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Lots of Calculations in General Relativity Susan Larsen Tuesday, August 20, 2019 1 http://physicssusan.mono.net [email protected] Content 1 Introduction ............................................................................................................................................... 2 1.1 Space-times ....................................................................................................................................... 4 2 The Metric tensor and Vector Transformations. ....................................................................................... 6 3 Four vectors and four velocity ................................................................................................................... 6 4 Christoffel Symbols, Geodesic Equation and Killing Vectors ..................................................................... 6 5 Covariant Derivative, Lie Derivative and Killings Equation ........................................................................ 7 6 The Riemann tensor .................................................................................................................................. 7 7 Cartan’s Structure Equations – a Shortcut to the Einstein equation ........................................................ 7 8 The Einstein Field Equations ...................................................................................................................... 7 8.1 The vacuum Einstein equations......................................................................................................... 7 8.2 The vacuum Einstein equations with a cosmological constant ......................................................... 7 8.3 General remarks on the Einstein equations with a cosmological constant ...................................... 8 8.4 Using the contracted Bianchi identities, prove that: = .................................................... 9 8.5 2+1 dimensions: Gravitational collapse of an inhomogeneous spherically symmetric dust cloud. 10 8.5.1 Find the components of the curvature tensor for the metric in 2+1 dimensions using Cartan’s structure equations ................................................................................................................................. 10 8.5.2 Find the components of the curvature tensor for the metric in 2+1 dimensions using Cartan’s structure equations – alternative solution .............................................................................................. 11 8.5.3 Find the components of the Einstein tensor in the coordinate basis for the metric in 2+1 dimensions............................................................................................................................................... 12 8.5.4 The Einstein equations of the metric in 2+1 dimensions. ....................................................... 15 8.6 Ricci rotation coefficients, Ricci scalar and Einstein equations for a general 4-dimensional metric: = − + , + , + , ........................................................................... 15 8.7 The de Sitter Spacetime................................................................................................................... 20 8.7.1 The Einstein equations ............................................................................................................ 20 8.7.2 Solving the Einstein equations................................................................................................. 21 8.8 The Anti-de Sitter Spacetime ........................................................................................................... 21 9 The Energy-Momentum Tensor .............................................................................................................. 22 9.1 The Einstein equation with source .................................................................................................. 22 9.2 Perfect Fluids ................................................................................................................................... 23 9.3 More examples on stress-energy tensors ....................................................................................... 24 9.3.1 Pure Matter ............................................................................................................................. 24 9.3.2 More complicated fluids .......................................................................................................... 24 9.3.3 The electromagnetic field ........................................................................................................ 24 9.4 The Gödel metric ............................................................................................................................. 27
Transcript
Page 1: Lots of Calculations in General Relativityphysicssusan.mono.net/upl/9111/LotsofCalculationsinGeneralRelativity8... · General relativity is the description of the smallest changes

Lots of Calculations in General Relativity

Susan Larsen Tuesday, August 20, 2019

1 http://physicssusan.mono.net [email protected]

Content 1 Introduction ............................................................................................................................................... 2

1.1 Space-times ....................................................................................................................................... 4

2 The Metric tensor and Vector Transformations. ....................................................................................... 6

3 Four vectors and four velocity ................................................................................................................... 6

4 Christoffel Symbols, Geodesic Equation and Killing Vectors ..................................................................... 6

5 Covariant Derivative, Lie Derivative and Killings Equation ........................................................................ 7

6 The Riemann tensor .................................................................................................................................. 7

7 Cartan’s Structure Equations – a Shortcut to the Einstein equation ........................................................ 7

8 The Einstein Field Equations ...................................................................................................................... 7

8.1 The vacuum Einstein equations ......................................................................................................... 7

8.2 The vacuum Einstein equations with a cosmological constant ......................................................... 7

8.3 General remarks on the Einstein equations with a cosmological constant ...................................... 8

8.4 Using the contracted Bianchi identities, prove that: 𝛁𝒃𝑮𝒂𝒃 = 𝟎 .................................................... 9

8.5 2+1 dimensions: Gravitational collapse of an inhomogeneous spherically symmetric dust cloud. 10

8.5.1 Find the components of the curvature tensor for the metric in 2+1 dimensions using Cartan’s structure equations ................................................................................................................................. 10

8.5.2 Find the components of the curvature tensor for the metric in 2+1 dimensions using Cartan’s structure equations – alternative solution .............................................................................................. 11

8.5.3 Find the components of the Einstein tensor in the coordinate basis for the metric in 2+1 dimensions............................................................................................................................................... 12

8.5.4 The Einstein equations of the metric in 2+1 dimensions. ....................................................... 15

8.6 Ricci rotation coefficients, Ricci scalar and Einstein equations for a general 4-dimensional metric: 𝒅𝒔𝟐 = −𝒅𝒕𝟐 + 𝑳𝟐𝒕, 𝒓𝒅𝒓𝟐 + 𝑩𝟐𝒕, 𝒓𝒅𝝓𝟐 +𝑴𝟐𝒕, 𝒓𝒅𝒛𝟐 ........................................................................... 15

8.7 The de Sitter Spacetime................................................................................................................... 20

8.7.1 The Einstein equations ............................................................................................................ 20

8.7.2 Solving the Einstein equations................................................................................................. 21

8.8 The Anti-de Sitter Spacetime ........................................................................................................... 21

9 The Energy-Momentum Tensor .............................................................................................................. 22

9.1 The Einstein equation with source .................................................................................................. 22

9.2 Perfect Fluids ................................................................................................................................... 23

9.3 More examples on stress-energy tensors ....................................................................................... 24

9.3.1 Pure Matter ............................................................................................................................. 24

9.3.2 More complicated fluids .......................................................................................................... 24

9.3.3 The electromagnetic field ........................................................................................................ 24

9.4 The Gödel metric ............................................................................................................................. 27

Page 2: Lots of Calculations in General Relativityphysicssusan.mono.net/upl/9111/LotsofCalculationsinGeneralRelativity8... · General relativity is the description of the smallest changes

Lots of Calculations in General Relativity

Susan Larsen Tuesday, August 20, 2019

2 http://physicssusan.mono.net [email protected]

9.5 The Einstein Cylinder ....................................................................................................................... 30

9.5.1 The line element ...................................................................................................................... 30

9.5.2 The Ricci tensor ....................................................................................................................... 30

9.5.3 The Einstein Equations ............................................................................................................ 32

9.5.4 The Einstein tensor with a cosmological constant .................................................................. 33

9.6 The Newtonian Approximation – The right hand side! ................................................................... 33

10 Null Tetrads and the Petrov Classification........................................................................................... 35

10.1 Weyl scalars and Petrov classification ............................................................................................. 35

10.2 Construct a null tetrad for the flat space-time Minkowski metric .................................................. 36

11 The Schwarzschild Solution ................................................................................................................. 37

12 Black Holes ........................................................................................................................................... 37

13 Cosmology ........................................................................................................................................... 37

14 Gravitational Waves ............................................................................................................................ 37

15 Appendix A: Tensor Calculus ............................................................................................................... 38

15.1 The Einstein Summation Convention .............................................................................................. 38

15.1.1 Repeated Indices. .................................................................................................................... 38

15.1.2 Substitutions ............................................................................................................................ 38

15.1.3 Double Sums ............................................................................................................................ 38

15.1.4 Kronecker Delta ....................................................................................................................... 38

16 Appendix B: Collection of results. ........................................................................................................ 40

Bibliografi ......................................................................................................................................................... 56

General relativity is the description of the smallest changes and the largest entities. The best way to describe this I found in a quote by Roger Penrose: Calculus is built from two basic ingredients: differentiation and integration. Differentiation is concerned with

velocities, accelerations, the slopes and curvature of curves and surfaces… These are the rates at which things change, and they are quantities defined locally, in terms of structure or behavior in the tiniest neigh-borhood of single points. Integration on the other hand, is concerned with areas and volumes, with centers of gravity…These are things which involves measures of totality … and they are not defined merely by what

is going on in the local or infinitesimal neighborhoods of individual points. (Penrose, 2004, s. 103)

1 Introduction We begin with a few important facts:

- The gravitational force is always attractive. - The gravitational force is a long range force without boundaries. - A gravitational field is created from all kinds of masses and (because 𝐸 = 𝑚𝑐2) all kinds of ener-

gies. - A mass/energy creates a curvature of the four-dimensional space-time, where masses (test

masses) moves along straight lines (geodesics).

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Lots of Calculations in General Relativity

Susan Larsen Tuesday, August 20, 2019

3 http://physicssusan.mono.net [email protected]

Calculus: 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 shifta:

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 geodesicb. 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∗c. Example: Killing vectors in the Schwarzschild metricd. The Killing vector that corresponds to the independence of the metric of 𝑡 is 𝜉 = (1,0,0,0) and of 𝜙 is 휂 =

(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 = 휂 ⋅ 𝑢 = 𝑔𝑎𝑏휂

𝑎𝑢𝑏 = 𝑔𝜙𝜙 ⋅ 1 ⋅𝑑𝜙

𝑑𝜏=

−𝑟2 sin2 휃𝑑𝜙

𝑑𝜏= −𝑟2

𝑑𝜙

𝑑𝜏 for 휃 =

𝜋

2

3. The Geodesic equation leads to equations of motion:

𝐾 =1

2𝑔𝑎𝑏�̇�

𝑎�̇�𝑏

𝜕𝐾

𝜕𝑥𝑎 =

𝑑

𝑑𝑠(𝜕𝐾

𝜕�̇�𝑎)

𝑑2𝑥𝑎

𝑑𝑠2+ Γ 𝑏𝑐

𝑎𝑑𝑥𝑏

𝑑𝑠

𝑑𝑥𝑐

𝑑𝑠 = 0

Example: Planetary orbitse 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 If 𝑛 = 4, 𝑅𝑎𝑏𝑐𝑑 has twenty independent component – ten of which are given by 𝑅𝑎𝑏 and the remaining ten by the Weyl tensorf.

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Lots of Calculations in General Relativity

Susan Larsen Tuesday, August 20, 2019

4 http://physicssusan.mono.net [email protected]

Example: The Friedmann equations A homogenous, isotropic and expanding universe described by the Robertson-Walker space-timeg, in this case the Einstein equations becomes the Friedmann equations:

8𝜋𝜌 =3

𝑎2(𝑘 + �̇�2) − Λ

−8𝜋𝑃 = 2

�̈�

𝑎+1

𝑎2(𝑘 + �̇�2) − Λ

1.1 Space-times This document includes many different space-time examples. In order to keep track of them I have made this list- in alphabetical order, so that you can see in which chapter you can find the space-time you are looking for.

Space-time Line-element Chap-

ter

Aichelburg-Sexl Solution 𝑑𝑠2 = 4𝜇 log(𝑥2 + 𝑦2) 𝑑𝑢2 + 2𝑑𝑢𝑑𝑟 − dx2 − 𝑑𝑦2 14

Anti de Sitter space-time 𝑑𝑠2 = −𝑑𝑡2 + cos2(𝑡) 𝑑𝑟2 + cos2(𝑡) sinh2(𝑟) 𝑑휃2

+ cos2(𝑡) sinh2(𝑟) sin2 휃 𝑑𝜙2 8

Brinkman space-time 𝑑𝑠2 = 𝐻(𝑢, 𝑥, 𝑦)𝑑𝑢2 + 2𝑑𝑢𝑑𝑣 − 𝑑𝑥2 − 𝑑𝑦2 14

Classically anti-de-Sitter space-time 𝑑𝑠2 = −cosh2(𝑟) 𝑑𝑡2 + 𝑑𝑟2 + sinh2(𝑟) 𝑑휃2

+ sinh2(𝑟) sin2 휃 𝑑𝜙2 4

Colliding gravitational waves 𝑑𝑠2 = 𝛿(𝑢)(𝑋2 − 𝑌2)𝑑𝑢2 + 2𝑑𝑢𝑑𝑟 − 𝑑𝑋2 − 𝑑𝑌2 14

Collision of a gravitational wave with a electromagnetic wave

𝑑𝑠2 = 2𝑑𝑢𝑑𝜈 − cos2 𝑎𝜈 (𝑑𝑥2 + 𝑑𝑦2) 14

Cylindrical coordinates 𝑑𝑠2 = 𝑑𝑟2 + 𝑟2𝑑𝜙2 + 𝑑𝑧2 4

De Sitter space-time 𝑑𝑠2 = −𝑑𝑡2 + 𝑎(𝑡)2(𝑑휃2

+ sin2 휃 (𝑑𝜙2 + sin2𝜙𝑑𝜓2)) 8

Eddington-Finkelstein coordinates 𝑑𝑠2 = −(1 −

2𝑚

𝑟)𝑑𝑣2 + 2𝑑𝑣𝑑𝑟

+ 𝑟2(𝑑휃2 + sin2 휃 𝑑𝜙2) 12

Egg geometry 𝑑𝑠2 = 𝑎2[(cos2 휃 + 4 sin2 휃)𝑑휃2 + sin2 휃 𝑑𝜙2] 2

Einstein cylinder 𝑑𝑠2 = −𝑑𝑡2 + (𝑎0)

2(𝑑휃2

+ sin2 휃 (𝑑𝜙2 + sin2𝜙𝑑𝜓2))

9 (4), 13

Ellipsoid 𝑑𝑠2 = 𝑎𝑑𝑥2 + 𝑏𝑑𝑦2 + 𝑐𝑑𝑧2 2

Example – three dimensional space 𝑑𝑠2 = (𝑢2 + 𝜈2)𝑑𝑢2 + (𝑢2 + 𝜈2)𝑑𝜈2 + 𝑢2𝜈2𝑑휃2 6

Example – three dimensional space 𝑑𝑠2 = 𝑑𝑥2 + 2𝑥𝑑𝑦2 + 2𝑦𝑑𝑧2 6

Example – two dimensional space 𝑑𝑠2 = 𝑦2 sin 𝑥 𝑑𝑥2 + 𝑥2 tan 𝑦 𝑑𝑦2 6

Example: Four-dimensional space-time

𝑑𝑠2 = −(1 − 𝐴𝑟2)2𝑑𝑡2 + (1 − 𝐴𝑟2)2𝑑𝑟2 + 𝑟2𝑑휃2

+ 𝑟2 sin2 휃 𝑑𝜙2 2

Example: Four-dimensional space-time

𝑑𝑠2 = −𝑑𝑡2 + 𝐿2(𝑡, 𝑟)𝑑𝑟2 + 𝐵2(𝑡, 𝑟)𝑑𝜙2

+𝑀2(𝑡, 𝑟)𝑑𝑧2 8

A Fifth dimension 𝑑𝑠2 = −𝑑𝑥2 + 𝑑𝑥2 + 𝑑𝑦2 + 𝑑𝑧2 + 𝑅2𝑑Ω2 2

Flat Minkowsky space-time 𝑑𝑠2 = −𝑑𝑡2 + 𝑑𝑥2 + 𝑑𝑦2 + 𝑑𝑧2 2

Flat Minkowsky space-time in polar coordinates

𝑑𝑠2 = −𝑑𝑡2 + 𝑑𝑟2 + 𝑟2𝑑휃2 + 𝑟2 sin2 휃 𝑑𝜙 2, 10

Flat space-time in Eddington Finkel-stein coordinates

𝑑𝑠2 = −𝑑𝑣2 + 2𝑑𝑣𝑑𝑟 + 𝑟2𝑑휃2 + 𝑟2 sin2 휃 𝑑𝜙 2, 12

General four-dimensional diagonal metric

𝑑𝑠2 = 𝑔00𝑑𝑥

0𝑑𝑥0 + 𝑔11𝑑𝑥1𝑑𝑥1 + 𝑔22𝑑𝑥

2𝑑𝑥2

+ 𝑔33𝑑𝑥3𝑑𝑥3

2

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Lots of Calculations in General Relativity

Susan Larsen Tuesday, August 20, 2019

5 http://physicssusan.mono.net [email protected]

Gödel metric 𝑑𝑠2 =1

2𝜔2((𝑑𝑡 + 𝑒𝑥𝑑𝑧)2 − 𝑑𝑥2 − 𝑑𝑦2 −

1

2𝑒2𝑥𝑑𝑧2) 2,9

Gravitationally collapse of an inho-mogeneous spherically symmetric dust cloud

𝑑𝑠2 = −𝑑𝑡2 + 𝑒2𝑏(𝑡,𝑟)𝑑𝑟2 + 𝑅2(𝑡, 𝑟)𝑑𝜙2 8

Homogenous closed universe 𝑑𝑆2 =1

1 − (𝑟𝑎)2 𝑑𝑟

2 + 𝑟2𝑑휃2 + 𝑟2 sin2 휃 𝑑𝜙2 2

Hyperbolic plane 𝑑𝑆2 = 𝑦−2(𝑑𝑥2 + 𝑑𝑦2) 4

Impulsive gravitational wave 𝑑𝑠2 = 2𝑑𝑢𝑑𝜈 − [1 − 𝜈Θ(𝜈)]2𝑑𝑥2 − [1 + 𝜈Θ(𝜈)]2𝑑𝑦2 14

Kerr Spinning black hole 𝑑𝑠2

= (1 −2𝑚𝑟

Σ)𝑑𝑡2 +

4𝑎𝑚𝑟 sin2 휃

Σ𝑑𝑡𝑑𝜙 −

Σ

Δ𝑑𝑟2

− Σ𝑑휃2

− (𝑟2 + 𝑎2

+2𝑎2𝑚𝑟 sin2 휃

Σ) sin2 휃 𝑑𝜙2

2, 12

Kruskal coordinates 𝑑𝑠2 =32𝑚3

𝑟𝑒−

𝑟2𝑚(𝑑𝑣2 − 𝑑𝑢2) − 𝑟2(𝑑휃 + sin2 휃 𝑑𝜙) 12

Linearized metric 𝑑𝑠2 = (휂𝑎𝑏 + 𝜖ℎ𝑎𝑏)𝑑𝑥𝑎𝑑𝑥𝑏 9, 14

Lorentz hyperboloid 𝑑𝑠2 = 𝑑𝜓2 + sinh2𝜓 𝑑휃2 + sinh2𝜓 sin2 휃 𝑑𝜙2 7, 13

Nariai space-time 𝑑𝑠2 = −Λ𝜈2𝑑𝑢2 + 2𝑑𝑢𝑑𝑣 −1

Ω2(𝑑𝑥2 + 𝑑𝑦2) 14

North pole 𝑑𝑠2 = (1 −𝑦2

3𝑎2)𝑑𝑥2 +

𝑥𝑦

3𝑎2𝑑𝑥𝑑𝑦 + (1 −

𝑥2

3𝑎2)𝑑𝑦2 2

Peanut geometry 𝑑𝑠2 = 𝑎2𝑑휃2 + 𝑎2𝑓2(휃)𝑑𝜙2𝑓(휃) 2

Penrose-Kahn metric 𝑑𝑠2 = 2𝑑𝑢𝑑𝑣 − (1 − 𝑢)2𝑑𝑥2 − (1 + 𝑢)2𝑑𝑦2 14

Plane waves: ℎ𝑎𝑏 = ℎ𝑎𝑏(𝑡 − 𝑧) 𝑑𝑠2 = (휂𝑎𝑏 + 𝜖ℎ𝑎𝑏)𝑑𝑥𝑎𝑑𝑥𝑏 14

Reissner-Nordström spacetime 𝑑𝑠2

= (1 −2𝑚

𝑟+𝑒2

𝑟2)𝑑𝑡2

− (1 −2𝑚

𝑟+𝑒2

𝑟2)

−1

𝑑𝑟2

− 𝑟2𝑑휃2 − 𝑟2 sin2 휃 𝑑𝜙2

11

Rindler metric 𝑑𝑠2 = 𝜉2𝑑𝜏2 − 𝑑𝜉2 7

Robertson Walker space-time 𝑑𝑠2 = −𝑑𝑡2 +𝑎2(𝑡)

1 − 𝑘𝑟2𝑑𝑟2 + 𝑎2(𝑡)𝑟2𝑑휃2

+ 𝑎2(𝑡)𝑟2 sin2 휃 𝑑𝜙2 13

Rosen line element 𝑑𝑠2 = 𝑑𝑈𝑑𝑉 − 𝑎2(𝑈)𝑑𝑥2 − 𝑏2(𝑈)𝑑𝑦2 14

Schwarzschild metric 𝑑𝑠2 = −(1 −2𝑚

𝑟)𝑑𝑡2 + (1 −

2𝑚

𝑟)−1

𝑑𝑟2 + 𝑟2𝑑휃2

+ 𝑟2 sin2 휃 𝑑𝜙2

11, 12

Schwarzschild metric: general solu-tion

𝑑𝑠2 = 𝑒2𝜈(𝑟)𝑑𝑡2 − 𝑒2𝜆(𝑟)𝑑𝑟2 − 𝑟2𝑑휃2

− 𝑟2 sin2 휃 𝑑𝜙2 11

Schwarzschild metric: general time dependent

𝑑𝑠2 = 𝑒2𝜈(𝑡,𝑟)𝑑𝑡2 − 𝑒2𝜆(𝑡,𝑟)𝑑𝑟2 − 𝑟2𝑑휃2

− 𝑟2 sin2 휃 𝑑𝜙2 11

Schwarzschild metric: The effect of the cosmological constant over the scale of the solar system

𝑑𝑠2 = (1 +1

3Λ𝑟2)𝑑𝑡2 −

𝑑𝑟2

1 +13Λ𝑟

2 11

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Schwarzschild metric: The general Schwarzschild metric with nonzero cosmological constant.

𝑑𝑠2 = 𝑓(𝑟)𝑑𝑡2 −

1

𝑓(𝑟)𝑑𝑟2 − 𝑟2𝑑휃2 − 𝑟2 sin2 휃 𝑑𝜙2

11

Schwarzschild metric: With two dif-ferent masses

𝑑𝑠2 = (1 −2𝑚1

𝑟) 𝑑𝑡2 − (1 −

2𝑚2

𝑟)−1

𝑑𝑟2 − 𝑟2𝑑휃2

− 𝑟2 sin2 휃 𝑑𝜙2

11

Schwarzschild metric: 휃 =𝜋

2 𝑑𝑠2 = −(1 −

2𝑚

𝑟)𝑑𝑡2 + (1 −

2𝑚

𝑟)−1

𝑑𝑟2 + 𝑟2𝑑𝜙2 11

Spatial part of a homogenous, iso-tropic metric

𝑑𝜎2 =𝑑𝑟2

1 − 𝑘𝑟2+ 𝑟2𝑑휃2 + 𝑟2 sin2 휃 𝑑𝜙 13

Static Weak field 𝑑𝑠2

= −(1 +2Φ(𝑥𝑖)

𝑐2) (𝑐𝑑𝑡)2

+ (1 −2Φ(𝑥𝑖)

𝑐2) (𝑑𝑥2 + 𝑑𝑦2

+ 𝑑𝑧2)

2

Taub-nut space-time 𝑑𝑠2 = −

𝑑𝑡2

𝑈2(𝑡)+ (2𝑙)2𝑈2(𝑡)(𝑑𝑟 + cos 휃 𝑑𝜙)2

+ 𝑉2(𝑡)(𝑑휃2 + sin2 휃 𝑑𝜙2)

7

Three dimensional flat space-time 𝑑𝑠2 = −(𝑐𝑑𝑡)2 + 𝑑𝑥2 + 𝑑𝑦2 3

Three-dimensional flat space in po-lar coordinates

𝑑𝑠2 = 𝑑𝑟2 + 𝑟2𝑑휃2 + 𝑟2 sin2 휃 𝑑𝜙2 2,7, 13

Three-dimensional flat space-time in polar coordinates

𝑑𝑠2 = −𝑑𝑡2 + 𝑑𝑟2 + 𝑟2𝑑𝜙2 4

Tolman-Bondi-de Sitter space-time 𝑑𝑠2 = 𝑑𝑡2 − 𝑒−2𝜓(𝑡,𝑟)𝑑𝑟2 − 𝑅2(𝑡, 𝑟)𝑑휃2

− 𝑅2(𝑡, 𝑟) sin2 휃 𝑑𝜙2 7

Two interacting waves 𝑑𝑠2 = 2𝑑𝑢𝑑𝜈 − cos2 𝑎𝜈 𝑑𝑥2 − cosh2 𝑎𝜈 𝑑𝑦2 14

Two-dimensional flat space 𝑑𝑆2 = 𝑑𝑥2 + 𝑑𝑦2 4

Two-dimensional flat space in polar coordinates

𝑑𝑆2 = 𝑑𝑟2 + 𝑟2𝑑𝜙2 2,5

Two-dimensional flat space-time 𝑑𝑠2 = −𝑋2𝑑𝑇2 + 𝑑𝑋2 2,3,4

Two-dimensional flat space-time 𝑑𝑠2 = −𝑑𝑡2 + 𝑑𝑥2 3

Two-dimensional sphere with ra-dius 𝑎

𝑑𝑠2 = 𝑎2𝑑휃2 + 𝑎2 sin2 휃 𝑑𝜙2 5

Two-dimensional unit sphere 𝑑𝑠2 = 𝑑휃2 + sin2 휃 𝑑𝜙2 7

Warp space-time 𝑑𝑠2 = −𝑑𝑡2 + [𝑑𝑥 − 𝑉𝑠(𝑡)𝑓(𝑟𝑠)𝑑𝑡]2 + 𝑑𝑦2 + 𝑑𝑧2 4

Worm hole geometry 𝑑𝑠2 = −𝑑𝑡2 + 𝑑𝑟2 + (𝑏2 + 𝑟2)(𝑑휃2 + sin2 휃 𝑑𝜙2) 4

2 The Metric tensor and Vector Transformations. See separate document

3 Four vectors and four velocity See separate document

4 Christoffel Symbols, Geodesic Equation and Killing Vectors See separate document

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5 Covariant Derivative, Lie Derivative and Killings Equation See separate document

6 The Riemann tensor See separate document

7 Cartan’s Structure Equations – a Shortcut to the Einstein equation See separate document

8 The Einstein Field Equations

8.1 hThe 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

𝐺𝑎𝑏 = 𝑅𝑎𝑏 −1

2𝑔𝑎𝑏𝑅

If 𝐺𝑎𝑏 = 𝜅𝑇𝑎𝑏 = 0:

⇒ 𝐺𝑎𝑏 = 𝑅𝑎𝑏 −1

2𝑔𝑎𝑏𝑅 = 0

⇒ 𝑅𝑎𝑏 =1

2𝑔𝑎𝑏𝑅

Contracting with 𝑔𝑎𝑏

⇒ 𝑔𝑎𝑏𝑅𝑎𝑏 =1

2𝑔𝑎𝑏𝑔𝑎𝑏𝑅

using the definition 𝑅 = 𝑔𝑎𝑏𝑅𝑎𝑏

and that in 4 dimensions 𝑔𝑎𝑏𝑔𝑎𝑏 = 4

⇒ 𝑅 =1

24𝑅 = 2𝑅

Now this can only be true if 𝑅𝑎𝑏 = 0

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

0 = 𝑅𝑎𝑏 −

1

2𝑔𝑎𝑏𝑅 + 𝑔𝑎𝑏Λ

⇒ 0 = 𝑔𝑎𝑏𝑅𝑎𝑏 −1

2𝑔𝑎𝑏𝑔𝑎𝑏𝑅 + 𝑔𝑎𝑏𝑔

𝑎𝑏Λ

= 𝑅 −

1

24𝑅 + 4Λ

⇒ 𝑅 = 4Λ Next we rewrite the Einstein equation

0 = 𝑅𝑎𝑏 −1

2𝑔𝑎𝑏𝑅 + 𝑔𝑎𝑏Λ

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= 𝑅𝑎𝑏 −

1

2𝑔𝑎𝑏(4Λ) + 𝑔𝑎𝑏Λ

= 𝑅𝑎𝑏 − 𝑔𝑎𝑏Λ ⇒ 𝑅𝑎𝑏 = 𝑔𝑎𝑏Λ Q.E.D.

