Lecture Notes On General Relativity

Øyvind Grøn

Oslo College, Department of engineering, Cort Adelers gt. 30, N-0254 Oslo, Norway and Department of Physics, University of Oslo, Box 1048 Blindern, N-0316, Norway May 8, 2006 Preface

These notes are a transcript of lectures delivered by Øyvind Grøn during the spring of 1997 at the University of Oslo. Two compendia, (Grøn and Flø 1984) and (Ravndal 1978) were provided by Grøn as additional reference material during the lectures. The present version of this document is an extended and corrected version of a set of Lecture Notes which were typesetted by S. Bard, Andreas O. Jaunsen, Frode Hansen and Ragnvald J. Irgens using LATEX2. Svend E. Hjelmeland has made many useful suggestions which have improved the text. While we hope that these typeset notes are of benefit particularly to stu- dents of general relativity and look forward to their comments, we welcome all interested readers and accept all feedback with thanks. All comment may be sent to the author either by e-mail or snail mail. Øyvind Grøn Fysisk Institutt Universitetet i Oslo P.O.Boks 1048, Blindern 0315 OSLO E-mail: [email protected]

Contents

List of Figures v

List of Definitions ix

List of Examples xi

1 Newton’s law of universal gravitation 1 1.1 The force law of gravitation ...... 1 1.2 Newton’s law of gravitation in its local form ...... 2 1.3 Tidal Forces ...... 5 1.4 The Principle of Equivalence ...... 9 1.5 The general principle of relativity ...... 10 1.6 The covariance principle ...... 11 1.7 Mach’s principle ...... 12

2 Vectors, Tensors and Forms 13 2.1 Vectors ...... 13 2.1.1 4-vectors ...... 14 2.1.2 Tangent vector fields and coordinate vectors ...... 17 2.1.3 Coordinate transformations ...... 20 2.1.4 Structure coefficients ...... 23 2.2 Tensors ...... 25 2.2.1 Transformation of tensor components ...... 27 2.2.2 Transformation of basis 1-forms ...... 27 2.2.3 The metric tensor ...... 28 2.3 Forms ...... 31

3 Accelerated Reference Frames 34 3.1 Rotating reference frames ...... 34 3.1.1 Space geometry ...... 34 3.1.2 Angular acceleration in the rotating frame ...... 38 3.1.3 Gravitational time dilation ...... 40 3.1.4 Path of photons emitted from axes in the rotating refer- ence frame (RF) ...... 41 3.1.5 The Sagnac effect ...... 42 3.2 Hyperbolically accelerated reference frames ...... 43

i 4 Covariant Differentiation 49 4.1 Differentiation of forms ...... 49 4.1.1 Exterior differentiation ...... 49 4.1.2 Covariant derivative ...... 51 4.2 The Christoffel Symbols ...... 53 4.3 Geodesic curves ...... 56 4.4 The covariant Euler-Lagrange equations ...... 57 4.5 Application of the Lagrangian formalism to free particles . . . . . 59 4.5.1 Equation of motion from Lagrange’s equation ...... 60 4.5.2 Geodesic world lines in spacetime ...... 61 4.5.3 Gravitational Doppler effect ...... 70 4.6 The Koszul connection ...... 71 α α 4.7 Connection coefficients Γ µν and structure coefficients c µν in ... . 74 4.8 Covariant differentiation of vectors, forms and tensors ...... 75 4.8.1 Covariant differentiation of a vector in an arbitrary basis . 75 4.8.2 Covariant differentiation of forms ...... 75 4.8.3 Generalization for tensors of higher rank ...... 77 4.9 The Cartan connection ...... 77

5 Curvature 81 5.1 The Riemann curvature tensor ...... 81 5.2 Differential geometry of surfaces ...... 86 5.2.1 Surface curvature, using the Cartan formalism ...... 89 5.3 The Ricci identity ...... 90 5.4 Bianchi’s 1st identity ...... 90 5.5 Bianchi’s 2nd identity ...... 91

6 Einstein’s Field Equations 93 6.1 Energy-momentum conservation ...... 93 6.1.1 Newtonian fluid ...... 93 6.1.2 Perfect fluids ...... 95 6.2 Einstein’s curvature tensor ...... 95 6.3 Einstein’s field equations ...... 96 6.4 The “geodesic postulate” as a consequence of the field equations . 98

7 The Schwarzschild spacetime 100 7.1 Schwarzschild’s exterior solution ...... 100 7.2 Radial free fall in Schwarzschild spacetime ...... 104 7.3 Light cones in Schwarzschild spacetime ...... 106 7.4 Analytical extension of the Schwarzschild spacetime ...... 108 7.5 Embedding of the Schwarzschild metric ...... 110 7.6 Deceleration of light ...... 110 7.7 Particle trajectories in Schwarzschild 3-space ...... 112 7.7.1 Motion in the equatorial plane ...... 114 7.8 Classical tests of Einstein’s general theory of relativity ...... 116 7.8.1 The Hafele-Keating experiment ...... 116

ii 7.8.2 Mercury’s perihelion precession ...... 118 7.8.3 Deflection of light ...... 119

8 Black Holes 121 8.1 ’Surface gravity’:gravitational acceleration on the horizon of a black hole ...... 121 8.2 Hawking radiation:radiation from a black hole (1973) ...... 122 8.3 Rotating Black Holes: The Kerr metric ...... 123 8.3.1 Zero-angular-momentum-observers (ZAMO’s) ...... 124 8.3.2 Does the Kerr space have a horizon? ...... 125

9 Schwarzschild’s Interior Solution 127 9.1 Newtonian incompressible star ...... 127 9.2 The pressure contribution to the gravitational mass of a static, spherical symmetric system ...... 129 9.3 The Tolman-Oppenheimer-Volkov equation ...... 130 9.4 An exact solution for incompressible stars - Schwarzschild’s inte- rior solution ...... 132

10 Cosmology 134 10.1 Comoving coordinate system ...... 134 10.2 Curvature isotropy - the Robertson-Walker metric ...... 135 10.3 Cosmic dynamics ...... 136 10.3.1 Hubbles law ...... 136 10.3.2 Cosmological redshift of light ...... 136 10.3.3 Cosmic fluids ...... 138 10.3.4 Isotropic and homogeneous universe models ...... 139 10.4 Some cosmological models ...... 142 10.4.1 Radiation dominated model ...... 142 10.4.2 Dust dominated model ...... 143 10.4.3 Friedmann-Lemaître model ...... 147 10.5 Inflationary Cosmology ...... 157 10.5.1 Problems with the Big Bang Models ...... 157 10.5.2 Cosmic Inflation ...... 160

Bibliography 165

iii

List of Figures

1.1 Newton’s law of universal gravitation ...... 1 1.2 Newton’s law of gravitation in its local form ...... 2 1.3 The definition of solid angle dΩ ...... 5 1.4 Tidal Forces ...... 6 1.5 A small Cartesian coordinate system at a distance R from a mass M...... 7 1.6 An elastic, circular ring falling freely in the Earth’s gravitational field ...... 8

2.1 Closed polygon (linearly dependent) ...... 13 2.2 Carriage at rest (top) and with velocity ~v (bottom) ...... 14 2.3 World-lines in a Minkowski diagram ...... 16 2.4 No position vectors ...... 17 2.5 Tangentplane ...... 17 2.6 Proper time ...... 20 2.7 Coordinate transformation, flat space...... 22 2.8 Basis-vectors ~e1 and ~e2 ...... 29 2.9 The covariant- and contravariant components of a vector . . . . . 30

3.1 Simultaneity in rotating frames ...... 36 3.2 Rotating system: Distance between points on the circumference . 36 3.3 Rotating system: Discontinuity in simultaneity ...... 37 3.4 Rotating system: Angular acceleration ...... 39 3.5 Rotating system: Distance increase ...... 39 3.6 Rotating system: Lorentz contraction ...... 40 3.7 The Sagnac effect ...... 42 3.8 Hyperbolic acceleration ...... 44 3.9 Simultaneity and hyperbolic acceleration ...... 46 3.10 The hyperbolically accelerated reference system ...... 48

4.1 Parallel transport ...... 55 4.2 Different world-lines connecting P1 and P2 in a Minkowski diagram 57 4.3 Geodesic on a flat surface ...... 59 4.4 Geodesic on a sphere ...... 59 4.5 Timelike geodesics ...... 61 4.6 Projectiles in 3-space ...... 63 4.7 Geodesics in rotating reference frames ...... 64

v 4.8 Coordinates on a rotating disc ...... 65 4.9 Projectiles in accelerated frames ...... 66 4.10 The twin “paradox” ...... 68 4.11 Rotating coordinate system ...... 72

5.1 Parallel transport of A~ ...... 81 5.2 Parallel transport of a vector along a triangle of angles 90◦ is rotated 90◦ ...... 82 5.3 Geometry of parallel transport ...... 83 5.4 Surface geometry ...... 86

7.1 Light cones in Schwarzschild spacetime ...... 107 7.2 Light cones in Schwarzschild spacetime ...... 107 7.3 Embedding of the Schwarzschild metric ...... 111 7.4 Deceleration of light ...... 111 7.5 Newtonian centrifugal barrier ...... 115 7.6 Gravitational collapse ...... 116 7.7 Deflection of light ...... 119

8.1 Static border and horizon of a Kerr black hole ...... 126

9.1 Hydrostatic equilibrium ...... 128

10.1 Schematic representation of cosmological redshift ...... 137 10.2 Expansion of a radiation dominated universe ...... 143 10.3 The size of the universe ...... 146 10.4 Expansion factor ...... 147 10.5 The expansion factor as function of cosmic time in units of the age of the universe...... 150 10.6 The Hubble parameter as function of cosmic time...... 150 10.7 ...... 151 10.8 The deceleration parameter as function of cosmic time...... 152 10.9 The ratio of the point of time when cosmic decelerations turn over to acceleration to the age of the universe...... 153 10.10The cosmic red shift of light emitted at the turnover time from deceleration to acceleration as function of the present relative density of vacuum energy...... 154 10.11The critical density in units of the constant density of the vacuum energy as function of time...... 154 10.12The relative density of the vacuum energy density as function of time...... 155 10.13The density of matter in units of the density of vacuum energy as function of time...... 156 10.14The relative density of matter as function of time...... 156 Ω ln( t ) ln( t ) = 10.15Rate of change of Λ as function of t0 . The value t0 40 corresponds to the cosmic point of time t 1s...... 158 − 0 ∼ 10.16The shape of the potential depends on the sign of µ2...... 161

vi 10.17The temperature dependence of a Higgs potential with a first order phase transition...... 162

vii

List of Definitions

1.2.1 Solid angle ...... 4 2.1.1 4-velocity ...... 15 2.1.2 4-momentum ...... 15 2.1.3 4-acceleration ...... 16 2.1.4 Reference frame ...... 17 2.1.5 Coordinate system ...... 18 2.1.6 Comoving coordinate system ...... 18 2.1.7 Orthonormal basis ...... 18 2.1.8 Coordinate basis vectors...... 18 2.1.9 Coordinate basis vectors...... 21 2.1.10Orthonormal basis ...... 22 2.1.11Commutators between vectors ...... 23 ρ 2.1.12Structure coefficients c µν ...... 24 2.2.1 Multilinear function, tensors ...... 26 2.2.2 Tensor product ...... 26 2.2.3 The metric tensor ...... 28 2.2.4 Contravariant components ...... 29 2.3.1 p-form ...... 32 3.2.1 Born-stiff motion ...... 44 4.3.1 Geodesic curves ...... 56 4.6.1 Koszul’s connection coeffecients in an arbitrary basis ...... 72 4.8.1 Covariant derivative of a vector ...... 75 4.8.2 Covariant directional derivative of a one-form field ...... 75 4.8.3 Covariant derivative of a one-form ...... 76 4.8.4 Covariant derivative of a tensor ...... 77 4.9.1 Exterior derivative of a basis vector ...... 77 ν 4.9.2 Connection forms Ω µ ...... 78 4.9.3 Scalar product between vector and 1-form ...... 78 5.5.1 Contraction ...... 91 7.1.1 Physical singularity ...... 104 7.1.2 Coordinate singularity ...... 104 8.3.1 Horizon ...... 125

ix

List of Examples

2.1.1 Photon clock ...... 14 2.1.2 Coordinate transformation ...... 21 2.1.3 Relativistic Doppler Effect ...... 23 2.1.4 Structure coefficients in planar polar coordinates ...... 25 2.2.1 Example of a tensor ...... 27 2.2.2 A mixed tensor of rank 3 ...... 28 2.2.3 Cartesian coordinates in a plane ...... 29 2.2.4 Basis-vectors in plane polar-coordinates ...... 29 2.2.5 Non-diagonal basis-vectors ...... 29 2.2.6 Cartesian coordinates in a plane ...... 31 2.2.7 Plane polar coordinates ...... 31 2.3.1 antisymmetric combinations ...... 32 2.3.2 antisymmetric combinations ...... 32 2.3.3 A 2-form in 3-space ...... 32 4.1.1 Outer product of 1-forms in 3-space ...... 50 4.1.2 The derivative of a vector field with rotation ...... 52 4.2.1 The Christoffel symbols in plane polar coordinates ...... 54 4.3.1 vertical motion of free particle in hyperb. acc. ref. frame ...... 56 4.5.1 How geodesics in spacetime can give parabolas in space ...... 61 4.5.2 Spatial geodesics described in the reference frame of a rotating disc. 62 4.5.3 Christoffel symbols in a hyperbolically accelerated reference frame . 65 4.5.4 Vertical projectile motion in a hyperbolically accelerated reference frame ...... 66 4.5.5 The twin “paradox” ...... 68 4.5.6 Measurements of gravitational Doppler effects (Pound and Rebka 1960) 71 4.6.1 The connection coefficients in a rotating reference frame...... 72 4.6.2 Acceleration in a non-rotating reference frame (Newton) ...... 73 4.6.3 The acceleration of a particle, relative to the rotating reference frame 73 4.9.1 Cartan-connection in an orthonormal basis field in plane polar coord. 79 6.1.1 Energy momentum tensor for a Newtonian fluid ...... 94 10.4.1Age-redshift relation for dust dominated universe with k = 0 . . . . 145

xi

Chapter 1

Newton’s law of universal gravitation

1.1 The force law of gravitation

m

˜r F˜

M

Figure 1.1: Newton’s law of universal gravitation states that the force between two masses is attractive, acts along the line joining them and is inversely pro- portional to the distance separating the masses.

M M F~ = mG ~r = mG ~e (1.1) − r3 − r2 r Let V be the potential energy of m (see figure 1.1). Then

∂V F~ = V (~r), Fi = (1.2) −∇ −∂xi

For a spherical mass distribution: V (~r) = mG M , with zero potential − r infinitely far from the center of M. Newton’s law of gravitation is valid for “small” velocities, i.e. velocities much smaller than the velocity of light and “weak” fields. Weak fields are fields in which the gravitational potential energy of a test particle is very small compared to its rest mass energy. (Note that here one is interested only in the absolute values of the above quantities and not their sign). M GM mG mc2 r . (1.3) r  ⇒  c2

1 2 Chapter 1. Newton’s law of universal gravitation

2GM The Schwarzschild radius for an object of mass M is Rs = c2 . Far outside the Schwarzschild radius we have a weak field. To get a feeling for magnitudes consider that Rs u 1 cm for the Earth which is to be compared with RE u 6400 km. That is, the gravitational field at the Earth’s surface can be said to be weak! This explains, in part, the success of the Newtonian theory.

1.2 Newton’s law of gravitation in its local form

i Let P be a point in the field (see figure 1.2) with position vector ~r = x ~ei and ~ i0 let the gravitating point source be at r0 = x ~ei0 . Newton’s law of gravitation for a continuous distribution of mass is

~ ~ ∞ ~ ~r r0 3 F = mG ρ(r0) − d r0 − r ~r r~ 3 (1.4) Z | − 0| = V (~r) −∇

See figure (1.2) for symbol definitions.

P ~r r~ − 0 ~r

~ r0

Figure 1.2: Newton’s law of gravitation in its local form. 1.2 Newton’s law of gravitation in its local form 3

Let’s consider equation (1.4) term by term. 1 ∂ 1 = ~ei ~ j j 1/2 ∇ ~r r0 ∂xi (x x 0 )(x x ) | − | − j − j0 ∂ 1/2  j j0  − = ~ei (x x )(xj xj0 ) ∂xi − − 1 h ∂xj i 3/2 k k0 − = ~ei − 2(xj xj0 ) (x x )(xk xk0 ) 2 − ∂xi − − h i (xj xj0 )δi (1.5) = ~e − j i 3/2 − [(xk xk0 )(x x )] − k − k0 (xi xi0 ) = ~ei − 3/2 − (xj xj0 )(x x ) − j − j0 ~r r~ = − 0  − ~r r~ 3 | − 0| Now equations (1.4) and (1.5) together ⇒ ~ ρ(r0) 3 V (~r) = mG d r0 (1.6) − ~r r~ Z | − 0| Gravitational potential at point P : ~ V (~r) ρ(r0) 3 φ(~r) = G d r0 ≡ m − ~r r~ Z | − 0| ~ ~ ~r r0 3 φ(~r) = G ρ(r0) − d r0 (1.7) ⇒ ∇ ~r r~ 3 Z | − 0| ~ 2 ~ ~r r0 3 φ(~r) = G ρ(r0) − d r0 ⇒ ∇ ∇· ~r r~ 3 Z | − 0| The above equation simplifies considerably if we calculate the divergence in the integrand. Note that “ ” ∇ operates on ~r ~ ~r r0 ~r ~ 1 only! − = ∇· + (~r r0) ∇· ~r r~ 3 ~r r~ 3 − · ∇ ~r r~ 3 | − 0| | − 0| | − 0| ~ 3 ~ 3(~r r0) = (~r r0) − ~r r~ 3 − − · ~r r~ 5 (1.8) | − 0| | − 0| 3 3 = ~r r~ 3 − ~r r~ 3 | − 0| | − 0| = 0 ~r = r~ ∀ 6 0 We conclude that the Newtonian gravitational potential at a point in a gravi- tational field outside a mass distribution satisfies Laplace’s equation

2φ = 0 (1.9) ∇ 4 Chapter 1. Newton’s law of universal gravitation

Digression 1.2.1 (Dirac’s delta function) The Dirac delta function has the following properties:

1. δ(~r r~ ) = 0 ~r = r~ − 0 ∀ 6 0 2. δ(~r r~ )d3r = 1 when ~r = r~ is contained in the integration domain. The − 0 0 0 integral is identically zero otherwise. R 3. f(r~ )δ(~r r~ )d3r = f(~r) 0 − 0 0 R

~r r~ 3 A calculation of the integral − 0 d r0 which is valid also in the case where ~r r~ 3 ∇·| − 0| the field point is inside the Rmass distribution is obtained through the use of Gauss’ integral theorem:

3 A~d r0 = A~ d~s, (1.10) ∇· · Zv Is where s is the boundary of v (s = ∂v is an area).

Definition 1.2.1 (Solid angle)

ds0 dΩ ⊥ (1.11) ≡ ~r r~ 2 | − 0| ~ where ds0 is the projection of the area ds0 normal to the line of sight. ds0 is the ⊥ ~ ⊥ component vector of ds0 along the line of sight which is equal to the normal vector of ds0 (see figure (1.3)). ⊥

Now, let’s apply Gauss’ integral theorem. ~ ~ ~r r0 3 ~r r0 ds0 − d r0 = − ds~ = ⊥ = dΩ (1.12) ~ 3 ~ 3 0 ~ 2 ∇· ~r r0 ~r r0 · ~r r0 Zv | − | Is | − | Is | − | Is So that, ~ ~r r0 3 4π if P is inside the mass distribution, − d r0 = (1.13) ~ 3 ∇· ~r r0 (0 if P is outside the mass distribution. Zv | − | The above relation is written concisely in terms of the Dirac delta function: ~ ~r r0 ~ − = 4πδ(~r r0) (1.14) ∇· ~r r~ 3 − | − 0| 1.3 Tidal Forces 5

~ ds0 ⊥

P ~r r~ − 0 ~r r~ ~r ds~ = − 0 ds~ 0 ~r r~ 0 dΩ ⊥ | − 0| · ~ r0

~ ds0 normal to bounding surface

Figure 1.3: The solid angle dΩ is defined such that the surface of a sphere subtends 4π at the center

We now have ~ 2 ~ ~r r0 3 φ(~r) = G ρ(r0) − d r0 ∇ ∇· ~r r~ 3 Z | − 0| 3 (1.15) = G ρ(r~ )4πδ(~r r~ )d r0 0 − 0 Z = 4πGρ(~r)

Newton’s theory of gravitation can now be expressed very succinctly indeed!

1. Mass generates gravitational potential according to

2φ = 4πGρ (1.16) ∇ 2. Gravitational potential generates motion according to

~g = φ (1.17) −∇ where ~g is the field strength of the gravitational field.

1.3 Tidal Forces

Tidal force is difference of gravitational force on two neighboring particles in a gravitational field. The tidal force is due to the inhomogeneity of a gravitational field. In figure 1.4 two points have a separation vector ζ~. The position vectors of 1 and 2 are ~r and ~r + ζ~, respectively, where ζ~ ~r . The gravitational forces on | |  | | 6 Chapter 1. Newton’s law of universal gravitation

2 ζ 1

F2

F1

Figure 1.4: Tidal Forces

a mass m at 1 and at 2 are F~ (~r) and F~ (~r + ζ~). By means of a Taylor expansion to lowest order in ζ~ we get for the i-component of the tidal force | |

∂Fi f = F (~r + ζ~) F (~r) = ζ . (1.18) i i − i j ∂xj  ~r The corresponding vector equation is

f~ = (ζ~ ) F~ . (1.19) · ∇ ~r Using that F~ = m φ, (1.20) − ∇ the tidal force may be expressed in terms of the gravitational potential according to f~ = m(ζ~ ) φ. (1.21) − · ∇ ∇ It follows that in a local Cartesian coordinate system, the i-coordinate of the relative acceleration of the particles is

2 2 d ζi ∂ φ = ζ . (1.22) dt2 − ∂xi∂xj j  ~r Let us look at a few simple examples. In the first one ζ~ has the same direction as ~g. Consider a small Cartesian coordinate system at a distance R from a mass M (see figure 1.5). If we place a particle of mass m at a point (0, 0, +z), it will, according to eq. (1.1) be acted upon by a force GM F (+z) = m (1.23) z − (R + z)2

while an identical particle at the origin will be acted upon by the force GM F (0) = m . (1.24) z − R2 1.3 Tidal Forces 7

z

¡ ¡ m

Fz(+z )

¨ ¤ ¤ ¤ ¦ ¦

¥ ¥ ¥ § §

¨ ¤ ¤ ¤ ¦ ¦

¥ ¥ ¥ § §

¨ ¤ ¤ ¤ ¦ ¦

¥ ¥ § §

¥ y

¨ ¤ ¤ ¤ ¦ ¦

¥ ¥ ¥ § §

¨ ¤ ¤ ¤ ¦ ¦

¥ ¥ ¥ § §

¨ ¤ ¤ ¦ ¦

¤ Fz(0)

¨

¨

¨

¨ ¨

¨ R

¨

¨

¨

¨

¨

¨

¨

¢ ¢ £ £ M

Figure 1.5: A small Cartesian coordinate system at a distance R from a mass M.

If this little coordinate system is falling freely towards M, an observer at the origin will say that the particle at (0, 0, +z) is acted upon by a force GM f = F (z) F (0) 2mz (1.25) z z − z ≈ R3 directed away from the origin, along the positive z-axis. We have assumed z R. This is the tidal force.  In the same way particles at the points (+x, 0, 0) and (0, +y, 0) are attracted towards the origin by tidal forces GM f = mx , (1.26) x − R3 GM f = my . (1.27) y − R3 Eqs. (1.25)–(1.27) have among others the following consequence: If an elastic, circular ring is falling freely in the Earth’s gravitational field, as shown in figure 1.6, it will be stretched in the vertical direction and compressed in the horizontal direction. In general, tidal forces cause changes of shape. The tidal forces from the Sun and the Moon cause flood and ebb on the Earth. Let us consider the effect due to the Moon. We then let M be the mass of the Moon, and choose a coordinate system with origin at the Earth’s center. The tidal force per unit mass at a point is the negative gradient of the tidal potential GM 1 1 GM φ(~r) = z2 x2 y2 = r2(3 cos2 θ 1), (1.28) − R3 − 2 − 2 − 2R3 −   8 Chapter 1. Newton’s law of universal gravitation

Figure 1.6: An elastic, circular ring falling freely in the Earth’s gravitational field

where we have introduced spherical coordinates, z = r cos θ, x2 + y2 = r2 sin2 θ, R is the distance between the Earth and the Moon, and the radius r of the spherical coordinate is equal to the radius of the Earth. The potential at a height h above the surface of the Earth has one term, mgh, due to the attraction of the Earth and one given by eq. (1.28), due to the attraction of the Moon. Thus,

GM Θ(r) = gh r2(3 cos2 θ 1). (1.29) − 2R3 −

At equilibrium, the surface of the Earth will be an equipotential surface, given by Θ = constant. The height of the water at flood, θ = 0 or θ = π, is therefore GM r 2 h = h + , (1.30) flood 0 gR R   π where h0 is an unknown constant. The height of the water at ebb (θ = 2 or 3π θ = 2 ) is 1 GM r 2 h = h . (1.31) ebb 0 − 2 gR R   The height difference between flood and ebb is therefore

3 GM r 2 ∆h = . (1.32) 2 gR R   For a numerical result we need the following values:

M = 7.35 1025g, g = 9.81m/s2, (1.33) Moon · R = 3.85 105km, r = 6378km. (1.34) · Earth With these values we find ∆h = 53cm, which is typical of tidal height differences. 1.4 The Principle of Equivalence 9

1.4 The Principle of Equivalence

Galilei investigated experimentally the motion of freely falling bodies. He found that they moved in the same way, regardless what sort of material they consisted of and what mass they had. In Newton’s theory of gravitation mass appears in two different ways; as gravitational mass, mG, in the law of gravitation, analogously to charge in Coulomb’s law, and as inertial mass, mI in Newton’s 2nd law. The equation of motion of a freely falling particle in the field of gravity from a spherical body with mass M then takes the form 2 d ~r mG M 2 = G 3 ~r. (1.35) dt − mI r The results of Galilei’s measurements imply that the quotient between gravita- tional and inertial mass must be the same for all bodies. With a suitable choice of units, we then obtain mG = mI. (1.36) Measurements performed by the Hungarian baron Eötvös around the turn of the century indicated that this equality holds with an accuracy better than 8 mI 13 10− . More recent experiments have given the result 1 < 9 10− . | mG − | · Einstein assumed the exact validity of eq.(1.52). He did not consider this as an accidental coincidence, but rather as an expression of a fundamental principle, called the principle of equivalence. A consequence of this principle is the possibility of removing the effect of a gravitational force by being in free fall. In order to clarify this, Einstein considered a homogeneous gravitational field in which the acceleration of gravity, g, is independent of the position. In a freely falling, non-rotating reference frame in this field, all free particles move according to d2~r m = (m m )~g = 0, (1.37) I dt2 G − I where eq. (1.36) has been used. This means that an observer in such a freely falling reference frame will say that the particles around him are not acted upon by forces. They move with constant velocities along straight paths. In other words, such a reference frame is inertial. Einstein’s heuristic reasoning suggests equivalence between inertial frames in regions far from mass distributions, where there are no gravitational fields, and inertial frames falling freely in a gravitational field. This equivalence between all types of inertial frames is so intimately connected with the equivalence between gravitational and inertial mass, that the term “principle of equivalence” is used whether one talks about masses or inertial frames. The equivalence of different types of inertial frames encompasses all types of physical phenomena, not only particles in free fall. The principle of equivalence has also been formulated in an “opposite” way. An observer at rest in a homogeneous gravitational field, and an observer in 10 Chapter 1. Newton’s law of universal gravitation

an accelerated reference frame in a region far from any mass distributions, will obtain identical results when they perform similar experiments. An inertial field caused by the acceleration of the reference frame, is equivalent to a field of gravity caused by a mass distribution, as far is tidal effects can be ignored.

