − 𝜆′) + 2

𝜈′

𝑟) 𝑒2𝜈−2𝜆 2

λ

r0 0

2λ

r(�� + ��(�� − ��)) 𝑒−2𝜈+2𝜆 − (𝜈′′ + 𝜈′(𝜈′ − 𝜆′) − 2

𝜆′

𝑟) 0 0

0 0 ((−𝜈′ + 𝜆′ )𝑟 − 1)𝑒−2𝜆 + 1 0

0 0 0 (((−𝜈′ + 𝜆′ )𝑟 − 1)𝑒−2𝜆 + 1) sin2 𝜃}






9.9 The Schwarzschild metric with nonzero cosmological constant.

9.9.1 138The Ricci rotation coefficients and Ricci tensor for the Schwarzschild metric with non-

zero cosmological constant.

The line element: 𝑑𝑠2 = −𝑓(𝑟)𝑑𝑡2 +1

𝑓(𝑟)𝑑𝑟2 + 𝑟2𝑑𝜃2 + 𝑟2 sin2 𝜃 𝑑𝜙2

Where 𝑓(𝑟) = 1 −2𝑚

𝑟−1

3Λ𝑟2

Now we can compare with the line element of the Schwarzschild metric with zero cosmological constant, where the primes should not be mistaken for the derivative 𝑑/𝑑𝑟.

𝑑𝑠2 = 𝑒2𝜈(𝑟′)𝑑𝑡′

2− 𝑒2𝜆(𝑟

′)𝑑𝑟′2− 𝑟′

2𝑑𝜃′

2− 𝑟′

2sin2 𝜃′ 𝑑𝜙′

2

And choose:

𝑒𝜈(𝑟′)𝑑𝑡′ = √𝑓(𝑟)𝑑𝑡

𝑒𝜆(𝑟′)𝑑𝑟′ =

1

√𝑓(𝑟)𝑑𝑟

𝑟′𝑑𝜃′ = 𝑟𝑑𝜃 𝑟′ sin 𝜃′ 𝑑𝜙′ = 𝑟 sin𝜃 𝑑𝜙

Comparing the two metrics we see: 𝜙′ = 𝜙, 𝜃′ = 𝜃, 𝑟′ = 𝑟, 𝑒𝜈(𝑟′) = √𝑓(𝑟), 𝜈 = −𝜆, 𝑡′ = 𝑡

Next we can use the former calculations of the Schwarzschild metric with zero cosmological constant to find the Ricci rotation coefficients and the Ricci tensor for the Schwarzschild metric with non-zero cosmo-logical constant.

But first we need to calculate 𝑑𝜈(𝑟′)

𝑑𝑟′ = 𝑒−𝜈(𝑟

′)𝑑

𝑑𝑟′(𝑒𝜈(𝑟

′)) = 𝑒−𝜈(𝑟′)𝑑

𝑑𝑟′(√𝑓(𝑟)

𝑑𝑡

𝑑𝑡′) = 𝑒−𝜈(𝑟

′)𝑑

𝑑𝑟(√𝑓(𝑟))

= 𝑒−𝜈(𝑟′)𝑑𝑓(𝑟)

𝑑𝑟

1

2√𝑓(𝑟)=

1

2𝑓(𝑟)

𝑑𝑓(𝑟)

𝑑𝑟

𝑑2𝜈(𝑟′)

𝑑𝑟′2 =

𝑑

𝑑𝑟′[1

2𝑓(𝑟)

𝑑𝑓(𝑟)

𝑑𝑟 ] =

𝑑

𝑑𝑟[1

2𝑓(𝑟)

𝑑𝑓(𝑟)

𝑑𝑟 ] =

−1

2𝑓2(𝑟)(𝑑𝑓(𝑟)

𝑑𝑟)

2

+1

2𝑓(𝑟)

𝑑2𝑓(𝑟)

𝑑𝑟2

𝑑𝑓(𝑟)

𝑑𝑟 =

𝑑

𝑑𝑟(1 −

2𝑚

𝑟−1

3Λ𝑟2) =

2𝑚

𝑟2−2

3Λ𝑟 = −

1

𝑟(𝑓(𝑟) + Λ𝑟2 − 1)

𝑑2𝑓(𝑟)

𝑑𝑟2 =

𝑑

𝑑𝑟(2𝑚

𝑟2−2

3Λ𝑟) = −

4𝑚

𝑟3−2

3Λ = −

2

𝑟(𝑑𝑓(𝑟)

𝑑𝑟+ Λ𝑟)

The Ricci rotation coefficients

Γ�� = Γ�� =

𝑑𝜈(𝑟′)

𝑑𝑟′𝑒−𝜆(𝑟

′) =1

2𝑓(𝑟)

𝑑𝑓(𝑟)

𝑑𝑟√𝑓(𝑟) =

1

2√𝑓(𝑟)

𝑑𝑓(𝑟)

𝑑𝑟=

2𝑚𝑟2

−23Λr

2√1 −2𝑚𝑟 −

13Λ𝑟

2

=𝑚 −

13Λ𝑟

3

√𝑟 − 2𝑚 −13Λ𝑟

3

=3𝑚 − Λ𝑟3

√9𝑟 − 18𝑚 − 3Λ𝑟3

138 (McMahon, 2006, pp. 231-32), quiz 10-3 and 10-4. The answer to quiz 10-3 is (b) and the answer to quiz 10-4 is (a)





Γ�� = Γ�� = −Γ�� = −Γ

�� = −

1

𝑟′𝑒−𝜆(𝑟

′) = −1

𝑟√𝑓(𝑟) = −

1

𝑟√1 −

2𝑚

𝑟−1

3Λ𝑟2

Γ�� = −Γ�� = −cot𝜃′

𝑟′= −

cot 𝜃

𝑟

The Ricci tensor

𝑅�� = [𝑑2𝜈

𝑑𝑟′2+ (

𝑑𝜈

𝑑𝑟′)2

− (𝑑𝜈

𝑑𝑟′) (𝑑𝜆

𝑑𝑟′) +

2

𝑟′𝑑𝜈

𝑑𝑟′] 𝑒−2𝜆(𝑟

′) (10.24)

= [𝑑2𝜈

𝑑𝑟′2+ 2(

𝑑𝜈

𝑑𝑟′)2

+2

𝑟′𝑑𝜈

𝑑𝑟′] 𝑒2𝜈(𝑟

′)

= [−1

2𝑓2(𝑟)(𝑑𝑓(𝑟)

𝑑𝑟)

2

+1

2𝑓(𝑟)

𝑑2𝑓(𝑟)

𝑑𝑟2+ 2(

1

2𝑓(𝑟)

𝑑𝑓(𝑟)

𝑑𝑟 )

2

+2

𝑟

1

2𝑓(𝑟)

𝑑𝑓(𝑟)

𝑑𝑟 ] 𝑓(𝑟)

= [1

2𝑓(𝑟)

𝑑2𝑓(𝑟)

𝑑𝑟2+2

𝑟

1

2𝑓(𝑟)

𝑑𝑓(𝑟)

𝑑𝑟 ] 𝑓(𝑟)

= [1

2𝑓(𝑟)(−

2

𝑟(𝑑𝑓(𝑟)

𝑑𝑟+ Λ𝑟)) +

2

𝑟

1

2𝑓(𝑟)

𝑑𝑓(𝑟)

𝑑𝑟 ] 𝑓(𝑟) = −Λ

𝑅�� = −[𝑑2𝜈

𝑑𝑟′2+ (

𝑑𝜈

𝑑𝑟′)2

− (𝑑𝜈

𝑑𝑟′) (𝑑𝜆

𝑑𝑟′) −

2

𝑟′𝑑𝜆


′) = −𝑅�� = Λ (10.25)

𝑅�� = 𝑅�� = [−

1

𝑟′𝑑𝜈

𝑑𝑟′+1

𝑟′𝑑𝜆


′) +1 − 𝑒−2𝜆(𝑟

′)

𝑟′2

= −2

𝑟′𝑑𝜈

𝑑𝑟′𝑒−2𝜆(𝑟

′) +1 − 𝑒−2𝜆(𝑟

′)

𝑟′2

(10.26)

= −2

𝑟

1

2𝑓(𝑟)

𝑑𝑓(𝑟)

𝑑𝑟𝑓(𝑟) +

1 − 𝑓(𝑟)

𝑟2= −

1

𝑟

𝑑𝑓(𝑟)

𝑑𝑟+1 − 𝑓(𝑟)

𝑟2

= −

1

𝑟(−

1

𝑟(𝑓(𝑟) + Λ𝑟2 − 1)) +

1 − 𝑓(𝑟)

𝑟2

=1

𝑟2((𝑓(𝑟) + Λ𝑟2 − 1)) +

1 − 𝑓(𝑟)

𝑟2

= Λ

Alternatively we could use the formula calculated earlier on page 138: 𝑅�� = 𝜂��Λ valid in vacuum sys-tems with a cosmological constant and positive signature, from which we immediately can see that −𝑅�� = 𝑅�� = 𝑅�� = 𝑅�� = Λ

9.9.2 139The general Schwarzschild metric in vacuum with a cosmological constant: The Ricci

scalar

The metric 𝑑𝑠2 = 𝑒2𝜈(𝑟)𝑑𝑡2 − 𝑒2𝜆(𝑟)𝑑𝑟2 − 𝑟2(𝑑𝜃2 + sin2 𝜃 𝑑𝜙2)

139 (McMahon, 2006, p. 277), quiz 12-1. And the answer to quiz 12-1 is (c)





In this case we can write the Einstein equation in the local frame (non-coordinate basis) – we name the

cosmological constant Ω140:

0 = 𝑅�� −1

2𝜂��𝑅 + 𝜂��Ω (6.6)

⇒ 0 = 𝜂��𝑅�� −1

2𝜂��𝜂��𝑅 + 𝜂

��𝜂��Ω

= 𝑅 −

1

24𝑅 + 4Ω

⇔ 𝑅 = 4Ω

9.9.3 141The general Schwarzschild metric in vacuum with a cosmological constant: Integration

constants

We know that 𝜆(𝑟) = ln𝑘 − 𝜈(𝑟)

⇒ 𝜆′(𝑟) = −𝜈′(𝑟)

and 𝑒−2𝜆(𝑟) =1

𝑘2𝑒2𝜈(𝑟)

We also need to find

(𝑟𝑒2𝜈(𝑟))′ = 𝑒2𝜈(𝑟) + 2𝑟𝜈′(𝑟)𝑒2𝜈(𝑟)

⇔ 𝜈′(𝑟) =(𝑟𝑒2𝜈(𝑟))

′

2𝑟𝑒2𝜈(𝑟)−1

2𝑟

As in quiz 12-1 we use the Einstein equation in the non-coordinate basis, but this time for the coordinate

𝜃

0 = 𝑅�� −1

2𝜂��𝑅 + 𝜂��Ω (6.6)

= 𝑅�� +1

2𝑅 − Ω

= 𝑅�� +1

24Ω − Ω

= 𝑅�� + Ω

⇔ 𝑅�� = −Ω Earlier (p.211) we calculated the Ricci tensor:

⇒ −Ω = −𝜈′


𝜆′


(1 − 𝑒−2𝜆(𝑟))

𝑟2

= −2𝜈′

𝑘2𝑟𝑒2𝜈(𝑟) +

1

𝑟2−𝑒2𝜈(𝑟)

𝑘2𝑟2

= −2((𝑟𝑒2𝜈(𝑟))

′

2𝑟𝑒2𝜈(𝑟)−1

2𝑟)𝑒2𝜈(𝑟)

𝑘2𝑟+1

𝑟2−𝑒2𝜈(𝑟)

𝑘2𝑟2

Renaming 𝑔(𝑟) = 𝑟𝑒2𝜈(𝑟)

⇒ −Ω = −(𝑔′(𝑟)

𝑔(𝑟)−1

𝑟)𝑔(𝑟)

𝑘2𝑟2+1

𝑟2−𝑔(𝑟)

𝑘2𝑟3

⇔ 𝑘2 + 𝑘2Ω𝑟2 = 𝑔′(𝑟)

140 If you compare this to quiz 10-3 and 10-4 page 231-32 you can conclude that Ω = −Λ. The reason is that the metric in the two cases changes signature, which implies that in the first case 𝑅 = 4Λ, and in the second 𝑅 = 4Ω = −4Λ. You might also check the proofs on page 138. 141 (McMahon, 2006, p. 277), quiz 12-2 and 12-3. The answers to quiz 12-2 is: 𝑟𝑒2𝜈(𝑟) = 𝐴 + 𝐵𝑟 +

1

3𝑘2Ω𝑟3 and quiz

12-3 is 𝐵 = 𝑘2.





We guess the solution (polynomials with exponents higher than 3 cannot contribute): 𝑔(𝑟) = 𝑟𝑒2𝜈(𝑟)

= 𝐴 + 𝐵𝑟 + 𝐶𝑟2 +𝐷𝑟3

⇒ 𝑔′(𝑟) = 𝐵 + 2𝐶𝑟 + 3𝐷𝑟2 ⇒ 𝑘2 + 𝑘2Ω𝑟2 = 𝐵 + 2𝐶𝑟 + 3𝐷𝑟2

Now comparing the coefficients we find 𝐵 = 𝑘2 𝐶 = 0

𝐷 =1

3𝑘2Ω

and we can conclude that

𝑟𝑒2𝜈(𝑟) = 𝐴 + 𝑘2𝑟 +1

3𝑘2Ω𝑟3

⇒ 𝑒2𝜈(𝑟) =𝐴

𝑟+ 𝑘2 +

1

3𝑘2Ω𝑟2

and the line element 𝑑𝑠2 = 𝑒2𝜈(𝑟)𝑑𝑡2 − 𝑒2𝜆(𝑟)𝑑𝑟2 − 𝑟2𝑑𝜃2 − 𝑟2 sin2 𝜃 𝑑𝜙2

becomes

𝑑𝑠2 = (𝐴

𝑟+ 𝑘2 +

1

3𝑘2Ω𝑟2)𝑑𝑡2 − (

𝐴

𝑟+ 𝑘2 +

1

3𝑘2Ω𝑟2)

−1

𝑑𝑟2 − 𝑟2𝑑𝜃2

− 𝑟2 sin2 𝜃 𝑑𝜙2 If 𝑘2 = 1 and Ω = 0 this should be identical to the ordinary Schwarzschild vacuum metric, which means that 𝐴 has to be equal to: 𝐴 = −2𝑚

9.9.4 142The general Schwarzschild metric in vacuum with a cosmological constant: The spatial

part of the line element.

The line element 𝑑𝑠2 = 𝑒2𝜈(𝑟)𝑑𝑡2 − 𝑒2𝜆(𝑟)𝑑𝑟2 − 𝑟2𝑑𝜃2 − 𝑟2 sin2 𝜃 𝑑𝜙2

can in Gaussian normal coordinates be written as 𝑑𝑠2 = 𝑑𝑡2 − 𝑎2(𝑡)𝑑𝜎2 In this case we want to find the spatial part of the line element 𝑑𝜎2 = 𝑔𝑖𝑗𝑑𝑥𝑖𝑑𝑥𝑗

and to do that we will use the method outlined on page 260, where the metric is found from the Ricci-tensor: 𝑅𝑖𝑗 = 2𝐾𝑔𝑗𝑙

⇔ 𝑔𝑖𝑗 =1

2𝐾𝑅𝑖𝑗

⇒ 𝑔𝑟𝑟 =1

2𝐾𝑅𝑟𝑟

=1

2𝐾(Λ 𝑟

�� )2𝑅��

=1

2𝐾(𝑒𝜆(𝑟))

2(−Ω)

= −Ω

2𝐾𝑒2𝜆(𝑟)

In the former quiz we found that

𝑒−2𝜆(𝑟) =1

𝑘2𝑒2𝜈(𝑟)

142 (McMahon, 2006, p. 277), quiz 12-4





=

1

𝑘21

𝑟(−2𝑚 + 𝑘2𝑟 +

1

3𝑘2Ω𝑟3)

= 1 −

2𝑚

𝑘2𝑟+1

3Ω𝑟2

⇒ 𝑔𝑟𝑟 = −Ω

2𝐾

1

1 −2𝑚𝑘2𝑟

+13Ω𝑟2

𝑔𝜃𝜃 =

1

2𝐾𝑅𝜃𝜃

=

1

2𝐾(Λ 𝜃

�� )2𝑅��

=

1

2𝐾(𝑟)2(−Ω)

= −

Ω

2𝐾𝑟2

𝑔𝜙𝜙 =1

2𝐾𝑅𝜙𝜙

=

1

2𝐾(Λ 𝜙

��)2

𝑅��

=

1

2𝐾(𝑟 sin𝜃)2(−Ω)

= −

Ω

2𝐾𝑟2 sin2 𝜃

⇒ 𝑑𝜎2 = −Ω

2𝐾

𝑑𝑟2

1 −2𝑚𝑘2𝑟

+13Ω𝑟

2−Ω

2𝐾𝑟2𝑑𝜃2 −

Ω

2𝐾𝑟2 sin2 𝜃 𝑑𝜙2

where we can omit the common factor = −Ω

2𝐾 and finally get if we choose 𝑘 = 1

𝑑𝜎2 =

𝑑𝑟2

1 −2𝑚𝑟 +

13Ω𝑟

2+ 𝑟2𝑑𝜃2 + 𝑟2 sin2 𝜃 𝑑𝜙2

9.9.5 143The effect of the cosmological constant over the scale of the solar system

To check the effect of the cosmological constant over the scale of the solar system we will look at the line element of a radially moving (i.e. 𝑑𝜙 = 0, 𝑑𝜃 = 0) light ray (𝑑𝑠2 = 0) in the solar system with a sun mass 𝑚 = 0 in a universe with a cosmological constant Λ proportional to the size of the Universe. The line element:

𝑑𝑠2 = 144 (1 +1

3Λ𝑟2)𝑑𝑡2 −

𝑑𝑟2

1 +13Λ𝑟

2

⇒ 𝑑𝑡 = (±)𝑑𝑟

1 +13Λ𝑟

2

⇒ ∫ 𝑑𝑡 = ∫𝑑𝑟

1 +13Λ𝑟

2

𝑅2

𝑅1

=3

Λ∫

𝑑𝑟

3Λ + 𝑟

2

𝑅2

𝑅1

143 (McMahon, 2006, p. 277), Quiz 12-5, answer (c) 144 Notice: If we set Λ = 0 we get the usual Minkowski line element.





= 145 [3

Λ√Λ

3tan−1(𝑟√

3

Λ)]

𝑅1

𝑅2

= √3

Λtan−1((𝑅2 − 𝑅1)√

3

Λ)

→ 0 𝑖𝑓 (𝑅2 − 𝑅1) ≪ Λ So the effect of the cosmological constant on a small scale is negligible.

9.10 146The Petrov type of the Schwarzschild spacetime

The metric tensor: 𝑔𝑎𝑏 =

{

1 −

2𝑚

𝑟

−1

(1 −2𝑚𝑟)

−𝑟2

−𝑟2 sin2 𝜃}


{

1

(1 −2𝑚𝑟)

−(1 −2𝑚

𝑟)

−1

𝑟2

−1

𝑟2 sin2 𝜃}

The basis one forms

𝜔�� = √1 −2𝑚

𝑟𝑑𝑡 𝑑𝑡

=1

√1 −2𝑚𝑟

𝜔��

𝜂𝑖𝑗 = {

1−1

−1−1

} 𝜔�� =

1

√1 −2𝑚𝑟

𝑑𝑟 𝑑𝑟 = √1 −

2𝑚

𝑟𝜔��


𝑟𝜔��




145 ∫

𝑑𝑥

𝑎2+𝑥2=

1

𝑎tan−1

𝑥

𝑎 (Spiegel, 1990) eq.( 14.162)

146 (McMahon, 2006, p. 232), quiz 10-5





Now we can use the basis one-forms to construct a orthonormal null tetrad (9.10)

(

𝑙𝑛𝑚��

) =1

√2(

1 1 0 01 −1 0 00 0 1 𝑖0 0 1 −𝑖

)

(

𝜔��

𝜔��

𝜔��

𝜔��)

=1

√2

(

𝜔�� +𝜔��

𝜔�� −𝜔��

𝜔�� + 𝑖𝜔��

𝜔�� − 𝑖𝜔��)

=1

√2

(

√1 −

2𝑚

𝑟𝑑𝑡 +

1

√1 −2𝑚𝑟

𝑑𝑟

√1 −2𝑚

𝑟𝑑𝑡 −

1

√1 −2𝑚𝑟

𝑑𝑟

𝑟𝑑𝜃 + 𝑖𝑟 sin𝜃 𝑑𝜙𝑟𝑑𝜃 − 𝑖𝑟 sin𝜃 𝑑𝜙 )


𝑙𝑎 =1

√2(√1 −

2𝑚

𝑟,

1

√1 −2𝑚𝑟

, 0, 0) 𝑛𝑎 =1

√2(√1 −

2𝑚

𝑟, −

1

√1 −2𝑚𝑟

, 0, 0)

𝑚𝑎 =1

√2(0, 0, 𝑟, 𝑖𝑟 sin𝜃) 𝑚𝑎 =

1

√2(0, 0, 𝑟, −𝑖𝑟 sin𝜃)


𝑙𝑡 = 𝑔𝑡𝑡𝑙𝑡 =1

(1 −2𝑚𝑟)

1

√2 √1 −

2𝑚

𝑟=1

√2

1

√1 −2𝑚𝑟

𝑙𝑟 = 𝑔𝑟𝑟𝑙𝑟 = −(1 −2𝑚

𝑟)1

√2

1

√1 −2𝑚𝑟

= −1

√2√1 −

2𝑚

𝑟

𝑙𝜃 = 𝑙𝜙 = 0

𝑛𝑡 = 𝑔𝑡𝑡𝑛𝑡 =1

(1 −2𝑚𝑟 )

1

√2√1 −

2𝑚

𝑟=1

√2

1

√1 −2𝑚𝑟

𝑛𝑟 = 𝑔𝑟𝑟𝑛𝑟 = −(1 −2𝑚

𝑟)1

√2(

−1

√1 −2𝑚𝑟 )

=1

√2√1 −

2𝑚

𝑟

𝑛𝜃 = 𝑛𝜙 = 0

𝑚𝑡 = 𝑚𝑟 = 0

𝑚𝜃 = 𝑔𝜃𝜃𝑚𝜃 = (−1

𝑟2)1

√2𝑟 = −

1

√2

1

𝑟

𝑚𝜙 = 𝑔𝜙𝜙𝑚𝜙 = −1

𝑟2 sin2 𝜃

1

√2𝑖𝑟 sin𝜃 = −𝑖

1

√2

1

𝑟 sin𝜃


𝑙𝑎 =1

√2(√1 −

2𝑚

𝑟,

1

√1 −2𝑚𝑟

, 0, 0) 𝑙𝑎 =1

√2(

1

√1 −2𝑚𝑟

, −√1 −2𝑚

𝑟, 0, 0)





𝑛𝑎 =1

√2(√1 −

2𝑚

𝑟, −

1

√1 −2𝑚𝑟

, 0, 0) 𝑛𝑎 =1

√2(

1

√1 −2𝑚𝑟

, √1 −2𝑚

𝑟, 0, 0)

𝑚𝑎 =1

√2(0, 0, 𝑟, 𝑖𝑟 sin𝜃) 𝑚𝑎 =

1

√2(0, 0, −

1

𝑟, −𝑖

1

𝑟 sin𝜃)

𝑚𝑎 =1

√2(0, 0, 𝑟, −𝑖𝑟 sin𝜃) ��𝑎 =

1

√2(0, 0, −

1

𝑟, 𝑖

1

𝑟 sin 𝜃)



𝑎𝑙𝑏 휀 =1


𝑎𝑙𝑏 − ∇𝑏𝑚𝑎��𝑎𝑙𝑏)

(9.15) 𝜈 = −∇𝑏𝑛𝑎��

𝑎𝑛𝑏 𝜏 = ∇𝑏𝑙𝑎𝑚𝑎𝑛𝑏 𝛾 =

1




𝑎��𝑏 𝛼 =1


𝑎��𝑏 − ∇𝑏𝑚𝑎��𝑎��𝑏)






𝜋 = −∇𝑏𝑛𝑎��𝑎𝑙𝑏

= −∇𝑡𝑛𝑎��𝑎𝑙𝑡 − ∇𝑟𝑛𝑎��

𝑎𝑙𝑟 = −∇𝑡𝑛𝜃��

𝜃𝑙𝑡 − ∇𝑟𝑛𝜃��𝜃𝑙𝑟 − ∇𝑡𝑛𝜙��

𝜙𝑙𝑡 − ∇𝑟𝑛𝜙��𝜙𝑙𝑟

= −(𝜕𝑡𝑛𝜃 − Γ 𝑡𝜃

𝑐 𝑛𝑐)��𝜃𝑙𝑡 − (𝜕𝑟𝑛𝜃 − Γ 𝑟𝜃

𝑐 𝑛𝑐)��𝜃𝑙𝑟 − (𝜕𝑡𝑛𝜙 − Γ 𝑡𝜙

𝑐 𝑛𝑐)��𝜙𝑙𝑡

− (𝜕𝑟𝑛𝜙 − Γ 𝑟𝜙𝑐 𝑛𝑐)��

𝜙𝑙𝑟

= 0 𝜈 = −∇𝑏𝑛𝑎��

𝑎𝑛𝑏 = 0 𝜆 = −∇𝑏𝑛𝑎��

𝑎��𝑏 = −∇𝜃𝑛𝑎��

𝑎��𝜃 − ∇𝜙𝑛𝑎��𝑎��𝜙

= −∇𝜃𝑛𝜃��𝜃��𝜃 − ∇𝜙𝑛𝜃��

𝜃��𝜙 − ∇𝜃𝑛𝜙��𝜙��𝜃 − ∇𝜙𝑛𝜙��

𝜙��𝜙

= −(𝜕𝜃𝑛𝜃 − Γ 𝜃𝜃

𝑐 𝑛𝑐)��𝜃��𝜃 − (𝜕𝜙𝑛𝜃 − Γ 𝜙𝜃

𝑐 𝑛𝑐)��𝜃��𝜙 − (𝜕𝜃𝑛𝜙 − Γ 𝜃𝜙

𝑐 𝑛𝑐)��𝜙��𝜃

− (𝜕𝜙𝑛𝜙 − Γ 𝜙𝜙𝑐 𝑛𝑐)��

𝜙��𝜙

= Γ 𝜃𝜃𝑟 𝑛𝑟��

𝜃��𝜃 + Γ 𝜙𝜙𝑟 𝑛𝑟��

𝜙��𝜙

= −(𝑟 − 2𝑚)𝑛𝑟 (1

√2(−

1

𝑟))

2

− (𝑟 − 2𝑚) sin2 𝜃 𝑛𝑟 (1

√2(𝑖

1

𝑟 sin 𝜃))

2

= 0 𝜇 = −∇𝑏𝑛𝑎��

𝑎𝑚𝑏 = 0 𝜅 = ∇𝑏𝑙𝑎𝑚

𝑎𝑙𝑏 = ∇𝑡𝑙𝑎𝑚

𝑎𝑙𝑡 + ∇𝑟𝑙𝑎𝑚𝑎𝑙𝑟

= ∇𝑡𝑙𝜃𝑚𝜃𝑙𝑡 + ∇𝑟𝑙𝜃𝑚

𝜃𝑙𝑟 + ∇𝑡𝑙𝜙𝑚𝜙𝑙𝑡 + ∇𝑟𝑙𝜙𝑚

𝜙𝑙𝑟

= (𝜕𝑡𝑙𝜃 − Γ 𝑡𝜃𝑐 𝑙𝑐)𝑚

𝜃𝑙𝑡 + (𝜕𝑟𝑙𝜃 − Γ 𝑟𝜃𝑐 𝑙𝑐)𝑚

𝜃𝑙𝑟 + (𝜕𝑡𝑙𝜙 − Γ 𝑡𝜙𝑐 𝑙𝑐)𝑚

𝜙𝑙𝑡 + (𝜕𝑟𝑙𝜙 − Γ 𝑟𝜙𝑐 𝑙𝑐)𝑚

𝜙𝑙𝑟

= 0





𝜏 = ∇𝑏𝑙𝑎𝑚𝑎𝑛𝑏

= 0 𝜌 = ∇𝑏𝑙𝑎𝑚

𝑎��𝑏 = ∇𝜃𝑙𝑎𝑚

𝑎��𝜃 + ∇𝜙𝑙𝑎𝑚𝑎��𝜙

= ∇𝜃𝑙𝜃𝑚𝜃��𝜃 + ∇𝜙𝑙𝜃𝑚

𝜃��𝜙 + ∇𝜃𝑙𝜙𝑚𝜙��𝜃 + ∇𝜙𝑙𝜙𝑚

𝜙��𝜙

= (𝜕𝜃𝑙𝜃 − Γ 𝜃𝜃

𝑐 𝑙𝑐)𝑚𝜃��𝜃 + (𝜕𝜙𝑙𝜃 − Γ 𝜙𝜃

𝑐 𝑙𝑐)𝑚𝜃��𝜙 + (𝜕𝜃𝑙𝜙 − Γ 𝜃𝜙

𝑐 𝑙𝑐)𝑚𝜙��𝜃

+ (𝜕𝜙𝑙𝜙 − Γ 𝜙𝜙𝑐 𝑙𝑐)𝑚

𝜙��𝜙

= −Γ 𝜃𝜃𝑟 𝑙𝑟𝑚

𝜃��𝜃 − Γ 𝜙𝜙𝑟 𝑙𝑟𝑚

𝜙��𝜙

= 𝑟 (1 −

2𝑚

𝑟)1

√2

1

√1 −2𝑚𝑟

(1

√2(−

1

𝑟))

2

+ 𝑟 (1 −2𝑚

𝑟) sin2 𝜃

1

√2

1

√1 −2𝑚𝑟

(1

√2(−𝑖

1

𝑟 sin 𝜃))(

1

√2𝑖

1

𝑟 sin𝜃)

=1

√2√(1 −

2𝑚

𝑟)1

𝑟

𝜎 = ∇𝑏𝑙𝑎𝑚𝑎𝑚𝑏


𝜃𝑚𝜃 − Γ 𝜙𝜙𝑟 𝑙𝑟𝑚

𝜙𝑚𝜙

= 0

휀 =1



=1

2(∇𝑡𝑙𝑎𝑛

𝑎𝑙𝑡 − ∇𝑡𝑚𝑎��𝑎𝑙𝑡) +

1

2(∇𝑟𝑙𝑎𝑛

𝑎𝑙𝑟 − ∇𝑟𝑚𝑎��𝑎𝑙𝑟)

=1

2(∇𝑡𝑙𝑡𝑛

𝑡𝑙𝑡 − ∇𝑡𝑚𝜃��𝜃𝑙𝑡) +

1

2(∇𝑟𝑙𝑡𝑛

𝑡𝑙𝑟 − ∇𝑟𝑚𝜃��𝜃𝑙𝑟) +

1

2(∇𝑡𝑙𝑟𝑛

𝑟𝑙𝑡 − ∇𝑡𝑚𝜙��𝜙𝑙𝑡)

+1

2(∇𝑟𝑙𝑟𝑛

𝑟𝑙𝑟 − ∇𝑟𝑚𝜙��𝜙𝑙𝑟)

=1

2((𝜕𝑡𝑙𝑡 − Γ 𝑡𝑡

𝑐 𝑙𝑐)𝑛𝑡𝑙𝑡 − (𝜕𝑡𝑚𝜃 − Γ 𝑡𝜃

𝑐 )��𝜃𝑙𝑡) +1

2((𝜕𝑟𝑙𝑡 − Γ 𝑟𝑡

𝑐 𝑙𝑐)𝑛𝑡𝑙𝑟 − (𝜕𝑟𝑚𝜃 − Γ 𝑟𝜃

𝑐 )��𝜃𝑙𝑟)

+1

2((𝜕𝑡𝑙𝑟 − Γ 𝑡𝑟

𝑐 𝑙𝑐)𝑛𝑟𝑙𝑡 − (𝜕𝑡𝑚𝜙 − Γ 𝑡𝜙

𝑐 )��𝜙𝑙𝑡)

+1

2((𝜕𝑟𝑙𝑟 − Γ 𝑟𝑟

𝑐 𝑙𝑐)𝑛𝑟𝑙𝑟 − (𝜕𝑟𝑚𝜙 − Γ 𝑟𝜙

𝑐 )��𝜙𝑙𝑟)

=1

2((−Γ 𝑡𝑡

𝑟 𝑙𝑟)𝑛𝑡𝑙𝑡) +

1


𝑡 𝑙𝑡)𝑛𝑡𝑙𝑟 − (𝜕𝑟𝑚𝜃 − Γ 𝑟𝜃

𝜃 𝑚𝜃)��𝜃𝑙𝑟) +

1

2((−Γ 𝑡𝑟

𝑡 𝑙𝑡)𝑛𝑟𝑙𝑡)

+1


𝑟 𝑙𝑟)𝑛𝑟𝑙𝑟 − (𝜕𝑟𝑚𝜙 − Γ 𝑟𝜙

𝜙𝑚𝜙) ��

𝜙𝑙𝑟)

=1

4((−Γ 𝑡𝑡

𝑟 𝑙𝑟)1

1 −2𝑚𝑟

)+1

4(−(𝜕𝑟𝑙𝑡 − Γ 𝑟𝑡

𝑡 𝑙𝑡) − (𝜕𝑟𝑚𝜃 − Γ 𝑟𝜃𝜃 𝑚𝜃) √1 −

2𝑚

𝑟

1

𝑟) +

1

4((−Γ 𝑡𝑟

𝑡 𝑙𝑡))

+1

4(−(𝜕𝑟𝑙𝑟 − Γ 𝑟𝑟

𝑟 𝑙𝑟) (1 −2𝑚

𝑟) + (𝜕𝑟𝑚𝜙 − Γ 𝑟𝜙

𝜙𝑚𝜙) √1 −

2𝑚

𝑟

𝑖

𝑟 sin𝜃)





=1

252

(

(

−𝑚

𝑟2(1 −

2𝑚

𝑟)

1

√1 −2𝑚𝑟 )

1

1 −2𝑚𝑟)

+1

252

(−(𝜕𝑟 (√1 −2𝑚

𝑟) −

𝑚

𝑟21

1 −2𝑚𝑟

√1 −2𝑚

𝑟) − (𝜕𝑟(𝑟) −

1

𝑟𝑟) √1 −

2𝑚

𝑟

1

𝑟)

+1

252

(−𝑚

𝑟21

1 −2𝑚𝑟

√1 −2𝑚

𝑟)

+1

25

2

(

−

(

𝜕𝑟

(

1

√1 −2𝑚𝑟 )

+𝑚

𝑟2

1

1 −2𝑚𝑟

1

√1 −2𝑚𝑟 )

(1 −2𝑚

𝑟)

+ (𝜕𝑟(𝑖𝑟 sin 𝜃) −1

𝑟𝑖𝑟 sin 𝜃) √1 −

2𝑚

𝑟

𝑖

𝑟 sin 𝜃

)

=1

252(

−𝑚

𝑟21

√1 −2𝑚𝑟 )

+1

252

(

−

(

𝑚

𝑟21

√1 −2𝑚𝑟

−𝑚

𝑟21

√1 −2𝑚𝑟 )

)

+1

252(

−𝑚

𝑟21

√1 −2𝑚𝑟 )

+1

252

(

−

(

−𝑚

𝑟21

(1 −2𝑚𝑟)

32

+𝑚

𝑟21

(1 −2𝑚𝑟)

32

)

(1 −

2𝑚

𝑟)

)

= −

1

232

𝑚

𝑟21

√1 −2𝑚𝑟

𝛾 =1



=1


𝑎𝑛𝑡 − ∇𝑡𝑚𝑎��𝑎𝑛𝑡) +

1


𝑎𝑛𝑟 − ∇𝑟𝑚𝑎��𝑎𝑛𝑟)

=1


𝑡𝑛𝑡) +1


𝑡𝑛𝑟) +1


𝑟𝑛𝑡) +1


𝑟𝑛𝑟)

=1

2(𝜕𝑡𝑙𝑡 − Γ 𝑡𝑡

𝑐 𝑙𝑐)𝑛𝑡𝑛𝑡 +

1

2(𝜕𝑟𝑙𝑡 − Γ 𝑟𝑡

𝑐 𝑙𝑐)𝑛𝑡𝑛𝑟 +

1

2(𝜕𝑡𝑙𝑟 − Γ 𝑡𝑟

𝑐 𝑙𝑐)𝑛𝑟𝑛𝑡 +

1

2(𝜕𝑟𝑙𝑟 − Γ 𝑟𝑟

𝑐 𝑙𝑐)𝑛𝑟𝑛𝑟

=1

2(−Γ 𝑡𝑡

𝑟 𝑙𝑟)𝑛𝑡𝑛𝑡 +

1

2(𝜕𝑟𝑙𝑡 − Γ 𝑟𝑡

𝑡 𝑙𝑡)𝑛𝑡𝑛𝑟 +

1

2(−Γ 𝑡𝑟

𝑡 𝑙𝑡)𝑛𝑟𝑛𝑡 +

1

2(𝜕𝑟𝑙𝑟 − Γ 𝑟𝑟

𝑟 𝑙𝑟)𝑛𝑟𝑛𝑟

= −

1

2Γ 𝑡𝑡𝑟 𝑙𝑟(𝑛

𝑡)2 +1

2𝜕𝑟𝑙𝑡𝑛

𝑡𝑛𝑟 − Γ 𝑟𝑡𝑡 𝑙𝑡𝑛

𝑡𝑛𝑟 +1

2𝜕𝑟𝑙𝑟(𝑛

𝑟)2 −1

2Γ 𝑟𝑟𝑟 𝑙𝑟(𝑛

𝑟)2

= −

1

252

1

√1 −2𝑚𝑟

𝑚

𝑟2+1

252

1

√1 −2𝑚𝑟

𝑚

𝑟2−1

232

1

√1 −2𝑚𝑟

𝑚

𝑟2−1

252

1

√1 −2𝑚𝑟

𝑚

𝑟2+1

252

1

√1 −2𝑚𝑟

𝑚

𝑟2

= −

1

232

1

√1 −2𝑚𝑟

𝑚

𝑟2





𝛼 =1


𝑎��𝑏 − ∇𝑏𝑚𝑎��𝑎��𝑏)

=1

2(∇𝜃𝑙𝑎𝑛

𝑎��𝜃 − ∇𝜃𝑚𝑎��𝑎��𝜃) +

1

2(∇𝜙𝑙𝑎𝑛

𝑎��𝜙 − ∇𝜙𝑚𝑎��𝑎��𝜙)

=1

2(∇𝜃𝑙𝑡𝑛

𝑡��𝜃 − ∇𝜃𝑚𝜃��𝜃��𝜃) +

1

2(∇𝜙𝑙𝑡𝑛

𝑡��𝜙 − ∇𝜙𝑚𝜃��𝜃��𝜙) +

1

2(∇𝜃𝑙𝑟𝑛

𝑟��𝜃 − ∇𝜃𝑚𝜙��𝜙��𝜃)

+1

2(∇𝜙𝑙𝑟𝑛

𝑟��𝜙 − ∇𝜙𝑚𝜙��𝜙��𝜙)

=1

2((𝜕𝜃𝑙𝑡 − Γ 𝜃𝑡

𝑐 𝑙𝑐)𝑛𝑡��𝜃 − (𝜕𝜃𝑚𝜃 − Γ 𝜃𝜃

𝑐 𝑚𝑐)��𝜃��𝜃)

+1

2((𝜕𝜙𝑙𝑡 − Γ 𝜙𝑡

𝑐 𝑙𝑐)𝑛𝑡��𝜙 − (𝜕𝜙𝑚𝜃 − Γ 𝜙𝜃

𝑐 𝑚𝑐)��𝜃��𝜙)

+1

2((𝜕𝜃𝑙𝑟 − Γ 𝜃𝑟

𝑐 𝑙𝑐)𝑛𝑟��𝜃 − (𝜕𝜃𝑚𝜙 − Γ 𝜃𝜙

𝑐 𝑚𝑐)��𝜙��𝜃)

+1

2((𝜕𝜙𝑙𝑟 − Γ 𝜙𝑟

𝑐 𝑙𝑐)𝑛𝑟��𝜙 − (𝜕𝜙𝑚𝜙 − Γ 𝜙𝜙

𝑐 𝑚𝑐)��𝜙��𝜙)

=1

2Γ 𝜙𝜃𝜙

𝑚𝜙��𝜃��𝜙 +

1

2(−(𝜕𝜃𝑚𝜙 − Γ 𝜃𝜙


𝜙��𝜃) +1

2Γ 𝜙𝜙𝜃 𝑚𝜃��

𝜙��𝜙

=1

252

1

𝑟cot 𝜃 +

1

2(−(𝜕𝜃(𝑖𝑟 sin𝜃) − cot 𝜃 (𝑖𝑟 sin 𝜃))��

𝜙��𝜃) +1

252

cos 𝜃 sin 𝜃1

𝑟

1

sin2 𝜃

=1

232

1

𝑟cot 𝜃

𝛽 =1



=1

2Γ 𝜙𝜃𝜙

𝑚𝜙��𝜃𝑚𝜙 +

1

2(−(𝜕𝜃𝑚𝜙 − Γ 𝜃𝜙


𝜙𝑚𝜃) +1

2Γ 𝜙𝜙𝜃 𝑚𝜃��

𝜙𝑚𝜙

= −1

252

1

𝑟cot 𝜃 +

1

2(−(𝜕𝜃(𝑖𝑟 sin 𝜃) − cot 𝜃 (𝑖𝑟 sin 𝜃))��

𝜙𝑚𝜃) −1

252

1

𝑟cot 𝜃

= −1

232

1

𝑟cot 𝜃


𝜋 = 0 𝜅 = 0 휀 = −1

232

1

√1 −2𝑚𝑟

𝑚

𝑟2

𝜈 = 0 𝜏 = 0 𝛾 = −1

232

1

√1 −2𝑚𝑟

𝑚

𝑟2

𝜆 = 0 𝜌 =1

√2√(1 −

2𝑚

𝑟)1

𝑟 𝛼 =

1

232

1

𝑟cot 𝜃

𝜇 = 0 𝜎 = 0 𝛽 = −1

232

1

𝑟cot 𝜃


Ψ0 = 𝐷𝜎 − 𝛿𝜅 − 𝜎(𝜌 + ��) − 𝜎(3휀 − 휀) + 𝜅(𝜋 − �� + �� + 3𝛽) (13.22)





Ψ1 = 𝐷𝛽 − 𝛿휀 − 𝜎(𝛼 + 𝜋) − 𝛽(�� − 휀) + 𝜅(𝜇 + 𝛾) + 휀(�� − ��) (13.23) Ψ2 = 𝛿𝜏 − Δ𝜌 − 𝜌�� − 𝜎𝜆 + 𝜏(�� − 𝛼 − ��) + 𝜌(𝛾 + ��) + 𝜅𝜈 − 2Λ (13.24)

Ψ3 = 𝛿𝛾 − Δα + 𝜈(𝜌 + 휀) − 𝜆(𝜏 + 𝛽) + 𝛼(�� − ��) + 𝛾(�� − ��) (13.25)

Ψ4 = 𝛿𝜈 − Δλ + 𝜆(𝜇 + ��) − 𝜆(3𝛾 − ��) + 𝜈(3𝛼 + �� + 𝜋 − ��) (13.26)


Ψ0 = 0 Ψ1 = 𝐷𝛽 − 𝛿휀 − 𝛽��

= 𝑙𝑎∇𝑎𝛽 −𝑚𝑎∇𝑎휀 − 𝛽��

= 𝑙𝑎𝜕𝑎𝛽 −𝑚𝑎𝜕𝑎휀 − 𝛽��

= (−1

√2√1 −

2𝑚

𝑟)𝜕𝑟 (−

1

232

1

𝑟cot 𝜃) − (−

1

232

1

𝑟cot 𝜃)(

1

√2√(1 −

2𝑚

𝑟)1

𝑟)

=1

4(√1 −

2𝑚

𝑟)cot 𝜃 [𝜕𝑟 (

1

𝑟) +

1

𝑟2]

= 0 Ψ2 = −Δ𝜌 + 2𝜌𝛾

= −𝑛𝑎∇𝑎𝜌 + 2𝜌𝛾 = −𝑛𝑎𝜕𝑎𝜌 + 2𝜌𝛾

= −(1

√2√1 −

2𝑚

𝑟)𝜕𝑟 (

1

√2√(1 −

2𝑚

𝑟)1

𝑟) + 2(

1

√2√(1 −

2𝑚

𝑟)1

𝑟)

(

−1

232

1

√1 −2𝑚𝑟

𝑚

𝑟2

)

= −(1

2√1 −

2𝑚

𝑟)

(

𝑚

𝑟31

√1 −2𝑚𝑟

− √(1 −2𝑚

𝑟)1

𝑟2

)

−1

2

𝑚

𝑟3

=1

2𝑟2−2𝑚

𝑟3

Ψ3 = 𝛿𝛾 − Δα = ��𝑎𝜕𝑎𝛾 − 𝑛

𝑎𝜕𝑎α

= −(1

√2√1 −

2𝑚

𝑟)𝜕𝑟 (

1

232

1

𝑟cot 𝜃)

= 1471

4cot 𝜃√1 −

2𝑚

𝑟

1

𝑟2

Ψ4 = 0 Ψ2 ≠ 0: This is a Petrov type D, which means there are two principal null directions. The Petrov type D is associated with the gravitational field of a star or a black hole. The two principal null directions correspond to ingoing and outgoing congruence of light rays.