In the non coordinate basis 𝑅�̂��̂� = 휂�̂��̂�Λ In the case of metrics with negative signature the Einstein equation in vacuum with a cosmological con-stant

0 = 𝑅𝑎𝑏 −1

2𝑔𝑎𝑏𝑅 − 𝑔𝑎𝑏Λ

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

8.3 jGeneral 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: ∇𝑏𝑇

𝑎𝑏 = 0

⇒ ∇𝑏(𝑅𝑎𝑏 + 𝜇𝑅𝑔𝑎𝑏 − Λ𝑔𝑎𝑏) = 0

⇒ ∇𝑏𝑅𝑎𝑏 + 𝜇∇𝑏(𝑅𝑔

𝑎𝑏) − Λ∇𝑏𝑔𝑎𝑏 = 0

⇒ 1 ∇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

⇒ 2 ∇𝑎𝑅𝑎𝑐 + ∇𝑏𝑅

𝑏𝑐 − ∇𝑐𝑅 = 0

⇒ 2 (∇𝑎𝑅𝑎𝑐 −

1

2∇𝑐𝑅) = 0

1 ∇𝑏𝑔

𝑎𝑏 = 0 2 𝑔𝑎𝑒𝑅 𝑒𝑐𝑎

𝑏 = 𝑔𝑎𝑒𝑅𝑒 𝑎𝑐 𝑏 = 𝑅 𝑎𝑐

𝑎𝑏 = 𝑅 𝑐𝑏

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⇒ 2𝑔𝑏𝑐 (∇𝑎𝑅𝑎𝑐 −

1

2∇𝑐𝑅) = 0

⇒ 2 (∇𝑎𝑔𝑏𝑐𝑅𝑎𝑐 −

1

2(∇𝑐𝑅)𝑔

𝑏𝑐) = 0

⇒ 2 (∇𝑎𝑅𝑎𝑏 −

1

2∇𝑎𝑅𝑔

𝑎𝑏) = 0

Now if we compare with ∇𝑏𝑅

𝑎𝑏 + 𝜇(∇𝑏𝑅)𝑔𝑎𝑏 = 0

we see that

𝜇 = −1

2

8.4 3Using the contracted Bianchi identities, prove that: 𝛁𝒃𝑮𝒂𝒃 = 𝟎

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

𝑎𝑏 If 0 = ∇𝑐𝑔

𝑎𝑏 ⇔ 0 = 𝑔𝑑𝑎𝑔𝑒𝑏∇𝑐𝑔

𝑎𝑏 ⇔ 0 = ∇𝑐𝑔𝑑𝑎𝑔𝑒𝑏𝑔

𝑎𝑏 ⇔ 0 = ∇𝑐𝑔𝑑𝑒 ⇔ 0 = ∇𝑐𝑔

𝑎𝑏 (S3) 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 𝑐𝑎 −

1

2∇𝑎𝑔 𝑐

𝑎 𝑅

use (2.15) = ∇𝑎𝑅 𝑐𝑎 −

1

2𝑔 𝑐𝑎 ∇𝑎𝑅

= ∇𝑎𝑅 𝑐𝑎 −

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 𝑐𝑎 + ∇𝑏𝑔

𝑎𝑒𝑅 𝑒𝑐𝑎𝑏 − ∇𝑐𝑅 𝑎

𝑎 ⇔ (use (4.47)) 0 = ∇𝑎𝑅 𝑐

𝑎 + ∇𝑏𝑔𝑎𝑒𝑅 𝑒𝑐𝑎

𝑏 − ∇𝑐𝑅

3 (McMahon, 2006, p. 152), (Kay, 1988, s. 120, 125)

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⇔(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 This is a very important result because it leads to the conservation laws of the right hand side of the Ein-stein equation, which we will look into later. ∇𝑎𝑇

𝑎𝑏 = 0

8.5 42+1 dimensions: Gravitational collapse of an inhomogeneous spherically symmetric dust cloud.

8.5.1 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

휂𝑖𝑗 = {−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 �̂�𝑐̂

�̂� 𝜔𝑐̂ (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�̂�

Summarizing the curvature one forms in a matrix:

Γ �̂��̂� =

{

0 �̇�𝜔�̂�

�̇�

𝑅𝜔�̂�

�̇�𝜔�̂� 0𝑅′

𝑅𝑒−𝑏(𝑡,𝑟)𝜔�̂�

�̇�

𝑅𝜔�̂� −

𝑅′

𝑅𝑒−𝑏(𝑡,𝑟)𝜔�̂� 0 }

Where �̂� refers to column and �̂� to row. The curvature two forms:

Ω �̂��̂� = 𝑑Γ �̂�

�̂� + Γ 𝑐̂�̂� ∧ Γ �̂�

𝑐̂ =1

2𝑅 �̂�𝑐̂�̂��̂� 𝜔𝑐̂ ∧ 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�̂�

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

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Γ 𝑐̂�̂� ∧ Γ �̂�

𝑐̂ = 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 𝑐̂�̂�∧ Γ �̂�

𝑐̂ = 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 𝑐̂�̂�∧ Γ �̂�

𝑐̂ = Γ �̂��̂�∧ Γ �̂�

�̂� + Γ �̂��̂�∧ Γ �̂�

�̂� + Γ �̂�

�̂�∧ Γ �̂�

�̂�= Γ

�̂��̂�∧ Γ �̂�

�̂� =�̇��̇�

𝑅𝜔�̂� ∧ 𝜔�̂�

⇒ Ω �̂��̂�

= −[(�̇�)′− 𝑅′�̇�]

𝑒−𝑏(𝑡,𝑟)

𝑅 𝜔�̂� ∧ 𝜔�̂� − ([𝑅′′ + 𝑅′𝑏′]

𝑒−2𝑏(𝑡,𝑟)

𝑅−�̇��̇�

𝑅)𝜔�̂� ∧ 𝜔�̂�

Summarized in a matrix:

Ω �̂��̂� =

{

0 − [�̈� + (�̇�)

2] 𝜔�̂� ∧ 𝜔�̂� −

�̈�

𝑅𝜔�̂� ∧ 𝜔�̂� − (

(�̇�)′

𝑅−𝑅′�̇�

𝑅) 𝑒−𝑏(𝑡,𝑟)𝜔�̂� ∧ 𝜔�̂�

𝑆 0 − [(�̇�)′− 𝑅′�̇�]

𝑒−𝑏(𝑡,𝑟)

𝑅 𝜔�̂� ∧ 𝜔�̂� − ([𝑅′′ + 𝑅′𝑏′]

𝑒−2𝑏(𝑡,𝑟)

𝑅−�̇��̇�

𝑅)𝜔�̂� ∧ 𝜔�̂�

𝑆 𝐴𝑆 0 }

Where �̂� refers to column and �̂� to row. Now we can find the independent elements of the Riemann tensor in the non-coordinate basis:

R �̂��̂��̂��̂� (𝐴) = −[�̈� + (�̇�)

2]𝜔�̂� ∧ 𝜔�̂� 𝑅

�̂��̂��̂�

�̂�(𝐵) = −

�̈�

𝑅

𝑅 �̂��̂��̂�

�̂� (𝐶) = −[(�̇�)′+ 𝑅′�̇�]

𝑒−𝑏(𝑡,𝑟)

𝑅

𝑅 �̂��̂��̂�

�̂�(𝐷) = −([𝑅′′ + 𝑅′𝑏′]

𝑒−2𝑏(𝑡,𝑟)

𝑅−�̇��̇�

𝑅)

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

8.5.2 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.

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

And chose: 𝑑𝑡′ = 𝑑𝑡

𝑒−𝜓(𝑡′,𝑟′)𝑑𝑟′ = 𝑒𝑏(𝑡,𝑟)𝑑𝑟

𝑅(𝑡′, 𝑟′)𝑑휃′ = 0 𝑅(𝑡′, 𝑟′) sin 휃′ 𝑑𝜙′ = 𝑅(𝑡, 𝑟)𝑑𝜙

Comparing the two metrics we see: 𝑑𝜙′ = 𝑑𝜙, 휃′ =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′)

𝑒2𝜓(𝑡,𝑟)

𝑅+�̇��̇�

𝑅]

𝑅 �̂��̂��̂�

�̂�= −

�̈�

𝑅 ⇒ 𝑅

�̂��̂��̂�

�̂�(𝐵) = −

�̈�

𝑅

𝑅 �̂��̂��̂�

�̂�= −[(�̇�)

′+ 𝑅′�̇�]

𝑒𝜓(𝑡,𝑟)

𝑅 ⇒ 𝑅

�̂��̂��̂�

�̂�(𝐶) = −[(�̇�)

′− 𝑅′�̇�]

𝑒−𝑏(𝑡,𝑟)

𝑅

𝑅 �̂��̂��̂�

�̂�= −[(𝑅′′ + 𝑅′𝜓′)

𝑒2𝜓(𝑡,𝑟)

𝑅+�̇��̇�

𝑅] ⇒ 𝑅

�̂��̂��̂�

�̂� (𝐷) = −[(𝑅′′ − 𝑅′𝑏′)𝑒−2𝑏(𝑡,𝑟)

𝑅−�̇��̇�

𝑅]

𝑅 �̂��̂��̂�

�̂�= [

1

𝑅2+(�̇�)

2

𝑅2−(𝑅′)2

𝑅2𝑒2𝜓(𝑡,𝑟)]

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

8.5.3 Find the components of the Einstein tensor in the coordinate basis for the metric in 2+1 dimensions.

The Ricci tensor:

𝑅�̂��̂� = 𝑅 �̂�𝑐̂�̂�𝑐̂ (4.46)

𝑅�̂��̂� = 𝑅 �̂�𝑐̂�̂�𝑐̂

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= 𝑅 �̂��̂��̂��̂� + 𝑅 �̂��̂��̂�

�̂� + 𝑅 �̂��̂��̂�

�̂�

= 𝑅 �̂��̂��̂��̂� + 𝑅

�̂��̂��̂�

�̂�

= −[�̈� + (�̇�)2] −

�̈�

𝑅

= 𝐴 + 𝐵

𝑅�̂��̂� = 𝑅 �̂�𝑐̂�̂�𝑐̂

= 𝑅 �̂��̂��̂��̂� + 𝑅 �̂��̂��̂�

�̂� + 𝑅 �̂��̂��̂�

�̂�

= 𝑅 �̂��̂��̂�

�̂�

= −[(�̇�)′− 𝑅′�̇�]

𝑒−𝑏(𝑡,𝑟)

𝑅

= 𝐶

𝑅�̂��̂� = 𝑅 �̂�𝑐̂�̂�𝑐̂ = 𝑅 �̂��̂��̂�

�̂� + 𝑅 �̂��̂��̂��̂� + 𝑅 �̂��̂��̂�

�̂� + 𝑅 �̂��̂��̂�

�̂�= 0

𝑅�̂��̂� = 𝑅 �̂�𝑐̂�̂�𝑐̂

= 𝑅 �̂��̂��̂��̂� + 𝑅 �̂��̂��̂�

�̂� + 𝑅 �̂��̂��̂�

�̂�

= −𝑅 �̂��̂��̂��̂� + 𝑅

�̂��̂��̂�

�̂�

= [�̈� + (�̇�)2] − [(𝑅′′ − 𝑅′𝑏′)

𝑒−2𝑏(𝑡,𝑟)

𝑅−�̇��̇�

𝑅]

= −𝐴 + 𝐷

𝑅�̂��̂� = 𝑅 �̂�𝑐̂�̂�𝑐̂

= 𝑅 �̂��̂��̂��̂� + 𝑅 �̂��̂��̂�

�̂� + 𝑅 �̂��̂��̂�

�̂�

= −𝑅 �̂��̂��̂�

�̂�+ 𝑅

�̂��̂��̂�

�̂�

=�̈�

𝑅− [(𝑅′′ − 𝑅′𝑏′)

𝑒−2𝑏(𝑡,𝑟)

𝑅−�̇��̇�

𝑅]

= −𝐵 + 𝐷 Summarized in a matrix:

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]}

Where �̂� refers to column and �̂� to row The Ricci scalar:

𝑅 = 휂�̂��̂�𝑅�̂��̂� (4.47)

𝑅 = 휂�̂��̂�𝑅�̂��̂� + 휂�̂��̂�𝑅�̂��̂� + 휂

�̂��̂�𝑅�̂��̂�

= −𝑅�̂��̂� + 𝑅�̂��̂� + 𝑅�̂��̂�

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

= −2𝑅 �̂��̂��̂��̂� − 2𝑅

�̂��̂��̂�

�̂�+ 2𝑅

�̂��̂��̂�

�̂�

= 2 [�̈� + (�̇�)2] + 2

�̈�

𝑅− 2 [(𝑅′′ − 𝑅′𝑏′)

𝑒−2𝑏(𝑡,𝑟)

𝑅−�̇��̇�

𝑅]

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The Einstein tensor:

𝐺�̂��̂� = 𝑅�̂��̂� −1

2휂�̂��̂�𝑅 (4.48)

𝐺�̂��̂� = 𝑅�̂��̂� −1

2휂�̂��̂�𝑅

= 𝑅�̂��̂� +1

2𝑅

= 𝐴 + 𝐵 +1

2(−2𝐴 − 2𝐵 + 2𝐷)

= 𝐷

= 𝑅 �̂��̂��̂�

�̂�

= −[(𝑅′′ − 𝑅′𝑏′)𝑒−2𝑏

𝑅−�̇��̇�

𝑅]

𝐺�̂��̂� = 𝑅�̂��̂� −1

2휂�̂��̂�𝑅 = 𝑅�̂��̂� = 𝑅 �̂��̂��̂�

�̂�= −[(�̇�)

′− 𝑅′�̇�]

𝑒−𝑏

𝑅

𝐺�̂��̂� = 𝑅�̂��̂� −1

2휂�̂��̂�𝑅 = 0

𝐺�̂��̂� = 𝑅�̂��̂� −1

2휂�̂��̂�𝑅

= 𝑅�̂��̂� −1

2𝑅

= −𝐴 + 𝐷 −1

2(−2𝐴 − 2𝐵 + 2𝐷)

= 𝐵

= 𝑅 �̂��̂��̂�

�̂�

= −�̈�

𝑅

𝐺�̂��̂� = 𝑅�̂��̂� −1

2휂�̂��̂�𝑅

= 𝑅�̂��̂� −1

2𝑅

= −𝐵 + 𝐷 −1

2(−2𝐴 − 2𝐵 + 2𝐷)

= 𝐴

= 𝑅 �̂��̂��̂��̂�

= −[�̈� + (�̇�)2]

Summarized in a matrix:

𝐺�̂��̂� =

{

−[(𝑅′′ − 𝑅′𝑏′)

𝑒−2𝑏

𝑅−�̇��̇�

𝑅] − [(�̇�)

′− 𝑅′�̇�]

𝑒−𝑏

𝑅0

𝑆 −�̈�

𝑅0

0 0 − [�̈� + (�̇�)2]}

Where �̂� refers to column and �̂� to row The Einstein tensor in the coordinate basis: The transformation:

𝐺𝑎𝑏 = Λ 𝑎𝑐̂ Λ 𝑏

�̂� 𝐺𝑐̂�̂� (6.34)

𝐺𝑡𝑡 = Λ 𝑡𝑐̂ Λ 𝑡

�̂� 𝐺𝑐̂�̂� = (Λ 𝑡�̂� )

2𝐺�̂��̂� = −[(𝑅′′ − 𝑅′𝑏′)

𝑒−2𝑏

𝑅−�̇��̇�

𝑅]

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15 http://physicssusan.mono.net [email protected]

𝐺𝑟𝑡 = Λ 𝑟𝑐̂ Λ 𝑡

�̂� 𝐺𝑐̂�̂� = Λ 𝑟�̂� Λ 𝑡

�̂� 𝐺�̂��̂� = −[(�̇�)

′− 𝑅′�̇�]

𝑅

𝐺𝑟𝑟 = Λ 𝑟𝑐̂ Λ 𝑟

�̂� 𝐺𝑐̂�̂� = (Λ 𝑟�̂� )

2𝐺�̂��̂� = −

�̈�

𝑅𝑒2𝑏

𝐺𝜙𝜙 = Λ 𝜙𝑐̂ Λ 𝜙

�̂� 𝐺𝑐̂�̂� = (Λ 𝜙�̂�)2

𝐺�̂��̂� = −𝑅2 [�̈� + (�̇�)2]

Summarized in a matrix:

𝐺𝑎𝑏 =

{

−[(𝑅′′ − 𝑅′𝑏′)

𝑒−2𝑏

𝑅−�̇��̇�

𝑅] −

[(�̇�)′− 𝑅′�̇�]

𝑅0

𝑆 −�̈�

𝑅𝑒2𝑏 0

0 0 −𝑅2 [�̈� + (�̇�)2]}

Where 𝑎 refers to column and 𝑏 to row

8.5.4 The Einstein equations of the metric in 2+1 dimensions. Given the Einstein equation ( if 𝑐 = 𝐺 = 1): 𝐺�̂��̂� + Λ휂�̂��̂� = 𝜅𝑇�̂��̂� (6.40) with Λ = −𝜆2 you get 𝐺�̂��̂� − 𝜆

2휂�̂��̂� = 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)

8.6 5Ricci 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

5 (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)

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𝜔�̂� = 𝑑𝑡

휂𝑖𝑗 = {

−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�̂�

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

𝑑𝜔�̂� = −Γ �̂��̂� ∧ 𝜔�̂� (5.9)

Γ �̂��̂� = Γ �̂�𝑐̂

�̂� 𝜔𝑐̂ (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�̂�

Summarizing the curvature one forms in a matrix:

Γ �̂��̂� =

{

0

�̇�

𝐿𝜔�̂�

�̇�

𝐵𝜔�̂�

�̇�

𝑀𝜔�̂�

�̇�

𝐿𝜔�̂� 0

𝐵′

𝐿𝐵𝜔�̂�

𝑀′

𝐿𝑀𝜔�̂�

�̇�

𝐵𝜔�̂� −

𝐵′

𝐿𝐵𝜔�̂� 0 0

�̇�

𝑀𝜔�̂� −

𝑀′

𝐿𝑀𝜔�̂� 0 0 }

Where �̂� refers to column and �̂� to row Now we can read off the Ricci rotation coefficients

Γ �̂��̂��̂� =

�̇�

𝐿 Γ �̂��̂�

�̂� =�̇�

𝐿 Γ

�̂��̂�

�̂� =

�̇�

𝐵 Γ �̂��̂�

�̂� =�̇�

𝑀

Γ �̂��̂��̂� =

�̇�

𝐵 Γ �̂��̂�

�̂� = −𝐵′

𝐿𝐵 Γ

�̂��̂�

�̂� =

𝐵′

𝐿𝐵 Γ �̂��̂�

�̂� =𝑀′

𝐿𝑀

Γ �̂��̂��̂� =

�̇�

𝑀 Γ �̂��̂�

�̂� = −𝑀′

𝐿𝑀

The curvature two forms:

Ω �̂��̂� = 𝑑Γ �̂�

�̂� + Γ 𝑐̂�̂� ∧ Γ �̂�

𝑐̂ =1

2𝑅 �̂�𝑐̂�̂��̂� 𝜔𝑐̂ ∧ 𝜔�̂� (5.27), (5.28)

Ω �̂��̂� : 𝑑Γ �̂�

�̂� = 𝑑 (�̇�

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

�̈�

𝐿𝜔�̂� ∧ 𝜔�̂�

Γ 𝑐̂�̂� ∧ Γ �̂�

𝑐̂ = 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 𝑐̂�̂�∧ Γ �̂�

𝑐̂ = Γ �̂��̂�∧ Γ �̂�

�̂� + Γ �̂��̂�∧ Γ �̂�

�̂� + Γ �̂�

�̂�∧ Γ

�̂��̂�+ Γ �̂�

�̂�∧ Γ �̂�

�̂� = Γ �̂��̂�∧ Γ �̂�

�̂� =𝐵´

𝐵𝐿𝜔�̂� ∧

�̇�

𝐿𝜔�̂�

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⇒ 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 𝑐̂�̂� ∧ Γ �̂�

𝑐̂ = 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 𝑐̂�̂�∧ Γ �̂�

𝑐̂ = 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 𝑐̂�̂� ∧ Γ �̂�

𝑐̂ = 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 𝑐̂

�̂� ∧ Γ �̂�𝑐̂

= Γ �̂��̂� ∧ Γ �̂�

�̂� + Γ �̂��̂� ∧ Γ �̂�

�̂� + Γ �̂��̂� ∧ Γ

�̂�

�̂�+ Γ �̂�

�̂� ∧ Γ �̂��̂�

= Γ �̂��̂� ∧ Γ �̂�

�̂� + Γ �̂��̂� ∧ Γ �̂�

�̂�

=�̇�

𝑀𝜔�̂� ∧

�̇�

𝐵𝜔�̂� +

𝑀′

𝐿𝑀𝜔�̂� ∧ (−

𝐵′

𝐿𝐵𝜔�̂�)

= (�̇��̇�

𝐵𝑀−𝐵′𝑀′

𝐿2𝐵𝑀)𝜔�̂� ∧ 𝜔�̂�

Summarized in a matrix:

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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 (�̇��̇�

𝐵𝑀−𝐵′𝑀′

𝐿2𝐵𝑀)𝜔�̂� ∧ 𝜔�̂�

𝑆 𝐴𝑆 𝐴𝑆 0 }

Where �̂� refers to column and �̂� to row. Now we can find the independent elements of the Riemann tensor in the non-coordinate basis:

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) =

�̇��̇�

𝐵𝑀−𝐵′𝑀′

𝐿2𝐵𝑀

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

𝑅�̂��̂� = 𝑅 �̂�𝑐̂�̂�𝑐̂ (4.46)