1.5 The general principle of relativity

The principle of equivalence led Einstein to a generalization of the special princi- ple of relativity. In his general theory of relativity Einstein formulated a general principle of relativity, which says that not only velocities are relative, but accel- erations, too. Consider two formulations of the special principle of relativity.

S1 All laws of Nature are the same (may be formulated in the same way) in all inertial frames.

S2 Every inertial observer can consider himself to be at rest.

These two formulations may be interpreted as different formulations of a single principle. But the generalization of S1 and S2 to the general case, which encompasses accelerated motion and non-inertial frames, leads to two different principles G1 and G2.

G1 The laws of Nature are the same in all reference frames.

G2 Every observer can consider himself to he at rest.

In the literature both G1 and G2 are mentioned as the general principle of relativity. But G2 is a stronger principle (i.e. stronger restriction on natural phenomena) than G1. Generally the course of events of a physical process in a certain reference frame, depends upon the laws of physics, the boundary conditions, the motion of the reference frame and the geometry of space-time. The two latter properties are described by means of a metrical tensor. By formulating the physical laws in a metric independent way, one obtains that G1 is valid for all types of physical phenomena. Even if the laws of Nature are the same in all reference frames, the course of events of a physical process will, as mentioned above, depend upon the motion of the reference frame. As to the spreading of light, for example, the law is that light follows null-geodesic curves (see ch. 4). This law implies that the path of a light particle is curved in non-inertial reference frames and straight in inertial frames. The question whether G2 is true in the general theory of relativity has been thoroughly discussed recently, and the answer is not clear yet. 1.6 The covariance principle 11

1.6 The covariance principle

The principle of relativity is a physical principle. It is concerned with physical phenomena. This principle motivates the introduction of a formal principle, called the covariance principle: The equations of a physical theory shall have the same form in every coordinate system. This principle is not concerned directly with physical phenomena. The principle may be fulfilled for every theory by writing the equations in a form- invariant i.e. covariant way. This may he done by using tensor (vector) quanti- ties, only, in the mathematical formulation of the theory. The covariance principle and the equivalence principle may be used to obtain a description of what happens in the presence of gravitation. We then start with the physical laws as formulated in the special theory of relativity. Then the laws are written in a covariant form, by writing them as tensor equations. They are then valid in an arbitrary, accelerated system. But the inertial field (“fictive force”) in the accelerated frame is equivalent to a gravitational field. So, starting with in a description referred to an inertial frame, we have obtained a description valid in the presence of a gravitational field. The tensor equations have in general a coordinate independent form. Yet, such form-invariant, or covariant, equations need not fulfill the principle of rel- ativity. This is due to the following circumstances. A physical principle, for example the principle of relativity, is concerned with observable relationships. Therefore, when one is going to deduce the observable consequences of an equation, one has to establish relations between the tensor-components of the equation and observable physical quantities. Such relations have to be defined; they are not determined by the covariance principle. From the tensor equations, that are covariant, and the defined relations between the tensor components and the observable physical quantities, one can deduce equations between physical quantities. The special principle of relativity, for example, demands that the laws which these equations express must be the same with reference to every inertial frame The relationships between physical quantities and tensors (vectors) are the- ory dependent. The relative velocity between two bodies, for example, is a vector within Newtonian kinematics. However, in the relativistic kinematics of four-dimensional space-time, an ordinary velocity, which has only three com- ponents, is not a vector. Vectors in space-time, so called 4-vectors, have four components. Equations between physical quantities are not covariant in general. For example, Maxwell’s equations in three-vector-form are not invariant un- der a Galilei transformation. However, if these equations are rewritten in tensor- form, then neither a Galilei transformation nor any other transformation will change the form of the equations. If all equations of a theory are tensor equations, the theory is said to be given a manifestly covariant form. A theory that is written in a manifestly covariant form, will automatically fulfill the covariance principle, but it need not fulfill the principle of relativity. 12 Chapter 1. Newton’s law of universal gravitation

1.7 Mach’s principle

Einstein gave up Newton’s idea of an absolute space. According to Einstein all motion is relative. This may sound simple, but it leads to some highly non-trivial and fundamental questions. Imagine that there are only two particles connected by a spring, in the universe. What will happen if the two particles rotate about each other? Will the spring be stretched due to centrifugal forces? Newton would have confirmed that this is indeed what will happen. However, when there is no longer any absolute space that the particles can rotate relatively to, the answer is not so obvious. If we, as observers, rotate around the particles, and they are at rest, we would not observe any stretching of the spring. But this situation is now kinematically equivalent to the one with rotating particles and observers at rest, which leads to stretching. Such problems led Mach to the view that all motion is relative. The motion of a particle in an empty universe is not defined. All motion is motion relatively to something else, i.e. relatively to other masses. According to Mach this implies that inertial forces must be due to a particle’s acceleration relatively to the great masses of the universe. If there were no such cosmic masses, there would not exist inertial forces, like the centrifugal force. In our example with two particles connected by a string, there would not be any stretching of the spring, if there were no cosmic masses that the particles could rotate relatively to. Another example may be illustrated by means of a turnabout. If we stay on this, while it rotates, we feel that the centrifugal forces lead us outwards. At the same time we observe that the heavenly bodies rotate. According to Mach identical centrifugal forces should appear if the turnabout is static and the heavenly bodies rotate. Einstein was strongly influenced by Mach’s arguments, which probably had some influence, at least with regards to motivation, on Einstein’s construction of his general theory of relativity. Yet, it is clear that general relativity does not fulfill all requirements set by Mach’s principle. For example there exist general relativistic, rotating cosmological models, where free particles will tend to rotate relative to the cosmic masses of the model. However, some Machian effects have been shown to follow from the equations of the general theory of relativity. For example, inside a rotating, massive shell the inertial frames, i.e. the free particles, are dragged on and tend to rotate in the same direction as the shell. This was discovered by Lense and Thirring in 1918 and is therefore called the Lense-Thirring effect. More recent investigations of this effect have, among others, lead to the following result (Brill and Cohen 1966): “A massive shell with radius equal to its Schwarzschild radius has often been used as an idealized model of our universe. Our result shows that in such models local inertial frames near the center cannot rotate relatively to the mass of the universe. In this way our result gives an explanation in accordance with Mach’s principle, of the fact that the “fixed stars” is at rest on heaven as observed from an inertial reference frame.” Chapter 2

Vectors, Tensors and Forms

2.1 Vectors

µ µ An expression on the form a ~eµ, where a , µ = 1, 2, ..., n are real numbers, is known as a linear combination of the vectors ~eµ. The vectors ~e1, ..., ~en are said to be linearly independent if there does not exist real numbers aµ = 0 such that aµ~e = 0. 6 µ

Figure 2.1: Closed polygon (linearly dependent)

Geometrical interpretation: A set of vectors are linearly independent if it is not possible to construct a closed polygon of the vectors (even by adjusting their lengths). A set of vectors ~e1, . . . , ~en are said to be maximally linearly independent if ~e , . . . , ~e , ~v are linearly dependent for all vectors ~v = ~e . We define the 1 n 6 µ dimension of a vector-space as the number of vectors in a maximally linearly independent set of vectors of the space. The vectors ~eµ in such a set are known

13 14 Chapter 2. Vectors, Tensors and Forms

as the basis-vectors of the space.

µ ~v + a ~eµ = 0

⇓ ~v = aµ~e (2.1) − µ

The components of ~v are the numbers vµ defined by vµ = aµ ~v = vµ~e . − ⇒ µ

2.1.1 4-vectors

4-vectors are vectors which exist in (4-dimensional) space-time. A 4-vector equation represents 4 independent component equations.

L c

L c v

∆ v t 2

Figure 2.2: Carriage at rest (top) and with velocity ~v (bottom)

Example 2.1.1 (Photon clock) Carriage at rest:

2L ∆t = 0 c 2.1 Vectors 15

Carriage with velocity ~v:

∆t 2 2 2 (v 2 ) + L ∆t = q c ⇓ c2∆t2 = v2∆t2 + 4L2

⇓ 2L 2L/c ∆t0 ∆t = = = (2.2) √c2 v2 1 v2/c2 1 v2/c2 − − − p p

The proper time-interval is denoted by dτ (above it was denoted ∆t0). The proper time-interval for a particle is measured with a standard clock which follows the particle.

Definition 2.1.1 (4-velocity)

dt dx dy dz U~ = c ~e + ~e + ~e + ~e , (2.3) dτ t dτ x dτ y dτ z where t is the coordinate time, measured with clocks at rest in the reference frame.

dxµ U~ = U µ~e = ~e , xµ = (ct, x, y, z), x0 ct µ dτ µ ≡ dt 1 = γ (2.4) dτ 1 v2 ≡ − c2 q U~ = γ(c, ~v), where ~v is the common 3-velocity of the particle.

Definition 2.1.2 (4-momentum)

P~ = m0U~ , (2.5) where m0 is the rest mass of the particle. ~ E P = ( c , p~), where p~ = γm0~v = m~v and E is the relativistic energy.

~ The 4-force or Minkowski-force F~ dP and the ’common force’ f~ = dp~ . ≡ dτ dt Then 1 F~ = γ( f~ ~v, f~) (2.6) c · 16 Chapter 2. Vectors, Tensors and Forms

ct

world line of a material particle

lightcone

tachyons, if they exist, should have v > c

y

x

Figure 2.3: World-lines in a Minkowski diagram

Definition 2.1.3 (4-acceleration)

dU~ A~ = (2.7) dτ The 4-velocity has the scalar value c so that

U~ U~ = c2 (2.8) · −

The 4-velocity identity eq. 2.8 gives U~ A~ = 0, in other words A~ U~ and A~ is · ⊥ space-like. The line element for Minkowski space-time (flat space-time) with Cartesian coordinates is ds2 = c2dt2 + dx2 + dy2 + dz2 (2.9) − In general relativity theory, gravitation is not considered a force. Gravitation is instead described as motion in a curved space-time. A particle in free fall, is in Newtonian gravitational theory said to be only influenced by the gravitational force. According to general relativity theory the particle is not influenced by any force. Such a particle has no 4-acceleration. A~ = 0 implies that the particle is not 6 in free fall. It is then influenced by non-gravitational forces. One has to distinguish between observed acceleration, ie. common 3-acceleration, and the absolute 4-acceleration. 2.1 Vectors 17

2.1.2 Tangent vector fields and coordinate vectors

In a curved space position vectors with finite length do not exist. (See figure 2.4).

N(North pole)

P

Figure 2.4: In curved space,vectors can only exist in tangent planes.The vectors in the tangent plane of N,do not contain the vector −N−→P (dashed line).

Different points in a curved space have different tangent planes. Finite vec- tors do only exist in these tangent planes (See figure 2.5). However, infinitesimal position vectors d~r do exist.

tangent plane of point P:

P

Figure 2.5: In curved space,vectors can only exist in tangent planes 18 Chapter 2. Vectors, Tensors and Forms

Definition 2.1.4 (Reference frame) A reference frame is defined as a continuum of non-intersecting timelike world lines in spacetime.

We can view a reference frame as a set of reference particles with a specified motion. An inertial reference frame is a non-rotating set of free particles.

Definition 2.1.5 (Coordinate system) A coordinate system is a continuum of 4-tuples giving a unique set of coordinates for events in spacetime.

Definition 2.1.6 (Comoving coordinate system) A comoving coordinate system in a frame is a coordinate system where the particles in the reference frame have constant spatial coordinates.

Definition 2.1.7 (Orthonormal basis) An orthonormal basis {~eµˆ} in spacetime is defined by

~etˆ ~etˆ = 1(c = 1) · − (2.10) ~e ~e = δ ˆi · ˆj ˆiˆj

where ˆi and ˆj are space indices.

Definition 2.1.8 (Coordinate basis vectors.) Temporary definition of coordinate basis vector: Assume any coordinate system xµ . { } ∂~r ~e (2.11) µ ≡ ∂xµ

A vector field is a continuum of vectors in a space, where the components are continuous and differentiable functions of the coordinates. Let ~v be a tangent vector to the curve ~r(λ):

d~r ~v = where ~r = ~r[xµ(λ)] (2.12) dλ 2.1 Vectors 19

The chain rule for differentiation yields: d~r ∂~r dxµ dxµ ~v = = = e~ = vµ~e (2.13) dλ ∂xµ dλ dλ µ µ Thus, the components of the tangent vector field along a curve, parameterised by λ, is given by: dxµ vµ = (2.14) dλ In the theory of relativity, the invariant parameter is often chosen to be the proper time. Tangent vector to the world line of a material particle: dxµ uµ = (2.15) dτ These are the components of the 4-velocity of the particle!

Digression 2.1.1 (Proper time of the photon.) Minkowski-space: ds2 = c2dt2 + dx2 − 1 dx = c2dt2 1 2 − − c2 dt (2.16) v2

= 1 c2dt2 − − c2 For a photon,v = c so:
lim ds2 = 0 v c (2.17) → Thus, the spacetime interval between two points on the world line of a photon, is zero! This also means that the proper time for the photon is zero!! (See example 2.1.2).

Digression 2.1.2 (Relationships between spacetime intervals, time and proper time.) Physical interpretation of the spacetime interval for a timelike interval:

ds2 = c2dτ 2 (2.18) − where dτ is the proper time interval between two events, measured on a clock moving in a way, such that it is present on both events (figure 2.6). v2 c2dτ 2 = c2 1 dt2 − − − c2 v2
dτ = 1 2 dt (2.19) ⇒ r − c 20 Chapter 2. Vectors, Tensors and Forms

ct

τ d P 2

P 1

x

Figure 2.6: P1 and P2 are two events in spacetime, separated by a proper time interval dτ.

The time interval between to events in the laboratory, is smaller measured on a moving clock than measured on a stationary one, because the moving clock is ticking slower!

2.1.3 Coordinate transformations

µ µ Given two coordinate systems x and x 0 . { } { } ∂~r ~eµ = (2.20) 0 ∂xµ0

Suppose there exists a coordinate transformation, such that the primed coor- dinates are functions of the unprimed, and vice versa. Then we can apply the chain rule:

∂~r ∂~r ∂xµ ∂xµ ~eµ = = = ~eµ (2.21) 0 ∂xµ0 ∂xµ ∂xµ0 ∂xµ0

µ This is the transformation equation for the basis vectors. ∂x are elements ∂xµ0 of the transformation matrix. Indices that are not sum-indices are called ’free indices’.

Rule: In all terms on each side in an equation, the free indices should behave identically (high or low), and there should be exactly the same indices in all terms! 2.1 Vectors 21

Applying this rule, we can now find the inverse transformation

∂xµ0 ~e = ~e µ µ0 ∂xµ ∂xµ µ0 µ µ0 ~v = v ~eµ = v ~eµ = v ~eµ 0 ∂xµ0 So, the transformation rules for the components of a vector becomes

∂xµ ∂xµ0 vµ = vµ0 ; vµ0 = vµ (2.22) ∂xµ0 ∂xµ

The directional derivative along a curve, parametrised by λ: d ∂ dxµ ∂ = = vµ (2.23) dλ ∂xµ dλ ∂xµ µ dxµ where v = dλ are the components of the tangent vector of the curve. Direc- tional derivative along a coordinate curve: ∂ ∂xµ ∂ ∂ λ = xν = δµ = (2.24) ∂xµ ∂xν ν ∂xµ ∂xν In the primed system: ∂ ∂xµ ∂ = (2.25) ∂xµ0 ∂xµ0 ∂xµ

Definition 2.1.9 (Coordinate basis vectors.) We define the coordinate basis vectors as:

∂ ~e = (2.26) µ ∂xµ

This definition is not based upon the existence of finite position vectors. It applies in curved spaces as well as in flat spaces.

Example 2.1.2 (Coordinate transformation) From Figure 2.7 we see that

x = r cos θ, y = r sin θ (2.27)

Coordinate basis vectors were defined by ∂ e~ (2.28) µ ≡ ∂xµ 22 Chapter 2. Vectors, Tensors and Forms

y eθ

er y

r ey

θ x x ex

Figure 2.7: Coordinate transformation, flat space.

This means that we have ∂ ∂ ∂ ∂ e~ = , e~ = , e~ = , e~ = x ∂x y ∂y r ∂r θ ∂θ (2.29) ∂ ∂x ∂ ∂y ∂ e~ = = + r ∂r ∂r ∂x ∂r ∂y

Using the chain rule and Equations (2.27) and (2.29) we get

e~r = cos θe~x + sin θe~y ∂ ∂x ∂ ∂y ∂ e~ = = + (2.30) θ ∂θ ∂θ ∂x ∂θ ∂y = r sin θe~ + r cos θe~ − x y But are the vectors in (2.30) also unit vectors?

e~ e~ = cos2θ + sin2θ = 1 (2.31) r · r So e~ is a unit vector, e~ = 1. r | r| e~ e~ = r2(cos2θ + sin2θ) = r2 (2.32) θ · θ and we see that e~ is not a unit vector, e~ = r. But we have that e~ e~ = 0 θ | θ| r · θ ⇒ e~ e~ . Coordinate basis vectors are not generally unit vectors. r⊥ θ 2.1 Vectors 23

Definition 2.1.10 (Orthonormal basis) An orthonormal basis is a vector basis consisting of unit vectors that are normal to each other. To show that we are using an orthonormal basis we will use ’hats’ over the indices, ~e . { µˆ}

Orthonormal basis associated with planar polar coordinates: 1 ~e = ~e , ~e = ~e (2.33) rˆ r θˆ r θ

Example 2.1.3 (Relativistic Doppler Effect) The Lorentz transformation is known from special relativity and relates the reference frames of two systems where one is moving with a constant velocity v with regard to the other,

x0 = γ(x vt) −vx t0 = γ(t ) − c2 According to the vector component transformation (2.22), the 4-momentum for a µ E particle moving in the x-direction, P = ( c , p, 0, 0) transforms as

∂xµ0 P µ0 = P µ, ∂xµ E0 = γ(E vp). − hν Using the fact that a photon has energy E = hν and momemtum p = c , where h is Planck’s constant and ν is the photon’s frequency, we get the equation for the frequency shift known as the relativistic Doppler effect,

v v 1 c ν ν0 = γ(ν ν) = − − c 1 v
1 + v − c c q

ν0 c v = − (2.34) ν c + v r

2.1.4 Structure coefficients Definition 2.1.11 (Commutators between vectors) The commutator between two vectors, ~u and ~v, is defined as

[~u , ~v] ~u~v ~v~u (2.35) ≡ − 24 Chapter 2. Vectors, Tensors and Forms

where ~u~v is defined as ∂ ∂ ~u~v uµe~ (vν e~ ) = uµ (vν ) (2.36) ≡ µ ν ∂xµ ∂xν

We can think of a vector as a linear combination of partial derivatives. We get: ∂vν ∂ ∂2 ~u~v = uµ + uµvν ∂xµ ∂xν ∂xµ∂xν (2.37) ∂vν ∂2 = uµ e~ + uµvν ∂xµ ν ∂xµ∂xν Due to the last term, ~u~v is not a vector. ∂ ∂ ~v~u = vν (uµ ) ∂xν ∂xµ ∂uµ ∂2 = vν e~ + vν uµ ∂xν µ ∂xν ∂xµ

∂vν ∂uµ (2.38) ~u~v ~v~u = uµ e~ vν e~ − ∂xµ ν − ∂xν µ

µ ∂uν v ∂xµ e~ν ∂vν | ∂u{zν } = (uµ vµ )e~ ∂xµ − ∂xµ ν Here we have used that ∂2 ∂2 = (2.39) ∂xµ∂xν ∂xν∂xµ The Einstein comma notation ⇒ ~u~v ~v~u = (uµvν vµuν )e~ (2.40) − ,µ − ,µ ν As we can see, the commutator between two vectors is itself a vector.

ρ Definition 2.1.12 (Structure coefficients c µν ) ρ The structure coefficients c µν in an arbitrary basis e~ are defined by: { µ} [e~ , e~ ] cρ e~ (2.41) µ ν ≡ µν ρ

Structure coefficients in a coordinate basis: ∂ ∂ [e~ , e~ ] = [ , ] µ ν ∂xµ ∂xν ∂ ∂ ∂ ∂ = ( ) ( ) (2.42) ∂xµ ∂xν − ∂xν ∂xµ ∂2 ∂2 = = 0 ∂xµ∂xν − ∂xν ∂xµ 2.2 Tensors 25

The commutator between two coordinate basis vectors is zero, so the structure coefficients are zero in coordinate basis.

Example 2.1.4 (Structure coefficients in planar polar coordinates) We will find the structure coefficients of an orthonormal basis in planar polar coor- dinates. In (2.33) we found that 1 ~e = ~e , ~e = ~e (2.43) rˆ r θˆ r θ We will now use this to find the structure coefficients. ∂ 1 ∂ [~e , ~e ] = [ , ] rˆ θˆ ∂r r ∂θ ∂ 1 ∂ 1 ∂ ∂ = ( ) ( ) ∂r r ∂θ − r ∂θ ∂r (2.44) 1 ∂ 1 ∂2 1 ∂2 = + −r2 ∂θ r ∂r∂θ − r ∂θ∂r 1 1 = ~e = ~e −r2 θ −r θˆ To find the structure coefficients in coordinate basis we must use [~e , ~e ] = 1~e . rˆ θˆ − r θˆ ρˆ [~eµˆ , ~eνˆ] = c µˆνˆ~eρˆ (2.45) Using (2.44) and (2.45) we get

ˆ 1 cθ = (2.46) rˆθˆ −r ρ From the definition of c µν ([~u , ~v] = [~v , ~u]) we see that the structure coefficients − are anti symmetric in their lower indices:

cρ = cρ (2.47) µν − νµ

ˆ 1 ˆ cθ = = cθ (2.48) θˆrˆ r − rˆθˆ

2.2 Tensors

A 1-form-basis ω1, . . ., ωn is defined by:

µ µ ω (e~ν ) = δ ν (2.49) An arbitrary 1-form can be expressed, in terms of its components, as a linear combination of the basis forms:

µ α = αµω (2.50) 26 Chapter 2. Vectors, Tensors and Forms

where αµ are the components of α in the given basis. Using eqs.(2.49) and (2.50), we find:

µ µ α(~eν ) = αµω (~eν) = αµδ ν = αν µ µ µ 1 2 (2.51) α(~v) = α(v ~eµ) = v α(~eµ) = v αµ = v α1 + v α2 + . . .

We will now look at functions of multiple variables.

Definition 2.2.1 (Multilinear function, tensors) A multilinear function is a function that is linear in all its arguments and maps one-forms and vectors into real numbers.

A covariant tensor only maps vectors. • A contravariant tensor only maps forms. • A mixed tensor maps both vectors and forms into R. • N A tensor of rank maps N one-forms and N 0 vectors into R. It is usual to N 0 say that a tensor is of rank (N + N ). A one-form, for example, is a covariant
0 tensor of rank 1:

µ α(~v) = v αµ (2.52)

Definition 2.2.2 (Tensor product) The basis of a tensor R of rank q contains a tensor product, . If T and S are ⊗ two tensors of rank m and n, the tensor product is defined by:

T S(u~ ,..., u~ , v~ ,..., v~ ) T (u~ ,..., u~ )S(v~ ,..., v~ ) (2.53) ⊗ 1 m 1 n ≡ 1 m 1 n where T and S are tensors of rank m and n, respectively. T S is a tensor of rank ⊗ (m + n).

Let R = T S. We then have ⊗ R = R ωµ1 ωµ2 ωµq (2.54) µ1,...,µq ⊗ ⊗ · · · ⊗ Notice that S T = T S. We get the components of a tensor (R) by using ⊗ 6 ⊗ the tensor on the basis vectors:

Rµ1,...,µq = R(e~µ1 ,..., e~µq ) (2.55)

The indices of the components of a contravariant tensor are written as upper indices, and the indices of a covariant tensor as lower indices. 2.2 Tensors 27

Example 2.2.1 (Example of a tensor) Let ~u and ~v be two vectors and α and β two 1-forms.