147 This should be 0. However, the result is reproduced in FE-12 concerning the Reissner-Nordström metric.





9.11 148The deflection of a light ray in a Schwarzschild metric with two different

masses The metric

𝑑𝑠2 = (1 −

2𝑚1

𝑟)𝑑𝑡2 − (1 −

2𝑚2

𝑟)−1


Copying the method pp 224-229

0 = (1 −2𝑚1

𝑟) ��2 − (1 −

2𝑚2

𝑟)−1

��2 − 𝑟2��2 (10.52’)

𝑒2 = 149 (1 −

2𝑚1

𝑟)2

��2 𝑙2 = 𝑟4𝜙2

⇒ 0 = (1 −2𝑚1

𝑟)−1

𝑒2 − (1 −2𝑚2

𝑟)−1

��2 −𝑙2

𝑟2 (10.53’)

�� = 𝑟′�� =

𝑟′𝑙

𝑟2= −𝑙𝑢′ if 𝑢′ = (

1

𝑟)′

= −1

𝑟2𝑟′

⇒ 0 = 𝑒2 − (1 −2𝑚1

𝑟) (1 −

2𝑚2

𝑟)−1

𝑙2𝑢′2− (1 −

2𝑚1

𝑟) 𝑙2𝑢2

= 𝑒2 −(1 − 2𝑚1𝑢)

(1 − 2𝑚2𝑢)𝑙2𝑢′

2− (1 − 2𝑚1𝑢)𝑙

2𝑢2

Differentiating with respect to 𝑟

⇒ 0 =

2𝑚1𝑢′

(1 − 2𝑚2𝑢)𝑙2𝑢′

2− 2𝑚2𝑢

′(1 − 2𝑚1𝑢)

(1 − 2𝑚2𝑢)2𝑙2𝑢′

2− 2𝑢′𝑢′′

(1 − 2𝑚1𝑢)

(1 − 2𝑚2𝑢)𝑙2 + 2𝑚1𝑢

′𝑙2𝑢2

− 2𝑢′𝑢(1 − 2𝑚1𝑢)𝑙2

= 𝑢′𝑙2 (2𝑚1

(1 − 2𝑚2𝑢)𝑢′2− 2𝑚2

(1 − 2𝑚1𝑢)

(1 − 2𝑚2𝑢)2𝑢′2− 2𝑢′′

(1 − 2𝑚1𝑢)

(1 − 2𝑚2𝑢)+ 2𝑚1𝑢

2

− 2𝑢(1 − 2𝑚1𝑢))

⇒ 0 = 𝑚1(1 − 2𝑚2𝑢)𝑢′2 −𝑚2(1 − 2𝑚1𝑢)𝑢

′2 − (1 − 2𝑚1𝑢)(1 − 2𝑚2𝑢)𝑢′′

+𝑚1(1 − 2𝑚2𝑢)2𝑢2 − (1 − 2𝑚1𝑢)(1 − 2𝑚2𝑢)

2𝑢

= (𝑚1 −𝑚2)𝑢′2 − (1 − 2𝑚1𝑢)(1 − 2𝑚2𝑢)𝑢

′′ +𝑚1(1 − 2𝑚2𝑢)2𝑢2

− (1 − 2𝑚1𝑢)(1 − 2𝑚2𝑢)2𝑢

= (𝑚1 −𝑚2)𝑢′2 − (1 − 2𝑚1𝑢)(1 − 2𝑚2𝑢)𝑢

′′ + (3𝑚1𝑢 − 1)(1 − 2𝑚2𝑢)2𝑢

In the limit where 𝑚𝑢 =𝑚

𝑟→ 0

⇒ 0 = (𝑚1 −𝑚2)𝑢′2 − 𝑢′′ + 𝑢

From which we can conclude, that the deflection is somehow dependent on (𝑚1 −𝑚2)

9.12 150The non-zero Weyl scalars of the Reissner-Nordström spacetime The Reissner-Nordström spacetime is a static solution to the Einstein-Maxwell field equations151, which

corresponds to the gravitational field of a charged non-rotating, spherically symmetric of mass 𝑚152.

The metric: 𝑑𝑠2 = (1 −2𝑚

𝑟+𝑒2

𝑟2)𝑑𝑡2 − (1 −

2𝑚

𝑟+𝑒2

𝑟2)

−1


148 (McMahon, 2006, p. 326) final exam 13 149 Killing vectors p.220 150 (McMahon, 2006, p. 325), final exam 12, If 𝑒2 = 0 the results from Quiz 10-5 are reproduced 151 http://en.wikipedia.org/wiki/Einstein-Maxwell_equations#Einstein.E2.80.93Maxwell_equations 152 http://en.wikipedia.org/wiki/Reissner%E2%80%93Nordstr%C3%B6m_metric


http://en.wikipedia.org/wiki/Einstein-Maxwell_equations#Einstein.E2.80.93Maxwell_equations

http://en.wikipedia.org/wiki/Reissner%E2%80%93Nordstr%C3%B6m_metric




The metric tensor:

𝑔𝑎𝑏 =

{

(1 −

2𝑚

𝑟+𝑒2

𝑟2)

−(1 −2𝑚

𝑟+𝑒2

𝑟2)

−1

−𝑟2

−𝑟2 sin2 𝜃}

and its in-verse:

𝑔𝑎𝑏 =

{

(1 −

2𝑚

𝑟+𝑒2

𝑟2)

−1

−(1 −2𝑚

𝑟+𝑒2

𝑟2)

−1

𝑟2

−1

𝑟2 sin2 𝜃}

The Christoffel symbols

To find the geodesic we use the Euler-Lagrange equation

0 =𝑑

𝑑𝑠(𝜕𝐹

𝜕��𝑎) −

𝜕𝐹

𝜕𝑥𝑎 (10.36)

where

𝐹 = (1 −2𝑚

𝑟+𝑒2

𝑟2) ��2 − (1 −

2𝑚

𝑟+𝑒2

𝑟2)

−1

��2 − 𝑟2��2 − 𝑟2 sin2 𝜃 ��2

𝑥𝑎 = 𝑡:

𝜕𝐹

𝜕𝑡 = 0

𝜕𝐹

𝜕�� = 2(1 −

2𝑚

𝑟+𝑒2

𝑟2) ��

𝑑

𝑑𝑠(𝜕𝐹

𝜕��) = 4(

𝑚𝑟 − 𝑒2

𝑟3) �� + 2(1 −

2𝑚

𝑟+𝑒2

𝑟2) ��

⇒ 0 = 4(𝑚𝑟 − 𝑒2

𝑟3) �� + 2(1 −

2𝑚

𝑟+𝑒2

𝑟2) ��

⇔ 0 = �� + 2(1 −2𝑚

𝑟+𝑒2

𝑟2)

−1

(𝑚𝑟 − 𝑒2

𝑟3) ��

𝑥𝑎 = 𝑟:

𝜕𝐹

𝜕𝑟 = (

2𝑚

𝑟2−2𝑒2

𝑟3) ��2 + (1 −

2𝑚

𝑟+𝑒2

𝑟2)

−2

(2𝑚

𝑟2−2𝑒2

𝑟3) ��2 − 2𝑟��2 − 2𝑟 sin2 𝜃 ��2

= 2(𝑚𝑟 − 𝑒2

𝑟3) ��2 + 2(1 −

2𝑚

𝑟+𝑒2

𝑟2)

−2

(𝑚𝑟 − 𝑒2

𝑟3) ��2 − 2𝑟��2 − 2𝑟 sin2 𝜃 ��2

𝜕𝐹

𝜕�� = −2(1 −

2𝑚

𝑟+𝑒2

𝑟2)

−1

��

𝑑

𝑑𝑠(𝜕𝐹

𝜕��) = −2(1 −

2𝑚

𝑟+𝑒2

𝑟2)

−1

�� + 2(1 −2𝑚

𝑟+𝑒2

𝑟2)

−2

(2𝑚

𝑟2−2𝑒2

𝑟3) ��2





= −2(1 −2𝑚

𝑟+𝑒2

𝑟2)

−1

�� + 4(1 −2𝑚

𝑟+𝑒2

𝑟2)

−2

(𝑚𝑟 − 𝑒2

𝑟3) ��2

⇒ 0

= −2(1 −2𝑚

𝑟+𝑒2

𝑟2)

−1

�� + 4(1 −2𝑚

𝑟+𝑒2

𝑟2)

−2

(𝑚𝑟 − 𝑒2

𝑟3) ��2 − 2(

𝑚𝑟 − 𝑒2

𝑟3) ��2

− 2(1 −2𝑚

𝑟+𝑒2

𝑟2)

−2

(𝑚𝑟 − 𝑒2

𝑟3) ��2 + 2𝑟��2 + 2𝑟 sin2 𝜃 ��2

= −2(1 −2𝑚

𝑟+𝑒2

𝑟2)

−1

�� + 2(1 −2𝑚

𝑟+𝑒2

𝑟2)

−2

(𝑚𝑟 − 𝑒2

𝑟3) ��2 − 2(

𝑚𝑟 − 𝑒2

𝑟3) ��2

+ 2𝑟��2 + 2𝑟 sin2 𝜃 ��2

⇔ 0 = �� − (1 −

2𝑚

𝑟+𝑒2

𝑟2)

−1

(𝑚𝑟 − 𝑒2

𝑟3) ��2 + (1 −

2𝑚

𝑟+𝑒2

𝑟2)(𝑚𝑟 − 𝑒2

𝑟3) ��2

− 𝑟 (1 −2𝑚

𝑟+𝑒2

𝑟2) ��2 − 𝑟 (1 −

2𝑚

𝑟+𝑒2

𝑟2)sin2 𝜃 ��2

𝑥𝑎 = 𝜃:

𝜕𝐹

𝜕𝜃 = −2𝑟2 cos 𝜃 sin𝜃 ��2

𝜕𝐹

𝜕�� = −2𝑟2��

𝑑

𝑑𝑠(𝜕𝐹

𝜕��) = −4𝑟�� − 2𝑟2��

⇒ 0 = −4𝑟�� − 2𝑟2�� + 2𝑟2 cos 𝜃 sin 𝜃 ��2

⇔ 0 = �� +2

𝑟�� − cos𝜃 sin𝜃 ��2 (10.39)

𝑥𝑎 = 𝜙:

𝜕𝐹

𝜕𝜙 = 0

𝜕𝐹

𝜕�� = −2𝑟2 sin2 𝜃 ��

𝑑

𝑑𝑠(𝜕𝐹


⇒ 0 = −4𝑟 sin2 𝜃 �� − 4𝑟2 cos 𝜃 sin 𝜃 �� − 2𝑟2 sin2 𝜃 ��

⇔ 0 = �� +2

𝑟�� + 2 cot 𝜃 �� (10.40)


0 = �� + 2(1 −2𝑚

𝑟+𝑒2

𝑟2)

−1

(𝑚𝑟 − 𝑒2

𝑟3) ��

0 = �� − (1 −

2𝑚

𝑟+𝑒2

𝑟2)

−1

(𝑚𝑟 − 𝑒2

𝑟3) ��2 + (1 −

2𝑚

𝑟+𝑒2

𝑟2)(𝑚𝑟 − 𝑒2

𝑟3) ��2 − 𝑟 (1 −

2𝑚

𝑟+𝑒2

𝑟2) ��2

− 𝑟 (1 −2𝑚

𝑟+𝑒2

𝑟2) sin2 𝜃 ��2

0 = �� +2


0 = �� +2

𝑟�� + 2 cot 𝜃 ��






Γ 𝑟𝑡𝑡 = (1 −

2𝑚

𝑟+𝑒2

𝑟2)

−1

(𝑚𝑟 − 𝑒2

𝑟3)

Γ 𝑟𝑟𝑟 = −(1 −

2𝑚

𝑟+𝑒2

𝑟2)

−1

(𝑚𝑟 − 𝑒2

𝑟3)

Γ 𝑡𝑡𝑟 = (1 −

2𝑚

𝑟+𝑒2

𝑟2)(𝑚𝑟 − 𝑒2

𝑟3)

Γ 𝜃𝜃𝑟 = −𝑟(1 −

2𝑚

𝑟+𝑒2

𝑟2)

Γ 𝜙𝜙𝑟 = −𝑟(1 −

2𝑚

𝑟+𝑒2

𝑟2) sin2 𝜃

Γ 𝑟𝜃𝜃 =

1

𝑟

Γ 𝜙𝜙𝜃 = −cos𝜃 sin𝜃

Γ 𝑟𝜙𝜙

=1

𝑟

Γ 𝜃𝜙𝜙

= cot 𝜃

The basis one forms

𝜔�� = (1 −2𝑚

𝑟+𝑒2

𝑟2)

12

𝑑𝑡 𝑑𝑡 = (1 −2𝑚

𝑟+𝑒2

𝑟2)

−12

𝜔��

𝜂𝑖𝑗 = {

1−1

−1−1

} 𝜔�� = (1 −

2𝑚

𝑟+𝑒2

𝑟2)

−12

𝑑𝑟 𝑑𝑟 = (1 −2𝑚

𝑟+𝑒2

𝑟2)

12

𝜔��


𝑟𝜔��









(

𝑙𝑛𝑚��

)

=1

√2(

1 1 0 01 −1 0 00 0 1 𝑖0 0 1 −𝑖

)

(

𝜔��

𝜔��

𝜔��

𝜔��)

=1

√2(

𝜔�� + 𝜔��

𝜔�� − 𝜔��

𝜔�� + 𝑖𝜔��

𝜔�� − 𝑖𝜔��)

=1

√2

(

(1 −

2𝑚

𝑟+𝑒2

𝑟2)

12

𝑑𝑡 + (1 −2𝑚

𝑟+𝑒2

𝑟2)

−12

𝑑𝑟

(1 −2𝑚

𝑟+𝑒2

𝑟2)

12

𝑑𝑡 − (1 −2𝑚

𝑟+𝑒2

𝑟2)

−12

𝑑𝑟

𝑟𝑑𝜃 + 𝑖𝑟 sin 𝜃 𝑑𝜙𝑟𝑑𝜃 − 𝑖𝑟 sin 𝜃 𝑑𝜙 )


𝑙𝑎 =1

√2((1 −

2𝑚

𝑟+𝑒2

𝑟2)

12

, (1 −2𝑚

𝑟+𝑒2

𝑟2)

−12

, 0, 0)

𝑛𝑎 =1

√2((1 −

2𝑚

𝑟+𝑒2

𝑟2)

12

, − (1 −2𝑚

𝑟+𝑒2

𝑟2)

−12

, 0, 0)

𝑚𝑎 =1

√2(0, 0, 𝑟, 𝑖𝑟 sin 𝜃)

𝑚𝑎 =1

√2(0, 0, 𝑟, −𝑖𝑟 sin 𝜃)


𝑙𝑡 = 𝑔𝑡𝑡𝑙𝑡 = (1 −2𝑚

𝑟+𝑒2

𝑟2)

−11

√2 (1 −

2𝑚

𝑟+𝑒2

𝑟2)

12

=1

√2(1 −

2𝑚

𝑟+𝑒2

𝑟2)

−12

𝑙𝑟 = 𝑔𝑟𝑟𝑙𝑟 = −(1 −2𝑚

𝑟+𝑒2

𝑟2)1

√2(1 −

2𝑚

𝑟+𝑒2

𝑟2)

−12

= −1

√2(1 −

2𝑚

𝑟+𝑒2

𝑟2)

12

𝑙𝜃 = 𝑙𝜙 = 0

𝑛𝑡 = 𝑔𝑡𝑡𝑛𝑡 = (1 −2𝑚

𝑟+𝑒2

𝑟2)

−11

√2(1 −

2𝑚

𝑟+𝑒2

𝑟2)

12

=1

√2(1 −

2𝑚

𝑟+𝑒2

𝑟2)

−12

𝑛𝑟 = 𝑔𝑟𝑟𝑛𝑟 = −(1 −2𝑚

𝑟+𝑒2

𝑟2)1

√2(−(1 −

2𝑚

𝑟+𝑒2

𝑟2)

−12

) =1

√2(1 −

2𝑚

𝑟+𝑒2

𝑟2)

12

𝑛𝜃 = 𝑛𝜙 = 0

𝑚𝑡 = 𝑚𝑟 = 0

𝑚𝜃 = 𝑔𝜃𝜃𝑚𝜃 = (−1

𝑟2)1

√2𝑟 = −

1

√2

1

𝑟

𝑚𝜙 = 𝑔𝜙𝜙𝑚𝜙 = −1

𝑟2 sin2 𝜃

1

√2𝑖𝑟 sin𝜃 = −𝑖

1

√2

1

𝑟 sin𝜃


𝑙𝑎 =1

√2((1 −

2𝑚

𝑟+𝑒2

𝑟2)

12

, (1 −2𝑚

𝑟+𝑒2

𝑟2)

−12

, 0, 0) 𝑙𝑎 =1

√2((1 −

2𝑚

𝑟+𝑒2

𝑟2)

−12

, − (1 −2𝑚

𝑟+𝑒2

𝑟2)

12

, 0, 0)

𝑛𝑎 =1

√2((1 −

2𝑚

𝑟+𝑒2

𝑟2)

12

, − (1 −2𝑚

𝑟+𝑒2

𝑟2)

−12

, 0, 0) 𝑛𝑎 =1

√2((1 −

2𝑚

𝑟+𝑒2

𝑟2)

−12

, (1 −2𝑚

𝑟+𝑒2

𝑟2)

12

, 0, 0)





𝑚𝑎 =1

√2(0, 0, 𝑟, 𝑖𝑟 sin𝜃) 𝑚𝑎 =

1

√2(0, 0, −

1

𝑟, −𝑖

1

𝑟 sin𝜃)

𝑚𝑎 =1

√2(0, 0, 𝑟, −𝑖𝑟 sin𝜃) ��𝑎 =

1

√2(0, 0, −

1

𝑟, 𝑖

1

𝑟 sin 𝜃)



𝑎𝑙𝑏 휀 =1



(9.15) 𝜈 = −∇𝑏𝑛𝑎��


1




𝑎��𝑏 𝛼 =1


𝑎��𝑏 − ∇𝑏𝑚𝑎��𝑎��𝑏)







= −∇𝑡𝑛𝑎��𝑎𝑙𝑡 − ∇𝑟𝑛𝑎��

𝑎𝑙𝑟 = −∇𝑡𝑛𝜃��

𝜃𝑙𝑡 − ∇𝑟𝑛𝜃��𝜃𝑙𝑟 − ∇𝑡𝑛𝜙��

𝜙𝑙𝑡 − ∇𝑟𝑛𝜙��𝜙𝑙𝑟

= −(𝜕𝑡𝑛𝜃 − Γ 𝑡𝜃

𝑐 𝑛𝑐)��𝜃𝑙𝑡 − (𝜕𝑟𝑛𝜃 − Γ 𝑟𝜃

𝑐 𝑛𝑐)��𝜃𝑙𝑟 − (𝜕𝑡𝑛𝜙 − Γ 𝑡𝜙

𝑐 𝑛𝑐)��𝜙𝑙𝑡

− (𝜕𝑟𝑛𝜙 − Γ 𝑟𝜙𝑐 𝑛𝑐)��

𝜙𝑙𝑟

= 0 𝜈 = −∇𝑏𝑛𝑎��

𝑎𝑛𝑏 = 0 𝜆 = −∇𝑏𝑛𝑎��

𝑎��𝑏 = −∇𝜃𝑛𝑎��

𝑎��𝜃 − ∇𝜙𝑛𝑎��𝑎��𝜙

= −∇𝜃𝑛𝜃��𝜃��𝜃 − ∇𝜙𝑛𝜃��

𝜃��𝜙 − ∇𝜃𝑛𝜙��𝜙��𝜃 − ∇𝜙𝑛𝜙��

𝜙��𝜙

= −(𝜕𝜃𝑛𝜃 − Γ 𝜃𝜃

𝑐 𝑛𝑐)��𝜃��𝜃 − (𝜕𝜙𝑛𝜃 − Γ 𝜙𝜃

𝑐 𝑛𝑐)��𝜃��𝜙 − (𝜕𝜃𝑛𝜙 − Γ 𝜃𝜙

𝑐 𝑛𝑐)��𝜙��𝜃

− (𝜕𝜙𝑛𝜙 − Γ 𝜙𝜙𝑐 𝑛𝑐)��

𝜙��𝜙

= Γ 𝜃𝜃𝑟 𝑛𝑟��

𝜃��𝜃 + Γ 𝜙𝜙𝑟 𝑛𝑟��

𝜙��𝜙

= −𝑟(1 −2𝑚

𝑟+𝑒2

𝑟2)𝑛𝑟 (

1

√2(−

1

𝑟))

2

− 𝑟 (1 −2𝑚

𝑟+𝑒2

𝑟2) sin2 𝜃 𝑛𝑟 (

1

√2(𝑖

1

𝑟 sin𝜃))

2

= 0 𝜇 = −∇𝑏𝑛𝑎��


𝑎𝑙𝑏 = ∇𝑡𝑙𝑎𝑚

𝑎𝑙𝑡 + ∇𝑟𝑙𝑎𝑚𝑎𝑙𝑟

= ∇𝑡𝑙𝜃𝑚𝜃𝑙𝑡 + ∇𝑟𝑙𝜃𝑚

𝜃𝑙𝑟 + ∇𝑡𝑙𝜙𝑚𝜙𝑙𝑡 + ∇𝑟𝑙𝜙𝑚

𝜙𝑙𝑟

= (𝜕𝑡𝑙𝜃 − Γ 𝑡𝜃𝑐 𝑙𝑐)𝑚

𝜃𝑙𝑡 + (𝜕𝑟𝑙𝜃 − Γ 𝑟𝜃𝑐 𝑙𝑐)𝑚

𝜃𝑙𝑟 + (𝜕𝑡𝑙𝜙 − Γ 𝑡𝜙𝑐 𝑙𝑐)𝑚

𝜙𝑙𝑡 + (𝜕𝑟𝑙𝜙 − Γ 𝑟𝜙𝑐 𝑙𝑐)𝑚

𝜙𝑙𝑟

= 0 𝜏 = ∇𝑏𝑙𝑎𝑚

𝑎𝑛𝑏 = 0 𝜌 = ∇𝑏𝑙𝑎𝑚

𝑎��𝑏 = ∇𝜃𝑙𝑎𝑚

𝑎��𝜃 + ∇𝜙𝑙𝑎𝑚𝑎��𝜙

= ∇𝜃𝑙𝜃𝑚𝜃��𝜃 + ∇𝜙𝑙𝜃𝑚

𝜃��𝜙 + ∇𝜃𝑙𝜙𝑚𝜙��𝜃 + ∇𝜙𝑙𝜙𝑚

𝜙��𝜙





= (𝜕𝜃𝑙𝜃 − Γ 𝜃𝜃

𝑐 𝑙𝑐)𝑚𝜃��𝜃 + (𝜕𝜙𝑙𝜃 − Γ 𝜙𝜃

𝑐 𝑙𝑐)𝑚𝜃��𝜙 + (𝜕𝜃𝑙𝜙 − Γ 𝜃𝜙

𝑐 𝑙𝑐)𝑚𝜙��𝜃

+ (𝜕𝜙𝑙𝜙 − Γ 𝜙𝜙𝑐 𝑙𝑐)𝑚

𝜙��𝜙


𝜃��𝜃 − Γ 𝜙𝜙𝑟 𝑙𝑟𝑚

𝜙��𝜙

= 𝑟 (1 −2𝑚

𝑟+𝑒2

𝑟2)1

√2(1 −

2𝑚

𝑟+𝑒2

𝑟2)

−12

(1

√2(−

1

𝑟))

2

+ 𝑟 (1 −2𝑚

𝑟+𝑒2

𝑟2) sin2 𝜃

1

√2(1 −

2𝑚

𝑟+𝑒2

𝑟2)

−12

(1

√2(−𝑖

1

𝑟 sin 𝜃))(

1

√2𝑖

1

𝑟 sin𝜃)

=1

√2(1 −

2𝑚

𝑟+𝑒2

𝑟2)

12 1

𝑟



𝜃𝑚𝜃 − Γ 𝜙𝜙𝑟 𝑙𝑟𝑚

𝜙𝑚𝜙

= 0

휀 =1



=1


𝑎𝑙𝑡 − ∇𝑡𝑚𝑎��𝑎𝑙𝑡) +

1


𝑎𝑙𝑟 − ∇𝑟𝑚𝑎��𝑎𝑙𝑟)

=1


𝑡𝑙𝑡 − ∇𝑡𝑚𝜃��𝜃𝑙𝑡) +

1


𝑡𝑙𝑟 − ∇𝑟𝑚𝜃��𝜃𝑙𝑟) +

1


𝑟𝑙𝑡 − ∇𝑡𝑚𝜙��𝜙𝑙𝑡)

+1


𝑟𝑙𝑟 − ∇𝑟𝑚𝜙��𝜙𝑙𝑟)

=1

2((𝜕𝑡𝑙𝑡 − Γ 𝑡𝑡

𝑐 𝑙𝑐)𝑛𝑡𝑙𝑡 − (𝜕𝑡𝑚𝜃 − Γ 𝑡𝜃

𝑐 )��𝜃𝑙𝑡) +1


𝑐 𝑙𝑐)𝑛𝑡𝑙𝑟 − (𝜕𝑟𝑚𝜃 − Γ 𝑟𝜃

𝑐 )��𝜃𝑙𝑟)

+1

2((𝜕𝑡𝑙𝑟 − Γ 𝑡𝑟

𝑐 𝑙𝑐)𝑛𝑟𝑙𝑡 − (𝜕𝑡𝑚𝜙 − Γ 𝑡𝜙

𝑐 )��𝜙𝑙𝑡)

+1


𝑐 𝑙𝑐)𝑛𝑟𝑙𝑟 − (𝜕𝑟𝑚𝜙 − Γ 𝑟𝜙

𝑐 )��𝜙𝑙𝑟)

=1

2((−Γ 𝑡𝑡


1


𝑡 𝑙𝑡)𝑛𝑡𝑙𝑟 − (𝜕𝑟𝑚𝜃 − Γ 𝑟𝜃

𝜃 𝑚𝜃)��𝜃𝑙𝑟) +

1

2((−Γ 𝑡𝑟


+1


𝑟 𝑙𝑟)𝑛𝑟𝑙𝑟 − (𝜕𝑟𝑚𝜙 − Γ 𝑟𝜙


𝜙𝑙𝑟)

=1

2((−Γ 𝑡𝑡


1

2((

1

√2𝜕𝑟 (1 −

2𝑚

𝑟+𝑒2

𝑟2)

12

− Γ 𝑟𝑡𝑡 𝑙𝑡)𝑛

𝑡𝑙𝑟 − (1

√2𝜕𝑟(𝑟) − Γ 𝑟𝜃

𝜃 𝑚𝜃) ��𝜃𝑙𝑟)

+1

2((−Γ 𝑡𝑟


+1

2((

1

√2𝜕𝑟 (1 −

2𝑚

𝑟+𝑒2

𝑟2)

−12

− Γ 𝑟𝑟𝑟 𝑙𝑟)𝑛

𝑟𝑙𝑟

− (1

√2𝜕𝑟(𝑖𝑟 sin𝜃) − Γ 𝑟𝜙


𝜙𝑙𝑟)





=1

2((−Γ 𝑡𝑡

𝑟 𝑙𝑟)𝑛𝑡𝑙𝑡)

+1

2((

1

√2(2𝑚

𝑟2−2𝑒2

𝑟3)1

2(1 −

2𝑚

𝑟+𝑒2

𝑟2)

−12

− Γ 𝑟𝑡𝑡 𝑙𝑡)𝑛

𝑡𝑙𝑟

− (1

√2− Γ 𝑟𝜃

𝜃 𝑚𝜃) ��𝜃𝑙𝑟)+

1

2((−Γ 𝑡𝑟


+1

2((

1

√2(2𝑚

𝑟2−2𝑒2

𝑟3)(−

1

2) (1 −

2𝑚

𝑟+𝑒2

𝑟2)

−32

− Γ 𝑟𝑟𝑟 𝑙𝑟)𝑛

𝑟𝑙𝑟

− (1

√2(𝑖 sin𝜃) − Γ 𝑟𝜙


𝜙𝑙𝑟)

=1

2((−Γ 𝑡𝑡

𝑟 𝑙𝑟)𝑛𝑡𝑙𝑡)

+1

2((

1

√2(𝑚𝑟 − 𝑒2

𝑟3)(1 −

2𝑚

𝑟+𝑒2

𝑟2)

−12

− (1 −2𝑚

𝑟+𝑒2

𝑟2)

−1

(𝑚𝑟 − 𝑒2

𝑟3)1

√2(1 −

2𝑚

𝑟+𝑒2

𝑟2)

12

)𝑛𝑡𝑙𝑟

− (1

√2−1

𝑟

1

√2𝑟) ��𝜃𝑙𝑟)+

1

2((−Γ 𝑡𝑟


+1

2((−

1

√2(𝑚𝑟 − 𝑒2

𝑟3)(1 −

2𝑚

𝑟+𝑒2

𝑟2)

−32

+ (1 −2𝑚

𝑟+𝑒2

𝑟2)

−1

(𝑚𝑟 − 𝑒2

𝑟3)1

√2(1 −

2𝑚

𝑟+𝑒2

𝑟2)

−12

)𝑛𝑟𝑙𝑟

− (1

√2(𝑖 sin𝜃) −

1

𝑟

1

√2𝑖𝑟 sin𝜃) ��𝜙𝑙𝑟)

=1

2((−Γ 𝑡𝑡


1

2((−Γ 𝑡𝑟


= −

1

2(Γ 𝑡𝑡𝑟 𝑙𝑟𝑛

𝑡 + Γ 𝑡𝑟𝑡 𝑙𝑡𝑛

𝑟)𝑙𝑡





= −1

2((1 −

2𝑚

𝑟+𝑒2

𝑟2)(𝑚𝑟 − 𝑒2

𝑟3)1

√2(1 −

2𝑚

𝑟+𝑒2

𝑟2)

−12 1

√2(1 −

2𝑚

𝑟+𝑒2

𝑟2)

−12

+ (1 −2𝑚

𝑟+𝑒2

𝑟2)

−1

(𝑚𝑟 − 𝑒2

𝑟3)1

√2(1 −

2𝑚

𝑟+𝑒2

𝑟2)

12 1

√2(1 −

2𝑚

𝑟+𝑒2

𝑟2)

12

) 𝑙𝑡

= −1

2(𝑚𝑟 − 𝑒2

𝑟3)1

√2(1 −

2𝑚

𝑟+𝑒2

𝑟2)

−12

= −1

232

(𝑚𝑟 − 𝑒2

𝑟3)(1 −

2𝑚

𝑟+𝑒2

𝑟2)

−12

𝛾 =1



= −1

2(Γ 𝑡𝑡𝑟 𝑙𝑟𝑛

𝑡 + Γ 𝑡𝑟𝑡 𝑙𝑡𝑛

𝑟)𝑛𝑡

= −1

232

(𝑚𝑟 − 𝑒2

𝑟3)(1 −

2𝑚

𝑟+𝑒2

𝑟2)

−12

𝛼 =1


𝑎��𝑏 − ∇𝑏𝑚𝑎��𝑎��𝑏)

=1

2(∇𝜃𝑙𝑎𝑛

𝑎��𝜃 − ∇𝜃𝑚𝑎��𝑎��𝜃) +

1

2(∇𝜙𝑙𝑎𝑛

𝑎��𝜙 − ∇𝜙𝑚𝑎��𝑎��𝜙)

=1

2(∇𝜃𝑙𝑡𝑛

𝑡��𝜃 − ∇𝜃𝑚𝜃��𝜃��𝜃) +

1

2(∇𝜙𝑙𝑡𝑛

𝑡��𝜙 − ∇𝜙𝑚𝜃��𝜃��𝜙) +

1

2(∇𝜃𝑙𝑟𝑛

𝑟��𝜃 − ∇𝜃𝑚𝜙��𝜙��𝜃)

+1

2(∇𝜙𝑙𝑟𝑛

𝑟��𝜙 − ∇𝜙𝑚𝜙��𝜙��𝜙)

=1

2((𝜕𝜃𝑙𝑡 − Γ 𝜃𝑡

𝑐 𝑙𝑐)𝑛𝑡��𝜃 − (𝜕𝜃𝑚𝜃 − Γ 𝜃𝜃

𝑐 𝑚𝑐)��𝜃��𝜃)

+1

2((𝜕𝜙𝑙𝑡 − Γ 𝜙𝑡

𝑐 𝑙𝑐)𝑛𝑡��𝜙 − (𝜕𝜙𝑚𝜃 − Γ 𝜙𝜃

𝑐 𝑚𝑐)��𝜃��𝜙)

+1

2((𝜕𝜃𝑙𝑟 − Γ 𝜃𝑟

𝑐 𝑙𝑐)𝑛𝑟��𝜃 − (𝜕𝜃𝑚𝜙 − Γ 𝜃𝜙

𝑐 𝑚𝑐)��𝜙��𝜃)

+1

2((𝜕𝜙𝑙𝑟 − Γ 𝜙𝑟

𝑐 𝑙𝑐)𝑛𝑟��𝜙 − (𝜕𝜙𝑚𝜙 − Γ 𝜙𝜙

𝑐 𝑚𝑐)��𝜙��𝜙)

=1

2(Γ 𝜙𝜃

𝜙𝑚𝜙��

𝜃��𝜙 − (𝜕𝜃𝑚𝜙 − Γ 𝜃𝜙𝜙

𝑚𝜙) ��𝜙��𝜃 + Γ 𝜙𝜙

𝜃 𝑚𝜃��𝜙��𝜙)

=1

2(cot 𝜃

1

√2𝑖𝑟 sin 𝜃

1

√2(−

1

𝑟) − (

1

√2𝜕𝜃(𝑖𝑟 sin𝜃) − cot 𝜃

1

√2𝑖𝑟 sin𝜃) ��𝜃

− cos 𝜃 sin𝜃1

√2𝑟1

√2𝑖

1

𝑟 sin 𝜃) ��𝜙

= −

1

2𝑖 cos 𝜃 ��𝜙

= −

1

2𝑖 cos 𝜃

1

√2𝑖

1

𝑟 sin 𝜃

=1

232

1

𝑟cot 𝜃

𝛽 =1







= −1

2𝑖 cos 𝜃𝑚𝜙

= −1

2𝑖 cos 𝜃

1

√2(−𝑖

1

𝑟 sin 𝜃)

= −1

232

1

𝑟cot 𝜃


𝜋 = 0 𝜅 = 0 휀 = −1

232

(𝑚𝑟 − 𝑒2

𝑟3)(1 −

2𝑚

𝑟+𝑒2

𝑟2)

−12

𝜈 = 0 𝜏 = 0 𝛾 = −1

232

(𝑚𝑟 − 𝑒2

𝑟3)(1 −

2𝑚

𝑟+𝑒2

𝑟2)

−12

𝜆 = 0 𝜌 =1

√2(1 −

2𝑚

𝑟+𝑒2

𝑟2)

12 1

𝑟 𝛼 =

1

232

1

𝑟cot 𝜃

𝜇 = 0 𝜎 = 0 𝛽 = −1

232

1

𝑟cot 𝜃



Ψ3 = 𝛿𝛾 − Δα + 𝜈(𝜌 + 휀) − 𝜆(𝜏 + 𝛽) + 𝛼(�� − ��) + 𝛾(�� − ��) (13.25)

Ψ4 = 𝛿𝜈 − Δλ + 𝜆(𝜇 + ��) − 𝜆(3𝛾 − ��) + 𝜈(3𝛼 + �� + 𝜋 − ��) (13.26)


Ψ0 = 0 Ψ1 = 𝐷𝛽 − 𝛿휀 − 𝛽(�� − 휀) + 휀�� = 𝑙𝑎∇𝑎𝛽 −𝑚

𝑎∇𝑎휀 − 𝛽�� + 𝛽휀 + 휀�� = 𝑙𝑡𝜕𝑡𝛽 + 𝑙

𝑟𝜕𝑟𝛽 −𝑚𝜃𝜕𝜃휀 −𝑚

𝜙 𝜕𝜙휀 − 𝛽��

= 𝑙𝑟𝜕𝑟𝛽 − 𝛽��

= −1

√2(1 −

2𝑚

𝑟+𝑒2

𝑟2)

12

𝜕𝑟 (−1

232

1


1

232

1

𝑟cot 𝜃)

1

√2(1 −

2𝑚

𝑟+𝑒2

𝑟2)

12 1

𝑟

= −1

√2(1 −

2𝑚

𝑟+𝑒2

𝑟2)

12

𝜕𝑟 (−1

232

1


1

232

1

𝑟cot 𝜃)

= 0 Ψ2 = −Δ𝜌 + 𝜌(𝛾 + ��) = −𝑛𝑡 ∂𝑡𝜌 − 𝑛

𝑟 ∂𝑟𝜌 + 2𝜌𝛾 = −𝑛𝑟 ∂𝑟𝜌 + 2𝜌𝛾





= −𝑛𝑟 ∂𝑟 (1

√2(1 −

2𝑚

𝑟+𝑒2

𝑟2)

12 1

𝑟)

+ 2(1

√2(1 −

2𝑚

𝑟+𝑒2

𝑟2)

12 1

𝑟)(−

1

232

(𝑚𝑟 − 𝑒2

𝑟3)(1 −

2𝑚

𝑟+𝑒2

𝑟2)

−12

)

= −1

√2(1 −

2𝑚

𝑟+𝑒2

𝑟2)

12 1

√2((1 −

2𝑚

𝑟+𝑒2

𝑟2)

12

(−1

𝑟2) + (

2𝑚

𝑟2−2𝑒2

𝑟3)1

2(1 −

2𝑚

𝑟+𝑒2

𝑟2)

−12 1

𝑟)

−1

2(1

𝑟(𝑚𝑟 − 𝑒2

𝑟3))

= −

1

2((1 −

2𝑚

𝑟+𝑒2

𝑟2)(−

1

𝑟2) + (

2𝑚

𝑟2−2𝑒2

𝑟3)1

2

1

𝑟) −

1

2(1

𝑟(𝑚𝑟 − 𝑒2

𝑟3))

= −

1

2

1

𝑟((−

1

𝑟+2𝑚

𝑟2−𝑒2

𝑟3) + (

𝑚

𝑟2−𝑒2

𝑟3) + (

𝑚𝑟 − 𝑒2

𝑟3))

=

1

2𝑟2−2𝑚

𝑟3+3𝑒2

2𝑟4

=𝑟2 − 4𝑟𝑚 + 3𝑒2

2𝑟4

Ψ3 = 𝛿𝛾 − Δα + +𝛼�� + 𝛾�� = ��𝑎𝜕𝑎𝛾 − 𝑛

𝑎𝜕𝑎α = ��𝜃𝜕𝜃𝛾 + ��

𝜙𝜕𝜙𝛾 − 𝑛𝑡𝜕𝑡α − 𝑛

𝑟𝜕𝑟α

= −(1

√2(1 −

2𝑚

𝑟+𝑒2

𝑟2)

12

)𝜕𝑟 (1

232

1

𝑟cot 𝜃)

=1

4cot 𝜃

1

𝑟2(1 −

2𝑚

𝑟+𝑒2

𝑟2)

12

Ψ4 = 0

10 Black Holes

10.1 153The Path of a Radially Infalling Particle154 Particle orbits in the Schwarzschild space time are described by

1 = (1 −

2𝑚

𝑟) (𝑑𝑡

𝑑𝜏)2

− (1 −2𝑚

𝑟)−1

(𝑑𝑟

𝑑𝜏)2

− 𝑟2 (𝑑𝜃

𝑑𝜏)2

− 𝑟2 sin2 𝜃 (𝑑𝜙

𝑑𝜏)2

(10.44)

where 𝜏 is the local particle time (proper time) and t can be described as a distant observers (our) time. For paths along radial lines we can set 𝑑𝜃 = 𝑑𝜙 = 0, and rearrange the Schwarzschild line element:

1 = (1 −2𝑚

𝑟) (𝑑𝑡

𝑑𝜏)2

− (1 −2𝑚

𝑟)−1

(𝑑𝑟

𝑑𝜏)2

153 (McMahon, 2006, p. 238) 154 A more thorough review of the physical interpretation of the equations can be found here: http://phys-icssusan.mono.net/upl/9129/Radiallyinfallingparticle.pdf


http://physicssusan.mono.net/upl/9129/Radiallyinfallingparticle.pdf

http://physicssusan.mono.net/upl/9129/Radiallyinfallingparticle.pdf




⇔ 1 −2𝑚

𝑟 = (1 −

2𝑚

𝑟)2

(𝑑𝑡

𝑑𝜏)2

− (𝑑𝑟

𝑑𝜏)2

On page 220 we investigated the Killing vector 𝜉 = (1,0,0,0) and found that (1 −2𝑚

𝑟)𝑑𝑡

𝑑𝜏 is a conserved

quantity and, since (1 −2𝑚

𝑟)𝑑𝑡

𝑑𝜏→ 1 𝑓𝑜𝑟 𝑟 → ∞, (1 −

2𝑚

𝑟)𝑑𝑡

𝑑𝜏 must have the overall value 1:

1 = (1 −2𝑚

𝑟) (𝑑𝑡

𝑑𝜏) ⇒ 1 = (1 −

2𝑚

𝑟)2

(𝑑𝑡

𝑑𝜏)2

and from the Schwarzschild line element above we can conclude

2𝑚

𝑟 = (

𝑑𝑟

𝑑𝜏)2

⇒ 𝑑𝑟

𝑑𝜏 = ±√

2𝑚

𝑟

Notice that for → ∞: 𝑑𝑡

𝑑𝜏= 1, and the particle proper time, 𝜏, and the time of the distant observer, 𝑡, are

equal. Notice also that for 𝑟 → ∞:𝑑𝑟

𝑑𝜏= 0, which means the velocity of the particle is zero in this limit. Also

notice that 𝑑𝑟

𝑑𝜏 has to be negative because 𝑑𝑟decreases as the particle moves inwards. Next we rearrange

the two equations into a differential equation:

𝑑𝑡

𝑑𝑟 = −

1

√2𝑚

√𝑟

(1 −2𝑚𝑟 )

(i)

We can solve this by integration from 𝑟0 (far out) to 𝑟 (in the vicinity of 2𝑚), and find the 𝑡(𝑟) which describes the particles path from our distant point of view, or more popular: what happens to the poor astronaut as he approaches the black holes event horizon from viewed from our distant position.