𝑅�̂��̂� = 𝑅 �̂�𝑐̂�̂�𝑐̂

= 𝑅 �̂��̂��̂��̂� + 𝑅 �̂��̂��̂�

�̂� + 𝑅 �̂��̂��̂�

�̂�+ 𝑅 �̂��̂��̂�

�̂�

= 𝑅 �̂��̂��̂��̂� + 𝑅

�̂��̂��̂�

�̂�+ 𝑅 �̂��̂��̂�

�̂�

= −�̈�

𝐿−�̈�

𝐵−�̈�

𝑀

= 𝐴 + 𝐵 + 𝐶

𝑅�̂��̂� = 𝑅 �̂�𝑐̂�̂�𝑐̂

= 𝑅 �̂��̂��̂��̂� + 𝑅 �̂��̂��̂�

�̂� + 𝑅 �̂��̂��̂�

�̂�+ 𝑅 �̂��̂��̂�

�̂�

= 𝑅 �̂��̂��̂�

�̂�+ 𝑅 �̂��̂��̂�

�̂�

=𝐵´�̇�

𝐵𝐿2−𝐵′̇

𝐵𝐿+𝑀´�̇�

𝑀𝐿2−�̇�′

𝑀𝐿

= 𝐷 + 𝐸

𝑅�̂��̂� = 𝑅 �̂�𝑐̂�̂�𝑐̂ = 𝑅 �̂��̂��̂�

�̂� + 𝑅 �̂��̂��̂��̂� + 𝑅

�̂��̂��̂�

�̂�+ 𝑅 �̂��̂��̂�

�̂� = 0

𝑅�̂��̂� = 𝑅 �̂�𝑐̂�̂�𝑐̂ = 𝑅 �̂��̂��̂�

�̂� + 𝑅 �̂��̂��̂��̂� + 𝑅

�̂��̂��̂�

�̂�+ 𝑅 �̂��̂��̂�

�̂� = 0

𝑅�̂��̂� = 𝑅 �̂�𝑐̂�̂�𝑐̂

= 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 �̂�𝑐̂�̂�𝑐̂

= 𝑅 �̂��̂��̂��̂� + 𝑅 �̂��̂��̂�

�̂� + 𝑅 �̂��̂��̂�

�̂�+ 𝑅 �̂��̂��̂�

�̂�

= −𝑅 �̂��̂��̂�

�̂�+ 𝑅

�̂��̂��̂�

�̂�+ 𝑅 �̂��̂��̂�

�̂�

=�̈�

𝐵+𝐵′𝐿′ − 𝐵′′𝐿

𝐵𝐿3+�̇��̇�

𝐵𝐿+�̇��̇�

𝐵𝑀−𝐵′𝑀′

𝐿2𝐵𝑀

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= −𝐵 + 𝐹 + 𝐻

𝑅�̂��̂� = 𝑅 �̂�𝑐̂�̂�𝑐̂

= 𝑅 �̂��̂��̂��̂� + 𝑅 �̂��̂��̂�

�̂� + 𝑅 �̂��̂��̂�

�̂�+ 𝑅 �̂��̂��̂�

�̂�

= −𝑅 �̂��̂��̂��̂� + 𝑅 �̂��̂��̂�

�̂� + 𝑅 �̂��̂��̂��̂�

=�̈�

𝑀+𝑀´𝐿´ − 𝑀´´𝐿

𝑀𝐿3+�̇��̇�

𝑀𝐿+�̇��̇�

𝐵𝑀−𝐵′𝑀′

𝐿2𝐵𝑀

= −𝐶 + 𝐺 +𝐻 Summarized in a matrix:

𝑅�̂��̂� =

{

�̈�

𝐿−�̈�

𝐵−�̈�

𝑀

𝐵´�̇�

𝐵𝐿2−𝐵′̇

𝐵𝐿+𝑀´�̇�

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+�̇��̇�

𝐵𝑀−𝐵′𝑀′

𝐿2𝐵𝑀0

0 0 0�̈�

𝑀+𝑀´𝐿´ − 𝑀´´𝐿

𝑀𝐿3+�̇��̇�

𝑀𝐿+�̇��̇�

𝐵𝑀−𝐵′𝑀′

𝐿2𝐵𝑀}

Where �̂� refers to column and �̂� to row The Ricci scalar:

𝑅 = 휂�̂��̂�𝑅�̂��̂� (4.47)

𝑅 = 휂�̂��̂�𝑅�̂��̂� + 휂�̂��̂�𝑅�̂��̂� + 휂

�̂��̂�𝑅�̂��̂� + 휂�̂��̂�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+�̇��̇�

𝐵𝐿+𝑀´𝐿´ − 𝑀´´𝐿

𝑀𝐿3+�̇��̇�

𝑀𝐿+�̇��̇�

𝐵𝑀−𝐵′𝑀′

𝐿2𝐵𝑀)

The Einstein tensor:

𝐺�̂��̂� = 𝑅�̂��̂� −1

2휂�̂��̂�𝑅 (4.48)

𝐺�̂��̂� = 𝑅�̂��̂� −1

2휂�̂��̂�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+�̇��̇�

𝐵𝐿+𝑀´𝐿´ − 𝑀´´𝐿

𝑀𝐿3+�̇��̇�

𝑀𝐿+�̇��̇�

𝐵𝑀−𝐵′𝑀′

𝐿2𝐵𝑀

𝐺�̂��̂� = 𝑅�̂��̂� −1

2휂�̂��̂�𝑅 = 𝑅�̂��̂� =

𝐵´�̇�

𝐵𝐿2−𝐵′̇

𝐵𝐿+𝑀´�̇�

𝑀𝐿2−�̇�′

𝑀𝐿

𝐺�̂��̂� = 𝑅�̂��̂� −1

2휂�̂��̂�𝑅 = 0

𝐺�̂��̂� = 𝑅�̂��̂� −1

2휂�̂��̂�𝑅 = 0

𝐺�̂��̂� = 𝑅�̂��̂� −1

2휂�̂��̂�𝑅

= 𝑅�̂��̂� −1

2𝑅

= −𝐴 + 𝐹 + 𝐺 −1

2(−2𝐴 − 2𝐵 − 2𝐶 + 2𝐹 + 2𝐺 + 2𝐻)

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= 𝐵 + 𝐶 + 𝐻

= 𝑅 �̂��̂��̂�

�̂�+ 𝑅 �̂�𝑧�̂�

�̂� + 𝑅 �̂��̂��̂��̂�

= −�̈�

𝐵−�̈�

𝑀+�̇��̇�

𝐵𝑀−𝐵′𝑀′

𝐿2𝐵𝑀

𝐺�̂��̂� = 𝑅�̂��̂� −1

2휂�̂��̂�𝑅 = 0

𝐺�̂��̂� = 𝑅�̂��̂� −1

2휂�̂��̂�𝑅 = 0

𝐺�̂��̂� = 𝑅�̂��̂� −1

2휂�̂��̂�𝑅

= 𝑅�̂��̂� −1

2𝑅

= −𝐵 + 𝐹 + 𝐻 −1

2(−2𝐴 − 2𝐵 − 2𝐶 + 2𝐹 + 2𝐺 + 2𝐻)

= 𝐴 + 𝐶 + 𝐺

= 𝑅 �̂��̂��̂��̂� + 𝑅 �̂�𝑧�̂�

�̂� + 𝑅 �̂��̂��̂��̂�

= −�̈�

𝐿−�̈�

𝑀

𝑀´𝐿´ − 𝑀´´𝐿

𝑀𝐿3+�̇��̇�

𝑀𝐿

𝐺�̂��̂� = 𝑅�̂��̂� −1

2휂�̂��̂�𝑅 = 0

𝐺�̂��̂� = 𝑅�̂��̂� −1

2휂�̂��̂�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

Summarized in a matrix:

𝐺�̂��̂� =

{

𝐵′𝐿′ − 𝐵′′𝐿

𝐵𝐿3+�̇��̇�

𝐵𝐿+𝑀´𝐿´ − 𝑀´´𝐿

𝑀𝐿3+�̇��̇�

𝑀𝐿+�̇��̇�

𝐵𝑀−𝐵′𝑀′

𝐿2𝐵𝑀

𝐵´�̇�

𝐵𝐿2−𝐵′̇

𝐵𝐿+𝑀´�̇�

𝑀𝐿2−�̇�′

𝑀𝐿0 0

𝑆 −�̈�

𝐵−�̈�

𝑀+�̇��̇�

𝐵𝑀−𝐵′𝑀′

𝐿2𝐵𝑀0 0

0 0 −�̈�

𝐿−�̈�

𝑀

𝑀´𝐿´ − 𝑀´´𝐿

𝑀𝐿3+�̇��̇�

𝑀𝐿0

0 0 0 −�̈�

𝐿−�̈�

𝐵+𝐵′𝐿′ − 𝐵′′𝐿

𝐵𝐿3+�̇��̇�

𝐵𝐿}

Where �̂� refers to column and �̂� to row

8.7 6The de Sitter Spacetime The line element 𝑑𝑠2 = −𝑑𝑡2 + 𝑎(𝑡)2(𝑑휃2 + sin2 휃 (𝑑𝜙2 + sin2𝜙𝑑𝜓2)) (IV.11.5)

8.7.1 The Einstein equations The de Sitter spacetime is an example of the Robertson Walker metric in vacuum, positive curvature and a cosmological constant. The solution is the Friedman equations7

0 =3

𝑎2(1 + �̇�2) − Λ

6 (Choquet-Bruhat, 2015, s. 96) Problem IV.11.2. 7 See the chapter named: The Einstein tensor and Friedmann-equations for the Robertson Walker metric

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0 = 2

�̈�

𝑎+1

𝑎2(1 + �̇�2) − Λ

We chose 𝑘2 =Λ

3

0 =1

𝑎2(1 + �̇�2) − 𝑘2

⇒ −1 = �̇�2 − 𝑘2𝑎2 (IV.11.7)

0 = 2�̈�

𝑎+1

𝑎2(1 + �̇�2) − 3𝑘2 = 2

�̈�

𝑎+ 𝑘2 − 3𝑘2 = 2

�̈�

𝑎− 2𝑘2

⇒ 0 = �̈� − 𝑘2𝑎 (IV.11.6)

8.7.2 Solving the Einstein equations These equations we can solve. The general solution to equation (IV.11.6) is 𝑎 = 𝐴𝑒𝑘𝑡 + 𝐵𝑒−𝑘𝑡 (IV.11.8)

⇒ 𝑎2 = (𝐴𝑒𝑘𝑡 + 𝐵𝑒−𝑘𝑡)2= 𝐴2𝑒2𝑘𝑡 + 𝐵2𝑒−2𝑘𝑡 + 2𝐴𝐵

⇒ �̇� = 𝐴𝑘𝑒𝑘𝑡 − 𝐵𝑘𝑒−𝑘𝑡

⇒ �̇�2 = (𝐴𝑘𝑒𝑘𝑡 − 𝐵𝑘𝑒−𝑘𝑡)2= 𝑘2(𝐴2𝑒2𝑘𝑡 +𝐵2𝑒−2𝑘𝑡 − 2𝐴𝐵)

�̈� = 𝐴𝑘2𝑒𝑘𝑡 + 𝐵𝑘2𝑒−𝑘𝑡 = 𝑘2𝑎 ⇒ 0 = �̈� − 𝑘2𝑎 = 𝑘2𝑎 − 𝑘2𝑎 = 0

⇒ −1 = �̇�2 − 𝑘2𝑎2 = 𝑘2(𝐴2𝑒2𝑘𝑡 + 𝐵2𝑒−2𝑘𝑡 − 2𝐴𝐵) − 𝑘2(𝐴2𝑒2𝑘𝑡 + 𝐵2𝑒−2𝑘𝑡 + 2𝐴𝐵)

= −4𝑘2𝐴𝐵

⇒ 1

𝑘2 = 4𝐴𝐵

We choose 𝐴 = 𝐵

⇒ 𝑎2 = (𝐴𝑒𝑘𝑡 + 𝐵𝑒−𝑘𝑡)2= 𝐴2(𝑒𝑘𝑡 + 𝑒−𝑘𝑡)

2=

1

4𝑘222 cosh2(𝑘𝑡) =

1

𝑘2cosh2(𝑘𝑡)

And find the de Sitter line element

𝑑𝑠2 = −𝑑𝑡2 +1

𝑘2cosh2(𝑘𝑡) (𝑑휃2 + sin2 휃 (𝑑𝜙2 + sin2𝜙𝑑𝜓2)) (IV.11.9)

8.8 8The Anti-de Sitter Spacetime The line element 𝑑𝑠2 = −𝑑𝑡2 + cos2(𝑡) 𝑑𝑟2 + cos2(𝑡) sinh2(𝑟) 𝑑휃2 + cos2(𝑡) sinh2(𝑟) sin2 휃 𝑑𝜙2 (IV.11.10) To show that this spacetime is a solution to the Einstein vacuum equation with cosmological constant Λ =−3 we can compare to the Tolman-Bondi de Sitter spacetime9 𝑑𝑠2 = 10𝑑𝑡2 − 𝑒−2𝜓(𝑡,𝑟)𝑑𝑟2 − 𝑅2(𝑡, 𝑟)𝑑휃2 − 𝑅2(𝑡, 𝑟) sin2 휃 𝑑𝜙2

Comparing the two metrics 𝑒−2𝜓(𝑡,𝑟) = cos2(𝑡)

𝑅2(𝑡, 𝑟) = cos2(𝑡) sinh2(𝑟) ⇒ 𝑅(𝑡, 𝑟) = ±cos(𝑡) sinh(𝑟)

We need

𝑑(𝑒−2𝜓(𝑡,𝑟))

𝑑𝑡 =

𝑑(cos2(𝑡))

𝑑𝑡= −2cos(𝑡) sin(𝑡) = −2�̇�𝑒−2𝜓(𝑡,𝑟)

⇒ �̇� =cos(𝑡) sin(𝑡)

𝑒−2𝜓(𝑡,𝑟)=cos(𝑡) sin(𝑡)

cos2(𝑡)= tan(𝑡)

8 (Choquet-Bruhat, 2015, s. 97) Problem IV.11.3.1 9 See the chapter named: The Einstein Tensor of the Tolman-Bondi de Sitter Metric 10 Notice the two spacetimes do not have the same signature. Since we are only going to use the Ricci scalar, and the Ricci scalar does not change sign, when the signature changes, this is not a problem.

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�̈� =

𝑑 tan(𝑡)

𝑑𝑡=

1

cos2(𝑡)

𝜓′ = 0

�̇� =𝑑(± cos(𝑡) sinh(𝑟))

𝑑𝑡= ∓ sin(𝑡) sinh(𝑟)

�̈� = ∓cos(𝑡) sinh(𝑟)

4�̈�

𝑅 = 4

∓cos(𝑡) sinh(𝑟)

± cos(𝑡) sinh(𝑟)= −4

𝑅′ = ±cos(𝑡) cosh(𝑟) 𝑅′′ = ±cos(𝑡) sinh(𝑟) = 𝑅(𝑡, 𝑟)

⇒ 2 [�̈� − (�̇�)2] = 2 [

1

cos2(𝑡)− tan2(𝑡)] = 2

1

cos2(𝑡)[1 − sin2(𝑡)] = 2

(𝑅′′ + 𝑅′𝜓′)𝑒2𝜓(𝑡,𝑟)

𝑅 = 𝑒2𝜓(𝑡,𝑟) =

1

cos2(𝑡)

�̇��̇�

𝑅 =

∓sin(𝑡) sinh(𝑟) tan(𝑡)

± cos(𝑡) sinh(𝑟)= − tan2(𝑡)

(�̇�)2

𝑅2 =

(∓ sin(𝑡) sinh(𝑟))2

cos2(𝑡) sinh2(𝑟)= tan2(𝑡)

(𝑅′)2

𝑅2𝑒2𝜓(𝑡,𝑟) =

(± cos(𝑡) cosh(𝑟))2

cos2(𝑡) sinh2(𝑟) cos2(𝑡)=

1

cos2(𝑡) tanh2(𝑟)

The Ricci scalar

𝑅 = 2 [�̈� − (�̇�)2] − 4

�̈�

𝑅+ 4 [(𝑅′′ + 𝑅′𝜓′)

𝑒2𝜓(𝑡,𝑟)

𝑅+�̇��̇�

𝑅] − 2 [

1

𝑅2+(�̇�)

2

𝑅2−(𝑅′)2

𝑅2𝑒2𝜓(𝑡,𝑟)]

= 2 − (−4) + 4 [1

cos2(𝑡)− tan2(𝑡)] − 2 [

1

cos2(𝑡) sinh2(𝑟)+ tan2(𝑡) −

1

cos2(𝑡) tanh2(𝑟)]

= 10 −

2

cos2(𝑡)[sin2(𝑡) +

1

sinh2(𝑟)(1 − cosh2(𝑟))]

= 10 −

2

cos2(𝑡)[sin2(𝑡) − 1]

= 12 The vacuum Einstein equation for a metric with negative signature and a cosmological constant is

0 = 𝐺𝑎𝑏 − 𝑔𝑎𝑏Λ = 𝑅𝑎𝑏 −1

2𝑔𝑎𝑏𝑅 − 𝑔𝑎𝑏Λ

⇒ 0 = 𝑔𝑎𝑏𝑅𝑎𝑏 −1

2𝑔𝑎𝑏𝑔𝑎𝑏𝑅 − 𝑔

𝑎𝑏𝑔𝑎𝑏Λ = 𝑅 −1

24𝑅 − 4Λ = −𝑅 − 4Λ

⇒ Λ = −𝑅

4= −

12

4= −3

9 The Energy-Momentum Tensor

9.1 11The Einstein equation with source You can show that in a spacetime of dimension 𝑛 + 1 > 2, the Einstein equation with sources is equivalent

to 𝑅𝑎𝑏 = 𝜅𝜌𝑎𝑏 with 𝜌𝑎𝑏 ≡ 𝑇𝑎𝑏 −𝑇𝑐𝑐

𝑛−1𝑔𝑎𝑏

The Einstein equation with sources is

𝐺𝑎𝑏 = 𝑅𝑎𝑏 −1

2𝑔𝑎𝑏𝑅 = 𝜅𝑇𝑎𝑏 (IV.2.4)

Contracting with 𝑔𝑎𝑏

11 (Choquet-Bruhat, 2015, s. 70) Exercise IV.2.1.

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⇒ 𝑔𝑎𝑏𝑅𝑎𝑏 −1

2𝑔𝑎𝑏𝑔𝑎𝑏𝑅 = 𝜅𝑔𝑎𝑏𝑇𝑎𝑏

⇒ 𝑅 −1

2(𝑛 + 1)𝑅 = 𝜅𝑇𝑎

𝑎

⇒ 1

2(1 − 𝑛)𝑅 = 𝜅𝑇𝑎

𝑎

⇒ 1

2𝑅 =

𝜅𝑇𝑎𝑎

(1 − 𝑛)

⇒ 1

2𝑔𝑎𝑏𝑅 = −

𝜅𝑇𝑎𝑎

(𝑛 − 1)𝑔𝑎𝑏

Inserting this into the Einstein equation

⇒ 𝑅𝑎𝑏 = 𝜅𝑇𝑎𝑏 +1

2𝑔𝑎𝑏𝑅

⇒ 𝑅𝑎𝑏 = 𝜅𝑇𝑎𝑏 −𝜅𝑇𝑎

𝑎

(𝑛 − 1)𝑔𝑎𝑏 = 𝜅 (𝑇𝑎𝑏 −

𝑇𝑎𝑎

(𝑛 − 1)𝑔𝑎𝑏) = 𝜅𝜌𝑎𝑏 (IV.2.5)

In 𝑛 + 1 dimensions with a cosmological constant Λ we have

𝜌𝑎𝑏 = 𝑇𝑎𝑏 +1

𝑛 − 1((𝑛 + 1)Λ − 𝑇𝑐

𝑐)𝑔𝑎𝑏 (IV.3.3)

9.2 12Perfect Fluids 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

휂�̂��̂� = {

1 0 0 00 −1 0 00 0 −1 00 0 0 −1

}

This we can use to find the constants 𝐴 and 𝐵

𝑇0̂0̂ = 𝐴𝑢0̂𝑢0̂ + 𝐵휂0̂0̂ = 𝐴 + 𝐵 = 𝜌

𝑇 �̂��̂� = 𝐴𝑢�̂�𝑢�̂� + 𝐵휂�̂��̂� = −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

휂�̂��̂� = {

−1 0 0 00 1 0 00 0 1 00 0 0 1

}

𝑇0̂0̂ = 𝐴𝑢0̂𝑢0̂ + 𝐵휂0̂0̂ = 𝐴 − 𝐵 = 𝜌

𝑇 �̂��̂� = 𝐴𝑢�̂�𝑢�̂� + 𝐵휂�̂��̂� = 𝐵 = {𝑃0

𝑖𝑓 𝑖 = 𝑗

12 (McMahon, 2006, p. 160), (Carroll, 2004, pp. 33-35)

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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)

9.3 More examples on stress-energy tensors

9.3.1 13Pure Matter In the case of pure matter with no pressure the stress-energy tensor is 𝑇𝑎𝑏 = 𝜌𝑢𝑎𝑢𝑏 (IV.2.7) Where 𝑢 is the unit flow velocity and 𝜌 is the rest-mass density. 14For a co-moving observer the four velocity reduces to 𝑢𝑎 = (1,0,0,0) And the energy-momentum tensor reduces to

𝑇𝑎𝑏 = {

𝜌 0 0 00 0 0 00 0 0 00 0 0 0

} (7.4)

15In the case of a stationary observer the dust particles have a four velocity 𝑢𝑎 = (𝛾, 𝛾𝑢𝑥 , 𝛾𝑢𝑦, 𝛾𝑢𝑧) And the energy-momentum tensor is

𝑇𝑎𝑏 = 𝜌𝛾2

{

1 𝑢𝑥 𝑢𝑦 𝑢𝑧

𝑢𝑥 (𝑢𝑥)2 𝑢𝑥𝑢𝑦 𝑢𝑥𝑢𝑧

𝑢𝑦 𝑢𝑦𝑢𝑥 (𝑢𝑦)2 𝑢𝑦𝑢𝑧

𝑢𝑧 𝑢𝑧𝑢𝑥 𝑢𝑧𝑢𝑦 (𝑢𝑧)2}

(7.5)

9.3.2 16More complicated fluids The most general form of a matter-stress-energy-tensor is the non-perfect fluid with viscosity and shear. 𝑇𝑎𝑏 = 𝜌(1 + 휀)𝑢𝑎𝑢𝑏 + (𝑃 − 휁휃)ℎ𝑎𝑏 − 2휂𝜎𝑎𝑏 + 𝑞𝑎𝑢𝑏 + 𝑞𝑏𝑢𝑎 Where the various quantities are defined as 휀 - Specific energy density of the fluid in its rest frame 𝑃 - Pressure ℎ𝑎𝑏 = 𝑢𝑎𝑢𝑏 + 𝑔𝑎𝑏 – The spatial projection tensor

휂 - shear viscosity 휁 - bulk viscosity 휃 = ∇𝑎𝑢

𝑎 – expansion 𝜎𝑎𝑏 =

1

2(∇𝑐𝑢

𝑎ℎ𝑐𝑏 + ∇𝑐𝑢𝑏ℎ𝑐𝑎) −

1

3휃ℎ𝑎𝑏 – shear tensor

𝑞𝑎 - energy flux tensor

9.3.3 17The electromagnetic field The stress energy tensor of an electromagnetic field is the Maxwell tensor 𝑇𝛼𝛽 = 𝐹𝛼

𝜆𝐹𝛽𝜆 −1

4휂𝛼𝛽𝐹

𝜇𝜈𝐹𝜇𝜈 (IV.2.10)

13 (Choquet-Bruhat, 2015, s. 71) 14 (McMahon, 2006, p. 158) 15 (McMahon, 2006, p. 159) 16 (McMahon, 2006, p. 164) 17 (Choquet-Bruhat, 2015, s. 71)

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9.3.3.1 18The Maxwell equations The electromagnetic field tensor 𝐹𝜇𝜈 is defined by

𝐹𝜇𝜈 = {

0 −𝐸1 −𝐸2 −𝐸3

𝐸1 0 −𝜖123𝐵3 −𝜖132𝐵2

𝐸2 −𝜖213𝐵3 0 −𝜖231𝐵1

𝐸3 −𝜖312𝐵2 −𝜖321𝐵1 0

} = {

0 −𝐸1 −𝐸2 −𝐸3

𝐸1 0 −𝐵3 𝐵2

𝐸2 𝐵3 0 −𝐵1

𝐸3 −𝐵2 𝐵1 0

}

i.e. 𝐹0𝑖 = −𝐸𝑖 𝐹𝑖𝑗 = −𝜖𝑖𝑗𝑘𝐵𝑘 Expressed by the vector potential19 𝐴𝜇 𝐹𝜇𝜈 = 𝜕𝜇𝐴𝜈 − 𝜕𝜈𝐴𝜇 Or �̅� = 20 − ∇̅ × �̅�

This we can use to find the four Maxwell equations a)Notice that ∇ ⋅ �̅� = −∇ ⋅ (∇̅ × �̅�)

= −∇ ⋅ {

𝜕2𝐴3 − 𝜕3𝐴

2

𝜕3𝐴1 − 𝜕1𝐴

3

𝜕1𝐴2 − 𝜕2𝐴

1

}

= 𝜕1(𝜕2𝐴3 − 𝜕3𝐴

2) + 𝜕2(𝜕3𝐴1 − 𝜕1𝐴

3) + 𝜕3(𝜕1𝐴2 − 𝜕2𝐴

1) = 0 (II) b)Also notice �̅� = 21 − 𝜕0�̅� − ∇̅𝐴0

⇒ ∇̅ × �̅� = ∇̅ × (−𝜕0�̅� − ∇̅𝐴0)