µ µ µ µ ~u = u ~eµ; ~v = v ~eµ; α = αµω ; β = βµω (2.56)

From these we can construct tensors of rank 2 through the relation R = ~u ~v as ⊗ follows: The components of R are

Rµ1µ2 = R(ωµ1 , ωµ2 ) = ~u ~v(ωµ1 , ωµ2 ) ⊗ = ~u(ωµ1 )~v(ωµ2 )

µ µ1 ν µ2 (2.57) = u ~eµ(ω )v ~eν(ω ) µ µ1 ν µ2 = u δ µ v δ ν = uµ1 vµ2

2.2.1 Transformation of tensor components We shall not limit our discussion to coordinate transformations. Instead, we will consider arbitrary transformations between bases, ~eµ ~eµ . The µ { } −→ 0 elements of transformation matrices are denoted by M such that µ0 

µ µ0 ~eµ = ~eµM and ~eµ = ~eµ M (2.58) 0 µ0 0 µ

µ0 where M µ are elements of the inverse transformation matrix. Thus, it follows that

M µ M µ0 = δµ (2.59) µ0 ν ν

If the transformation is a coordinate transformation, the elements of the matrix become

∂xµ0 M µ0 = (2.60) µ ∂xµ

2.2.2 Transformation of basis 1-forms

µ0 µ0 µ ω = M µ ω (2.61) ωµ = M µ ωµ0 µ0

The components of a tensor of higher rank transform such that every con- travariant index (upper) transforms as a basis 1-form and every covariant index (lower) as a basis vector. Also, all elements of the transformation matrix are multiplied with one another. 28 Chapter 2. Vectors, Tensors and Forms

Example 2.2.2 (A mixed tensor of rank 3)

T α0 = M α0 M µ M ν T α (2.62) µ0ν0 α µ0 ν0 µν

The components in the primed basis are linear combinations of the components in the unprimed basis.

Tensor transformation of components means that tensors have a basis in- dependent existence. That is, if a tensor has non-vanishing components in a given basis then it has non-vanishing components in all bases. This means that tensor equations have a basis independent form. Tensor equations are invariant. A basis transformation might result in the vanishing of one or more tensor components. Equations in component form may differ from one basis to another. But an equation expressed in tensor components can be transformed from one basis to another using the tensor component transformation rules. An equation that is expressed only in terms of tensor components is said to be covariant.

2.2.3 The metric tensor Definition 2.2.3 (The metric tensor) The scalar product of two vectors ~u and ~v is denoted by g(~u,~v) and is defined as a symmetric linear mapping which for each pair of vectors gives a scalar g(~v,~u) = g(~u,~v).

The value of the scalar product g(~u,~v) is given by specifying the scalar products of each pair of basis-vectors in a basis. g is a symmetric covariant tensor of rank 2. This tensor is known as the metric tensor. The components of this tensor are

g(~eµ, ~eν ) = g µν (2.63)

~u ~v = g(~u, ~v) = g(uµ~e , vν~e ) = uµvν g(~e , ~e ) = uµuνg (2.64) · µ ν µ ν µν Usual notation:

~u ~v = g uµvν (2.65) · µν

The absolute value of a vector:

~v = g(~v, ~v) = g vµvν (2.66) | | | µν | p q 2.2 Tensors 29

e 2

Θ e 1

Figure 2.8: Basis-vectors ~e1 and ~e2

Example 2.2.3 (Cartesian coordinates in a plane)

~e ~e = 1, ~e ~e = 1, ~e ~e = ~e ~e = 0 x · x y · y x · y y · x g xx = g yy = 1, g xy = g yx = 0 (2.67) 1 0 g = µν 0 1  

Example 2.2.4 (Basis-vectors in plane polar-coordinates)

~e ~e = 1, ~e ~e = r2, ~e ~e = 0, (2.68) r · r θ · θ r · θ The metric tensor in plane polar-coordinates: 1 0 g = (2.69) µν 0 r2  

Example 2.2.5 (Non-diagonal basis-vectors)

~e ~e = 1, ~e ~e = 1, ~e ~e = cos θ = ~e ~e 1 · 1 2 · 2 1 · 2 2 · 1 1 cos θ g g (2.70) g = = 11 12 µν cos θ 1 g g    21 22

Definition 2.2.4 (Contravariant components) The contravariant components gµα of the metric tensor are defined as: gµαg δµ gµν = w~ µ w~ ν , (2.71) αν ≡ ν · where w~ µ is defined by w~ µ w~ δµ . (2.72) · ν ≡ ν µν g is the inverse matrix of g µν . 30 Chapter 2. Vectors, Tensors and Forms

ω 2

ω2 A2 A2 x2 = constant e 2

2 1 A e 2 A x = constant

e 1 1 A e 1 A1

A ω1 1 ω 1

Figure 2.9: The covariant- and contravariant components of a vector

It is possible to define a mapping between tensors of different type (eg. covariant on contravariant) using the metric tensor. We can for instance map a vector on a 1-form: α α α vµ = g(~v, ~eµ) = g(v ~eα, ~eµ) = v g(~eα, ~eµ) = v g αµ (2.73) This is known as lowering of an index. Raising of an index becomes : µ µα v = g vα (2.74) The mixed components of the metric tensor becomes: µ µα µ g ν = g g αν = δ ν (2.75) We now define distance along a curve. Let the curve be parameterized by λ (proper-time τ for time-like curves). Let ~v be the tangent vector-field of the curve. The squared distance ds2 between the points along the curve is defined as:

ds2 g(~v, ~v)dλ2 (2.76) ≡ gives 2 µ ν 2 ds = g µν v v dλ . (2.77) 2.3 Forms 31

µ dxµ The tangent vector has components v = dλ , which gives:

2 µ ν ds = g µνdx dx (2.78)

The expression ds2 is known as the line-element.

Example 2.2.6 (Cartesian coordinates in a plane)

x y g xx = g yy = 1, g y = g x = 0 (2.79) ds2 = dx2 + dy2

Example 2.2.7 (Plane polar coordinates)

2 g rr = 1, g θθ = r (2.80) ds2 = dr2 + r2dθ2

Cartesian coordinates in the (flat) Minkowski space-time :

ds2 = c2dt2 + dx2 + dy2 + dz2 (2.81) − In an arbitrary curved space, an orthonormal basis can be adopted in any point. If ~etˆ is tangent vector to the world line of an observer, then ~etˆ = ~u where ~u is the 4-velocity of the observer. In this case, we are using what we call the comoving orthonormal basis of the observer. In a such basis, we have the Minkowski-metric: 2 µˆ νˆ ds = ηµˆνˆdx dx (2.82)

2.3 Forms

An antisymmetric tensor is a tensor whose sign changes under an arbitrary exchange of two arguments.

A( , ~u, , ~v, ) = A( , ~v, , ~u, ) (2.83) · · · · · · · · · − · · · · · · · · · The components of an antisymmetric tensor change sign under exchange of two indices.

A µ ν = A ν µ (2.84) ··· ··· ··· − ··· ··· ··· 32 Chapter 2. Vectors, Tensors and Forms

Definition 2.3.1 (p-form) A p-form is defined to be an antisymmetric, covariant tensor of rank p. An antisymmetric tensor product is defined by: ∧ (p + q)! ω[µ1 ωµp] ω[ν1 ωνq] ω[µ1 ωνq] (2.85) ⊗ · · · ⊗ ∧ ⊗ · · · ⊗ ≡ p!q! ⊗ · · · ⊗

where [ ] denotes antisymmetric combinations defined by: 1 ω[µ1 ωµp] (the sum of terms with ⊗ · · · ⊗ ≡ p! · all possible permutations (2.86) of indices with, “+” for even and “-” for odd permutations)

Example 2.3.1 (antisymmetric combinations)

1 ω[µ1 ωµ2] = (ωµ1 ωµ2 ωµ2 ωµ1 ) (2.87) ⊗ 2 ⊗ − ⊗

Example 2.3.2 (antisymmetric combinations)

ω[µ1 ωµ2 ωµ3] = ⊗ ⊗ 1 (ωµ1 ωµ2 ωµ3 + ωµ3 ωµ1 ωµ2 + ωµ2 ωµ3 ωµ1 3! ⊗ ⊗ ⊗ ⊗ ⊗ ⊗ ωµ2 ωµ1 ωµ3 ωµ3 ωµ2 ωµ1 ωµ1 ωµ3 ωµ2 ) − ⊗ ⊗ − ⊗ ⊗ − ⊗ ⊗ 1 =  (ωµi ωµj ωµk ) (2.88) 3! ijk ⊗ ⊗

Example 2.3.3 (A 2-form in 3-space)

α = α ω1 ω2 +α ω2 ω1 +α ω1 ω3 +α ω3 ω1 +α ω2 ω3 +α ω3 ω2 12 ⊗ 21 ⊗ 13 ⊗ 31 ⊗ 23 ⊗ 32 ⊗ (2.89)

Now the antisymmetry of α means that

+α = α ; +α = α ; +α = α (2.90) 21 − 12 31 − 13 32 − 23 2.3 Forms 33

α =α (ω1 ω2 ω2 ω1) 12 ⊗ − ⊗ 1 3 3 1 + α13(ω ω ω ω ) ⊗ − ⊗ (2.91) + α (ω2 ω3 ω3 ω2) 23 ⊗ − ⊗ [µ ν] = α µν 2ω ω | | ⊗ where µν means summation only for µ < ν (see (Misner, Thorne and Wheeler | | 1973)). We now use the definition of with p = q = 1. This gives ∧ µ ν α = α µν ω ω | | ∧ ωµ ων is the ∧ We can also write form basis. 1 α = α ωµ ων 2 µν ∧

A tensor of rank 2 can always be split up into a symmetric and an anti- symmetric part. (Note that tensors of higher rank can not be split up in this way.) 1 1 Tµν = (Tµν Tνµ) + (Tµν + Tνµ) 2 − 2 (2.92) = Aµν + Sµν

We thus have: 1 S Aµν = (T + T )(T µν T νµ) µν 4 µν νµ − 1 = (T T µν T T νµ + T T µν T T νµ) (2.93) 4 µν − µν νµ − νµ = 0

In general, summation over indices of a symmetric and an antisymmetric quan- µν µν tity vanishes. In a summation TµνA where A is antisymmetric and Tµν has no symmetry, only the antisymmetric part of Tµν contributes. So that, in 1 α = α ωµ ων (2.94) 2 µν ∧ only the antisymmetric elements α = α , contribute to the summation. νµ − µν These antisymmetric elements are the form components Forms are antisymmetric covariant tensors. Because of this antisymmetry a form with two identical components must be a null form (= zero). e.g. α = α α = 0 131 − 131 ⇒ 131 In an n-dimensional space all p-forms with p > n are null forms. Chapter 3

Accelerated Reference Frames

3.1 Rotating reference frames

3.1.1 Space geometry 0 0 Let ~e0ˆ be the 4-velocity field (x = ct, c = 1, x = t) of the reference particles in a reference frame R. A set of simultanous events in R, defines a 3 dimensional space called ’3-space’ in R. This space is orthogonal to ~e0ˆ. We are going to find the metric tensor γij in this space, expressed by the metric tensor gµν of spacetime. In an arbitrary coordinate basis {~eµ}, {~ei} is not necessarily orthogonal to ~e0. We choose ~e0 ~eˆ. Let ~e i be the component of ~ei orthogonal to ~e0, that k 0 ⊥ is:~e i ~e0 = 0. The metric tensor of space is defined by: ⊥ ·

γij = ~e i ~e j, γi0 = 0, γ00 = 0 ⊥ · ⊥

~e i = ~ei ~e i ⊥ − k ~ei ~e0 gi0 ~e i = · ~e0 = ~e0 k ~e ~e g 0 · o 00 γij = (~ei ~e i) (~ej ~e j) − k · − k gi0 gj0 = (~ei ~e0) (~ej ~e0) − g00 · − g00 gj0 gi0 gi0gj0 = ~ei ~ej ~e0 ~ei ~e0 ~ej + 2 ~e0 ~e0 · − g00 · − g00 · g00 · gi0gj0 gi0gj0 gi0gj0 = gij + − g00 − g00 g00

gi0gj0 γij = gij (3.1) ⇒ − g00

(Note:g = g γ = γ ) ij ji ⇒ ij ji The line element in space:

2 i j gi0gj0 i j dl = γijdx dx = gij dx dx (3.2) − g00
34 3.1 Rotating reference frames 35 gives the geometry of a “simultaneity space” in a reference frame where the metric tensor of spacetime in a comoving coordinate system is gµν . The line element for spacetime can be expressed as:

ds2 = dtˆ2 + dl2 (3.3) − It follows that dtˆ = 0 represents the simultaneity defining the 3-space with metric γij.

dtˆ2 = dl2 ds2 = (γ g )dxµdxν − µν − µν = (γ g )dxidxj + 2(γ g )dxidx0 + (γ g )dx0dx0 ij − ij i0 − i0 00 − 00 gi0gj0 i j i 0 0 2 = (gij gij)dx dx 2gi0dx dx g00(dx ) − g00 − − − g g g = g (dx0)2 + 2 i0 dx0dxi + i0 j0 dxidxj − 00 g g2  00 00  g 2 = ( g )1/2(dx0 + i0 dxi) − 00 g  00  So finally we get

1/2 0 gi0 i dtˆ = ( g00) (dx + dx ) (3.4) − g00

The 3-space orthogonal to the world lines of the reference particles in R, dtˆ = 0, corresponds to a coordinate time interval dt = gi0 dxi. This is not an exact − g00 differential , that is , the line integral of dt around a closed curve is in general not equal to 0. Hence you can not in general define simultaneity (given by dtˆ = 0) around closed curves. It can, however, be done if the spacetime metric is diagonal, gi0 = 0. The condition dtˆ = 0 means simultaneity on Einstein synchronized clocks . Conclusion:It is in general (g = 0) not possible to i0 6 Einstein synchronize clocks around closed curves. In particular, it is not possible to Einstein-synchronize clocks around a closed curve in a rotating reference frame. If this is attempted, contradictory bound- ary conditions in the non-rotating lab frame will arise, due to the relativity of simultaneity. (See figure 3.1) The distance in the laboratory frame between two points is: 2πr L = (3.5) 0 n

Lorentz transformation from the instantaneous rest frame (x0, t0) to the labo- ratory system (x, t):

v 1 ∆t = γ(∆t0 + 2 ∆x0), γ = c r2ω2 1 2 (3.6) − c ∆x = γ(∆x0 + v∆t0) q 36 Chapter 3. Accelerated Reference Frames

t+2∆ t S2 3 t+∆ t 2 S ω 1

t & t+n ∆ t (discontinuity)

r

Sn

n t+(n-1) ∆ t

Sn-1 n-1

Figure 3.1: Events simultanous in the rotating reference frame. 1 comes before 2, before 3, etc. . . Note the discontinuity at t.

v = rω L0

Figure 3.2: The distance between two points on the circumference is L0.

3.1 Rotating reference frames 37

¡£¢¥¤§¦©¨ ¨



¢¥¤§¦¨ ¨

Figure 3.3: Discontinuity in simultaneity.

Since we for simultaneous events in the rotating reference frame have ∆t0 = 0, and proper distance ∆x0 = γL0, we get in the laboratory frame rω rω 2πr ∆t = γ2 L = γ2 (3.7) c2 0 c2 n The fact that ∆t = 0 and ∆t = 0 is an expression of the relativity of simul- 0 6 taneity. Around the circumference this is accumulated to

2πr2ω n∆t = γ2 (3.8) c2 and we get a discontinuity in simultaneity, as shown in figure 3.3. Let IF be an inertial frame with cylinder coordinates (T, R, Θ, Z). The line element is then given by

ds2 = dT 2 + dR2 + R2dΘ2 + dZ2 (c = 1) (3.9) − In a rotating reference frame, RF, we have cylinder coordinates (t, r, θ, z). We then have the following coordinate transformation :

t = T, r = R, θ = Θ ωT, z = Z (3.10) − The line element in the co-moving coordinate system in RF is then

ds2 = dt2 + dr2 + r2(dθ + ωdt)2 + dz2 − (3.11) = (1 r2ω2)dt2 + dr2 + r2dθ2 + dz2 + 2r2ωdθdt (c = 1) − − 38 Chapter 3. Accelerated Reference Frames

The metric tensor have the following components:

g = (1 r2ω2), g = 1, g = r2, g = 1 tt − − rr θθ zz 2 (3.12) gθt = gtθ = r ω

dt = 0 gives

ds2 = dr2 + r2dθ2 + dz2 (3.13)

This represents the Euclidean geometry of the 3-space (simultaneity space, t = T ) in IF. The spatial geometry in the rotating system is given by the spatial line element:

2 gi0gj0 i j dl = (gij )dx dx − g00 γrr = grr = 1, γzz = gzz = 1, 2 gθ0 γθθ = gθθ − g00 (r2ω)2 r2 = r2 = − (1 r2ω2) 1 r2ω2 − − −

r2dθ2 dl2 = dr2 + + dz2 (3.14) ⇒ 1 r2ω2 − So we have a non Euclidean spatial geometry in RF. The circumference of a circle with radius r is 2πr lθ = > 2πr (3.15) √1 r2ω2 − We see that the quotient between circumference and radius > 2π which means that the spatial geometry is hyperbolic. (For spherical geometry we have lθ < 2πr.)

3.1.2 Angular acceleration in the rotating frame We will now investigate what happens when we give RF an angular acceleration. Then we use a rotating circle made of standard measuring rods, as shown in Figure 3.4. All points on a circle are accelerated simultaneously in IF (the laboratory system). We let the angular velocity increase from ω to ω + dω, measured in IF. Lorentz transformation to an instantaneous rest frame for a point on the circumference then gives an increase in velocity in this system: rdω rdω0 = , (3.16) 1 r2ω2 − where we have used that the initial velocity in this frame is zero. 3.1 Rotating reference frames 39

"nail"

Standard measuring rod

Figure 3.4: The standard measuring rods are fastened with nails in one end. We will see what then happens when we have an angular acceleration.

The time difference for the accelerations of the front and back ends of the rods (the front end is accelerated first) in the instantaneous rest frame is:

rωL0 ∆t0 = (3.17) √1 r2ω2 − where L0 is the distance between points on the circumference when at rest (= the 2πr length of the rods when at rest), L0 = n . In IF all points on the circumference are accelerated simultaneously. In RF, however, this is not the case. Here the distance between points on the circumference will increase, see Figure 3.5. The rest distance increases by

2 r ωL0dω dL0 = rdω0∆t0 = . (3.18) (1 r2ω2)3/2 −

The increase of the distance during the acceleration (in an instantaneous

t'+ ∆t' t' dv'

Figure 3.5: In RF two points on the circumference are accelerated at different times. Thus the distance between them is increased. 40 Chapter 3. Accelerated Reference Frames

rest frame) is ω 2 ωdω 1 L0 = r L0 = ( 1)L0. (3.19) 2 2 3/2 2 2 0 (1 r ω ) √1 r ω − Z − − Hence, after the acceleration there is a proper distance L0 between the rods. In the laboratory system (IF) the distance between the rods is

2 2 2 2 1 2 2 L = 1 r ω L0 = 1 r ω ( 1)L0 = L0 L0 1 r ω , − − √1 r2ω2 − − − p p − p (3.20)

where L is the rest length of the rods and L √1 r2ω2 is their Lorentz con- 0 0 − tracted length. We now have the situation shown in Figure 3.6.

Standard measuring rod, Lorenz contracted

Figure 3.6: The standard measuring rods have been Lorentz contracted.

Thus, there is room for more standard rods around the periphery the faster the disk rotates. This means that the measured length of the periphery (number of standard rods) gets larger with increasing angular velocity.

3.1.3 Gravitational time dilation

r2ω2 ds2 = (1 )c2dt2 + dr2 + r2dθ2 + dz2 + 2r2ωdθdt (3.21) − − c2 We now look at standard clocks with constant r and z. r2ω2 r2 dθ r2ω dθ ds2 = c2dt2[ (1 ) + ( )2 + 2 ] (3.22) − − c2 c2 dt c2 dt 3.1 Rotating reference frames 41

Let dθ θ˙ be the angular velocity of the clock in RF. The proper time interval dt ≡ measured by the clock is then

ds2 = c2dτ 2 (3.23) − From this we see that

r2ω2 r2θ˙2 r2ωθ˙ dτ = dt 1 2 (3.24) s − c2 − c2 − c2

A non-moving standard clock in RF: θ˙ = 0 ⇒ r2ω2 dτ = dt 1 2 (3.25) r − c Seen from IF, the non-rotating laboratory system, (3.25) represents the velocity dependent time dilation from the special theory of relativity. But how is (3.25) interpreted in RF? The clock does not move relative to an observer in this system, hence what happens can not bee interpreted as a velocity dependent phenomenon. According to Einstein, the fact that standard clocks slow down the farther away from the axis of rotation they are, is due to a gravitational effect. We will now find the gravitational potential at a distance r from the axis. The sentripetal acceleration is v2/r, v = rω so: r r 1 Φ = g(r)dr = rω2dr = r2ω2 − − −2 Z0 Z0 We then get:

r2ω2 2Φ dτ = dt 1 2 = dt 1 + 2 (3.26) r − c r c In RF the position dependent time dilation is interpreted as a gravitational time dilation: Time flows slower further down in a gravitational field.

3.1.4 Path of photons emitted from axes in the rotating refer- ence frame (RF) We start with description in the inertial frame (IF). In IF photon paths are radial. Consider a photon path with Θ = 0, R = T with light source at R = 0. Transforming to RF:

t = T, r = R, θ = Θ ωT − (3.27) r = t, θ = ωt ⇒ − The orbit equation is thus θ = ωr which is the equation for an Archimedean − r spiral. The time used by a photon out to distance r from axis is t = c . 42 Chapter 3. Accelerated Reference Frames

3.1.5 The Sagnac effect IF description: Here the velocity of light is isotropic, but the emitter/receiver moves due to the disc’s rotation as shown in Figure 3.7. Photons are emitted/received in/from opposite directions. Let t1 be the travel time of photons which move with the rotation. Emitter/Receiver X

r + ω

Figure 3.7: The Sagnac effect demonstrates the anisotropy of the speed of light when measured in a rotating reference frame.

Then 2πr + rωt = ct ⇒ 1 1 2πr (3.28) t = ⇒ 1 c rω − Let t2 be the travel time for photons moving against the rotation of the disc. A is the area The difference in travel time is enclosed by the 1 1 photon path or ∆t = t t = 2πr 1 − 2 c rω − c + rω orbit.  −  2πr2rω = (3.29) c2 r2ω2 4−Aω = γ2 c2

RF description: ds2 = 0 along the world line of a r2ω2 photon ds2 = 1 c2dt2 + r2dθ2 + 2r2ωdθdt − − c2   dθ let θ˙ = dt r2θ˙2 + 2r2ωθ˙ (c2 r2ω2) = 0 − − 2 4 2 2 2 4 2 ˙ r ω (r ω + r c r ω ) θ = −  2 − p r 3.2 Hyperbolically accelerated reference frames 43

rc θ˙ = ω −  r2 c (3.30) = ω −  r

The speed of light: v = rθ˙ = rω c. We see that in the rotating frame RF,  −  the measured (coordinate) velocity of light is NOT isotropic. The difference in the travel time of the two beams is

2πr 2πr ∆t = c rω − c + rω (3.31) −4Aω = γ2 c2

(See Phil. Mag. series 6, vol. 8 (1904) for Michelson’s article)

3.2 Hyperbolically accelerated reference frames

Consider a particle moving along a straight line with velocity u and acceleration du a = dT . Rest acceleration is aˆ.

3/2 a = 1 u2/c2 a.ˆ (3.32) ⇒ −
Assume that the particle has constant rest acceleration aˆ = g. That is

du 3/2 = 1 u2/c2 g. (3.33) dT −
Which on integration with u(0) = 0 gives

gT dX u = = 1/2 g2T 2 dT 1 + c2   1/2 c2 g2T 2 X = 1 + + k ⇒ g c2  

c4 = (X k)2 c2T 2 (3.34) ⇒ g2 − −

In its final form the above equation describes a hyperbola in the Minkowski diagram as shown in figure(3.8). 44 Chapter 3. Accelerated Reference Frames

cT

X

Figure 3.8: Hyperbolically accelerated reference frames are so called because the loci of particle trajectories in space-time are hyperbolae.

The proper time interval as measured by a clock which follows the particle:

u2 1/2 dτ = 1 dT (3.35) − c2   Substitution for u(T ) and integration with τ(0) = 0 gives c gT τ = arcsinh g c c gτ  or T = sinh (3.36) g c c2  gτ and X = cosh + k g c   We now use this particle as the origin of space in an hyperbolically accelerated reference frame.

Definition 3.2.1 (Born-stiff motion) Born-stiff motion of a system is motion such that every element of the system has constant rest length. We demand that our accelerated reference frame is Born-stiff.

Let the inertial frame have coordinates (T, X, Y, Z) and the accelerated frame have coordinates (t, x, y, z). We now denote the X-coordinate of the “origin particle” by X0.

gX0 gτ0 1 + = cosh (3.37) c2 c 3.2 Hyperbolically accelerated reference frames 45

c2 where τ0 is the proper time for this particle and k is set to −g . (These are Møller coordinates. Setting k = 0 gives Rindler coordinates). Let us denote the accelerated frame by Σ. The coordinate time at an ar- bitrary point in Σ is defined by t = τ0. That is coordinate clocks in Σ run identically with the standard clock at the “origin particle”. Let X~ 0 be the posi- tion 4-vector of the “origin particle”. Decomposed in the laboratory frame, this becomes

c2 gt c2 gt X~ = sinh , cosh 1 , 0, 0 (3.38) 0 g c g c −    

P is chosen such that P and P0 are simultaneous in the accelerated frame Σ. The distance (see figure(3.9)) vector from P0 to P , decomposed into an orthonormal comoving basis of the “origin particle” is Xˆ = (0, xˆ, yˆ, zˆ) where x,ˆ yˆ and zˆ are physical distances measured simultaneously in Σ. The space coordinates in Σ are defined by

x x,ˆ y yˆ, z zˆ. (3.39) ≡ ≡ ≡

ˆ The position vector of P is X~ = X~ 0 +X~ . The relationship between basis vectors in IF and the comoving orthonormal basis is given by a Lorentz transformation in the x-direction.