⇒ 𝑡 − 𝑡0 =1

√2𝑚∫

√𝑟

(1 −2𝑚𝑟)

𝑟0

𝑟

𝑑𝑟

=1

√2𝑚∫

𝑟3/2

(𝑟 − 2𝑚)𝑑𝑟

𝑟0

𝑟

=1

√2𝑚∫

(𝑥 + 2𝑚)3/2

𝑥𝑑𝑥

𝑟0

𝑟

𝑥 = 𝑟 − 2𝑚

= 1551

√2𝑚([2(𝑥 + 2𝑚)3/2

3]𝑟

𝑟0

+ 2𝑚∫√(𝑥 + 2𝑚)

𝑥𝑑𝑥

𝑟0

𝑟

)

=1

√2𝑚([2(𝑥 + 2𝑚)3/2

3+ 2𝑚2√(𝑥 + 2𝑚)]

𝑟

𝑟0

+ (2𝑚)2∫𝑑𝑥

𝑥√𝑥 + 2𝑚

𝑟0

𝑟

)

= 1561

√2𝑚[2(𝑥 + 2𝑚)3/2

3+ 4𝑚√(𝑥 + 2𝑚) +

4𝑚2

√2𝑚ln√𝑥 + 2𝑚 − √2𝑚

√𝑥 + 2𝑚 + √2𝑚 ]𝑟

𝑟0

=1

√2𝑚[2𝑟3/2

3+ 4𝑚√𝑟 + (2𝑚)3/2 ln

√𝑟 − √2𝑚

√𝑟 + √2𝑚 ]𝑟

𝑟0

𝑟 = 𝑥 + 2𝑚

=2

3√2𝑚[𝑟3/2 + 6𝑚√𝑟]

𝑟

𝑟0+ [2𝑚 ln

√𝑟 − √2𝑚

√𝑟 + √2𝑚 ]𝑟

𝑟0

=2

3√2𝑚(𝑟0

3/2 − 𝑟3/2 + 6𝑚√𝑟0 − 6𝑚√𝑟) + 2𝑚 ln√𝑟0 − √2𝑚

√𝑟0 + √2𝑚 √𝑟 + √2𝑚

√𝑟 − √2𝑚

𝑖𝑓 𝑟0 → 𝑟 → 2𝑚: We set 휀0 = 1 −𝑟0

2𝑚→ 0, 휀 = 1 −

𝑟

2𝑚→ 0

155 ∫

(𝑎𝑥+𝑏)𝑛/2

𝑥𝑑𝑥 =

2(𝑎𝑥+𝑏)𝑛/2

𝑛+ 𝑏 ∫

(𝑎𝑥+𝑏)(𝑛−2)/2

𝑥𝑑𝑥 (14.102) (Spiegel, 1990)

156 ∫𝑑𝑥

𝑥√𝑎𝑥+𝑏=

1

√𝑏ln

√𝑎𝑥+𝑏−√𝑏

√𝑎𝑥+𝑏+√𝑏 𝑖𝑓 𝑏 > 0 (14.87) (Spiegel, 1990)





𝑡 − 𝑡0 =2

3√2𝑚(𝑟0

32 − 𝑟

32 + 6𝑚√𝑟0 − 6𝑚√𝑟) + 2𝑚 ln

√𝑟0 − √2𝑚

√𝑟0 + √2𝑚 √𝑟 + √2𝑚

√𝑟 − √2𝑚

→ 2𝑚 ln√𝑟0 − √2𝑚

√𝑟0 + √2𝑚 √𝑟 + √2𝑚

√𝑟 − √2𝑚

= 2𝑚 ln√𝑟02𝑚 − 1

√𝑟02𝑚

+ 1

√𝑟2𝑚 + 1

√𝑟2𝑚

− 1

→ 2𝑚 ln√1 − 휀0 − 1

√1 − 휀0 + 1 √1 − 휀 + 1

√1 − 휀 − 1

→ 2𝑚 ln1 −

12휀0 − 1

1 −12휀0 + 1

1 −

12휀 + 1

1 −12휀 − 1

= 2𝑚 ln−12휀0

2 −12 휀0

2 −

12휀

−12 휀

→ 2𝑚 ln휀0휀

= 2𝑚 ln1 −

𝑟02𝑚

1 −𝑟2𝑚

= 2𝑚 ln2𝑚 − 𝑟02𝑚 − 𝑟

⇔ 2𝑚 − 𝑟 = 157(2𝑚 − 𝑟0) exp(−1

2𝑚(𝑡 − 𝑡0))

Another way to come around this is to insert 휀 = 1 −𝑟

2𝑚→ 0 in the differential equation (i) and solve it.

𝑑𝑡

𝑑𝑟 = −

1

√2𝑚

√𝑟

(1 −2𝑚𝑟 )

= −√𝑟

2𝑚

1

(1 −2𝑚𝑟 )

→ −√1 − 휀1

1 − (1

1 + 휀)

→ −√1 − 휀1

1 − (1 − 휀)

= √1 − 휀1

휀

→ (1 −1

2휀)1

휀

→1

휀 𝑓𝑜𝑟 휀 → 0

⇒ 𝑑𝑡

𝑑𝑟 =

1

1 −𝑟2𝑚

157 (d'Inverno, 1992, p. 219)





⇒ 𝑡0 − 𝑡 = ∫1

1 −𝑟2𝑚

𝑑𝑟𝑟0

𝑟

= −2𝑚∫1

𝑥𝑑𝑥

1−2𝑚𝑥0

1−2𝑚𝑥

𝑥 = 1 − 𝑟/2𝑚

= −2𝑚[ln 𝑥]1−2𝑚𝑥1−2𝑚𝑥0

= −2𝑚 [ln (1 −𝑟

2𝑚)]𝑟

𝑟0

= −2𝑚(ln (1 −𝑟02𝑚

) − ln (1 −𝑟

2𝑚))

= −2𝑚 ln(1 −

𝑟02𝑚)

(1 −𝑟2𝑚

)

= −2𝑚 ln(𝑟0 − 2𝑚)

(𝑟 − 2𝑚)

⇒ 𝑟 − 2𝑚 = (𝑟0 − 2𝑚)𝑒−𝑡−𝑡02𝑚

10.2 158The Schwarzschild metric in Kruskal Coordinates. The Kruskal coordinates 𝒓 > 2𝒎

𝑢 = 𝑒

𝑟4𝑚 √

𝑟

2𝑚− 1 cosh

𝑡

4𝑚

(11.5)

𝑣 = 𝑒𝑟4𝑚 √

𝑟

2𝑚− 1 sinh

𝑡

4𝑚

where 𝑢2 − 𝑣2 = 𝑒

𝑟2𝑚 (

𝑟

2𝑚− 1) (11.8)

We calculate

𝜕𝑢

𝜕𝑡 =

1

4𝑚𝑒𝑟4𝑚 √

𝑟

2𝑚− 1 sinh

𝑡

4𝑚=

𝑣

4𝑚

𝜕𝑣

𝜕𝑡 =

1


𝑟

2𝑚− 1 cosh

𝑡

4𝑚=

𝑢

4𝑚

𝜕𝑢

𝜕𝑟 =

1


𝑟

2𝑚− 1 cosh

𝑡

4𝑚+ 𝑒

𝑟4𝑚

1

2𝑚

1

2√𝑟2𝑚 − 1

cosh𝑡

4𝑚=

𝑢

4𝑚(

1

1 −2𝑚𝑟

)

𝜕𝑣

𝜕𝑟 =

1


𝑟

2𝑚− 1 sinh

𝑡

4𝑚+ 𝑒

𝑟4𝑚

1

2𝑚

1

2√𝑟2𝑚 − 1

sinh𝑡

4𝑚=

𝑣

4𝑚(

1

1 −2𝑚𝑟

)

Now we can use the chain rule

𝑑𝑢 =𝜕𝑢

𝜕𝑡𝑑𝑡 +

𝜕𝑢

𝜕𝑟𝑑𝑟

𝑑𝑣 =𝜕𝑣

𝜕𝑡𝑑𝑡 +

𝜕𝑣

𝜕𝑟𝑑𝑟

Written as a matrix

{𝑑𝑢𝑑𝑣} = {

𝜕𝑢

𝜕𝑡

𝜕𝑢

𝜕𝑟𝜕𝑣

𝜕𝑡

𝜕𝑣

𝜕𝑟

} {𝑑𝑡𝑑𝑟}

158 (McMahon, 2006, p. 242)





With the inverse

{𝑑𝑡𝑑𝑟} = 159{

𝜕𝑢

𝜕𝑡

𝜕𝑢

𝜕𝑟𝜕𝑣

𝜕𝑡

𝜕𝑣

𝜕𝑟

}

−1

{𝑑𝑢𝑑𝑣} =

1

𝜕𝑢𝜕𝑡𝜕𝑣𝜕𝑟−𝜕𝑢𝜕𝑟𝜕𝑣𝜕𝑡

{

𝜕𝑣

𝜕𝑟−𝜕𝑢

𝜕𝑟

−𝜕𝑣

𝜕𝑡

𝜕𝑢

𝜕𝑡

} {𝑑𝑢𝑑𝑣}

=1

(𝑣4𝑚)

2(

1

1 −2𝑚𝑟

)− (𝑢4𝑚)

2(

1

1 −2𝑚𝑟

){

𝑣

4𝑚(

1

1 −2𝑚𝑟

) −𝑢

4𝑚(

1

1 −2𝑚𝑟

)

−𝑢

4𝑚

𝑣

4𝑚 }

{𝑑𝑢𝑑𝑣}

=4𝑚

𝑣2 − 𝑢2{

𝑣 −𝑢

−𝑢 (1 −2𝑚

𝑟) 𝑣 (1 −

2𝑚

𝑟)} {

𝑑𝑢𝑑𝑣}

⇒ 𝑑𝑡 =4𝑚

𝑣2 − 𝑢2(𝑣𝑑𝑢 − 𝑢𝑑𝑣)

⇒ 𝑑𝑡2 =16𝑚2

(𝑣2 − 𝑢2)2(𝑣2𝑑𝑢2 + 𝑢2𝑑𝑣2 − 2𝑢𝑣𝑑𝑢𝑑𝑣)

⇒ 𝑑𝑟 =4𝑚

𝑣2 − 𝑢2(1 −

2𝑚

𝑟) (−𝑢𝑑𝑢 + 𝑣𝑑𝑣)

⇒ 𝑑𝑟2 =16𝑚2

(𝑣2 − 𝑢2)2(1 −

2𝑚

𝑟)2

(𝑢2𝑑𝑢2 + 𝑣2𝑑𝑣2 − 2𝑢𝑣𝑑𝑢𝑑𝑣)

Next we find

⇒ (1 −2𝑚

𝑟)𝑑𝑡2 = (1 −

2𝑚

𝑟)

16𝑚2

(𝑣2 − 𝑢2)2(𝑣2𝑑𝑢2 + 𝑢2𝑑𝑣2 − 2𝑢𝑣𝑑𝑢𝑑𝑣 )

⇒

(1

−2𝑚

𝑟)−1

𝑑𝑟2 = (1 −

2𝑚

𝑟)

16𝑚2

(𝑣2 − 𝑢2)2(𝑢2𝑑𝑢2 + 𝑣2𝑑𝑣2 − 2𝑢𝑣𝑑𝑢𝑑𝑣)

Inserting into the Schwarzschild metric

𝑑𝑠2 = (1 −2𝑚

𝑟)𝑑𝑡2 − (1 −

2𝑚

𝑟)−1

𝑑𝑟2 − 𝑟2(𝑑𝜃 + sin2 𝜃 𝑑𝜙)

= (1 −2𝑚

𝑟)

16𝑚2

(𝑣2 − 𝑢2)2(𝑣2𝑑𝑢2 + 𝑢2𝑑𝑣2 − (𝑢2𝑑𝑢2 + 𝑣2𝑑𝑣2))

−𝑟2(𝑑𝜃 + sin2 𝜃 𝑑𝜙)

= (1 −2𝑚

𝑟)

16𝑚2

(𝑢2 − 𝑣2)(𝑑𝑣2 − 𝑑𝑟2) − 𝑟2(𝑑𝜃 + sin2 𝜃 𝑑𝜙)

= 16𝑚2 (1 −2𝑚

𝑟) 𝑒−

𝑟2𝑚 (

𝑟

2𝑚− 1)

−1

(𝑑𝑢2 − 𝑑𝑣2) − 𝑟2(𝑑𝜃 + sin2 𝜃 𝑑𝜙)

=32𝑚3

𝑟𝑒−

𝑟2𝑚(𝑑𝑣2 − 𝑑𝑢2) − 𝑟2(𝑑𝜃 + sin2 𝜃 𝑑𝜙)

(11.7)

The Kruskal coordinates 𝒓 < 2𝒎

𝑢 = 𝑒

𝑟4𝑚 √1 −

𝑟

2𝑚sinh

𝑡

4𝑚

(11.6)

159 {𝑎 𝑏𝑐 𝑑

}−1

=1

𝑎𝑑−𝑏𝑐{𝑑 −𝑏−𝑐 𝑎

}





𝑣 = 𝑒

𝑟4𝑚 √1 −

𝑟

2𝑚cosh

𝑡

4𝑚

where 𝑢2 − 𝑣2 = 𝑒

𝑟2𝑚 (

𝑟

2𝑚− 1) (11.8)

We calculate

𝜕𝑢

𝜕𝑡 =

1

4𝑚𝑒𝑟4𝑚 √1 −

𝑟

2𝑚cosh

𝑡

4𝑚=

𝑣

4𝑚

𝜕𝑣

𝜕𝑡 =

1

4𝑚𝑒𝑟4𝑚 √1 −

𝑟

2𝑚sinh

𝑡

4𝑚=

𝑢

4𝑚

𝜕𝑢

𝜕𝑟 =

1

4𝑚𝑒𝑟4𝑚 √1 −

𝑟

2𝑚sinh

𝑡

4𝑚− 𝑒

𝑟4𝑚

1

2𝑚

1

2√1 −𝑟2𝑚

sinh𝑡

4𝑚=

𝑢

4𝑚 (

1

2𝑚𝑟− 1

)

𝜕𝑣

𝜕𝑟 =

1

4𝑚𝑒𝑟4𝑚 √1 −

𝑟

2𝑚cosh

𝑡

4𝑚− 𝑒

𝑟4𝑚

1

2𝑚

1

2√1 −𝑟2𝑚

cosh𝑡

4𝑚=

𝑣

4𝑚(

1

2𝑚𝑟− 1

)

As before we find

{𝑑𝑡𝑑𝑟} = {

𝜕𝑢

𝜕𝑡

𝜕𝑢

𝜕𝑟𝜕𝑣

𝜕𝑡

𝜕𝑣

𝜕𝑟

}

−1

{𝑑𝑢𝑑𝑣} =

1

𝜕𝑢𝜕𝑡𝜕𝑣𝜕𝑟−𝜕𝑢𝜕𝑟𝜕𝑣𝜕𝑡

{

𝜕𝑣

𝜕𝑟−𝜕𝑢

𝜕𝑟

−𝜕𝑣

𝜕𝑡

𝜕𝑢

𝜕𝑡

} {𝑑𝑢𝑑𝑣}

=1

(𝑣4𝑚)

2(

12𝑚𝑟 − 1

) − (𝑢4𝑚)

2(

12𝑚𝑟 − 1

){

𝑣

4𝑚(

1

2𝑚𝑟 − 1

) −𝑢

4𝑚(

1

2𝑚𝑟 − 1

)

−𝑢

4𝑚

𝑣

4𝑚 }

{𝑑𝑢𝑑𝑣}

=4𝑚

𝑣2 − 𝑢2{

𝑣 −𝑢

−𝑢 (2𝑚

𝑟− 1) 𝑣 (

2𝑚

𝑟− 1)} {

𝑑𝑢𝑑𝑣}

⇒ 𝑑𝑡 =4𝑚

𝑣2 − 𝑢2(𝑣𝑑𝑢 − 𝑢𝑑𝑣)

⇒ 𝑑𝑡2 =16𝑚2

(𝑣2 − 𝑢2)2(𝑣2𝑑𝑢2 + 𝑢2𝑑𝑣2 − 2𝑢𝑣𝑑𝑢𝑑𝑣)

⇒ 𝑑𝑟 =4𝑚

𝑣2 − 𝑢2(2𝑚

𝑟− 1) (−𝑢𝑑𝑢 + 𝑣𝑑𝑣)

⇒ 𝑑𝑟2 =16𝑚2

(𝑣2 − 𝑢2)2(2𝑚

𝑟− 1)

2

(𝑢2𝑑𝑢2 + 𝑣2𝑑𝑣2 − 2𝑢𝑣𝑑𝑢𝑑𝑣)

Next we find

⇒ (1 −2𝑚

𝑟)𝑑𝑡2 = (1 −

2𝑚

𝑟)

16𝑚2

(𝑣2 − 𝑢2)2(𝑣2𝑑𝑢2 + 𝑢2𝑑𝑣2 − 2𝑢𝑣𝑑𝑢𝑑𝑣 )

⇒

(1

−2𝑚

𝑟)−1

𝑑𝑟2 = (1 −

2𝑚

𝑟)

16𝑚2

(𝑣2 − 𝑢2)2(𝑢2𝑑𝑢2 + 𝑣2𝑑𝑣2 − 2𝑢𝑣𝑑𝑢𝑑𝑣)

Inserting into the Schwarzschild metric we find as before

𝑑𝑠2 = (1 −2𝑚

𝑟)𝑑𝑡2 − (1 −

2𝑚

𝑟)−1

𝑑𝑟2 − 𝑟2(𝑑𝜃 + sin2 𝜃 𝑑𝜙)





=32𝑚3

𝑟𝑒−

𝑟2𝑚(𝑑𝑣2 − 𝑑𝑢2) − 𝑟2(𝑑𝜃 + sin2 𝜃 𝑑𝜙)

(11.7)

10.3 The Kerr metric

10.3.1 160The Kerr-Newman geometry

A more general metric is the Kerr-Newman geometry, corresponding to a simultaneously rotating and electrically charged black hole of mass 𝑚, charge 𝑄 and angular momentum 𝑆.

𝑑𝑠2 =Δ

Σ(𝑑𝑡 − 𝑎 sin2 𝜃 𝑑𝜙)2 −

sin2 𝜃

Σ((𝑟2 + 𝑎2)𝑑𝜙 − 𝑎𝑑𝑡)

2−Σ

Δ𝑑𝑟2 − Σ𝑑𝜃2 (33.2)

=Δ

Σ[𝑑𝑡2 + (𝑎 sin2 𝜃)2𝑑𝜙2 − 2𝑎 sin2 𝜃 𝑑𝑡𝑑𝜙]

−sin2 𝜃

Σ[𝑎2𝑑𝑡2 + (𝑟2 + 𝑎2)2𝑑𝜙2 − 2𝑎(𝑟2 + 𝑎2)𝑑𝑡𝑑𝜙] −

Σ

Δ𝑑𝑟2 − Σ𝑑𝜃2

=1

Σ[Δ − 𝑎2 sin2 𝜃]𝑑𝑡2 +

1

Σ[−Δ2𝑎 sin2 𝜃 + 2 sin2 𝜃 𝑎(𝑟2 + 𝑎2)]𝑑𝑡𝑑𝜙 −

Σ


+1

Σ[Δ(𝑎 sin2 𝜃)2 − sin2 𝜃 (𝑟2 + 𝑎2)2]𝑑𝜙2

=1

Σ[Δ − 𝑎2 sin2 𝜃]𝑑𝑡2 +

2𝑎 sin2 𝜃

Σ[−Δ + (𝑟2 + 𝑎2)]𝑑𝑡𝑑𝜙 −

Σ


+sin2 𝜃

Σ[Δ𝑎2 sin2 𝜃 − (𝑟2 + 𝑎2)2]𝑑𝜙2

=1

Σ[(Σ − 2𝑚𝑟 + 𝑎2 sin2 𝜃 + 𝑄2) − 𝑎2 sin2 𝜃]𝑑𝑡2

+2𝑎 sin2 𝜃

Σ[−(Σ − 2𝑚𝑟 + 𝑎2 sin2 𝜃) + (Σ + 𝑎2 sin2 𝜃)]𝑑𝑡𝑑𝜙 −

Σ


+sin2 𝜃

Σ[(Σ − 2𝑚𝑟 + 𝑎2 sin2 𝜃)𝑎2 sin2 𝜃 − (Σ + 𝑎2 sin2 𝜃)2]𝑑𝜙2

= (1 −

2𝑚𝑟

Σ+𝑄2

Σ)𝑑𝑡2 +

4𝑎𝑚𝑟 sin2 𝜃

Σ𝑑𝑡𝑑𝜙 −

Σ


+sin2 𝜃

Σ[Σ𝑎2 sin2 𝜃 − 2𝑚𝑟𝑎2 sin2 𝜃 + (𝑎2 sin2 𝜃)2

− (Σ2 + (𝑎2 sin2 𝜃)2 + 2Σ𝑎2 sin2 𝜃)]𝑑𝜙2

= (1 −2𝑚𝑟

Σ+𝑄2

Σ)𝑑𝑡2 +



Σ


+sin2 𝜃

Σ[−Σ𝑎2 sin2 𝜃 − 2𝑚𝑟𝑎2 sin2 𝜃 − Σ2]𝑑𝜙2

= (1 −

2𝑚𝑟

Σ+𝑄2

Σ)𝑑𝑡2 +



Σ


− sin2 𝜃 [𝑎2 sin2 𝜃 +2𝑚𝑟𝑎2 sin2 𝜃

Σ+ Σ]𝑑𝜙2

= (1 −

2𝑚𝑟

Σ+𝑄2

Σ)𝑑𝑡2 +



Σ


− (𝑟2 + 𝑎2 +2𝑎2𝑚𝑟 sin2 𝜃

Σ) sin2 𝜃 𝑑𝜙2

Δ = 𝑟2 − 2𝑚𝑟 + 𝑎2 + 𝑄2 Σ = 𝑟2 + 𝑎2 cos2 𝜃

160 (C.W.Misner, 1973) chapter 33





= 𝑟2 + 𝑎2 − 𝑎2 sin2 𝜃 = Δ − 𝑄2 + 2𝑚𝑟 − 𝑎2 sin2 𝜃

𝑎 =𝑆

𝑚

We look at three special cases:

10.3.1.1 𝑸 = 𝟎 In the case of 𝑄 = 0 we see immediately that the Kerr-Newman geometry reduces to the Kerr geometry describing a non-charged rotating black hole.

𝑑𝑠2 = (1 −

2𝑚𝑟

Σ)𝑑𝑡2 +



Σ


− (𝑟2 + 𝑎2 +2𝑎2𝑚𝑟 sin2 𝜃


Δ = 𝑟2 − 2𝑚𝑟 + 𝑎2

Σ = 𝑟2 + 𝑎2 cos2 𝜃 = 𝑟2 + 𝑎2 − 𝑎2 sin2 𝜃 = Δ + 2𝑚𝑟 − 𝑎2 sin2 𝜃

𝑎 =𝑆

𝑚

10.3.1.2 𝑺 = 𝟎 In the case of 𝑆 = 0 we see immediately that the Kerr-Newman geometry reduces to the Reissner-Nordstrøm geometry describing a charged non-rotating black hole.

𝑑𝑠2

= (1 −2𝑚𝑟

Σ+𝑄2

Σ)𝑑𝑡2 +



Σ


− (𝑟2 + 𝑎2 +2𝑎2𝑚𝑟 sin2 𝜃


= (1 −2𝑚𝑟

Σ+𝑄2

Σ)𝑑𝑡2 −

Σ

Δ𝑑𝑟2 − Σ𝑑𝜃2 − 𝑟2 sin2 𝜃 𝑑𝜙2

= (1 −2𝑚

𝑟+𝑄2

𝑟2)𝑑𝑡2 −

1

1 −2𝑚𝑟 +

𝑄2

𝑟2

𝑑𝑟2 − 𝑟2𝑑𝜃2 − 𝑟2 𝑠𝑖𝑛2 𝜃 𝑑𝜙2

Δ = 𝑟2 − 2𝑚𝑟 + 𝑄2 Σ = 𝑟2

10.3.1.3 𝑸 = 𝟎 and 𝑺 = 𝟎 In the case of 𝑄 = 0 and 𝑆 = 0 we see immediately that the Kerr-Newman geometry reduces to the Schwarzschild geometry describing a non-charged non-rotating black hole.

𝑑𝑠2 = (1 −

2𝑚𝑟

Σ)𝑑𝑡2 +



Σ


− (𝑟2 + 𝑎2 +2𝑎2𝑚𝑟 sin2 𝜃


= (1 −2𝑚

𝑟)𝑑𝑡2 −

1

1 −2𝑚𝑟

𝑑𝑟2 − 𝑟2𝑑𝜃2 − 𝑟2 𝑠𝑖𝑛2 𝜃 𝑑𝜙2

Δ = 𝑟2 − 2𝑚𝑟 Σ = 𝑟2





10.3.2 161The inverse metric of the Kerr Spinning Black Hole

The Kerr metric of a spinning black hole with mass 𝑚 and angular momentum 𝑆.

𝑑𝑠2 = (1 −2𝑚𝑟

Σ)𝑑𝑡2 +



Σ

Δ𝑑𝑟2 − Σ𝑑𝜃2 − (𝑟2 + 𝑎2 +

2𝑎2𝑚𝑟 sin2 𝜃

Σ) sin2 𝜃 𝑑𝜙2 (11.9)

where Δ = 𝑟2 − 2𝑚𝑟 + 𝑎2 Σ = 𝑟2 + 𝑎2 cos2 𝜃

= 𝑟2 + 𝑎2 − 𝑎2 sin2 𝜃 = Δ + 2𝑚𝑟 − 𝑎2 sin2 𝜃

𝑎 =𝑆

𝑚

the metric tensor

𝑔𝑎𝑏 =

{

(1 −

2𝑚𝑟

Σ)


Σ

−Σ

Δ−Σ


Σ−(𝑟2 + 𝑎2 +


Σ) sin2 𝜃

}

with the inverse

𝑔𝑎𝑏 =

{

𝑔𝑡𝑡 𝑔𝑡𝜙

−Δ

Σ

−1

Σ𝑔𝜙𝑡 𝑔𝜙𝜙}

=

{

1

ΣΔ((𝑟2 + 𝑎2)2 − Δ𝑎2 sin2 𝜃)

2𝑎𝑚𝑟

ΣΔ

−Δ

Σ

−1

Σ2𝑎𝑚𝑟

ΣΔ−(Δ − 𝑎2 sin2 𝜃)

ΣΔ sin2 𝜃 }

where we can calculate 𝑔𝑡𝑡, 𝑔𝑡𝜙, 𝑔𝜙𝜙 from the inverse162

𝑔𝑎𝑏 =1

𝑔𝑡𝑡𝑔𝜙𝜙 − (𝑔𝜙𝑡)2 {𝑔𝜙𝜙 −𝑔𝑡𝜙−𝑔𝜙𝑡 𝑔𝑡𝑡

}

First we calculate the common factor

1

𝑔𝑡𝑡𝑔𝜙𝜙 − (𝑔𝜙𝑡)2 = (−(1 −

2𝑚𝑟

Σ)(𝑟2 + 𝑎2 +


Σ) sin2 𝜃 − (


Σ)

2

)

−1

= (−(1 −2𝑚𝑟

Σ)(𝑟2 + 𝑎2 +


Σ) − (

2𝑎𝑚𝑟

Σ)2

sin2 𝜃)

−11

sin2 𝜃

= (−(𝑟2 + 𝑎2 +2𝑎2𝑚𝑟 sin2 𝜃

Σ) +

2𝑚𝑟

Σ(𝑟2 + 𝑎2 +


Σ) − (

2𝑎𝑚𝑟

Σ)2

sin2 𝜃)

−11

sin2 𝜃

161 (McMahon, 2006, p. 246)

162{𝑎 𝑏𝑐 𝑑

}−1

=1

𝑎𝑑−𝑏𝑐{𝑑 −𝑏−𝑐 𝑎

}





= (−(𝑟2 + 𝑎2 +2𝑎2𝑚𝑟 sin2 𝜃

Σ) +

2𝑚𝑟

Σ(𝑟2 + 𝑎2))

−11

sin2 𝜃

= (−(𝑟2 + 𝑎2 +2𝑚𝑟

Σ𝑎2 sin2 𝜃) +

2𝑚𝑟

Σ(𝑟2 + 𝑎2))

−11

sin2 𝜃

= (−(𝑟2 + 𝑎2 +2𝑚𝑟

Σ𝑎2 sin2 𝜃) +

2𝑚𝑟

Σ(𝑟2 + 𝑎2 cos2 𝜃 + 𝑎2 sin2 𝜃))

−11

sin2 𝜃

= (−(𝑟2 + 𝑎2) +

2𝑚𝑟

Σ(𝑟2 + 𝑎2 cos2 𝜃))

−11

sin2 𝜃

= (−(𝑟2 + 𝑎2) +2𝑚𝑟

ΣΣ)

−1 1

sin2 𝜃

= (−(𝑟2 + 𝑎2) + 2𝑚𝑟)−11

sin2 𝜃

= −1

Δ sin2 θ

Now we can calculate the inverse metric

𝑔𝑡𝑡 =𝑔𝜙𝜙

𝑔𝑡𝑡𝑔𝜙𝜙 − (𝑔𝜙𝑡)2

= −𝑔𝜙𝜙

Δ sin2 𝜃

= −

−(𝑟2 + 𝑎2 +2𝑎2𝑚𝑟 sin2 𝜃

Σ ) sin2 𝜃

Δ sin2 𝜃

=1

Δ(𝑟2 + 𝑎2 +


Σ)

=1

ΣΔ(Σ(𝑟2 + 𝑎2) + 2𝑎2𝑚𝑟 sin2 𝜃)

=1

ΣΔ((𝑟2 + 𝑎2 cos2 𝜃)(𝑟2 + 𝑎2) + (𝑟2 + 𝑎2 − Δ)𝑎2 sin2 𝜃)

=1

ΣΔ((𝑟2 + 𝑎2 cos2 𝜃)(𝑟2 + 𝑎2) + (𝑟2 + 𝑎2)𝑎2(1 − cos2 𝜃) − Δ𝑎2 sin2 𝜃)

=1

ΣΔ((𝑟2 + 𝑎2)2 − Δ𝑎2 sin2 𝜃)

(11.13)

𝑔𝑡𝜙 = −𝑔𝑡𝜙


=𝑔𝑡𝜙

Δ sin2 𝜃

=1

Δ sin2 𝜃


Σ

=2𝑎𝑚𝑟

ΣΔ

(11.13)

𝑔𝜙𝜙 =𝑔𝑡𝑡


= −𝑔𝑡𝑡

Δ sin2 𝜃

= −1

Δ sin2 𝜃(1 −

2𝑚𝑟

Σ)

= −1

ΣΔ sin2 𝜃(Σ − 2𝑚𝑟)





= −1

ΣΔ sin2 𝜃(Δ + 2𝑚𝑟 − 𝑎2 sin2 𝜃 − 2𝑚𝑟)

= −

(Δ − 𝑎2 sin2 𝜃)

ΣΔ sin2 𝜃

(11.13)

11 Cosmology

11.1 163Light travelling in the Universe Light travelling in the universe can be described by the line element 𝑑𝑠2 = 𝑑𝑡2 − 𝑎2(𝑡)𝑑𝑥2, where 𝑑𝑥2 =

𝑑𝑥12 + 𝑑𝑥2

2 + 𝑑𝑥32 and the 𝑥𝑖 are commoving coordinates. Light travel along null geodesic i.e. 𝑑𝑠2 = 0.

We can now write ∫𝑑𝑡

𝑎(𝑡)

𝑡0𝑡

for the total commoving distance light emitted at time 𝑡 can travel by time 𝑡0. If

we multiply this by the value of the scale factor 𝑎(𝑡0) at time 𝑡0, then we will have calculated the physical

distance that the light has traveled in this time interval. This algorithm can be widely used to calculate how

far light can travel in any given time interval, revealing whether to points in space , for example are in causal

contact. As you can see, for accelerated expansion, even for arbitrarily large 𝑡0 the integral is bounded, show-

ing that the light will never reach arbitrarily distant commoving locations. Thus, in a universe with accelerated

expansion, there are locations with which we can never communicate.

11.2 164Spaces of Positive, Negative, and Zero Curvature According to (12.5) the spatial part of a homogenous, isotropic metric is

𝑑𝜎2 =

𝑑𝑟2

1 − 𝑘𝑟2+ 𝑟2𝑑𝜃2 + 𝑟2 sin2 𝜃 𝑑𝜙 (12.5)

rewriting it in a more general form 𝑑𝜎2 = 𝑑𝜒2 + 𝑟2(𝜒)𝑑𝜃2 + 𝑟2(𝜒) sin2 𝜃 𝑑𝜙2 (12.6) we see that

𝑑𝜒2 =

𝑑𝑟2

1 − 𝑘𝑟2

⇒ 𝑑𝜒 = ±𝑑𝑟

√1 − 𝑘𝑟2

and in order to identify the metric for the different 𝑘-values we solve the latter differential equation. 𝑘 = 0: ∫ 𝑑𝜒 = ±∫ 𝑑𝑟 ⇒ 𝜒 = ±𝑟 ⇒ 𝑟 = ±𝜒 ⇒ 𝑟2(𝜒) = 𝜒2 and the metric 𝑑𝜎2 = 𝑑𝜒2 + 𝜒2𝑑𝜃2 + 𝜒2 sin2 𝜃 𝑑𝜙2 𝑘 > 0:

∫ 𝑑𝜒 = ±∫

𝑑𝑟

√1 − 𝑘𝑟2

⇒ 𝜒 = ±∫𝑑𝑟

√1 − 𝑘𝑟2= ±

1

√𝑘∫

𝑑𝑥

√1 − 𝑥2= 165 ±

1

√𝑘sin−1 𝑥 𝑥 = √𝑘𝑟

163 (Greene, s. 516) note 10 164 (McMahon, 2006, p. 262) 165 ∫

𝑑𝑥

√𝑎2−𝑥2= sin−1

𝑥

𝑎 (14.237) (Spiegel, 1990)





= ±1

√𝑘sin−1 √𝑘𝑟

⇒ 𝑟 =1

√𝑘sin(±√𝑘𝜒) = ±

1

√𝑘sin(√𝑘𝜒)

⇒ 𝑟2(𝜒) =1

𝑘sin2(√𝑘𝜒)

if 𝑘 = 1 we get 𝑟2(𝜒) = sin2(𝜒) and the metric 𝑑𝜎2 = 𝑑𝜒2 + sin2(𝜒) 𝑑𝜃2 + sin2(𝜒) sin2 𝜃 𝑑𝜙2 𝑘 < 0:

⇒ ∫ 𝑑𝜒 = ±∫𝑑𝑟

√1 + 𝐾𝑟2 𝐾 = −𝑘

⇒ 𝜒 = ±∫

𝑑𝑟

√1 + (√𝐾𝑟)2

= ±1

√𝐾∫

𝑑𝑥

√1 + 𝑥2 𝑥 = √𝐾𝑟 = √−𝑘𝑟

= 166 ±1

√𝐾ln (𝑥 + √𝑥2 + 1)

= 167 ±1

√𝐾sinh−1 𝑥

= ±1

√−𝑘sinh−1(√−𝑘𝑟)

⇒ 𝑟 =1

√−𝑘sinh(±√−𝑘𝜒)

= ±1

√−𝑘sinh(√−𝑘𝜒)

⇒ 𝑟2(𝜒) =1

−𝑘sinh2(√−𝑘𝜒)

if 𝑘 = −1 we get

⇒ 𝑟2(𝜒) = sinh2(𝜒) and the metric 𝑑𝜎2 = 𝑑𝜒2 + sinh2(𝜒) 𝑑𝜃2 + sinh2(𝜒) sin2 𝜃 𝑑𝜙2

Note: We have omitted the constants of integration because of symmetry reasons. The metrics have to fulfill the requirement of homogeneity and isotropy

11.3 New – The critical density168 The critical density

𝜌𝑐 =3

8𝜋

𝐻02

𝐺 (12.8)

can be found from some simple Newtonian considerations169.