= −∇̅ × {

𝜕0𝐴1 + 𝜕1𝐴0𝜕0𝐴2 + 𝜕2𝐴0𝜕0𝐴3 + 𝜕3𝐴0

}

= −{

𝜕2(𝜕0𝐴3 + 𝜕3𝐴0) − 𝜕3(𝜕0𝐴2 + 𝜕2𝐴0)

𝜕3(𝜕0𝐴1 + 𝜕1𝐴0) − 𝜕1(𝜕0𝐴3 + 𝜕3𝐴0)

𝜕1(𝜕0𝐴2 + 𝜕2𝐴0) − 𝜕2(𝜕0𝐴1 + 𝜕1𝐴0)}

= −{

𝜕2(𝜕0𝐴3) − 𝜕3(𝜕0𝐴2)

𝜕3(𝜕0𝐴1) − 𝜕1(𝜕0𝐴3)

𝜕1(𝜕0𝐴2) − 𝜕2(𝜕0𝐴1)}

= −𝜕0 {

𝜕2(𝐴3) − 𝜕3(𝐴2)

𝜕3(𝐴1) − 𝜕1(𝐴3)

𝜕1(𝐴2) − 𝜕2(𝐴1)}

= −𝜕0(∇̅ × �̅�)

18 (Michael E. Peskin, 1995, s. 33) 19 Definition: Vector potential: A function �̅� such that �̅� ≡ ∇ × �̅�. The most common use of a vector potential is the representation of a magnetic field. If a vector field has zero divergence, it may be represented by a vector potential http://mathworld.wolfram.com/VectorPotential.html

20 𝜖𝑖𝑗𝑘𝐵𝑘 = −𝐹𝑖𝑗 = −(𝜕𝑖𝐴𝑗– 𝜕𝑗𝐴𝑖) ⇒ �̅� = {𝐵1

𝐵2

𝐵3} = − {

𝐹23

−𝐹13

𝐹12} = − {

𝜕2𝐴3– 𝜕3𝐴2

−(𝜕1𝐴3– 𝜕3𝐴1)

𝜕1𝐴2– 𝜕2𝐴1} = − {

𝜕2𝐴3– 𝜕3𝐴2

𝜕3𝐴1– 𝜕1𝐴3

𝜕1𝐴2– 𝜕2𝐴1} =

−{

−𝜕2𝐴3 + 𝜕3𝐴

2

−𝜕3𝐴1 + 𝜕1𝐴

3

−𝜕1𝐴2 + 𝜕2𝐴

1

} = {

𝜕2𝐴3 − 𝜕3𝐴

2

𝜕3𝐴1 − 𝜕1𝐴

3

𝜕1𝐴2 − 𝜕2𝐴

1

} = ∇̅ × �̅�

21𝐸𝑖 = −𝐹0𝑖 = −(𝜕0𝐴𝑖– 𝜕𝑖𝐴0) = −(𝜕0𝐴𝑖 + 𝜕𝑖𝐴0) ⇒ �̅� = −𝜕0�̅�− ∇̅𝐴0

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= −𝜕0�̅� (III) c) The Lagrangian

ℒ = −1

4𝐹𝜇𝜈𝐹

𝜇𝜈

The Euler-Lagrange equation

0 = 𝜕𝜇 (𝜕ℒ

𝜕(𝜕𝜇𝐴𝜈)) −

𝜕ℒ

𝜕𝐴𝜈 (2.3)

𝜕ℒ

𝜕𝐴𝜈 = 0

𝜕ℒ

𝜕(𝜕𝜇𝐴𝜈) =

𝜕 (−14𝐹𝜌𝛿𝐹

𝜌𝛿)

𝜕(𝜕𝜇𝐴𝜈)

= −

1

4[𝐹𝜌𝛿

𝜕(𝐹𝜌𝛿)

𝜕(𝜕𝜇𝐴𝜈)+ 𝐹𝜌𝛿

𝜕(𝐹𝜌𝛿)

𝜕(𝜕𝜇𝐴𝜈)]

= −

1

4[𝐹𝜌𝛿

𝜕(𝐹𝜌𝛿)

𝜕(𝜕𝜇𝐴𝜈)+ 𝐹𝜌𝛿

𝜕(𝐹𝜌𝛿)

𝜕(𝜕𝜇𝐴𝜈)]

= −

1

2𝐹𝜌𝛿

𝜕(𝐹𝜌𝛿)

𝜕(𝜕𝜇𝐴𝜈)

= −

1

2𝐹𝜌𝛿 (

𝜕(𝜕𝜌𝐴𝛿 − 𝜕𝛿𝐴𝜌)

𝜕(𝜕𝜇𝐴𝜈))

= −

1

2𝐹𝜇𝜈 (

𝜕(𝜕𝜇𝐴𝜈 − 𝜕𝜈𝐴𝜇)

𝜕(𝜕𝜇𝐴𝜈)) −

1

2𝐹𝜈𝜇 (

𝜕(𝜕𝜈𝐴𝜇 − 𝜕𝜇𝐴𝜈)

𝜕(𝜕𝜇𝐴𝜈))

= 22 −

1

2𝐹𝜇𝜈 +

1

2𝐹𝜈𝜇

= −

1

2𝐹𝜇𝜈 −

1

2𝐹𝜇𝜈

= −𝐹𝜇𝜈 ⇒ 𝜕𝜇(𝐹

𝜇𝜈) = 0

if 𝜈 is space-like we get 𝜕𝜇(𝐹

𝜇𝜈) = 𝜕𝜇(𝐹𝜇𝑖) = 𝜕0(𝐹

0𝑖) + 𝜕𝑗(𝐹𝑗𝑖) = −𝜕0𝐸

𝑖 + 𝜕𝑗(−𝜖𝑗𝑖𝑘𝐵𝑘) = 0

⇒ 0 = {

−𝜕0𝐸1 + 𝜕𝑗(−𝜖

𝑗1𝑘𝐵𝑘)

−𝜕0𝐸2 + 𝜕𝑗(−𝜖

𝑗2𝑘𝐵𝑘)

−𝜕0𝐸3 + 𝜕𝑗(−𝜖

𝑗3𝑘𝐵𝑘)

}

= {

−𝜕0𝐸1 + 𝜕3(−𝜖

312𝐵2) + 𝜕2(−𝜖213𝐵3)

−𝜕0𝐸2 + 𝜕3(−𝜖

321𝐵1) + 𝜕1(−𝜖123𝐵3)

−𝜕0𝐸3 + 𝜕2(−𝜖

231𝐵1) + 𝜕1(−𝜖132𝐵2)

}

= −𝜕0�̅� + {

−𝜕3(𝐵2) + 𝜕2(𝐵

3)

𝜕3(𝐵1) − 𝜕1(𝐵

3)

−𝜕2(𝐵1) + 𝜕1(𝐵

2)

}

= −𝜕0�̅� + ∇̅ × �̅� ⇒ ∇̅ × �̅� = −𝜕0�̅� (IV) d) If 𝜇 is space-like we get and 𝜈 is space-like we get 𝜕𝜇(𝐹

𝜇𝜈) = 𝜕𝑖(𝐹𝑖0) = −𝜕𝑖𝐸

𝑖 = −div �̅� = 0 (I)

Collecting the results

22 𝐹𝜈𝜇 = 𝜕𝜈𝐴𝜇 − 𝜕𝜇𝐴𝜈 = −(𝜕𝜇𝐴𝜈 − 𝜕𝜈𝐴𝜇) = −𝐹𝜇𝜈

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∇ ⋅ �̅� = 230 (I)

∇ ⋅ �̅� = 0 (II)

∇ × �̅� = −𝜕�̅�

𝜕𝑡 (III)

∇ × �̅� = 24

𝜕�̅�

𝜕𝑡 (IV)

9.4 25The 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.26 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

⇒ 27 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

23 Recall, if a charge is present the equation is Gauss law: ∇ ⋅ �̅� =

𝜌

0

24 Recall, if a charge is present the equation is Amperes law: ∇ × �̅� = 𝜇0𝑗̅ + 𝜇0휀0𝜕�̅�

𝜕𝑡

25 (McMahon, 2006, p. 326), final exam 14, the answer to Final Exam quiz 14 is (a). 26 http://en.wikipedia.org/wiki/G%C3%B6del_metric 27 𝑔𝑎𝑏𝑇𝑎𝑏 =

𝜌

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

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The Basis one forms

𝜔�̂� =1

√2𝜔(𝑑𝑡 + 𝑒𝑥𝑑𝑧) 𝑑𝑡 = √2𝜔𝜔�̂� − 2𝜔𝜔�̂�

휂𝑖𝑗 = {

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�̂� = 𝑑

1

√2𝜔(𝑑𝑡 + 𝑒𝑥𝑑𝑧)

=

𝑒𝑥

√2𝜔𝑑𝑥 ∧ 𝑑𝑧

=

𝑒𝑥

√2𝜔(√2𝜔𝜔𝑥) ∧ (2𝜔𝑒−𝑥𝜔�̂�)

= 2𝜔𝜔𝑥 ∧ 𝜔�̂� = −𝜔𝜔 �̂� ∧ 𝜔𝑥 +𝜔𝜔�̂� ∧ 𝜔�̂� = −Γ �̂�

�̂� ∧ 𝜔𝑥 − Γ �̂��̂� ∧ 𝜔�̂� − Γ �̂�

�̂� ∧ 𝜔 �̂�

𝑑𝜔𝑥 = 0 𝑑𝜔�̂� = 0

𝑑𝜔�̂� = 𝑑 (1

2𝜔𝑒𝑥𝑑𝑧)

=

1

2𝜔𝑒𝑥𝑑𝑥 ∧ 𝑑𝑧

=

1

2𝜔𝑒𝑥(√2𝜔𝜔�̂�) ∧ (2𝜔𝑒−𝑥𝜔�̂�)

= √2𝜔𝜔𝑥 ∧ 𝜔�̂�

Summarizing the curvature one forms in a matrix:

Γ �̂��̂� =

{

0 𝜔𝜔�̂� 0 −𝜔𝜔𝑥

𝜔𝜔 �̂� 0 0 √2𝜔𝜔�̂�

0 0 0 0

−𝜔𝜔�̂� −√2𝜔𝜔�̂� 0 0 }

Where �̂� refers to column and �̂� to row. The curvature two forms and the Riemann tensor:

Ω �̂��̂� = 𝑑Γ �̂�

�̂� + Γ 𝑐̂�̂� ∧ Γ �̂�

𝑐̂ =1

2𝑅 �̂�𝑐̂�̂��̂� 𝜔𝑐̂ ∧ 𝜔�̂� (5.27), (5.28)

Ω �̂��̂� : 𝑑Γ �̂�

�̂� = 0

Γ 𝑐̂�̂� ∧ Γ �̂�

𝑐̂ = 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(√2𝜔𝜔𝑥) ∧ 2𝜔𝑒−𝑥𝜔�̂�

= √2𝜔2𝜔�̂� ∧ 𝜔�̂�

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Γ 𝑐̂𝑥 ∧ Γ �̂�

𝑐̂ = 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 𝑐̂�̂�∧ Γ �̂�

𝑐̂ = Γ �̂��̂�∧ Γ �̂�

�̂� + Γ �̂��̂�∧ Γ �̂�

𝑥 + Γ �̂��̂�∧ Γ �̂�

�̂�+ Γ �̂�

�̂�∧ Γ �̂�

�̂� = 0

⇒ Ω �̂��̂�

= 0

Ω �̂��̂� : 𝑑Γ �̂�

�̂� = 𝑑(−𝜔𝜔𝑥) = 𝑑 (−𝜔1

√2𝜔𝑑𝑥)

Γ 𝑐̂�̂� ∧ Γ �̂�

𝑐̂ = Γ �̂��̂� ∧ Γ �̂�

�̂� + Γ �̂��̂� ∧ Γ �̂�

𝑥 + Γ �̂��̂� ∧ Γ �̂�

�̂�+ Γ �̂�

�̂� ∧ Γ �̂��̂� = 0

⇒ Ω �̂��̂� = 0

Ω 𝑥𝑥 : 𝑑Γ �̂�

𝑥 = 0

Γ 𝑐̂𝑥 ∧ Γ �̂�

𝑐̂ = Γ �̂�𝑥 ∧ Γ �̂�

�̂� + Γ �̂��̂� ∧ Γ �̂�

𝑥 + Γ �̂�𝑥 ∧ Γ �̂�

�̂�+ Γ �̂�

𝑥 ∧ Γ �̂� �̂� = 0

⇒ Ω 𝑥𝑥 = 0

Ω �̂�𝑥 : 𝑑Γ �̂�

𝑥 = 0

Γ 𝑐̂𝑥 ∧ Γ �̂�

𝑐̂ = 0

⇒ Ω �̂�𝑥 = 0

Ω �̂��̂�: 𝑑Γ �̂�

�̂� = 0

Γ 𝑐̂�̂�∧ Γ �̂�

𝑐̂ = Γ �̂��̂�∧ Γ �̂�

�̂� + Γ �̂��̂�∧ Γ �̂�

𝑥 + Γ �̂��̂�∧ Γ �̂�

�̂�+ Γ �̂�

�̂�∧ Γ �̂�

�̂� = 0

⇒ Ω �̂��̂�

= 0

Ω 𝑥�̂� : 𝑑Γ �̂�

�̂� = 𝑑(√2𝜔𝜔�̂�)

= 𝑑 (1

√2𝑒𝑥𝑑𝑥 ∧ 𝑑𝑧)

=1

√2𝑒𝑥(√2𝜔𝜔𝑥) ∧ (2𝜔𝑒−𝑥𝜔�̂�)

= 2𝜔2𝜔�̂� ∧ 𝜔�̂�

Γ 𝑐̂�̂� ∧ Γ �̂�

𝑐̂ = Γ �̂��̂� ∧ Γ �̂�

�̂� + Γ �̂��̂� ∧ Γ �̂�

𝑥 + Γ �̂��̂� ∧ Γ �̂�

�̂�+ Γ �̂�

�̂� ∧ Γ �̂��̂�

= Γ �̂��̂� ∧ Γ �̂�

�̂�

= (−𝜔𝜔𝑥) ∧ (𝜔𝜔�̂�)

= −𝜔2𝜔�̂� ∧ 𝜔�̂�

⇒ Ω 𝑥�̂� = 𝜔2𝜔�̂� ∧ 𝜔�̂�

Summarized in a matrix:

Ω �̂��̂� = {

0 0 0 00 0 0 𝜔2𝜔�̂� ∧ 𝜔 �̂�

0 0 0 00 −𝜔2𝜔𝑥 ∧ 𝜔�̂� 0 0

}

Where �̂� refers to column and �̂� to row

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 �̂�𝑐̂�̂�

𝑐̂ (4.46)

𝑅𝑥𝑥 = 𝑅 �̂�𝑐̂�̂�𝑐̂ = 𝑅 �̂��̂�𝑥

�̂� = 𝑅 �̂�𝑥�̂�𝑥 = −𝜔2

𝑅�̂��̂� = 𝑅 �̂�𝑐̂�̂�𝑐̂ = 𝑅 �̂�𝑥�̂�

𝑥 = −𝜔2

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Summarized in a matrix:

𝑅�̂��̂� = {

0 0 0 00 −𝜔2 0 00 0 0 00 0 0 −𝜔2

}

Where �̂� refers to column and �̂� to row The Ricci scalar: 𝑅 = 휂�̂��̂�𝑅�̂��̂� (4.47)

𝑅 = 휂�̂��̂�𝑅�̂��̂� + 휂�̂��̂�𝑅�̂��̂� + 휂

�̂��̂�𝑅�̂��̂� + 휂�̂��̂�𝑅�̂��̂� = −𝑅�̂��̂� − 𝑅�̂��̂� = 2𝜔

2

Now we can find Λ

Λ = −1

4(8𝜋𝐺𝜌 + 𝑅) = −

1

4(8𝜋𝐺𝜌 + 2𝜔2)

If we use geometrized units28 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.

9.5 29The Einstein Cylinder

9.5.1 The line element The Einstein cylinder has the line element 𝑑𝑠2 = −𝑑𝑡2 + (𝑎0)

2(𝑑휃2 + sin2 휃 (𝑑𝜙2 + sin2𝜙𝑑𝜓2)) (IV.11.1)

Make the coordinate transformation 𝑟 = sin휃 ⇒ 𝑑𝑟 = 𝑑(sin휃) = cos 휃 𝑑휃

⇒ 𝑑𝑠2 = −𝑑𝑡2 + (𝑎0)2 (

𝑑𝑟2

cos2 휃+ 𝑟2(𝑑𝜙2 + sin2𝜙𝑑𝜓2))

= −𝑑𝑡2 + (𝑎0)

2 (𝑑𝑟2

1 − sin2 휃+ 𝑟2(𝑑𝜙2 + sin2𝜙𝑑𝜓2))

= −𝑑𝑡2 + (𝑎0)

2 (𝑑𝑟2

1 − 𝑟2+ 𝑟2(𝑑𝜙2 + sin2𝜙𝑑𝜓2)) (IV.11.2)

9.5.2 The Ricci tensor To find the Ricci tensor we can compare with calculations made for the Robertson Walker and find a proper transformation of the coordinates. The Einstein cylinder

𝑑𝑠2 = −𝑑𝑡2 +(𝑎0)

2

1 − 𝑟2𝑑𝑟2 + (𝑎0𝑟)

2𝑑𝜙2 + (𝑎0𝑟)2 sin2𝜙𝑑𝜓2

The Robertson Walker line element

𝑑𝑠2 = −𝑑𝑡′2+

𝑎2(𝑡′)

1 − 𝑘𝑟′2𝑑𝑟′

2+ 𝑎2(𝑡′)𝑟′

2𝑑휃′

2+ 𝑎2(𝑡′)𝑟′

2sin2 휃′ 𝑑𝜙′

2

Comparing the line elements 𝑑𝑡 = 𝑑𝑡′

28 http://en.wikipedia.org/wiki/Geometrized_unit_system 29 (Choquet-Bruhat, 2015, s. 95) Problem IV.11.1.

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𝑎0

√1 − 𝑟2 𝑑𝑟 =

𝑎(𝑡′)

√1 − 𝑘𝑟′2𝑑𝑟′

𝑎0𝑟𝑑𝜙 = 𝑎(𝑡′)𝑟′𝑑휃′ 𝑎0𝑟 sin𝜙 𝑑𝜓 = 𝑎(𝑡′)𝑟′ sin휃′ 𝑑𝜙′ A proper transformation would be 𝑡′ = 𝑡 𝑎(𝑡′) = 𝑎0 𝑘 = 1 𝑟′ = 𝑟 휃′ = 𝜙 𝜙′ = 𝜓 The Ricci tensor in the non-coordinate basis is30

𝑅�̂��̂� = −3�̈�

𝑎= 0

𝑅�̂��̂� =

�̈�

𝑎+ 2

(�̇�2 + 𝑘)

𝑎2=2

𝑎02

𝑅�̂��̂� =

�̈�

𝑎+ 2

(�̇�2 + 𝑘)

𝑎2=2

𝑎02

𝑅�̂��̂� =

�̈�

𝑎+ 2

(�̇�2 + 𝑘)

𝑎2=2

𝑎02

The Ricci scalar

𝑅 = 휂�̂��̂�𝑅�̂��̂� = 휂�̂��̂�𝑅�̂��̂� + 휂

�̂��̂�𝑅�̂��̂� + 휂�̂��̂�𝑅�̂��̂� + 휂

�̂��̂�𝑅�̂��̂� = 3 ⋅2

𝑎02

To find the Ricci tensor in the coordinate basis we need the basis one-forms

𝜔�̂� = 𝑑𝑡

𝜔�̂� =𝑎0

√1 − 𝑟2𝑑𝑟 𝑑𝑟 =

√1 − 𝑟2

𝑎0𝜔�̂�

𝜔�̂� = 𝑎0𝑟𝑑𝜙 𝑑𝜙 =1

𝑎0𝑟𝜔�̂� 휂𝑖𝑗 = {

−11

11

}

𝜔�̂� = 𝑎0𝑟 sin𝜙 𝑑𝜓 𝑑𝜓 =1

𝑎0𝑟 sin𝜙𝜔�̂�

The transformation 𝑅𝑎𝑏 = Λ 𝑎

𝑐̂ Λ 𝑏�̂� 𝑅𝑐̂�̂�

⇒ 𝑅𝑡𝑡 = 0

𝑅𝑟𝑟 = Λ 𝑟𝑐̂ Λ 𝑟

�̂� 𝑅𝑐̂�̂� = (Λ 𝑟�̂� )

2𝑅�̂��̂� =

𝑎02

1 − 𝑟22

𝑎02 =

2

1 − 𝑟2

𝑅𝜙𝜙 = Λ 𝜙

𝑐̂ Λ 𝜙�̂� 𝑅𝑐̂�̂� = (Λ 𝜙

�̂�)2

𝑅�̂��̂� = (𝑎0𝑟)22

𝑎02 = 2𝑟

2

𝑅𝜓𝜓 = Λ 𝜓

𝑐̂ Λ 𝜓�̂� 𝑅𝑐̂�̂� = (Λ 𝜓

�̂�)2

𝑅�̂��̂� = (𝑎0𝑟 sin𝜙)22

𝑎02 = 2𝑟

2 sin2𝜙

Summarized in a matrix

30 The chapter is named: The Robertson Walker metric

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𝑅𝑎𝑏 =

{

0 0 0 0

02

1 − 𝑟20 0

0 0 2𝑟2 00 0 0 2𝑟2 sin2𝜙}

(IV.11.3)

Where 𝑎 refers to column and 𝑏 to row

9.5.3 31The Einstein Equations To find the Einstein tensor we once more copy the result from the Robertson Walker metric

𝐺�̂��̂� = 3�̇�2 + 𝑘

𝑎2=3

𝑎02

𝐺�̂��̂� = −(2

�̈�

𝑎+𝑎2̇ + 𝑘

𝑎2) = −

1

𝑎02

𝐺�̂��̂� = −(2

�̈�

𝑎+�̇�2 + 𝑘

𝑎2) = −

1

𝑎02

𝐺�̂��̂� = −(2

�̈�

𝑎+�̇�2 + 𝑘

𝑎2) = −

1

𝑎02

Summarized in a matrix

𝐺�̂��̂� =

{

3

𝑎02 0 0 0

0 −1

𝑎02 0 0

0 0 −1

𝑎02 0

0 0 0 −1

𝑎02}

Where 𝑎 refers to column and 𝑏 to row In case of a perfect fluid, the stress energy tensor is 𝑇�̂��̂� = 𝜇𝑢�̂�𝑢�̂� + 𝑝(휂�̂��̂� + 𝑢�̂�𝑢�̂�) (IV.2.8) = 𝜇휂�̂��̂�𝑢

�̂�𝑢�̂� + 𝑝(휂�̂��̂� + 휂�̂��̂�𝑢�̂�𝑢�̂�)

⇒ 𝑇�̂��̂� = 32𝜇휂�̂��̂�𝑢�̂�𝑢�̂� + 𝑝(휂�̂��̂� + 휂�̂��̂�𝑢

�̂�𝑢�̂�)

= 𝜇(−1)(−1) + 𝑝((−1) + (−1)(−1))

= 𝜇 𝑇�̂��̂� = 𝜇휂�̂��̂�𝑢

�̂�𝑢�̂� + 𝑝(휂�̂��̂� + 휂�̂��̂�𝑢�̂�𝑢�̂�) = 𝑝

𝑇�̂��̂� = 𝜇휂�̂��̂�𝑢�̂�𝑢�̂� + 𝑝(휂�̂��̂� + 휂�̂��̂�𝑢

�̂�𝑢�̂�) = 𝑝

𝑇�̂��̂� = 𝜇휂�̂��̂�𝑢�̂�𝑢�̂� + 𝑝(휂�̂��̂� + 휂�̂��̂�𝑢

�̂�𝑢�̂�) = 𝑝

Summarized in a matrix

𝑇�̂��̂� = {

𝜇 0 0 00 𝑝 0 00 0 𝑝 00 0 0 𝑝

}

Where 𝑎 refers to column and 𝑏 to row

31 (Choquet-Bruhat, 2015) Problem IV.11.2 32 𝑢�̂� = (1,0,0,0), 𝑢

�̂� = 휂�̂��̂�𝑢�̂� = (−1,0,0,0)

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⇒ 𝑝 = −1

𝑎02 < 0

(IV.11.4)

𝜇 =3

𝑎02

9.5.4 The Einstein tensor with a cosmological constant A negative pressure is problematic in classical physics and we include a cosmological constant. The Einstein equation with a cosmological constant 𝐺�̂��̂� = 𝑇�̂��̂� − 휂�̂��̂�Λ (IV.3.2)

{

3

𝑎02 0 0 0

0 −1

𝑎02 0 0

0 0 −1

𝑎02 0

0 0 0 −1

𝑎02}

= {

𝜇 0 0 00 𝑝 0 00 0 𝑝 00 0 0 𝑝

} − Λ{

−11

11

}

⇒ 𝜇 =3

𝑎02 − Λ

𝑝 = −

1

𝑎02 + Λ

9.6 33The Newtonian Approximation – The right hand side! The newtionian approximation is characterized by a weak gravitational field34 and bodies of low masses and velocities.