∂xµ ~e = ~e µˆ µ ∂xµˆ cosh θ sinh θ 0 0 (3.40) sinh θ cosh θ 0 0 = (~eT , ~eX , ~eY , ~eZ , )   0 0 1 0  0 0 0 1     where θ is the rapidity defined by

U0 tanh θ (3.41) ≡ c

U0 being the velocity of the “origin particle”.

dX gt U = 0 = c tanh 0 dT c 0 (3.42) gt θ = ∴ c 46 Chapter 3. Accelerated Reference Frames

X~ P

cT b ~etˆ P0 ~eXˆ

X~

X~ 0

X Figure 3.9: Simultaneity in hyperbolically accelerated reference frames. The vector X~ lies along the “simultaneity line” which makes the same angle with the X-axis as does ~etˆ with the cT-axis. b

So the basis vectors can be written as follows

gt gt ~e = ~e cosh + ~e sinh tˆ T c X c gt gt ~e = ~e sinh + ~e cosh xˆ T c X c (3.43) ~eyˆ = ~eY

~ezˆ = ~eZ

ˆ The equation X~ = X~ 0 + X~ can now be decomposed in IF:

cT~eT + X~eX + Y ~eY + Z~eZ = c gt c2 gt x gt gt sinh ~e + cosh 1 ~e + sinh ~e + x cosh ~e + y~e + z~e g c T g c − X c c T c X Y Z   (3.44) 3.2 Hyperbolically accelerated reference frames 47

This then, gives the coordinate transformations c gt x gt T = sinh + sinh g c c c c2 gt gt X = cosh 1 + x cosh g c − c   Y = y Z = z gT gx gt = 1 + sinh ⇒ c c2 c gX  gx gt 1 + = 1 + cosh c2 c2 c   Now dividing the last two of the above equations we get

gT gX gt = 1 + tanh (3.45) c c2 c   showing that the coordinate curves t = constant are straight lines in the T,X- 2 frame passing through the point T = 0, X = c . Using the identity cosh2 θ − g − sinh2 θ = 1 we get

gX 2 gT 2 gx 2 1 + = 1 + (3.46) c2 − c c2       showing that the coordinate curves x = constant are hypebolae in the T,X- diagram. The line element (the metric) gives : ds2 is an invariant ds2 = c2dT 2 + dX2 + dY 2 + dZ2 quantity − gx = (1 + )2c2dt2 + dx2 + dy2 + dz2 (3.47) − c2 Note: When the metric is diagonal the unit vectors are orthogonal. Clocks as rest in the accelerated system:

dx = dy = dz = 0, ds2 = c2dτ 2 −

⇓ gx c2dτ 2 = (1 + )2c2dt2 − − c2

⇓ gx dτ = (1 + )dt (3.48) c2 Here dτ is the proper time and dt the coordinate time. An observer in the accelerated system Σ experiences a gravitational field in the negative x-direction. When x < 0 then dτ < dt. The coordinate clocks 48 Chapter 3. Accelerated Reference Frames

cT

horizon

light t=constant

x=constant

X -c 2 g

Figure 3.10: The hyperbolically accelerated reference system

tick equally fast independently of their position. This implies that time passes slower further down in a gravitational field. Consider a standard clock moving in the x-direction with velocity v = dx/dt. Then gx 2 c2dτ 2 = 1 + c2dt2 + dx2 − − c2  gx 2 v2 = 1 + c2dt2 (3.49) − c2 − c2    Hence gx 2 v2 dτ = 1 + 2 2 dt (3.50) r c − c   This expresses the combined effect of the gravitational- and the kinematic time dilation. Chapter 4

Covariant Differentiation

4.1 Differentiation of forms

We must have a method of differentiation that maintains the anti symmetry, thus making sure that what we end up with after differentiation is still a form.

4.1.1 Exterior differentiation The exterior derivative of a 0-form, i.e. a scalar function, f, is given by: ∂f df = ωµ = f ωµ (4.1) ∂xµ ,µ where ωµ are coordinate basis forms: ∂ ωµ( ) = δµ (4.2) ∂xν ν We then (in general) get:

∂xµ ωµ = δµ ων = ων = dxµ (4.3) ν ∂xν In coordinate basis we can always write the basis forms as exterior derivatives of the coordinates. The differential dxµ is given by

dxµ(d~r) = dxµ (4.4) where d~r is an infinitesimal position vector. dxµ are not infinitesimal quantities. In coordinate basis the exterior derivative of a p-form

1 µ1 µp α = αµ1 µp dx dx (4.5) p! ··· ∧ · · · ∧ will have the following component form:

1 µ0 µ1 µp d α = αµ1 µp,µ0 dx dx dx (4.6) p! ··· ∧ ∧ · · · ∧

49 50 Chapter 4. Covariant Differentiation

∂ where , µ µ . The exterior derivative of a p-form is a (p + 1)-form. 0 ≡ ∂x 0 Consider the exterior derivative of a p-form α.

1 µ0 µp dα = αµ1 µp,µ0 dx dx . (4.7) p! ··· ∧ · · · ∧

Let (dα)µ0 µp be the form components of dα. They must, by definition, be ··· antisymmetric under an arbitrary interchange of indices.

1 µ0 µp dα = (dα)µ0 µp dx dx (p + 1)! ··· ∧ · · · ∧

1 µ0 µp which, by (4.7) = α[αµ µp,µ ]dx dx ⇒ p! 1··· 0 ∧ · · · ∧

∴ (dα)µ0 µp = (p + 1)α[µ1 µp,µ0] (4.8) ··· ···

The form equation dα = 0 in component form is

α[µ1 µp,µ0] = 0 (4.9) ···

Example 4.1.1 (Outer product of 1-forms in 3-space)

i i α = αidx x = (x, y, z) (4.10) dα = α dxj dxi i,j ∧ Also, assume that dα = 0. The corresponding component equation is

α = 0 α α = 0 [i,j] ⇒ i,j − j,i ∂α ∂α ∂α ∂α ∂α ∂α (4.11) x y = 0, x z = 0, y z = 0 ⇒ ∂y − ∂x ∂z − ∂x ∂z − ∂y

which corresponds to

α~ = 0 (4.12) ∇× The outer product of an outer product!

d2α d(dα) ≡ (4.13) 2 1 ν2 ν1 µp d α = αµ1 µp,ν1ν2 dx dx dx p! ··· ∧ ∧ · · · ∧

∂2 ,ν ν (4.14) 1 2 ≡ ∂xν1 ∂xν2 4.1 Differentiation of forms 51

Since ∂2 ∂2 ,ν ν =,ν ν (4.15) 1 2 ≡ ∂xν1 ∂xν2 2 1 ≡ ∂xν2 ∂xν1 summation over ν1 and ν2 which are symmetric in αµ1 µp,ν1ν2 and antisymmetric ··· in the basis we get Poincaré’s lemma (valid only for scalar fields)

d2α = 0 (4.16)

This corresponds to the vector equation

( A~) = 0 (4.17) ∇ · ∇×

Let α be a p-form and β be a q-form. Then

d(α β) = dα β + ( 1)pα dβ (4.18) ∧ ∧ − ∧ 4.1.2 Covariant derivative The general theory of relativity contains a covariance principle which states that all equations expressing laws of nature must have the same form irrespective of the coordinate system in which they are derived. This is achieved by writing all equations in terms of tensors. Let us see if the partial derivative of vector µ components transform as tensor components. Given a vector A~ = A ~eµ = µ0 A ~eµ0 with the transformation of basis given by ∂ ∂xν ∂ ν = ν ν (4.19) ∂x 0 ∂x 0 ∂x So that ∂ µ0 µ0 A ,ν ν A 0 ≡ ∂x 0 ∂xν ∂  µ0 = ν ν A ∂x 0 ∂x ∂xν ∂  ∂xµ = 0 Aµ ∂xν ∂xν ∂xµ 0   ν µ ν 2 µ0 ∂x ∂x 0 µ ∂x µ ∂ x = ν µ A ,ν + ν A ν µ (4.20) ∂x 0 ∂x ∂x 0 ∂x ∂x The first term corresponds to a tensorial transformation. The existence of the µ last term shows that A ,ν does not, in general, transform as the components of µ a tensor. Note that A ,ν will transform as a tensor under linear transformations such as the Lorentz transformations. The partial derivative must be generalized such as to ensure that when it is applied to tensor components it produces tensor components. 52 Chapter 4. Covariant Differentiation

Example 4.1.2 (The derivative of a vector field with rotation) We have a vector field:

A~ = kr~eθ

The chain rule for derivation gives:

d ∂ dxν ∂ = = uν dτ ∂xν · dτ ∂xν

dA~ = uν (Aµ~e ) dτ µ ,ν ν µ µ = u A ,ν~eµ + A ~eµ,ν

The change of the vector field with a displacement
along a coordinate-curve is expressed by:

∂A~ = A~ = Aµ ~e + Aµ~e ∂xν ,ν ,ν µ µ,ν

The change in A~ with the displacement in the θ-direction is:

∂A~ = Aµ ~e + Aµ~e ∂θ ,θ µ µ,θ For our vector field, with Ar = 0, we get

∂A~ = Aθ ~e + Aθ~e ∂θ ,θ θ θ,θ =0

θ θ |{z} and since A ,θ = 0 because A = kr we end up with

∂A~ = Aθ~e = kr~e ∂θ θ,θ θ,θ 4.2 The Christoffel Symbols 53

We now need to calculate the derivative of ~eθ. We have:

x = r cos θ y = r sin θ

∂ Using ~eµ = ∂xµ we can write: ∂ ∂x ∂ ∂y ∂ ~e = = + θ ∂θ ∂θ ∂x ∂θ ∂y = r sin θ~e + r cos θ~e − x y

∂ ~e = = cos θ~e + sin θ~e r ∂r x y Gives: ~e = r cos θ~e r sin θ~e θ,θ − x − y = r(cos θ~e + sin θ~e ) = r~e − x y − r This gives us finally:

∂A~ = kr2~e ∂θ − r

~ Thus ∂A = 0 even if A~ = Aθ~e and Aθ = 0. ∂θ 6 θ ,θ

4.2 The Christoffel Symbols

The covariant derivative was introduced by Christoffel to be able to differenti- ate tensor fields. It is defined in coordinate basis by generalizing the partially µ µ derivative A ,ν to a derivative written as A ;ν and which transforms tensorially,

µ0 ν µ0 ∂x ∂x µ A ;ν A ;ν. (4.21) 0 ≡ ∂xµ · ∂xν0 The covariant derivative of the contravariant vector components are written as:

Aµ Aµ + AαΓµ (4.22) ;ν ≡ ,ν αν µ This equation defines the Christoffel symbols Γ αν, which are also called the “connection coefficients in coordinate basis”. From the transformation formulae for the two first terms follows that the Christoffel symbols transform as:

ν µ α0 α0 2 α α0 ∂x ∂x ∂x α ∂x ∂ x Γ µ ν = Γ µν + (4.23) 0 0 ∂xν0 ∂xµ0 ∂xα ∂xα ∂xµ0 ∂xν0 The Christoffel symbols do not transform as tensor components. It is possible to cancel all Christoffel symbols by transforming into a locally Cartesian coordinate 54 Chapter 4. Covariant Differentiation

system which is co-moving in a locally non-rotating reference frame in free fall. Such coordinates are known as Gaussian coordinates. In general relativity theory an inertial frame is defined as a non-rotating frame in free fall. The Christoffel symbols are 0 (zero) in a locally Cartesian coordinate system which is co-moving in a local inertial frame. Local Gaussian coordinates are indicated with a bar over the indices, giving

α¯ Γ µ¯ν¯ = 0 (4.24)

A transformation from local Gaussian coordinates to any coordinates leads to:

α0 2 α¯ α0 ∂x ∂ x Γ µ ν = (4.25) 0 0 ∂xα¯ ∂xµ0 ∂xν0 This equation shows that the Christoffel symbols are symmetric in the two lower indices, ie.

Γα0 = Γα0 µ0ν0 ν0µ0 (4.26)

Example 4.2.1 (The Christoffel symbols in plane polar coordinates)

x = r cos θ, y = r sin θ y r = x2 + y2, θ = arctan x p ∂x ∂x ∂r x ∂r = cos θ, = r sin θ = = cos θ, = sin θ ∂r ∂θ − ∂x r ∂y ∂y ∂y ∂θ sin θ ∂θ cos θ = sin θ, = r cos θ = , = ∂r ∂θ ∂x − r ∂y r

∂r ∂2x ∂r ∂2y Γr = + θθ ∂x ∂θ2 ∂y ∂θ2 = cos θ( r cos θ) + sin θ( r sin θ) − − = r(cos θ2 + sin θ2) = r − −

∂θ ∂2x ∂θ ∂2y Γθ = Γθ = + rθ θr ∂x ∂θ∂r ∂y ∂θ∂r sin θ cos θ = ( sin θ) + (cos θ) − r − r 1 = r 4.2 The Christoffel Symbols 55

The geometrical interpretation of the covariant derivative was given by Levi- Civita. Consider a curve S in any (eg. curved) space. It is parameterized by λ, ie. xµ = xµ(λ). λ is invariant and chosen to be the curve length. µ The tangent vector field of the curve is ~u = (dx /dλ)~eµ. The curve passes through a vector field A~. The covariant directional derivative of the vector field along the curve is defined as:

dA~ dxν A~ = Aµ ~e = Aµ uν~e (4.27) ∇u~ dλ ≡ ;ν dλ µ ;ν µ

The vectors in the vector field are said to be connected by parallel transport along the curve if µ ν A ;νu = 0

A( λ+∆λ) B

λ+∆λ A ( λ+∆λ ) Q A( λ) u

λ P

µ ν Figure 4.1: Parallel transport from P to Q. The vector B~ = A ;νu ∆λ~eµ

dxµ ~u = ~e (4.28) dλ µ According to the geometrical interpretation of Levi-Civita, the covariant direc- tional derivative is:

A~ (λ + ∆λ) A~(λ) ~ µ ν u~ A = A ;νu ~eµ = lim k − (4.29) ∇ ∆λ 0 ∆λ → where A~ (λ + ∆λ) means the vector A~ parallel transported from Q to P . k 56 Chapter 4. Covariant Differentiation

4.3 Geodesic curves Definition 4.3.1 (Geodesic curves) A geodesic curve is defined in such a way that,the vectors of the tangent vector field of the curve is connected by parallell transport.

This definition says that geodesic curves are ’as straight as possible’. If vectors in a vector field A~(λ) are connected by parallell transport by a dis- µ ν placement along a vector ~u , we have A ;νu = 0. For geodesic curves, we then have: µ ν u ;νu = 0 (4.30)

which is the geodesic equation.

µ µ α ν (u ,ν + Γ αν u )u = 0 (4.31)

d dxν ∂ ν ∂ Then we are using that ν = u ν : dλ ≡ dλ ∂x ∂x duµ ∂uµ = uν = uνuµ (4.32) dλ ∂xν ,ν

The geodesic equation can also be written as:

duµ + Γµ uαuν = 0 (4.33) dλ αν

d Usual notation: ˙ = dλ dxµ uµ = = x˙ µ (4.34) dλ

µ µ α ν x¨ + Γ αν x˙ x˙ = 0 (4.35)

By comparing eq.4.35 with the equation of motion(4.53) for a free particle (which we deduced from the Lagrangian equations) , we find the equations to be iden- tical. Conclusion:Free particles follow geodesic curves in spacetime.

Example 4.3.1 (vertical motion of free particle in hyperb. acc. ref. frame) x gx Inserting the Christoffel symbols Γ tt = (1 + c2 )g from example 4.5.3 into the geodesic equation for a vertical geodesic curve in a hyperbolically accelerated refer- ence frame, we get: gx x¨ + (1 + )gt˙2 = 0 c2 4.4 The covariant Euler-Lagrange equations 57

4.4 The covariant Euler-Lagrange equations

Geodesic curves can also be defined as curves with an extremal distance between two points. Let a particle have a world-line (in space-time) between two points (events) P1 and P2. Let the curves be described by an invariant parameter λ (proper time τ is used for particles with a rest mass). The Lagrange-function is a function of coordinates and their derivatives, dxµ L = L(xµ, x˙ µ), x˙ µ . (4.36) ≡ dλ (Note: if λ = τ then x˙ µ are the 4-velocity components) The action-integral is S = L(xµ, x˙ µ)dλ. The principle of extremal action (Hamiltons-principle): The world-line of a particle is determined by the condi- R tion that S shall be extremal for all infinitesimal variations of curves which keep P1 and P2 rigid, ie.

λ2 δ L(xµ, x˙ µ)dλ = 0, (4.37) Zλ1 where λ1 and λ2 are the parameter-values at P1 and P2. For all the variations ct

P2

P1 x

Figure 4.2: Different world-lines connecting P1 and P2 in a Minkowski diagram the following condition applies:

µ µ δx (λ1) = δx (λ2) = 0 (4.38)

We write Eq. (4.37) as

λ2 λ2 ∂L ∂L δ Ldλ = δxµ + δx˙ µ dλ (4.39) ∂xµ ∂x˙ µ Zλ1 Zλ1   58 Chapter 4. Covariant Differentiation

Partial integration of the last term

λ2 ∂L ∂L λ2 λ2 d ∂L δx˙ µdλ = δxµ δxµdλ (4.40) ∂x˙ µ ∂x˙ µ − dλ ∂x˙ µ Zλ1  λ1 Zλ1   µ µ Due to the conditions δx (λ1) = δx (λ2) = 0 the first term becomes zero. Then we have : λ2 ∂L d ∂L δS = δxµdλ (4.41) ∂xµ − dλ ∂x˙ µ Zλ1    The world-line the particle follows is determined by the condition δS = 0 for any variation δxµ. Hence, the world-line of the particle must be given by

∂L d ∂L = 0 (4.42) ∂xµ − dλ ∂x˙ µ   These are the covariant Euler-Lagrange equations. µ The canonical momentum pµ conjugated to a coordinate x is defined as ∂L p (4.43) µ ≡ ∂x˙ µ The Lagrange-equations can now be written as

dpµ ∂L ∂L = or p˙ = . (4.44) dλ ∂xµ µ ∂xµ

A coordinate which the Lagrange-function does not depend on is known as a ∂L cyclic coordinate. Hence, ∂xµ = 0 for a cyclic coordinate. From this follows:

The canonical momentum conjugated to a cyclic coordinate is a constant of motion

µ ie. pµ = C (constant) if x is cyclic. A free particle in space-time (curved space-time includes gravitation) has the Lagrange function 1 1 1 L = ~u ~u = x˙ x˙ µ = g x˙ µx˙ ν (4.45) 2 · 2 µ 2 µν An integral of the Lagrange-equations is obtained readily from the 4-velocity identity: µ 2 x˙ µx˙ = c for a particle with rest-mass µ − (4.46) (x˙ µx˙ = 0 for light The line-element is:

2 µ ν µ ν 2 2 ds = gµν dx dx = gµν x˙ x˙ dλ = 2Ldλ . (4.47)

Thus the Lagrange function of a free particle is obtained from the line-element. 4.5 Application of the Lagrangian formalism to free particles 59

flat surface: Q

P

Figure 4.3: On a flat surface, the geodesic curve is the minimal distance between P and Q

sphere:

Q

P

Figure 4.4: On a sphere, the geodesic curves are great circles.

4.5 Application of the Lagrangian formalism to free particles

To describe the motion of a free particle, we start by setting up the line element of the space-time in the chosen coordinate system. There are coordinates on which the metric does not depend. For example, given axial symmetry we may choose the angle θ which is a cyclic coordinate here and the conjugate (covariant) impulse Pθ is a constant of the motion (the orbital spin of the particle). If, in addition, the metric is time independent (stationary metric) then t is also cyclic and pt is a constant of the motion (the mechanical energy of the particle). A static metric is time-independent and unchanged under time reversal (i.e. t t). A stationary metric changed under time reversal. Examples → − of static metrics are Minkowski and hyperbolically accelerated frames. The rotating cylindrical coordinate system is stationary. 60 Chapter 4. Covariant Differentiation

4.5.1 Equation of motion from Lagrange’s equation

The Lagrange function for a free particle is:

1 L = g x˙ µx˙ ν 2 µν (4.48)

λ where g µν = g µν (x ). And the Lagrange equations are

∂L d ∂L = 0, ∂xβ − dτ ∂x˙ β   ∂L = g x˙ ν, (4.49) ∂x˙ β βν ∂L 1 = g x˙ µx˙ ν. ∂xβ 2 µν,β

d ∂L ∂L • = g˙ x˙ ν + g x¨ν dτ ∂x˙ β ≡ ∂x˙ β βν βν     (4.50) µ ν ν = g βν,µx˙ x˙ + g βνx¨ .

Now, (4.50) and (4.49) together give:

1 g x˙ µx˙ ν g x˙ µx˙ ν g x¨ν = 0. (4.51) 2 µν,β − βν,µ − βν

The second term on the left hand side of (4.51) may be rewritten making use of the fact that x˙ µx˙ ν is symmetric in µν, as as follows

µ ν 1 µ ν g βν,µx˙ x˙ = (g βµ,ν + g βν,µ)x˙ x˙ 2 (4.52) 1 g x¨ν + (g + g g )x˙ µx˙ ν = 0. ⇒ βν 2 βµ,ν βν,µ − µν,β

Finally, since we are free to multiply (4.52) through by gαβ, we can isolate x¨α to get the equation of motion in a particularly elegant and simple form:

α α µ ν x¨ + Γ µν x˙ x˙ = 0 (4.53)

α where the Christoffel symbols Γ µν in (4.53) are defined by

1 Γα gαβ(g + g g ). (4.54) µν ≡ 2 βµ,ν βν,µ − µν,β

Equation(4.53) describes a geodesic curve . 4.5 Application of the Lagrangian formalism to free particles 61

4.5.2 Geodesic world lines in spacetime Consider two timelike curves between two events in spacetime. In fig.4.5 they are drawn in a Minkowski diagram which refers to an inertial reference frame.

cT

t1 P

non-geodetic curve between O and P

geodetic curve v(t)

O t 0 X Figure 4.5: Timelike curves in spacetime.

The general interpretation of the line-element for a time-like interval is; The spacetime distance between O and P (See figure 4.5) equals the proper time interval between two events O and P measured on a clock moving in a such way, that it is present both at O and P.

ds2 = c2dτ 2 (4.55) − which gives T1 v2(T ) τ0 1 = 1 dT (4.56) − − c2 ZT0 r We can see that τ0 1 is maximal along the geodesic curve with v(T ) = 0. Time- − like geodesic curves in spacetime have maximal distance between two points.

Example 4.5.1 (How geodesics in spacetime can give parabolas in space) A geodesic curve between two events O and P has maximal proper time. Consider the last expression in Section 3.2 of the propertime interval of a particle with position x and velocity v in a gravitational field with acceleration of gravity g.

2 gx 2 v dτ = dt 1 + 2 2 r c − c
62 Chapter 4. Covariant Differentiation

This expression shows that the proper time of the particle proceeds faster the higher up in the field the particle is, and it proceeds slower the faster the particle moves. Consider figure 4.6. The path a free particle follows between the events O and P is a compromize between moving as slowly as possible in space, in order to keep the velocity dependent time dilation small, and moving through regions high up in the gravitational field, in order to prevent the slow proceeding of proper time far down. However if the particle moves too high up, its velocity becomes so large that it proceeds slower again. The compromise between kinematic and gravitational time dilation which gives maximal proper time between O and P is obtained for the thick curve in fig. 4.6. This is the curve followed by a free particle between the events O and P. We shall now deduce the mathematical expression of what has been said above. Timelike geodesic curves are curves with maximal proper time, i.e.

τ1 τ = g x˙ µx˙ νdτ − µν Z0 p is maximal for a geodesic curve. However the action

τ1 τ1 J = 2 Ldτ = g x˙ µx˙ νdτ − − µν Z0 Z0 is maximal for the same curves and this gives an easier calculation. In the case of a vertical curve in a hyperbolically accelerated reference frame the Lagrangian is 1 gx 2 x˙ 2 L = 1 + t˙2 + (4.57) 2 − c2 c2     Using the Euler-Lagrange equations now gives gx x¨ + (1 + )gt˙2 = 0 c2 which is the equation of the geodesic curve in example 4.3.1. Since spacetime is flat, the equation represents straight lines in spacetime. The projection of such curves into the three space of arbitrary inertial frames gives straight paths in 3-space, in accordance with Newton’s 1st law. However projecting it into an accelerated frame where the particle also has a horizontal motion, and taking the Newtonian limit, one finds the parabolic path of projectile motion.

Example 4.5.2 (Spatial geodesics described in the reference frame of a rotating disc.) In Figure 4.7, we see a rotating disc. We can see two geodesic curves between P1 and P2. The dashed line is the geodesic for the non-rotating disc. The other curve is a geodesic for the 3-space of a rotating reference frame. We can see that the geodesic is curved inward when the disc is rotating. The curve has to curve inward since the measuring rods are longer there (because of Lorentz-contraction). Thus, the minimum distance between P1 and P2 will be achieved by an inwardly bent curve. 4.5 Application of the Lagrangian formalism to free particles 63

The path of the particle g

O P Figure 4.6: The particle moves between two events O and P at fixed points in time. The path chosen by the particle between O and P is such that the proper time taken by the particle betweem these two events is as large as possible. Thus the goal of the particle is to follow a path such that its comoving standard clocks goes as fast as possible. If the particle follows the horizontal line between O and P it goes as slowly as possible and the kinematical time dilation is as small as possible. Then the slowing down of its comoving standard clocks due to the kinematical time dilation is as small as possible, but the particle is far down in the gravitational field and its proper time goes slowly for that reason. Paths further up leads to a greater rate of proper time. But above the curve drawn as a thick line, the kinematical time dilation will dominate, and the proper time proceeds more slowly. 64 Chapter 4. Covariant Differentiation

We will show this mathematically, using the Lagrangian equations. The line element for the space dt = dz = 0 of the rotating reference frame is

P 1

ω

P 2

Figure 4.7: Geodesic curves on a non-rotating (dashed line) and rotating (solid line) disc.

r2dθ2 dl2 = dr2 + 1 r2ω2 − c2 Lagrangian function: 1 1 r2θ˙2 L = r˙2 + 2 2 1 r2ω2 − c2 We will also use the identity: r2θ˙2 r˙2 + = 1 (4.58) 1 r2ω2 − c2 (We got this from using ~u ~u = 1) We see that θ is cyclic ( ∂L = 0), implying: · ∂θ ∂L r2θ˙ pθ = = = constant ∂θ˙ 1 r2ω2 − c2 This gives: 2 2 2 r ω pθ pθ ω pθ θ˙ = 1 = (4.59) − c2 r2 r2 − c2 Inserting 4.59 into 4.58:
ω2p2 p2 r˙2 = 1 + θ θ (4.60) c2 − r2 This gives us the equation of the geodesic curve between P1 and P2:

2 2 2 2 ω pθ pθ r˙ dr r 1 + 2 2 = = c − r (4.61) θ˙ dθ qp 1 r2ω2 θ − c2
4.5 Application of the Lagrangian formalism to free particles 65

Boundary conditions:

P 2

r θ

r 0

geodesic

P1 Figure 4.8: Geodesic curves on a rotating disc,coordinates

r˙ = 0, r = r0, for θ = 0

Inserting this into 4.60 gives:

2 2 pθ pθω = 1 + 2 (4.62) r0 s c

Rearranging 4.61,using 4.62 gives:

dr ω2 rdr dθ 2 = r r2 r2 − c r2 r2 r0 − 0 − 0 p p Integrating this yields:

r ω2 r θ = 0 r2 r2 arccos 0  c2 − 0 ∓ r q

Example 4.5.3 (Christoffel symbols in a hyperbolically accelerated reference frame) The Christoffel symbols were defined in Equation (4.53).