166 ∫

𝑑𝑥

√𝑥2−𝑎2= ln(𝑥 + √𝑥2 − 𝑎2) (14.210) (Spiegel, 1990)

167 sinh−1 𝑥 = ln(𝑥 + √𝑥2 + 1) (8.55) (Spiegel, 1990) 168 (McMahon, 2006, p. 267) 169 (Weinberg, 1979, s. 157)





Consider a galaxy with mass 𝑚 on the surface of a sphere with radius 𝑅, density 𝜌 and an expansion rate corresponding to the Hubble expansion 𝐻0. The velocity of the galaxy is given by the Hubble law 𝑣 = 𝐻0𝑅. The potential energy of the galaxy is 𝐸𝑝𝑜𝑡 = −

4𝜋

3𝑅3𝜌 ∗

𝑚𝐺

𝑅

= −

4𝜋𝑅2𝜌𝑚𝐺

3

The kinetic energy of the galaxy is

𝐸𝑘𝑖𝑛 =1

2𝑚𝑣2

=1

2𝑚𝐻0

2𝑅2

The total energy 𝐸𝑡𝑜𝑡 = 𝐸𝑘𝑖𝑛 + 𝐸𝑝𝑜𝑡

=1

2𝑚𝐻0

2𝑅2 −4𝜋𝑅2𝜌𝑚𝐺

3

= 𝑚𝑅2 (

1

2𝐻02 −

4𝜋𝜌𝐺

3)

Setting the total energy to zero defines the critical density

0 = 𝑚𝑅2 (1

2𝐻02 −

4𝜋𝜌𝑐𝐺

3)

⇒ 𝜌𝑐 =3𝐻0

2

8𝜋𝐺

11.4 170The Robertson-Walker metric

11.4.1 171Find the components of the Riemann tensor of the Robertson-Walker metric (Homog-

enous, isotropic and expanding universe) using Cartan’s structure equations

The metric: 𝑑𝑠2 = −𝑑𝑡2 +𝑎2(𝑡)

1 − 𝑘𝑟2𝑑𝑟2 + 𝑎2(𝑡)𝑟2𝑑𝜃2 + 𝑎2(𝑡)𝑟2 sin2 𝜃 𝑑𝜙2

The Basis one forms

𝜔�� = 𝑑𝑡

𝜔�� =𝑎(𝑡)

√1 − 𝑘𝑟2𝑑𝑟 𝑑𝑟 =

√1 − 𝑘𝑟2

𝑎(𝑡)𝜔��

𝜔�� = 𝑎(𝑡)𝑟𝑑𝜃 𝑑𝜃 =1

𝑎(𝑡)𝑟𝜔�� 𝜂𝑖𝑗 = {

−11

11

}

𝜔�� = 𝑎(𝑡)𝑟 sin𝜃 𝑑𝜙 𝑑𝜙 =1

𝑎(𝑡)𝑟 sin𝜃𝜔��

Cartan’s First Structure equation and the calculation of the curvature one forms:

170 (McMahon, 2006, p. 161), example 7-2 171 (McMahon, 2006, p. 116), example 5-3





𝑑𝜔�� = −Γ �� ∧ 𝜔��

𝑑𝜔�� = 0

𝑑𝜔�� = 𝑑 (𝑎(𝑡)

√1 − 𝑘𝑟2𝑑𝑟) =

��

√1 − 𝑘𝑟2𝑑𝑡 ∧ 𝑑𝑟 = −

��

𝑎𝜔�� ∧ 𝜔��

𝑑𝜔�� = 𝑑(𝑎(𝑡)𝑟𝑑𝜃) = ��𝑟𝑑𝑡 ∧ 𝑑𝜃 + 𝑎𝑑𝑟 ∧ 𝑑𝜃 =��

𝑎𝜔�� ∧ 𝜔�� +

√1 − 𝑘𝑟2

𝑎𝑟𝜔�� ∧ 𝜔��

𝑑𝜔�� = 𝑑(𝑎(𝑡)𝑟 sin𝜃 𝑑𝜙) = ��𝑟 sin𝜃 𝑑𝑡 ∧ 𝑑𝜙 + 𝑎 sin 𝜃 𝑑𝑟 ∧ 𝑑𝜙 + 𝑎𝑟 cos 𝜃 𝑑𝜃 ∧ 𝑑𝜙

= ��𝑟 sin𝜃 𝜔�� ∧

1

𝑎𝑟 sin𝜃𝜔�� + 𝑎 sin 𝜃

√1 − 𝑘𝑟2

𝑎𝜔�� ∧

1

𝑎𝑟 sin𝜃𝜔�� + 𝑎𝑟 cos 𝜃

1

𝑎𝑟𝜔��

∧1

𝑎𝑟 sin 𝜃𝜔��

=��

𝑎𝜔�� ∧ 𝜔�� +

√1 − 𝑘𝑟2

𝑎𝑟𝜔�� ∧ 𝜔�� +

cot 𝜃

𝑎𝑟𝜔�� ∧ 𝜔��

The curvature one-forms summerized in a matrix

Γ ��

=

{

0

��

𝑎𝜔��

��

𝑎𝜔��

��

𝑎𝜔��

��

𝑎𝜔�� 0

√1 − 𝑘𝑟2

𝑎𝑟𝜔��

√1 − 𝑘𝑟2

𝑎𝑟𝜔��

��

𝑎𝜔�� −

√1 − 𝑘𝑟2

𝑎𝑟𝜔�� 0

cot 𝜃

𝑎𝑟𝜔��

��

𝑎𝜔�� −

√1 − 𝑘𝑟2

𝑎𝑟𝜔�� −

cot 𝜃

𝑎𝑟𝜔�� 0 }

Where �� refers to the column and �� the row

Curvature two forms:

Ω �� = 𝑑Γ ��

�� + Γ 𝑐�� ∧ Γ ��

𝑐 =1

2𝑅 ��𝑐�� 𝜔𝑐 ∧ 𝜔�� (5.27), (5.28)

Ω �� : 𝑑Γ ��

�� = 𝑑 (��

𝑎𝜔��) = 𝑑 (

��

𝑎

𝑎

√1 − 𝑘𝑟2𝑑𝑟) = 𝑑 (

��

√1 − 𝑘𝑟2𝑑𝑟) =

��

√1 − 𝑘𝑟2𝑑𝑡 ∧ 𝑑𝑟

=��

√1 − 𝑘𝑟2𝜔�� ∧

√1 − 𝑘𝑟2

𝑎𝜔�� =

��

𝑎𝜔�� ∧ 𝜔��

Γ 𝑐�� ∧ Γ ��

𝑐 = Γ �� ∧ Γ ��

�� + Γ �� ∧ Γ ��

�� + Γ �� ∧ Γ ��

�� + Γ �� ∧ Γ

��= 0

⇒ Ω �� =

��

𝑎𝜔�� ∧ 𝜔��

Ω �� : 𝑑Γ ��

�� = 𝑑 (��

𝑎𝜔��) = 𝑑 (

��

𝑎𝑎𝑟𝑑𝜃) = 𝑑(��𝑟𝑑𝜃) = ��𝑟𝑑𝑡 ∧ 𝑑𝜃 + ��𝑑𝑟 ∧ 𝑑𝜃

= ��𝑟𝜔�� ∧1

𝑎𝑟𝜔�� + ��

√1 − 𝑘𝑟2

𝑎𝜔�� ∧

1

𝑎𝑟𝜔�� =

��

𝑎𝜔�� ∧ 𝜔�� + ��

√1 − 𝑘𝑟2

𝑎2𝑟𝜔�� ∧ 𝜔��





Γ 𝑐�� ∧ Γ ��

𝑐 = Γ �� ∧ Γ ��

�� + Γ �� ∧ Γ ��

�� + Γ �� ∧ Γ ��

�� ++Γ �� ∧ Γ

��= Γ ��

�� ∧ Γ ��

=��√1 − 𝑘𝑟2

𝑎2𝑟𝜔�� ∧ 𝜔��

⇒ Ω �� =

��

𝑎𝜔�� ∧ 𝜔��

Ω ��: 𝑑Γ

��

= 𝑑(��

𝑎𝜔��) = 𝑑 (

��

𝑎𝑎𝑟 sin𝜃 𝑑𝜙) = 𝑑(��𝑟 sin𝜃 𝑑𝜙)

= ��𝑟 sin𝜃 𝑑𝑡 ∧ 𝑑𝜙 + �� sin𝜃 𝑑𝑟 ∧ 𝑑𝜙 + ��𝑟 cos 𝜃 𝑑𝜃 ∧ 𝑑𝜙

=��

𝑎𝜔�� ∧ 𝜔�� + ��

√1 − 𝑘𝑟2

𝑎2𝑟𝜔�� ∧ 𝜔�� + ��

cot 𝜃

𝑎2𝑟𝜔�� ∧ 𝜔��

Γ 𝑐��∧ Γ ��

𝑐 = Γ ��∧ Γ ��

�� + Γ ��∧ Γ ��

�� + Γ ��

��∧ Γ ��

�� + Γ 𝜙��∧ Γ

��= Γ ��

��∧ Γ ��

�� + Γ ��

��∧ Γ ��

��

=��√1 − 𝑘𝑟2

𝑎2𝑟𝜔�� ∧ 𝜔�� +

�� cot 𝜃

𝑎2𝑟𝜔�� ∧ 𝜔��

⇒ Ω ��

=��

𝑎𝜔�� ∧ 𝜔��

Ω �� : 𝑑Γ ��

�� = 𝑑 (√1 − 𝑘𝑟2

𝑎𝑟𝜔��) = 𝑑 (

√1 − 𝑘𝑟2

𝑎𝑟𝑎𝑟𝑑𝜃) = 𝑑 (√1 − 𝑘𝑟2𝑑𝜃)

=−𝑘𝑟

√1 − 𝑘𝑟2𝑑𝑟 ∧ 𝑑𝜃 =

−𝑘

𝑎2𝜔�� ∧ 𝜔��

Γ 𝑐�� ∧ Γ ��

𝑐 = Γ �� ∧ Γ ��

�� + Γ �� ∧ Γ ��

�� + Γ �� ∧ Γ ��

�� + Γ �� ∧ Γ ��

��= Γ ��

�� ∧ Γ ��

= (��

𝑎)2

𝜔�� ∧ 𝜔��

⇒ Ω �� =

−(��2 + 𝑘)

𝑎2𝜔�� ∧ 𝜔��

Ω ��: 𝑑Γ ��

�� = 𝑑 (

√1 − 𝑘𝑟2

𝑎𝑟𝜔��) = 𝑑 (√1 − 𝑘𝑟2 sin𝜃 𝑑𝜙)

=−𝑘𝑟

√1 − 𝑘𝑟2sin𝜃 𝑑𝑟 ∧ 𝑑𝜙 +√1 − 𝑘𝑟2 cos 𝜃 𝑑𝜃 ∧ 𝑑𝜙

=−𝑘𝑟

𝑎2𝑟𝜔�� ∧ 𝜔�� +

√1 − 𝑘𝑟2 cot 𝜃

𝑎2𝑟2𝜔�� ∧ 𝜔��

Γ 𝑐��∧ Γ ��

𝑐 = Γ ��∧ Γ ��

�� + Γ ��∧ Γ ��

�� + Γ ��

��∧ Γ ��

�� + Γ ��

��∧ Γ ��

��= Γ

��∧ Γ ��

�� + Γ ��

��∧ Γ ��

��

= −��2

𝑎2𝜔�� ∧ 𝜔�� −

√1 − 𝑘𝑟2 cot 𝜃

𝑎2𝑟2𝜔�� ∧ 𝜔��

⇒ Ω ��

=−(��2 + 𝑘)

𝑎2𝜔�� ∧ 𝜔��

Ω ��

��: 𝑑Γ

��

�� = 𝑑 (

cot 𝜃

𝑎𝑟𝜔��) = 𝑑 (

cot 𝜃

𝑎𝑟𝑎𝑟 sin𝜃 𝑑𝜙) = 𝑑(cos 𝜃 𝑑𝜙) = −sin 𝜃 𝑑𝜃 ∧ 𝑑𝜙

= −1

𝑎2𝑟2𝜔�� ∧ 𝜔��

Γ 𝑐��∧ Γ ��

𝑐 = Γ ��∧ Γ ��

�� + Γ ��∧ Γ ��

�� + Γ ��

��∧ Γ ��

�� + Γ ��

��∧ Γ

��

��= Γ

��∧ Γ ��

�� + Γ ��∧ Γ ��

��

= (��

𝑎)2

𝜔�� ∧ 𝜔�� −1 − 𝑘𝑟2

(𝑎𝑟)2𝜔�� ∧ 𝜔��

⇒ Ω ��

�� =

(��2 + 𝑘)

𝑎2𝜔�� ∧ 𝜔��

Summarized in a matrix





Ω �� =

{

0

��

𝑎𝜔�� ∧ 𝜔��

��

𝑎𝜔�� ∧ 𝜔��

��

𝑎𝜔�� ∧ 𝜔��

𝑆 0−(��2 + 𝑘)

𝑎2𝜔�� ∧ 𝜔��

−(��2 + 𝑘)

𝑎2𝜔�� ∧ 𝜔��

𝑆 𝐴𝑆 0(��2 + 𝑘)

𝑎2𝜔�� ∧ 𝜔��


Now we can read off the elements in the Riemann tensor in the non-coordinate basis

𝑅 �� = −

��

𝑎 𝑅 ��

�� = −��

𝑎 𝑅

��

�� = −

��

𝑎

𝑅 �� =

(��2 + 𝑘)

𝑎2 𝑅

��

�� =

(��2 + 𝑘)

𝑎2

Γ

��

�� =

��2 + 𝑘

𝑎2

11.4.2 The Einstein tensor and Friedmann-equations for the Robertson Walker metric

The Ricci scalar:

𝑅 = 𝜂��𝑅�� = −𝑅�� + 𝑅�� + 𝑅�� + 𝑅�� = −𝑅 ��𝑐��𝑐 + 𝑅 ��𝑐��

𝑐 + 𝑅 ��𝑐��𝑐 + 𝑅 ��𝑐��

𝑐

= −𝑅 �� − 𝑅 ��

�� − 𝑅 ��

��+ 𝑅 ��

�� + 𝑅 �� + 𝑅

��

��+ 𝑅 ��

�� + 𝑅 �� + 𝑅

��

��+ 𝑅 ��

�� + 𝑅 ��

+𝑅 ��

= −2𝑅 �� − 2𝑅 ��

�� − 2𝑅 ��

��+ 2𝑅 ��

�� + 2𝑅 ��

��+ 2𝑅

��

��

= 2��

𝑎+ 2

��

𝑎+ 2

��

𝑎+ 2

(��2 + 𝑘)

𝑎2+ 2

(��2 + 𝑘)

𝑎2+ 2

(��2 + 𝑘)

𝑎2= (

a

a+(��2 + k)

a2)


𝐺�� = 𝑅�� −1

2𝜂��𝑅 (4.48)

𝐺�� = 𝑅�� −1

2𝜂��𝑅

= 𝑅 �� + 𝑅 ��

�� + 𝑅 ��

��+1

2(−2𝑅 ��

�� − 2𝑅 �� − 2𝑅

��

��+ 2𝑅 ��

�� + 2𝑅 ��

��+ 2𝑅

��

��)

= 𝑅 �� + 𝑅

��

��+ 𝑅

��

��= 3(

(��2 + k)

a2)

𝐺�� = 𝑅�� −1

2𝜂��𝑅 = 𝑅 ��𝑐��

𝑐 = 0

𝐺�� = 𝐺�� = 𝐺�� = 𝐺�� = 𝐺�� = 𝐺�� = 𝐺�� = 𝐺�� = 𝐺�� = 𝐺�� = 𝐺�� = 0

𝐺�� = 𝑅�� −1

2𝜂��𝑅

= 𝑅 �� + 𝑅 ��

�� + 𝑅 ��

��−1

2(−2𝑅 ��

�� − 2𝑅 �� − 2𝑅

��

��+ 2𝑅 ��

�� + 2𝑅 ��

��+ 2𝑅

��

��)

= 𝑅 �� + 𝑅 ��

�� + 𝑅 ��

��+ 𝑅 ��

�� + 𝑅 �� + 𝑅

��

��− 𝑅 ��

�� − 𝑅 ��

��− 𝑅

��

��

= 𝑅 �� + 𝑅

��

��− 𝑅

��

��= −

��

𝑎−��

𝑎−��2 + 𝑘

𝑎2= −(

2��

𝑎+��2 + 𝑘

𝑎2)





𝐺�� = 𝑅�� −1

2𝜂��𝑅

= 𝑅 �� + 𝑅 ��

�� + 𝑅 ��

��−1

2(−2𝑅 ��

�� − 2𝑅 �� − 2𝑅

��

��+ 2𝑅 ��

�� + 2𝑅 ��

��+ 2𝑅

��

��)

= 𝑅 �� + 𝑅 ��

�� + 𝑅 ��

��+ 𝑅 ��

�� + 𝑅 �� + 𝑅

��

��− 𝑅 ��

�� − 𝑅 ��

��− 𝑅

��

��

= 𝑅 �� + 𝑅

��

��− 𝑅

��

��= −

��

𝑎−��

𝑎−��2 + 𝑘

𝑎2= −(

2��

𝑎+��2 + 𝑘

𝑎2)

𝐺�� = 𝑅�� −1

2𝜂��𝑅

= 𝑅 �� + 𝑅 ��

�� + 𝑅 �� −

1

2(−2𝑅 ��

�� − 2𝑅 �� − 2𝑅

��

��+ 2𝑅 ��

�� + 2𝑅 ��

��+ 2𝑅

��

��)

= 𝑅 �� + 𝑅 ��

�� + 𝑅 �� + 𝑅 ��

�� + 𝑅 �� + 𝑅

��

��− 𝑅 ��

�� − 𝑅 ��

��− 𝑅

��

��

= 𝑅 �� + 𝑅 ��

�� − 𝑅 �� = −

��

𝑎−��

𝑎−��2 + 𝑘

𝑎2= −(

2��

𝑎+��2 + 𝑘

𝑎2)


𝐺�� =

{

3(

(��2 + k)

a2) 0 0 0

0 −(2��

𝑎+��2 + 𝑘

𝑎2) 0 0

0 0 −(2��

𝑎+��2 + 𝑘

𝑎2) 0

0 0 0 −(2��

𝑎+��2 + 𝑘

𝑎2)}


The Friedmann equations:

Given the Einstein equation ( if 𝑐 = 𝐺 = 1): 𝐺�� − Λ𝜂�� = 8𝜋𝑇�� (7.14)

and the stress-energy tensor: 𝑇�� = 8𝜋{

𝜌 0 0 00 𝑃 0 00 0 𝑃 00 0 0 𝑃

} (7.16)

You can find the Friedmann- equations

{

3(

(��2 + k)

a2) 0 0 0

0 −(2��

𝑎+��2 + 𝑘

𝑎2) 0 0

0 0 −(2��

𝑎+��2 + 𝑘

𝑎2) 0

0 0 0 −(2��

𝑎+��2 + 𝑘

𝑎2)}

− Λ{

−11

11

} = 8𝜋{

𝜌 0 0 00 𝑃 0 00 0 𝑃 00 0 0 𝑃

}





⇒ 3

𝑎2(𝑘 + ��2) + Λ = 8𝜋𝜌 (7.17)

2��

𝑎+1

𝑎2(𝑘 + ��2) + Λ = −8𝜋𝑃 (8.18)

11.4.3 The Einstein tensor for the Robertson Walker metric – Alternative version.

The line element: 𝑑𝑠2 = −𝑑𝑡2 +𝑎2(𝑡)

1 − 𝑘𝑟2𝑑𝑟2 + 𝑎2(𝑡)𝑟2𝑑𝜃2 + 𝑎2(𝑡)𝑟2 sin2 𝜃 𝑑𝜙2

Now we can compare with the Tolman-Bondi – de Sitter line element, where the primes should not be mistaken for the derivative 𝑑/𝑑𝑟.

𝑑𝑠2 = 𝑑𝑡′2 − 𝑒−2𝜓(𝑡′,𝑟′)𝑑𝑟′2 − 𝑅2(𝑡′, 𝑟′)𝑑𝜃′2 − 𝑅2(𝑡′, 𝑟′) sin2 𝜃′ 𝑑𝜙′2

And chose: 𝑑𝑡′ = 𝑑𝑡

𝑒−𝜓(𝑡′,𝑟′)𝑑𝑟′ =

𝑎(𝑡)

√1 − 𝑘𝑟2𝑑𝑟

𝑅(𝑡′, 𝑟′)𝑑𝜃′ = 𝑎(𝑡)𝑟𝑑𝜃 𝑅(𝑡′, 𝑟′) sin𝜃′ 𝑑𝜙′ = 𝑎(𝑡)𝑟 sin 𝜃 𝑑𝜙

Comparing the two metrics we see: 𝑑𝜙′ = 𝑑𝜙, 𝑑𝜃′ = 𝑑𝜃, 𝜃′ = 𝜃, 𝑅(𝑡′, 𝑟′) = 𝑎(𝑡)𝑟, 𝑑𝑡′ = 𝑑𝑡 Next we can use the former calculations of the Tolman-Bondi – de Sitter metric to find the Einstein ten-sor for the Robertson-Walker metric.

But first we need to find

�� =𝑑𝜓(𝑡′, 𝑟′)

𝑑𝑡′= 𝑒−𝜓(𝑡

′,𝑟′)𝑑

𝑑𝑡′(𝑒𝜓(𝑡

′,𝑟′)) =𝑎(𝑡)

√1 − 𝑘𝑟2

𝑑𝑟

𝑑𝑟′𝑑

𝑑𝑡(√1 − 𝑘𝑟2

𝑎(𝑡)

𝑑𝑟′

𝑑𝑟) = −

��(𝑡)

𝑎(𝑡)

�� =𝑑

𝑑𝑡′(−

��(𝑡)

𝑎(𝑡)) =

𝑑

𝑑𝑡(−

��(𝑡)

𝑎(𝑡)) = −

(��(𝑡)𝑎(𝑡) − ��(𝑡)��(𝑡))

𝑎(𝑡)2= (

��(𝑡)

𝑎(𝑡))

2

−��(𝑡)

𝑎(𝑡)

𝜓′ =𝑑𝜓(𝑡′, 𝑟′)

𝑑𝑟′= 𝑒−𝜓(𝑡

′,𝑟′)𝑑

𝑑𝑟′(𝑒𝜓(𝑡

′,𝑟′)) = 𝑒−𝜓(𝑡′,𝑟′)

𝑑𝑟

𝑑𝑟′𝑑

𝑑𝑟(√1 − 𝑘𝑟2

𝑎(𝑡)

𝑑𝑟′

𝑑𝑟)

= −𝑘𝑟

𝑎(𝑡)√1 − 𝑘𝑟2𝑒−𝜓(𝑡

′,𝑟′)

�� =𝑑𝑅(𝑡′, 𝑟′)

𝑑𝑡′=𝑑𝑎(𝑡)𝑟

𝑑𝑡= ��(𝑡)𝑟

�� =𝑑��(𝑡)𝑟

𝑑𝑡′=𝑑��(𝑡)𝑟

𝑑𝑡= ��(𝑡)𝑟

𝑅′ =𝑑𝑅(𝑡′, 𝑟′)

𝑑𝑟′=𝑑𝑟

𝑑𝑟′𝑑

𝑑𝑟(𝑎(𝑡)𝑟) =

√1 − 𝑘𝑟2

𝑎(𝑡)𝑒−𝜓(𝑡

′,𝑟′)𝑎(𝑡) = √1 − 𝑘𝑟2𝑒−𝜓(𝑡′,𝑟′)

𝑅′′ =𝑑

𝑑𝑟′(√1 − 𝑘𝑟2𝑒−𝜓(𝑡

′,𝑟′)) =𝑑𝑟

𝑑𝑟′𝑑

𝑑𝑟(√1 − 𝑘𝑟2

𝑎(𝑡)

√1 − 𝑘𝑟2

𝑑𝑟

𝑑𝑟′) = 0

��′ =𝑑

𝑑𝑟′(��(𝑡)𝑟) =

𝑑𝑟

𝑑𝑟′𝑑

𝑑𝑟(��(𝑡)𝑟) =

��(𝑡)√1 − 𝑘𝑟2

𝑎(𝑡)𝑒−𝜓(𝑡

′,𝑟′)






Tolman –Bondi – de Sitter Robertson-Walker

𝐺�� =1

𝑅2[1 − 2𝑅�� + (��)

2

− (2𝑅𝑅′′ + 2𝑅𝑅′𝜓′ + (𝑅′)2)𝑒2𝜓(𝑡,𝑟)] ⇒ 𝐺�� = 3

��(𝑡)2 + 𝑘

𝑎(𝑡)2

𝐺�� = −2 [(��)′+ 𝑅′��]

𝑒𝜓(𝑡,𝑟)

𝑅 ⇒ 𝐺�� = 0

𝐺�� =1

𝑅2[(𝑅′)2𝑒2𝜓(𝑡,𝑟) − 2𝑅�� − 1 − (��)

2] ⇒ 𝐺�� = −(2

��(𝑡)

𝑎(𝑡)+��(𝑡)2 + 𝑘

𝑎(𝑡)2)

𝐺�� = [�� − (��)2] +

1

𝑅[(𝑅′′ + 𝑅′𝜓′)𝑒2𝜓(𝑡,𝑟) + �� − ��] ⇒ 𝐺�� = −(2

��(𝑡)

𝑎(𝑡)+��(𝑡)2 + 𝑘

𝑎(𝑡)2)

𝐺�� = [�� − (��)2] +

1

𝑅[(𝑅′′ + 𝑅′𝜓′)𝑒2𝜓(𝑡,𝑟) + �� − ��] ⇒ 𝐺�� = −(2

��(𝑡)

𝑎(𝑡)+��(𝑡)2 + 𝑘

𝑎(𝑡)2)

11.5 172Manipulating the Friedmann equations. Show that the two Friedman equations

3

𝑎2(𝑘 + ��2) + Λ = 8𝜋𝜌 (7.17)

2��

𝑎+1

𝑎2(𝑘 + ��2) + Λ = −8𝜋𝑃 (8.18)

can be manipulated into: 𝑑

𝑑𝑡(𝜌𝑎3) + 𝑃

𝑑

𝑑𝑡(𝑎3) = 0

Rewriting (7.17):

8𝜋𝜌 =3

𝑎2(𝑘 + ��2) + Λ

⇒ 8𝜋𝜌𝑎3 = 3𝑎(𝑘 + ��2) + Λa3

⇒ 8𝜋𝑑

𝑑𝑡(𝜌𝑎3) =

𝑑

𝑑𝑡(3𝑎(𝑘 + ��2) + Λa3) = 3��(𝑘 + ��2) + 6𝑎�� + 3Λ𝑎2��

Rewriting (7.18):

−8𝜋𝑃 = 2��

𝑎+1

𝑎2(𝑘 + ��2) + Λ

⇒ −8𝜋𝑃𝑑

𝑑𝑡(𝑎3) = (2

��

𝑎+1

𝑎2(𝑘 + ��2) + Λ)

𝑑

𝑑𝑡(𝑎3) = (2

��

𝑎+1

𝑎2(𝑘 + ��2) + Λ)3𝑎2��

= 6𝑎�� + 3��(𝑘 + ��2) + 3Λ𝑎2��

⇔ 8𝜋𝑃𝑑

𝑑𝑡(𝑎3) = −(6𝑎�� + 3��(𝑘 + ��2) + 3Λ𝑎2��)

Now adding

8𝜋𝑑

𝑑𝑡(𝜌𝑎3) + 8𝜋𝑃

𝑑

𝑑𝑡(𝑎3) = 3��(𝑘 + ��2) + 6𝑎�� + 3Λ𝑎2�� − (6𝑎�� + 3��(𝑘 + ��2) + 3Λ𝑎2��)

= 0 Q.E.D.

11.6 173Parameters in an flat universe with positive cosmological constant: Start-

ing with ��𝟐 =𝑪

𝒂+

𝚲

𝟑𝒂𝟐 use a change of variables 𝒖 =

𝟐𝚲

𝟑𝑪𝒂𝟑

We have

172 (McMahon, 2006, p. 165), quiz 7-3 173 (McMahon, 2006, p. 278), quiz 12-6. The answer to quiz 12-6 is (a)





��2 =𝐶

𝑎+Λ

3𝑎2

⇒ (𝑎

𝑎) 2

=𝐶

𝑎3+Λ

3

and

𝑢 =2Λ

3𝐶𝑎3

⇒ �� =2Λ

𝐶��𝑎2

= 3𝑢

𝑎��

⇒ ��

𝑢 = 3

��

𝑎

Rearranging we get

(1

3

��

𝑢)2

=2Λ

3𝑢+Λ

3

⇒ �� = 3√2Λ

3𝑢 +

Λ

3𝑢2

= √3Λ√𝑢√𝑢 + 2

⇒ ∫ 𝑑𝑡 = ∫𝑑𝑢

√3Λ√𝑢√𝑢 + 2

⇒ 𝑡 − 𝑡0 = 1741

√3Λ2 ln(√𝑢 + 2 + √𝑢)

=1

√3Λln(√𝑢 + 2 + √𝑢)

2

=1

√3Λln(2𝑢 + 2 + 2√𝑢√𝑢 + 2)

=1

√3Λln (2𝑢 + 2 + 2√𝑢2 + 2𝑢)

=1

√3Λln (2𝑢 + 2 + 2√(𝑢 + 1)2 − 1)

=1

√3Λln 2 (𝑢 + 1 + √(𝑢 + 1)2 − 1)

=1

√3Λ(ln 2 + ln (𝑢 + 1 + √(𝑢 + 1)2 − 1))

= 1751

√3Λ(ln 2 + cosh−1(𝑢 + 1))

⇒ 𝑢 = cosh(√3Λ(𝑡 − 𝑡0) − ln 2) − 1

⇒ 𝑎3 =3C

2Λ[cosh(√3Λ(𝑡 − 𝑡0) − ln 2) − 1]

Leaving out the constants of integration −√3Λ𝑡0 − ln2 we get

𝑎3 =3C

2Λ[cosh(√3Λ𝑡) − 1]

174 ∫

𝑑𝑥

√(𝑎𝑥+𝑏)(𝑝𝑥+𝑞) =

2

√𝑎𝑝ln(√𝑎(𝑝𝑥 + 𝑞) + 𝑝(𝑎𝑥 + 𝑏)) (14.280) (Spiegel, 1990)

175 cosh−1 𝑥 = ln(𝑥 + √𝑥2 − 1) (8.56) (Spiegel, 1990)





12 Gravitational Waves

12.1 176Gauge transformation - The Einstein Gauge Requiring that 𝑅 𝑏𝑐𝑑

𝑎 , 𝑅𝑎𝑏 and 𝑅 are unchanged under a gauge-transformation of first order in 휀, show that this is fulfilled by the coordinate transformations

𝑥𝑎 → 𝑥𝑎′= 𝑥𝑎 + 휀𝜙𝑎 (13.11)

ℎ𝑎𝑏 → ℎ𝑎𝑏′ = ℎ𝑎𝑏 − 𝜙𝑎,𝑏 − 𝜙𝑏,𝑎

𝜓 𝑏,𝑎𝑎 → 𝜓′ 𝑏,𝑎

𝑎= 𝜓 𝑏,𝑎

𝑎 − □𝜙𝑏

where 𝜙𝑎 is a function of position and |𝜙 ,𝑏𝑎 | ≪ 1. We have

𝑅 𝑏𝑐𝑑𝑎 =

1

2휀𝜂𝑎𝑒 (−

𝜕2ℎ𝑏𝑑𝜕𝑥𝑐𝜕𝑥𝑓

+𝜕2ℎ𝑑𝑓

𝜕𝑥𝑐𝜕𝑥𝑏+

𝜕2ℎ𝑏𝑐𝜕𝑥𝑑𝜕𝑥𝑓

−𝜕2ℎ𝑐𝑓

𝜕𝑥𝑑𝜕𝑥𝑏) (13.4)

𝑅𝑎𝑏 =1

2휀 (

𝜕2ℎ 𝑎𝑐

𝜕𝑥𝑏𝜕𝑥𝑐+𝜕2ℎ 𝑏

𝑐

𝜕𝑥𝑎𝜕𝑥𝑐−𝑊ℎ𝑎𝑏 −

𝜕2ℎ

𝜕𝑥𝑎𝜕𝑥𝑏) (13.5)

𝑅 = 휀 (𝜕2ℎ𝑐𝑑

𝜕𝑥𝑐𝜕𝑥𝑑−𝑊ℎ) (13.6)

𝜓𝑎𝑏 = ℎ𝑎𝑏 −1

2𝜂𝑎𝑏ℎ (13.8)

The Einstein gauge transformation is a coordinate transformation that leaves 𝑅 𝑏𝑐𝑑𝑎 , 𝑅𝑎𝑏 and 𝑅 unchanged.

The coordinate transformation that will do this is 𝑥𝑎 → 𝑥𝑎´ = 𝑥𝑎 + 휀𝜙𝑎 (13.11)

In order to show this you only have to convince yourself that the line element is unchanged. Checking

𝑑𝑠2 = 𝑔𝑎𝑏𝑑𝑥𝑎𝑑𝑥𝑏

= (𝜂𝑎𝑏 + 휀ℎ𝑎𝑏)𝑑𝑥𝑎𝑑𝑥𝑏

𝑑𝑠´2 = 𝑔𝑎´𝑏´𝑑𝑥𝑎´𝑑𝑥𝑏´

= (𝜂´𝑎𝑏 + 휀ℎ´𝑎𝑏)𝑑𝑥𝑎´𝑑𝑥𝑏´ 𝜂´𝑎𝑏 = 𝜂𝑎𝑏

= (𝜂𝑎𝑏 + 휀ℎ´𝑎𝑏)𝑑(𝑥𝑎 + 휀𝜙𝑎)𝑑(𝑥𝑏 + 휀𝜙𝑏)

= (𝜂𝑎𝑏 + 휀ℎ´𝑎𝑏) (

𝜕𝑥𝑎

𝜕𝑥𝑐𝑑𝑥𝑐 + 휀

𝜕𝜙𝑎

𝜕𝑥𝑐𝑑𝑥𝑐)(

𝜕𝑥𝑏

𝜕𝑥𝑑𝑑𝑥𝑑 + 휀

𝜕𝜙𝑏

𝜕𝑥𝑑𝑑𝑥𝑑)

= (𝜂𝑎𝑏 + 휀ℎ´𝑎𝑏)(𝛿𝑐𝑎𝑑𝑥𝑐 + 휀𝜙 ,𝑐

𝑎 𝑑𝑥𝑐)(𝛿𝑑𝑏𝑑𝑥𝑑 + 휀𝜙 ,𝑑

𝑏 𝑑𝑥𝑑)

= (𝜂𝑎𝑏 + 휀ℎ´𝑎𝑏)(𝛿𝑐𝑎𝛿𝑑

𝑏𝑑𝑥𝑐𝑑𝑥𝑑 + 휀𝜙 ,𝑐𝑎 𝛿𝑑

𝑏𝑑𝑥𝑐𝑑𝑥𝑑 + 휀𝜙 ,𝑑𝑏 𝛿𝑐

𝑎𝑑𝑥𝑐𝑑𝑥𝑑)

+ 휀2…

= (𝜂𝑎𝑏 + 휀ℎ´𝑎𝑏)(𝑑𝑥𝑎𝑑𝑥𝑏 + 휀𝜙 ,𝑐

𝑎 𝑑𝑥𝑐𝑑𝑥𝑏 + 휀𝜙 ,𝑑𝑏 𝑑𝑥𝑎𝑑𝑥𝑑)

= (𝜂𝑎𝑏 + 휀ℎ´𝑎𝑏)(𝑑𝑥𝑎𝑑𝑥𝑏 + 휀𝜂𝑎𝑒𝜙𝑒,𝑐𝑑𝑥

𝑐𝑑𝑥𝑏 + 휀𝜂𝑏𝑓𝜙𝑓,𝑑 𝑑𝑥𝑎𝑑𝑥𝑑)

= (𝜂𝑎𝑏 + 휀ℎ´𝑎𝑏)𝑑𝑥𝑎𝑑𝑥𝑏 + 𝜂𝑎𝑏(휀𝜂

𝑎𝑒𝜙𝑒,𝑐𝑑𝑥𝑐𝑑𝑥𝑏 + 휀𝜂𝑏𝑓𝜙𝑓,𝑑 𝑑𝑥

𝑎𝑑𝑥𝑑)

+ 휀2…

= (𝜂𝑎𝑏 + 휀ℎ´𝑎𝑏)𝑑𝑥𝑎𝑑𝑥𝑏 + (휀𝛿𝑏

𝑒𝜙𝑒,𝑐𝑑𝑥𝑐𝑑𝑥𝑏 + 휀𝛿𝑎

𝑓𝜙𝑓,𝑑 𝑑𝑥

𝑎𝑑𝑥𝑑)

= (𝜂𝑎𝑏 + 휀ℎ´𝑎𝑏)𝑑𝑥𝑎𝑑𝑥𝑏 + (휀𝜙𝑏,𝑐𝑑𝑥

𝑐𝑑𝑥𝑏 + 휀𝜙𝑎,𝑑 𝑑𝑥𝑎𝑑𝑥𝑑)

Renaming the dummy variables = (𝜂𝑎𝑏 + 휀ℎ´𝑎𝑏)𝑑𝑥

𝑎𝑑𝑥𝑏 + (휀𝜙𝑏,𝑎𝑑𝑥𝑎𝑑𝑥𝑏 + 휀𝜙𝑎,𝑏 𝑑𝑥

𝑎𝑑𝑥𝑏)

= (𝜂𝑎𝑏 + 휀ℎ´𝑎𝑏 + 휀𝜙𝑏,𝑎 + 휀𝜙𝑎,𝑏 )𝑑𝑥𝑎𝑑𝑥𝑏

= (𝜂𝑎𝑏 + 휀ℎ𝑎𝑏)𝑑𝑥𝑎𝑑𝑥𝑏

if ℎ´𝑎𝑏 = ℎ𝑎𝑏 − 𝜙𝑏,𝑎 −𝜙𝑎,𝑏 Q.E.D

176 (McMahon, 2006, p. 286)





Next we are going to investigate the transformation of the derivative of the trace reverse 𝜓 𝑏,𝑎𝑎 →

𝜓′ 𝑏,𝑎𝑎

= 𝜓 𝑏,𝑎𝑎 − □𝜙𝑏


= 𝜂𝑎𝑐𝜓𝑐𝑏,𝑎′

= 𝜂𝑎𝑐 (ℎ′𝑐𝑏,𝑎 −1

2𝜂𝑐𝑏ℎ,𝑎

′ )


2𝜂𝑐𝑏ℎ 𝑑,𝑎

′𝑑 )


2𝜂𝑐𝑏𝜂

𝑒𝑑ℎ 𝑒𝑑,𝑎′ )

= 𝜂𝑎𝑐(ℎ𝑐𝑏,𝑎 − 𝜙𝑏,𝑐𝑎 − 𝜙𝑐,𝑏𝑎 ) −1

2𝜂𝑎𝑐𝜂𝑐𝑏𝜂

𝑒𝑑(ℎ𝑒𝑑,𝑎 − 𝜙𝑑,𝑒𝑎 − 𝜙𝑒,𝑑𝑎 )

= 𝜂𝑎𝑐 (ℎ𝑐𝑏,𝑎 −1

2𝜂𝑐𝑏𝜂

𝑒𝑑ℎ𝑒𝑑,𝑎) − 𝜂𝑎𝑐 (𝜙𝑏,𝑐𝑎 + 𝜙𝑐,𝑏𝑎 −

1

2𝜂𝑐𝑏𝜂

𝑒𝑑(𝜙𝑑,𝑒𝑎 +𝜙𝑒,𝑑𝑎 ))


2𝜂𝑐𝑏ℎ 𝑑,𝑎

𝑑 ) − 𝜂𝑎𝑐𝜙𝑏,𝑐𝑎 − 𝜂𝑎𝑐 (𝜙𝑐,𝑏𝑎 −

1

2𝜂𝑐𝑏𝜂

𝑒𝑑(𝜙𝑑,𝑒𝑎 + 𝜙𝑒,𝑑𝑎 ))


2𝜂𝑐𝑏ℎ,𝑎) − □𝜙𝑏

− (𝜂𝑎𝑐𝜙𝑐,𝑏𝑎 −1

2𝜂𝑎𝑐𝜂𝑐𝑏𝜂

𝑒𝑑(𝜙𝑑,𝑒𝑎 + 𝜙𝑒,𝑑𝑎 ))

= 𝜂𝑎𝑐𝜓𝑐𝑏,𝑎 − □𝜙𝑏 − (𝜂𝑎𝑐𝜙𝑐,𝑏𝑎 −

1

2𝛿𝑏𝑎𝜂𝑒𝑑(𝜙𝑑,𝑒𝑎 + 𝜙𝑒,𝑑𝑎 ))

= 𝜓 𝑏,𝑎𝑎 − □𝜙𝑏 − (𝜙 ,𝑏𝑎

𝑎 −1

2𝛿𝑏𝑎(𝜙 ,𝑒𝑎

𝑒 + 𝜙 ,𝑑𝑎 𝑑 ))

= 𝜓 𝑏,𝑎𝑎 − □𝜙𝑏 − (𝜙 ,𝑏𝑎

𝑎 −1

2(𝜙 ,𝑒𝑏

𝑒 + 𝜙 ,𝑑𝑏 𝑑 ))

Renaming the dummy variables

= 𝜓 𝑏,𝑎𝑎 − □𝜙𝑏 − (𝜙 ,𝑏𝑎

𝑎 −1

2(𝜙 ,𝑎𝑏

𝑎 + 𝜙 ,𝑎𝑏 𝑎 ))


= 𝜓 𝑏,𝑎𝑎 − □𝜙𝑏 Q.E.D.