The Einstein space-time in a weak gravitational field can be described by the linearized metric 𝑑𝑠2 = 𝑔𝑎𝑏𝑑𝑥

𝑎𝑑𝑥𝑏 𝑔𝑎𝑏 = 휂𝑎𝑏 + 𝜖ℎ𝑎𝑏 𝜖 ≪ 1 (13.1)35

We look at a particle moving along 𝑥1 with velocity 𝑣 =𝑑𝑥1

𝑑𝑥0≪ 𝑐

⇒ 𝑑𝑠2 = 𝑔𝑜𝑜𝑑𝑥0 + (𝑔01 + 𝑔10)𝑑𝑥1𝑑𝑡 + 𝑔11𝑑𝑥1

2

⇒ (𝑑𝑠

𝑑𝑥0)2

= 𝑔𝑜𝑜 + (𝑔01 + 𝑔10) (𝑑𝑥1𝑑𝑥0

) + 𝑔11 (𝑑𝑥1𝑑𝑥0

)2

= 𝑔𝑜𝑜 + (𝑔01 + 𝑔10)𝑣 + 𝑔11𝑣2

= (−1 + 𝜖ℎ00) + (𝜖ℎ01 + 𝜖ℎ10)𝑣 + (1 + 𝜖ℎ11)𝑣2

→ {−1 + 𝜖ℎ00 𝑖𝑓 𝑣 → 0

𝜖(ℎ00 + ℎ01 + ℎ10 + ℎ11) 𝑖𝑓 𝑣 → 1

(I)

i.e. in the approximation where 𝑣 ≪ 𝑐, the time component ℎ00 is dominant with respect to the space-components (ℎ01, ℎ10, ℎ11).

36The Ricci tensor: In this linearized theory we have the Christoffel symbols

33 (Choquet-Bruhat, 2015, s. 75) Exercise IV.5.1. 34 For a detailed calculation of the Christoffel symbols, the Riemann and Ricci tensors, the Ricci scalar and the Einstein equation (The left hand side!) see a later chapter named “Linearized metric” 35 (McMahon, 2006, p. 280) 36 (Choquet-Bruhat, 2015, s. 75) Chapter IV.5.1.

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Γ 𝑏𝑐𝑎 =

1

2𝜖휂𝑎𝑑 (

𝜕ℎ𝑏𝑑𝜕𝑥𝑐

+𝜕ℎ𝑐𝑑𝜕𝑥𝑏

−𝜕ℎ𝑏𝑐𝜕𝑥𝑑

)

(13.3)37

𝑅𝑎𝑏 = 𝜕𝑐Γ 𝑎𝑏𝑐 − 𝜕𝑏Γ 𝑎𝑐

𝑐 38 ⇒ 𝑅00 = 𝜕𝑐Γ 00

𝑐 − 𝜕0Γ 0𝑐𝑐

Assuming39 the time derivatives 𝜕0ℎ𝑎𝑏 are small compared to the space derivatives 𝜕𝑖ℎ𝑎𝑏 ⇒ 𝑅00 = 𝜕𝑐Γ 00

𝑐

= 𝜕0Γ 000 + ∑ 𝜕𝑖Γ 00

𝑖

1,2,3

𝑖

= ∑ 𝜕𝑖 (1

2𝜖휂𝑖𝑑 (

𝜕ℎ0𝑑𝜕𝑥0

+𝜕ℎ0𝑑𝜕𝑥0

−𝜕ℎ00𝜕𝑥𝑑

))

1,2,3

𝑖

=1

2𝜖 ∑ −

𝜕2ℎ00𝜕𝑥𝑖𝜕𝑥𝑖

1,2,3

𝑖

= 40 −

1

2𝜖∇2ℎ00

Notice

Γ 00𝑖 = −

1

2𝜖𝜕ℎ00𝜕𝑥𝑖

In another chapter41 we found out that

𝑅𝑎𝑏 = 𝜅 (𝑇𝑎𝑏 −1

2𝑔𝑎𝑏𝑇𝑎𝑏𝑔𝑎𝑏) = 𝜅𝜌𝑎𝑏 (IV.2.5)

The stress-energy tensor is 𝑇𝑎𝑏 = 𝜇𝑢𝑎𝑢𝑏

In the simplest case, pure matter no pressure42, 𝑢𝑎 = (𝑑𝑥0

𝑑𝑠,𝑑𝑥𝑖

𝑑𝑠) = (−1,0,0,0) and the only non-zero ele-

ment in the stress energy tensor is 𝑇00 = 𝜇(𝑢0)

2 = 𝜇 And

𝑇 = 𝑔𝑎𝑏𝑇𝑎𝑏 = 𝑔00𝜇

⇒ 𝜌00 = 𝑇00 −1

2𝑇𝑔00 = 43𝜇 −

1

2𝜇𝑔00𝑔00 =

1

2𝜇

(IV.5.2)

Now, because

𝑅00 = −1

2𝜖∇2ℎ00 = 𝜅𝜌00 =

1

2𝜅𝜇

⇒ 𝜖∇2ℎ00 = 44 −1

2𝐺𝐸𝜇

45Which is the equivalent of Poisson’s equation for gravity

∇2𝜙(𝑟) = 46 − 4𝜋𝐺𝑁𝜌

37 (McMahon, 2006, p. 282) 38 (McMahon, 2006) 39 I don’t know how to prove this except to state, that we in the Newtonian approximation work in a regime where frequencies are low and wavelengths long.

40 ∇2= ∑𝜕2

𝜕𝑥𝑖𝜕𝑥𝑖1,2,3𝑖

41 ”The Einstein equation with source.” 42 Chapter “Pure matter” 43 𝑔00𝑔00 ≈ 1 because the off-diagonal elements are ≪ 1. 44 Renaming 𝜅 = 𝐺𝐸 45 http://mathworld.wolfram.com/PoissonsEquation.html 46 If 𝜖ℎ00 = 𝜙 and 𝐺𝐸 = 8𝜋𝐺𝑁

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The equation of motion: We can also find the equation of motion in this approximation. In GR a test particle follows geodesics described by

𝑑2𝑥𝑎

𝑑𝑠2+ Γ 𝑏𝑐

𝑎𝑑𝑥𝑏

𝑑𝑠

𝑑𝑥𝑐

𝑑𝑠 = 0

(4.34)47

⇒ 𝑑2𝑥𝑎

𝑑𝑠2 = −Γ 𝑏𝑐

𝑎𝑑𝑥𝑏

𝑑𝑠

𝑑𝑥𝑐

𝑑𝑠

⇒ 48 𝑑2𝑥𝑖

(𝑑𝑥0)2 = −Γ 00

𝑖 (𝑑𝑥0

𝑑𝑠)

2

≅ 49 − Γ 00𝑖 (−1)2 =

1

2𝜖𝜕ℎ00𝜕𝑥𝑖

(IV.5.4)

Again we see the equivalence to Newton’s laws, where the left side represent the force and the right side the gradient of the potential. Recall

�̅� = ∇𝑈 with

𝑈 = −𝐺𝑀𝑚

𝑟

leads to Newton’s famous equation

𝐹 =𝐺𝑀𝑚

𝑟2

10 Null Tetrads and the Petrov Classification

10.1 Weyl scalars and Petrov classification Petrov Classifica-tion50

Weyl scalars51 PND52

Type I: Alg. gen-eral

[1,1,1,1] Ψ0, Ψ1, Ψ2, Ψ3, Ψ4≠ 0

Type II [2,1,1] Ψ0, Ψ1 = 0 Ψ2, Ψ3, Ψ4 ≠ 0

𝑙𝑎 , 𝑙𝑎

Type D or de-gener-ate

[2,2] Ψ0, Ψ1,Ψ3, Ψ4 = 0

Ψ2 ≠ 0 𝑙𝑎 , 𝑙𝑎 , 𝑛𝑎, 𝑛𝑎 Gravitational field of a star or black hole

(Schwarzschild or Kerr vacuum). The two princi-pal null directions correspond to ingoing and outgoing congruences of light rays. 53The fact that the spacetime contains Ψ2 and not Ψ4 or Ψ0 indicates that this spacetime de-scribes electromagnetic fields and not gravita-tional radiation. This spacetime represents a vacuum universe that contains electromagnetic fields with no matter.

47 (McMahon, 2006, p. 82)

48 𝑑2𝑥𝑖

𝑑𝑠2=

𝑑

𝑑𝑠(𝑑𝑥𝑖

𝑑𝑠) =

𝑑

𝑑𝑠(𝑑𝑥𝑖

𝑑𝑥0

𝑑𝑥0

𝑑𝑠) = −

𝑑

𝑑𝑠(𝑑𝑥𝑖

𝑑𝑥0) = −

𝑑

𝑑𝑥0(𝑑𝑥𝑖

𝑑𝑠) = −

𝑑

𝑑𝑥0(𝑑𝑥𝑖

𝑑𝑥0

𝑑𝑥0

𝑑𝑠) =

𝑑

𝑑𝑥0(𝑑𝑥𝑖

𝑑𝑥0)

49 Eq. (I) 50 (Scolarpedia, n.d.), (McMahon, 2006) 51 If 𝑛 = 4, 𝑅𝑎𝑏𝑐𝑑 has twenty independent component – ten of which are given by 𝑅𝑎𝑏 and the remaining ten by the Weyl tensor (d'Inverno, 1992, p. 87) 52 Principal Null Directions 53 (McMahon, p. 321)

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Type III [3,1] Ψ0, Ψ1, Ψ2 = 0 Ψ3, Ψ4 ≠ 0

𝑙𝑎 , 𝑙𝑎 , 𝑙𝑎 Longitudinal gravity waves with shear due to tidal effects.

Type IV or N

[4] Ψ0, Ψ1, Ψ2, Ψ3= 0 Ψ4 ≠ 0

𝑙𝑎,𝑙𝑎 , 𝑙𝑎 , 𝑙𝑎 Transverse gravity waves, single principal null di-rection of multiplicity 4

54Ψ1, Ψ2, Ψ3, Ψ4= 0 Ψ0 ≠ 0

𝑛𝑎,𝑛𝑎, 𝑛𝑎 , 𝑛𝑎 Gravitational wave region III

Type O Ψ0, Ψ1, Ψ2, Ψ3, Ψ4= 0

In the context of gravitational radiation, we have the following interpretations: Ψ0 transverse wave component in the 𝑛𝑎 direction Ψ1 longitudinal wave component in the 𝑛𝑎 direc-

tion Ψ2 denotes a coulomb component Ψ3 transverse wave component in the 𝑙𝑎 direction Ψ4 longitudinal wave component in the 𝑙𝑎 direc-

tion

10.2 55Construct a null tetrad for the flat space-time Minkowski metric The line element:

𝑑𝑠2 = 𝑑𝑡2 − 𝑑𝑟2 − 𝑟2𝑑휃2 − 𝑟2 sin2 휃 𝑑𝜙2 The metric tensor:

𝑔𝑎𝑏 = {

1−1

−𝑟2

−𝑟2 sin2 휃

}

and its inverse:

𝑔𝑎𝑏 =

{

1

−1

−1

𝑟2

−1

𝑟2 sin2 휃}

The basis one forms

𝜔�̂� = 𝑑𝑡

𝜔�̂� = 𝑑𝑟

𝜔�̂� = 𝑟𝑑휃

𝜔�̂� = 𝑟 sin휃 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

𝑟𝑑휃 + 𝑖𝑟 sin 휃 𝑑𝜙𝑟𝑑휃 − 𝑖𝑟 sin 휃 𝑑𝜙

) (9.10)

Written in terms of the coordinate basis

54 http://www-staff.lboro.ac.uk/~majbg/jbg/book/chap3.pdf 55 (McMahon, 2006, p. 186), example 9-3 and 9-4

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𝑙𝑎 =1

√2(1, 1, 0, 0) 𝑛𝑎 =

1

√2(1, −1, 0, 0)

𝑚𝑎 =1

√2(0, 0, 𝑟, 𝑖𝑟 sin휃) 𝑚𝑎̅̅ ̅̅ =

1

√2(0, 0, 𝑟, −𝑖𝑟 sin휃)

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 휃) ⋅𝑖𝑟 sin 휃

√2= −𝑖

1

𝑟 sin휃 √2

Collecting the results

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 sin휃) 𝑚𝑎 =

1

√2(0, 0, −

1

𝑟, −

𝑖

𝑟 sin휃)

𝑚𝑎̅̅ ̅̅ =1

√2(0, 0, 𝑟, −𝑖𝑟 sin휃) 𝑚𝑎̅̅ ̅̅ =

1

√2(0, 0, −

1

𝑟,

𝑖

𝑟 sin휃)

11 The Schwarzschild Solution See separate document

12 Black Holes See separate document

13 Cosmology See separate document

14 Gravitational Waves See separate document

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15 Appendix A: Tensor Calculus

15.1 56The Einstein Summation Convention

15.1.1 Repeated Indices. (1.15) 𝑎𝑖𝑏𝑖 (𝑛 = 6) = 𝑎1𝑏1 + 𝑎2𝑏2 + 𝑎3𝑏3 + 𝑎4𝑏4 + 𝑎4𝑏4 + 𝑎6𝑏6

(1.16) 𝑅 𝑗𝑘𝑖𝑖 (𝑛 = 4) = 𝑅 𝑗𝑘1

1 + 𝑅 𝑗𝑘22 + 𝑅 𝑗𝑘3

3 + 𝑅 𝑗𝑘44

= 𝑅 1𝑘1

1 + 𝑅 1𝑘22 + 𝑅 1𝑘3

3 + 𝑅 1𝑘44 + 𝑅 2𝑘1

1 + 𝑅 2𝑘22 + 𝑅 2𝑘3

3 + 𝑅 2𝑘44 + 𝑅 3𝑘1

1 + 𝑅 3𝑘22

+ 𝑅 3𝑘33 + 𝑅 3𝑘4

4 + 𝑅 4𝑘11 + 𝑅 4𝑘2

2 + 𝑅 4𝑘33 + 𝑅 4𝑘4

4 = 𝑅 111

1 + 𝑅 1122 + 𝑅 113

3 + 𝑅 1144 + 𝑅 211

1 + 𝑅 2122 + 𝑅 213

3 + 𝑅 2144 + 𝑅 311

1 + 𝑅 3122

+ 𝑅 3133 + 𝑅 314

4 + 𝑅 4111 + 𝑅 412

2 + 𝑅 4133 + 𝑅 414

4 + 𝑅 1211 + 𝑅 122

2

+ 𝑅 1233 + 𝑅 124

4 + 𝑅 2211 + 𝑅 222

2 + 𝑅 2233 + 𝑅 224

4 + 𝑅 3211 + 𝑅 322

2

+ 𝑅 3233 + 𝑅 324

4 + 𝑅 4211 + 𝑅 422

2 + 𝑅 4233 + 𝑅 424

4 + 𝑅 1311 + 𝑅 132

2

+ 𝑅 1333 + 𝑅 134

4 + 𝑅 2311 + 𝑅 232

2 + 𝑅 2333 + 𝑅 234

4 + 𝑅 3311 + 𝑅 332

2

+ 𝑅 3333 + 𝑅 334

4 + 𝑅 4311 + 𝑅 432

2 + 𝑅 4333 + 𝑅 434

4 + 𝑅 1411 + 𝑅 142

2

+ 𝑅 1433 + 𝑅 4𝑘4

4 + 𝑅 2411 + 𝑅 242

2 + 𝑅 2433 + 𝑅 244

4 + 𝑅 3411 + 𝑅 342

2

+ 𝑅 3433 + 𝑅 344

4 + 𝑅 4411 + 𝑅 442

2 + 𝑅 4433 + 𝑅 444

4 Dummy index: 𝑖. Free indices: 𝑗, 𝑘. Number of summations: 4 ⋅ 42 = 64 (1.19) 𝑎13𝑏13 + 𝑎23𝑏23 + 𝑎33𝑏33 = 𝑎𝑖3𝑏𝑖3 (𝑛 = 3 )

(1.20) 𝑎11(𝑥1)

2 + 𝑎12𝑥1𝑥2 + 𝑎13𝑥1𝑥3 + 𝑎21𝑥2𝑥1 + 𝑎22(𝑥2)2 + 𝑎23𝑥2𝑥3 + 𝑎31𝑥3𝑥1 + 𝑎32𝑥3𝑥2

+ 𝑎33(𝑥33)2

= 𝑎𝑖𝑗𝑥𝑖𝑥𝑗 (𝑛 = 3)

Dummy indices : 𝑖, 𝑗. Number of summations: 33 = 9

15.1.2 Substitutions (1.5) 𝑦𝑖 = 𝑎𝑖𝑗𝑥𝑗

⇒ 𝑔𝑖𝑗𝑦𝑖𝑦𝑗 = 𝑔𝑖𝑗(𝑎𝑖𝑘𝑥𝑘)(𝑎𝑗𝑙𝑥𝑙) = 𝑔𝑖𝑗𝑎𝑖𝑘𝑎𝑗𝑙𝑥𝑙𝑥𝑘

(1.27a) 𝑦𝑖 = 𝑇 𝑖𝑗𝑗

⇒ 𝑏 𝑗𝑖 𝑦𝑖 = 𝑏 𝑗

𝑖 𝑇 𝑖𝑘𝑘

(1.27b) 𝑦𝑖 = 𝑏𝑖𝑗𝑥𝑗

⇒ 𝑎𝑖𝑗𝑦𝑗 = 𝑎𝑖𝑗𝑏𝑗𝑘𝑥𝑗𝑘

(1.27c) 𝑦𝑖 = 𝑏𝑖𝑗𝑥𝑗

⇒ 𝑎𝑖𝑗𝑘𝑦𝑖𝑦𝑗𝑦𝑘 = 𝑎𝑖𝑗𝑘(𝑏𝑖𝑙𝑥𝑙)(𝑏𝑗𝑚𝑥𝑚)(𝑏𝑘𝑛𝑥𝑛) = 𝑎𝑖𝑗𝑘𝑏𝑖𝑙𝑏𝑗𝑚𝑏𝑘𝑛𝑥𝑙𝑥𝑚𝑥𝑛

15.1.3 Double Sums (1.21) 𝑦1 = 𝑐11𝑥1 + 𝑐12𝑥2 𝑦2 = 𝑐21𝑥1 + 𝑐22𝑥2 ⇒ 𝑦𝑖 = 𝑐𝑖𝑗𝑥𝑗 (𝑛 = 2)

15.1.4 Kronecker Delta

(1.8) 𝜕

𝜕𝑥𝑘(𝑎𝑖𝑗𝑥𝑖𝑥𝑗) =

𝜕

𝜕𝑥𝑘(𝑎𝑘𝑗𝑥𝑘𝑥𝑗) +

𝜕

𝜕𝑥𝑘(𝑎𝑖𝑘𝑥𝑖𝑥𝑘)

= 𝑎𝑘𝑗𝑥𝑗 + 𝑎𝑖𝑘𝑥𝑖 = 𝑎𝑘𝑖𝑥𝑖 + 𝑎𝑖𝑘𝑥𝑖 = (𝑎𝑘𝑖 + 𝑎𝑖𝑘)𝑥𝑖

(1.10) 𝜕

𝜕𝑥𝑘(𝑎𝑖𝑗𝑥𝑖𝑥𝑗) = 𝑎𝑖𝑗𝑥𝑗

𝜕𝑥𝑖𝜕𝑥𝑘

+ 𝑎𝑖𝑗𝑥𝑖𝜕𝑥𝑗

𝜕𝑥𝑘

56 (Kay, 1988, s. 1)

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= 𝑎𝑖𝑗𝑥𝑗𝛿𝑖𝑘 + 𝑎𝑖𝑗𝑥𝑖𝛿𝑗𝑘

= 𝑎𝑘𝑗𝑥𝑗 + 𝑎𝑖𝑘𝑥𝑖 = 𝑎𝑘𝑖𝑥𝑖 + 𝑎𝑖𝑘𝑥𝑖 = (𝑎𝑘𝑖 + 𝑎𝑖𝑘)𝑥𝑖

(1.11) 𝜕2

𝜕𝑥𝑘𝜕𝑥𝑙(𝑎𝑖𝑗𝑥𝑖𝑥𝑗) =

𝜕

𝜕𝑥𝑙((𝑎𝑘𝑖 + 𝑎𝑖𝑘)𝑥𝑖)

=𝜕

𝜕𝑥𝑙((𝑎𝑘𝑙 + 𝑎𝑙𝑘)𝑥𝑙)

= (𝑎𝑘𝑙 + 𝑎𝑙𝑘) = 2𝑎𝑘𝑙

(1.11’) 𝜕2

𝜕𝑥𝑘𝜕𝑥𝑙(𝑎𝑖𝑗𝑥𝑖𝑥𝑗) =

𝜕

𝜕𝑥𝑙((𝑎𝑘𝑖 + 𝑎𝑖𝑘)𝑥𝑖)

= (𝑎𝑘𝑖 + 𝑎𝑖𝑘)𝜕𝑥𝑖𝜕𝑥𝑙

= (𝑎𝑘𝑖 + 𝑎𝑖𝑘)𝛿𝑙𝑖

= (𝑎𝑘𝑙 + 𝑎𝑙𝑘) = 2𝑎𝑘𝑙

(1.17) 𝛿𝑗 𝑖𝑥𝑖 = 𝑥𝑗

(1.18a) 𝛿𝑖𝑖 = 𝑛 (1.18b) 𝛿𝑖𝑗𝛿𝑖𝑗 = 𝛿𝑖𝑗 = 𝛿𝑖𝑖 = 𝑛

(1.18c) 𝛿𝑖𝑗𝛿𝑘𝑗𝑐𝑖𝑘 = 𝛿𝑖𝑗𝑐𝑖𝑗 = 𝑐𝑖𝑖 = 𝑐11 + 𝑐22 +⋯+ 𝑐𝑛𝑛

(1.22) 𝜕

𝜕𝑥𝑘(𝑎11𝑥1 + 𝑎12𝑥2 + 𝑎13𝑥3) =

𝜕

𝜕𝑥𝑘(𝑎1𝑖𝑥𝑖) (𝑛 = 3)

=𝜕

𝜕𝑥𝑘(𝑎1𝑘𝑥𝑘) (𝑛 = 3)

= 𝑎1𝑘 (𝑛 = 3) = 𝑎11 + 𝑎12 + 𝑎13

(1.22’) 𝜕

𝜕𝑥𝑘(𝑎11𝑥1 + 𝑎12𝑥2 + 𝑎13𝑥3) =

𝜕

𝜕𝑥𝑘(𝑎1𝑖𝑥𝑖) (𝑛 = 3)

= 𝑎1𝑖𝜕𝑥𝑖𝜕𝑥𝑘

(𝑛 = 3)

= 𝑎1𝑖𝛿𝑘𝑖 (𝑛 = 3)

= 𝑎1𝑘 (𝑛 = 3) = 𝑎11 + 𝑎12 + 𝑎13

(1.23) 𝜕

𝜕𝑥𝑘(𝑎𝑖𝑗𝑥𝑗) =

𝜕

𝜕𝑥𝑘(𝑎𝑖𝑘𝑥𝑘) = 𝑎𝑖𝑘

(1.23’) 𝜕

𝜕𝑥𝑘(𝑎𝑖𝑗𝑥𝑗) = 𝑎𝑖𝑗

𝜕𝑥𝑗

𝜕𝑥𝑘= 𝑎𝑖𝑗𝛿𝑘

𝑗= 𝑎𝑖𝑘

(1.24) 𝜕

𝜕𝑥𝑘[𝑎𝑖𝑗𝑥𝑖(𝑥𝑗)

2] =

𝜕

𝜕𝑥𝑘[𝑎𝑘𝑗𝑥𝑘(𝑥𝑗)

2] +

𝜕

𝜕𝑥𝑘[𝑎𝑖𝑘𝑥𝑖(𝑥𝑘)

2]

= 𝑎𝑘𝑗(𝑥𝑗)2+ 𝑎𝑖𝑘𝑥𝑖 [

𝜕

𝜕𝑥𝑘(𝑥𝑘)

2]

= 𝑎𝑘𝑗(𝑥𝑗)2+ 2𝑎𝑖𝑘𝑥𝑖𝑥𝑘

= 𝑎𝑘𝑗(𝑥𝑗)2+ 2𝑎𝑗𝑘𝑥𝑗𝑥𝑘

= 𝑎𝑘𝑗 [(𝑥𝑗)2+ 2𝑥𝑗𝑥𝑘]

(1.24’) 𝜕

𝜕𝑥𝑘[𝑎𝑖𝑗𝑥𝑖(𝑥𝑗)

2] = 𝑎𝑖𝑗(𝑥𝑗)