1 Γα gαβ(g + g g ). µν ≡ 2 βµ,ν βν,µ − µν,β

In this example

gx 2 g = 1 + c2, g = g = g = 1 tt − c2 xx yy zz   66 Chapter 4. Covariant Differentiation

ct

LOW HIGH

g P

non-geodetic curve between O and P geodetic curve 2 1

O x Figure 4.9: Vertical throw in the accelerated referenceframe.

∂gtt α and only the term ∂x contributes to Γµν. Thus the only non-vanishing Christoffel symbols are

1 ∂g Γt = Γt = gtt tt xt tx 2 ∂x   1 ∂g = tt 2gtt ∂x 2 1 + gx g = c2 gx 2 2 2 1 + c2
c 1 g
= gx 2 1 + c2 c 1 ∂g Γx = gxx
tt tt −2 ∂x   1 gx g = 2 1 + c2 −2 − c2 c2 ngx   o = 1 + g c2   4.5 Application of the Lagrangian formalism to free particles 67

Example 4.5.4 (Vertical projectile motion in a hyperbolically accelerated reference frame)

gx 2 ds2 = 1 + c2dt2 + dx2 + dy2 + dz2 (4.63) − c2   Vertical motion implies that dy = dz = 0 and the Lagrange function becomes 1 L = g x˙ µx˙ ν 2 µν 1 gx 2 1 = 1 + c2t˙2 + x˙ 2 −2 c2 2   where the dots imply differentiation w.r.t the particle’s proper time, τ. And the initial conditions are: x(0) = 0, x˙(0) = (u0, ux, 0, 0) = γ(c, v, 0, 0), 1/2 where, γ = 1 v2/c2 − . − What is the maximum height, hreached by
the particle? 2 Newtonian description: 1 mv2 = mgh h = v . 2 ⇒ 2g Relativistic description: t is a cyclic coordinate x0 = ct is cyclic and p = ⇒ 0 constant.

∂L 1 ∂L gx 2 p0 = = = c 1 + t˙ (4.64) ∂x˙ 0 c ∂t˙ − c2   Now the 4-velocity identity is

~u ~u = g x˙ µx˙ ν = c2 (4.65) · µν − so 1 gx 2 1 1 1 + c2t˙2 + x˙ 2 = c2 (4.66) −2 c2 2 −2   and given that the maximum height h is reached when x˙ = 0 we get gh 2 1 + t˙2 = 1. (4.67) c2 x=h   Now, since p0 is a constant of the motion, it preserves its initial value throughout the flight (i.e. p = ct˙(0) = γc) and particularly at x = h, 0 − − gh 2 (4.64) p = γc = c 1 + t˙ (4.68) ⇒ 0 − − c2 x=h   Finally, dividing equation (4.67) by equation (4.68) and substituting back in equation (4.67) gives c2 h = (γ 1) (4.69) g − 68 Chapter 4. Covariant Differentiation

In the Newtonian limit (4.69) becomes

c2 1 c2 1 v2 v2 h = 1 u 1 + 1 h u g (1 v2/c2)1/2 − g 2 c2 − ⇒ 2g  −   

Example 4.5.5 (The twin “paradox”) Eva travels to Alpha Centauri, 4 light years from the Earth, with a velocity v = 0.8c (γ = 1/0.6). The trip takes 5 years out and 5 years back. This means that Eli, who remains at Earth is 10 years older when she meets Eva at the end of her journey. Eva, on the other hand, is 10(1 v2/c2)1/2 = 10(0.6) = 6 years older. − T

Eli Eva

X Figure 4.10: The twins Eli and Eva each travel between two fixed events in space- time

According to the theory of relativity, Eva can consider herself as being stationary and Eli as the one whom undertakes the long journey. In this picture it seems that Eva and Eli must be 10 and 6 years older respectively upon their return. Let us accept the principle of general relativity as applied to accelerated reference frames and review the twin “paradox” in this light. Eva’s description of the trip when she sees herself as stationary is as follows. Eva perceives a Lorentz contracted distance between the Earth and Alpha Cen- tauri, namely, 4 light years 1/γ = 2.4light years. The Earth and Eli travel with × 2.4light years v = 0.8c. Her travel time in one direction is then 0.8c = 3 years. So the round trip takes 6 years according to Eva. That is Eva is 6 years older when they meet again. This is in accordance with the result arrived at by Eli. According to 4.5 Application of the Lagrangian formalism to free particles 69

Eva, Eli ages by only 6 years 1/γ = 3.6 years during the round trip, not 10 years × as Eli found. On turning about Eva experiences a force which reduces her velocity and acceler- ates her towards the Earth and Eli. This means that she experiences a gravitational force directed away from the Earth. Eli is higher up in this gravitational field and ages faster than Eva, because of the gravitational time dilation. We assume that Eva has constant proper acceleration and is stationary in a hyperbolically accelerated frame as she turns about. The canonical momentum pt for Eli is then(see Equation (4.64))

gx 2 p = 1 + ct˙ t − c2   Inserting this into the 4-velocity identity gives

gx 2 gx 2 p2 c2 1 + = 1 + x˙ 2, (4.70) t − c2 c2     or 1 + gx dτ = c2 dx p2 c2 1 + gx 2 t − c2 q
Now, since x˙ = 0 for x = x2 (x2 is Eli’s turning point according to Eva), we have that gx p = c 1 + 2 t c2   Let x1 be Eli’s position according to Eva just as Eva begins to notice the gravitational field. That is when Eli begins to slow down in Eva’s frame. Integration from x1 to x2 and inserting the value of pt gives

c gx2 2 gx1 2 τ1 2 = 1 + 1 + − g c2 − c2 r    1 2 2 lim τ1 2 = x2 x1. ⇒ g − c − →∞ q Now setting x2 = 4 and x1 = 2.4 light years respectively we get

lim τ1 2 = 3.2 years g − →∞ Eli’s aging as she turns about is, according to Eva,

∆τEli = 2 lim τ1 2 = 6.4 years. g − →∞

So Eli’s has aged by a total of τEli = 3.6 + 6.4 = 10 years, according to Eva, which is just what Eli herself found. 70 Chapter 4. Covariant Differentiation

4.5.3 Gravitational Doppler effect This concerns the frequency shift of light traversing up or down in a gravitational field. The 4-momentum of a particle with relativistic energy E and spatial velocity w~ (3-velocity) is given by:

P~ = E(1, w~) (c = 1) (4.71)

Let U~ be the 4-velocity of an observer. In a co-moving orthonormal basis of the observer we have U~ = (1, 0, 0, 0). This gives

U~ P~ = Eˆ (4.72) · − The energy of a particle with 4-momentum P~ measured by an observer with 4-velocity U~ is Eˆ = U~ P~ (4.73) − · Let E = (U~ P~ ) and E = (U~ P~ ) be the energy of a photon, measured S − · S a − · a locally by observers in rest in the transmitter and receiver positions, respectively. This gives1 ES Ea = (4.74) (U~ P~ ) (U~ P~ ) · S · a Let the angular frequency of the light, measured by the transmitter and receiver, be ws and wa, respectively. We then have

ES Ea w = , w = , (4.75) s h a h which gives:

(U~ P~ )a wa = · ws (4.76) (U~ P~ ) · s For an observer at rest in a time-independent orthogonal metric we have dt U~ P~ = U tP = P (4.77) · t dτ t

where Pt is a constant of motion (since t is a cyclic coordinate) for photons and hence has the same value in transmitter and receiver positions. The line-element is 2 2 i 2 ds = gttdt + gii(dx ) (4.78) Using the physical interpretation (4.55) of the line-element for a time like inter- val, we obtain for the proper time of an observer at rest

dτ 2 = g dt2 dτ = √ g dt (4.79) − tt ⇒ − tt 1 0 1 0 0 1 1 A~ B~ = A0B + A1B + ... = g00A B + g11A B + ..., an orthonormal basis gives · A~ B~ = A0B0 + A1B1 + ... · − 4.6 The Koszul connection 71

Hence dt 1 = , (4.80) dτ √ g − tt which gives 1 U~ P~ = P . (4.81) · √ g t − tt Inserting this into the expression for angular frequency (4.76) gives

(gtt)s wa = ws (4.82) s(gtt)a

Note: we have assumed an orthogonal and time-independent metric, i.e. Pt1 =

Pt2 . Inserting the metric of a hyperbolically accelerated reference system with gx g = (1 + )2 (4.83) tt − c2 gives gxs 1 + c2 wa = gxa ws, (4.84) 1 + c2 or gxs g w w 1 + 2 2 (xs xa) g a s c c − − = gxa 1 = gxa 2 H, (4.85) ws 1 + c2 − 1 + c2 ≈ c where H = x x is the difference in height between transmitter and receiver. s − a

Example 4.5.6 (Measurements of gravitational Doppler effects (Pound and Rebka 1960))

H 20m, g = 10m/s2 ≈ gives ∆w 200 15 = = 2.2 10− . w 9 1016 × × This effect was measured by Pound and Rebka in 1960.

4.6 The Koszul connection

The covariant directional derivative of a scalar field f in the direction of a vector ~u is defined as: f ~u(f) (4.86) ∇u~ ≡ Here the vector ~u should be taken as a differensial operator. (In coordinate µ ∂ basis, ~u = u ∂xµ ) The directional derivative along a basis vector ~eν is written as: (4.87) ∇ν ≡ ∇~eν Hence ( ) = ( ) = ~e ( ) ∇µ ∇~eµ µ 72 Chapter 4. Covariant Differentiation

Definition 4.6.1 (Koszul’s connection coeffecients in an arbitrary basis) In an aribitrary basis the Koszul connection coefficients are defined by

~e Γα ~e (4.88) ∇ν µ ≡ µν α

α α which may also be written ~eν(~eµ) = Γµν~eα. In coordinate basis , Γ µν is reduced α to Christoffel symbols and one often writes ~eµ,ν = Γ µν~eα. In an arbitrary basis α , Γ µν has no symmetry.

Example 4.6.1 (The connection coefficients in a rotating reference frame.) Coordinate transformation: (T, R, Θ are coordinates in the non-rotating reference frame, t, r, θ in the rotating.) Corresponding Cartesian coordinates:X, Y and x, y.

t = T, r = R, θ = Θ ωT − X = R cos Θ, Y = R sin Θ X = r cos(θ + ωt), Y = r sin(θ + ωt)

Y

(x,y) P x (X,Y)

y

ωt θ+ωt X

Figure 4.11: The non-rotating coordinate system (X,Y) and the rotating system (x,y),rotating with angular velocity ω

∂ ∂X ∂ ∂Y ∂ ∂T ∂ ~e = = + + t ∂t ∂t ∂X ∂t ∂Y ∂t ∂T 4.6 The Koszul connection 73 gives: ~e = rω sin(θ + ωt)~e + rω cos(θ + ωt)~e + ~e t − X Y T ∂X ∂ ∂Y ∂ ~e = + r ∂r ∂X ∂r ∂Y = cos(θ + ωt)~eX + sin(θ + ωt)~eY ∂X ∂ ∂Y ∂ ~e = + θ ∂θ ∂X ∂θ ∂Y = r sin(θ + ωt)~e + r cos(θ + ωt)~e − X Y We are going to find the Christoffel symbols, which involves differentiation of ba- sis vectors. This coordinate transformation makes this easy, since ~eX , ~eY , ~eT are constant. Differentiation gives ~e = rω2 cos(θ + ωt)~e rω2 sin(θ + ωt)~e (4.89) ∇t t − X − Y The connection coefficients are (see eq. 4.88) ~e Γα ~e (4.90) ∇ν µ ≡ µν α α So, to calculate Γ µν , the right hand side of eq.4.89 has to by expressed by the basis that we are differentiating. By inspection, the right hand side is rω2~e . − r That is ~e = rω2~e giving Γr = rω2. ∇t t − r tt − The other nonzero Christoffel symbols are ω 1 Γθ = Γθ = , Γθ = Γθ = rt tr r θr rθ r Γr = Γr = rω, Γr = r θt tθ θθ −

Example 4.6.2 (Acceleration in a non-rotating reference frame (Newton))

¨ ˙ i i j k ~r = ~v = (v˙ + Γ jkv v )~ei, where d . i, j, and k are space indices. Inserting the Christoffel symbols for · ≡ dt plane polar coordinates (see example 4.2.1), gives: 2 ~a = (r¨ rθ˙2)~e + (θ¨ + r˙θ˙)~e inert − r r θ

Example 4.6.3 (The acceleration of a particle, relative to the rotating reference frame) Inserting the Christoffel symbols from example 4.6.1: 2 ~a = (r¨ rθ˙2 rω2 + Γr θ˙t˙ + Γr t˙θ˙)~e + (θ¨ + r˙θ˙ + Γθ r˙t˙ + Γθ t˙r˙)~e rot − − θt tθ r r rt tr θ = (r¨ rθ˙2 rω2 2rωθ˙)~e + (rθ¨ + 2r˙θ˙ + 2r˙ω)~e − − − rˆ θˆ = ~a (rω2 2rωθ˙)~e + 2r˙ω~e inert − − rˆ θˆ 74 Chapter 4. Covariant Differentiation

The angular velocity of the reference frame, is ω~ = ω~ez. We also introduce ~r = r~er. The velocity relative to the rotating reference frame is then:

~r˙ = r˙~er + r~e˙r

Furthermore d~e ∂~e dxi ~e˙ = r = r = vi~e r dt ∂xi dt r,i Using definition 4.6.1 in a coordinate basis, this may be written ˙ i j ~er = v Γ ri~ej Using the expressions of the Christoffel symbols in example 4.6.1, we get 1 ~e˙ = vθΓθ ~e = θ˙ ~e = θ˙~e r rθ θ r θ θˆ Hence ˙ ˙ ~v = ~r = r˙~erˆ + rθ~eθˆ Inserting this into the expression for the acceleration, gives:

~r¨ = ~r¨ + ω~ (ω~ ~r) + 2ω~ ~v rot inert × × × We can see that the centrifugal acceleration (the term in the middle) and the coriolis acceleration (last term) is contained in the expression for the covariant derivative.

α 4.7 Connection coefficients Γ µν and structure coeffi- α cients c µν in a Riemannian (torsion free) space

The commutator of two vectors, ~u and ~v, expressed by covariant directional derivatives is given by: [~u, ~v] = ~v ~u (4.91) ∇u~ − ∇~v Let ~u = e~µ and ~v = e~ν. We then have:

[e~ , e~ ] = e~ e~ . (4.92) µ ν ∇µ ν − ∇ν µ Using the definitions of the connection and structure coefficients we get:

cα e~ = (Γα Γα )e~ (4.93) µν α νµ − µν α Thus in a torsion free space

cα = Γα Γα (4.94) µν νµ − µν In coordinate basis we have ∂ ∂ e~ = , e~ = (4.95) µ ∂xµ ν ∂xν 4.8 Covariant differentiation of vectors, forms and tensors 75

And therefore: ∂ ∂ [e~ , e~ ] = [ , ] µ ν ∂xµ ∂xν ∂ ∂ ∂ ∂ = ( ) ( ) (4.96) ∂xµ ∂xν − ∂xν ∂xµ ∂2 ∂2 = = 0 ∂xµ∂xν − ∂xν ∂xµ α Equation (4.96) shows that c µν = 0, and that the connection coefficients in Equation (4.94) therefore are symmetrical in a coordinate basis: α α Γ νµ = Γ µν (4.97)

4.8 Covariant differentiation of vectors, forms and ten- sors

4.8.1 Covariant differentiation of a vector in an arbitrary basis

µ νA~ = ν(A e~µ) ∇ ∇ (4.98) = Aµe~ + Aα e~ ∇ν µ ∇ν α ∂ Aµ = e~ (Aµ) , e~ = M µ , (4.99) ∇ν ν ν ν ∂xµ µ where M ν are the elements of a transformation matrix between a coordinate ∂ basis ∂xµ and an arbitrary basis e~ν . ( If e~ν had been a coordinate basis { } µ { } ∂ µ µ vector, we would have gotten e~ν(A ) = ∂xν (A ) = A ,ν). A~ = [e~ (Aµ) + AαΓµ ]e~ (4.100) ∇ν ν αν µ

Definition 4.8.1 (Covariant derivative of a vector) The covariant derivative of a vector in an arbitrary basis is defined by: A~ Aµ ~e (4.101) ∇ν ≡ ;ν µ So: µ µ α µ A;ν = ~eν(A ) + A Γ αν (4.102) where ~e Γµ ~e ∇ν α ≡ αν µ

4.8.2 Covariant differentiation of forms Definition 4.8.2 (Covariant directional derivative of a one-form field) Given a vector field A~ and a one-form field α, the covariant directional derivative of α in the direction of the vector ~u is defined by: ( α)(A~ ) [α(A~ )] α( A~) (4.103) ∇u~ ≡ ∇u~ − ∇u~ µ αµA | {z } 76 Chapter 4. Covariant Differentiation

Let α = ωµ (basis form), ωµ(e~ ) δµ and let A~ = e~ and ~u = e~ . We then ν ≡ ν ν λ have: ( ωµ)(e~ ) = [ωµ(e~ )] ωµ( e~ ) (4.104) ∇λ ν ∇λ ν − ∇λ ν µ δ ν The covariant directional derivative of a constant scalar field is zero, δµ = ∇λ | {z } ∇λ ν 0. We therefore get: ( ωµ)(e~ ) = ωµ( e~ ) ∇λ ν − ∇λ ν = ωµ(Γα e~ ) − νλ α = Γα ωµ(e~ ) (4.105) − νλ α = Γα δµ − νλ α = Γµ − νλ The contraction between a one-form and a basis vector gives the components of the one-form, α(e~ν ) = αν . Equation (4.105) tells us that the ν-component of ωµ is equal to Γµ , and that we therefore have ∇λ − νλ ωµ = Γµ ων (4.106) ∇λ − νλ Equation (4.106) gives the directional derivatives of the basis forms. Using the product of differentiation gives α = (α ωµ) ∇λ ∇λ µ = (α )ωµ + α ωµ (4.107) ∇λ µ µ∇λ = e~ (α )ωµ α Γµ ων λ µ − µ νλ

Definition 4.8.3 (Covariant derivative of a one-form) µ The covariant derivative of a one-form α = αµω is therefore given by Equation (4.108) below, when we let µ ν in the first term on the right hand side in (4.107): → α = [e~ (α ) α Γµ ]ων (4.108) ∇λ λ ν − µ νλ

The covariant derivative of the one-form components αµ are denoted by αν;λ and are defined by α α ων (4.109) ∇λ ≡ ν;λ It then follows that α = e~ (α ) α Γµ (4.110) ν;λ λ ν − µ νλ

µ It is worth to note that Γ νλ in Equation (4.110) are not Christoffel symbols. In coordinate basis we get: µ αν;λ = αν,λ αν Γ λν (4.111) µ µ − where Γ λν = Γ νλ are Christoffel symbols. 4.9 The Cartan connection 77

4.8.3 Generalization for tensors of higher rank Definition 4.8.4 (Covariant derivative of a tensor) Let A and B be two tensors of arbitrary rank. The covariant directional derivative along a basis vector e~ of a tensor A B of arbitrary rank is defined by: α ⊗ (A B) ( A) B + A ( B) (4.112) ∇λ ⊗ ≡ ∇λ ⊗ ⊗ ∇λ

We will use (4.112) to find the formula for the covariant derivative of the com- ponents of a tensor of rank 2:

µ ν αS = α(Sµν ω ω ) ∇ ∇ µ⊗ ν µ ν µ ν = ( αSµν )ω ω + Sµν( αω ) ω + Sµν ω ( αω ) (4.113) ∇ ⊗ ∇ ⊗ ⊗ ∇ = (S S Γβ S Γβ )ωµ ων µν,α − βν µα − µβ να ⊗ where Sµν,α = ~eα(Sµν). Defining the covariant derivative Sµν;α by

S = S ωµ ων (4.114) ∇α µν;α ⊗ we get

S = S S Γβ S Γβ (4.115) µν;α µν,α − βν µα − µβ να For the metric tensor we get

g = g g Γβ g Γβ (4.116) µν;α µν,α − βν µα − µβ να From

g = e~ e~ (4.117) µν µ · ν we get:

g = ( e~ ) e~ + e~ ( e~ ) µν,α ∇α µ · ν µ ∇α ν β β = Γ e~ e~ν + e~µ Γ e~ (4.118) µα β · · να β β β = gβν Γ µα + gµβΓ να

This means that gµν;α = 0 (4.119) So the metric tensor is a (covariant) constant tensor.

4.9 The Cartan connection Definition 4.9.1 (Exterior derivative of a basis vector)

d~e Γν ~e ωα (4.120) µ ≡ µα ν ⊗ 78 Chapter 4. Covariant Differentiation

Exterior derivative of a vector field:

dA~ = d(~e Aµ) = ~e dAν + Aµd~e (4.121) µ ν ⊗ µ In arbitrary basis: ν ν λ dA = ~eλ(A )ω (4.122) ν ∂ ν ν (In coordinate basis, ~eλ(A ) = ∂xλ (A ) = A,λ) giving:

~ ν λ µ ν λ dA = ~eν [~eλ(A )ω ] + A Γ µλ~eν ω ⊗ ⊗ (4.123) = (~e (Aν ) + AµΓν )~e ωλ λ µλ ν ⊗

dA~ = Aν ~e ωλ (4.124) ;λ ν ⊗

ν Definition 4.9.2 (Connection forms Ω µ) ν The connection forms Ω µ are 1-forms, defined by:

ν d~eµ ~eν Ω µ ≡ ⊗ (4.125) Γν ~e ωα = ~e Γν ωα = ~e Ων µα ν ⊗ ν ⊗ µα ν ⊗ µ

ν ν α Ω µ = Γ µαω (4.126)

The exterior derivatives of the components of the metric tensor:

dg = d(~e ~e ) = ~e d~e + ~e d~e (4.127) µν µ · ν µ · ν ν · µ where the meaning of the dot is defined as follows:

Definition 4.9.3 (Scalar product between vector and 1-form) µ ν The scalar product between a vector ~u and a (vectorial) one form A = A ν~e ω µ ⊗ is defined by: ~u A uαAµ (~e ~e )ων (4.128) · ≡ ν α · µ

Using this definition, we get:

dg = (~e ~e )Ωλ + (~e ~e )Ωγ µν µ · λ ν ν · γ µ λ γ (4.129) = gµλΩ ν + gνγ Ω µ 4.9 The Cartan connection 79

Lowering an index gives

dgµν = Ωµν + Ωνµ (4.130)

In an orthonormal basis field there is Minkowski-metric:

gµˆνˆ = ηµˆνˆ (4.131) which is constant. Then we have :

dg = 0 Ω = Ω (4.132) µˆνˆ ⇒ νˆµˆ − µˆνˆ where we write Ω = Γ ωαˆ. It follows that Γ = Γ . νˆµˆ νˆµˆαˆ νˆµˆαˆ − µˆνˆαˆ It also follows that ˆ Γtˆ = Γ = Γ = Γi ˆiˆj − tˆˆiˆj ˆitˆˆj tˆˆj ˆ ˆ (4.133) Γi = Γj ˆjkˆ − ˆikˆ Cartans 1st structure equation (without proof): 1 dωρ = cρ ωµ ων −2 µν ∧ 1 = (Γρ Γρ )ωµ ων −2 νµ − µν ∧ (4.134) = Γρ ωµ ων − νµ ∧ = Ωρ ων − ν ∧ dωρ = Ωρ ων and dωρ = Γρ ωµ ων (4.135) − ν ∧ µν ∧ In coordinate basis, we have ωρ = dxρ. Thus, dωρ = d2xρ = 0. ρ We also have c µν = 0 , and C1 is reduced to an identity.This formalism can not be used in coordinate basis!