P.288: The choice of 𝜓′ 𝑏,𝑎𝑎

= 0 leads to177


= 𝜂𝑎𝑐𝜓′𝑐𝑏,𝑎


2𝜂𝑐𝑏ℎ

′,𝑎)

= 𝜂𝑎𝑐ℎ′𝑐𝑏,𝑎 −1

2𝛿𝑏𝑎ℎ′,𝑎

= ℎ′𝑎𝑏,𝑎 −

1

2ℎ′,𝑏 = 0

12.2 Plane waves

12.2.1 178The Riemann tensor of a plane wave

Here we want to show that the Riemann tensor only depends on ℎ𝑥𝑥, ℎ𝑥𝑦, ℎ𝑦𝑥 and ℎ𝑦𝑦. For symmetry

reasons it is only necessary to show that the Riemann tensor does not depend on ℎ𝑡𝑡 and ℎ𝑡𝑥. The Riemann tensor

177 However I don’t know how to show that the Riemann-tensor keeps the same form if we make this choice 178 (McMahon, 2006, pp. 288,13)





𝑅 𝑏𝑐𝑑𝑎 =

1

2휀𝜂𝑎𝑓 (

𝜕2ℎ𝑑𝑓

𝜕𝑥𝑐𝜕𝑥𝑏−𝜕2ℎ𝑏𝑑𝜕𝑥𝑐𝜕𝑥𝑓

+𝜕2ℎ𝑏𝑐𝜕𝑥𝑑𝜕𝑥𝑓

−𝜕2ℎ𝑐𝑓

𝜕𝑥𝑑𝜕𝑥𝑏) (13.4)

For plane waves we have ℎ𝑎𝑏 = ℎ𝑎𝑏(𝑡 − 𝑧)

⇒ 𝜕ℎ𝑎𝑏𝜕𝑥

=𝜕ℎ𝑎𝑏𝜕𝑦

= 0

We also need 𝜕ℎ𝑎𝑏

𝜕𝑧 =

𝜕(𝑡 − 𝑧)

𝜕𝑧

𝜕ℎ𝑎𝑏𝜕(𝑡 − 𝑧)

= −𝜕ℎ𝑎𝑏

𝜕(𝑡 − 𝑧)

𝜕ℎ𝑎𝑏𝜕𝑡

=𝜕(𝑡 − 𝑧)

𝜕𝑡

𝜕ℎ𝑎𝑏𝜕(𝑡 − 𝑧)

=𝜕ℎ𝑎𝑏

𝜕(𝑡 − 𝑧)

⇒ 𝜕ℎ𝑎𝑏𝜕𝑧

= −𝜕ℎ𝑎𝑏𝜕𝑡

𝜕2ℎ𝑎𝑏𝜕𝑧2

=𝜕2ℎ𝑎𝑏𝜕𝑡2

and

ℎ𝑎𝑏 =

(

ℎ𝑡𝑡 ℎ𝑡𝑥 ℎ𝑡𝑦 ℎ𝑡𝑧ℎ𝑥𝑡 ℎ𝑥𝑥 ℎ𝑥𝑦 ℎ𝑥𝑧ℎ𝑦𝑡 ℎ𝑦𝑥 ℎ𝑦𝑦 ℎ𝑦𝑧ℎ𝑧𝑡 ℎ𝑧𝑥 ℎ𝑧𝑦 ℎ𝑧𝑧)

=

(

ℎ𝑡𝑡 ℎ𝑡𝑥 ℎ𝑡𝑦 −1

2(ℎ𝑡𝑡 + ℎ𝑧𝑧)

ℎ𝑡𝑥 ℎ𝑥𝑥 ℎ𝑥𝑦 −ℎ𝑡𝑥ℎ𝑡𝑦 ℎ𝑥𝑦 −ℎ𝑥𝑥 −ℎ𝑡𝑦

−1

2(ℎ𝑡𝑡 + ℎ𝑧𝑧) −ℎ𝑡𝑥 −ℎ𝑡𝑦 ℎ𝑧𝑧 )

(13.16)

The Minkowski

𝜂𝑎𝑏 = (

1−1

−1−1

)

The dependence on ℎ𝑡𝑡 𝑑 = 𝑓 = 𝑡 (⇒ 𝑎 = 𝑡):

𝑅 𝑏𝑐𝑡𝑡 =

1

2휀𝜂𝑡𝑡 (

𝜕2ℎ𝑡𝑡𝜕𝑥𝑐𝜕𝑥𝑏

−𝜕2ℎ𝑏𝑡𝜕𝑥𝑐𝜕𝑡

+𝜕2ℎ𝑏𝑐𝜕2𝑡

−𝜕2ℎ𝑐𝑡𝜕𝑡𝜕𝑥𝑏

)

𝑏 = 𝑡:

𝑅 𝑡𝑐𝑡𝑡 =

1

2휀 (𝜕2ℎ𝑡𝑡𝜕𝑥𝑐𝜕𝑡

−𝜕2ℎ𝑡𝑡𝜕𝑥𝑐𝜕𝑡

+𝜕2ℎ𝑡𝑐𝜕2𝑡

−𝜕2ℎ𝑐𝑡𝜕2𝑡

) = 0

𝑏 = 𝑥:

𝑅 𝑥𝑐𝑡𝑡 =

1

2휀 (

𝜕2ℎ𝑡𝑡𝜕𝑥𝑐𝜕𝑥

−𝜕2ℎ𝑥𝑡𝜕𝑥𝑐𝜕𝑡

+𝜕2ℎ𝑥𝑐𝜕2𝑡

−𝜕2ℎ𝑐𝑡𝜕𝑡𝜕𝑥

) =1

2휀 (−

𝜕2ℎ𝑥𝑡𝜕𝑥𝑐𝜕𝑡

+𝜕2ℎ𝑥𝑐𝜕2𝑡

)

𝑐 = 𝑡:

𝑅 𝑥𝑡𝑡𝑡 =

1

2휀 (−

𝜕2ℎ𝑥𝑡𝜕2𝑡

+𝜕2ℎ𝑥𝑡𝜕2𝑡

) = 0

𝑐 = 𝑥:

𝑅 𝑥𝑥𝑡𝑡 =

1

2휀 (−

𝜕2ℎ𝑥𝑡𝜕𝑥𝜕𝑡

+𝜕2ℎ𝑥𝑥𝜕2𝑡

) =1

2휀𝜕2ℎ𝑥𝑥𝜕2𝑡

𝑐 = 𝑦:





𝑅 𝑥𝑦𝑡𝑡 =

1

2휀 (−

𝜕2ℎ𝑥𝑡𝜕𝑦𝜕𝑡

+𝜕2ℎ𝑥𝑦

𝜕2𝑡) =

1

2휀𝜕2ℎ𝑥𝑦

𝜕2𝑡

𝑐 = 𝑧:

𝑅 𝑥𝑧𝑡𝑡 =

1

2휀 (−

𝜕2ℎ𝑥𝑡𝜕𝑧𝜕𝑡

+𝜕2ℎ𝑥𝑧𝜕2𝑡

) =1

2휀 (𝜕2ℎ𝑥𝑡𝜕2𝑡

−𝜕2ℎ𝑡𝑥𝜕2𝑡

) = 0

𝑏 = 𝑦:

𝑅 𝑦𝑐𝑡𝑡 =

1

2휀 (

𝜕2ℎ𝑡𝑡𝜕𝑥𝑐𝜕𝑦

−𝜕2ℎ𝑦𝑡

𝜕𝑥𝑐𝜕𝑡+𝜕2ℎ𝑦𝑐

𝜕2𝑡−𝜕2ℎ𝑐𝑡𝜕𝑡𝜕𝑦

) =1

2휀 (−

𝜕2ℎ𝑦𝑡

𝜕𝑥𝑐𝜕𝑡+𝜕2ℎ𝑦𝑐

𝜕2𝑡)

𝑐 = 𝑡:

𝑅 𝑦𝑡𝑡𝑡 =

1

2휀 (−

𝜕2ℎ𝑦𝑡

𝜕2𝑡+𝜕2ℎ𝑦𝑡

𝜕2𝑡) = 0

𝑐 = 𝑥:

𝑅 𝑦𝑥𝑡𝑡 =

1

2휀 (−

𝜕2ℎ𝑦𝑡

𝜕𝑥𝜕𝑡+𝜕2ℎ𝑦𝑥

𝜕2𝑡) =

1

2휀𝜕2ℎ𝑦𝑥

𝜕2𝑡

𝑐 = 𝑦:

𝑅 𝑦𝑦𝑡𝑡 =

1

2휀 (−

𝜕2ℎ𝑦𝑡

𝜕𝑦𝜕𝑡+𝜕2ℎ𝑦𝑦

𝜕2𝑡) =

1

2휀𝜕2ℎ𝑦𝑦

𝜕2𝑡

𝑐 = 𝑧:

𝑅 𝑦𝑧𝑡𝑡 =

1

2휀 (−

𝜕2ℎ𝑦𝑡

𝜕𝑧𝜕𝑡+𝜕2ℎ𝑦𝑧

𝜕2𝑡) =

1

2휀 (𝜕2ℎ𝑦𝑡

𝜕2𝑡−𝜕2ℎ𝑦𝑡

𝜕2𝑡) = 0

𝑏 = 𝑧:

𝑅 𝑧𝑐𝑡𝑡 =

1

2휀 (

𝜕2ℎ𝑡𝑡𝜕𝑥𝑐𝜕𝑧

−𝜕2ℎ𝑧𝑡𝜕𝑥𝑐𝜕𝑡

+𝜕2ℎ𝑧𝑐𝜕2𝑡

−𝜕2ℎ𝑐𝑡𝜕𝑡𝜕𝑧

)

𝑐 = 𝑡:

𝑅 𝑧𝑡𝑡𝑡 =

1

2휀 (𝜕2ℎ𝑡𝑡𝜕𝑡𝜕𝑧

−𝜕2ℎ𝑧𝑡𝜕2𝑡

+𝜕2ℎ𝑧𝑡𝜕2𝑡

−𝜕2ℎ𝑡𝑡𝜕𝑡𝜕𝑧

) = 0

𝑐 = 𝑥:

𝑅 𝑧𝑥𝑡𝑡 =

1

2휀 (𝜕2ℎ𝑡𝑡𝜕𝑥𝜕𝑧

−𝜕2ℎ𝑧𝑡𝜕𝑥𝜕𝑡

+𝜕2ℎ𝑧𝑥𝜕2𝑡

−𝜕2ℎ𝑥𝑡𝜕𝑡𝜕𝑧

) =1

2휀 (−

𝜕2ℎ𝑡𝑥𝜕2𝑡

+𝜕2ℎ𝑥𝑡𝜕2𝑡

) = 0

𝑐 = 𝑦:

𝑅 𝑧𝑦𝑡𝑡 =

1

2휀 (𝜕2ℎ𝑡𝑡𝜕𝑦𝜕𝑧

−𝜕2ℎ𝑧𝑡𝜕𝑦𝜕𝑡

+𝜕2ℎ𝑧𝑦

𝜕2𝑡−𝜕2ℎ𝑦𝑡

𝜕𝑡𝜕𝑧) =

1

2휀 (−

𝜕2ℎ𝑡𝑦

𝜕2𝑡+𝜕2ℎ𝑦𝑡

𝜕2𝑡) = 0

𝑐 = 𝑧:

𝑅 𝑧𝑧𝑡𝑡 =

1

2휀 (𝜕2ℎ𝑡𝑡𝜕2𝑧

−𝜕2ℎ𝑧𝑡𝜕𝑧𝜕𝑡

+𝜕2ℎ𝑧𝑧𝜕2𝑡

−𝜕2ℎ𝑧𝑡𝜕𝑡𝜕𝑧

)

=1

2휀

(

𝜕2ℎ𝑡𝑡𝜕2𝑡

+

𝜕2 (−12(ℎ𝑡𝑡 + ℎ𝑧𝑧))

𝜕2𝑡+𝜕2ℎ𝑧𝑧𝜕2𝑡

+

𝜕2 (−12(ℎ𝑡𝑡 + ℎ𝑧𝑧))

𝜕2𝑡

)

= 0 𝑏 = 𝑑 = 𝑡:

𝑅 𝑡𝑐𝑡𝑎 =

1

2휀𝜂𝑎𝑓 (

𝜕2ℎ𝑡𝑓

𝜕𝑥𝑐𝜕𝑡−

𝜕2ℎ𝑡𝑡𝜕𝑥𝑐𝜕𝑥𝑓

+𝜕2ℎ𝑡𝑐𝜕𝑡𝜕𝑥𝑓

−𝜕2ℎ𝑐𝑓

𝜕2𝑡)

𝑎 = 𝑥(⇒ 𝑓 = 𝑥):





𝑅 𝑡𝑐𝑡𝑥 =

1

2휀𝜂𝑥𝑥 (

𝜕2ℎ𝑡𝑥𝜕𝑥𝑐𝜕𝑡

−𝜕2ℎ𝑡𝑡𝜕𝑥𝑐𝜕𝑥

+𝜕2ℎ𝑡𝑐𝜕𝑡𝜕𝑥

−𝜕2ℎ𝑐𝑥𝜕2𝑡

) = −1

2휀 (𝜕2ℎ𝑡𝑥𝜕𝑥𝑐𝜕𝑡

−𝜕2ℎ𝑐𝑥𝜕2𝑡

)

𝑐 = 𝑥:

𝑅 𝑡𝑥𝑡𝑥 = −

1

2휀 (𝜕2ℎ𝑡𝑥𝜕𝑥𝜕𝑡

−𝜕2ℎ𝑥𝑥𝜕2𝑡

) =1


𝑐 = 𝑦:

𝑅 𝑡𝑦𝑡𝑥 = −

1

2휀 (𝜕2ℎ𝑡𝑥𝜕𝑦𝜕𝑡

−𝜕2ℎ𝑦𝑥

𝜕2𝑡) =

1


𝜕2𝑡

𝑐 = 𝑧:

𝑅 𝑡𝑧𝑡𝑥 = −

1

2휀 (𝜕2ℎ𝑡𝑥𝜕𝑧𝜕𝑡

−𝜕2ℎ𝑧𝑥𝜕2𝑡

) = −1

2휀 (−

𝜕2ℎ𝑡𝑥𝜕2𝑡

+𝜕2ℎ𝑡𝑥𝜕2𝑡

) = 0

𝑎 = 𝑦(⇒ 𝑓 = 𝑦):

𝑅 𝑡𝑐𝑡𝑦

=1

2휀𝜂𝑦𝑦 (

𝜕2ℎ𝑡𝑦

𝜕𝑥𝑐𝜕𝑡−𝜕2ℎ𝑡𝑡𝜕𝑥𝑐𝜕𝑦

+𝜕2ℎ𝑡𝑐𝜕𝑡𝜕𝑦

−𝜕2ℎ𝑐𝑦

𝜕2𝑡) = −

1

2휀 (𝜕2ℎ𝑡𝑦

𝜕𝑥𝑐𝜕𝑡−𝜕2ℎ𝑐𝑦

𝜕2𝑡)

𝑐 = 𝑥:

𝑅 𝑡𝑥𝑡𝑦

= −1


𝜕𝑥𝜕𝑡−𝜕2ℎ𝑥𝑦

𝜕2𝑡) =

1


𝜕2𝑡

𝑐 = 𝑦:

𝑅 𝑡𝑦𝑡𝑦

= −1


𝜕𝑦𝜕𝑡−𝜕2ℎ𝑦𝑦

𝜕2𝑡) =

1


𝜕2𝑡

𝑐 = 𝑧:

𝑅 𝑡𝑧𝑡𝑦

= −1


𝜕𝑧𝜕𝑡−𝜕2ℎ𝑧𝑦

𝜕2𝑡) = −

1

2휀 (−

𝜕2ℎ𝑡𝑦

𝜕2𝑡+𝜕2ℎ𝑡𝑦

𝜕2𝑡) = 0

𝑎 = 𝑧(⇒ 𝑓 = 𝑧):

𝑅 𝑡𝑐𝑡𝑧

=1

2휀𝜂𝑧𝑧 (

𝜕2ℎ𝑡𝑧𝜕𝑥𝑐𝜕𝑡

−𝜕2ℎ𝑡𝑡𝜕𝑥𝑐𝜕𝑧

+𝜕2ℎ𝑡𝑐𝜕𝑡𝜕𝑧

−𝜕2ℎ𝑐𝑧𝜕2𝑡

)

= −1

2휀 (𝜕2ℎ𝑡𝑧𝜕𝑥𝑐𝜕𝑡

−𝜕2ℎ𝑡𝑡𝜕𝑥𝑐𝜕𝑧

+𝜕2ℎ𝑡𝑐𝜕𝑡𝜕𝑧

−𝜕2ℎ𝑐𝑧𝜕2𝑡

)

𝑐 = 𝑥:

𝑅 𝑡𝑥𝑡𝑧 = −

1

2휀 (𝜕2ℎ𝑡𝑧𝜕𝑥𝜕𝑡

−𝜕2ℎ𝑡𝑡𝜕𝑥𝜕𝑧

+𝜕2ℎ𝑡𝑥𝜕𝑡𝜕𝑧

−𝜕2ℎ𝑥𝑧𝜕2𝑡

) = 0

𝑐 = 𝑦:

𝑅 𝑡𝑦𝑡𝑧 = −

1

2휀 (𝜕2ℎ𝑡𝑧𝜕𝑦𝜕𝑡

−𝜕2ℎ𝑡𝑡𝜕𝑦𝜕𝑧

+𝜕2ℎ𝑡𝑦

𝜕𝑡𝜕𝑧−𝜕2ℎ𝑦𝑧

𝜕2𝑡) = 0

𝑐 = 𝑧:

𝑅 𝑡𝑧𝑡𝑧 = −

1

2휀 (𝜕2ℎ𝑡𝑧𝜕𝑧𝜕𝑡

−𝜕2ℎ𝑡𝑡𝜕2𝑧

+𝜕2ℎ𝑡𝑧𝜕𝑡𝜕𝑧

−𝜕2ℎ𝑧𝑧𝜕2𝑡

) = 0

The dependence on ℎ𝑡𝑥 𝑑 = 𝑡, 𝑓 = 𝑥 (⇒ 𝑎 = 𝑥):

𝑅 𝑏𝑐𝑡𝑥 =

1

2휀𝜂𝑥𝑥 (

𝜕2ℎ𝑡𝑥𝜕𝑥𝑐𝜕𝑥𝑏

−𝜕2ℎ𝑏𝑡𝜕𝑥𝑐𝜕𝑥

+𝜕2ℎ𝑏𝑐𝜕𝑡𝜕𝑥

−𝜕2ℎ𝑐𝑥𝜕𝑡𝜕𝑥𝑏

) = −1

2휀 (

𝜕2ℎ𝑡𝑥𝜕𝑥𝑐𝜕𝑥𝑏

−𝜕2ℎ𝑐𝑥𝜕𝑡𝜕𝑥𝑏

)

𝑏 = 𝑥: 𝑅 𝑥𝑐𝑡

𝑥 = 0 𝑏 = 𝑦: 𝑅 𝑦𝑐𝑡

𝑥 = 0

𝑏 = 𝑧:





𝑅 𝑧𝑐𝑡𝑥 = −

1

2휀 (𝜕2ℎ𝑡𝑥𝜕𝑥𝑐𝜕𝑧

−𝜕2ℎ𝑐𝑥𝜕𝑡𝜕𝑧

)

𝑏 = 𝑡, 𝑑 = 𝑥:

𝑅 𝑡𝑐𝑥𝑎 =

1

2휀𝜂𝑎𝑓 (

𝜕2ℎ𝑥𝑓


𝜕2ℎ𝑡𝑥𝜕𝑥𝑐𝜕𝑥𝑓

+𝜕2ℎ𝑡𝑐𝜕𝑥𝜕𝑥𝑓

−𝜕2ℎ𝑐𝑓

𝜕𝑥𝜕𝑡) =

1

2휀𝜂𝑎𝑓 (

𝜕2ℎ𝑥𝑓


𝜕2ℎ𝑡𝑥𝜕𝑥𝑐𝜕𝑥𝑓

)

𝑎 = 𝑡(⇒ 𝑓 = 𝑡):

𝑅 𝑡𝑐𝑥𝑡 =

1

2휀𝜂𝑡𝑡 (

𝜕2ℎ𝑥𝑡𝜕𝑥𝑐𝜕𝑡

−𝜕2ℎ𝑡𝑥𝜕𝑥𝑐𝜕𝑡

) = 0

𝑎 = 𝑥(⇒ 𝑓 = 𝑥):

𝑅 𝑡𝑐𝑥𝑥 =

1

2휀𝜂𝑥𝑥 (

𝜕2ℎ𝑥𝑥𝜕𝑥𝑐𝜕𝑡

−𝜕2ℎ𝑡𝑥𝜕𝑥𝑐𝜕𝑥

) = −1

2휀𝜕2ℎ𝑥𝑥𝜕𝑥𝑐𝜕𝑡

𝑎 = 𝑦(⇒ 𝑓 = 𝑦):

𝑅 𝑡𝑐𝑥𝑦

=1

2휀𝜂𝑦𝑦 (

𝜕2ℎ𝑥𝑦

𝜕𝑥𝑐𝜕𝑡−𝜕2ℎ𝑡𝑥𝜕𝑥𝑐𝜕𝑦

) = −1


𝜕𝑥𝑐𝜕𝑡

𝑎 = 𝑧(⇒ 𝑓 = 𝑧):

𝑅 𝑡𝑐𝑥𝑧 =

1

2휀𝜂𝑧𝑧 (

𝜕2ℎ𝑥𝑧𝜕𝑥𝑐𝜕𝑡

−𝜕2ℎ𝑡𝑥𝜕𝑥𝑐𝜕𝑧

) =1

2휀𝜂𝑧𝑧 (−

𝜕2ℎ𝑡𝑥𝜕𝑥𝑐𝜕𝑡

+𝜕2ℎ𝑡𝑥𝜕𝑥𝑐𝜕𝑡

) = 0

𝑏 = 𝑡, 𝑐 = 𝑥:

𝑅 𝑡𝑥𝑑𝑎 =

1

2휀𝜂𝑎𝑓 (

𝜕2ℎ𝑑𝑓

𝜕𝑥𝜕𝑡−𝜕2ℎ𝑡𝑑𝜕𝑥𝜕𝑥𝑓

+𝜕2ℎ𝑡𝑥𝜕𝑥𝑑𝜕𝑥𝑓

−𝜕2ℎ𝑥𝑓

𝜕𝑥𝑑𝜕𝑡) =

1

2휀𝜂𝑎𝑓 (

𝜕2ℎ𝑡𝑥𝜕𝑥𝑑𝜕𝑥𝑓

−𝜕2ℎ𝑥𝑓

𝜕𝑥𝑑𝜕𝑡)

𝑎 = 𝑡(⇒ 𝑓 = 𝑡):

𝑅 𝑡𝑥𝑑𝑡 =

1

2휀𝜂𝑡𝑡 (

𝜕2ℎ𝑡𝑥𝜕𝑥𝑑𝜕𝑡

−𝜕2ℎ𝑥𝑡𝜕𝑥𝑑𝜕𝑡

) = 0

𝑎 = 𝑥(⇒ 𝑓 = 𝑥):

𝑅 𝑡𝑥𝑑𝑥 =

1

2휀𝜂𝑥𝑥 (

𝜕2ℎ𝑡𝑥𝜕𝑥𝑑𝜕𝑥

−𝜕2ℎ𝑥𝑥𝜕𝑥𝑑𝜕𝑡

) =1

2휀𝜕2ℎ𝑥𝑥𝜕𝑥𝑑𝜕𝑡

𝑎 = 𝑦(⇒ 𝑓 = 𝑦):

𝑅 𝑡𝑥𝑑𝑦

=1

2휀𝜂𝑦𝑦 (

𝜕2ℎ𝑡𝑥𝜕𝑥𝑑𝜕𝑦

−𝜕2ℎ𝑥𝑦

𝜕𝑥𝑑𝜕𝑡) =

1


𝜕𝑥𝑑𝜕𝑡

𝑎 = 𝑧(⇒ 𝑓 = 𝑧):

𝑅 𝑡𝑥𝑑𝑧 =

1

2휀𝜂𝑧𝑧 (

𝜕2ℎ𝑡𝑥𝜕𝑥𝑑𝜕𝑧

−𝜕2ℎ𝑥𝑧𝜕𝑥𝑑𝜕𝑡

) = −1

2휀 (−

𝜕2ℎ𝑡𝑥𝜕𝑥𝑑𝜕𝑡

+𝜕2ℎ𝑡𝑥𝜕𝑥𝑑𝜕𝑡

) = 0

𝑐 = 𝑡, 𝑓 = 𝑥(⇒ 𝑎 = 𝑥):

𝑅 𝑏𝑡𝑑𝑥 =

1

2휀𝜂𝑥𝑥 (

𝜕2ℎ𝑑𝑥𝜕𝑡𝜕𝑥𝑏

−𝜕2ℎ𝑏𝑑𝜕𝑡𝜕𝑥

+𝜕2ℎ𝑏𝑡𝜕𝑥𝑑𝜕𝑥

−𝜕2ℎ𝑡𝑥𝜕𝑥𝑑𝜕𝑥𝑏

) = −1

2휀 (𝜕2ℎ𝑑𝑥𝜕𝑡𝜕𝑥𝑏

−𝜕2ℎ𝑡𝑥𝜕𝑥𝑑𝜕𝑥𝑏

)

The nonzero calculated elements of the Riemann tensor, from which we can conclude that the Rie-

mann tensor only depends on ℎ𝑥𝑥, ℎ𝑥𝑦, ℎ𝑦𝑥 and ℎ𝑦𝑦 .

𝑅 𝑥𝑥𝑡𝑡 =

1


𝑅 𝑡𝑥𝑡𝑥 =

1


𝑅 𝑡𝑥𝑡𝑦

=1


𝜕2𝑡

𝑅 𝑥𝑦𝑡𝑡 =

1


𝜕2𝑡 𝑅 𝑡𝑥𝑧

𝑥 = −1


𝑅 𝑡𝑥𝑧𝑦

= −1


𝜕2𝑡

𝑅 𝑦𝑥𝑡𝑡 =

1


𝜕2𝑡 𝑅 𝑡𝑦𝑡

𝑥 =1


𝜕2𝑡 𝑅 𝑡𝑦𝑡

𝑦 =

1


𝜕2𝑡





𝑅 𝑦𝑦𝑡𝑡 =

1


𝜕2𝑡 𝑅 𝑡𝑧𝑥

𝑥 =1


𝑅 𝑡𝑧𝑥𝑦

=1


𝜕2𝑡

𝑅 𝑧𝑥𝑡𝑥 = −

1


𝑅 𝑧𝑦𝑡𝑥 = −

1


𝜕2𝑡

12.2.2 179The line element of a plane wave in the Einstein gauge

The perturbation

ℎ𝑎𝑏 =

(

ℎ𝑡𝑡 ℎ𝑡𝑥 ℎ𝑡𝑦 −1

2(ℎ𝑡𝑡 + ℎ𝑧𝑧)

ℎ𝑡𝑥 ℎ𝑥𝑥 ℎ𝑥𝑦 −ℎ𝑡𝑥ℎ𝑡𝑦 ℎ𝑥𝑦 −ℎ𝑥𝑥 −ℎ𝑡𝑦

−1

2(ℎ𝑡𝑡 + ℎ𝑧𝑧) −ℎ𝑡𝑥 −ℎ𝑡𝑦 ℎ𝑧𝑧 )

(13.16)

the perturbation in the Einstein gauge

ℎ′𝑎𝑏 = (

0 0 0 00 ℎ′𝑥𝑥 ℎ′𝑥𝑦 0

0 ℎ′𝑥𝑦 −ℎ′𝑥𝑥 0

0 0 0 0

) (13.17)

with the transformation ℎ´𝑎𝑏 = ℎ𝑎𝑏 − 𝜙𝑏,𝑎 −𝜙𝑎,𝑏 where we can assume that 𝜙𝑎 = 𝜙𝑎(𝑡 − 𝑧) How to prove (13.17)180:

1) ℎ𝑥𝑥 and ℎ𝑥𝑦 are unchanged by the transformation

ℎ´𝑥𝑥 = ℎ𝑥𝑥 −𝜙𝑥,𝑥 −𝜙𝑥,𝑥 = ℎ𝑥𝑥 − 2

𝜕𝜙𝑥𝜕𝑥

= ℎ𝑥𝑥

ℎ´𝑥𝑦 = ℎ𝑥𝑦 − 𝜙𝑦,𝑥 − 𝜙𝑥,𝑦 = ℎ𝑥𝑦 −

𝜕𝜙𝑦

𝜕𝑥−𝜕𝜙𝑥𝜕𝑦

= ℎ𝑥𝑦

2) Choosing the remaining elements ℎ𝑎𝑏′ = 0 leaves ℎ𝑥𝑥 and ℎ𝑥𝑦 unchanged

ℎ´𝑡𝑡 = ℎ𝑡𝑡 − 𝜙𝑡,𝑡 − 𝜙𝑡,𝑡 = ℎ𝑡𝑡 − 2

𝜕𝜙𝑡𝜕𝑡

= 0

⇔ ℎ𝑡𝑡 = 2𝜕𝜙𝑡𝜕𝑡

ℎ´𝑡𝑥 = ℎ𝑡𝑥 −𝜙𝑥,𝑡 − 𝜙𝑡,𝑥 = ℎ𝑡𝑥 −

𝜕𝜙𝑥𝜕𝑡

−𝜕𝜙𝑡𝜕𝑥

= ℎ𝑡𝑥 −𝜕𝜙𝑥𝜕𝑡

= 0

⇔ ℎ𝑡𝑥 =𝜕𝜙𝑥𝜕𝑡

ℎ´𝑡𝑦 = ℎ𝑡𝑦 − 𝜙𝑦,𝑡 − 𝜙𝑡,𝑦 = ℎ𝑡𝑦 −𝜕𝜙𝑦

𝜕𝑡−𝜕𝜙𝑦

𝜕𝑥= ℎ𝑡𝑦 −

𝜕𝜙𝑦

𝜕𝑡= 0

⇔ ℎ𝑡𝑦 =𝜕𝜙𝑦

𝜕𝑡

ℎ´𝑡𝑧 = ℎ𝑡𝑧 − 𝜙𝑧,𝑡 − 𝜙𝑡,𝑧 = ℎ𝑡𝑧 −

𝜕𝜙𝑧𝜕𝑡

−𝜕𝜙𝑡𝜕𝑧

= 0

ℎ´𝑧𝑡 = ℎ𝑧𝑡 − 𝜙𝑡,𝑧 −𝜙𝑧,𝑡 = ℎ𝑧𝑡 −𝜕𝜙𝑡𝜕𝑧

−𝜕𝜙𝑧𝜕𝑡

= 0

⇔ ℎ𝑡𝑧 = ℎ𝑧𝑡 =𝜕𝜙𝑧𝜕𝑡

+𝜕𝜙𝑡𝜕𝑧

179 (McMahon, 2006, pp. 290,12) 180 (d'Inverno, 1992, pp. 277-278)





ℎ´𝑧𝑧 = ℎ𝑧𝑧 − 𝜙𝑧,𝑧 − 𝜙𝑧,𝑧 = ℎ𝑧𝑧 − 2

𝜕𝜙𝑧𝜕𝑧

= 0

⇔ ℎ𝑧𝑧 = 2𝜕𝜙𝑧𝜕𝑧

12.2.3 181The line element of a plane wave

With

ℎ𝑎𝑏 = (

0 0 0 00 ℎ𝑥𝑥 ℎ𝑥𝑦 0

0 ℎ𝑥𝑦 −ℎ𝑥𝑥 0

0 0 0 0

) (13.17)

we find the line element 𝑑𝑠2 = 𝑔𝑎𝑏𝑑𝑥

𝑎𝑑𝑥𝑏 = (𝜂𝑎𝑏 + 휀ℎ𝑎𝑏)𝑑𝑥

𝑎𝑑𝑥𝑏 = (𝜂𝑡𝑡 + 휀ℎ𝑡𝑡)𝑑𝑡

2 + (𝜂𝑡𝑥 + 휀ℎ𝑡𝑥)𝑑𝑡𝑑𝑥 + (𝜂𝑡𝑦 + 휀ℎ𝑡𝑦)𝑑𝑡𝑑𝑦 + (𝜂𝑡𝑧 + 휀ℎ𝑡𝑧)𝑑𝑡𝑑𝑧

+ (𝜂𝑥𝑡 + 휀ℎ𝑥𝑡)𝑑𝑥𝑑𝑡 + (𝜂𝑥𝑥 + 휀ℎ𝑥𝑥)𝑑𝑥2 + (𝜂𝑥𝑦 + 휀ℎ𝑥𝑦)𝑑𝑥𝑑𝑦

+ (𝜂𝑥𝑧 + 휀ℎ𝑥𝑧)𝑑𝑥𝑑𝑧 + (𝜂𝑦𝑡 + 휀ℎ𝑦𝑡)𝑑𝑦𝑑𝑡 + (𝜂𝑦𝑥 + 휀ℎ𝑦𝑥)𝑑𝑦𝑑𝑥

+ (𝜂𝑦𝑦 + 휀ℎ𝑦𝑦)𝑑𝑦2 + (𝜂𝑦𝑧 + 휀ℎ𝑦𝑧)𝑑𝑦𝑑𝑧 + (𝜂𝑧𝑡 + 휀ℎ𝑧𝑡)𝑑𝑧𝑑𝑡

+ (𝜂𝑧𝑥 + 휀ℎ𝑧𝑥)𝑑𝑧𝑑𝑥 + (𝜂𝑧𝑦 + 휀ℎ𝑧𝑦)𝑑𝑧𝑑𝑦 + (𝜂𝑧𝑧 + 휀ℎ𝑧𝑧)𝑑𝑦2

= 𝑑𝑡2 − (1 − 휀ℎ𝑥𝑥)𝑑𝑥2 + 휀ℎ𝑥𝑦𝑑𝑥𝑑𝑦 + 휀ℎ𝑦𝑥𝑑𝑦𝑑𝑥 − (1 + 휀ℎ𝑥𝑥)𝑑𝑦

2 − 𝑑𝑧2

= 𝑑𝑡2 − (1 − 휀ℎ𝑥𝑥)𝑑𝑥2 − (1 + 휀ℎ𝑥𝑥)𝑑𝑦

2 − 𝑑𝑧2 + 2휀ℎ𝑥𝑦𝑑𝑥𝑑𝑦

𝒉𝒙𝒚 = 𝟎:

𝑑𝑠2 = 𝑑𝑡2 − (1 − 휀ℎ𝑥𝑥)𝑑𝑥2 − (1 + 휀ℎ𝑥𝑥)𝑑𝑦

2 − 𝑑𝑧2 (13.18) 𝒉𝒙𝒙 = 𝟎: 𝑑𝑠2 = 𝑑𝑡2 − 𝑑𝑥2 − 𝑑𝑦2 − 𝑑𝑧2 + 2휀ℎ𝑥𝑦𝑑𝑥𝑑𝑦 (13.19)

Considering the following transformation

𝑑𝑥′ =𝑑𝑥 − 𝑑𝑦

√2 𝑑𝑦′ =

𝑑𝑥 + 𝑑𝑦

√2

⇒ 𝑑𝑥 =1

√2(𝑑𝑥′ + 𝑑𝑦′) 𝑑𝑦 =

−1

√2(𝑑𝑥′ − 𝑑𝑦′)

⇒ 𝑑𝑥2 =1

2𝑑𝑥′2 +

1

2𝑑𝑦′2 + 𝑑𝑥′𝑑𝑦′ 𝑑𝑦2 =

1

2𝑑𝑥′2 +

1

2𝑑𝑦′2 − 𝑑𝑥′𝑑𝑦′

⇒ 𝑑𝑥𝑑𝑦 = −1

2(𝑑𝑥′2 − 𝑑𝑦′2) 𝑑𝑥2 + 𝑑𝑦2 = 𝑑𝑥′2 + 𝑑𝑦′2

we can rewrite the line element 𝑑𝑠2 = 𝑑𝑡2 − 𝑑𝑥2 − 𝑑𝑦2 − 𝑑𝑧2 + 2휀ℎ𝑥𝑦𝑑𝑥𝑑𝑦

= 𝑑𝑡2 − 𝑑𝑥′2 − 𝑑𝑦′2 − 𝑑𝑧2 − 휀ℎ𝑥𝑦(𝑑𝑥′2 − 𝑑𝑦′2)

= 𝑑𝑡2 − (1 + 휀ℎ𝑥𝑦)𝑑𝑥′2 − (1 − 휀ℎ𝑥𝑦)𝑑𝑦

′2 − 𝑑𝑧2 (13.20)

12.2.4 182The Rosen line element

The line element: 𝑑𝑠2 = 𝑑𝑈𝑑𝑉 − 𝑎2(𝑈)𝑑𝑥2 − 𝑏2(𝑈)𝑑𝑦2

181 (McMahon, 2006, p. 291) 182 (McMahon, 2006, p. 298)





The metric tensor: 𝑔𝑎𝑏 =

{

1

21

2−𝑎2(𝑈)

−𝑏2(𝑈)}

The basis one forms

Finding the basis one forms is not so obvious, we write: 𝑑𝑠2 = 𝑑𝑈𝑑𝑉 − 𝑎2(𝑈)𝑑𝑥2 − 𝑏2(𝑈)𝑑𝑦2

= (𝜔0)2− (𝜔1)

2− (𝜔2)

2− (𝜔3)

2

= (𝜔0 +𝜔1)(𝜔0 −𝜔1) − (𝜔2)2− (𝜔3)

2

⇒ 𝑑𝑈 = 𝜔0 +𝜔1

𝑑𝑉 = 𝜔0 −𝜔1

⇒ 𝜔0 =1

2(𝑑𝑈 + 𝑑𝑉) 𝑑𝑈 = 𝜔0 +𝜔1

𝜂𝑖𝑗 = {

1−1

−1−1

}

𝜔1 =1

2(𝑑𝑈 − 𝑑𝑉) 𝑑𝑉 = 𝜔0 −𝜔1

𝜔2 = 𝑎(𝑈)𝑑𝑥 𝑑𝑥 =1

𝑎(𝑈)𝜔2

𝜔3 = 𝑏(𝑈)𝑑𝑦 𝑑𝑦 =1

𝑏(𝑈)𝜔3


𝑑𝜔�� = −Γ �� ∧ 𝜔�� (5.9)

𝑑𝜔0 = 𝑑 (1

2(𝑑𝑈 + 𝑑𝑉)) = 0

𝑑𝜔1 = 𝑑 (1

2(𝑑𝑈 − 𝑑𝑉)) = 0

𝑑𝜔2 = 𝑑(𝑎(𝑈)𝑑𝑥) =𝑑𝑎

𝑑𝑈𝑑𝑈 ∧ 𝑑𝑥 =

𝑑𝑎

𝑑𝑈(𝜔0 +𝜔1) ∧

1

𝑎(𝑈)𝜔2 = −

1

𝑎

𝑑𝑎

𝑑𝑈𝜔2 ∧ (𝜔0 +𝜔1)

𝑑𝜔3 = 𝑑(𝑏(𝑈)𝑑𝑦) =𝑑𝑏

𝑑𝑈𝑑𝑈 ∧ 𝑑𝑦 =

𝑑𝑏

𝑑𝑈(𝜔0 +𝜔1) ∧

1

𝑏(𝑈)𝜔3 = −

1

𝑏

𝑑𝑏

𝑑𝑈𝜔3 ∧ (𝜔0 +𝜔1)






Γ �� =

{

0 0

1

𝑎

𝑑𝑎

𝑑𝑈𝜔2(𝐴)

1

𝑏

𝑑𝑏

𝑑𝑈𝜔3(𝐵)

0 01

𝑎

𝑑𝑎

𝑑𝑈𝜔2(𝐴)

1

𝑏

𝑑𝑏

𝑑𝑈𝜔3(𝐵)

1

𝑎

𝑑𝑎

𝑑𝑈𝜔2(𝐴) −

1

𝑎

𝑑𝑎

𝑑𝑈𝜔2(−𝐴) 0 0

1

𝑏

𝑑𝑏

𝑑𝑈𝜔3(𝐵) −

1

𝑏

𝑑𝑏

𝑑𝑈𝜔3(−𝐵) 0 0 }

Where �� refers to column and �� to row and A and B will be used later, to make the calculations easier


Ω �� = 𝑑Γ ��

�� + Γ 𝑐�� ∧ Γ ��

𝑐 =1

2𝑅 ��𝑐�� 𝜔𝑐 ∧ 𝜔�� (5.27), (5.28)

First we will calculate

𝑑𝐴 = 𝑑 (1

𝑎

𝑑𝑎

𝑑𝑈𝜔2) = 𝑑 (

𝑑𝑎

𝑑𝑈𝑑𝑥) =

𝑑2𝑎

𝑑𝑈2𝑑𝑈 ∧ 𝑑𝑥 =

𝑑2𝑎

𝑑𝑈2(𝜔0 +𝜔1) ∧

1

𝑎𝜔2

=1

𝑎

𝑑2𝑎

𝑑𝑈2(𝜔0 ∧ 𝜔2 +𝜔1 ∧ 𝜔2)

𝑑𝐵 = 𝑑 (1

𝑏

𝑑𝑏

𝑑𝑈𝜔3) = 𝑑 (

𝑑𝑏

𝑑𝑈𝑑𝑦) =

𝑑2𝑏

𝑑𝑈2𝑑𝑈 ∧ 𝑑𝑦 =

𝑑2𝑏

𝑑𝑈2(𝜔0 +𝜔1) ∧

1

𝑏𝜔3

=1

𝑏

𝑑2𝑏

𝑑𝑈2(𝜔0 ∧ 𝜔3 +𝜔1 ∧ 𝜔3)