2 𝜕𝑥𝑖𝜕𝑥𝑘

+ 2𝑎𝑖𝑗𝑥𝑖𝑥𝑗𝜕𝑥𝑗

𝜕𝑥𝑘

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= 𝑎𝑖𝑗(𝑥𝑗)2𝛿𝑘𝑖 + 2𝑎𝑖𝑗𝑥𝑖𝑥𝑗𝛿𝑘

𝑗

= 𝑎𝑘𝑗(𝑥𝑗)2+ 2𝑎𝑖𝑘𝑥𝑖𝑥𝑘

= 𝑎𝑘𝑗(𝑥𝑗)2+ 2𝑎𝑗𝑘𝑥𝑗𝑥𝑘

= 𝑎𝑘𝑗 [(𝑥𝑗)2+ 2𝑥𝑗𝑥𝑘]

(1.25) 𝜕

𝜕𝑥𝑙(𝑎𝑖𝑗𝑘𝑥𝑖𝑥𝑗𝑥𝑘) =

𝜕

𝜕𝑥𝑙(𝑎𝑙𝑗𝑘𝑥𝑙𝑥𝑗𝑥𝑘) +

𝜕

𝜕𝑥𝑙(𝑎𝑖𝑙𝑘𝑥𝑖𝑥𝑙𝑥𝑘) +

𝜕

𝜕𝑥𝑙(𝑎𝑖𝑗𝑙𝑥𝑖𝑥𝑗𝑥𝑙)

= 𝑎𝑙𝑗𝑘𝑥𝑗𝑥𝑘 + 𝑎𝑖𝑙𝑘𝑥𝑖𝑥𝑘 + 𝑎𝑖𝑗𝑙𝑥𝑖𝑥𝑗

= 𝑎𝑙𝑖𝑘𝑥𝑖𝑥𝑘 + 𝑎𝑖𝑙𝑘𝑥𝑖𝑥𝑘 + 𝑎𝑖𝑘𝑙𝑥𝑖𝑥𝑘 = (𝑎𝑙𝑖𝑘 + 𝑎𝑖𝑙𝑘 + 𝑎𝑖𝑘𝑙)𝑥𝑖𝑥𝑘

(1.25’) 𝜕

𝜕𝑥𝑙(𝑎𝑖𝑗𝑘𝑥𝑖𝑥𝑗𝑥𝑘) = 𝑎𝑖𝑗𝑘𝑥𝑗𝑥𝑘

𝜕𝑥𝑖𝜕𝑥𝑙

+ 𝑎𝑖𝑗𝑘𝑥𝑖𝑥𝑘𝜕𝑥𝑗

𝜕𝑥𝑙+ 𝑎𝑖𝑗𝑘𝑥𝑖𝑥𝑗

𝜕𝑥𝑘𝜕𝑥𝑙

= 𝑎𝑖𝑗𝑘𝑥𝑗𝑥𝑘𝛿𝑙𝑖 + 𝑎𝑖𝑗𝑘𝑥𝑖𝑥𝑘𝛿𝑙

𝑗+ 𝑎𝑖𝑗𝑘𝑥𝑖𝑥𝑗𝛿𝑙

𝑘

= 𝑎𝑙𝑗𝑘𝑥𝑗𝑥𝑘 + 𝑎𝑖𝑙𝑘𝑥𝑖𝑥𝑘 + 𝑎𝑖𝑗𝑙𝑥𝑖𝑥𝑗

= 𝑎𝑙𝑖𝑘𝑥𝑖𝑥𝑘 + 𝑎𝑖𝑙𝑘𝑥𝑖𝑥𝑘 + 𝑎𝑖𝑘𝑙𝑥𝑖𝑥𝑘 = (𝑎𝑙𝑖𝑘 + 𝑎𝑖𝑙𝑘 + 𝑎𝑖𝑘𝑙)𝑥𝑖𝑥𝑘

16 Appendix B: Collection of results.

1) 57General Surfaces.

a) The line element: 𝑑𝑠2 = 𝑎2(𝑑휃2 + 𝑓2(휃)𝑑𝜙2)

i) Since the line element is the same for all 𝜙, it corresponds to a surface that is axi-symmetric

about an axis.

ii) The circumference 𝐶(휃) of a circle of constant 휃 is 𝐶(휃) = ∮𝑑𝑠 = ∫ 𝑎𝑓(휃)𝑑𝜙2𝜋

0= 2𝜋𝑎𝑓(휃)

iii) The distance from pole to pole is 𝑑 = 𝑎 ∫ 𝑑휃𝜋

0= 𝜋𝑎

2) The Surface of a Sphere.

a) The line element: 𝑑𝑠2 = 𝑎2(𝑑휃2 + sin2 휃 𝑑𝜙2)

i) 𝑓(휃) = sin휃

ii) 𝐶(휃) = 2𝜋𝑎 sin휃 , 𝑑 = 𝜋𝑎

3) The Surface of a peanut.

i) 𝑓(휃) = sin휃 (1 −3

4sin2 휃)

ii) 𝐶(휃) = 2𝜋𝑎 sin휃 (1 −3

4sin2 휃)

4) The Surface of an ellipsoid.

a) The line element: 𝑑𝑠2 = [cos2 휃 (𝑎2 cos2𝜙 + 𝑏2 sin2𝜙) + 𝑐2 sin2 휃]𝑑휃2 + sin2 휃 (𝑎2 sin2𝜙 +

𝑏2 cos2𝜙)𝑑𝜙2 + 2(𝑏2 − 𝑎2)(cos𝜙 cos휃 sin𝜙 sin휃)𝑑휃𝑑𝜙

5) 58The Surface of an Egg.

a) The line element: 𝑑𝑠2 = 𝑎2[(cos2 휃 + 4 sin2 휃)𝑑휃2 + sin2 휃 𝑑𝜙2] 𝑖𝑓 𝑎 = 𝑏 =1

2𝑐

6) 59Cylindindrical Coordinates.

57 (Hartle, Gravity - An introduction to Einstein's General Relativity, 2003, p. 26) 58 (Hartle, Gravity An Introduction to Einstein's General Relativity, 2003) problem 2-8 59 (McMahon, p. 83)

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a) The line element: 𝑑𝑠2 = 𝑑𝑟2 + 𝑟2𝑑𝜙2 + 𝑑𝑧2

b) Geodesic equations.

i) �̈� − 𝑟�̇�2 = 0

ii) �̈� +1

𝑟�̇��̇� +

1

𝑟�̇��̇� = 0

iii) �̈� = 0

7) General non-diagonal space-time.

a) The metric tensor: 𝑔𝑎𝑏 =

{

𝑔12

𝑔12𝑔33

𝑔44}

b) Ricci constant: 𝑅 = 4𝑔12𝑅 1323 + 4𝑔12𝑅 142

4 + 2𝑔33𝑅 3434

8) 60Metric Example 1.

a) Line element: 𝑑𝑠2 = 𝑦2 sin 𝑥 𝑑𝑥2 + 𝑥2 tan 𝑦 𝑑𝑦2

b) Ricci scalar: 𝑅 = (𝑥 cos𝑥 tan2 𝑦+𝑦 sin2 𝑥+𝑦 sin2 𝑥 tan2 𝑦

𝑥2𝑦2 sin2 𝑥 tan2 𝑦)

9) 61Metric Example 2.

a) Line element: 𝑑𝑠2 = 𝑑𝜓2 + sinh2𝜓 𝑑휃2 + sinh2𝜓 sin2 휃 𝑑𝜙2

b) The basis one-forms: 𝜔�̂� = 𝑑𝜓, 𝜔�̂� = sinh𝜓𝑑휃, 𝜔�̂� = sinh𝜓 sin휃 𝑑𝜙

c) Ricci rotation coefficients: Γ �̂��̂�

�̂�= −coth𝜓, Γ �̂��̂�

�̂� = coth𝜓, Γ �̂��̂�

�̂�=

coth𝜓

sin𝜃, Γ �̂��̂�

�̂�= −

coth𝜓

sin𝜃,

Γ �̂��̂��̂� = −

cot𝜃

sinh2𝜓, Γ �̂��̂�

�̂�=

cot𝜃

sinh2𝜓

10) 62Metric Example 3.

a) The line element: 𝑑𝑠2 = (𝑢2 + 𝜈2)𝑑𝑢2 + (𝑢2 + 𝜈2)𝑑𝜈2 + 𝑢2𝜈2𝑑휃2

b) The Riemann tensor: 𝑅𝑎𝑏𝑐𝑑 = 0

11) 63General 4-dimensional space-time.

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

b) The basis one-forms: 𝜔�̂� = 𝑑𝑡, 𝜔�̂� = 𝐿(𝑡, 𝑟)𝑑𝑟, 𝜔�̂� = 𝐵(𝑡, 𝑟)𝑑𝜙, 𝜔�̂� = 𝑀(𝑡, 𝑟)𝑑𝑧

c) The Einstein tensor: 𝐺�̂��̂� =𝐵′𝐿′−𝐵′′𝐿

𝐵𝐿3+�̇��̇�

𝐵𝐿+𝑀´𝐿´−𝑀´´𝐿

𝑀𝐿3+�̇��̇�

𝑀𝐿+�̇��̇�

𝐵𝑀−

𝐵′𝑀′

𝐿2𝐵𝑀, 𝐺�̂��̂� =

𝐵´�̇�

𝐵𝐿2−

𝐵′̇

𝐵𝐿+

𝑀´�̇�

𝑀𝐿2−

�̇�′

𝑀𝐿,

𝐺�̂��̂� = −�̈�

𝐵−�̈�

𝑀+�̇��̇�

𝐵𝑀−

𝐵′𝑀′

𝐿2𝐵𝑀, 𝐺�̂��̂� = −

�̈�

𝐿−�̈�

𝑀

𝑀´𝐿´−𝑀´´𝐿

𝑀𝐿3+�̇��̇�

𝑀𝐿, 𝐺�̂��̂� = −

�̈�

𝐿−�̈�

𝐵+𝐵′𝐿′−𝐵′′𝐿

𝐵𝐿3+�̇��̇�

𝐵𝐿

12) 64The Plane in polar coordinates.

a) The line element: 𝑑𝑆2 = 𝑑𝑟2 + 𝑟2𝑑𝜙2

b) The geodesic equations:

i) �̈� = 𝑟�̇�2

ii) �̈� = −2

𝑟�̇��̇�

13) 65Mathematical singularity: The two-dimensional plane in polar coordinates 𝑑𝑆2 = 𝑑𝑟2 + 𝑟2𝑑𝜙2 can

blow up in a singularity by making the transformation 𝑟 = 𝑎2/𝑟′ for some constant 𝑎 𝑑𝑆2 =

60 (McMahon, p. 92) 61 (McMahon, 2006, p. 325) 62 (McMahon, 2006, p. 324) 63 (McMahon, p. 153) 64 (Hartle, Gravity An Introduction to Einstein's General Relativity, 2003, p. 171) Example 8-1 65 (Hartle, Gravity An Introduction to Einstein's General Relativity, 2003, p. 136)

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𝑎4

𝑟′4(𝑑𝑟′2 + 𝑟′2𝑑𝜙2). This singularity arises because the transformation 𝑟 = 𝑎2/𝑟′ map all the points at

𝑟 = ∞ into 𝑟 = 0.

14) 66The Hyperbolic Plane.

a) The line element: 𝑑𝑆2 = 𝑦−2(𝑑𝑥2 + 𝑑𝑦2)

b) Conservation equations:

i) (𝑥 − 𝑥0)2 + 𝑦2 =

1

𝐾12

ii) 𝑥 =1

𝐾1tanh(𝑆)

iii) 𝑦 =1

𝐾1 cosh(𝑆)

15) 672-sphere With Radius 𝑎.

a) Line element: 𝑑𝑠2 = 𝑎2𝑑휃2 + 𝑎2 sin2 휃 𝑑𝜙2.

b) Basis one-forms: 𝜔�̂� = 𝑑, 𝜔�̂� = sin 휃 𝑑𝜙

c) Killing vector: 𝑋 = (𝑋𝜃

𝑋𝜙) = (

𝐴′ sin𝜙 + 𝐵′ cos𝜙

cot 휃 (𝐴′ cos𝜙 − 𝐵′ sin𝜙) + 𝐶′); 𝑋 = 𝑋𝑎𝜕𝑎 = 𝐴𝐿𝑥 + 𝐵𝐿𝑦 + 𝐶𝐿𝑧

i) 𝐿𝑥 = cos𝜙𝜕

𝜕𝜃− cot 휃 sin𝜙

𝜕

𝜕𝜙

ii) 𝐿𝑦 = sin𝜙𝜕

𝜕𝜃+ cot 휃 cos𝜙

𝜕

𝜕𝜙

iii) 𝐿𝑧 =𝜕

𝜕𝜙

d) Ricci scalar: 𝑅 = 2

16) 68Three-dimensional Flat Space in Spherical Polar Coordinates.

a) Line element: 𝑑𝑠2 = 𝑑𝑟2 + 𝑟2𝑑휃2 + 𝑟2 sin2 휃 𝑑𝜙2

b) Christoffel symbols: Γ 𝜃𝜃𝑟 = −𝑟, Γ 𝑟𝜃

𝜃 = Γ 𝜃𝑟𝜃 =

1

𝑟, Γ 𝜙𝜙

𝑟 = −𝑟 sin2 휃, Γ 𝑟𝜙𝜙

= Γ 𝜙𝑟𝜙

=1

𝑟, Γ 𝜙𝜙

𝜃 =

−sin휃 cos휃, Γ 𝜃𝜙𝜙

= Γ 𝜙𝜃𝜙

= cot 휃

17) 69Flat Minkowski space-time: Flat Minkowski space-time is the mathematical setting in which Einstein’s

special theory of relativity is most conveniently formulated.

a) Line element:

i) 𝑑𝑠2 = −𝑑𝑡2 + 𝑑𝑟2 + 𝑟2𝑑휃2 + 𝑟2 sin2 휃 𝑑𝜙2 (Spherical polar coordinates 70)

ii) 𝑑𝑠2 = −𝑑𝑡2 + 𝑑𝑥2 + 𝑑𝑦2 + 𝑑𝑧2

(1) 𝑑𝑠2 < 0, time-like, inside the light cone

(2) 𝑑𝑠 = 0: null-vector, on the light cone

(3) 𝑑𝑠2 > 0, space-like, outside the light cone.

b) Basis one-forms: 𝜔�̂� = 𝑑𝑡, 𝜔�̂� = 𝑑𝑟, 𝜔�̂� = 𝑟𝑑휃, 𝜔�̂� = 𝑟 sin휃 𝑑𝜙

c) Orthonormal tetrad

i) 𝑙𝑎 =1

√2(1, 1, 0, 0), 𝑙𝑎 =

1

√2(1, −1, 0, 0)

ii) 𝑛𝑎 =1

√2(1, −1, 0, 0), 𝑛𝑎 =

1

√2(1, 1, 0, 0)

66 (Hartle, Gravity An Introduction to Einstein's General Relativity, 2003, p. 184) Problem 8-12 67 (McMahon, p. 170) 68 (McMahon, 2006, p. 91) 69 (McMahon, p. 186) 70 (McMahon, p. 91)

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iii) 𝑚𝑎 =1

√2(0, 0, 𝑟, 𝑖𝑟 sin 휃), 𝑚𝑎 =

1

√2(0, 0, −

1

𝑟, −

𝑖

𝑟 sin𝜃)

iv) 𝑚𝑎̅̅ ̅̅ =1

√2(0, 0, 𝑟, −𝑖𝑟 sin휃), 𝑚𝑎̅̅ ̅̅ =

1

√2(0, 0, −

1

𝑟,

𝑖

𝑟 sin𝜃)

d) Spin coefficients: 𝜋 = 𝜈 = 𝜆 = 𝜇 = 𝜅 = 𝜏 = 𝜌 = 𝜎 = 휀 = 𝛾 = 0, 𝛼 =cot𝜃

𝑟, 𝛽 = −

cot𝜃

𝑟

e) 71Rotating the axes (𝑡, 𝑟) by 45°:

i) Coordinate transformation:

(1) 3𝑢 ≡ 𝑡 − 𝑟

(2) 𝑣 ≡ 𝑡 + 𝑟

ii) Line element: 𝑑𝑠2 = −𝑑𝑢𝑑𝑣 +1

4(𝑢 − 𝑣)2(𝑑휃2 + sin2 휃 𝑑𝜙2)

f) Make a further transformation, that map the coordinates from −∞ < 𝑡 < +∞ and 0 < 𝑟 < +∞

into a finite range: 𝑢′ ≡ tan− 𝑢 ≡ 𝑡′ − 𝑟′, 𝑣′ ≡ tan−1 𝑣 ≡ 𝑡′ + 𝑟′

18) 72Three-dimensional Flat Space-time.

a) Line element: 𝑑𝑠2 = −𝑑𝑡2 + 𝑑𝑟2 + 𝑟2𝑑𝜙2

b) Conservation equations: Light rays moves on straight lines in curved space. From our point of view

the tip of the light cone (𝑡, 𝑟) moves along a hyperbolic path.

i) (𝐾2

𝐾1)2= 𝑟2 − (𝑡 − 𝑡0)

2

ii) 𝑡 − 𝑡0 =𝐾2

𝐾1tan(𝜙 − 𝜙0)

iii) 𝑟 = ± 𝐾2

𝐾1√tan2(𝜙 − 𝜙0) + 1

c) Coordinate transformation:

i) 𝑡 = 𝑡′ − 𝑥′

ii) 𝑥 = 𝑡′

(1) Line element:73 𝑑𝑠2 = −𝑑𝑡2 + 2𝑑𝑥𝑑𝑡 + 𝑑𝑦2 + 𝑑𝑧2

19) 74Two-dimensional Flat Space-time.

a) Coordinate transformation:

i) 𝑡 = 𝑋 sinh(𝑇)

ii) 𝑥 = 𝑋 cosh(𝑇)

b) Line element: 𝑑𝑠2 = −𝑋2𝑑𝑇2 + 𝑑𝑋2

c) Conservation equation: 𝑋2𝑑𝑇

𝑑𝑆= 𝐾

d) Geodesic equation: 𝑋(𝑇) = −𝐾

sinh(𝑇−𝑇∗)

20) Conformally Flat Space:

a) Line element: 𝑔𝑎𝑏 = 𝑓(𝑥)휂𝑎𝑏

b) Weyl tensor: 𝐶𝑎𝑏𝑐𝑑 = 0

21) 75Static 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

71 (Hartle, Gravity An Introduction to Einstein's General Relativity, 2003, p. 137) 72 (Hartle, Gravity - An introduction to Einstein's General Relativity, 2003) Problem 8-11 73 (Hartle, Gravity - An introduction to Einstein's General Relativity, 2003, p. 164), problem 2 74 (Hartle, Gravity An Introduction to Einstein's General Relativity, 2003, p. 143) 75 (Hartle, Gravity An Introduction to Einstein's General Relativity, 2003, p. 126)

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mechanics for motion in a gravitational potential for nonrelativistic velocities. Φ(𝑥𝑖) is a function of

position satisfying the Newtonian field equation76 ∇2Φ(�⃗�) = 4𝜋𝐺𝜇(�⃗�)and assumed to vanish at infin-

ity. For example outside Earth Φ(𝑟) = −𝐺𝑀⊕

𝑟. This line element is predicted by general relativity for

small curvatures produced by time-independent weak sources, and it is a good approximation to the

curved spacetime geometry produced by the Sun.

a) Line element: 𝑑𝑠2 = −(1 +2Φ(𝑥𝑖)

𝑐2) (𝑐𝑑𝑡)2 + (1 −

2Φ(𝑥𝑖)

𝑐2) (𝑑𝑥2 + 𝑑𝑦2 + 𝑑𝑧2)

77Δ𝜏𝐵~(1 +Φ𝐵−Φ𝐴

𝑐2)Δ𝜏𝐴 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. 22) 78Rindler metric: The Rindler coordinate system or frame describes a uniformly accelerating frame of

reference in Minkowski space.

a) Line element: 𝑑𝑠2 = 𝜉2𝑑𝜏2 − 𝑑𝜉2

b) Basis one-forms: 𝜔�̂� = 𝑑𝜉, 𝜔�̂� = 𝜉𝑑𝜏

c) Geodesic equations:

i) �̈� + 𝜉�̇�2 = 0

ii) �̈� +2

𝜉�̇��̇� = 0

23) 79The Einstein Cylinder:

a) Line element: 𝑑𝑠2 = −𝑑𝑡2 + (𝑎0)2(𝑑휃2 + sin2 휃 (𝑑𝜙2 + sin2𝜙𝑑𝜓2))

b) Coordinate transformation: 𝑟 = sin 휃

i) Line element: 𝑑𝑠2 = −𝑑𝑡2 + (𝑎0)2 (

𝑑𝑟2

1−𝑟2+ 𝑟2(𝑑𝜙2 + sin2𝜙𝑑𝜓2))

c) Ricci scalar: 𝑅 = 3 ⋅2

𝑎02

d) Einstein equations:

i) 𝜇 =3

𝑎02 − Λ

ii) 𝑝 = −1

𝑎02 + Λ

24) 80Classical Anti-de Sitter Space-time

a) Line element: 𝑑𝑠2 = −cosh2(𝑟) 𝑑𝑡2 + 𝑑𝑟2 + sinh2(𝑟) 𝑑휃2 + sinh2(𝑟) sin2 휃 𝑑𝜙2

b) Coordinate transformation: cosh(𝑟) =1

cos𝜓

i) Line element: 𝑑𝑠2 = −1

cos2𝜓(𝑑𝑡2 + 𝑑𝜓2 + sin2𝜓𝑑휃2 + sin2𝜓 sin2 휃 𝑑𝜙2) - which is confor-

mally related to the Einstein cylinder.

c) Killing vectors and conservation equations

i) 𝝃 = (𝜉𝑡 , 𝜉𝑟, 𝜉𝜃, 𝜉𝜙) = (1,0,0,0) ⇒ �̇� =𝐾1

cosh2(𝑟)

76 (Hartle, Gravity An Introduction to Einstein's General Relativity, 2003, p. 40) eq. (3.18) 77 (Hartle, Gravity - An introduction to Einstein's General Relativity, 2003, p. 127) 78 http://en.wikipedia.org/wiki/Rindler_coordinates, (McMahon, p. 84) 79 (Choquet-Bruhat, 2015, s. 95) 80 (Choquet-Bruhat, 2015, s. 97)

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ii) 𝜻 = (휁𝑡, 휁𝑟, 휁𝜃, 휁𝜙) = (0,0,0,1) ⇒ �̇� =𝐾2

sinh2(𝑟) sin2 𝜃

d) Geodesic equations:

i) 0 = cosh2(𝑟) �̈� + 2 sinh(𝑟) �̇��̇�

ii) 0 = �̈� + sinh(𝑟) �̇�2 − cosh(𝑟) 휃̇2 − cosh(𝑟) sin2 휃 �̇�2

iii) 0 = 2 cosh(𝑟) �̇�휃̇ + sinh2(𝑟) 휃̈ − sinh2(𝑟) cos휃 �̇�2

iv) 0 = 2 cosh(𝑟) sin2 휃 �̇� + 2 sinh2(𝑟) cos 휃 �̇� + sinh2(𝑟) sin2 휃 �̈�

e) The coordinates 𝑡 and 𝑟: �̇� = cosh(𝑟) ⇒ 𝑡 − 𝑡0 =sinh(𝑟)

2 cosh2(𝑟)+ tan−1(𝑒𝑟) − (

sinh(𝑟0)

2 cosh2(𝑟0)+

tan−1(𝑒𝑟0)) →sinh(𝑟)

2cosh2(𝑟)+ tan−1(𝑒𝑟) − 𝐾3 (if 𝑟0 → 0) → 𝐾4 (if 𝑟 → ∞). Interpreting this means,

that no matter how far the light travels in this space-time from 𝑟 = 0 to 𝑟 → ∞ this happens within

a limited time.