Example 4.9.1 (Cartan-connection in an orthonormal basis field in plane polar coord.)

ds2 = dr2 + r2dθ2

Introducing basis forms in an orthonormal basis field (where the metric is grˆrˆ = gθˆθˆ = 1): ˆ ˆ ˆ ˆ ds2 = g ωrˆ ωrˆ + g ωθ ωθ = ωrˆ ωrˆ + ωθ ωθ rˆrˆ ⊗ θˆθˆ ⊗ ⊗ ⊗ ˆ ωrˆ = dr, ωθ = rdθ ⇒ Exterior differentiation gives:

ˆ 1 ˆ dωrˆ = d2r = 0, dωθ = dr dθ = ωrˆ ωθ ∧ r ∧ 80 Chapter 4. Covariant Differentiation

C1:

dωµˆ = Ωµˆ ωνˆ − νˆ ∧ ˆ = Ωµˆ ωrˆ Ωµˆ ωθ − rˆ ∧ − θˆ ∧ We have that dωrˆ = 0 , which gives:

ˆ Ωrˆ = Γrˆ ωθ θˆ θˆθˆ (4.136)

θˆ θˆ rˆ since ω ω = 0. (Ω rˆ=0 because of the antisymmetry Ωνˆµˆ = Ωµˆνˆ.) ∧ ˆ ˆ − We also have: dωθ = 1 ωθ ωrˆ. C1: − r ∧ ˆ ˆ ˆ ˆ dωθ = Ωθ ωrˆ Ωθ ωθ − rˆ ∧ − θˆ ∧ =0

ˆ ˆ ˆ |{z}ˆ Ωθ = Γθ ωθ + Γθ ωrˆ rˆ rˆθˆ rˆrˆ (4.137) Γθˆ = 1 giving rˆθˆ r . ˆ We have: Ωrˆ = Ωθ . Using equations 4.136 and 4.137 we get: θˆ − rˆ

θˆ Γ rˆrˆ = 0 1 Γrˆ = ⇒ θˆθˆ −r ˆ ˆ giving Ωrˆ = Ωθ = 1 ωθ. θˆ − rˆ − r Chapter 5

Curvature

5.1 The Riemann curvature tensor

A~(λ + ∆λ) 

~ ~vA∆λ1 ∇ 

M A~QP (λ + ∆λ) A~(λ) ~v * Q λ + ∆λ

P λ

Figure 5.1: Parallel transport of A~

The covariant directional derivative of a vector field A~ along a vector ~u was defined and interpreted geometrically in section 4.2, as follows

dA~ A~ = = Aµ vν~e ∇~v dλ ;ν µ (5.1) A~ (λ + ∆λ) A~(λ) = lim QP − ∆λ 0 ∆λ →

Let A~QP be the parallel transported of A~ from Q to P. Then to first order in ∆λ we have: A~ = A~ + ( A~) ∆λ and QP P ∇~v P

81 82 Chapter 5. Curvature

 

3 A~ k 6

: - A~

Figure 5.2: Parallel transport of a vector along a triangle of angles 90◦ is rotated 90◦

A~ = A~ ( A~) ∆λ (5.2) P Q Q − ∇~v Q To second order in ∆λ we have: 1 A~ = (1 ∆λ + (∆λ)2)A~ (5.3) P Q − ∇~v 2∇~v∇~v Q

If A~P Q is parallel transported further on to R we get

1 A~ = (1 ∆λ + (∆λ)2) P QR u~ 2 u~ u~ − ∇ ∇ ∇ (5.4) 1 (1 ∆λ + (∆λ)2)A~ · − ∇~v 2∇~v∇~v R

where A~Q is replaced by A~R because the differential operator always shall be applied to the vector in the first position. If we parallel transport A~ around the whole polygon in figure 5.3 we get: 1 A~ = (1 + ∆λ + (∆λ)2) P QRST P ∇u~ 2∇u~ ∇u~ 1 (1 + ∆λ + (∆λ)2) ~v 2 ~v ~v · ∇ ∇ ∇ (5.5) 1 (1 (∆λ)2) (1 ∆λ + (∆λ)2) · − ∇[u~,~v] · − ∇u~ 2∇u~ ∇u~ 1 (1 ∆λ + (∆λ)2)A~ · − ∇~v 2∇~v∇~v P 5.1 The Riemann curvature tensor 83

1 S 6] ( u~∆λ~v ~v∆λ~u)∆λ ∇ −=∇[~u, ~v]∆λ2 ~v(T )∆λ ~v∆λ ∇u~∆λ T -  ~vP T ∆λ  ~u∆λ ~u(P )∆λ ~uP Q∆λ ∇~v∆λ j 3 R ~ ~vA 1 ~ ∇ OAQ I I ~u(Q)∆λ A~P A~P Q

- P ~v(P )∆λ Q λ λ + ∆λ

Figure 5.3: Geometrically implied curvature from non-zero differences between vectors along a curve (parameterized by λ) and their parallel transported equiv- alents

Calculating to 2. order in ∆λ gives:

A~ = A~ + ([ , ] )(∆λ)2A~ (5.6) P QRST P P ∇u~ ∇~v − ∇[u~,~v] P There is a variation of the vector under parallel transport around the closed polygon:

δA~ = A~ A~ = ([ , ] )A~ (∆λ)2 (5.7) P QRST P − P ∇u~ ∇~v − ∇[u~,~v] P We now introduce the Riemann’s curvature tensor as:

R( , A~, ~u, ~v) ([ , ] )(A~) (5.8) ≡ ∇u~ ∇~v − ∇[u~,~v] The components of the Riemann curvature tensor is defined by applying the tensor on basis vectors,

Rµ ~e ([ , ] )(~e ) (5.9) ναβ µ ≡ ∇α ∇β − ∇[~eα,~eβ] ν Anti-symmetry follows from the definition:

Rµ = Rµ (5.10) νβα − ναβ

The expression for the variation of A~ under parallel transport around the poly- 84 Chapter 5. Curvature

gon, Eq. (5.7), can now be written as: δA~ = R( , A,~ ~u, ~v)(∆λ)2 ν α β 2 = R( , A ~eν, u ~eα, v ~eβ)(∆λ) = ~e Rµ Aν uαvβ (∆λ)2 (5.11) µ ναβ · 1 = ~e Rµ Aν(uαvβ uβvα)(∆λ)2 2 µ ναβ − The area of the parallellogram defined by the vectors ~u∆λ and ~v∆λ is

∆S~ = ~u ~v(∆λ)2 × 6 ~u∆λ  ∆S

- ~v∆λ

∆S~ = ~u ~v(∆λ)2. × Using that (~u ~v)αβ = uαvβ uβvα . × − we can write Eq. (5.11) as:

1 δA~ = AνRµ ∆Sαβ~e . (5.12) 2 ναβ µ

The components of the Riemann tensor expressed by the connection- and structure- coefficients are given below: ~e Rµ = [ , ]~e ~e µ ναβ ∇α ∇β ν − ∇[~eα,~eβ] ν = ( cρ )~e ∇α∇β − ∇β∇α − αβ∇ρ ν = ~e ~e cρ ~e ∇α∇β ν − ∇β∇α ν − αβ∇ρ ν µ µ ρ µ (Kozul-connection) = αΓ νβ~eµ βΓ να~eµ c αβΓ νρ~eµ ∇ − ∇ − (5.13) = ( Γµ )~e + Γµ ~e ∇α νβ µ νβ∇α µ ( Γµ )~e Γµ ~e cρ Γµ ~e − ∇β να µ − να∇β µ − αβ νρ µ µ ρ µ = ~eα(Γ νβ)~eµ + Γ νβΓ ρα~eµ ~e (Γµ )~e Γρ Γµ ~e cρ Γµ ~e . − β να µ − να ρβ µ − αβ νρ µ This gives (in arbitrary basis): µ µ µ R ναβ = ~eα(Γ νβ) ~eβ(Γ να) − (5.14) + Γρ Γµ Γρ Γµ cρ Γµ . νβ ρα − να ρβ − αβ νρ 5.1 The Riemann curvature tensor 85

In coordinate basis eq. (5.14) is reduced to:

Rµ = Γµ Γµ + Γρ Γµ Γρ Γµ , (5.15) ναβ νβ,α − να,β νβ ρα − να ρβ

µ µ where Γ νβ = Γ βν are the Christoffel symbols. Due to the antisymmetry (5.10) we can define a matrix of curvature-forms

1 Rµ = Rµ ωα ωβ (5.16) ν 2 ναβ ∧

Inserting the components of the Riemann tensor from eq. (5.14) gives

1 Rµ = (~e (Γµ ) + Γρ Γµ cρ Γµ )ωα ωβ (5.17) ν α νβ νβ ρα − 2 αβ νρ ∧

The connection forms:

µ µ α Ω ν = Γ ναω (5.18)

Exterior derivatives of basis forms:

1 dωρ = cρ ωα ωβ (5.19) −2 αβ ∧

Exterior derivatives of connection forms (C1: dωρ = Ωρ ωα) : − α ∧

dΩµ = dΓµ ωβ + Γµ dωρ ν νβ ∧ νρ 1 (5.20) = ~e (Γµ ωα ωβ cρ Γµ ωα ωβ α νβ ∧ − 2 αβ νρ ∧

The curvature forms can now be written as:

Rµ = dΩµ + Ωµ Ωλ (5.21) ν ν λ ∧ ν

This is Cartans 2nd structure equation. 86 Chapter 5. Curvature

5.2 Differential geometry of surfaces

N

λ eν

eµ

u

Figure 5.4: The geometry of a surface. We see the normal vector and the unit vectors of the tangent plane of a point on the surface.

Imagine an arbitrary surface embedded in an Euclidian 3 dimensional space. (See figure 5.4). Coordinate vectors on the surface :

∂ ∂ ~e = , ~e = (5.22) u ∂u v ∂v where u and v are coordinates on the surface. Line element on the surface:

2 µ ν ds = gµν dx dx (5.23)

with x1 = u and x2 = v. (1st fundamental form) The directional derivatives of the basis vectors are written

α ~ ~eµ,ν = Γ µν~eα + Kµν N, α = 1, 2 (5.24)

Greek indices run through the surface coordinates, N~ is a unit vector orthogonal to the surface. 5.2 Differential geometry of surfaces 87

The equation above is called Gauss’ equation. We have: Kµν = ~eµ,ν N~ . In ∂2 ∂2 · coordinate basis, we have ~eµ,ν = ∂xµ∂xν = ∂xν ∂xµ = ~eν,µ. It follows that

Kµν = Kνµ (5.25)

Let ~u be the unit tangent vector to a curve on the surface, parametrised by λ. Differentiating ~u along the curve: d~u = uµ uν~e + K uµuν N~ (5.26) dλ ;ν µ µν 2nd fundamental form

We define κg and κN by: | {z } d~u = κ ~e + κ N~ (5.27) dλ g N

κg is called geodesic curvature. κN is called normal curvature (external curva- ture). κg = 0 for geodesic curves on the surface.

κ ~e = uµ uν~e = ~u g ;ν µ ∇u~ µ ν κN = Kµν u u (5.28) d~u And :κ = N~ N dλ ·

We also have that ~u N~ = 0 along the whole curve. Differentiation gives: · d~u dN~ N~ + ~u = 0 (5.29) dλ · · dλ gives: dN~ κ = ~u (5.30) N − · dλ which is called Weingarten’s equation. κg and κN together give a complete description of the geometry of a surface in a flat 3 dimensional space. We are now going to consider geodesic curves through a point on the surface. Tangent vector ~u = uµ~e with ~u ~u = g uµuν = µ · µν 1. Directions with maximum and minimum values for the normal curvatures are µ ν found, by extremalizing κN under the condition gµν u u = 1. We then solve the variation problem δF = 0 for arbitrary uµ, where F = K uµuν k(g uµuν 1). µν − µν − Here k is the Lagrange multiplicator. Variation with respect to uµ gives:

δF = 2(K kg )uνδuµ µν − µν δF = 0 for arbitrary δuµ demands:

(K kg )uν = 0 (5.31) µν − µν For this system of equations to have nonzero solutions, we must have:

det(K kg ) = 0 (5.32) µν − µν 88 Chapter 5. Curvature

K kg K kg 11 − 11 12 − 12 = 0 (5.33) K kg K kg 21 − 21 22 − 22

This gives the following quadratic equation for k:

k2det(g ) (g K 2g K + g K )k + det(K ) = 0 (5.34) µν − 11 22 − 12 12 22 11 µν (K symmetric K12 = K21)

The equation has two solutions, k1 and k2. These are the extremal values of k. To find the meaning of k, we multiply eq.5.31 by uµ:

0 = (K kg )uµuν µν − µν = K uµuν kg uµuν (5.35) µν − µν = κ k k = κ N − ⇒ N

The extremal values of κN are called the principal curvatures of the surface. Let the directions of the geodesics with extreme normal curvature be given by the tangent vectors ~u and ~v.Eq.5.31 gives:

ν ν Kµν u = kgµν u (5.36)

We then get:

ν µ ν µ Kµν u v = k1gµν u v = k u vµ = k (~u ~v) 1 µ 1 · K vνuµ = k g vν uµ = k (~u ~v) µν 2 µν 2 ·

gives (k k )(~u ~v) = K (uν vµ vνuµ) 1 − 2 · µν − [ν µ] (5.37) = 2Kµν u v

K is symmetric in µ and ν. So we get:(k k )(~u ~v) = 0. For k = k we µν 1 − 2 · 1 6 2 have to demand ~u ~v = 0. So the geodesics with extremal normal curvature, are · orthogonal to each other. The Gaussian curvature (at a point) is defined as:

K = κ κ (5.38) N1 · N2

Since κN1 and κN2 are solutions of the quadratic equation above, we get:

det(Kµν ) K = (5.39) det(gµν ) 5.2 Differential geometry of surfaces 89

5.2.1 Surface curvature, using the Cartan formalism In each point on the surface we have an orthonormal set of basis vectors. Greek indices run through the surface coordinates (two dimensional) and Latin indices through the space coordinates (three dimensional): e~ = (e~ , e~ , N~ ) , e~ = e~ , e~ (5.40) a 1 2 µ { 1 2} Using the exterior derivative and form formalism, we find how the unit vectors on the surface change: a de~ν = e~a Ω ν ⊗ (5.41) = e~ Ωα + N~ Ω3 , α ⊗ ν ⊗ ν µ µ α where Ω ν = Γ ναω are the connection forms on the surface, i.e. the intrinsic connection forms. The extrinsic connection forms are 3 α µ µ α Ω ν = Kναω , Ω 3 = K αω (5.42) We let the surface be embedded in an Euclidean (flat) 3-dimensional space. This means that the curvature forms of the 3-dimensional space are zero: Ra = 0 = d Ωa + Ωa Ωk (5.43) 3b b k ∧ b which gives: µ µ µ α µ 3 R 3ν = 0 = d Ω ν + Ω α Ω ν + Ω 3 Ω ν ∧ ∧ (5.44) = Rµ + Ωµ Ω3 , ν 3 ∧ ν µ where R ν are the curvature forms of the surface. We then have: 1 Rµ ωα ωβ = Ωµ Ω3 (5.45) 2 ναβ ∧ − 3 ∧ ν Inserting the components of the extrinsic connection forms, we get: (when we µ remember the anti symmetry of α and β in R ναβ) Rµ = Kµ K Kµ K (5.46) ναβ α νβ − β να We now lower the first index: R = K K K K (5.47) µναβ µα νβ − µβ να Rµναβ are the components of a curvature tensor which only refer to the dimen- sions of the surface. In particular we have: R = K K K K = det K (5.48) 1212 11 22 − 12 21 We then have the following connection between this component of the Riemann curvature tensor of the surface and the Gaussian curvature of the surface: det Kµν R1212 K = κN1 κN2 = = (5.49) · det gµν det gµν Since the right hand side refers to the intrinsic curvature and the metric on the surface, we have proved that the Gaussian curvature of a surface is an intrinsic quantity. It can be measured by observers on the surface without embedding the surface in a three-dimensional space. This is the contents of Gauss’ theorema egregium. 90 Chapter 5. Curvature

5.3 The Ricci identity

e~ Rµ Aν = ( )(A~) (5.50) µ ναβ ∇α∇β − ∇β∇α − ∇[e~α,e~β] In coordinate basis this is reduced to

e~ Rµ Aν = (Aµ Aµ )e~ , (5.51) µ ναβ ;βα − ;αβ µ where

Aµ (Aµ ) (5.52) ;αβ ≡ ;β ;α The Ricci identity on component form is:

Aν Rµ = Aµ Aµ (5.53) ναβ ;βα − ;αβ We can write this as: 1 d2A~ = Rµ Aν e~ ωα ωβ (5.54) 2 ναβ µ ⊗ ∧ This shows us that the 2nd exterior derivative of a vector is equal to zero only in a flat space. Equations (5.53) and (5.54) both represents the Ricci identity.

5.4 Bianchi’s 1st identity

Cartan’s 1st structure equation:

d ωµ = Ωµ ων (5.55) − ν ∧ Cartan’s 2nd structure equation:

Rµ = d Ωµ + Ωµ Ωλ (5.56) ν ν λ ∧ ν

Exterior differentiation of (5.55) and use of P oincare´0s lemma (4.16) gives: (d2 ωµ = 0)

0 = d Ωµ ων Ωµ d ωλ (5.57) ν ∧ − λ ∧ Use of (5.55) gives:

d Ωµ ων + Ωµ Ωλ ων = 0 (5.58) ν ∧ λ ∧ ν ∧ From this we see that

(d Ωµ + Ωµ Ωλ ) ων = 0 (5.59) ν λ ∧ ν ∧ We now get Bianchi’s 1st identity:

Rµ ων = 0 (5.60) ν ∧ 5.5 Bianchi’s 2nd identity 91

On component form Bianchi’s 1st identity is 1 Rµ ωα ωβ ων = 0 (5.61) 2 ναβ ∧ ∧ µ R ν | {z } The component equation is: (remember the anti symmetry in α and β)

µ R [ναβ] = 0 (5.62) or

µ µ µ R ναβ + R αβν + R βνα+ = 0 (5.63) where the anti symmetry Rµ = Rµ has been used. Without this anti ναβ − νβα symmetry we would have gotten six, and not three, terms in this equation.

5.5 Bianchi’s 2nd identity

Exterior differentiation of (5.56) ⇒ µ µ λ µ ρ λ µ λ µ λ ρ d R ν = R Ω ν Ω ρ Ω Ω ν Ω R ν + Ω Ω ρ Ω ν λ ∧ − ∧ λ ∧ − λ ∧ λ ∧ ∧ (5.64) = Rµ Ωλ Ωµ Rλ λ ∧ ν − λ ∧ ν We now have Bianchi’s 2nd identity as a form equation:

d Rµ + Ωµ Rλ Rµ Ωλ = 0 (5.65) ν λ ∧ ν − λ ∧ ν

As a component equation Bianchi’s 2nd identity is given by

µ R ν[αβ;γ] = 0 (5.66)

Definition 5.5.1 (Contraction) ‘Contraction’ is a tensor operation defined by

R Rµ (5.67) νβ ≡ νµβ We must here have summation over µ. What we do, then, is constructing a new tensor from another given tensor, with a rank 2 lower than the given one.

The tensor with components Rνβ is called the Ricci curvature tensor. µ Another contraction gives the Ricci curvature scalar, R = R µ. 92 Chapter 5. Curvature

Riemann curvature tensor has four symmetries. The definition of the Rie- µ µ mann tensor implies that R ναβ = R νβα µ − Bianchi’s 1st identity: R [ναβ] = 0 From Cartan’s 2nd structure equation follows

λ R µν = dΩ µν + Ωµλ Ω ν ∧ (5.68) R = R ⇒ µναβ − νµαβ By choosing a locally Cartesian coordinate system in an inertial frame we get the following expression for the components of the Riemann curvature tensor: 1 R = (g g + g g ) (5.69) µναβ 2 µβ,να − µα,νβ να,µβ − νβ,µα

from which it follows that Rµναβ = Rαβµν . Contraction of µ and α leads to:

α α R ναβ = R βαν (5.70) R = R ⇒ νβ βν i.e. the Ricci tensor is symmetric. In 4-D the Ricci tensor has 10 independent components. Chapter 6

Einstein’s Field Equations

6.1 Energy-momentum conservation

6.1.1 Newtonian fluid Energy-momentum conservation for a Newtonian fluid in terms of the divergence of the energy momentum tensor can be shown as follows. The total derivative of a velocity field is D~v ∂~v + (~v ~ )~v (6.1) Dt ≡ ∂t · ∇ ∂~v ∂t is the local derivative which gives the change in ~v as a function of time at a given point in space. (~v ~ )~v is called the convective derivative of ~v. It · ∇ represents the change of ~v for a moving fluid particle due to the inhomogeneity of the fluid velocity field. In component notation the above become Dvi ∂vi ∂vi + vj (6.2) Dt ≡ ∂t ∂xj The continuity equation ∂ρ ∂ρ ∂(ρvi) + (ρ~v) = 0 or + = 0 (6.3) ∂t ∇ · ∂t ∂xi Euler’s equation of motion (ignoring gravity)

D~v ∂vi ∂vi ∂p ρ = ~ p or ρ + vj = (6.4) Dt −∇ ∂t ∂xj −∂xi   The energy momentum tensor is a symmetric tensor of rank 2 that describes material characteristics. T 00 T 01 T 02 T 03 T 10 T 11 T 12 T 13 T µν =   (6.5) T 20 T 21 T 22 T 23 T 30 T 31 T 32 T 33     c 1 ≡

93 94 Chapter 6. Einstein’s Field Equations

T 00 represents energy density.

T i0 represents momentum density.

T ii represents pressure (T ii > 0).

T ii represents stress (T ii < 0).

T ij represents shear forces (i = j). 6

Example 6.1.1 (Energy momentum tensor for a Newtonian fluid)

T 00 = ρ T i0 = ρvi (6.6) T ij = ρvivj + pδij

where p is pressure, assumed isotropic here. We choose a locally Cartesian coordinate system in an inertial frame such that the covariant derivatives are reduced to partial µν derivatives. The divergence of the momentum energy tensor, T ;ν has 4 components, one for each value of µ. The zeroth component is

0ν 0ν 00 0i T ;ν = T ,ν = T ,0 + T ,i ∂ρ ∂(ρvi) (6.7) = + ∂t ∂xi 0ν which by comparison to Newtonian hydrodynamics implies that T ;ν = 0 is the continuity equation. This equation represents the conservation of energy. The ith component of the divergence is

iν i0 ij T ,ν = T ,0 + T ,j ∂(ρvi) ∂(ρvivj + pδij) = + ∂t ∂xj (6.8) ∂vi ∂ρ ∂ρvj ∂vi ∂p = ρ + vi + vi + ρvj + ∂t ∂t ∂xj ∂xj ∂xi now, according to the continuity equation ∂(ρvi) ∂ρ = ∂xi − ∂t i i iν ∂v i ∂ρ i ∂ρ j ∂v ∂p T ,ν = ρ + v v + ρv j + i ⇒ ∂t ∂t − ∂t ∂x ∂x (6.9) Dvi ∂p = ρ + Dt ∂xi Dvi ∂p T iν = 0 ρ = ∴ ;ν ⇒ Dt −∂xi which is Euler’s equation of motion. It expresses the conservation of momentum. 6.2 Einstein’s curvature tensor 95

µν The equations T ;ν = 0 are general expressions for energy and momentum conservation.

6.1.2 Perfect fluids A perfect fluid is a fluid with no viscosity and is given by the energy-momentum tensor p T = (ρ + )u u + pg (6.10) µν c2 µ ν µν where ρ and p are the mass density and the stress, respectively, measured in the fluids rest frame, uµ are the components of the 4-velocity of the fluid. In a comoving orthonormal basis the components of the 4-velocity are uµˆ = (c, 0, 0, 0). Then the energy-momentum tensor is given by ρc2 0 0 0 0 p 0 0 Tµˆνˆ =   (6.11) 0 0 p 0  0 0 0 p     where p > 0 is pressure and p < 0 is tension. There are three different types of perfect fluids that are useful. 1. dust or non-relativistic gas is given by p = 0 and the energy-momemtum tensor Tµν = ρuµuν. 2. radiation or ultra-relativistic gas is given by a traceless energy-momemtum µ 1 2 tensor, i.e. T µ = 0. It follows that p = 3 ρc . 3. vacuum energy: If we assume that no velocity can be measured relatively to vacuum, then all the components of the energy-momentum tensor must be Lorentz-invariant. It follows that T g . If vacuum is defined as a µν ∝ µν perfect fluid we get p = ρc2 so that T = pg = ρc2g . − µν µν − µν 6.2 Einstein’s curvature tensor

The field equations are assumed to have the form:

space-time curvature momentum-energy tensor ∝ Also, it is demanded that energy and momentum conservation should follow as a consequence of the field equation. This puts the following constraints on the curvature tensor: It must be a symmetric, divergence free tensor of rank 2. Bianchi’s 2nd identity: µ µ µ R ναβ;σ + R νσα;β + R νβσ;α = 0 (6.12) contraction of µ and α ⇒ µ µ µ R νµβ;σ R νµσ;β + R νβσ;µ = 0 − (6.13) R R + Rµ = 0 νβ;σ − νσ;β νβσ;µ 96 Chapter 6. Einstein’s Field Equations

further contraction of ν and σ gives

Rσ Rσ + Rσµ = 0 β;σ − σ;β σβ;µ Rσ R + Rσ = 0 (6.14) β;σ − ;β β;σ σ ∴ 2R β;σ = R ;β

Thus, we have calculated the divergence of the Ricci tensor, 1 Rσ = R (6.15) β;σ 2 ;β Now we use this expression together with the fact that the metric tensor is co- variant and divergence free to construct a new divergence free curvature tensor. 1 Rσ R = 0 (6.16) β;σ − 2 ;β

σ σ β Keeping in mind that (g βR);σ = g βR;σ we multiply (6.16) by g α to get

1 gβ Rσ gβ R = 0 α β;σ − α 2 ;β (6.17) β σ 1 β g αR β g αR = 0 ;σ − 2 ;β     interchanging σ and β in the first term of the last equation:

σ β 1 β g αR σ g αR = 0 ;β − 2 ;β   1  (6.18) Rβ δβ R = 0 ⇒ α − 2 α  ;β

σ β σ β β β 1 β since g R σ=δ R σ=R α. So that R α δ αR is the divergence free curvature α α − 2 tensor desired. This tensor is called the Einstein tensor and its covariant components are denoted by Eαβ. That is

1 E = R g R (6.19) αβ αβ − 2 αβ

µν NOTE THAT: E ;ν = 0 4 equations, giving only 6 equations from E , → µν which secures a free choice of coordinate system.