Now we are ready to calculate the curvature two-forms

Ω 00 = 𝑑Γ 0

0 + Γ 𝑐0 ∧ Γ 0

𝑐 = Γ 00 ∧ Γ 0

0 + Γ 10 ∧ Γ 0

1 + Γ 20 ∧ Γ 0

2 + Γ 30 ∧ Γ 0

3 = 0

Ω 01 = 𝑑Γ 0

1 + Γ 𝑐1 ∧ Γ 0

𝑐 = Γ 01 ∧ Γ 0

0 + Γ 11 ∧ Γ 0

1 + Γ 21 ∧ Γ 0

2 + Γ 31 ∧ Γ 0

3 = 0

Ω 02 = 𝑑Γ 0

2 + Γ 𝑐2 ∧ Γ 0

𝑐 = 𝑑Γ 02 + Γ 0

2 ∧ Γ 0 0 + Γ 1

2 ∧ Γ 0 1 + Γ 2

2 ∧ Γ 0 2 + Γ 3

2 ∧ Γ 0 3

=

1

𝑎

𝑑2𝑎

𝑑𝑈2(𝜔0 ∧ 𝜔2 +𝜔1 ∧ 𝜔2)

Ω 03 = 𝑑Γ 0

3 + Γ 𝑐3 ∧ Γ 0

𝑐 = 𝑑Γ 03 + Γ 0

3 ∧ Γ 0 0 + Γ 1

3 ∧ Γ 0 1 + Γ 2

3 ∧ Γ 0 2 + Γ 3

3 ∧ Γ 0 3

=

1

𝑏

𝑑2𝑏

𝑑𝑈2(𝜔0 ∧ 𝜔3 +𝜔1 ∧ 𝜔3)

Ω 11 = 𝑑Γ 1

1 + Γ 𝑐1 ∧ Γ 1

𝑐 = Γ 01 ∧ Γ 1

0 + Γ 11 ∧ Γ 1

1 + Γ 21 ∧ Γ 1

2 + Γ 31 ∧ Γ 1

3 = 0

Ω 12 = 𝑑Γ 1

2 + Γ 𝑐2 ∧ Γ 1

𝑐 = 𝑑Γ 12 + Γ 0

2 ∧ Γ 1 0 + Γ 1

2 ∧ Γ 1 1 + Γ 2

2 ∧ Γ 1 2 + Γ 3

2 ∧ Γ 1 3

=

1

𝑎

𝑑2𝑎

𝑑𝑈2(𝜔0 ∧ 𝜔2 +𝜔1 ∧ 𝜔2)

Ω 13 = 𝑑Γ 1

3 + Γ 𝑐3 ∧ Γ 1

𝑐 = 𝑑Γ 13 + Γ 0

3 ∧ Γ 1 0 + Γ 1

3 ∧ Γ 1 1 + Γ 2

3 ∧ Γ 1 2 + Γ 3

3 ∧ Γ 1 3

=1

𝑏

𝑑2𝑏

𝑑𝑈2(𝜔0 ∧ 𝜔3 +𝜔1 ∧ 𝜔3)

Ω 22 = 𝑑Γ 2

2 + Γ 𝑐2 ∧ Γ 2

𝑐 = Γ 02 ∧ Γ 2

0 + Γ 12 ∧ Γ 2

1 + Γ 22 ∧ Γ 2

2 + Γ 32 ∧ Γ 2

3 = 0

Ω 23 = 𝑑Γ 2

3 + Γ 𝑐3 ∧ Γ 2

𝑐 = Γ 03 ∧ Γ 2

0 + Γ 13 ∧ Γ 2

1 + Γ 23 ∧ Γ 2

2 + Γ 33 ∧ Γ 2

3

=1

𝑏

𝑑𝑏

𝑑𝑈𝜔3 ∧

1

𝑎

𝑑𝑎

𝑑𝑈𝜔2 +

1

𝑏

𝑑𝑏

𝑑𝑈𝜔3 ∧ (−

1

𝑎

𝑑𝑎

𝑑𝑈𝜔2) = 0





Ω 33 = 𝑑Γ 3

3 + Γ 𝑐3 ∧ Γ 3

𝑐 = Γ 03 ∧ Γ 3

0 + Γ 13 ∧ Γ 3

1 + Γ 23 ∧ Γ 3

2 + Γ 33 ∧ Γ 3

3 = 0


Ω �� =

{

0 0

1

𝑎

𝑑2𝑎

𝑑𝑈2(𝜔0 ∧ 𝜔2 +𝜔1 ∧ 𝜔2)

1

𝑏

𝑑2𝑏

𝑑𝑈2(𝜔0 ∧ 𝜔3 +𝜔1 ∧ 𝜔3)

0 01

𝑎

𝑑2𝑎

𝑑𝑈2(𝜔0 ∧ 𝜔2 +𝜔1 ∧ 𝜔2)

1

𝑏

𝑑2𝑏

𝑑𝑈2(𝜔0 ∧ 𝜔3 +𝜔1 ∧ 𝜔3)

𝑆 𝐴𝑆 0 0𝑆 𝐴𝑆 0 0 }



R 0202 = −

1

𝑎

𝑑2𝑎

𝑑𝑈2 R 030

3 = −1

𝑏

𝑑2𝑏

𝑑𝑈2

R 0212 = −

1

𝑎

𝑑2𝑎

𝑑𝑈2 R 031

3 = −1

𝑏

𝑑2𝑏

𝑑𝑈2

R 1212 = −

1

𝑎

𝑑2𝑎

𝑑𝑈2 R 131

3 = −1

𝑏

𝑑2𝑏

𝑑𝑈2

The Ricci tensor:

𝑅�� = 𝑅 ��𝑐��𝑐 (4.46)

𝑅00 = 𝑅 0𝑐0𝑐 = 𝑅 000

0 + 𝑅 0101 + 𝑅 020

2 + 𝑅 0303 = −(

1

𝑎

𝑑2𝑎

𝑑𝑈2+1

𝑏

𝑑2𝑏

𝑑𝑈2)

𝑅10 = 𝑅 1𝑐0𝑐 = 𝑅 100

0 + 𝑅 1101 + 𝑅 120

2 + 𝑅 1303 = −(

1

𝑎

𝑑2𝑎

𝑑𝑈2+1

𝑏

𝑑2𝑏

𝑑𝑈2)

𝑅20 = 𝑅 2𝑐0𝑐 = 𝑅 200

0 + 𝑅 2101 + 𝑅 220

2 + 𝑅 2303 = 0

𝑅30 = 𝑅 3𝑐0𝑐 = 𝑅 300

0 + 𝑅 3101 + 𝑅 320

2 + 𝑅 3303 = 0

𝑅01 = 𝑅 0𝑐1𝑐 = 𝑅 001

0 + 𝑅 0111 + 𝑅 021

2 + 𝑅 0313 = −(

1

𝑎

𝑑2𝑎

𝑑𝑈2+1

𝑏

𝑑2𝑏

𝑑𝑈2)

𝑅11 = 𝑅 1𝑐1𝑐 = 𝑅 101

0 + 𝑅 1111 + 𝑅 121

2 + 𝑅 1313 = −(

1

𝑎

𝑑2𝑎

𝑑𝑈2+1

𝑏

𝑑2𝑏

𝑑𝑈2)

𝑅12 = 𝑅 1𝑐2𝑐 = 𝑅 102

0 + 𝑅 1121 + 𝑅 122

2 + 𝑅 1323 = 0

𝑅13 = 𝑅 1𝑐3𝑐 = 𝑅 103

0 + 𝑅 1131 + 𝑅 123

2 + 𝑅 1333 = 0

𝑅02 = 𝑅 0𝑐2𝑐 = 𝑅 002

0 + 𝑅 0121 + 𝑅 022

2 + 𝑅 0323 = 0

𝑅12 = 𝑅 1𝑐2𝑐 = 𝑅 102

0 + 𝑅 1121 + 𝑅 122

2 + 𝑅 1323 = 0

𝑅22 = 𝑅 2𝑐2𝑐 = 𝑅 202

0 + 𝑅 2121 + 𝑅 222

2 + 𝑅 2323 =

1

𝑎

𝑑2𝑎

𝑑𝑈2−1

𝑎

𝑑2𝑎

𝑑𝑈2= 0

𝑅32 = 𝑅 3𝑐2𝑐 = 0

𝑅03 = 𝑅 0𝑐3𝑐 = 𝑅 003

0 + 𝑅 0131 + 𝑅 023

2 + 𝑅 0333 = 0

𝑅13 = 𝑅 1𝑐3𝑐 = 𝑅 103

0 + 𝑅 1131 + 𝑅 123

2 + 𝑅 1333 = 0





𝑅33 = 𝑅 3𝑐3𝑐 = 𝑅 303

0 + 𝑅 3131 + 𝑅 323

2 + 𝑅 3333 =

1

𝑏

𝑑2𝑏

𝑑𝑈2−1

𝑏

𝑑2𝑏

𝑑𝑈2= 0


𝑅�� =

{

−(

1

𝑎

𝑑2𝑎

𝑑𝑈2+1

𝑏

𝑑2𝑏

𝑑𝑈2) −(

1

𝑎

𝑑2𝑎

𝑑𝑈2+1

𝑏

𝑑2𝑏

𝑑𝑈2) 0 0

−(1

𝑎

𝑑2𝑎

𝑑𝑈2+1

𝑏

𝑑2𝑏

𝑑𝑈2) −(

1

𝑎

𝑑2𝑎

𝑑𝑈2+1

𝑏

𝑑2𝑏

𝑑𝑈2) 0 0

0 0 0 00 0 0 0}


12.3 183Colliding gravity waves - coordinate transformation The metric of a plane gravitational wave

𝑑𝑠2 = 𝛿(𝑢)(𝑋2 − 𝑌2)𝑑𝑢2 + 2𝑑𝑢𝑑𝑟 − 𝑑𝑋2 − 𝑑𝑌2 (13.42) can be written in terms of the null coordinates 𝑢 and 𝜈 by using the following coordinate transfor-mation184

𝑢 = 𝑢

𝑟 = 𝜈 −1

2Θ(𝑢)(1 − 𝑢)𝑥2 +

1

2Θ(𝑢)(1 + 𝑢)𝑦2

𝑋 = (1 − 𝑢Θ(𝑢))𝑥

𝑌 = (1 + 𝑢Θ(𝑢))𝑦

𝑑𝑢 = 𝑑𝑢

𝑑𝑟 =𝜕𝑟

𝜕𝜈𝑑𝜈 +

𝜕𝑟

𝜕𝑢𝑑𝑢 +

𝜕𝑟

𝜕𝑥𝑑𝑥 +

𝜕𝑟

𝜕𝑦𝑑𝑦

= 𝑑𝜈 − (1

2(Θ′(𝑢)(1 − 𝑢) − Θ(𝑢))𝑥2 −

1

2(Θ′(𝑢)(1 + 𝑢) + Θ(𝑢))𝑦2)𝑑𝑢

− Θ(𝑢)(1 − 𝑢)𝑥𝑑𝑥 + Θ(𝑢)(1 + 𝑢)𝑦𝑑𝑦

= 185𝑑𝜈 − (1

2(𝛿(𝑢) − Θ(𝑢))𝑥2 −

1

2(𝛿(𝑢) + Θ(𝑢))𝑦2)𝑑𝑢 − Θ(𝑢)(1 − 𝑢)𝑥𝑑𝑥

+ Θ(𝑢)(1 + 𝑢)𝑦𝑑𝑦

= 𝑑𝜈 −1

2(𝛿(𝑢)(𝑥2 − 𝑦2) − Θ(𝑢)(𝑥2 + 𝑦2))𝑑𝑢 − Θ(𝑢)(1 − 𝑢)𝑥𝑑𝑥 + Θ(𝑢)(1 + 𝑢)𝑦𝑑𝑦

𝑑𝑋 = (−Θ(𝑢) − 𝑢Θ′(𝑢))𝑥𝑑𝑢 + (1 − 𝑢Θ(𝑢))𝑑𝑥

= −Θ(𝑢)𝑥𝑑𝑢 + (1 − 𝑢Θ(𝑢))𝑑𝑥

𝑑𝑌 = (Θ(𝑢) + 𝑢Θ′(𝑢))𝑦𝑑𝑢 + (1 + 𝑢Θ(𝑢))𝑑𝑦

= Θ(𝑢)𝑦𝑑𝑢 + (1 + 𝑢Θ(𝑢))𝑑𝑦

𝑋2 − 𝑌2 = (1 − 𝑢Θ(𝑢))2𝑥2 − (1 + 𝑢Θ(𝑢))

2𝑦2

= (1 + 𝑢2Θ2(𝑢))(𝑥2 − 𝑦2) − 2𝑢Θ(𝑢)(𝑥2 + 𝑦2)

𝛿(𝑢)(𝑋2

− 𝑌2) = 𝛿(𝑢)(1 + 𝑢2Θ2(𝑢))(𝑥2 − 𝑦2) − 𝛿(𝑢)2𝑢Θ(𝑢)(𝑥2 + 𝑦2)

= 𝛿(𝑢)(𝑥2 − 𝑦2)

𝑑𝑋2 = (−Θ(𝑢)𝑥𝑑𝑢 + (1 − 𝑢Θ(𝑢))𝑑𝑥)2

183 (McMahon, 2006, p. 304) 184 http://www-staff.lboro.ac.uk/~majbg/jbg/book/chap3.pdf 185 𝑢Θ′(𝑢) = 𝑢𝛿(𝑢) = 0


http://www-staff.lboro.ac.uk/~majbg/jbg/book/chap3.pdf




= Θ2(𝑢)𝑥2𝑑𝑢2 + (1 − 𝑢Θ(𝑢))2𝑑𝑥2 − 2Θ(𝑢)(1 − 𝑢Θ(𝑢))𝑥𝑑𝑢𝑑𝑥

𝑑𝑌2 = (Θ(𝑢)𝑦𝑑𝑢 + (1 + 𝑢Θ(𝑢))𝑑𝑦)2

= Θ2(𝑢)𝑦2𝑑𝑢2+(1 + 𝑢Θ(𝑢))2𝑑𝑦2 + 2Θ(𝑢)(1 + 𝑢Θ(𝑢))𝑦𝑑𝑢𝑑𝑦

𝑑𝑋2

+ 𝑑𝑌2

= Θ2(𝑢)𝑥2𝑑𝑢2 + (1 − 𝑢Θ(𝑢))2𝑑𝑥2 − 2Θ(𝑢)(1 − 𝑢Θ(𝑢))𝑥𝑑𝑢𝑑𝑥 + Θ2(𝑢)𝑦2𝑑𝑢2

+ (1 + 𝑢Θ(𝑢))2𝑑𝑦2 + 2Θ(𝑢)(1 + 𝑢Θ(𝑢))𝑦𝑑𝑢𝑑𝑦

= Θ2(𝑢)(𝑥2 + 𝑦2)𝑑𝑢2 + (1 − 𝑢Θ(𝑢))

2𝑑𝑥2 − 2Θ(𝑢)(1 − 𝑢Θ(𝑢))𝑥𝑑𝑢𝑑𝑥

+ (1 + 𝑢Θ(𝑢))2𝑑𝑦2 + 2Θ(𝑢)(1 + 𝑢Θ(𝑢))𝑦𝑑𝑢𝑑𝑦

𝑑𝑠2 = 𝛿(𝑢)(𝑋2 − 𝑌2)𝑑𝑢2 + 2𝑑𝑢𝑑𝑟 − 𝑑𝑋2 − 𝑑𝑌2 = 𝛿(𝑢)(𝑥2 − 𝑦2)𝑑𝑢2

+ 2𝑑𝑢 (𝑑𝜈 −1

2(𝛿(𝑢)(𝑥2 − 𝑦2) − Θ(𝑢)(𝑥2 + 𝑦2))𝑑𝑢 − Θ(𝑢)(1 − 𝑢)𝑥𝑑𝑥

+ Θ(𝑢)(1 + 𝑢)𝑦𝑑𝑦)

− (Θ2(𝑢)(𝑥2 + 𝑦2)𝑑𝑢2 + (1 − 𝑢Θ(𝑢))2𝑑𝑥2 − 2Θ(𝑢)(1 − 𝑢Θ(𝑢))𝑥𝑑𝑢𝑑𝑥

+ (1 + 𝑢Θ(𝑢))2𝑑𝑦2 + 2Θ(𝑢)(1 + 𝑢Θ(𝑢))𝑦𝑑𝑢𝑑𝑦)

= 2𝑑𝑢𝑑𝜈 + 2𝑑𝑢(−Θ(𝑢)(1 − 𝑢)𝑥𝑑𝑥 + Θ(𝑢)(1 + 𝑢)𝑦𝑑𝑦)

− ((1 − 𝑢Θ(𝑢))2𝑑𝑥2 − 2Θ(𝑢)(1 − 𝑢Θ(𝑢))𝑥𝑑𝑢𝑑𝑥 + (1 + 𝑢Θ(𝑢))

2𝑑𝑦2

+ 2Θ(𝑢)(1 + 𝑢Θ(𝑢))𝑦𝑑𝑢𝑑𝑦)

= 1862𝑑𝑢𝑑𝜈 − (1 − 𝑢Θ(𝑢))2𝑑𝑥2 − (1 + 𝑢Θ(𝑢))

2𝑑𝑦2 (13.43)

12.4 187The delta – 𝜹(𝒖) and heavy-side –𝚯(𝒖) functions: prove that 𝒖𝜹(𝒖) = 𝟎 Definitions

𝛿(𝑢) = {

+∞ 𝑖𝑓 𝑢 = 00 𝑖𝑓 𝑢 ≠ 0

; ∫ 𝛿(𝑢)𝑑𝑢∞

−∞

= 1

Θ(𝑢) = {

0 𝑖𝑓 𝑢 ≤ 01 𝑖𝑓 𝑢 > 0

; 𝑑Θ(𝑢)

𝑑𝑢 = 188Θ′(𝑢) =𝛿(𝑢);

Θ(𝑢) = ∫ 𝛿(𝑢)𝑑𝑢𝑢

−∞

We calculate

∫ 𝑓(𝑢)𝛿(𝑢)𝑑𝑢∞

−∞

= ∫ 𝑓(𝑢)Θ′(𝑢)𝑑𝑢∞

−∞

= [𝑓(𝑢)Θ(𝑢)]−∞

∞ −∫ 𝑓′(𝑢)Θ(𝑢)𝑑𝑢∞

−∞

= 𝑓(∞) − ∫ 𝑓′(𝑢)𝑑𝑢

∞

0

= 𝑓(∞) − (𝑓(∞) − 𝑓(0))

= 𝑓(0) if 𝑓(𝑢) = 𝑢 we find

∫ 𝑓(𝑢)𝛿(𝑢)𝑑𝑢∞

−∞

= ∫ 𝑢𝛿(𝑢)𝑑𝑢∞

−∞

= 0

Next we assume that 𝑢𝛿(𝑢) = 0. Multiplying both sides with a test function 𝑓(𝑢) and integrating we get

∫ 𝑓(𝑢)𝑢𝛿(𝑢)𝑑𝑢∞

−∞

= ∫ 𝑓(𝑢) ⋅ 0𝑑𝑢∞

−∞

186 Θ2(𝑢) = Θ(𝑢) 187 (McMahon, 2006, p. 304) 188 (13.44)





⇔ 0 ⋅ 𝑓(0) = 0 which is consistent with our initial assumption and we can therefore conclude that 𝑢𝛿(𝑢) = 0 Next we calculate

∫ 𝑓(𝑢)𝛿′(𝑢)𝑑𝑢∞

−∞

= [𝑓(𝑢)𝛿(𝑢)]−∞∞ −∫ 𝑓′(𝑢)𝛿(𝑢)𝑑𝑢

∞

−∞

= [𝑓(𝑢)𝛿(𝑢)]−∞

∞ −∫ 𝑓′(𝑢)Θ′(𝑢)𝑑𝑢∞

−∞

= [𝑓(𝑢)𝛿(𝑢)]−∞

∞ − ([𝑓′(𝑢)Θ(𝑢)]−∞∞ −∫ 𝑓′′(𝑢)Θ(𝑢)𝑑𝑢

∞

−∞

)

= 0 − (𝑓′(∞) − ∫ 𝑓′′(𝑢)𝑑𝑢

∞

0

)

= −(𝑓′(∞) − (𝑓′(∞) − 𝑓′(0)))

= 189 − 𝑓′(0) if 𝑓(𝑢) = 𝑢 we find

∫ 𝑓(𝑢)𝛿′(𝑢)𝑑𝑢∞

−∞

= ∫ 𝑢𝛿′(𝑢)𝑑𝑢∞

−∞

= −1

Next we assume that −𝛿(𝑢) = 𝑢𝛿′(𝑢). Multiplying both sides with a test function 𝑓(𝑢) and integrating we get

∫ −𝑓(𝑢)𝛿(𝑢)𝑑𝑢∞

−∞

= ∫ 𝑓(𝑢)𝑢𝛿′(𝑢)𝑑𝑢∞

−∞

⇔ −𝑓(0) = −(𝑓(𝑢)𝑢)′(𝑢 = 0) = −(𝑓′(𝑢) ⋅ 𝑢 + 𝑓(𝑢))(𝑢 = 0)

= −𝑓(0) which is consistent with our initial assumption and we can therefore conclude that −𝛿(𝑢) = 𝑢𝛿′(𝑢)

12.5 190Impulsive gravitational wave Region III The line element: 𝑑𝑠2 = 2𝑑𝑢𝑑𝜈 − [1 − 𝜈Θ(𝜈)]2𝑑𝑥2 − [1 + 𝜈Θ(𝜈)]2𝑑𝑦2



0 =𝑑

𝑑𝑠(𝜕𝐹

𝜕��𝑎) −

𝜕𝐹

𝜕𝑥𝑎 (10.36)

where 𝐹 = 2�� − [1 − 𝜈Θ(𝜈)]2��2 − [1 + 𝜈Θ(𝜈)]2��2 𝑥𝑎 = 𝑢:

𝜕𝐹

𝜕𝑢 = 0

𝜕𝐹

𝜕�� = 2��

𝑑

𝑑𝑠(𝜕𝐹

𝜕��) = 2��

189 The general formula is ∫ 𝑓(𝑢)𝛿(𝑛)(𝑢)𝑑𝑢 = (−1)𝑛𝑓(𝑛)(0)

∞

−∞






⇒ 0 = �� 𝑥𝑎 = 𝜈:

𝜕𝐹

𝜕𝜈 = 2Θ(𝜈)[1 − 𝜈Θ(𝜈)]��2 − 2Θ(𝜈)[1 + 𝜈Θ(𝜈)]��2

𝜕𝐹

𝜕�� = 2��

𝑑

𝑑𝑠(𝜕𝐹

𝜕��) = 2��

⇒ 0 = �� − Θ(𝜈)[1 − 𝜈Θ(𝜈)]��2 + Θ(𝜈)[1 + 𝜈Θ(𝜈)]��2 𝑥𝑎 = 𝑥:

𝜕𝐹

𝜕𝑥 = 0

𝜕𝐹

𝜕�� = −2[1 − 𝜈Θ(𝜈)]2��

𝑑

𝑑𝑠(𝜕𝐹

𝜕��) = −2(−��Θ(𝜈) − 𝜈��

𝑑Θ(𝜈)

𝑑𝜈) 2[1 − 𝜈Θ(𝜈)]�� − 2[1 − 𝜈Θ(𝜈)]2��

= 1914(Θ(𝜈) + 𝜈𝛿(𝜈))[1 − 𝜈Θ(𝜈)]�� − 2[1 − 𝜈Θ(𝜈)]2��

⇒ 0 = 192[1 − 𝜈Θ(𝜈)]2�� − 2Θ(𝜈)[1 − 𝜈Θ(𝜈)]��

⇔ 0 = �� −2Θ(𝜈)

[1 − 𝜈Θ(𝜈)]��

𝑥𝑎 = 𝑦:

𝜕𝐹

𝜕𝑦 = 0

𝜕𝐹

𝜕�� = −2[1 + 𝜈Θ(𝜈)]2��

𝑑

𝑑𝑠(𝜕𝐹

𝜕��) = −2(��Θ(𝜈) + 𝜈��

𝑑Θ(𝜈)

𝑑𝜈)2[1 + 𝜈Θ(𝜈)]�� − 2[1 + 𝜈Θ(𝜈)]2��

= −4(Θ(𝜈) + 𝜈𝛿(𝜈))[1 + 𝜈Θ(𝜈)]�� − 2[1 + 𝜈Θ(𝜈)]2��

⇒ 0 = [1 + 𝜈Θ(𝜈)]2�� + 2Θ(𝜈)[1 + Θ(𝜈)]��

⇔ 0 = �� +2Θ(𝜈)

[1 + 𝜈Θ(𝜈)]��

Collecting the results 0 = �� 0 = �� − Θ(𝜈)[1 − 𝜈Θ(𝜈)]��2 + Θ(𝜈)[1 + 𝜈Θ(𝜈)]��2

0 = �� −2Θ(𝜈)

[1 − 𝜈Θ(𝜈)]��

0 = �� +2Θ(𝜈)

[1 + 𝜈Θ(𝜈)]��


Γ 𝑥𝑥𝑢 = −Θ(𝜈)[1 − 𝜈Θ(𝜈)] Γ 𝜈𝑥

𝑥 = −Θ(𝜈)

[1 − 𝜈Θ(𝜈)] Γ 𝑦𝜈

𝑦 =

Θ(𝜈)

[1 + 𝜈Θ(𝜈)]

Γ 𝑦𝑦𝑢 = Θ(𝜈)[1 + 𝜈Θ(𝜈)]

191

𝑑Θ(𝜈)

𝑑𝜈= 𝛿(𝜈)

192 𝜈𝛿(𝜈) = 0





The Petrov type

The line element 𝑑𝑠2 = 2𝑑𝑢𝑑𝜈 − [1 − 𝜈Θ(𝜈)]2𝑑𝑥2 − [1 + 𝜈Θ(𝜈)]2𝑑𝑦2


11

−[1 − 𝜈Θ(𝜈)]2

−[1 + 𝜈Θ(𝜈)]2

}


{

11

−1

[1 − 𝜈Θ(𝜈)]2

−1

[1 + 𝜈Θ(𝜈)]2}

The basis one forms

Finding the basis one forms is not so obvious, we write: 𝑑𝑠2 = 2𝑑𝑢𝑑𝜈 − [1 − 𝜈Θ(𝜈)]2𝑑𝑥2 − [1 + 𝜈Θ(𝜈)]2𝑑𝑦2

= (𝜔��)2− (𝜔��)

2− (𝜔𝑥)

2− (𝜔��)

2

= (𝜔�� +𝜔��)(𝜔�� −𝜔��) − (𝜔𝑥)2− (𝜔��)

2

⇒ √2𝑑𝑢 = (𝜔�� +𝜔��)

√2𝑑𝜈 = (𝜔�� −𝜔��)

⇒ 𝜔�� =1

√2(𝑑𝑢 + 𝑑𝜈) 𝑑𝑢 =

1

√2(𝜔�� +𝜔��)

𝜂𝑖𝑗 = {

1−1

−1−1

} 𝜔�� =

1

√2(𝑑𝑢 − 𝑑𝜈) 𝑑𝜈 =

1

√2(𝜔�� −𝜔��)

𝜔�� = (1 − 𝜈Θ(𝜈))𝑑𝑥 𝑑𝑥 =1

1 − 𝜈Θ(𝜈)𝜔𝑥

𝜔�� = (1 + 𝜈Θ(𝜈))𝑑𝑦 𝑑𝑦 =1

1 + 𝜈Θ(𝜈)𝜔��







(

𝑙𝑛𝑚��

)

=1

√2(

1 1 0 01 −1 0 00 0 1 𝑖0 0 1 −𝑖

)(

𝜔��

𝜔��

𝜔𝑥

𝜔��

) =1

√2(

𝜔�� +𝜔��

𝜔�� −𝜔��

𝜔𝑥 + 𝑖𝜔��

𝜔𝑥 − 𝑖𝜔��

)

=1

√2

(

√2𝑑𝑢

√2𝑑𝜈

(1 − 𝜈Θ(𝜈))𝑑𝑥 + 𝑖(1 + 𝜈Θ(𝜈))𝑑𝑦

(1 − 𝜈Θ(𝜈))𝑑𝑥 − 𝑖(1 + 𝜈Θ(𝜈))𝑑𝑦)

Written in terms of the coordinate basis 𝑙𝑎 = (1, 0, 0, 0) 𝑛𝑎 = (0, 1, 0, 0)

𝑚𝑎 =1

√2(0, 0, (1 − 𝜈Θ(𝜈)), 𝑖(1 + 𝜈Θ(𝜈))) 𝑚𝑎 =

1

√2(0, 0, (1 − 𝜈Θ(𝜈)), −𝑖(1 + 𝜈Θ(𝜈)))


𝜈𝑢𝑙𝜈 = 1 ⋅ 0 = 0 𝑙𝜈 = 𝑔𝑎𝜈𝑙𝑎 = 𝑔

𝑢𝜈𝑙𝑢 = 1 ⋅ 1 = 1 𝑙𝑥 = 𝑙𝑦 = 0 𝑛𝑢 = 𝑔𝑎𝑢𝑛𝑎 = 𝑔

𝜈𝑢𝑛𝜈 = 1 ⋅ 1 = 1 𝑛𝜈 = 𝑔𝑎𝜈𝑛𝑎 = 𝑔

𝑢𝜈𝑛𝑢 = 1 ⋅ 0 = 0 𝑛𝑥 = 𝑛𝑦 = 0 𝑚𝜈 = 𝑚𝑢 = 0

𝑚𝑥 = 𝑔𝑥𝑥𝑚𝑥 = −1

[1 − 𝜈Θ(𝜈)]2⋅1

√2⋅ (1 − 𝜈Θ(𝜈)) = −

1

√2

1

(1 − 𝜈Θ(𝜈))

𝑚𝑦 = 𝑔𝑦𝑦𝑚𝑦 = −1

[1 + 𝜈Θ(𝜈)]21

√2⋅ 𝑖(1 + 𝜈Θ(𝜈)) = −𝑖

1

√2

1

(1 + 𝜈Θ(𝜈))

Collecting the results 𝑙𝑎 = (1, 0, 0, 0) 𝑙𝑎 = (0, 1, 0, 0) 𝑛𝑎 = (0, 1, 0, 0) 𝑛𝑎 = (1, 0, 0, 0)

𝑚𝑎 =1

√2(0, 0, (1 − 𝜈Θ(𝜈)), 𝑖(1 + 𝜈Θ(𝜈))) 𝑚𝑎 =

1

√2(0, 0, −

1

(1 − 𝜈Θ(𝜈)), −𝑖

1

(1 + 𝜈Θ(𝜈)))

𝑚𝑎 =1

√2(0, 0, (1 − 𝜈Θ(𝜈)), −𝑖(1 + 𝜈Θ(𝜈))) ��𝑎 =

1

√2(0, 0, −

1

(1 − 𝜈Θ(𝜈)), 𝑖

1

(1 + 𝜈Θ(𝜈)))

The spin coefficients calculated from the orthonormal tetrad


𝑎𝑙𝑏 휀 =1



(9.15) 𝜈 = −∇𝑏𝑛𝑎��


1




𝑎��𝑏 𝛼 =1


𝑎��𝑏 − ∇𝑏𝑚𝑎��𝑎��𝑏)










𝜋 = −∇𝑏𝑛𝑎��𝑎𝑙𝑏 = −∇ν𝑛𝑎��

𝑎𝑙𝜈 = −∇𝜈𝑛𝑥��𝑥𝑙𝜈 − ∇𝜈𝑛𝑦��

𝑦𝑙𝜈 = 0


𝑎𝑛𝑢 = −∇𝑢𝑛𝑥��𝑥𝑛𝑢 − ∇𝑢𝑛𝑦��

𝑦𝑛𝑢 = 0

𝜆 = −∇𝑏𝑛𝑎��𝑎��𝑏 = −∇𝑥𝑛𝑎��

𝑎��𝑥 − ∇𝑦𝑛𝑎��𝑎��𝑦

= −∇𝑥𝑛𝑥��𝑥��𝑥 − ∇𝑦𝑛𝑥��

𝑥��𝑦 − ∇𝑥𝑛𝑦��𝑦��𝑥 − ∇𝑦𝑛𝑦��

𝑦��𝑦

= −(𝜕𝑥𝑛𝑥 − Γ 𝑥𝑥

𝑐 𝑛𝑐)��𝑥��𝑥 − (𝜕𝑦𝑛𝑥 − Γ 𝑦𝑥

𝑐 𝑛𝑐)��𝑥��𝑦 − (𝜕𝑥𝑛𝑦 − Γ 𝑥𝑦

𝑐 𝑛𝑐)��𝑦��𝑥

− (𝜕𝑦𝑛𝑦 − Γ 𝑦𝑦𝑐 𝑛𝑐)��

𝑦��𝑦

= Γ 𝑥𝑥𝑢 𝑛𝑢��

𝑥��𝑥 + Γ 𝑦𝑦𝑢 𝑛𝑢��

𝑦��𝑦

= 0

𝜇 = −∇𝑏𝑛𝑎��𝑎𝑚𝑏 = Γ 𝑥𝑥

𝑢 𝑛𝑢��𝑥𝑚𝑥 + Γ 𝑦𝑦

𝑢 𝑛𝑢��𝑦𝑚𝑦 = 0

𝜅 = ∇𝑏𝑙𝑎𝑚𝑎𝑙𝑏 = ∇𝜈𝑙𝑎𝑚

𝑎𝑙𝜈 = ∇𝜈𝑙𝑥𝑚𝑥𝑙𝜈 + ∇𝜈𝑙𝑦𝑚

𝑦𝑙𝜈 = 0


𝑎𝑛𝑢 = ∇𝑢𝑙𝑥𝑚𝑥𝑛𝑢 + ∇𝑢𝑙𝑦𝑚

𝑢𝑛𝜈 = 0

𝜌 = ∇𝑏𝑙𝑎𝑚𝑎��𝑏

= ∇𝑥𝑙𝑎𝑚𝑎��𝑥 + ∇𝑦𝑙𝑎𝑚

𝑎��𝑦

= ∇𝑥𝑙𝑥𝑚𝑥��𝑥 + ∇𝑦𝑙𝑥𝑚

𝑥��𝑦 + ∇𝑥𝑙𝑦𝑚𝑦��𝑥 + ∇𝑦𝑙𝑦𝑚

𝑦��𝑦

= (𝜕𝑥𝑙𝑥 − Γ 𝑥𝑥

𝑐 𝑙𝑐)𝑚𝑥��𝑥 + (𝜕𝑦𝑙𝑥 − Γ 𝑦𝑥

𝑐 𝑙𝑐)𝑚𝑥��𝑦 + (𝜕𝑥𝑙𝑦 − Γ 𝑥𝑦

𝑐 𝑙𝑐)𝑚𝑦��𝑥

+ (𝜕𝑦𝑙𝑦 − Γ 𝑦𝑦𝑐 𝑙𝑐)𝑚

𝑦��𝑦

= −(Γ 𝑥𝑥𝑢 𝑙𝑢𝑚

𝑥��𝑥 + Γ 𝑦𝑦𝑢 𝑙𝑢𝑚

𝑦��𝑦)

= −(−Θ(𝜈)[1 − 𝜈Θ(𝜈)]1

2 (−

1

(1 − 𝜈Θ(𝜈)))(−

1

(1 − 𝜈Θ(𝜈)))

+ Θ(𝜈)[1 + 𝜈Θ(𝜈)]1

2(−𝑖

1

(1 + 𝜈Θ(𝜈)))(𝑖

1

(1 + 𝜈Θ(𝜈))))

=Θ(𝜈)

2(

1

(1 − 𝜈Θ(𝜈))−

1

(1 + 𝜈Θ(𝜈))) = 193

𝜈Θ(𝜈)

(1 + 𝜈Θ(𝜈))(1 − 𝜈Θ(𝜈))


=Θ(𝜈)

2(

1

(1 − 𝜈Θ(𝜈))+

1

(1 + 𝜈Θ(𝜈)))

=Θ(𝜈)

(1 + 𝜈Θ(𝜈))(1 − 𝜈Θ(𝜈))

휀 =1



=1

2(∇𝜈𝑙𝑎𝑛

𝑎𝑙𝜈 − ∇𝜈𝑚𝑎��𝑎𝑙𝜈)

=1

2(∇𝜈𝑙𝑢𝑛

𝑢𝑙𝜈 − ∇𝜈𝑚𝑥��𝑥𝑙𝜈 − ∇𝜈𝑚𝑦��

𝑦𝑙𝜈)

= −

1

2((𝜕𝜈𝑚𝑥 − Γ 𝜈𝑥

𝑐 𝑚𝑐)��𝑥 + (𝜕𝜈𝑚𝑦 − Γ 𝜈𝑦

𝑐 𝑚𝑐)��𝑦)

= −

1

2((𝜕𝜈(1 − 𝜈Θ(𝜈)) − Γ 𝜈𝑥

𝑥 𝑚𝑥)��𝑥𝑛𝜈 + (𝜕𝜈 (𝑖(1 + 𝜈Θ(𝜈))) − Γ 𝜈𝑦

𝑦𝑚𝑦) ��

𝑦𝑛𝜈)

193 Θ2(𝜈) = Θ(𝜈)





= −1

2((−Θ(𝜈) − 𝜈𝛿(𝜈) − (−

Θ(𝜈)

[1 − 𝜈Θ(𝜈)]) (1 − 𝜈Θ(𝜈))) ��𝑥𝑛𝜈

+ (𝑖(Θ(𝜈) + 𝜈𝛿(𝜈)) −Θ(𝜈)

[1 + 𝜈Θ(𝜈)]𝑖(1 + 𝜈Θ(𝜈))) ��𝑦𝑛𝜈)

= 0

𝛾 =1



=1

2(∇u𝑙𝑎𝑛

𝑎𝑛𝑢 − ∇𝑢𝑚𝑎��𝑎𝑛𝑢)

=1

2(∇u𝑙𝑢𝑛

𝑢𝑛𝑢 − ∇𝑢𝑚𝑥��𝑥𝑛𝑢 − ∇𝑢𝑚𝑦��

𝑦𝑛𝑢)

= 0

𝛼 =1


𝑎��𝑏 − ∇𝑏𝑚𝑎��𝑎��𝑏)

=1


𝑎��𝑥 − ∇𝑥𝑚𝑎��𝑎��𝑥) +

1


𝑎��𝑦 − ∇𝑦𝑚𝑎��𝑎��𝑦)

=1

2(∇𝑥𝑙𝑢𝑛

𝑢��𝑥 − ∇𝑥𝑚𝑥��𝑥��𝑥 − ∇𝑥𝑚𝑦��

𝑦��𝑥 + ∇𝑦𝑙𝑢𝑛𝑢��𝑦 − ∇𝑦𝑚𝑥��

𝑥��𝑦 − ∇𝑦𝑚𝑦��𝑦��𝑦)

= 0

𝛽 =1



Collecting the results 𝜋 = 0 𝜅 = 0 휀 = 0 𝜈 = 0 𝜏 = 0 𝛾 = 0

𝜆 = 0 𝜌 =𝜈Θ(𝜈)

(1 + 𝜈Θ(𝜈))(1 − 𝜈Θ(𝜈)) 𝛼 = 0

𝜇 = 0 𝜎 =Θ(𝜈)

(1 + 𝜈Θ(𝜈))(1 − 𝜈Θ(𝜈)) 𝛽 = 0



Ψ3 = 𝛿𝛾 − Δα + 𝜈(𝜌 + 휀) − 𝜆(𝜏 + 𝛽) + 𝛼(�� − ��) + 𝛾(�� − ��) (13.25)

Ψ4 = 𝛿𝜈 − Δλ + 𝜆(𝜇 + ��) − 𝜆(3𝛾 − ��) + 𝜈(3𝛼 + �� + 𝜋 − ��) (13.26)


Ψ0 = 𝐷𝜎 − 𝛿𝜅 − 𝜎(𝜌 + ��) − 𝜎(3휀 − 휀) + 𝜅(𝜋 − �� + �� + 3𝛽) = 𝐷𝜎 − 2𝜎𝜌

= 𝑙𝑎∇𝑎 (Θ(𝜈)

(1 + 𝜈Θ(𝜈))(1 − 𝜈Θ(𝜈))) − 2(

Θ(𝜈)

(1 + 𝜈Θ(𝜈))(1 − 𝜈Θ(𝜈)))

𝜈Θ(𝜈)

(1 + 𝜈Θ(𝜈))(1 − 𝜈Θ(𝜈))





= 𝑙𝜈 ∂ν (

Θ(𝜈)

(1 − 𝜈Θ(𝜈))(1 + 𝜈Θ(𝜈))) − 2

𝜈Θ(𝜈)

(1 − 𝜈2Θ(𝜈))2

= ∂ν (

Θ(𝜈)

1 − 𝜈2Θ(𝜈)) − 2

𝜈Θ(𝜈)

(1 − 𝜈2Θ(𝜈))2

=𝛿(𝜈)(1 − 𝜈2Θ(𝜈)) − Θ(𝜈)(−2𝜈Θ(𝜈) − 𝜈2𝛿(𝜈))

(1 − 𝜈2Θ(𝜈))2 − 2

𝜈Θ(𝜈)

(1 − 𝜈2Θ(𝜈))2

= 𝛿(𝜈) Ψ1 = 0 Ψ2 = 𝛿𝜏 − Δ𝜌 − 𝜌�� − 𝜎𝜆 + 𝜏(�� − 𝛼 − ��) + 𝜌(𝛾 + ��) + 𝜅𝜈 − 2Λ = −Δ𝜌 = −𝑛𝑎∇𝑎𝜌 = −𝑛

𝑢∇𝑢𝜌 = 0

Ψ3 = 0 Ψ4 = 0

194Ψ0 ≠ 0: This is a Petrov type N, which means there is a single principal null direction of multiplicity 4. This corresponds to transverse gravity waves in region III.