25) 81The de Sitter Space-time: The de Sitter space-time is an example of the Robertson Walker metric in

vacuum, positive curvature and a cosmological constant.

a) Line-element: 𝑑𝑠2 = −𝑑𝑡2 + 𝑎(𝑡)2(𝑑휃2 + sin2 휃 (𝑑𝜙2 + sin2𝜙𝑑𝜓2)) = −𝑑𝑡2 +1

𝑘2cosh2(𝑘𝑡) (𝑑휃2 + sin2 휃 (𝑑𝜙2 + sin2𝜙𝑑𝜓2))

b) Einstein equations:

i) 0 =3

𝑎2(1 + �̇�2) − Λ

ii) 0 = 2�̈�

𝑎+

1

𝑎2(1 + �̇�2) − Λ

iii) 𝑎2 =1

𝑘2cosh2(𝑘𝑡)

26) 82Tolman-Bondi de Sitter: Spherical dust with a cosmological constant

a) Line element: 𝑑𝑠2 = 𝑑𝑡2 − 𝑒−2𝜓(𝑡,𝑟)𝑑𝑟2 − 𝑅2(𝑡, 𝑟)𝑑휃2 − 𝑅2(𝑡, 𝑟) sin2 휃 𝑑𝜙2

b) Basis one-forms: 𝜔�̂� = 𝑑𝑡, 𝜔�̂� = 𝑒−𝜓(𝑡,𝑟)𝑑𝑟, 𝜔�̂� = 𝑅(𝑡, 𝑟)𝑑휃, 𝜔�̂� = 𝑅(𝑡, 𝑟) sin휃 𝑑𝜙

c) The Einstein tensor.

i) 𝐺𝑡𝑡 =1

𝑅2[1 − 2𝑅�̇��̇� + (�̇�)

2− (2𝑅𝑅′′ + 2𝑅𝑅′𝜓′ + (𝑅′)2)𝑒2𝜓]

ii) 𝐺𝑟𝑡 = −2 [(�̇�)′

𝑅+𝑅′�̇�

𝑅]

iii) 𝐺𝑟𝑟 =1

𝑅2[(𝑅′)2 − (2𝑅�̈� + 1 + (�̇�)

2) 𝑒−2𝜓]

iv) 𝐺𝜃𝜃 = 𝑅2 ([�̈� − (�̇�)

2] +

1

𝑅[(𝑅′′ + 𝑅′𝜓′)𝑒2𝜓 + �̇��̇� − �̈�])

v) 𝐺𝜙𝜙 = 𝑅2 sin2 휃 ([�̈� − (�̇�)

2] +

1

𝑅[(𝑅′′ + 𝑅′𝜓′)𝑒2𝜓 + �̇��̇� − �̈�])

27) 83The Anti-de Sitter Space-time: This spacetime is a solution to the Einstein vacuum equation with cos-

mological constant Λ = −3.

a) Line element: 𝑑𝑠2 = −𝑑𝑡2 + cos2(𝑡) 𝑑𝑟2 + cos2(𝑡) sinh2(𝑟) 𝑑휃2 + cos2(𝑡) sinh2(𝑟) sin2 휃 𝑑𝜙2

b) Ricci scalar: R = 12

28) 842+1 dimensions: Gravitational Collapse of an Inhomogenous Spherically Symmetric Dust Cloud with

Nonzero Cosmological Constant.

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

81 (Choquet-Bruhat, 2015, s. 96) 82 (McMahon, p. 121) 83 (Choquet-Bruhat, 2015, s. 97) 84 (McMahon, p. 150)

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b) Basis one-forms: 𝜔�̂� = 𝑑𝑡, 𝜔�̂� = 𝑒𝑏(𝑡,𝑟)𝑑𝑟, 𝜔�̂� = 𝑅(𝑡, 𝑟)𝑑𝜙

c) The stress tensor: 𝑇𝑎𝑏 = 𝜌𝑢𝑎𝑢𝑏, 𝑢𝑎 = (𝑢𝑡 , 𝑢𝑟, 𝑢𝜙) = (1,0,0)

d) The Einstein equations.

i) −[(𝑅′′ − 𝑅′𝑏′)𝑒−2𝑏

𝑅−�̇��̇�

𝑅] + 𝜆2 = 𝜅𝜌

ii) (�̇�)′− 𝑅′�̇� = 0

iii) �̈� + 𝜆2𝑅 = 0

iv) −[�̈� + (�̇�)2] − 𝜆2 = 0

29) 85Robertson-Walker: Homogenous, isotropic and expanding universe. The constant 𝑘 = −1,0,1 de-

pending on whether the universe is open, flat or closed.

a) Line element: 𝑑𝑠2 = −𝑑𝑡2 +𝑎2(𝑡)

1−𝑘𝑟2𝑑𝑟2 + 𝑎2(𝑡)𝑟2𝑑휃2 + 𝑎2(𝑡)𝑟2 sin2 휃 𝑑𝜙2

b) Basis one-forms: 𝜔�̂� = 𝑑𝑡, 𝜔�̂� =𝑎(𝑡)

√1−𝑘𝑟2𝑑𝑟, 𝜔�̂� = 𝑎(𝑡)𝑟𝑑휃, 𝜔�̂� = 𝑎(𝑡)𝑟 sin휃 𝑑𝜙

c) Einstein tensor.

i) 𝐺�̂��̂� = 3((�̇�2+k)

a2)

ii) 𝐺�̂��̂� = 𝐺�̂��̂� = 𝐺�̂��̂� = −(2�̈�(𝑡)

𝑎(𝑡)+�̇�(𝑡)2+𝑘

𝑎(𝑡)2)

d) Stress tensor: 𝑇�̂��̂� = 𝑑𝑖𝑎𝑔(𝜌, 𝑃, 𝑃, 𝑃)

e) Einstein equations (The Friedmann equations).

i) 8𝜋𝜌 =3

𝑎2(𝑘 + �̇�2) + Λ

ii) 8𝜋𝑃 = 2�̈�

𝑎+

1

𝑎2(𝑘 + �̇�2) + Λ

f) Solutions.

i) 86Matter, no radiation 𝑃 = 0: 𝜌 ∝ 𝑎−3

ii) 87No curvature,𝑘 = 0, no cosmological constant, Λ = 0: 𝑎(𝑡) ∝ 𝑡2

3

iii) 88Matter and radiation, late universe 𝑃 =1

3𝜌: 𝜌 ∝ 𝑎−4

iv) 89No matter, radiation, early universe: 𝑎(𝑡) ∝ √𝑡

v) 90No matter, no radiation, no curvature 𝑘 = 0, cosmological constant: 𝑎(𝑡) = 𝐶𝑒√Λ

3𝑡

vi) 91 A particle of mass 𝑚, sitting on a surface of a ball of radius 𝑅 and mass density 𝜌, experiences

an acceleration, 𝑑2𝑅

𝑑𝑡2 given by

4𝜋

3𝑅3𝐺𝜌

𝑅2, and so

1

𝑅

𝑑2𝑅

𝑑𝑡2=

4𝜋

3𝐺𝜌. If we formally identify 𝑅 with the

radius of the Universe, and 𝜌 with the mass density of the Universe, this is Einstein’s equation

for how the size of the Universe evolves, assuming the absence of pressure.

85 (McMahon, p. 161) 86 (McMahon, p. 270) 87 (McMahon, p. 274) 88 (McMahon, p. 271) 89 (McMahon, p. 272) 90 (McMahon, p. 273) 91 (Greene, 2004, s. 515) note 6

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vii) 92 It is not difficult to see how accelerated expansion arises. One of Einstein’s equations

is: �̈�(𝑡)

𝑎(𝑡)= −

4𝜋

3(𝜌 + 3𝑝). Notice that if the right-hand side of this equation is positive, the scale

factor will grow at an increasing rate: the Universe’s rate of growth will accelerate with time.

30) 93Light Travelling in the Universe: 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 con-

tact. As you can see, for accelerated expansion, even for arbitrarily large 𝑡0 the integral is bounded,

showing 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.

31) 94The Schwarzschild Space-time.

a) Line element: 𝑑𝑠2 = (1 −2𝑚

𝑟)𝑑𝑡2 −

𝑑𝑟2

1−2𝑚

𝑟

− 𝑟2(𝑑휃2 + sin2 휃 𝑑𝜙2)

b) Basis one-forms: 𝜔�̂� = √1 −2𝑚

𝑟𝑑𝑡, 𝜔�̂� =

1

√1−2𝑚

𝑟

𝑑𝑟, 𝜔�̂� = 𝑟𝑑휃, 𝜔�̂� = 𝑟 sin휃 𝑑𝜙

c) Killing vectors and conservation laws.

i) 95The Killing vector that corresponds to the independence of the metric of 𝑡 is 𝜉 = (1,0,0,0).

The conserved energy per unit rest mass: 𝑒 = −𝜉 ⋅ 𝑢 = −𝑔𝑎𝑏𝜉𝑎𝑢𝑏 = −𝑔𝑡𝑡 ⋅ 1 ⋅

𝑑𝑡

𝑑𝜏=

−(1 −2𝑚

𝑟)𝑑𝑡

𝑑𝜏

ii) and of 𝜙 is 휂 = (0,0,0,1). The conserved angular momentum per unit rest mass: 𝑙 = 휂 ⋅ 𝑢 =

𝑔𝑎𝑏휂𝑎𝑢𝑏 = 𝑔𝜙𝜙 ⋅ 1 ⋅

𝑑𝜙

𝑑𝜏= −𝑟2 sin2 휃

𝑑𝜙

𝑑𝜏= −𝑟2

𝑑𝜙

𝑑𝜏 for 휃 =

𝜋

2

d) Geodesic Equations.

i) �̈� +2𝑚

𝑟(𝑟−2𝑚)�̇��̇� = 0

ii) �̈� +𝑚

𝑟3(𝑟 − 2𝑚)�̇�2 −

𝑚

𝑟(𝑟−2𝑚)�̇�2 − (𝑟 − 2𝑚)(휃̇2 + sin2 휃 �̇�2) = 0

iii) 휃̈ +2

𝑟�̇�휃̇ − sin 휃 cos 휃 �̇�2 = 0

iv) �̈� + 2 cot 휃 휃̇�̇� +2

𝑟�̇��̇� = 0

e) Ricci, Einstein and Stress Tensor: 𝑅𝑎𝑏 = 𝐺𝑎𝑏 = 𝑇𝑎𝑏 = 0

f) Ortonormal tetrad.

i) 𝑙𝑎 =1

√2(√1 −

2𝑚

𝑟,

1

√1−2𝑚

𝑟

, 0, 0), 𝑙𝑎 =1

√2(

1

√1−2𝑚

𝑟

, −√1 −2𝑚

𝑟, 0, 0)

ii) 𝑛𝑎 =1

√2(√1 −

2𝑚

𝑟, −

1

√1−2𝑚

𝑟

, 0, 0), 𝑛𝑎 =1

√2(

1

√1−2𝑚

𝑟

, √1 −2𝑚

𝑟, 0, 0)

92 (Greene, s. 520) note 9 93 (Greene, s. 516) note 10 94 (McMahon, p. 215) 95 (McMahon, p. 220)

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iii) 𝑚𝑎 =1

√2(0, 0, 𝑟, 𝑖𝑟 sin 휃), 𝑚𝑎 =

1

√2(0, 0, −

1

𝑟, −𝑖

1

𝑟 sin𝜃)

iv) 𝑚𝑎̅̅ ̅̅ =1

√2(0, 0, 𝑟, −𝑖𝑟 sin휃), �̅�𝑎 =

1

√2(0, 0, −

1

𝑟, 𝑖

1

𝑟 sin𝜃)

g) Spin Coefficients.

i) 𝜋 = 𝜈 = 𝜆 = 𝜇 = 𝜅 = 0

ii) Expansion: 𝜌 =1

√2√(1 −

2𝑚

𝑟)1

𝑟 (no twist)

iii) 𝜎 = 0 (no shear)

iv) 𝜏 = 0 (the null rays defined by 𝑙𝑎 are parallel.

v) 휀 = −1

232

1

√1−2𝑚

𝑟

𝑚

𝑟2, 𝛾 = −

1

232

1

√1−2𝑚

𝑟

𝑚

𝑟2, 𝛼 =

1

232

1

𝑟cot 휃, 𝛽 = −

1

232

1

𝑟cot 휃

h) Weyl scalars.

i) Ψ0 = Ψ1 = 96Ψ3 = Ψ4 = 0

ii) Ψ2 =1

2𝑟2−2𝑚

𝑟3

i) Petrov Classification: Ψ2 ≠ 0: This is a Petrov type D, which means there are two principal null di-

rections. 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.

j) Particle (planetary) orbits: Choose coordinates so that the particle moves initially in a plane 휃 =𝜋

2.

i) Conservation equation: 97These values correspond to circular orbits, and we can use the bino-

mial expansion of the term under the square root to rewrite the term as. 𝑟1,2 =

𝑙2

2𝑚(1 ± √1 −

12𝑚2

𝑙2) ≅

𝑙2

2𝑚[1 ± (1 −

6𝑚2

𝑙2)] ⇒ Stable orbit: 𝑟1 ≅

𝑙2

𝑚 unstable orbit: 𝑟2 ≅ 3𝑚

ii) Geodesic equations.

(1) 98�̈� + 2𝑑𝜈

𝑑𝑟�̇��̇� = 0

(2) �̈� +𝑑𝜆

𝑑𝑟�̇�2 + 𝑒2(𝜈(𝑟)−𝜆(𝑟))

𝑑𝜈

𝑑𝑟�̇�2 − 𝑟𝑒−2𝜆(𝑟)�̇�2 = 0

(3) �̈� +2

𝑟�̇��̇� = 0

(4) ⇒ (𝑑𝑢

𝑑𝜙)2+ 𝑢2 =

𝑘2−1

ℎ2+2𝑚

ℎ2𝑢 + 2𝑚𝑢3 Which can be interpreted in terms of elliptic func-

tions, 𝑢 =1

𝑟, and h and k are constants of integration.

k) 99The Advance of Perihelion: 𝑢 =𝑚

ℎ2(1 + 𝑒 cos(𝜙 − 𝜔)) corresponds to the perihelion 𝜔 advances

a fraction of a revolution equal to 𝛿𝜔

𝜙=

12𝜋2𝑎2

𝑐2𝑇2(1−𝑒2)

l) 100The Deflection of Light Rays: 휃 =𝜋

2, 𝑑𝑠2 = 0 ⇒ (1 −

2𝑚

𝑟)2�̇�2 − (1 −

2𝑚

𝑟)−1�̇�2 − 𝑟2�̇�2 = 0.

101Deflection angle, in the weak field perturbation: 𝛼 =4𝐺𝑀

𝑐2𝑏, where b is closest approach.

96 Ψ3 =1

4cot 휃 √1 −

2𝑚

𝑟

1

𝑟2= 0 if 휃 =

𝜋

2

97 (McMahon, 2006, p. 222) 98 (A.S.Eddington, pp. 85-86) 99 (A.S.Eddington, p. 88 100 (McMahon, p. 224) 101 (Carroll, p. 291)

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m) 102Time Delay: 휃 =𝜋

2, 𝑑𝑠2 = 0, 𝑟0 = 𝑟 sin𝜙 ⇒ 𝑑𝑡2 =

𝑑𝑟2(1−2𝑚𝑟0

2

𝑟3)

(1−𝑟02

𝑟2)(1−

2𝑚

𝑟)2. To consider the travel time of

light between Earth and a planet in the solar system, we integrate between 𝑟0 (distance of closets

approach) to 𝑟𝑣 (planet orbit radius) and 𝑟0 to 𝑟𝑒 (Earth orbit radius): 𝑡𝑑𝑒𝑙𝑎𝑦 =

𝑚𝐺

𝑐3[2 ln (

(𝑟𝑣+√𝑟𝑣2−𝑟0

2)(𝑟𝑒+√𝑟𝑒2−𝑟0

2)

𝑟02 ) −

√𝑟𝑣2−𝑟0

2

𝑟𝑣−√𝑟𝑒

2−𝑟02

𝑟𝑒]. The ordinary flat space term is given by

√𝑟𝑣2 − 𝑟0

2 +√𝑟𝑒2 − 𝑟0

2.

n) 103Gravitational Red Shift: 𝑑𝜏 = √1 −2𝑚

𝑟𝑑𝑡. Light emitted upward in a gravitational field, from an

observer located at some inner radius 𝑟1 to an observer positioned at some outer radius 𝑟2. 𝛼 =

√1−2𝑚

𝑟2

√1−2𝑚

𝑟1

o) 104The Path of a Radially Infalling Particle.

i) Line element: From infinity with vanishing initial velocity: 𝑑휃 = 𝑑𝜙 = 0 ⇒ 1 −2𝑚

𝑟=

(1 −2𝑚

𝑟)2(𝑑𝑡

𝑑𝜏)2− (

𝑑𝑟

𝑑𝜏)2

ii) Killing equation: From Killings equation we know that (1 −2𝑚

𝑟)𝑑𝑡

𝑑𝜏 is a constant ⇒

(1 −2𝑚

𝑟)𝑑𝑡

𝑑𝜏= 1 and (

𝑑𝑟

𝑑𝜏)2=

2𝑚

𝑟 ⇒ 𝑡 − 𝑡0 =

2

3√2𝑚(𝑟0

3

2 − 𝑟3

2 + 6𝑚√𝑟0 − 6𝑚√𝑟) +

2𝑚 ln√𝑟0−√2𝑚

√𝑟0+√2𝑚 √𝑟+√2𝑚

√𝑟−√2𝑚

32) 105Schwarzschild Space-time with 휃 =𝜋

2.

a) Line element: 𝑑𝑠2 = −(1 −2𝑚

𝑟) 𝑑𝑡2 + (1 −

2𝑚

𝑟)−1𝑑𝑟2 + 𝑟2𝑑𝜙2

b) Geodesic equations:

i) �̈� +2𝑚

𝑟(𝑟−2𝑚)�̇��̇� = 0

ii) �̈� +𝑚

𝑟3(𝑟 − 2𝑚)�̇�2 −

𝑚

𝑟(𝑟−2𝑚)�̇�2 − (𝑟 − 2𝑚)�̇�2 = 0

iii) �̈� +2

𝑟�̇��̇� = 0

33) 106The Schwarzschild Metric in Kruskal Coordinates.

a) Line element: 𝑑𝑠2 =32𝑚3

𝑟𝑒−

𝑟

2𝑚(𝑑𝑣2 − 𝑑𝑢2) − 𝑟2(𝑑휃 + sin2 휃 𝑑𝜙)

i) 𝑟 > 2𝑚:

(1) 𝑢 = 𝑒𝑟

4𝑚 √𝑟

2𝑚− 1cosh

𝑡

4𝑚

(2) 𝑣 = 𝑒𝑟

4𝑚 √𝑟

2𝑚− 1 sinh

𝑡

4𝑚

ii) 𝑟 < 2𝑚:

102 (McMahon, p. 229) 103 (McMahon, p. 234) 104 (McMahon, p. 238) 105 (Hartle, Gravity - An introduction to Einstein's General Relativity, 2003, p. 183) problem 8-3 106 (McMahon, 2006, p. 242)

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(1) 𝑢 = 𝑒𝑟

4𝑚 √1 −𝑟

2𝑚sinh

𝑡

4𝑚

(2) 𝑣 = 𝑒𝑟

4𝑚 √1 −𝑟

2𝑚cosh

𝑡

4𝑚

b) Ricci scalar: 𝑅 = 0

34) 107The General Schwarzschild Metric: a static, spherically symmetric space-time.

a) Line element: 𝑑𝑠2 = 𝑒2𝜈(𝑟)𝑑𝑡2 − 𝑒2𝜆(𝑟)𝑑𝑟2 − 𝑟2𝑑휃2 − 𝑟2 sin2 휃 𝑑𝜙2

b) Basis one-forms: 𝜔�̂� = 𝑒𝜈(𝑟)𝑑𝑡, 𝜔�̂� = 𝑒𝜆(𝑟)𝑑𝑟, 𝜔�̂� = 𝑟𝑑휃, 𝜔�̂� = 𝑟 sin휃 𝑑𝜙

c) Geodesic equations.

i) �̈� + 2𝑑𝜈

𝑑𝑟�̇��̇� = 0

ii) �̈� +𝑑𝜆

𝑑𝑟�̇�2 + 𝑒2(𝜈(𝑟)−𝜆(𝑟))

𝑑𝜈

𝑑𝑟�̇�2 − 𝑟𝑒−2𝜆(𝑟)휃̇2 − 𝑟 sin2 휃 𝑒−2𝜆(𝑟)�̇�2 = 0

iii) 휃̈ +2

𝑟�̇�휃̇ − cos 휃 sin휃 �̇�2 = 0

iv) �̈� +2

𝑟�̇��̇� + 2 cot 휃 휃̇�̇� = 0

d) Ricci tensor.

i) 𝑅�̂��̂� = (𝜈′′ + 𝜈′(𝜈′ − 𝜆′) + 2

𝜈′

𝑟) 𝑒−2𝜆(𝑟)

ii) 𝑅�̂��̂� = −(𝜈′′ + 𝜈′(𝜈′ − 𝜆′) − 2

𝜆′

𝑟) 𝑒−2𝜆(𝑟)

iii) 𝑅�̂��̂� = 𝑅�̂��̂� = −𝜈′

𝑟𝑒−2𝜆(𝑟) +

𝜆′

𝑟𝑒−2𝜆(𝑟) +

(1−𝑒−2𝜆(𝑟))

𝑟2

35) 108General Time Dependent Schwarzschild Metric.

a) Line element: 𝑑𝑠2 = 𝑒2𝜈(𝑡,𝑟)𝑑𝑡2 − 𝑒2𝜆(𝑡,𝑟)𝑑𝑟2 − 𝑟2𝑑휃2 − 𝑟2 sin2 휃 𝑑𝜙2

b) Basis one-forms: 𝜔�̂� = 𝑒𝜈(𝑡,𝑟)𝑑𝑡, 𝜔�̂� = 𝑒𝜆(𝑡,𝑟)𝑑𝑟, 𝜔�̂� = 𝑟𝑑휃, 𝜔�̂� = 𝑟 sin 휃 𝑑𝜙

c) Ricci tensor.

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

𝜈′

𝑟) 𝑒2𝜈(𝑡,𝑟)−2𝜆(𝑡,𝑟)

ii) 𝑅𝑟𝑡 = 2�̇�

𝑟

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

𝜆′

𝑟)

iv) 𝑅𝜃𝜃 = ((−𝜈′ + 𝜆′ )𝑟 − 1)𝑒−2𝜆(𝑡,𝑟) + 1

v) 𝑅𝜙𝜙 = (((−𝜈′ + 𝜆′ )𝑟 − 1)𝑒−2𝜆(𝑡,𝑟) + 1) sin2 휃

36) 109The General Schwarzschild Metric: A Static, Spherically Symmetric Space-time With Cosmological

Constant, the Spatial Line Element.

a) Line element: 𝑑𝜎2 =𝑑𝑟2

1−2𝑚

𝑟+1

3Ω𝑟2

+ 𝑟2𝑑휃2 + 𝑟2 sin2 휃 𝑑𝜙2

b) Ricci scalar: 𝑅 = 4Ω

37) 110The Reissner-Nordström Space-time: A static solution to the Einstein-Maxwell field equations, which

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

a) The line element: 𝑑𝑠2 = (1 −2𝑚

𝑟+𝑒2

𝑟2) 𝑑𝑡2 − (1 −

2𝑚

𝑟+𝑒2

𝑟2)−1

𝑑𝑟2 − 𝑟2𝑑휃2 − 𝑟2 sin2 휃 𝑑𝜙2

107 (McMahon, p. 211) 108 (McMahon, p. 231) 109 (McMahon, p. 277) 110 (McMahon, p. 325)

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b) Basis one-forms: 𝜔�̂� = (1 −2𝑚

𝑟+𝑒2

𝑟2)

1

2𝑑𝑡, 𝜔�̂� = (1 −

2𝑚

𝑟+𝑒2

𝑟2)−1

2𝑑𝑟, 𝜔�̂� = 𝑟𝑑휃, 𝜔�̂� = 𝑟 sin휃 𝑑𝜙

c) Geodesic equations.

i) �̈� + 2 (1 −2𝑚

𝑟+𝑒2

𝑟2)−1

(𝑚𝑟−𝑒2

𝑟3) �̇��̇� = 0

ii) �̈� − (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) 휃̇2 −

𝑟 (1 −2𝑚

𝑟+𝑒2

𝑟2) sin2 휃 �̇�2 = 0

iii) 휃̈ +2

𝑟�̇�휃̇ − cos 휃 sin휃 �̇�2 = 0

iv) �̈� +2

𝑟�̇��̇� + 2 cot 휃 휃̇�̇� = 0

d) Spin Coefficients: 𝜌, 휀, 𝛾, 𝛼, 𝛽 ≠ 0

e) Petrov Classification: Ψ2 (and Ψ3) ≠ 0

38) 111The Kerr Black Hole (a Spinning Black Hole):

a) The line element: 𝑑𝑠2 = (1 −2𝑚𝑟

Σ) 𝑑𝑡2 +

4𝑎𝑚𝑟 sin2 𝜃

Σ𝑑𝑡𝑑𝜙 −

Σ

Δ𝑑𝑟2 − Σ𝑑휃2 − (𝑟2 + 𝑎2 +

2𝑎2𝑚𝑟 sin2 𝜃

Σ) sin2 휃 𝑑𝜙2

i) Δ = 𝑟2 − 2𝑚𝑟 + 𝑎2

ii) Σ = 𝑟2 + 𝑎2 cos2 휃

b) Killing vectors and conservation laws:

i) The Killing vector that corresponds to the independence of the metric of 𝑡 is 𝑋 = (1,0,0,0) ⇒

𝑋 ⋅ 𝑢 = 𝑔𝑎𝑏𝑋𝑎𝑢𝑏 = 𝑔𝑡𝑡𝑋

𝑡𝑢𝑡 + 𝑔𝑡𝜙𝑋𝑡𝑢𝜙 = (1 −

2𝑚𝑟

Σ)𝑑𝑡

𝑑𝜏+2𝑎𝑚𝑟 sin2 𝜃

Σ 𝑑𝜙

𝑑𝜏= 𝑐𝑜𝑛𝑠𝑡

ii) The Killing vector that corresponds to the independence of the metric of 𝜙 is 𝑌 = (0,0,0,1) ⇒

𝑌 ⋅ 𝑢 = 𝑔𝑎𝑏𝑌𝑎𝑢𝑏 = 𝑔𝜙𝜙𝑌

𝜙𝑢𝜙 + 𝑔𝜙𝑡𝑌𝜙𝑢𝑡 = (𝑟2 + 𝑎2 +

2𝑎2𝑚𝑟 sin2 𝜃

Σ) sin2 휃

𝑑𝜙

𝑑𝜏+

2𝑎𝑚𝑟 sin2 𝜃

Σ 𝑑𝑡

𝑑𝜏= 𝑐𝑜𝑛𝑠𝑡

c) Ricci scalar: 𝑅 = 0

d) The Static limit: 𝑟 = 𝑟𝑠 = 2𝑚,𝑅2 = 6𝑚2

i) 𝑑𝜙

𝑑𝑡=

𝑎

3𝑚2 Light that is emitted in the same direction in which the black hole is spinning

ii) 𝑑𝜙

𝑑𝑡= 0 The light is emitted in the direction opposite to that of a black hole’s rotation; that is,

the light is completely stationary.