6.3 Einstein’s field equations

Einstein’s field equations:

Eµν = κTµν (6.20) 6.3 Einstein’s field equations 97 or 1 R g R = κT (6.21) µν − 2 µν µν Contraction gives: 1 R 4R = κT , where T T µ − 2 ≡ µ (6.22) R = κT −

1 R = g ( κT ) + κT , (6.23) µν 2 µν − µν Thus the field equations may be written in the form 1 R = κ(T g T ) (6.24) µν µν − 2 µν In the Newtonian limit the metric may be written

2φ ds2 = 1 + dt2 + (1 + h )(dx2 + dy2 + dz2) (6.25) − c2 ii   where the Newtonian potential φ c2. We also have T T and T T . | |  00  kk ≈ − 00 Then the 00-component of the field equations becomes κ R T (6.26) 00 ≈ 2 00 Furthermore we have

µ i R00 = R 0µ0 = R 0i0 = Γi Γi 00,i − 0i,0 ∂Γk 1 = 00 = 2φ (6.27) ∂xk c2 ∇

Since T ρc2 eq.(6.26) can be written 2φ = 1 κc4ρ. Comparing this equation 00 ≈ ∇ 2 with the Newtonian law of gravitation on local form: 2φ = 4πGρ, we see that 8πG ∇ κ = c4 . In classical vacuum we have : Tµν = 0, which gives

Eµν = 0 or Rµν = 0 . (6.28)

These are the “vacuum field equations”. Note that Rµν = 0 does not imply Rµναβ = 0. 98 Chapter 6. Einstein’s Field Equations

Digression 6.3.1 (Lagrange (variation principle)) It was shown by Hilbert that the field equations may be deduced from a variation principle with action R√ gd4x , (6.29) − Z where R√ g is the Lagrange density. One may also include a so-called cosmological − constant Λ: (R + 2Λ)√ gd4x (6.30) − Z The field equations with cosmological constant are 1 R g R + Λg = κT (6.31) µν − 2 µν µν µν

6.4 The “geodesic postulate” as a consequence of the field equations

The principle that free particles follow geodesic curves has been called the “geodesic postulate”. We shall now show that the “geodesic postulate” follows as a consequence of the field equations. Consider a system of free particles in curved space-time. This system can be regarded as a pressure-free gas. Such a gas is called dust. It is described by an energy-momentum tensor

T µν = ρuµuν (6.32)

where ρ is the rest density of the dust as measured by an observer at rest in the dust and uµ are the components of the four-velocity of the dust particles. Einstein’s field equations as applied to space-time filled with dust, take the form 1 Rµν gµν R = κρuµuν (6.33) − 2 Because the divergence of the left hand side is zero, the divergence of the right hand side must be zero, too

µ ν (ρu u );ν = 0 (6.34)

or

ν µ (ρu u );ν = 0 (6.35)

we now regard the quantity in the parenthesis as a product of ρuν and uµ. By the rule for differentiating a product we get

ν µ ν µ (ρu );νu + ρu u ;ν = 0 (6.36) 6.4 The “geodesic postulate” as a consequence of the field equations 99

Since the four-velocity of any object has a magnitude equal to the velocity of light we have

u uµ = c2 (6.37) µ − Differentiation gives

µ (uµu );ν = 0 (6.38)

Using, again, the rule for differentiating a product, we get

µ µ uµ;νu + uµu ;ν = 0 (6.39)

From the rule for raising an index and the freedom of changing a summation index from α to µ, say, we get

µ µ µα µα α µ u µ;νu = u uµ;ν = g uαuµ;ν = uαg uµ;ν = uαu ;ν = uµu ;ν (6.40)

Thus the two terms of eq.(6.39) are equal. It follows that each of them are equal to zero. So we have

µ uµu ;ν = 0 (6.41)

Multiplying eq.(6.36) by uµ, we get

ν µ ν µ (ρu );νuµu + ρu uµu ;ν = 0 (6.42)

Using eq.(6.37) in the first term, and eq.(6.41) in the last term, which then vanishes, we get

ν (ρu );ν = 0 (6.43)

Thus the first term in eq.(6.36) vanishes and we get

ν µ ρu u ;ν = 0 (6.44)

Since ρ = 0 we must have 6 ν µ u u ;ν = 0 (6.45)

This is just the geodesic equation. Conclusion: It follows from Einstein’s field equations that free particles move along paths corresponding to geodesic curves of space-time. Chapter 7

The Schwarzschild spacetime

7.1 Schwarzschild’s exterior solution

This is a solution of the vacuum field equations Eµν = 0 for a static spherically symmetric spacetime. One can then choose the following form of the line element (employing units so that c=1),

ds2 = e2α(r)dt2 + e2β(r)dr2 + r2dΩ2 − (7.1) dΩ2 = dθ2 + sin2 θdφ2

These coordinates are chosen so that the area of a sphere with radius r is 4πr 2. Physical distance in radial direction, corresponding to a coordinate distance β(r) dr, is dlr = √grrdr = e dr. Here follows a stepwise algorithm to determine the components of the Ein- stein tensor by using the Cartan formalism:

1. Using orthonormal basis (ie. solving Eµˆνˆ = 0) we find

ˆ ˆ ˆ ωt = eα(r)dt , ωrˆ = eβ(r)dr , ωθ = rdθ , ωφ = r sin θdφ (7.2)

2. Computing the connection forms by applying Cartan’s 1. structure equa- tions dωµˆ = Ωµˆ ωνˆ (7.3) − νˆ ∧

tˆ α dω = e α0dr dt ∧ α β rˆ α tˆ = e α0e− ω e− ω ∧ (7.4) β tˆ rˆ = e− α0ω ω − ∧ = Ωtˆ ωrˆ − rˆ ∧

tˆ β tˆ rˆ ∴ Ω rˆ = e− α0ω + f1ω (7.5)

100 7.1 Schwarzschild’s exterior solution 101

3. To determine the f-functions we apply the anti-symmetry

Ω = Ω (7.6) µˆνˆ − νˆµˆ This gives:

rˆ φˆ 1 β φˆ Ω = Ω = e− ω φˆ − rˆ −r θˆ φˆ 1 φˆ Ω ˆ = Ω ˆ = cot θω φ − θ −r (7.7) tˆ rˆ β tˆ Ω rˆ = +Ω tˆ = e− α0ω rˆ θˆ 1 β θˆ Ω = Ω = e− ω θˆ − rˆ −r

4. We then proceed to determine the curvature forms by applying Cartan’s 2nd structure equations

Rµˆ = dΩµˆ + Ωµˆ Ωαˆ (7.8) νˆ νˆ αˆ ∧ νˆ which gives:

tˆ 2β 2 tˆ rˆ R = e− (α00 + α0 α0β0)ω ω rˆ − − ∧ tˆ 1 2β tˆ θˆ R = e− α0ω ω θˆ −r ∧ tˆ 1 2β tˆ φˆ R = e− α0ω ω φˆ −r ∧ (7.9) rˆ 1 2β rˆ θˆ R = e− β0ω ω θˆ r ∧ rˆ 1 2β rˆ φˆ R = e− β0ω ω φˆ r ∧ θˆ 1 2β θˆ φˆ R = (1 e− )ω ω φˆ r2 − ∧

5. By applying the following relation

1 ˆ Rµˆ = Rµˆ ωαˆ ωβ (7.10) νˆ 2 νˆαˆβˆ ∧ we find the components of Riemann’s curvature tensor.

6. Contraction gives the components of Ricci’s curvature tensor

R Rαˆ (7.11) µˆνˆ ≡ µˆαˆνˆ

7. A new contraction gives Ricci’s curvature scalar

R Rµˆ (7.12) ≡ µˆ 102 Chapter 7. The Schwarzschild spacetime

8. The components of the Einstein tensor can then be found 1 E = R η R , (7.13) µˆνˆ µˆνˆ − 2 µˆνˆ where η = diag( 1, 1, 1, 1). We then have: µˆνˆ −

2 2β 1 2β E = e− β0 + (1 e− ) tˆtˆ r r2 − 2 2β 1 2β E = e− α0 (1 e− ) (7.14) rˆrˆ r − r2 − 2β 2 α0 β0 E = E = e− (α00 + α0 α0β0 + ) θˆθˆ φˆφˆ − r − r

We want to solve the equations Eµˆνˆ = 0. We get only 2 independent equations, and choose to solve those:

Etˆtˆ = 0 and Erˆrˆ = 0 (7.15)

By adding the 2 equations we get:

Etˆtˆ + Erˆrˆ = 0

2 2β e− (β0 + α0) = 0 ⇒ r

(α + β)0 = 0 α + β = K (const) (7.16) ⇒ ⇒ 1 We now have: ds2 = e2αdt2 + e2βdr2 + r2dΩ2 (7.17) − By choosing a suitable coordinate time, we can achieve

K = 0 α = β 1 ⇒ − Since we have ds2 = e2αdt2 + e 2αdr2 + r2dΩ2, this means that g = − − rr 1 . We still have to solve one more equation to get the complete solution, − gtt and choose the equation Etˆtˆ = 0, which gives

2 2β 1 2β e− β0 + (1 e− ) = 0 r r2 − This equation can be written:

1 d 2β [r(1 e− )] = 0 r2 dr − (7.18) 2β r(1 e− ) = K (const) ∴ − 2

If we choose K2 = 0 we get β = 0 giving α = 0 and

ds2 = dt2 + dr2 + r2dΩ2 , (7.19) − 7.1 Schwarzschild’s exterior solution 103

which is the Minkowski space-time described in spherical coordinates. In general, K = 0 and 1 e 2β = K2 K , giving 2 6 − − r ≡ r

2α 2β K e = e− = 1 − r and K dr2 ds2 = (1 )dt2 + + r2dΩ2 (7.20) − − r 1 K − r We can find K by going to the Newtonian limit. We calculate the gravita- tional acceleration ( that is, the acceleration of a free particle instantanously at rest ) in the limit of a weak field of a particle at a distance r from a spherical mass M. Newtonian: d2r GM g = = (7.21) dt2 − r2 We anticipate that r >> K. Then the proper time τ of a particle will be approximately equal to the coordinate time, since dτ = 1 K dt − r The acceleration of a particle in 3-space, is given by theqgeodesic equation:

2 µ d x µ α β 2 + Γ αβu u = 0 dτ (7.22) dxα uα = dτ For a particle instantanously at rest in a weak field, we have dτ dt. Using ≈ uµ = (1, 0, 0, 0), we get: d2r g = = Γr (7.23) dt2 − tt r This equation gives a physical interpretation of Γ tt as the gravitational acceler- ation. This is a mathematical way to express the principle of equivalence: The gravitational acceleration can be transformed to 0, since the Christoffel symbols always can be transformed to 0 locally, in a freely falling non-rotating frame, i.e. a local inertial frame.

1 ∂g ∂g ∂g Γr = grα αt + αt tt tt 2 ∂t ∂t −∂xα 1 grα =0 =0
|{z}1 ∂g = | {ztt } | {z } −2grr ∂r K ∂gtt K g = (1 ) , = (7.24) tt − − r ∂r −r2 K GM g = Γr = = − tt −2r2 − r2 gives K = 2GM 2GM or with c: K = c2 104 Chapter 7. The Schwarzschild spacetime

Then we have the line element of the exterior Schwarzschild metric: 2GM dr2 ds2 = (1 )c2dt2 + + r2dΩ2 (7.25) − − c2r 1 2GM − c2r 2GM R 2 is the Schwarzschild radius of a mass M. S ≡ c Weak field: r >> RS. For the earth: R 0.9cm S ∼ For the sun: R 3km S ∼ A standard clock at rest in the Schwarzschild spacetime shows a proper time τ:

RS dτ = 1 dt (7.26) r − r So the coordinate clocks showing t, are ticking with the same rate as the stan- dard clocks far from M. Coordinate clocks are running equally fast no matter where they are. If they hadn’t, the spatial distance between simultanous events with given spatial coordinates, would depend on the time of the measuring of the distance. Then the metric would be time dependent. Time is not running at the Schwarzschild radius.

Definition 7.1.1 (Physical singularity) A physical singularity is a point where the curvature is infinitely large.

Definition 7.1.2 (Coordinate singularity) A coordinate singularity is a point (or a surface) where at least one of the components of the metric tensor is infinitely large, but where the curvature of spacetime is finite.

µναβ Kretschmann’s curvature scalar is RµναβR . From the Schwarzschild metric, we get: 48G2M 2 R Rµναβ = (7.27) µναβ r8 which diverges only at the origin. Since there is no physical singularity at r = RS, the singularity here is just a coordinate singularity, and can be re- moved by a transformation to a coordinate system falling inward. (Eddington - Finkelstein coordinates, Kruskal - Szekers analytical extension of the description of Schwarzschild spacetime to include the area inside RS).

7.2 Radial free fall in Schwarzschild spacetime

The Lagrangian function of a particle moving radially in Schwarzschild space- time 2 1 RS 2 2 1 r˙ d L = (1 )c t˙ + , · (7.28) −2 − r 2 (1 RS ) ≡ dτ − r 7.2 Radial free fall in Schwarzschild spacetime 105 where τ is the time measured on a standard clock which the particle is carrying. The momentum conjugate pt of the cyclic coordinate t, is a constant of motion.

∂L RS 2 pt = = (1 )c t˙ (7.29) ∂t˙ − − r

4-velocity identity: u uµ = c2: µ − 2 RS r˙ (1 )c2t˙2 + = c2 (7.30) − − r 1 RS − − r Inserting the expression for t˙ gives:

2 p RS r˙2 t = (1 )c2 (7.31) − c2 − − r

Boundary conditions: the particle is falling from rest at r = r0.

2 RS c RS 2 pt = (1 ) = 1 c (7.32) r R r − − 0 1 S −r − 0 − r0 q t˙(r=r0) | {z } giving

dr RS r0 r r˙ = = c − (7.33) dτ r r − r 0 r dr RS = c τ (7.34) r0 r − r0 Z r− r q Integration with τ = 0 for r = 0 gives:

r0 r0 r r r τ = (arcsin 1 ) (7.35) − c R r − r − r r S r 0 r 0 r 0 τ is the proper time that the particle spends on the part of the fall which is r from r to r=0. To the lowest order in r0 we get:

2 r r τ = (7.36) −3 R c r S

Travelling time from r = RS to r = 0 for RS = 2km is then:

2 RS 6 τ(R ) = = 4 10− s (7.37) | S | 3 c × 106 Chapter 7. The Schwarzschild spacetime

7.3 Light cones in Schwarzschild spacetime

The Schwarzschild line-element (with c = 1) is 2 RS dr ds2 = (1 )dt2 + + r2dΩ2 (7.38) − − r (1 RS ) − r We will look at radially moving photons (ds2 = dΩ2 = 0). We then get

1 dr RS r 2 dr √r RS = 1 dt = −1 dt 1 RS r − r ⇔ √r RS  r 2 − r − (7.39) q rdr = dt r R  − S with + for outward motion and for inward motion. For inwardly moving − photons, integration yields r r + t + RS ln 1 = k = constant (7.40) |RS − |

We now introduce a new time coordinate t0 such that the equation of motion for photons moving inwards takes the following form dr r + t0 = k = 1 ⇒ dt0 − r (7.41) ∴ t0 = t + RS ln 1 |RS − |

The coordinate t0 is called an ingoing Eddington-Finkelstein coordinate. The photons here always move with the local velocity of light, c. For photons moving outwards we have r r + RS ln 1 = t + k (7.42) |RS − | r Making use of t = t0 RS ln 1 we get − | RS − | r r + 2RS ln 1 = t0 + k |RS − | dr 2RS dr r + RS dr + = 1 = 1 (7.43) ⇒ dt r R dt ⇔ r R dt 0 − S 0 − S 0 dr r R = − S ⇔ dt0 r + RS Making use of ordinary Schwarzschild coordinates we would have gotten the following coordinate velocities for inn- and outwardly moving photons:

dr RS = (1 ) (7.44) dt  − r which shows us how light is decelerated in a gravitational field. Figure 7.1 shows how this is viewed by a non-moving observer located far away from the mass. In Figure 7.2 we have instead used the alternative time coordinate t0. The special theory of relativity is valid locally, and all material particles thus have to remain inside the light cone. 7.3 Light cones in Schwarzschild spacetime 107

t

Collaps of light cone

at the horizon, r = R S ?

Trajectory of transmitter

horizon Light cone

r

RS

Figure 7.1: At a radius r = RS the light cones collapse, and nothing can any longer escape, when we use the Schwarzschild coordinate time.

t'

Trajectory of transmitter horizon

r

RS

Figure 7.2: Using the ingoing Eddingto n-Finkelstein time coordinate there is no collapse of the light cone at r = RS. Instead we get a collapse at the singularity at r = 0. The angle between the left part of the light cone and the t0-axis is always 45 degrees. We also see that once the transmitter gets inside the horizon at r = RS, no particles can escape. 108 Chapter 7. The Schwarzschild spacetime

7.4 Analytical extension of the Schwarzschild space- time

The Schwarzschild coordinates are comoving with a static reference frame out- side a spherical mass distribution. If the mass has collapsed to a black hole there exist a horizon at the Schwarzschild radius. As we have seen in section 7.3 there do not exist static observers at finite radii inside the horizon. Hence, the Schwarzschild coordinates are well defined only outside the horizon. Also the rr-component of the metric tensor has a coordinate singularity at the Schwarzschild radius. The curvature of spacetime is finite here. Kruskal and Szekeres have introduced new coordinates that are well defined inside as well as outside the Schwarzschild radius, and with the property that the metric tensor is non-singular for all r > 0. In order to arrive at these coordinates we start by considering a photon moving radially inwards. From eq. (7.40) we then have r t = r R ln 1 + v (7.45) − − S R − S

where v is a constant along the world line of the photon. We introduce a new

radial coordinate r r∗ r + R ln 1 (7.46) ≡ S R − S

Then the equation of the worldline of the photon tak es the form

t + r∗ = v (7.47) The value of the constant v does only depend upon the point of time when the photon was emitted. We may therefore use v as a new time coordinate. For an outgoing photon we get in the same way

t r∗ = u (7.48) − where u is a constant of integration, which may be used as a new time coordinate for outgoing photons. The coordinates u and v are the generalization of the light cone coordinates of Minkowski spacetime to the Schwarzchild spacetime. From eqs. (7.47) and (7.48) we get 1 dt = (dv + du) (7.49) 2 1 dr∗ = (dv du) (7.50) 2 − and from eq. (7.46) Rs dr = 1 dr∗ (7.51) − r   Inserting these differentials into eq. (7.38) we arrive at a new form of the Schwarzschild line-element,

Rs ds2 = 1 du dv + r2dΩ2 (7.52) − − r   7.4 Analytical extension of the Schwarzschild spacetime 109

The metric is still not well behaved at the horizon. Introducing the coordinates

u U = e− 2Rs (7.53) − v V = e 2Rs (7.54) gives v u r r − ∗ Rs UV = e 2Rs = e Rs = 1 e Rs (7.55) − − − r −

and dU dV du dv = 4R 2 (7.56) − s UV The line-element (7.52) then takes the form

3 2 4Rs r 2 2 ds = e− Rs dUdV + r dΩ (7.57) − r This is the first form of the Kruskal-Szekeres line-element. Here is no coordinate singularity, only a physical singularity at r = 0. We may furthermore introduce two new coordinates

1 r r t T = (V + U) = 1 e 2Rs sinh (7.58) 2 R − 2R  s  s 1 r r t Z = (V U) = 1 e 2Rs cosh (7.59) 2 − R − 2R  s  s Hence

V = T + Z (7.60) U = T Z (7.61) − giving dUdV = dT 2 dZ2 (7.62) − Inserting this into eq. (7.57) we arrive at the second form of the Kruskal-Szekeres line-element 3 2 4Rs r 2 2 2 2 ds = e− Rs dT dZ + r dΩ (7.63) − r − The inverse transformations of eqs. (7.58) and
(7.59) is

r r 2 2 1 e Rs = T Z (7.64) − R −  s  t T tanh = (7.65) 2Rs Z Note from eq. (7.63) that with the Kruskal-Szekeres coordinates T and Z the equation of the radial null geodesics has the same form as in flat spacetime

Z = T + constant (7.66)  110 Chapter 7. The Schwarzschild spacetime

7.5 Embedding of the Schwarzschild metric

We will now look at a static, spherically symmetric space. A curved simultaneity plane (dt = 0) through the equatorial plane (dθ = 0) has the line element

2 2 2 2 ds = grrdr + r dφ (7.67)

with a radial coordinate such that a circle with radius r has a circumference of length 2πr. We now embed this surface in a flat 3-dimensional space with cylinder co- ordinates (z, r, φ) and line element

ds2 = dz2 + dr2 + r2dφ2 (7.68)

The surface described by the line element in (7.67) has the equation z = z(r). The line element in (7.68) is therefore written as dz ds2 = [1 + ( )2]dr2 + r2dφ2 (7.69) dr Demanding that (7.69) is in agreement with (7.67) we get dz dz g = 1 + ( )2 = g 1 (7.70) rr dr ⇔ dr  rr − p Choosing the positive solution gives

dz = g 1dr (7.71) rr − p In the Schwarzschild spacetime we have 1 grr = (7.72) 1 RS − r Making use of this we find z: r dr z = = 4RS(r RS) (7.73) R − ZRS 1 S − r p q This is shown in Figure 7.3.

7.6 Deceleration of light

The radial speed of light in Schwarzschild coordinates found by putting ds2 = dΩ2 = 0 in eq. (7.38) is

RS c¯ = 1 (7.74) − r To measure this effect one can look at how long it takes for light to get from Mercury to the Earth. This is illustrated in Figure 7.4. The travel time from 7.6 Deceleration of light 111

Figure 7.3: Embedding of the Schwarzschild metric.

z

Mercury

z2 >0 r 2

Sun b x

r 1

z1 <0 Earth

Figure 7.4: General relativity predicts that light traveling from Mercury to the Earth will be delayed due to the effect of the Suns gravity field on the speed of light. This effect has been measured. 112 Chapter 7. The Schwarzschild spacetime

z1 to z2 is z2 dz z2 R z2 R ∆t = (1 + S )dz = (1 + S )dz RS 2 2 z1 1 ≈ z1 r z1 √b + z Z − r Z Z (7.75) 2 2 √z2 + b + z2 = z2 + z1 + RS ln | | √z 2 + b2 z 1 − | 1| where RS is the Schwarzschild radius of the Sun. The deceleration is greatest when Earth and Mercury (where the light is reflected) are on nearly opposite sides of the Sun. The impact parameter b is then small. A series expansion to the lowest order of b/z gives

4 z1 z2 ∆t = z + z + R ln | | (7.76) 2 | 1| S b2 The last term represents the extra traveling time due to the effect of the Suns gravity field on the speed of light. The journey takes longer time:

R = the Schwarzschild radius of the Sun 2km S ∼ z = the radius of Earth’s orbit = 15 1010m | 1| × z = the radius of Mercury’s orbit = 5.8 1010m 2 × b = R = 7 108m ×

give a delay of 1.1 10 4s. In addition to this one must also, of course, take × − into account among other things the effects of the curvature of spacetime near the Sun and atmospheric effects on Earth.

7.7 Particle trajectories in Schwarzschild 3-space

1 L = g X˙ µX˙ ν 2 µν 1 R 1 r˙2 1 1 (7.77) = 1 s t˙2 + 2 + r2θ˙2 + r2 sin2 θφ˙2 −2 − r 1 Rs 2 2   − r Since t is a cyclic coordinate

∂L Rs pt = = 1 t˙ = constant = E (7.78) − − ∂t˙ − r   where E is the particle’s energy as measured by an observer "far away" (r R ).  s Also φ is a cyclic coordinate so that

∂L 2 2 pφ = = r sin θφ˙ = constant (7.79) ∂φ˙

where pφ is the particle’s orbital angular momentum. 7.7 Particle trajectories in Schwarzschild 3-space 113

Making use of the 4-velocity identity U~ 2 = g X˙ µX˙ ν = 1 we transform µν − the above to get

2 Rs r˙ 1 t˙2 + + r2θ˙2 + r2 sin2 θφ˙2 = 1 (7.80) − − r 1 Rs −   − r

E ˙ pφ which on substitution for t˙ = and φ = 2 becomes 1 Rs r2 sin θ − r

2 2 2 E r˙ pφ + + r2θ˙2 + = 1 (7.81) −1 Rs 1 Rs r2 sin2 θ − − r − r Now, refering back to the Lagrange equation

d ∂L ∂L = 0 (7.82) dτ ˙ µ − ∂Xµ ∂X  we get, for θ

2 2 2 (r θ˙)• = r sin θ cos θφ˙ p2 cos θ (7.83) = φ r2 sin3 θ

Multiplying this by r2θ˙ we get

˙ 2 2 cos θθ 2 (r θ˙)(r θ˙)• = p (7.84) sin3 θ φ which, on integration, gives

pφ 2 (r2θ˙)2 = k (7.85) − sin θ   where k is the constant of integration. Because of the spherical geometry we are free to choose a coordinate system such that the particle moves in the equatorial plane and along the equator at a π ˙ given time t = 0. That is θ = 2 and θ = 0 at time t = 0. This determines the 2 constant of integration and k = pφ such that

1 (r2θ˙)2 = p2 1 (7.86) φ − sin2 θ   The RHS is negative for all θ = π . It follows that the particle cannot deviate 6 2 from its original (equatorial) trajectory. Also, since this particular choice of trajectory was arbitrary we can conclude, quite generally, that any motion of free particles in a spherically symmetric gravitational field is planar motion. 114 Chapter 7. The Schwarzschild spacetime

7.7.1 Motion in the equatorial plane

2 2 2 E r˙ pφ + + = 1 (7.87) −1 Rs 1 Rs r2 − − r − r that is

2 Rs pφ r˙2 = E2 1 1 + (7.88) − − r r2   ! This corresponds to an energy equation with an effective potential V (r) given by

R p2 V 2(r) = 1 s 1 + φ − r r2   ! r˙2 + V 2(r) = E2

2 2 (7.89) r p Rsp V = 1 s + φ φ ⇒ s − r r2 − r3 2 1 Rs 1 pφ u 1 + − 2 r 2 r2

Newtonian potential VN is defined by using the last expression in

2 GM pφ V = V 1 V = + (7.90) N − ⇒ N − r 2r2 The possible trajectories of particles in the Schwarzschild 3-space are shown schematically in Figure 7.5 as functions of position and energy of the particle in the Newtonian limit. To take into account the relativistic effects the above picture must be mod- ified. We introduce dimensionless variables

r pφ X = and k = (7.91) GM GMm

The potential V 2(r) now take the form

2 k2 2k2 1/2 V = 1 + (7.92) − X X2 − X3   For r equal to the Schwarzschild radius (X = 2) we have

k2 2k2 V (2) = 1 1 + = 0 (7.93) r − 4 − 8 For k2 < 12 particles will fall in towards r = 0. 7.7 Particle trajectories in Schwarzschild 3-space 115

Figure 7.5: Newtonian particle trajectories are functions of the position and energy of the particle. Note the centrifugal barrier. Due to this particles with p = 0 cannot arrive at r = 0. φ 6

An orbit equation is one which connects r and φ. So for motion in the equatorial plane for weaks fields we have

dφ pφ d pφ d = • = (7.94) dt mr2 ≡ dτ mr2 dφ Introducing the new radial coordinate u 1 our equations transform to ≡ r du 1 dr 1 mr2 dr m = = = r˙ dφ −r2 dφ −r2 p dt −p φ φ (7.95) p du r˙ = φ ⇒ − m dφ Substitution from above for r˙ in the energy equation yields the orbit equation, du 2 m2 E2 + (1 2GMu) u2 + = . (7.96) dφ − p2 p2   φ ! φ Differentiating this, we find 2 2 d u GMm 2 2 + u = 2 + 3GMu (7.97) dφ pφ The last term on the RHS is a relativistic correction term. 116 Chapter 7. The Schwarzschild spacetime

1.05

20 40 60 80 100 0.95

0.9

Figure 7.6: When relativistic effects are included there is no longer a limit to the values that r can take and collapse to a singularity is "possible". Note that V 2 is plotted here.