12.6 195Two interacting waves

The line element: 𝑑𝑠2 = 2𝑑𝑢𝑑𝜈 − cos2 𝑎𝜈 𝑑𝑥2 − cosh2 𝑎𝜈 𝑑𝑦2



0 =𝑑

𝑑𝑠(𝜕𝐹

𝜕��𝑎) −

𝜕𝐹

𝜕𝑥𝑎 (10.36)

where 𝐹 = 2�� − cos2 𝑎𝜈 ��2 − cosh2 𝑎𝜈 ��2 𝑥𝑎 = 𝜈:

𝜕𝐹

𝜕𝜈 = 2𝑎 cos 𝑎𝜈 sin𝑎𝜈 ��2 − 2𝑎 cosh𝑎𝜈 sinh𝑎𝜈 ��2

𝜕𝐹

𝜕�� = 2��

𝑑

𝑑𝑠(𝜕𝐹

𝜕��) = 2��

⇒ 0 = �� − 𝑎 cos𝑎𝜈 sin𝑎𝜈 ��2 + 𝑎 cosh𝑎𝜈 sinh𝑎𝜈 ��2 𝑥𝑎 = 𝑢:

𝜕𝐹

𝜕𝑢 = 0

𝜕𝐹

𝜕�� = 2��

𝑑

𝑑𝑠(𝜕𝐹

𝜕��) = 2��

⇒ 0 = �� 𝑥𝑎 = 𝑥:

𝜕𝐹

𝜕𝑥 = 0

194 http://www-staff.lboro.ac.uk/~majbg/jbg/book/chap3.pdf 195 (McMahon, 2006, p. 313), example 13-2


http://www-staff.lboro.ac.uk/~majbg/jbg/book/chap3.pdf




𝜕𝐹

𝜕�� = −2cos2 𝑎𝜈 ��

𝑑

𝑑𝑠(𝜕𝐹

𝜕��) = 4𝑎 cos 𝑎𝜈 sin𝑎𝜈 �� − 2 cos2 𝑎𝜈 ��

⇒ 0 = 2𝑎 cos 𝑎𝜈 sin𝑎𝜈 �� − cos2 𝑎𝜈 ��

⇔ 0 = �� − 2𝑎 tan𝑎𝜈 �� 𝑥𝑎 = 𝑦:

𝜕𝐹

𝜕𝑦 = 0

𝜕𝐹

𝜕�� = −2cosh2 𝑎𝜈 ��

𝑑

𝑑𝑠(𝜕𝐹

𝜕��) = −4𝑎 cosh𝑎𝜈 sinh𝑎𝜈 �� − 2 cos2 𝑎𝜈 ��

⇒ 0 = −2𝑎 cosh𝑎𝜈 sinh𝑎𝜈 �� − cosh2 𝑎𝜈 �� ⇔ 0 = �� + 2𝑎 tanh 𝑎𝜈 ��

Collecting the results 0 = �� 0 = �� − 𝑎 cos𝑎𝜈 sin𝑎𝜈 ��2 + 𝑎 cosh𝑎𝜈 sinh𝑎𝜈 ��2 0 = �� − 2𝑎 tan𝑎𝜈 �� 0 = �� + 2𝑎 tanh 𝑎𝜈 ��


Γ 𝑥𝑥𝑢 = −𝑎 cos 𝑎𝜈 sin 𝑎𝜈 Γ 𝑦𝑦

𝑢 = 𝑎 cosh𝑎𝜈 sinh𝑎𝜈

Γ 𝑥𝜈𝑥 = −𝑎 tan𝑎𝜈 Γ 𝜈𝑥

𝑥 = −𝑎 tan𝑎𝜈 Γ 𝑦𝜈𝑦

= 𝑎 tanh𝑎𝜈 Γ 𝜈𝑦𝑦

= 𝑎 tanh𝑎𝜈

The Petrov type

The line element 𝑑𝑠2 = 2𝑑𝑢𝑑𝜈 − cos2 𝑎𝜈 𝑑𝑥2 − cosh2 𝑎𝜈 𝑑𝑦2


11

−cos2 𝑎𝜈− cosh2 𝑎𝜈

}


{

11

−1

cos2 𝑎𝜈

−1

cosh2 𝑎𝜈}

The basis one forms

Finding the basis one forms is not so obvious, we write: 𝑑𝑠2 = 2𝑑𝑢𝑑𝜈 − cos2 𝑎𝜈 𝑑𝑥2 − cosh2 𝑎𝜈 𝑑𝑦2





= (𝜔��)2− (𝜔��)

2− (𝜔𝑥)

2− (𝜔��)

2

= (𝜔�� +𝜔��)(𝜔�� −𝜔��) − (𝜔𝑥)2− (𝜔��)

2

⇒ √2𝑑𝑢 = (𝜔�� +𝜔��)

√2𝑑𝜈 = (𝜔�� −𝜔��)

𝜔�� = cos 𝑎𝜈 𝑑𝑥

𝜔�� = cosh𝑎𝜈 𝑑𝑦

⇒ 𝜔�� =1

√2(𝑑𝑢 + 𝑑𝜈) 𝑑𝑢 =

1

√2(𝜔�� +𝜔��)

𝜂𝑖𝑗 = {

1−1

−1−1

} 𝜔�� =

1

√2(𝑑𝑢 − 𝑑𝜈) 𝑑𝜈 =

1

√2(𝜔�� −𝜔��)

𝜔�� = cos 𝑎𝜈 𝑑𝑥 𝑑𝑥 =1

cos 𝑎𝜈𝜔𝑥

𝜔�� = cosh𝑎𝜈 𝑑𝑦 𝑑𝑦 =1

cosh𝑎𝜈𝜔��



(

𝑙𝑛𝑚��

) =1

√2(

1 1 0 01 −1 0 00 0 1 𝑖0 0 1 −𝑖

)(

𝜔��

𝜔��

𝜔��

𝜔��

) =1

√2(

𝜔�� +𝜔��

𝜔�� −𝜔��

𝜔�� + 𝑖𝜔 ��

𝜔�� − 𝑖𝜔 ��

) =1

√2(

√2𝑑𝑢

√2𝑑𝜈cos 𝑎𝜈 𝑑𝑥 + 𝑖 cosh𝑎𝜈 𝑑𝑦cos 𝑎𝜈 𝑑𝑥 − 𝑖 cosh𝑎𝜈 𝑑𝑦)

Written in terms of the coordinate basis 𝑙𝑎 = (1, 0, 0, 0) 𝑛𝑎 = (0, 1, 0, 0)

𝑚𝑎 =1

√2(0, 0, cos𝑎𝜈 , 𝑖 cosh𝑎𝜈) 𝑚𝑎 =

1

√2(0, 0, cos 𝑎𝜈 , −𝑖 cosh𝑎𝜈)


𝜈𝑢𝑙𝜈 = 1 ⋅ 0 = 0 𝑙𝜈 = 𝑔𝑎𝜈𝑙𝑎 = 𝑔

𝑢𝜈𝑙𝑢 = 1 ⋅ 1 = 1 𝑙𝑥 = 𝑙𝑦 = 0 𝑛𝑢 = 𝑔𝑎𝑢𝑛𝑎 = 𝑔

𝜈𝑢𝑛𝜈 = 1 ⋅ 1 = 1 𝑛𝜈 = 𝑔𝑎𝜈𝑛𝑎 = 𝑔

𝑢𝜈𝑛𝑢 = 1 ⋅ 0 = 0 𝑛𝑥 = 𝑛𝑦 = 0 𝑚𝜈 = 𝑚𝑢 = 0


cos2 𝑎𝜈⋅1

√2⋅ cos𝑎𝜈 = −

1

√2

1

cos 𝑎𝜈


cosh2 𝑎𝜈

1

√2⋅ 𝑖 cosh𝑎𝜈 = −𝑖

1

√2

1

cosh𝑎𝜈

Collecting the results 𝑙𝑎 = (1, 0, 0, 0) 𝑙𝑎 = (0, 1, 0, 0) 𝑛𝑎 = (0, 1, 0, 0) 𝑛𝑎 = (1, 0, 0, 0)

𝑚𝑎 =1

√2(0, 0, cos𝑎𝜈 , 𝑖 cosh𝑎𝜈) 𝑚𝑎 =

1

√2(0, 0, −

1

cos𝑎𝜈, −𝑖

1

cosh𝑎𝜈)





𝑚𝑎 =1

√2(0, 0, cos𝑎𝜈 , −𝑖 cosh𝑎𝜈) ��𝑎 =

1

√2(0, 0, −

1

cos𝑎𝜈, 𝑖

1

cosh𝑎𝜈)



𝑎𝑙𝑏 휀 =1



(9.15) 𝜈 = −∇𝑏𝑛𝑎��


1




𝑎��𝑏 𝛼 =1


𝑎��𝑏 − ∇𝑏𝑚𝑎��𝑎��𝑏)






𝜋 = −∇𝑏𝑛𝑎��𝑎𝑙𝑏 = −∇𝜈𝑛𝑎��

𝑎𝑙𝜈 = −∇𝜈𝑛𝑥��𝑥𝑙𝜈 − ∇ν𝑛𝑦��

𝑦𝑙𝜈 = 0



𝑦𝑛𝑢 = 0

𝜆 = −∇𝑏𝑛𝑎��𝑎��𝑏

= −∇𝑥𝑛𝑎��𝑎��𝑥 − ∇𝑦𝑛𝑎��

𝑎��𝑦

= −∇𝑥𝑛𝑥��𝑥��𝑥 − ∇𝑦𝑛𝑥��

𝑥��𝑦 − ∇𝑥𝑛𝑦��𝑦��𝑥 − ∇𝑦𝑛𝑦��

𝑦��𝑦

= 0 𝜇 = −∇𝑏𝑛𝑎��


𝑎𝑙𝑏 = ∇ν𝑙𝑎𝑚𝑎𝑙ν = ∇ν𝑙𝑥𝑚

𝑥𝑙ν + ∇ν𝑙𝑦𝑚𝑦𝑙ν = 0



𝑦𝑛𝑢 = 0



𝑎��𝑦



𝑦��𝑦






𝑦��𝑦



𝑦��𝑦)

= −(−𝑎 cos 𝑎𝜈 sin𝑎𝜈 (−1

√2 cos 𝑎𝜈)2

+ 𝑎 cosh𝑎𝜈 sinh𝑎𝜈 (−𝑖1

√2 cosh𝑎𝜈) (𝑖

1

√2 cosh𝑎𝜈))

=𝑎

2(tan𝑎𝜈 − tanh 𝑎𝜈)

𝜎 = ∇𝑏𝑙𝑎𝑚𝑎𝑚𝑏 = −(Γ 𝑥𝑥

𝑢 𝑙𝑢𝑚𝑥𝑚𝑥 + Γ 𝑦𝑦

𝑢 𝑙𝑢𝑚𝑦𝑚𝑦) =

𝑎

2(tan 𝑎𝜈 + tanh𝑎𝜈)

휀 =1



=1

2(∇ν𝑙𝑎𝑛

𝑎𝑙ν − ∇ν𝑚𝑎��𝑎𝑙ν)

=1

2(∇ν𝑙𝑢𝑛

𝑢𝑙ν − ∇ν𝑚𝑥��𝑥𝑙ν − ∇ν𝑚𝑦��

𝑦𝑙ν)

= −

1


𝑐 𝑚𝑐)��𝑥𝑙ν + (𝜕𝜈𝑚𝑦 − Γ 𝜈𝑦

𝑐 𝑚𝑐)��𝑦𝑙ν)

= −

1

2((𝜕𝜈

1

√2cos 𝑎𝜈 − Γ 𝜈𝑥

𝑥 𝑚𝑥) ��𝑥𝑙ν + (𝜕𝜈𝑖

1

√2cosh𝑎𝜈 − Γ 𝜈𝑦


𝑦𝑙ν)





= −

1

2((−𝑎

1

√2sin𝑎𝜈—𝑎 tan𝑎𝜈)

1

√2cos 𝑎𝜈) ��𝑥 + (𝑖𝑎

1

√2sinh𝑎𝜈 − 𝑎 tanh 𝑎𝜈 𝑖

1

√2cosh𝑎𝜈) ��𝑦

= 0

𝛾 =1



=1



=1



𝑦𝑛𝑢)

= 0

𝛼 =1


𝑎��𝑏 − ∇𝑏𝑚𝑎��𝑎��𝑏)

=1


𝑎��𝑥 − ∇𝑥𝑚𝑎��𝑎��𝑥) +

1


𝑎��𝑦 − ∇𝑦𝑚𝑎��𝑎��𝑦)

= 0

𝛽 =1



Collecting the results 𝜋 = 0 𝜅 = 0 휀 = 0 𝜈 = 0 𝜏 = 0 𝛾 = 0

𝜆 = 0 𝜌 =𝑎

2(tan𝑎𝜈 − tanh 𝑎𝜈) 𝛼 = 0

𝜇 = 0 𝜎 =𝑎

2(tan𝑎𝜈 + tanh 𝑎𝜈) 𝛽 = 0



Ψ3 = 𝛿𝛾 − Δα + 𝜈(𝜌 + 휀) − 𝜆(𝜏 + 𝛽) + 𝛼(�� − ��) + 𝛾(�� − ��) (13.25)

Ψ4 = 𝛿𝜈 − Δλ + 𝜆(𝜇 + ��) − 𝜆(3𝛾 − ��) + 𝜈(3𝛼 + �� + 𝜋 − ��) (13.26)


Ψ0 = 𝐷𝜎 − 𝜎(𝜌 + ��) = 𝐷𝜎 − 2𝜎𝜌 = 𝑙𝑎∇𝑎𝜎 − 2𝜎𝜌 = 𝑙𝜈𝜕𝜈𝜎 − 2𝜎𝜌

= 𝑙𝜈𝜕𝜈 (𝑎

2(tan 𝑎𝜈 + tanh 𝑎𝜈)) − 2(

𝑎

2(tan 𝑎𝜈 + tanh 𝑎𝜈))(

𝑎

2(tan 𝑎𝜈 − tanh 𝑎𝜈))

= (𝑎2

2(1 + tan2 𝑎𝜈 + 1 − tanh2 𝑎𝜈)) − (

𝑎2

2(tan2 𝑎𝜈 − tanh2 𝑎𝜈))

= 𝑎2 Ψ1 = 0





Ψ2 = −Δ𝜌 = −𝑛𝑎∇𝑎𝜌 = −𝑛𝑢 ∂𝑢𝜌 = 0

Ψ3 = 0 Ψ4 = 0 Ψ0 ≠ 0: This is a Petrov type N, which means there is a single principal null direction (𝑛𝑎) of multiplicity 4.

12.7 196The Nariai spacetime

The line element: 𝑑𝑠2 = −Λ𝜈2𝑑𝑢2 + 2𝑑𝑢𝑑𝑣 −1

Ω2(𝑑𝑥2 + 𝑑𝑦2)

Ω = 1 +Λ

2(𝑥2 + 𝑦2)

The metric tensor and its inverse:

𝑔𝑎𝑏 =

{

−Λ𝜈2 11

−1

Ω2

−1

Ω2}

𝑔𝑎𝑏 = {

11 Λν2

−Ω2

−Ω2

}


Γ𝑎𝑏𝑐 =1



Γ𝑢𝑢𝜈 =1

2𝜕𝜈(𝑔𝑢𝑢) = −Λ𝜈 ⇒ Γ 𝑢𝜈

𝜈 = 𝑔𝜈𝑢Γ𝑢𝑢𝜈 = −Λ𝜈


2𝜕𝜈(𝑔𝑢𝑢) = Λ𝜈 ⇒ Γ 𝑢𝑢

𝜈 = 𝑔𝜈𝜈Γ𝜈𝑢𝑢 = Λ2𝜈3

⇒ Γ 𝑢𝑢𝑢 = 𝑔𝑢𝜈Γ𝜈𝑢𝑢 = Λ𝜈

Γ𝑥𝑥𝑥 =1

2𝜕𝑥(𝑔𝑥𝑥) =

Λ𝑥

Ω3 ⇒ Γ 𝑥𝑥

𝑥 = 𝑔𝑥𝑥Γ𝑥𝑥𝑥 = −Λ𝑥

Ω

Γ𝑥𝑥𝑦 =1

2𝜕𝑦(𝑔𝑥𝑥) =

Λ𝑦

Ω3 ⇒ Γ 𝑥𝑦

𝑥 = 𝑔𝑥𝑥Γ𝑥𝑥𝑦 = −Λ𝑦

Ω

Γ𝑦𝑥𝑥 = −1

2𝜕𝑦(𝑔𝑥𝑥) = −

Λ𝑦

Ω3 ⇒ Γ 𝑥𝑥

𝑦 = 𝑔𝑦𝑦Γ𝑦𝑥𝑥 =

Λ𝑦

Ω

Γ𝑦𝑦𝑦 =1

2𝜕𝑦(𝑔𝑦𝑦) =

Λ𝑦

Ω3 ⇒ Γ 𝑦𝑦

𝑦 = 𝑔𝑦𝑦Γ𝑦𝑦𝑦 = −

Λ𝑦

Ω

Γ𝑦𝑦𝑥 =1

2𝜕𝑥(𝑔𝑦𝑦) =

Λ𝑥

Ω3 ⇒ Γ 𝑦𝑥

𝑦 = 𝑔𝑦𝑦Γ𝑦𝑦𝑥 = −

Λ𝑥

Ω

Γ𝑥𝑦𝑦 = −1

2𝜕𝑥(𝑔𝑦𝑦) = −

Λ𝑥

Ω3 ⇒ Γ 𝑦𝑦

𝑥 = 𝑔𝑥𝑥Γ𝑥𝑦𝑦 =Λ𝑥

Ω


Γ 𝑢𝜈𝜈 = −Γ 𝑢𝑢

𝑢 = −Λ𝜈

Γ 𝑢𝑢𝜈 = Λ2𝜈3

Γ 𝑥𝑥𝑥 = Γ 𝑦𝑥

𝑦= −Γ 𝑦𝑦

𝑥 = −Λ𝑥

Ω

Γ 𝑥𝑦𝑥 = −Γ 𝑥𝑥

𝑦= Γ 𝑦𝑦

𝑦= −

Λ𝑦

Ω






The basis one forms


𝑑𝑠2 = −Λ𝜈2𝑑𝑢2 + 2𝑑𝑢𝑑𝑣 −1

Ω2(𝑑𝑥2 + 𝑑𝑦2) = (𝜔��)

2− (𝜔��)

2− (𝜔��)

2− (𝜔��)

2

⇒ 𝑑𝑢[−Λ𝜈2𝑑𝑢 + 2𝑑𝑣] −1

Ω2𝑑𝑥2 −

1

Ω2𝑑𝑦2 = (𝜔�� +𝜔��)(𝜔�� −𝜔��) − (𝜔��)

2− (𝜔��)

2

⇒ 𝜔�� +𝜔�� = 𝑑𝑢 𝜔�� −𝜔�� = −Λ𝜈2𝑑𝑢 + 2𝑑𝑣

𝜔𝑥 =1

Ω𝑑𝑥

𝜔�� =1

Ω𝑑𝑦

⇒ 𝜔�� =1

2(−Λ𝜈2 + 1)𝑑𝑢 + 𝑑𝑣 𝑑𝑢 = 𝜔�� +𝜔��

𝜔�� =1

2(Λ𝜈2 + 1)𝑑𝑢 − 𝑑𝑣 𝑑𝑣 =

1

2(1 + Λ𝜈2)𝜔�� −

1

2(1 − Λ𝜈2)𝜔��

𝜔�� =1

Ω𝑑𝑥 𝑑𝑥 = Ω𝜔𝑥

𝜔�� =1

Ω𝑑𝑦 𝑑𝑦 = Ω𝜔��

𝜂𝑖𝑗 = {

1−1

−1−1

}


Now we can use the basis one-forms to construct a orthonormal null tetrad

(

𝑙𝑛𝑚��

) =1

√2(

1 1 0 01 −1 0 00 0 1 𝑖0 0 1 −𝑖

)(

𝜔��

𝜔��

𝜔��

𝜔��

) =1

√2(

𝜔�� +𝜔��

𝜔�� −𝜔��

𝜔�� + 𝑖𝜔 ��

𝜔�� − 𝑖𝜔 ��

) =1

√2

(

𝑑𝑢−Λ𝜈2𝑑𝑢 + 2𝑑𝑣1

Ω𝑑𝑥 + 𝑖

1

Ω𝑑𝑦

1

Ω𝑑𝑥 − 𝑖

1

Ω𝑑𝑦 )

(9.10)


𝑙𝑎 =1

√2(1, 0, 0, 0) 𝑛𝑎 =

1

√2(−Λ𝜈2, 2, 0, 0)

𝑚𝑎 =1

√2(0, 0,

1

Ω, 𝑖

1

Ω) 𝑚𝑎 =

1

√2(0, 0,

1

Ω, −𝑖

1

Ω)


𝑙𝑢 = 𝑔𝑎𝑢𝑙𝑎 = 𝑔𝑣𝑢𝑙𝑣 = 1 ⋅ 0 = 0

𝑙𝑣 = 𝑔𝑎𝑣𝑙𝑎 = 𝑔𝑢𝑣𝑙𝑢 + 𝑔

𝑣𝑣𝑙𝑣 = 1 ⋅ (1

√2) + Λ𝜈2 ⋅ 0 =

1

√2

𝑙𝑥 = 𝑙𝑦 = 0

𝑛𝑢 = 𝑔𝑎𝑢𝑛𝑎 = 𝑔𝑣𝑢𝑛𝑣 = 1 ⋅ (

2

√2) = √2





𝑛𝑣 = 𝑔𝑎𝑣𝑛𝑎 = 𝑔𝑢𝑣𝑛𝑢 + 𝑔

𝑣𝑣𝑛𝑣 = 1 ⋅ (1

√2(−Λ𝜈2)) + Λ𝜈2 ⋅ (

2

√2) =

1

√2Λ𝜈2

𝑛𝑥 = 𝑛𝑦 = 0 𝑚𝑢 = 𝑚𝑣 = 0

𝑚𝑥 = 𝑔𝑎𝑥𝑚𝑎 = 𝑔𝑥𝑥𝑚𝑥 = (−Ω

2) ⋅1

√2

1

Ω= −

Ω

√2

𝑚𝑦 = 𝑔𝑎𝑦𝑚𝑎 = 𝑔𝑦𝑦𝑚𝑦 = (−Ω

2) ⋅ 𝑖1

√2

1

Ω= −𝑖

Ω

√2


𝑙𝑎 =1

√2(1, 0, 0, 0) 𝑙𝑎 =

1

√2(0, 1, 0, 0)

𝑛𝑎 =1

√2(−Λ𝜈2, 2, 0, 0) 𝑛𝑎 =

1

√2(2, Λ𝜈2, 0, 0)

𝑚𝑎 =1

√2(0, 0,

1

Ω, 𝑖

1

Ω) 𝑚𝑎 =

1

√2(0, 0, −Ω, −𝑖Ω)

��𝑎 =1

√2(0, 0,

1

Ω, −𝑖

1

Ω) ��𝑎 =

1

√2(0, 0, −Ω, 𝑖Ω)



𝑎𝑙𝑏 휀 =1



(9.15) 𝜈 = −∇𝑏𝑛𝑎��


1




𝑎��𝑏 𝛼 =1


𝑎��𝑏 − ∇𝑏𝑚𝑎��𝑎��𝑏)







= −∇𝜈𝑛𝑎��𝑎𝑙𝜈

= −∇𝜈𝑛𝑥��𝑥𝑙𝜈 − ∇ν𝑛𝑦��

𝑦𝑙𝜈

= −(𝜕𝜈𝑛𝑥 − Γc𝜈𝑥𝑛𝑐)��

𝑥𝑙𝜈 − (𝜕𝜈𝑛𝑦 − Γc𝜈𝑦𝑛𝑐)��

𝑦𝑙𝜈

= 0



𝑦𝑛𝑢 = 0


= −∇𝑥𝑛𝑎��𝑎��𝑥 − ∇𝑦𝑛𝑎��

𝑎��𝑦

= −∇𝑥𝑛𝑥��𝑥��𝑥 − ∇𝑦𝑛𝑥��

𝑥��𝑦 − ∇𝑥𝑛𝑦��𝑦��𝑥 − ∇𝑦𝑛𝑦��

𝑦��𝑦

= 0 𝜇 = −∇𝑏𝑛𝑎��


𝑎𝑙𝑏 = ∇ν𝑙𝑎𝑚𝑎𝑙ν = ∇ν𝑙𝑥𝑚

𝑥𝑙ν + ∇ν𝑙𝑦𝑚𝑦𝑙ν = 0



𝑦𝑛𝑢 = 0







𝑎��𝑦



𝑦��𝑦






𝑦��𝑦

= 0 𝜎 = ∇𝑏𝑙𝑎𝑚

𝑎𝑚𝑏 = 0

휀 =1



=1

2(∇ν𝑙𝑎𝑛


=1

2(∇ν𝑙𝑢𝑛


𝑦𝑙ν)

= 0

𝛾 =1



=1



=1



𝑦𝑛𝑢)

=1

2(𝜕𝑢𝑙𝑢 − Γ 𝑢𝑢

𝑐 𝑙𝑐)𝑛𝑢𝑛𝑢

= −

1

2Γ 𝑢𝑢𝑢 𝑙𝑢𝑛

𝑢𝑛𝑢

= −

1

2Λ𝜈

1

√2(2

√2)2

= −

1

√2Λ𝜈

𝛼 =1


𝑎��𝑏 − ∇𝑏𝑚𝑎��𝑎��𝑏)

=1


𝑎��𝑥 − ∇𝑥𝑚𝑎��𝑎��𝑥) +

1


𝑎��𝑦 − ∇𝑦𝑚𝑎��𝑎��𝑦)

= −

1

2(∇𝑥𝑚𝑥��

𝑥��𝑥 + ∇𝑦𝑚𝑥��𝑥��𝑦 + ∇𝑥𝑚𝑦��

𝑦��𝑥 + ∇𝑦𝑚𝑦��𝑦��𝑦)

= −

1

2([𝜕𝑥𝑚𝑥 − Γ 𝑥𝑥

𝑐 𝑚𝑐]��𝑥��𝑥 + [𝜕𝑦𝑚𝑥 − Γ 𝑦𝑥

𝑐 𝑚𝑐]��𝑥��𝑦 + [𝜕𝑥𝑚𝑦 − Γ 𝑥𝑦

𝑐 𝑚𝑐]��𝑦��𝑥

+ [𝜕𝑦𝑚𝑦 − Γ 𝑦𝑦𝑐 𝑚𝑐]��

𝑦��𝑦)

= −

1

2([𝜕𝑥 (

1

√2Ω) − Γ 𝑥𝑥

𝑥 𝑚𝑥 − Γ 𝑥𝑥𝑦

𝑚𝑦] ��𝑥��𝑥 + [𝜕𝑦 (

1

√2Ω) − Γ 𝑦𝑥

𝑥 𝑚𝑥 − Γ 𝑦𝑥𝑦

𝑚𝑦] ��𝑥��𝑦

+ [𝜕𝑥 (𝑖

√2Ω) − Γ 𝑥𝑦

𝑥 𝑚𝑥 − Γ 𝑥𝑦𝑦

𝑚𝑦] ��𝑦��𝑥

+ [𝜕𝑦 (𝑖

√2Ω) − Γ 𝑦𝑦

𝑥 𝑚𝑥 − Γ 𝑦𝑦𝑦

𝑚𝑦] ��𝑦��𝑦)

= −

1

2([(−

Λ𝑥

√2Ω2) − (−

Λ𝑥

Ω)(

1

√2Ω) − (

Λ𝑦

Ω) (𝑖

1

√2Ω)] ��𝑥��𝑥

+ [(−Λ𝑦

√2Ω2) − (−

Λ𝑦

Ω)(

1

√2Ω) − (−

Λ𝑥

Ω)(𝑖

1

√2Ω)] ��𝑥��𝑦

+ [(−𝑖Λ𝑥

√2Ω2) − (−

Λ𝑦

Ω)(

1

√2Ω) − (−

Λ𝑥

Ω)(𝑖

1

√2Ω)] ��𝑦��𝑥

+ [(−𝑖Λ𝑦

√2Ω2) − (

Λ𝑥

Ω) (

1

√2Ω) − (−

Λ𝑦

Ω)(𝑖

1

√2Ω)] ��𝑦��𝑦)





= −

1

2([− (

Λ𝑦

Ω) (𝑖

1

√2Ω)] ��𝑥��𝑥 + [−(−

Λ𝑥

Ω)(𝑖

1

√2Ω)] ��𝑥��𝑦 + [−(−

Λ𝑦

Ω)(

1

√2Ω)] ��𝑦��𝑥

+ [−(Λ𝑥

Ω)(

1

√2Ω)] ��𝑦��𝑦)

= −1

2((−𝑖

Λ𝑦

√2Ω2) (−

Ω

√2)2

+ (𝑖Λ𝑥

√2Ω2) (−

Ω

√2) (𝑖

Ω

√2) + (

Λ𝑦

√2Ω2) (𝑖

Ω

√2) (−

Ω

√2)

+ (−Λ𝑥

√2Ω2) (𝑖

Ω

√2)2

)

= −

1

2((−𝑖

Λ𝑦

2√2) + (

Λ𝑥

2√2) − (𝑖

Λ𝑦

2√2) + (

Λ𝑥

2√2Ω2))

=

Λ

2√2(−𝑥 + 𝑖𝑦)

𝛽 =1



= −1

2((−

Λ𝑦

Ω)(

𝑖

√2Ω) ��𝑥𝑚𝑥 + (

Λ𝑥

Ω) (

𝑖

√2Ω) ��𝑥𝑚𝑦 + (

Λ𝑦

Ω)(

1

√2Ω) ��𝑦𝑚𝑥 − (

Λ𝑥

Ω)(

1

√2Ω) ��𝑦𝑚𝑦)

= −1

2((−

Λ𝑦

Ω)(

𝑖

√2Ω) (−

Ω

√2)2

+ (Λ𝑥

Ω)(

𝑖

√2Ω) (−

Ω

√2) (−

𝑖Ω

√2) + (

Λ𝑦

Ω) (

1

√2Ω) (𝑖Ω

√2) (−

Ω

√2)

− (Λ𝑥

Ω) (

1

√2Ω) (𝑖Ω

√2) (−

𝑖Ω

√2))

= −1

2(−

𝑖Λ𝑦

2√2−Λ𝑥

2√2−𝑖Λ𝑦

2√2−Λ𝑥

2√2)

=Λ

2√2(𝑥 + 𝑖𝑦)


𝜈 = 0 𝜏 = 0 𝛾 = −1

√2Λ𝜈

𝜆 = 0 𝜌 = 0 𝛼 = −Λ

2√2(𝑥 − 𝑖𝑦)

𝜇 = 0 𝜎 = 0 𝛽 =Λ

2√2(𝑥 + 𝑖𝑦)

Newman-Penrose identities

𝐷𝛾 − Δ휀 = 𝛼(𝜏 + ��) + 𝛽(�� + 𝜋) − 𝛾(휀 + 휀) + 𝜏𝜋 − 𝜈𝜅 + Ψ2 − ΛNP +Φ11 (13.58)

𝛿𝛼 − 𝛿𝛽 = 𝜌𝜇 − 𝜎𝜆 + 𝛼�� + 𝛽�� − 2𝛼𝛽 + 𝛾(𝜌 − ��) + 휀(𝜇 − ��) − Ψ2 + ΛNP +Φ11 (13.59)

Δ𝜌 − 𝛿𝜏 = −𝜌�� − 𝜎𝜆 + 𝜏(�� − 𝛼 − ��) + 𝜌(𝛾 + ��) + 𝜅𝜈 − Ψ2 − 2ΛNP (13.60)






Reduces to 𝐷𝛾 = Ψ2 − ΛNP +Φ11

𝛿𝛼 − 𝛿𝛽 = 𝛼�� + 𝛽�� − 2𝛼𝛽 − Ψ2 + ΛNP +Φ11 0 = −Ψ2 − 2ΛNP These we can solve

Ψ2 − ΛNP +Φ11 = 𝐷𝛾 = 𝑙𝑎∇𝑎 (−1

√2Λ𝜈) = 𝑙𝜈𝜕𝜈 (−

1

√2Λ𝜈) =

1

√2(−

1

√2Λ) = −

1

2Λ

Ψ2 − ΛNP −Φ11 = 𝛼�� + 𝛽�� − 2𝛼𝛽 − 𝛿𝛼 + 𝛿𝛽

= (−Λ

2√2(𝑥 − 𝑖𝑦))(−

Λ

2√2(𝑥 + 𝑖𝑦)) + (

Λ

2√2(𝑥 + 𝑖𝑦))(

Λ

2√2(𝑥 − 𝑖𝑦))

− 2(−Λ

2√2(𝑥 − 𝑖𝑦))(

Λ

2√2(𝑥 + 𝑖𝑦))

−𝑚𝑎∇𝑎 (−Λ

2√2(𝑥 − 𝑖𝑦)) + ��𝑎∇𝑎 (

Λ

2√2(𝑥 + 𝑖𝑦))

=Λ2

8(𝑥2 + 𝑦2) +

Λ2

8(𝑥2 + 𝑦2) + 2

Λ2

8(𝑥2 + 𝑦2) − 𝑚𝑥𝜕𝑥 (−

Λ

2√2(𝑥 − 𝑖𝑦))

−𝑚𝑦𝜕𝑦 (−Λ

2√2(𝑥 − 𝑖𝑦)) + ��𝑥𝜕𝑥 (

Λ

2√2(𝑥 + 𝑖𝑦))

+ ��𝑦𝜕𝑦 (Λ

2√2(𝑥 + 𝑖𝑦))

=Λ2

2(𝑥2 + 𝑦2) +

Ω

√2(−

Λ

2√2) + 𝑖

Ω

√2(𝑖

Λ

2√2) + (−

Ω

√2) (

Λ

2√2)

+ (𝑖Ω

√2) (𝑖

Λ

2√2)

=

Λ2

2(𝑥2 + 𝑦2) −

ΩΛ

4−ΩΛ

4−ΩΛ

4−ΩΛ

4

=

Λ2

2(𝑥2 + 𝑦2) − ΩΛ

=

Λ2

2(𝑥2 + 𝑦2) − (1 +

Λ

2(𝑥2 + 𝑦2))Λ

= −Λ Collecting the results

−1

2Λ = Ψ2 − ΛNP +Φ11

−Λ = Ψ2 − ΛNP −Φ11 0 = −Ψ2 − 2ΛNP

⇒ Ψ2 = −1

2Λ (13.65)

ΛNP =

1

4Λ (13.64)

Φ11 =

1

4Λ (13.65)

Checking Φ11 and 𝑅𝑎𝑏 = −𝑔𝑎𝑏Λ





Φ11 = −1

4𝑅𝑎𝑏(𝑙

𝑎𝑛𝑏 +𝑚𝑎��𝑏) (9.22)

=1

4𝑔𝑎𝑏Λ(𝑙

𝑎𝑛𝑏 +𝑚𝑎��𝑏) 𝑅𝑎𝑏 = −𝑔𝑎𝑏Λ

=1

4Λ(𝑔𝑎𝑏𝑙

𝑎𝑛𝑏 + 𝑔𝑎𝑏𝑚𝑎��𝑏)

=1

4Λ(𝑔𝑢𝑢𝑙

𝑢𝑛𝑢 + 𝑔𝑢𝜈𝑙𝑢𝑛𝜈 + 𝑔𝜈𝑢𝑙

𝜈𝑛𝑢 + 𝑔𝑥𝑥𝑚𝑥��𝑥 + 𝑔𝑦𝑦𝑚

𝑦��𝑦)

=1

4Λ(𝑔𝜈𝑢𝑙

𝜈𝑛𝑢 + 𝑔𝑥𝑥𝑚𝑥��𝑥 + 𝑔𝑦𝑦𝑚

𝑦��𝑦)

=1

4Λ(𝑙𝜈𝑛𝑢 −

1

Ω2(𝑚𝑥��𝑥 +𝑚𝑦��𝑦))

=1

4Λ(

1

√2⋅2

√2−1

Ω2((−

1

√2Ω)

2

+ (−𝑖1

√2Ω) 𝑖

1

√2Ω))

=1

4Λ(1 − 1)

= 0 And we can conclude that 𝑅𝑎𝑏 ≠ −𝑔𝑎𝑏Λ

Instead we will look at a generalized Nariai spacetime

The line element: 𝑑𝑠2 = −𝐴Λ𝜈2𝑑𝑢2 + 𝐵𝑑𝑢𝑑𝑣 −𝐶

Ω2(𝑑𝑥2 + 𝑑𝑦2)

Ω = 𝐷 + 𝐸Λ(𝑥2 + 𝑦2)


𝑑𝑠2 = −𝐴Λ𝜈2𝑑𝑢2 + 𝐵𝑑𝑢𝑑𝑣 −𝐶

Ω2(𝑑𝑥2 + 𝑑𝑦2) = (𝜔��)

2− (𝜔��)

2− (𝜔��)

2− (𝜔��)

2

⇒ 𝑑𝑢[−𝐴Λ𝜈2𝑑𝑢 + 𝐵𝑑𝑣] −𝐶

Ω2𝑑𝑥2 −

𝐶

Ω2𝑑𝑦2 = (𝜔�� +𝜔��)(𝜔�� −𝜔��) − (𝜔��)

2− (𝜔��)

2

⇒ 𝜔�� +𝜔�� = 𝑑𝑢 𝜔�� −𝜔�� = −𝐴Λ𝜈2𝑑𝑢 + 𝐵𝑑𝑣

𝜔�� =√𝐶

Ω𝑑𝑥

𝜔�� =√𝐶

Ω𝑑𝑦

The basis one-forms

𝜔�� =1

2(−𝐴Λ𝜈2 + 1)𝑑𝑢 +

1

2𝐵𝑑𝑣 𝑑𝑢 = 𝜔�� +𝜔��

𝜔�� =1

2(AΛ𝜈2 + 1)𝑑𝑢 −

1

2𝐵𝑑𝑣 𝑑𝑣 =

1

𝐵(1 + 𝐴Λ𝜈2)𝜔�� −

1

𝐵(1 − 𝐴Λ𝜈2)𝜔��

𝜔�� =√𝐶

Ω𝑑𝑥 𝑑𝑥 =

Ω

√𝐶𝜔𝑥

𝜔 �� =√𝐶

Ω𝑑𝑦 𝑑𝑦 =

Ω

√𝐶𝜔��





𝜂𝑖𝑗 = {

1−1

−1−1

}


𝑑𝜔�� = −Γ �� ∧ 𝜔�� (5.9)

𝑑𝜔�� = 𝑑 (1

2(−𝐴Λ𝜈2 + 1)𝑑𝑢 +

1

2𝐵𝑑𝑣)

= −𝐴Λ𝜈𝑑𝜈 ∧ 𝑑𝑢

= −𝐴Λ𝜈 (1

𝐵(1 + 𝐴Λ𝜈2)𝜔�� −

1

𝐵(1 − 𝐴Λ𝜈2)𝜔��) ∧ (𝜔�� +𝜔��)

= −2𝐴

𝐵Λ𝜈𝜔�� ∧ 𝜔��

𝑑𝜔�� = 𝑑 (1

2(AΛ𝜈2 + 1)𝑑𝑢 −

1

2𝐵𝑑𝑣)

= 𝐴Λ𝜈𝑑𝜈 ∧ 𝑑𝑢

= 𝐴Λ𝜈 (1

𝐵(1 + 𝐴Λ𝜈2)𝜔�� −

1

𝐵(1 − 𝐴Λ𝜈2)𝜔��) ∧ (𝜔�� +𝜔��)

= −2𝐴

𝐵Λ𝜈𝜔�� ∧ 𝜔��

𝑑𝜔𝑥 = 𝑑 (√𝐶

Ω𝑑𝑥) = 𝑑 (

√𝐶

𝐷 + 𝐸Λ(𝑥2 + 𝑦2)𝑑𝑥) =

−2𝑦Λ𝐸√𝐶

Ω2𝑑𝑦 ∧ 𝑑𝑥 =

2𝑦Λ𝐸

√𝐶𝜔𝑥 ∧ 𝜔��

𝑑𝜔�� = 𝑑 (√𝐶

Ω𝑑𝑦) = 𝑑 (

√𝐶

𝐷 + 𝐸Λ(𝑥2 + 𝑦2)𝑑𝑦) =

−2𝑥Λ𝐸√𝐶

Ω2𝑑𝑥 ∧ 𝑑𝑦 =

2𝑥Λ𝐸

√𝐶𝜔�� ∧ 𝜔𝑥


Γ �� =

{

0

2𝐴

𝐵Λ𝜈(𝜔�� +𝜔��) 0 0

2𝐴

𝐵Λ𝜈(𝜔�� +𝜔��) 0 0 0

0 0 0 −2Λ𝐸

√𝐶(𝑥𝜔�� − 𝑦𝜔𝑥)

0 02Λ𝐸

√𝐶(𝑥𝜔�� − 𝑦𝜔��) 0

}



Ω �� = 𝑑Γ ��

�� + Γ 𝑐�� ∧ Γ ��

𝑐 =1

2𝑅 ��𝑐�� 𝜔𝑐 ∧ 𝜔�� (5.27), (5.28)





First we see that Γ 𝑐�� ∧ Γ ��

𝑐 = 0 for all combinations

Ω 𝜈 �� = 𝑑Γ ��

��

= 𝑑 (2𝐴

𝐵Λ𝜈(𝜔�� +𝜔��))

= 𝑑 (2𝐴

𝐵Λ𝜈𝑑𝑢)

=2𝐴

𝐵𝛬𝑑𝜈 ∧ 𝑑𝑢

=2𝐴

𝐵𝛬 (1

𝐵(1 + 𝐴Λ𝜈2)𝜔�� −

1

𝐵(1 − 𝐴Λ𝜈2)𝜔��) ∧ (𝜔�� +𝜔��)

=4𝐴

𝐵2𝛬𝜔�� ∧ 𝜔��

Ω �� = 𝑑Γ ��

�� = 𝑑Γ �� = −

4𝐴

𝐵2𝛬𝜔�� ∧ 𝜔��

Ω 𝑦 𝑥 = 𝑑Γ ��

𝑥

= 𝑑 (2Λ𝐸

√𝐶(𝑥𝜔�� − 𝑦𝜔𝑥))

= 𝑑 (2Λ𝐸

Ω(𝑥𝑑𝑦 − 𝑦𝑑𝑥))

= (2Λ𝐸

Ω−4Λ2𝐸2

Ω2𝑥2)𝑑𝑥 ∧ 𝑑𝑦 + (−

2Λ𝐸

Ω+4Λ2𝐸2

Ω2𝑦2)𝑑𝑦 ∧ 𝑑𝑥

= (2Λ𝐸Ω

𝐶−4Λ2𝐸2

𝐶𝑥2)𝜔�� ∧ 𝜔�� + (−

2Λ𝐸Ω

𝐶+4Λ2𝐸2

𝐶𝑦2)𝜔�� ∧ 𝜔��

= (4Λ𝐸Ω

𝐶−4Λ2𝐸2

𝐶(𝑥2 + 𝑦2))𝜔�� ∧ 𝜔��

=4Λ𝐸

𝐶(Ω − Λ𝐸(𝑥2 + 𝑦2))𝜔�� ∧ 𝜔��

=4Λ𝐸

𝐶(𝐷 + 𝐸Λ(𝑥2 + 𝑦2) − Λ𝐸(𝑥2 + 𝑦2))𝜔�� ∧ 𝜔 ��

=4Λ𝐸𝐷

𝐶𝜔𝑥 ∧ 𝜔��

Ω ��

= 𝑑Γ ��= −𝑑Γ ��

𝑥 =4Λ𝐸𝐷

𝐶𝜔�� ∧ 𝜔𝑥


Ω �� =

{

0 −

4𝐴

𝐵2𝛬𝜔�� ∧ 𝜔�� 0 0

4𝐴

𝐵2𝛬𝜔�� ∧ 𝜔�� 0 0 0

0 0 04Λ𝐸𝐷

𝐶𝜔�� ∧ 𝜔𝑥

0 04Λ𝐸𝐷

𝐶𝜔𝑥 ∧ 𝜔�� 0 }






R �� = −

4𝐴

𝐵2Λ R ��

𝑥 =4Λ𝐸𝐷

𝐶

R �� =

4𝐴

𝐵2Λ R ��

�� =

4Λ𝐸𝐷

𝐶

The Ricci tensor:

𝑅�� = 𝑅 ��𝑐��𝑐 (4.46)

𝑅�� = 𝑅 ��𝑐��𝑐 = 𝑅 ��

�� + 𝑅 �� + 𝑅 ��

�� + 𝑅 ��

= 𝑅 �� = −

4𝐴

𝐵2Λ

𝑅�� = 𝑅 ��𝑐��𝑐 = 𝑅 ��

�� + 𝑅 �� + 𝑅 ��

�� + 𝑅 ��

= 0

𝑅�� = 0 𝑅�� = 0

𝑅�� = 𝑅 ��𝑐��𝑐 = 𝑅 ��

�� + 𝑅 �� + 𝑅 ��

�� + 𝑅 ��

= 𝑅 �� =

4𝐴

𝐵2Λ

𝑅𝑥�� = 0 𝑅�� = 0

𝑅𝑥�� = 𝑅 ��𝑐��𝑐 = 𝑅 ��

�� + 𝑅 �� + 𝑅 ��

�� + 𝑅 ��

= 𝑅 ��

=4Λ𝐸𝐷

𝐶

𝑅��𝑥 = 𝑅 ��𝑐��𝑐 = 𝑅 ��

�� + 𝑅 �� + 𝑅 ��

�� + 𝑅 ��

= 0

𝑅�� = 𝑅 ��𝑐��𝑐 = 𝑅 ��

�� + 𝑅 �� + 𝑅 ��

�� + 𝑅 ��

= 𝑅 �� =

4Λ𝐸𝐷

𝐶


𝑅�� =

{

−

4𝐴

𝐵2Λ 0 0 0

04𝐴

𝐵2Λ 0 0

0 04Λ𝐸𝐷

𝐶0

0 0 04Λ𝐸𝐷

𝐶 }

= {

−Λ 0 0 00 Λ 0 00 0 Λ 00 0 0 Λ

} = −𝜂��Λ

Where �� refers to column and �� to row Compared with197 𝑅�� = −𝜂��Λ we can see the that we can choose the coefficients are 𝐴 = 1;𝐵 =

2; 𝐶 = 1,𝐷 = 1 and 𝐸 =1

4, which corresponds to a Nariai line element consistent with 𝑅𝑎𝑏 = −𝑔𝑎𝑏Λ

𝑑𝑠2 = −Λ𝜈2𝑑𝑢2 + 2𝑑𝑢𝑑𝑣 −1

Ω2(𝑑𝑥2 + 𝑑𝑦2)

Ω = 1 +Λ

4(𝑥2 + 𝑦2)

So let’s copy the Christoffel, spin coefficient and Newman-Penrose identity calculations with this new

Ω = 1 +Λ

4(𝑥2 + 𝑦2). The null tetrad is unchanged.