39) 112The Spatial Part of a Homogenous, Isotropic Metric.

a) Line element: 𝑑𝜎2 =𝑑𝑟2

1−𝑘𝑟2+ 𝑟2𝑑휃2 + 𝑟2 sin2 휃 𝑑𝜙

i) 𝑘 = 0: 𝑑𝜎2 = 𝑑𝜒2 + 𝜒2𝑑휃2 + 𝜒2 sin2 휃 𝑑𝜙2

ii) 𝑘 = 1: 𝑑𝜎2 = 𝑑𝜒2 + sin2(𝜒) 𝑑휃2 + sin2(𝜒) sin2 휃 𝑑𝜙2

iii) 𝑘 = −1: 𝑑𝜎2 = 𝑑𝜒2 + sinh2(𝜒) 𝑑휃2 + sinh2(𝜒) sin2 휃 𝑑𝜙2

40) 113Linearized Theory.

a) Line element: 𝑑𝑠2 = 𝑔𝑎𝑏𝑑𝑥𝑎𝑑𝑥𝑏, 𝑔𝑎𝑏 = 휂𝑎𝑏 + 𝜖ℎ𝑎𝑏, 𝜖 ≪ 1

b) Γ 𝑏𝑐𝑎 =

𝜖

2휂𝑎𝑑 (

𝜕ℎ𝑏𝑑

𝜕𝑥𝑐+𝜕ℎ𝑐𝑑

𝜕𝑥𝑏−𝜕ℎ𝑏𝑐

𝜕𝑥𝑑)

111 (McMahon, p. 244) 112 (McMahon, p. 262) 113 (McMahon, 2006, p. 280)

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c) 𝑅 𝑏𝑐𝑑𝑎 =

𝜖

2휂𝑎𝑒 (

𝜕ℎ𝑑𝑒

𝜕𝑥𝑏𝜕𝑥𝑐−

𝜕ℎ𝑏𝑑

𝜕𝑥𝑐𝜕𝑥𝑒−

𝜕ℎ𝑐𝑒

𝜕𝑥𝑏𝜕𝑥𝑑+

𝜕ℎ𝑏𝑐

𝜕𝑥𝑑𝜕𝑥𝑒)

d) 𝑅𝑎𝑏 =𝜖

2(𝜕ℎ 𝑏

𝑐

𝜕𝑥𝑎𝜕𝑥𝑐−𝑊ℎ𝑎𝑏 −

𝜕ℎ

𝜕𝑥𝑎𝜕𝑥𝑏+

𝜕ℎ 𝑎𝑐

𝜕𝑥𝑏𝜕𝑥𝑐)

e) 𝑅 = 𝜖 (𝜕ℎ𝑎𝑏

𝜕𝑥𝑎𝜕𝑥𝑏−𝑊ℎ)

f) 𝐺𝑎𝑏 =𝜖

2(𝜕ℎ 𝑏

𝑐

𝜕𝑥𝑎𝜕𝑥𝑐−𝑊ℎ𝑎𝑏 −

𝜕ℎ

𝜕𝑥𝑎𝜕𝑥𝑏+

𝜕ℎ 𝑎𝑐

𝜕𝑥𝑏𝜕𝑥𝑐− 휂𝑎𝑏

𝜕ℎ𝑐𝑑

𝜕𝑥𝑐𝜕𝑥𝑐+ 휂𝑎𝑏𝑊ℎ)

g) 114The Newtonian Approximation: Weak gravitational field and bodies of low masses and velocities.

𝑣 =𝑑𝑥1

𝑑𝑥0≪ 𝑐 ⇒ (

𝑑𝑠

𝑑𝑥0)2→ −1+ 𝜖ℎ00

i) Christoffel symbol: Γ 00𝑖 = −

1

2𝜖𝜕ℎ00

𝜕𝑥𝑖

ii) Geodesic equations: 𝑑2𝑥𝑖

(𝑑𝑥0)2=

1

2𝜖𝜕ℎ00

𝜕𝑥𝑖

iii) Ricci tensor: 𝑅00 = −1

2𝜖∇2ℎ00

iv) Stress tensor and Einstein equation: Pure matter no pressure: 𝑇00 = 𝜇 ⇒ 𝜖∇2ℎ00 = −1

2𝐺𝐸𝜇

41) 115Plane Gravitational Waves.

a) Line element: 𝑑𝑠2 = 𝑔𝑎𝑏𝑑𝑥𝑎𝑑𝑥𝑏 = (휂𝑎𝑏 + ℎ𝑎𝑏)𝑑𝑥

𝑎𝑑𝑥𝑏

b) Plane waves: ℎ𝑎𝑏 = ℎ𝑎𝑏(𝑡 − 𝑧) ⇒

i) 𝜕ℎ𝑎𝑏

𝜕𝑥=

𝜕ℎ𝑎𝑏

𝜕𝑦= 0

ii) 𝜕ℎ𝑎𝑏

𝜕𝑧= −

𝜕ℎ𝑎𝑏

𝜕𝑡

iii) 𝜕2ℎ𝑎𝑏

𝜕𝑧2=

𝜕2ℎ𝑎𝑏

𝜕𝑡2

c) Riemann tensor: 𝑅 𝑏𝑐𝑑𝑎 =

1

2휀휂𝑎𝑓 (

𝜕2ℎ𝑑𝑓

𝜕𝑥𝑐𝜕𝑥𝑏−

𝜕2ℎ𝑏𝑑

𝜕𝑥𝑐𝜕𝑥𝑓+

𝜕2ℎ𝑏𝑐

𝜕𝑥𝑑𝜕𝑥𝑓−

𝜕2ℎ𝑐𝑓

𝜕𝑥𝑑𝜕𝑥𝑏)

42) The Einstein Gauge.

a) Line element: 𝑑𝑠2 = 𝑑𝑡2 − (1 − 휀ℎ𝑥𝑥)𝑑𝑥2 − (1 + 휀ℎ𝑥𝑥)𝑑𝑦

2 − 𝑑𝑧2 + 2휀ℎ𝑥𝑦𝑑𝑥𝑑𝑦

i) 116+-polarization, ℎ𝑥𝑦 = 0: 𝑑𝑠2 = 𝑑𝑡2 − (1 − 휀ℎ𝑥𝑥)𝑑𝑥

2 − (1 + 휀ℎ𝑥𝑥)𝑑𝑦2 − 𝑑𝑧2

ii) 117x-polarization,ℎ𝑥𝑥 = 0: 𝑑𝑠2 = 𝑑𝑡2 − 𝑑𝑥2 − 𝑑𝑦2 − 𝑑𝑧2 + 2휀ℎ𝑥𝑦𝑑𝑥𝑑𝑦

b) Coordinate Transformation:

i) 𝑑𝑥 =1

√2(𝑑𝑥′ + 𝑑𝑦′)

ii) 𝑑𝑦 =−1

√2(𝑑𝑥′ − 𝑑𝑦′)

iii) Line element: 𝑑𝑠2 = 𝑑𝑡2 − (1 + 휀ℎ𝑥𝑦)𝑑𝑥′2 − (1 − 휀ℎ𝑥𝑦)𝑑𝑦

′2 − 𝑑𝑧2

43) 118The Rosen Line Element.

a) Line Element: 𝑑𝑠2 = 𝑑𝑈𝑑𝑉 − 𝑎2(𝑈)𝑑𝑥2 − 𝑏2(𝑈)𝑑𝑦2, 𝑈 = 𝑡 − 𝑧, 𝑎2(𝑈) = 1 − ℎ𝑥𝑥(𝑈), 𝑏(𝑈) =

1 + ℎ𝑥𝑥(𝑈)

b) Basis one-forms: 𝜔0̂ =1

2(𝑑𝑈 + 𝑑𝑉), 𝜔1̂ =

1

2(𝑑𝑈 − 𝑑𝑉), 𝜔2̂ = 𝑎(𝑈)𝑑𝑥, 𝜔3̂ = 𝑏(𝑈)𝑑𝑦

c) Einstein Tensor: 𝐺00 = −1

4(1

𝑎

𝑑2𝑎

𝑑𝑈2+1

𝑏

𝑑2𝑏

𝑑𝑈2)

114 (Choquet-Bruhat, 2015, s. 75) Exercise IV.5.1. 115 (McMahon, p. 288) 116 (McMahon, p. 291) 117 (McMahon, p. 293) 118 (McMahon, p. 298)

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44) 119Kahn- Penrose Metric: Colliding gravitational waves.

a) Line element: 𝑑𝑠2 = 2𝑑𝑢𝑑𝑣 − (1 − 𝑢)2𝑑𝑥2 − (1 + 𝑢)2𝑑𝑦2

b) Ricci tensor: 𝑅 = 0

45) 120Brinkmann Metric: Plane gravitational waves.

a) Line element: 𝑑𝑠2 = 𝐻(𝑢, 𝑥, 𝑦)𝑑𝑢2 + 2𝑑𝑢𝑑𝑣 − 𝑑𝑥2 − 𝑑𝑦2(𝑢, 𝑣, 𝑥, 𝑦)

b) Principal null direction: 𝑢, 𝑙𝑎 = (0,1,0,0)

c) Basis one-forms: 𝜔�̂� =1

2(𝐻 + 1)𝑑𝑢 + 𝑑𝑣, 𝜔�̂� =

1

2(1 − 𝐻)𝑑𝑢 − 𝑑𝑣, 𝜔𝑥 = 𝑑𝑥, 𝜔 �̂� = 𝑑𝑦

d) Ricci and Einstein tensor: 121𝑅𝑢𝑢 = 𝐺𝑢𝑢 =1

2

𝜕2𝐻

𝜕𝑥2+1

2

𝜕2𝐻

𝜕𝑦2

e) Ortonormal tetrad.

i) 𝑙𝑎 =1

√2(1, 0, 0, 0), 𝑙𝑎 =

1

√2(0, 1, 0, 0)

ii) 𝑛𝑎 =1

√2(𝐻, 2, 0, 0), 𝑛𝑎 =

1

√2(2, −𝐻, 0, 0)

iii) 𝑚𝑎 =1

√2(0, 0, 1, 𝑖), 𝑚𝑎 =

1

√2(0, 0, −1, −𝑖)

iv) 𝑚𝑎̅̅ ̅̅ =1

√2(0, 0, 1, −𝑖), �̅�𝑎 =

1

√2(0, 0, −1, 𝑖)

f) Spin coefficients.

i) 𝜋 = 𝜆 = 𝜇 = 𝜅 = 휀 = 𝛾 = 𝛼 = 𝛽 = 0

ii) 𝜌 = 0 (no expansion, no twist)

iii) 𝜎 = 0 (no shear)

iv) 𝜏 = 0 (the null rays defined by 𝑙𝑎 are parallel.

v) 𝜈 =1

√2(−

𝜕𝐻

𝜕𝑥+ 𝑖

𝜕𝐻

𝜕𝑦)

g) Weyl scalars.

i) Ψ0 = Ψ1 = Ψ2 = Ψ3

ii) Ψ4 =1

2[𝜕2𝐻

𝜕𝑥2−𝜕2𝐻

𝜕𝑦2− 2𝑖

𝜕2𝐻

𝜕𝑥𝜕𝑦]

h) Petrov classification: 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.

i) 122General Plane wave: 𝐻(𝑢, 𝑥, 𝑦) = 𝑎(𝑢)(𝑥2 − 𝑦2) + 2𝑏(𝑢)𝑥𝑦 + 𝑐(𝑢)(𝑥2 + 𝑦2), where 𝑎 and 𝑏

describe the polarization states, and 𝑐 represents waves of other types of radiation.

i) Einstein tensor: 𝐺𝑢𝑢 = 2𝑐(𝑢)

j) 123Plane Wave in vacuum: 𝐻(𝑢, 𝑥, 𝑦) = 𝑎(𝑢)(𝑥2 − 𝑦2) + 2𝑏(𝑢)𝑥𝑦

i) Einstein tensor: 𝐺𝑢𝑢 = 0

k) 124Constant Linear Polarization in Vacuum: 𝐻(𝑢, 𝑥, 𝑦) = ℎ(𝑢)[cos 𝛼 (𝑥2 − 𝑦2) + 2 sin𝛼 𝑥𝑦, where

𝛼 is the angle between the polarization vector and the 𝑥-axis.

i) Einstein tensor: 𝐺𝑢𝑢 = 0

46) 125The Aichelburg-Sexl Solution: A black hole passing near by.

119 (McMahon, 2006, p. 92) 120 (McMahon, p. 195) 121 According to the Weyl scalar calculation the sign is wrong 122 (McMahon, p. 301) 123 (McMahon, p. 301) 124 (McMahon, p. 303) 125 (McMahon, p. 303)

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a) Line element: 𝑑𝑠2 = 4𝜇 log(𝑥2 + 𝑦2) 𝑑𝑢2 + 𝑑𝑢𝑑𝑟 − 𝑑𝑥2 − 𝑑𝑦2, 𝐻(𝑢, 𝑥, 𝑦) = 4𝜇 log(𝑥2 + 𝑦2)

47) 126Colliding Gravity Waves.

a) Line element: 𝑑𝑠2 = 2𝑑𝑢𝑑𝜈 − [1 − 𝑢Θ(𝑢)]2𝑑𝑥2 − [1 + 𝑢Θ(𝑢)]2𝑑𝑦2

48) 127Impulsive Gravitational wave region III.

a) Line element: 𝑑𝑠2 = 2𝑑𝑢𝑑𝜈 − [1 − 𝜈Θ(𝜈)]2𝑑𝑥2 − [1 + 𝜈Θ(𝜈)]2𝑑𝑦2

b) principal null direction 𝜈, 𝑛𝑎 = (0,1,0,0)

c) Basis one-forms: 𝜔�̂� =1

√2(𝑑𝑢 + 𝑑𝜈), 𝜔�̂� =

1

√2(𝑑𝑢 − 𝑑𝜈), 𝜔𝑥 = (1 − 𝜈Θ(𝜈))𝑑𝑥, 𝜔�̂� =

(1 + 𝜈Θ(𝜈))𝑑𝑦

d) Geodesic equations.

i) �̈� = 0

ii) �̈� − Θ(𝜈)[1 − 𝜈Θ(𝜈)]�̇�2 + Θ(𝜈)[1 + 𝜈Θ(𝜈)]�̇�2 = 0

iii) �̈� −2Θ(𝜈)

[1−𝜈Θ(𝜈)]�̇��̇� = 0

iv) �̈� +2Θ(𝜈)

[1+𝜈Θ(𝜈)]�̇��̇� = 0

e) orthonormal tetrad (𝑢, 𝜈, 𝑥, 𝑦)

i) 𝑙𝑎 = (1, 0, 0, 0), 𝑙𝑎 = (0, 1, 0, 0)

ii) 𝑛𝑎 = (0, 1, 0, 0), 𝑛𝑎 = (1, 0, 0, 0)

iii) 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)))

iv) 𝑚𝑎̅̅ ̅̅ =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)))

f) Spin coefficients.

i) 𝜋 = 𝜈 = 𝜆 = 𝜇 = 𝜅 = 휀 = 𝛾 = 𝛼 = 𝛽 = 0

ii) 𝜌 =𝜈Θ(𝜈)

(1+𝜈Θ(𝜈))(1−𝜈Θ(𝜈))≠ 0 (expansion)

iii) 𝜎 =Θ(𝜈)

(1+𝜈Θ(𝜈))(1−𝜈Θ(𝜈))≠ 0 (shear)

iv) 𝜏 = 0, the null rays defined by 𝑛𝑎 are parallel

g) Weyl scalars.

i) Ψ1 = Ψ2 = Ψ3 = Ψ4

ii) Ψ0 = 𝛿(𝜈) − 4𝜈Θ(𝜈)

(1−𝜈2Θ(𝜈))2

h) Petrov classification: 128Ψ0 ≠ 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.

49) 129Two Interacting Waves: A null congruence begins in vacuum. 𝜈 < 0: flat region of spacetime. Defin-

ing a plane wave by 𝜈 = 𝑐𝑜𝑛𝑠𝑡, we choose the null vector 𝑙𝑎 to point along 𝜈. The null hypersurface is

given by 𝜈 = 0. In the region past 𝜈 = 0, an opposing wave is encountered.

a) Line element: 𝑑𝑠2 = 2𝑑𝑢𝑑𝜈 − cos2 𝑎𝜈 𝑑𝑥2 − cosh2 𝑎𝜈 𝑑𝑦2 (𝑢, 𝜈, 𝑥, 𝑦)

126 (McMahon, p. 304) 127 (McMahon, p. 305) 128 http://www-staff.lboro.ac.uk/~majbg/jbg/book/chap3.pdf 129 (McMahon, p. 313)

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b) Basis one-forms: 𝜔�̂� =1

√2(𝑑𝑢 + 𝑑𝜈), 𝜔�̂� =

1

√2(𝑑𝑢 − 𝑑𝜈), 𝜔𝑥 = cos 𝑎𝜈 𝑑𝑥, 𝜔�̂� = cosh𝑎𝜈 𝑑𝑦

c) Geodesic equations:

i) 0 = �̈� − 𝑎 cos 𝑎𝜈 sin 𝑎𝜈 �̇�2 + 𝑎 cosh𝑎𝜈 sinh𝑎𝜈 �̇�2

ii) 0 = �̈�

iii) 0 = �̈� − 2𝑎 tan 𝑎𝜈 �̇��̇�

iv) 0 = �̈� + 2𝑎 tanh𝑎𝜈 �̇��̇�

d) Orthonormal tetrad (𝑢, 𝜈, 𝑥, 𝑦)

i) 𝑙𝑎 = (1, 0, 0, 0), 𝑙𝑎 = (0, 1, 0, 0)

ii) 𝑛𝑎 = (0, 1, 0, 0), 𝑛𝑎 = (1, 0, 0, 0)

iii) 𝑚𝑎 =1

√2(0, 0, cos 𝑎𝜈 , 𝑖 cosh𝑎𝜈), 𝑚𝑎 =

1

√2(0, 0, −

1

cos𝑎𝜈, −𝑖

1

cosh𝑎𝜈)

iv) 𝑚𝑎̅̅ ̅̅ =1

√2(0, 0, cos 𝑎𝜈 , −𝑖 cosh𝑎𝜈), �̅�𝑎 =

1

√2(0, 0, −

1

cos𝑎𝜈, 𝑖

1

cosh𝑎𝜈)

e) Spin coefficients.

i) 𝜋 = 𝜈 = 𝜆 = 𝜇 = 𝜅 = 𝜏 = 휀 = 𝛾 = 𝛼 = 𝛽 = 0

ii) 𝜌 =𝑎

2(tan 𝑎𝜈 − tanh𝑎𝜈),

iii) 𝜎 =𝑎

2(tan 𝑎𝜈 + tanh𝑎𝜈)

f) Petrov classification: 130Ψ0 ≠ 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

50) 131The Narai Space-time: A solution to the vacuum field equations with positive cosmological constant.

a) Line element: 𝑑𝑠2 = −Λ𝜈2𝑑𝑢2 + 2𝑑𝑢𝑑𝑣 −1

Ω2(𝑑𝑥2 + 𝑑𝑦2), Ω = 1 +

Λ

4(𝑥2 + 𝑦2)

b) Basis one-forms: 𝜔�̂� =1

2(−Λ𝜈2 + 1)𝑑𝑢 + 𝑑𝑣, 𝜔�̂� =

1

2(Λ𝜈2 + 1)𝑑𝑢 − 𝑑𝑣, 𝜔�̂� =

1

Ω𝑑𝑥, 𝜔�̂� =

1

Ω𝑑𝑦,

휂𝑖𝑖 = (1,−1,−1,−1)

c) Ricci tensor: 𝑅𝑢𝑢 = Λ2𝜈2, 𝑅𝑢𝜈 = −Λ, 𝑅𝑥𝑥 = 𝑅𝑦𝑦 =

Λ

Ω2

d) Ricci scalar: 𝑅 = −4Λ

e) Spin coefficients: 𝛾 = −1

√2Λ𝜈, 𝛼 = −

Λ

4√2(𝑥 − 𝑖𝑦), 𝛽 =

Λ

4√2(𝑥 + 𝑖𝑦)

f) Weyl scalars: 132Ψ2 = −1

3Λ, ΛNP =

1

6Λ, Φ11 = 0

g) Petrov classification: With Ψ2 ≠ 0 we can conclude that this is a Petrov type D. This means there

are two principal null directions, each doubly repeated. The fact that this spacetime contain Ψ2 and

not Ψ4 or Ψ0 indicates that this spacetime describes electromagnetic fields and not gravitational

radiation. This spacetime represents a vacuum universe that contains electromagnetic fields with

no matter.

51) 133Collision of a Gravitational Wave with an Electromagnetic Wave.

a) Line element: 𝑑𝑠2 = 2𝑑𝑢𝑑𝜈 − cos2 𝑎𝜈 (𝑑𝑥2 + 𝑑𝑦2)

b) Basis one-forms: 𝜔�̂� =1

√2(𝑑𝑢 + 𝑑𝜈), 𝜔�̂� =

1

√2(𝑑𝑢 − 𝑑𝜈), 𝜔𝑥 = cos 𝑎𝜈 𝑑𝑥, 𝜔�̂� = cos 𝑎𝜈 𝑑𝑦

c) Spin coefficient: 𝜌 = 𝑎 tan𝑎𝜈

130 http://www-staff.lboro.ac.uk/~majbg/jbg/book/chap3.pdf 131 (McMahon, p. 318) 132 (McMahon, p. 321) 133 (McMahon, p. 322)

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52) 134The 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 con-

stant.135

a) Line element: 𝑑𝑠2 =1

2𝜔2((𝑑𝑡 + 𝑒𝑥𝑑𝑧)2 − 𝑑𝑥2 − 𝑑𝑦2 −

1

2𝑒2𝑥𝑑𝑧2)

b) Ricci scalar: 𝑅 = 2𝜔2

53) 136Warp-Drive Space-time.

a) Line element: 𝑑𝑠2 = −𝑑𝑡2 + [𝑑𝑥 − 𝑉𝑠(𝑡)𝑓(𝑟𝑠)𝑑𝑡2]2 + 𝑑𝑦2 + 𝑑𝑧2

54) 137Worm Hole Geometry.

a) Line element: 𝑑𝑠2 = −𝑑𝑡2 + 𝑑𝑟2 + (𝑏2 + 𝑟2)(𝑑휃2 + sin2 휃 𝑑𝜙2)

b) Conservation equation: The travel time through a worm-hole: Δ𝜏 =2𝑅

𝑈

c) Geodesic equations.

i) �̈� = 0

ii) �̈� = 𝑟휃̇2 + 𝑟 sin2 휃 �̇�2

iii) 휃̈ = sin휃 cos 휃 �̇�2 −2𝑟

(𝑏2+𝑟2)�̇�휃̇

iv) �̈� = −2𝑟

(𝑏2+𝑟2)�̇��̇� − 2 cot 휃 휃̇�̇�

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. Choquet-Bruhat, Y. (2015). Introduction to General Relativity, Black Holes and Cosmology. Oxford: Oxford

University Press. 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. Hartle, J. B. (2003). Gravity An Introduction to Einstein's General Relativity. San Fransisco, CA: Addison

Wesley. Kay, D. C. (1988). Tensor Calculus. McGraw-Hill. McMahon, D. (2006). Relativity Demystified. McGraw-Hill Companies, Inc. Penrose, R. (2004). The Road to Reality. New York: Vintage Books. Scolarpedia. (n.d.). Retrieved from http://www.scholarpedia.org/article/Spin-coefficient_formalism. 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.

a (McMahon, p. 234)

134 (McMahon, 2006, p. 326) Final exam 14 135 http://en.wikipedia.org/wiki/G%C3%B6del_metric 136 (Hartle, Gravity An Introduction to Einstein's General Relativity, 2003, p. 144) Example 7.4 137 (Hartle, Gravity An Introduction to Einstein's General Relativity, 2003) eq (7.39)

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Lots of Calculations in General Relativity

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57 http://physicssusan.mono.net [email protected]

b (McMahon, p. 168) c (Carroll, 2004) d (McMahon, p. 220) e (A.S.Eddington, pp. 85-86) f (d'Inverno, p. 87) g (McMahon, p. 161) h (McMahon, 2006, p. 138) i An excellent qualitative explanation of the cosmological constant, you can find in (Greene, 2004, s. 273-279) j (d'Inverno, 1992, p. 172)


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