197 Page 138





Γ𝑎𝑏𝑐 =1



Γ𝑢𝑢𝜈 =1

2𝜕𝜈(𝑔𝑢𝑢) = −Λ𝜈 ⇒ Γ 𝑢𝜈

𝜈 = 𝑔𝜈𝑢Γ𝑢𝑢𝜈 = −Λ𝜈


2𝜕𝜈(𝑔𝑢𝑢) = Λ𝜈 ⇒ Γ 𝑢𝑢

𝜈 = 𝑔𝜈𝜈Γ𝜈𝑢𝑢 = Λ2𝜈3

⇒ Γ 𝑢𝑢𝑢 = 𝑔𝑢𝜈Γ𝜈𝑢𝑢 = Λ𝜈

Γ𝑥𝑥𝑥 =1

2𝜕𝑥(𝑔𝑥𝑥) =

Λ𝑥

2Ω3 ⇒ Γ 𝑥𝑥

𝑥 = 𝑔𝑥𝑥Γ𝑥𝑥𝑥 = −Λ𝑥

2Ω

Γ𝑥𝑥𝑦 =1

2𝜕𝑦(𝑔𝑥𝑥) =

Λ𝑦

2Ω3 ⇒ Γ 𝑥𝑦

𝑥 = 𝑔𝑥𝑥Γ𝑥𝑥𝑦 = −Λ𝑦

2Ω

Γ𝑦𝑥𝑥 = −1

2𝜕𝑦(𝑔𝑥𝑥) = −

Λ𝑦

2Ω3 ⇒ Γ 𝑥𝑥

𝑦 = 𝑔𝑦𝑦Γ𝑦𝑥𝑥 =

Λ𝑦

2Ω

Γ𝑦𝑦𝑦 =1

2𝜕𝑦(𝑔𝑦𝑦) =

Λ𝑦

2Ω3 ⇒ Γ 𝑦𝑦

𝑦 = 𝑔𝑦𝑦Γ𝑦𝑦𝑦 = −

Λ𝑦

2Ω

Γ𝑦𝑦𝑥 =1

2𝜕𝑥(𝑔𝑦𝑦) =

Λ𝑥

2Ω3 ⇒ Γ 𝑦𝑥

𝑦 = 𝑔𝑦𝑦Γ𝑦𝑦𝑥 = −

Λ𝑥

2Ω

Γ𝑥𝑦𝑦 = −1

2𝜕𝑥(𝑔𝑦𝑦) = −

Λ𝑥

2Ω3 ⇒ Γ 𝑦𝑦

𝑥 = 𝑔𝑥𝑥Γ𝑥𝑦𝑦 =Λ𝑥

2Ω


Γ 𝑢𝜈𝜈 = −Γ 𝑢𝑢

𝑢 = −Λ𝜈

Γ 𝑢𝑢𝜈 = Λ2𝜈3

Γ 𝑥𝑥𝑥 = Γ 𝑦𝑥

𝑦= −Γ 𝑦𝑦

𝑥 = −Λ𝑥

2Ω

Γ 𝑥𝑦𝑥 = −Γ 𝑥𝑥

𝑦= Γ 𝑦𝑦

𝑦= −

Λ𝑦

2Ω



𝑎𝑙𝑏 휀 =1



(9.15) 𝜈 = −∇𝑏𝑛𝑎��


1




𝑎��𝑏 𝛼 =1


𝑎��𝑏 − ∇𝑏𝑚𝑎��𝑎��𝑏)







= −∇𝜈𝑛𝑎��𝑎𝑙𝜈

= −∇𝜈𝑛𝑥��𝑥𝑙𝜈 − ∇ν𝑛𝑦��

𝑦𝑙𝜈

= −(𝜕𝜈𝑛𝑥 − Γc𝜈𝑥𝑛𝑐)��

𝑥𝑙𝜈 − (𝜕𝜈𝑛𝑦 − Γc𝜈𝑦𝑛𝑐)��

𝑦𝑙𝜈

= 0



𝑦𝑛𝑢 = 0


= −∇𝑥𝑛𝑎��𝑎��𝑥 − ∇𝑦𝑛𝑎��

𝑎��𝑦

= −∇𝑥𝑛𝑥��𝑥��𝑥 − ∇𝑦𝑛𝑥��

𝑥��𝑦 − ∇𝑥𝑛𝑦��𝑦��𝑥 − ∇𝑦𝑛𝑦��

𝑦��𝑦

= 0





𝜇 = −∇𝑏𝑛𝑎��𝑎𝑚𝑏 = 0

𝜅 = ∇𝑏𝑙𝑎𝑚𝑎𝑙𝑏 = ∇ν𝑙𝑎𝑚

𝑎𝑙ν = ∇ν𝑙𝑥𝑚𝑥𝑙ν + ∇ν𝑙𝑦𝑚

𝑦𝑙ν = 0



𝑦𝑛𝑢 = 0



𝑎��𝑦



𝑦��𝑦






𝑦��𝑦

= 0 𝜎 = ∇𝑏𝑙𝑎𝑚

𝑎𝑚𝑏 = 0

휀 =1



=1

2(∇ν𝑙𝑎𝑛


=1

2(∇ν𝑙𝑢𝑛


𝑦𝑙ν)

= 0

𝛾 =1



=1



=1



𝑦𝑛𝑢)

=1

2(𝜕𝑢𝑙𝑢 − Γ 𝑢𝑢

𝑐 𝑙𝑐)𝑛𝑢𝑛𝑢

= −

1

2Γ 𝑢𝑢𝑢 𝑙𝑢𝑛

𝑢𝑛𝑢

= −

1

2Λ𝜈

1

√2(2

√2)2

= −

1

√2Λ𝜈

𝛼 =1


𝑎��𝑏 − ∇𝑏𝑚𝑎��𝑎��𝑏)

=1


𝑎��𝑥 − ∇𝑥𝑚𝑎��𝑎��𝑥) +

1


𝑎��𝑦 − ∇𝑦𝑚𝑎��𝑎��𝑦)

= −

1

2(∇𝑥𝑚𝑥��

𝑥��𝑥 + ∇𝑦𝑚𝑥��𝑥��𝑦 + ∇𝑥𝑚𝑦��

𝑦��𝑥 + ∇𝑦𝑚𝑦��𝑦��𝑦)

= −

1

2((𝜕𝑥𝑚𝑥 − Γ 𝑥𝑥

𝑐 𝑚𝑐)��𝑥��𝑥 + (𝜕𝑦𝑚𝑥 − Γ 𝑦𝑥

𝑐 𝑚𝑐)��𝑥��𝑦 + (𝜕𝑥𝑚𝑦 − Γ 𝑥𝑦

𝑐 𝑚𝑐)��𝑦��𝑥

+ (𝜕𝑦𝑚𝑦 − Γ 𝑦𝑦𝑐 𝑚𝑐)��

𝑦��𝑦)

= −

1

2((𝜕𝑥 (

1

√2Ω) − Γ 𝑥𝑥

𝑥 𝑚𝑥 − Γ 𝑥𝑥𝑦

𝑚𝑦) ��𝑥��𝑥 + (𝜕𝑦 (

1

√2Ω) − Γ 𝑦𝑥

𝑥 𝑚𝑥 − Γ 𝑦𝑥𝑦

𝑚𝑦) ��𝑥��𝑦

+ (𝜕𝑥 (𝑖

√2Ω) − Γ 𝑥𝑦

𝑥 𝑚𝑥 − Γ 𝑥𝑦𝑦

𝑚𝑦) ��𝑦��𝑥

+ (𝜕𝑦 (𝑖

√2Ω) − Γ 𝑦𝑦

𝑥 𝑚𝑥 − Γ 𝑦𝑦𝑦

𝑚𝑦) ��𝑦��𝑦)





= −

1

2([(−

Λ𝑥

2√2Ω2) + (

Λ𝑥

2Ω) (

1

√2Ω) − (

Λ𝑦

2Ω) (𝑖

1

√2Ω)] ��𝑥��𝑥

+ [(−Λ𝑦

2√2Ω2) + (

Λ𝑦

2Ω) (

1

√2Ω) − (

Λ𝑥

2Ω) (𝑖

1

√2Ω)] ��𝑥��𝑦

+ [(−𝑖Λ𝑥

2√2Ω2) + (

Λ𝑦

2Ω) (

1

√2Ω) + (

Λ𝑥

2Ω) (𝑖

1

√2Ω)] ��𝑦��𝑥

+ [(−𝑖Λ𝑦

2√2Ω2) − (

Λ𝑥

2Ω) (

1

√2Ω) + (

Λ𝑦

2Ω) (𝑖

1

√2Ω)] ��𝑦��𝑦)

= −

1

2([(−

Λ𝑦

2Ω) (𝑖

1

√2Ω)] ��𝑥��𝑥 + [(

Λ𝑥

2Ω) (𝑖

1

√2Ω)] ��𝑥��𝑦 + [(

Λ𝑦

2Ω) (

1

√2Ω)] ��𝑦��𝑥

+ [−(Λ𝑥

2Ω) (

1

√2Ω)] ��𝑦��𝑦)

= −1

2((−𝑖

Λ𝑦

2√2Ω2) (−

Ω

√2)2

+ (𝑖Λ𝑥

2√2Ω2) (−

Ω

√2) (𝑖

Ω

√2) + (

Λ𝑦

2√2Ω2) (𝑖

Ω

√2) (−

Ω

√2)

− (Λ𝑥

2√2Ω2) (𝑖

Ω

√2)2

)

= −

1

2(−(𝑖

Λ𝑦

4√2) + (

Λ𝑥

4√2) − (𝑖

Λ𝑦

4√2) + (

Λ𝑥

4√2Ω2))

= −

Λ

4√2(𝑥 − 𝑖𝑦)

𝛽 =1



= −

1

2([(−

Λ𝑦

2Ω) (𝑖

1

√2Ω)] ��𝑥𝑚𝑥 + [(

Λ𝑥

2Ω) (𝑖

1

√2Ω)] ��𝑥𝑚𝑦 + [(

Λ𝑦

2Ω) (

1

√2Ω)] ��𝑦𝑚𝑥

+ [−(Λ𝑥

2Ω) (

1

√2Ω)] ��𝑦𝑚𝑦)

= −1

2((−

Λ𝑦

2Ω) (

𝑖

√2Ω) (−

Ω

√2)2

+ (Λ𝑥

2Ω) (

𝑖

√2Ω) (−

Ω

√2) (−

𝑖Ω

√2) + (

Λ𝑦

2Ω) (

1

√2Ω) (𝑖Ω

√2) (−

Ω

√2)

− (Λ𝑥

2Ω) (

1

√2Ω) (𝑖Ω

√2) (−

𝑖Ω

√2))

= −1

2(−

𝑖Λ𝑦

4√2−Λ𝑥

4√2−𝑖Λ𝑦

4√2−Λ𝑥

4√2)

=Λ

4√2(𝑥 + 𝑖𝑦)


𝜈 = 0 𝜏 = 0 𝛾 = −1

√2Λ𝜈

𝜆 = 0 𝜌 = 0 𝛼 = −Λ

4√2(𝑥 − 𝑖𝑦)

𝜇 = 0 𝜎 = 0 𝛽 =Λ

4√2(𝑥 + 𝑖𝑦)





Newman-Penrose identities

𝐷𝛾 − Δ휀 = 𝛼(𝜏 + ��) + 𝛽(�� + 𝜋) − 𝛾(휀 + 휀) + 𝜏𝜋 − 𝜈𝜅 + Ψ2 − ΛNP +Φ11 (13.58)

𝛿𝛼 − 𝛿𝛽 = 𝜌𝜇 − 𝜎𝜆 + 𝛼�� + 𝛽�� − 2𝛼𝛽 + 𝛾(𝜌 − ��) + 휀(𝜇 − ��) − Ψ2 + ΛNP +Φ11 (13.59)

Δ𝜌 − 𝛿𝜏 = −𝜌�� − 𝜎𝜆 + 𝜏(�� − 𝛼 − ��) + 𝜌(𝛾 + ��) + 𝜅𝜈 − Ψ2 − 2ΛNP (13.60)

Where 𝐷 = 𝑙𝑎∇𝑎 Δ = 𝑛𝑎∇𝑎 𝛿 = 𝑚𝑎∇𝑎 𝛿 = ��𝑎∇𝑎 (9.13) Reduces to 𝐷𝛾 = Ψ2 − ΛNP +Φ11

𝛿𝛼 − 𝛿𝛽 = 𝛼�� + 𝛽�� − 2𝛼𝛽 − Ψ2 + ΛNP +Φ11 0 = −Ψ2 − 2ΛNP These we can solve

Ψ2 − ΛNP +Φ11 = 𝐷𝛾 = 𝑙𝑎∇𝑎 (−

1

√2Λ𝜈) = 𝑙𝜈𝜕𝜈 (−

1

√2Λ𝜈) =

1

√2(−

1

√2Λ) = −

1

2Λ

Ψ2 − ΛNP −Φ11 = 𝛼�� + 𝛽�� − 2𝛼𝛽 − 𝛿𝛼 + 𝛿𝛽 = (−

Λ

4√2(𝑥 − 𝑖𝑦))(−

Λ

4√2(𝑥 + 𝑖𝑦)) + (

Λ

4√2(𝑥 + 𝑖𝑦))(

Λ

4√2(𝑥 − 𝑖𝑦))

−2(−Λ

4√2(𝑥 − 𝑖𝑦))(

Λ

4√2(𝑥 + 𝑖𝑦)) −𝑚𝑎∇𝑎 (−

Λ

4√2(𝑥 − 𝑖𝑦)) + ��𝑎∇𝑎 (

Λ

4√2(𝑥 + 𝑖𝑦))

=Λ2

32(𝑥2 + 𝑦2) +

Λ2

32(𝑥2 + 𝑦2) + 2

Λ2

32(𝑥2 + 𝑦2) − 𝑚𝑥𝜕𝑥 (−

Λ

4√2(𝑥 − 𝑖𝑦))

−𝑚𝑦𝜕𝑦 (−Λ

4√2(𝑥 − 𝑖𝑦)) + ��𝑥𝜕𝑥 (

Λ

4√2(𝑥 + 𝑖𝑦)) + ��𝑦𝜕𝑦 (

Λ

4√2(𝑥 + 𝑖𝑦))

=Λ2

8(𝑥2 + 𝑦2) +

Ω

√2(−

Λ

4√2) + 𝑖

Ω

√2(𝑖

Λ

4√2) + (−

Ω

√2) (

Λ

4√2) + (𝑖

Ω

√2) (𝑖

Λ

4√2)

=Λ2

8(𝑥2 + 𝑦2) −

ΩΛ

8−ΩΛ

8−ΩΛ

8−ΩΛ

8

=Λ2

8(𝑥2 + 𝑦2) −

ΩΛ

2

=Λ2

8(𝑥2 + 𝑦2) − (1 +

Λ

4(𝑥2 + 𝑦2))

Λ

2

= −

1

2Λ


−1

2Λ = Ψ2 − ΛNP +Φ11

−1

2Λ = Ψ2 − ΛNP −Φ11

0 = −Ψ2 − 2ΛNP

⇒ Ψ2 = −1

3Λ (13.65)

ΛNP =

1

6Λ (13.64)

Φ11 = 0 (13.65)





Which is consistent with the former calculation of Φ11 = 0, 𝑎nd we can conclude that if 𝑅𝑎𝑏 = −𝑔𝑎𝑏Λ so

Ω should rightfully be Ω = 1 +Λ

4(𝑥2 + 𝑦2)

12.8 198Collision of a gravitational wave with an electromagnetic wave – The non-

zero spin coefficients The line element in region 𝜈 ≥ 0: 𝑑𝑠2 = 2𝑑𝑢𝑑𝜈 − cos2 𝑎𝜈 (𝑑𝑥2 + 𝑑𝑦2)


11

− cos2 𝑎𝜈− cos2 𝑎𝜈

}


{

11

−1

cos2 𝑎𝜈

−1

cos2 𝑎𝜈}

The Christoffel symbols: To find the Christoffel symbols we calculate the geodesic from the Euler-Lagrange equation

0 =𝑑

𝑑𝑠(𝜕𝐹

𝜕��𝑎) −

𝜕𝐹

𝜕𝑥𝑎 (10.36)

where 𝐹 = 2�� − cos2 𝑎𝜈 (��2 + ��2) 𝑥𝑎 = 𝑢:

𝜕𝐹

𝜕𝑢 = 0

𝜕𝐹

𝜕�� = 2��

𝑑

𝑑𝑠(𝜕𝐹

𝜕��) = 2��

⇒ 0 = 2�� 𝑥𝑎 = 𝜈:

𝜕𝐹

𝜕𝜈 = 2𝑎 cos 𝑎𝜈 sin𝑎𝜈 (��2 + ��2)

𝜕𝐹

𝜕�� = 2��

𝑑

𝑑𝑠(𝜕𝐹

𝜕��) = 2��

⇒ 0 = �� − 𝑎 cos𝑎𝜈 sin𝑎𝜈 (��2 + ��2) 𝑥𝑎 = 𝑥:

𝜕𝐹

𝜕𝑥 = 0

𝜕𝐹

𝜕�� = −2cos2 𝑎𝜈 ��

𝑑

𝑑𝑠(𝜕𝐹


⇒ 0 = 2𝑎 cos 𝑎𝜈 sin𝑎𝜈 �� − cos2 𝑎𝜈 ��

⇔ 0 = �� − 2𝑎 tan𝑎𝜈 ��

198 (McMahon, 2006, p. 322), quiz 13-1. The answer to quiz 13-1 is (a)





𝑥𝑎 = 𝑦:

𝜕𝐹

𝜕𝑦 = 0

𝜕𝐹

𝜕�� = −2cos2 𝑎𝜈 ��

𝑑

𝑑𝑠(𝜕𝐹


⇒ 0 = 2𝑎 cos 𝑎𝜈 sin𝑎𝜈 �� − cos2 𝑎𝜈 �� ⇔ 0 = �� − 2𝑎 tan 𝑎𝜈 ��

Collecting the results 0 = �� 0 = �� − 𝑎 cos𝑎𝜈 sin𝑎𝜈 (��2 + ��2) 0 = �� − 2𝑎 tan𝑎𝜈 �� 0 = �� − 2𝑎 tan 𝑎𝜈 ��


Γ 𝑥𝑥𝑢 = −𝑎 cos 𝑎𝜈 sin 𝑎𝜈 Γ 𝑦𝑦

𝑢 = −𝑎 cos 𝑎𝜈 sin 𝑎𝜈 Γ 𝑥𝜈𝑥 = −𝑎 tan𝑎𝜈 Γ 𝜈𝑥

𝑥 = −𝑎 tan𝑎𝜈 Γ 𝑦𝜈𝑦

= −𝑎 tan𝑎𝜈 Γ 𝜈𝑦𝑦

= −𝑎 tan𝑎𝜈

The basis one forms: 𝑑𝑠2 = 2𝑑𝑢𝑑𝜈 − cos2 𝑎𝜈 (𝑑𝑥2 + 𝑑𝑦2)

= (𝜔��)2− (𝜔��)

2− (𝜔𝑥)

2− (𝜔��)

2

= (𝜔�� +𝜔��)(𝜔�� −𝜔��) − (𝜔𝑥)2− (𝜔��)

2

⇒ √2𝑑𝑢 = (𝜔�� +𝜔��)

√2𝑑𝜈 = (𝜔�� −𝜔��)

𝜔�� = cos 𝑎𝜈 𝑑𝑥 𝜔�� = cos 𝑎𝜈 𝑑𝑦

⇒ 𝜔�� =1

√2(𝑑𝑢 + 𝑑𝜈) 𝑑𝑢 =

1

√2(𝜔�� +𝜔��)

𝜂𝑖𝑗 = {

1−1

−1−1

}

𝜔�� =

1

√2(𝑑𝑢 − 𝑑𝜈) 𝑑𝜈 =

1

√2(𝜔�� −𝜔��)

𝜔�� = cos 𝑎𝜈 𝑑𝑥 𝑑𝑥 =

1

cos 𝑎𝜈𝜔𝑥

𝜔�� = cos 𝑎𝜈 𝑑𝑦 𝑑𝑦 =

1

cos 𝑎𝜈𝜔��

The orthonormal null tetrad: Now we can use the basis one-forms to construct a orthonormal null tetrad





(

𝑙𝑛𝑚��

) =1

√2(

1 1 0 01 −1 0 00 0 1 𝑖0 0 1 −𝑖

)(

𝜔��

𝜔��

𝜔��

𝜔��

) =1

√2(

𝜔�� +𝜔��

𝜔�� −𝜔��

𝜔�� + 𝑖𝜔 ��

𝜔�� − 𝑖𝜔 ��

) =1

√2(

√2𝑑𝑢

√2𝑑𝜈cos 𝑎𝜈 𝑑𝑥 + 𝑖 cos 𝑎𝜈 𝑑𝑦cos 𝑎𝜈 𝑑𝑥 − 𝑖 cos 𝑎𝜈 𝑑𝑦)

(9.10)


𝑙𝑎 = (1, 0, 0, 0) 𝑛𝑎 = (0, 1, 0, 0)

𝑚𝑎 =1

√2(0, 0, cos𝑎𝜈 , 𝑖 cos 𝑎𝜈) 𝑚𝑎 =

1

√2(0, 0, cos 𝑎𝜈 , −𝑖 cos 𝑎𝜈)


𝑙𝑢 = 𝑔𝑎𝑢𝑙𝑎 = 𝑔𝜈𝑢𝑙𝜈 = 1 ⋅ 0 = 0

𝑙𝜈 = 𝑔𝑎𝜈𝑙𝑎 = 𝑔𝑢𝜈𝑙𝑢 = 1 ⋅ 1 = 1

𝑙𝑥 = 𝑙𝑦 = 0 𝑛𝑢 = 𝑔𝑎𝑢𝑛𝑎 = 𝑔

𝜈𝑢𝑛𝜈 = 1 ⋅ 1 = 1

𝑛𝜈 = 𝑔𝑎𝜈𝑛𝑎 = 𝑔𝑢𝜈𝑛𝑢 = 1 ⋅ 0 = 0

𝑛𝑥 = 𝑛𝑦 = 0 𝑚𝜈 = 𝑚𝑢 = 0


cos2 𝑎𝜈⋅1

√2⋅ cos𝑎𝜈 = −

1

√2

1

cos 𝑎𝜈


cos2 𝑎𝜈

1

√2⋅ 𝑖 cos𝑎𝜈 = −𝑖

1

√2

1

cos 𝑎𝜈

Collecting the results: 𝑙𝑎 = (1, 0, 0, 0) 𝑙𝑎 = (0, 1, 0, 0) 𝑛𝑎 = (0, 1, 0, 0) 𝑛𝑎 = (1, 0, 0, 0)

𝑚𝑎 =1

√2(0, 0, cos𝑎𝜈 , 𝑖 cos 𝑎𝜈) 𝑚𝑎 =

1

√2(0, 0, −

1

cos𝑎𝜈, −𝑖

1

cos 𝑎𝜈)

𝑚𝑎 =1

√2(0, 0, cos𝑎𝜈 , −𝑖 cos 𝑎𝜈) ��𝑎 =

1

√2(0, 0, −

1

cos𝑎𝜈, 𝑖

1

cos 𝑎𝜈)

The spin coefficients calculated from the orthonormal tetrad:


𝑎𝑙𝑏 휀 =1



(9.15) 𝜈 = −∇𝑏𝑛𝑎��


1




𝑎��𝑏 𝛼 =1


𝑎��𝑏 − ∇𝑏𝑚𝑎��𝑎��𝑏)






𝜋 = −∇𝑏𝑛𝑎��𝑎𝑙𝑏 = −∇𝜈𝑛𝑎��

𝑎𝑙𝜈 = −∇𝜈𝑛𝑥��𝑥𝑙𝜈 − ∇ν𝑛𝑦��

𝑦𝑙𝜈 = 0



𝑦𝑛𝑢 = 0


= −∇𝑥𝑛𝑎��𝑎��𝑥 − ∇𝑦𝑛𝑎��

𝑎��𝑦

= −∇𝑥𝑛𝑥��𝑥��𝑥 − ∇𝑦𝑛𝑥��

𝑥��𝑦 − ∇𝑥𝑛𝑦��𝑦��𝑥 − ∇𝑦𝑛𝑦��

𝑦��𝑦

= 0 𝜇 = −∇𝑏𝑛𝑎��

𝑎𝑚𝑏 = 0





𝜅 = ∇𝑏𝑙𝑎𝑚𝑎𝑙𝑏 = ∇ν𝑙𝑎𝑚

𝑎𝑙ν = ∇ν𝑙𝑥𝑚𝑥𝑙ν + ∇ν𝑙𝑦𝑚

𝑦𝑙ν = 0



𝑦𝑛𝑢 = 0



𝑎��𝑦



𝑦��𝑦






𝑦��𝑦



𝑦��𝑦)


√2 cos 𝑎𝜈)2

− 𝑎 cos𝑎𝜈 sin𝑎𝜈 (−𝑖1

√2 cos 𝑎𝜈) (𝑖

1

√2 cos𝑎𝜈))

= 𝑎 tan𝑎𝜈 𝜎 = ∇𝑏𝑙𝑎𝑚

𝑎𝑚𝑏 = −(Γ 𝑥𝑥

𝑢 𝑙𝑢𝑚𝑥𝑚𝑥 + Γ 𝑦𝑦

𝑢 𝑙𝑢𝑚𝑦𝑚𝑦)


√2 cos 𝑎𝜈)2

− 𝑎 cos𝑎𝜈 sin𝑎𝜈 (−𝑖1

√2 cos 𝑎𝜈)2

)

= 0

휀 =1



=1

2(∇ν𝑙𝑎𝑛


=1

2(∇ν𝑙𝑢𝑛


𝑦𝑙ν)

= −

1


𝑐 𝑚𝑐)��𝑥𝑙ν + (𝜕𝜈𝑚𝑦 − Γ 𝜈𝑦

𝑐 𝑚𝑐)��𝑦𝑙ν)

= −

1

2((𝜕𝜈

1

√2cos 𝑎𝜈 − Γ 𝜈𝑥

𝑥 𝑚𝑥) ��𝑥𝑙ν + (𝜕𝜈𝑖

1

√2cosh𝑎𝜈 − Γ 𝜈𝑦


𝑦𝑙ν)

= −

1

2((−𝑎

1

√2sin𝑎𝜈 − (−𝑎 tan𝑎𝜈)

1

√2cos 𝑎𝜈) ��𝑥 + (−𝑖𝑎

1

√2sin 𝑎𝜈 + 𝑎 tan𝑎𝜈 𝑖

1

√2cos 𝑎𝜈) ��𝑦)

= 0

𝛾 =1



=1



=1



𝑦𝑛𝑢)

= 0

𝛼 =1


𝑎��𝑏 − ∇𝑏𝑚𝑎��𝑎��𝑏)

=1


𝑎��𝑥 − ∇𝑥𝑚𝑎��𝑎��𝑥) +

1


𝑎��𝑦 − ∇𝑦𝑚𝑎��𝑎��𝑦)

= 0

𝛽 =1








𝜈 = 0 𝜏 = 0 𝛾 = 0 𝜆 = 0 𝜌 = 𝑎 tan𝑎𝜈 𝛼 = 0 𝜇 = 0 𝜎 = 0 𝛽 = 0

This means that –𝑅𝑒(𝜌) ≠ 0 and there is expansion (or pure focusing=divergence).

12.9 199The Aichelburg-Sexl Solution – The passing of a black hole The line element 𝑑𝑠2 = 4𝜇 log(𝑥2 + 𝑦2) 𝑑𝑢2 + 2𝑑𝑢𝑑𝑟 − dx2 − 𝑑𝑦2 Comparing with the Brinkmann metric 𝑑𝑠2 = 𝐻(𝑢, 𝑥, 𝑦)𝑑𝑢2 + 2𝑑𝑢𝑑𝑣 − 𝑑𝑥2 − 𝑑𝑦2 We see that we can copy the results from the Brinkmann calculations p.195 if 𝐻(𝑢, 𝑥, 𝑦) =4𝜇 log(𝑥2 + 𝑦2)

The only non-zero spin-coefficient is:

𝜈 = −1

2√2(𝜕𝐻

𝜕𝑥+ 𝑖

𝜕𝐻

𝜕𝑦) (9.30)

= −

1

2√2(𝜕(4𝜇 log(𝑥2 + 𝑦2))

𝜕𝑥+ 𝑖

𝜕(4𝜇 log(𝑥2 + 𝑦2))

𝜕𝑦)

= −

4𝜇

2√2(

2𝑥

𝑥2 + 𝑦2+ 𝑖

2𝑦

𝑥2 + 𝑦2)

= −2√2𝜇 (

𝑥

𝑥2 + 𝑦2+ 𝑖

𝑦

𝑥2 + 𝑦2)

12.10 Observations: The Future Gravitational Wave detectors200. GWs are faint deformations of the space-time geometry, propagating at the speed of light and generated by cata-strophic events in the Universe, in which strong gravita-tional fields and sudden acceleration of asymmetric distri-bution of large masses are involved. GWs have a quadrupo-lar nature and have two polarizations, ℎ+ and ℎ𝑥, where ℎ

is the so called space-time strain ℎ =|𝐿−𝐿0|

𝐿0=

𝛿𝐿

𝐿0, the rela-

tive dimensional8distortion of an extended mass distribu-tion. The effect of these two polarizations on a circular mass distribution is shown in the figure. GWs are created by ac-celerating masses, but because gravity is the weakest of the four fundamental forces, GWs are extremely small. For this reason, only extremely massive and compact objects having intense and asymmetric gravitational fields, like neutron star and black hole binary systems, are expected to be able to generate detectable GW emission. The direct detection of GWs is still missing and it is quite easy to understand why. For example, the expected amplitude on Earth of the GW emitted by a coalescing binary system of neutron star located in the Virgo cluster is of the order of ℎ~10−22. This means that a detector having a dimension of a meter experiences an oscillating deformation of 10−22𝑚, an astonishingly small quantity. In the 1960s, the first GW detectors were based on a (multi)-ton resonant bar, that should resonate when excited by the passage of a GW. These detectors evolved, operating at cryogenic temperature to minimize the disturbance of the thermal Brownian vibration and being read by

199 (McMahon, 2006, p. 322), quiz 13-2. The answer to quiz 13-2 is (b) 200 This is an extract of the article: Opening a New Window on the Universe – The Future Gravitational Wave detectors http://www.europhysicsnews.org/articles/epn/abs/2013/02/epn2013442p16/epn2013442p16.html (Michele Punturo - 2013).


http://www.europhysicsnews.org/articles/epn/abs/2013/02/epn2013442p16/epn2013442p16.html




very low noise transducers. These detectors reached a sensitivity of the order of a few × 10−21 around the main resonant mode frequency, which is of the order of one kHz. Although two of these detectors are still operating, it is worth stating that their era has ended due to the realization of a new kind of GW de-tector: giant interferometers, operating since the first years of the 2000 decade. These instruments profit from two key elements of the GW; (i) the tidal nature of a GW: the expected metric deformation 𝛿𝐿 of a body traversed by a GW is proportional to its size 𝛿𝐿~ℎ × 𝐿0. Hence, if the expected space-time defor-mation ℎ is of the order of 10−22, the effect on a multi-km detector will be a deformation 𝛿𝐿~10−19 −10−18𝑚. (ii) the quadrupolar nature of the GW. A Michelson interferometer is sensitive to the difference in optical path length between its two arms, and it can match the metric deformation imposed by the GW. The first operative GW interferometric detector has been the Japanese TAMA, a 300m Michelson interfer-ometer that opened the path to this new family of instruments, but had a sensitivity limited by its reduced length and by its location, in the center of Tokyo, affected by too high environmental disturbance. In Eu-rope two interferometric GW detectors have been realized; GEO600, a 600 m Michelson interferometer, built close to Hannover and Virgo, a Michelson interferometer having Fabry-Perot resonating cavities in-serted in the 3 km long arms, built close to Pisa. The longest interferometric GW detectors in the World are the two ‘Laser Interferometer Gravitational wave Observatory’ (LIGO) detectors, having 4 km long arms, realized in Louisiana and Washington State (USA) with a topology similar to Virgo. Thanks to the long Fabry-Perot cavities in the arms of the Virgo and LIGO detectors, the photons are forced to bounce back-and forth between the suspended mirrors, thus squeezing a hundreds km long optical path in the few km long detector infrastructure, increasing the sensitivity to the space-time deformation. The length limita-tion, dictated by technical constrains and affecting the terrestrial GW detector infrastructures, will obvi-ously disappear in the space-based observatories, like the eLISA/NGO detector. This project of a multi-million km GW interferometer is to be launched at the end of the 2020 decade, in a heliocentric orbit, and

is devoted to the observation of ultra-low-frequency sources (10−5 − 10−3𝐻𝑧), like hypermassive black holes. Virgo, GEO600 and LIGO detectors operated in a network during the second half of the 2000 decade, ‘listening’to the sky in the 10-10000 Hz frequency range. Even though, no detection of GW signal has been obtained so far but, relevant scientific targets have been reached, putting constraints on potential GW emission by some astrophysical source. For example, thanks to the joint LIGO and Virgo data, an upper limit has been set to the possible GW emission of the Vela and Crab pulsars. These pulsars, remnants of supernovae explosions, are compact neutron stars rotating about 11 and 30 times per second, respec-tively. The pulsars are expected to emit GWs at a frequency double their rotation rate, and at an amplitude depending on many (unknown) parameters characterizing these stars. Through the radio signal, it is well known that these pulsars are slowing down because of emission of energy, due to several possible mech-anisms. LIGO and Virgo have been able to set an upper limit to the fraction of that energy emission due to GWs, stating that no more than few per cent of the energy loss can be due to GW radiation. The Virgo and LIGO detectors are currently offline, being upgraded toward the 2nd generation. In the period 2011-2015 several parts of the detectors will be replaced to improve the sensitivity by a factor of ten. An improvement by a factor of ten in sensitivity corresponds to an increase by a factor of a thousand in detection rate: in one year of operation of the advanced detectors at the nominal sensitivity, about 40 coalescences of neu-tron star binary systems are expected to be detected. The advanced detectors’ capability to detect a coa-lescence of a binary neutron star system at a distance of about 140 Mpc, and a coalescence of a binary systems of black holes at a distance of about 1 Gpc, will open up a gravitational-wave astrophysics era. It will be possible, for example, to compare the signal detected from the coalescence of a binary system of neutron stars with the general relativity prediction. Or it will be possible to investigate the nature of an isolated neutron star by looking at its GW emission. Few years later the completion of the Advanced Virgo detector in Europe and of the Advanced LIGO detectors in USA, new nodes will enter the network of GW observatories: a very innovative 3km interferometer (KAGRA), underground and cryogenic, is under con-struction in Japan. Furthermore, a 3rd Advanced LIGO site is under evaluation in India. European scientists are attempting to drive the evolution of this research field and the conceptual design of a 3rd generation





GW observatory has been realized, able to compete and collaborate with the most sensitive optical tele-scopes: the Einstein GW Telescope (ET). This new infrastructure, aimed to be operative in the 2020 decade, will test the cosmological model of the universe using GW signals, thanks to its capability to see many sources at large red-shift; ET will be a wonderful proofing tool of the general relativity predictions in all radiative processes involving intense gravitational fields, like in the presence of intermediate-mass black holes (𝑀~10 − 1000𝑀𝑆𝑢𝑛). It will allow detailed investigations of the nature of isolated neutron stars looking both to the continuous emission of the pulsars and to the explosion of supernovae.

Bibliografi A.S.Eddington. (1924). The Mathematical Theory of Relativity. Cambridge: At the University Press.

C.W.Misner, K. a. (1973). Gravitation. New York: W.H.Freeman and Company.

Carroll, S. M. (2004). An Introduction to General Relativity, Spacetime and Geometry. San Fransisco, CA:

Addison Wesley.

d'Inverno, R. (1992). Introducing Einstein's Relativity. Oxford: Clarendon Press.

Greene, B. (2004). The Fabric of the Cosmos. Penguin books.

Hartle, J. B. (2003). Gravity - An introduction to Einstein's General Relativity. Addison Wesley.

McMahon, D. (2006). Relativity Demystified. McGraw-Hill Companies, Inc.

Spiegel, M. R. (1990). SCHAUM'S OUTLINE SERIES: Mathematical Handbook of FORMULAS and TABLES.

McGraw-Hill Publishing Company.

Weinberg, S. (1979). De første tre minutter. Gyldendal.


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