7.8 Classical tests of Einstein’s general theory of rel- ativity

7.8.1 The Hafele-Keating experiment

Hafele and Keating measured the difference in time shown on moving and sta- tionary atomic clocks. This was done by flying around the Earth in the East- West direction comparing the time on the clock in the plane with the time on a clock on the ground. i dxi The proper time interval measured on a clock moving with a velocity v = dt in an arbitrary coordinate system with metric tensor gµν is given by

gµν µ ν 1 0 dτ = ( dx dx ) 2 , dx = cdt − c2 i 2 v v 1 = ( g 2g ) 2 dt (7.98) − 00 − i0 c − c2 v2 g vivj ≡ ij 7.8 Classical tests of Einstein’s general theory of relativity 117

For a diagonal metric tensor (gi0 = 0) we get

2 v 1 2 i 2 dτ = ( g ) 2 dt , v = g (v ) (7.99) − 00 − c2 ii

We now look at an idealized situation where a plane flies at constant altitude and with constant speed along the equator.

2 RS v 1 dτ = (1 ) 2 dt , r = R + h (7.100) − r − c2

2 RS v To the lowest order in r and c2 we get

2 RS 1 v dτ = (1 )dt (7.101) − 2r − 2 c2

The speed of the moving clock is

v = (R + h)Ω + u (7.102) where Ω is the angular velocity of the Earth and u is the speed of the plane. A series expansion and use of this value for v gives

GM 1 R2Ω2 gh 2RΩu + u2 GM ∆τ = (1 + )∆t , g = RΩ2 − Rc2 − 2 c2 c2 − 2c2 R2 − (7.103) u > 0 when flying in the direction of the Earth’s rotation, i.e. eastwards. For a clock that is left on the airport (stationary, h = u = 0) we get

GM 1 R2Ω2 ∆τ = (1 )∆t (7.104) 0 − Rc2 − 2 c2

To the lowest order the relative difference in travel time is

2 ∆τ ∆τ0 gh 2RΩu + u k = − = 2 2 (7.105) ∆τ0 ∼ c − 2c

Measurements: Travel time: ∆τ = 1.2 105s (a little over 24h) 0 × Traveling eastwards: k = 1.0 10 12 e − × − Traveling westwards: k = 2.1 10 12 w × − (∆τ ∆τ ) = 1.2 10 7s 120ns − 0 e − × − ≈ − (∆τ ∆τ ) = 2.5 10 7s 250ns − 0 w × − ≈ 118 Chapter 7. The Schwarzschild spacetime

7.8.2 Mercury’s perihelion precession The orbit equation for a planet orbiting a star of mass M is given by equation (7.97), 2 2 d u GMm 2 2 + u = 2 + ku (7.106) dφ pφ where k = 3GM. We will be slightly more general, and allow k to be a theory- or situation dependent term. This equation has a circular solution, such that 2 GMm 2 u0 = 2 + ku0 (7.107) pφ

With a small perturbation from the circular motion u is changed by u1, where u u . To lowest order in u we have 1  0 1 2 2 d u1 GMm 2 2 + u0 + u1 = 2 + ku0 + 2ku0u1 (7.108) dφ pφ or 2 2 d u1 d u1 + u = 2ku u + (1 2ku )u = 0 (7.109) dφ2 1 0 1 ⇔ dφ2 − 0 1 For ku 1 the equilibrium orbit is stable and we get a periodic solution: 0  u = u cos[ 1 2ku (φ φ )] (7.110) 1 0 − 0 − 0 where  and φ0 are integration constanp ts.  is the eccentricity of the orbit. We can choose φ0 = 0 and then have 1 = u = u + u = u [1 +  cos( 1 2ku φ) (7.111) r 0 1 0 − 0 Let f √1 2ku p ≡ − 0 ⇒ 1 1 = (1 +  cos fφ) (7.112) r r0 For f = 1 (k = 0, no relativistic term) this expression describes a non-precessing elliptic orbit (a Kepler-orbit). For f < 1 (k > 0) the ellipse is not closed. To give the same value for r as 2π on a given starting point, φ has to increase by f > 2π. The extra angle per rotation is 2π( 1 1) = ∆φ . f − 1 1 ∆φ1 = 2π( 1) 2πku0 (7.113) √1 2ku − ≈ − 0 Using general relativity we get for Mercury GMm2 k = 3GM ∆φ = 6πGMu0 6πGM 2 (7.114) ⇒ ≈ pφ

GMm ∆φ = 6π( )2per orbit. (7.115) pφ

which in Mercury’s case amounts to (∆φ)century = 4300 7.8 Classical tests of Einstein’s general theory of relativity 119

7.8.3 Deflection of light The orbit equation for a free particle with mass m = 0 is d2u + u = ku2 (7.116) dφ2 If light is not deflected it will follow the straight line b cos φ = = b u (7.117) r 0 where b is the impact parameter of the path. This is the horizontal dashed line in Figure 7.7. The 0’th order solution (7.117) fullfills

2 d u0 + u = 0 (7.118) dφ2 0 Hence it is a solution of (7.116) with k = 0. photon ∆θ 2

b r

φ M, GM<

Figure 7.7: Light traveling close to a massive object is deflected.

The perturbed solution is

u = u + u , u u (7.119) 0 1 | 1|  0 Inserting this into the orbit equation gives

2 2 d u0 d u1 + + u + u = ku 2 + 2ku u + ku 2 (7.120) dφ2 dφ2 0 1 0 0 1 1 The first and third term at the left hand side cancel each other due to eq. (7.118] and the last term at the right hand side is small to second order in u1 and will be neglected. Hence we get

2 d u1 + u = ku 2 + 2ku u (7.121) dφ2 1 0 0 1 120 Chapter 7. The Schwarzschild spacetime

The last term at the right hand side is much smaller then the first, and will also be neglected. Inserting for u0 from (7.117) we then get

2 d u1 k + u = cos2 φ (7.122) dφ2 1 b2

This equation has a particular solution of the form

2 u1p = A + B cos φ (7.123)

Inserting this into (7.122) we find 2k k A = , B = (7.124) 3b2 −3b2 Hence k u = 2 cos2 φ (7.125) 1p 3b2 −
giving 1 cos φ k = u = u + u = + 2 cos2 φ (7.126) r 0 1 b 3b2 −
The deflection of the light ∆θ is assumed to be small. We therefore put φ = n + ∆θ where ∆θ π, (see Figure 7.7). Hence 2 2  n ∆θ ∆θ ∆θ cos φ = cos + = sin (7.127) 2 2 − 2 ≈ 2   Thus, the term cos2 φ in (7.126) can be neglected. Furthermore, the deflection of the light is found by letting r , i. e. u 0. Then we get → ∞ → 4k ∆θ = (7.128) 3b

3 For motion in the Schwarzschild spacetime outside the Sun, k = 2 RS where RS is the Schwarzschild radius of the Sun. And for light passing the surface of the Sun b = R where R is the actual radius of the Sun. The deflection is then

RS ∆θ = 2 = 1.7500 (7.129) R

Chapter 8

Black Holes

8.1 ’Surface gravity’:gravitational acceleration on the horizon of a black hole

Surface gravity is denoted by κ1 and is defined by

a µ κ = lim a = aµa (8.1) r r+ t → u p t where r+ is the horizon radius, r+ = RS for the Schwarzschild spacetime, u is the time component of the 4-velocity. The 4-velocity of a free particle instantanously at rest in the Schwarzschild spacetime: t dt 1 ~et ~u = u ~et = ~et = ~et = (8.2) dτ √ gtt 1 RS − − r q The only component of the 4-acceleration different from zero, is ar. The µ ν µ µ α ν 4-acceleration:~a = ~u = u ;νu ~e = (u ,ν + Γ ανu )u ~e . ∇u~ µ µ α ν ar = (ur,ν + Γrαν u )u ν t 2 = ur,νu +Γrtt(u ) =0 Γ = | {zrtt} 1 RS − r 1 ∂g R (8.3) Γ = tt = S rtt −2 ∂r −2r2 RS 2r2 ar = 1 RS − r r rr ar RS RS a = g ar = = (1 )ar = 2 grr − r 2r

RS √ r 2r2 The acceleration scalar: a = ara = R (measured with standard instru- 1 S − r q 121 122 Chapter 8. Black Holes

ments: at the horizon, time is not running).

a RS = (8.4) ut 2r2 With c: 2 a c RS GM = = (8.5) ut 2r2 r2 a 1 1 κ = lim t = = (8.6) r RS u 2R 4GM → S c2 Including c the expression is κ = 4GM . On the horizon of a black hole with one 13 m solar mass, we get κ = 2 10 2 . × s

8.2 Hawking radiation:radiation from a black hole (1973)

The radiation from a black hole has a thermal spectrum. We are going to ’find’ the temperature of a Schwarzschild black hole of mass M. The Planck spectrum has an intensity maximum at a wavelength given by Wien’s displacement law.

N~c Λ = where k is the Boltzmann constant, and N=0.2014 kT For radiation emitted from a black hole, Hawking derived the following expres- sion for the wavelength at a maximum intensity 8πNGM Λ = 4πNR = (8.7) S c2 Inserting Λ from Wien’s displacement law, gives:

~c3 ~c T = = κ (8.8) 8πGkM 2πk

Inserting values for ~, c and k gives:

4 2 10− m T × K (8.9) ≈ RS

7 For a black hole with one solar mass,we have T 10− . When the mass is ≈ decreasing because of the radiation, the temperature is increasing.So a black hole has a negative heat capacity. The energy loss of a black hole because of radiation, is given by the Stefan-Boltzmann law: dM A = σT 4 (8.10) − dt c2 where A is the surface of the horizon. 16πG2M 2 A = 4πR2 = (8.11) S c4 8.3 Rotating Black Holes: The Kerr metric 123 gives:

dM 1 ~c6 Q = − dt 15360π G2M 2 ≡ M 2 (8.12) M(t) = (M 3 3Qt)1/3, M = M(0) 0 − 0

A black hole with mass M0 early in the history of the universe which is about to explode now, had to have a starting mass

M = (3Qt )1/3 1012kg (8.13) 0 0 ≈ about the mass of a mountain. They are called ’mini black holes’.

8.3 Rotating Black Holes: The Kerr metric

This solution was found by Roy Kerr in 1963. A time-independent, time-orthogonal metric is known as a static metric. A time-independent metric is known as a stationary metric. A stationary metric allows rotation. Consider a stationary metric which describes a axial-symmetric space

ds2 = e2ν dt2 + e2µdr2 + e2ψ(dφ ωdt)2 + e2λdθ2 , (8.14) − − where ν, µ, ψ, λ and ω are functions of r and θ. By solving the vacuum field equations for this line-element, Kerr found the solution: ρ2∆ ρ2 Σ2 e2ν = , e2µ = , e2ψ = sin2 θ , e2λ = ρ2 , Σ2 ∆ ρ2 2Mar ω = , where ρ2 = r2 + a2 cos2 θ Σ2 ∆ = r2 + a2 2Mr − Σ2 = (r2 + a2)2 a2∆ sin2 θ − (8.15)

This is the Kerr solution expressed in Boyer-Lindquist coordinates. The function ω is the angular-velocity. The Kerr-solution is the metric for space-time outside a rotating mass-distribution. The constant a is spin per mass-unit for the mass- distribution and M is its mass. Line-element: 2Mr ρ2 4Mar ds2 = (1 )dt2 + dr2 sin2 θdtdφ + ρ2dθ2 − − ρ2 ∆ − ρ2 (8.16) 2Ma2r + (r2 + a2 + sin2 θ) sin2 θdφ2 ρ2

(Here M is a measure of the mass so that M = G mass, ie. G = 1) · 124 Chapter 8. Black Holes

Light emitted from the surface, r = r0, where g tt = 0 is infinitely redshifted further out. Observed from the outside time stands still. 2 2 2 2 ρ = 2Mr0 r0 + a cos θ = 2Mr0 ⇒ (8.17) r = M M 2 a2 cos2 θ 0  − This is the equation for the surface which represenp ts infinite redshift.

8.3.1 Zero-angular-momentum-observers (ZAMO’s) π The Lagrange function of a free particle in the equator plane, θ = 2 1 1 1 1 L = (e2ν ω2e2ψ)t˙2 + e2µr˙2 + e2ψφ˙2 + e2λθ˙2 ωe2ψt˙φ˙ (8.18) −2 − 2 2 2 −

Here θ˙ = 0. The momentum pφ of the cyclic coordinates φ:

∂L 2ψ dt dφ pφ = e (φ˙ ωt˙) , t˙ = , φ˙ = (8.19) ≡ ∂φ˙ − dτ dτ The angular speed of the particle relative to the coordinate system: dφ φ˙ Ω = = , φ˙ = Ωt˙ dt t˙ (8.20) p = e2ψt˙(Ω ω) ⇒ φ − pφ is conserved during the movement. 2Mar ω = , (r2 + a2)2 a2(r2 + a2 2Mr) − − (8.21) ω 0 when r → → ∞ When studying the Kerr metric one finds that Kerr Minkowski for large → r. The coordinate clocks in the Kerr space-time show the same time as the standard-clocks at rest in the asymptotic Minkowski space-time. A ZAMO is per definition a particle or observer with pφ = 0. Consider a far away observer who let a stone fall with vanishing initial velocity. pφ is a constant of motion, so the stone remains a ZAMO during the movement. A local reference frame which coincides with the stone is a local inertial frame. dφ p = 0 Ω = = ω (8.22) φ ⇒ dt That is, the local inertial frame obtains an angular speed relative to the BL- system (Boyer-Lindquist system). Since the Kerr metric is time independent, the BL-system is stiff. The distant observer has no motion relative to the BL-system. To this observer the BL-system will appear stiff and non-rotating. The observer will observe that a is spin the local inertial system of the stone obtains an angular speed per mass dφ 2Mar = ω = (8.23) unity and dt (r2 + a2)2 a2(r2 + a2 2Mr) Ma is spin − − In other words, inertial systems at finite distances from the rotating mass M are dragged with it in the same direction. This is known as inertial dragging or the Lense-Thirring effect (about 1920). 8.3 Rotating Black Holes: The Kerr metric 125

8.3.2 Does the Kerr space have a horizon? Definition 8.3.1 (Horizon) a surface one can enter, but not exit.

Consider a particle in an orbit with constant r and θ. It’s 4-velocity is: d~x dt d~x ~u = = dτ dτ dt (8.24) 2 1 dφ = ( g 2g Ω g Ω )− 2 (1, Ω) , where Ω = − tt − tφ − φφ dt To have stationary orbits the following must be true

2 g φφΩ + 2g tφΩ + g tt < 0 (8.25)

This implies that Ω must be in the interval

Ωmin < Ω < Ωmax , (8.26)

2 g tt 2 g tt where Ωmin = ω ω g , Ωmax = ω + ω g since g tφ = ωg φφ. − − φφ − φφ − Outside the surfaceq with infinite redshift gqtt < 0. That is Ω can be negative, zero and positive. Inside the surface r = r0 with infinite redshift g tt > 0. Here Ωmin > 0 and static particles, Ω = 0, cannot exist. This is due to the inertial dragging effect. The surface r = r0 is therefore known as “the static border”. The interval of Ω, where stationary orbits are allowed, is reduced to zero 2 g tt 2 when Ωmin = Ωmax, that is ω = g g tt = ω g φφ (equation for the horizon). φφ ⇒ For the Kerr metric we have:

g = ω2g e2ν (8.27) tt φφ − Therefore the horizon equation becomes

e2ν = 0 ∆ = 0 r2 2mr + a2 = 0 (8.28) ⇒ ∴ − The largest solution is r = M + √M 2 a2 and this is the equation for a + − spherical surface. The static border is r = M + √M 2 a2 cos θ. 0 − 126 Chapter 8. Black Holes

θ = 0 horizon

r0

r+

π Ω > 0 θ = 2 2M

Ω = 0  K

M static border ergo-sphere stationary paths

Figure 8.1: Static border and horizon of a Kerr black hole Chapter 9

Schwarzschild’s Interior Solution

9.1 Newtonian incompressible star

2φ = 4πGρ, φ = φ(r) ∇ 1 d dφ (9.1) (r2 ) = 4πGρ r2 dr dr Assuming ρ = constant. dφ d(r2 ) = 4πGρr2dr dr dφ 4π r2 = Gρr3 + K (9.2) dr 3 = M(r) + K

Gravitational acceleration: ~g = φ = dφ~e −∇ − dr r M(r) K1 4π K1 g = + = Gρr + (9.3) r2 r2 3 r2

Finite g in r = 0 demands K1 = 0. 4π dφ 4π g = Gρr, = Gρr (9.4) 3 dr 3 Assume that the massdistribution has a radius R. 2π φ = Gρr2 + K (9.5) 3 2 Demands continuous potensial at r = R.

2π 2 M(R) 4π 2 GρR + K2 = = GρR 3 R − 3 (9.6) K = 2πGρR2 ⇒ 2 − (with zero level at infinite distance). Gives the potensial inside the mass distri- bution: 2π φ = Gρ(r2 3R2) (9.7) 3 −

127 128 Chapter 9. Schwarzschild’s Interior Solution

The star is in hydrostatic equilibrium, that is, the pressure forces are in equi- librium with the gravitational forces.

dm=4 πρr 2dr

4π r 3ρ 3

Figure 9.1: The shell with thickness dr, is affected by both gravitational and pressure forces.

Consider figure 9.1. The pressure forces on the shell is 4πr2dp. Gravitational forces on the shell:

mass inside shell mass of the shell G · r2 (9.8) 4π ρr3 4πρr2dr =G 3 · r2

Equilibrium:

4π 4πr2dp = G ρr3 4πρdr − 3 · 4π dp = Gρ2rdr − 3 ⇓ 2πG (9.9) p = K ρ2r2 3 − 3 2πG P (R) = 0 gives : K = ρ2R2 3 3 2πG p(r) = ρ2(R2 r2) 3 −

No matter how massive the star is, it is possible for the pressure forces to keep the equilibrium with gravity. In Newtonian theory, gravitational collapse is not a necessity. 9.2 The pressure contribution to the gravitational mass of a static, spherical symmetric system 129

9.2 The pressure contribution to the gravitational mass of a static, spherical symmetric system

We now give a new definition of the gravitational acceleration (not equivalent to (7.23)) a g = + , a = a aµ (9.10) ut µ p We have the line element: ds2 = e2α(r)dt2 + e2β(r)dr2 + r2dΩ2 − (9.11) g = e2α , g = e2β tt − rr gives (because of the gravitational acceleration)

α β g = +e − α0 (9.12)

E E E E From the expressions for tˆtˆ, rˆrˆ, θˆθˆ, φˆφˆ follow (see Section 7.1)

tˆ rˆ θˆ φˆ 2β 2α0 2 E E E E = 2e ( + α00 + α0 α0β0) . (9.13) tˆ − rˆ − θˆ − φˆ − r − We also have

2 α β 2 α β 2α0 2 (r e − α0)0 = r e − ( + α00 + α0 α0β0) , (9.14) r − which gives

1 ˆ ˆ ˆ g = + (Et Erˆ Eθ Eφ )r2eα+βdr . (9.15) 2r2 tˆ − rˆ − θˆ − φˆ Z By applying Einstein’s field equations

µˆ µˆ E νˆ = 8πGT νˆ (9.16) we get

4πG ˆ ˆ ˆ g = + (T t T rˆ T θ T φ)r2eα+βdr . (9.17) r2 tˆ − rˆ − θˆ − φˆ Z This is the Tolman-Whittaker expression for gravitational acceleration. The corresponding Newtonian expression is : 4πG g = ρr2dr (9.18) N − r2 Z The relativistic gravitational mass density is therefore defined as

ˆ ˆ ˆ ρ = T t + T rˆ + T θ + T φ (9.19) G − tˆ rˆ θˆ φˆ 130 Chapter 9. Schwarzschild’s Interior Solution

For an isotropic fluid with

ˆ ˆ ˆ T t = ρ , T rˆ = T θ = T φ = p (9.20) tˆ − rˆ θˆ φˆ

we get ρG = ρ + 3p (with c = 1), which becomes

3p ρ = ρ + (9.21) G c2

It follows that in relativity, pressure has a gravitational effect. Greater pressure gives increasing gravitational attraction. Strain (p < 0) decreases the gravita- tional attraction. In the Newtonian limit, c , pressure has no gravitational effect. → ∞

9.3 The Tolman-Oppenheimer-Volkov equation

With spherical symmetry the spacetime line-element may be written

ds2 = e2α(r)dt2 + e2β(r)dr2 + r2dΩ2 − (9.22) E = 8πGT , T µˆ = diag( ρ, p, p, p) tˆtˆ tˆtˆ νˆ − E From tˆtˆ we get

1 d 2β [r(1 e− )] = 8πGρ r2 dr − r (9.23) 2β 2 r(1 e− ) = 2G 4πρr dr , − Z0 r 2 where m(r) = 0 4πρr dr giving R 2β 2Gm(r) 1 e− = 1 = (9.24) − r g rr

From E rˆrˆ we have

E rˆrˆ = 8πGT rˆrˆ 2 dα 2β 1 2β (9.25) e− (1 e− ) = 8πGp r dr − r2 − We get 2 dα 2Gm(r) 2Gm(r) (1 ) = 8πGp r dr − r − r3

dα m(r) + 4πr3p(r) = G (9.26) dr r(r 2Gm(r)) − 9.3 The Tolman-Oppenheimer-Volkov equation 131

rˆνˆ The relativistic generalized equation for hydrostatic equilibrium is T ;νˆ = 0, giving

rˆνˆ νˆ rˆαˆ rˆ αˆνˆ T ,νˆ + Γ αˆνˆT + Γ αˆνˆT = 0

rˆνˆ rˆrˆ 1 ∂p T ,νˆ = T ,rˆ = p ,rˆ = √g rr ∂r rˆνˆ β dp (9.27) T = e− ,νˆ dr νˆ rˆαˆ νˆ tˆ αˆ Γ αˆνˆT = Γ rˆνˆp = Γ rˆtˆp + Γ rˆαˆp rˆ αˆνˆ rˆ νˆνˆ rˆ rˆ Γ αˆνˆT = Γ νˆνˆT = Γ tˆtˆρ + Γ αˆαˆp In orthonormal basis we have

Ω νˆµˆ = Ω µˆνˆ Γ µˆνˆαˆ = Γ νˆµˆαˆ − ⇒ − (9.28) Γαˆ = Γ = Γ = Γrˆ rˆαˆ αˆrˆαˆ − rˆαˆαˆ − αˆαˆ rˆνˆ T ;νˆ = 0 now takes the form:

β dp tˆ rˆ e− + Γ p + Γ ρ (9.29) dr rˆtˆ tˆtˆ We have

ˆ Γt = Γ = Γ = Γrˆ (9.30) rˆtˆ − tˆrˆtˆ rˆtˆtˆ tˆtˆ Γrˆ = e β dα and we also have tˆtˆ − dr , giving: dp dα + (p + ρ) = 0 (9.31) dr dr Inserting Equation 9.26 into Equation 9.31 gives

dp m(r) + 4πr3p(r) = G(ρ + p) (9.32) dr − r(r 2Gm(r)) −

This is the Tolman-Oppenheimer-Volkov (TOV) equation. The component gtt = e2α(r) may now be calculated as follows − dp = dα , ρ = constant ρ + p − ln(ρ + p) = K α (9.33) − α α ρ + p = K e− , p = K e− ρ 1 1 − Hence

α α(R) p 1 e = e (1 + )− (9.34) ρ where R is the radius of the mass distribution. 132 Chapter 9. Schwarzschild’s Interior Solution

9.4 An exact solution for incompressible stars - Schwarzschild’s interior solution

The mass inside a radius r for an incompressiable star is 4 m(r) = πρr3 (9.35) 3

2 2β 2Gm(r) r e− = 1 1 (9.36) − r ≡ − a2 where 3 r3 r2 a2 = , m(r) = , r = 2Gm = r (9.37) 8πGρ 2Ga2 s a2 TOV equation: dp 4 πρr3 + 4πr3p(r) = G 3 (ρ + p(r)) dr − r(r 2G 4 πρr3) − 3 4 ρ + 3p(r) = G π r(ρ + p(r)) − 3 1 G 8 πρr2 − 3 1 ρ + 3p(r) = 2 r(ρ + p(r)) (9.38) −2a2ρ 1 r − a2 p dp 1 r r = dr 2 r2 ⇒ 0 (ρ + 3p)(ρ + p −2a ρ R 1 Z Z − a2 p + ρ a2 R2 = − 3p + ρ a2 r2 r − So the relativistic pressure distribution is √a2 r2 √a2 R2 p(r) = − − − ρ, r R (9.39) 3√a2 R2 √a2 r2 ∀ ≤ − − − also 2 2 3 a r a = , 2 = > 1 a > r (9.40) 8πGρ r rs ⇒ To satisfy the condition for hydrostatic equilibrium we must have p > 0 or p(0) > 0 which gives a √a2 R2 p(0) pc = − − > 0 (9.41) ≡ 3√a2 R2 a − − in which the numerator is positive so that 3 a2 R2 > a − 9pa2 9R2 > a2 − 8 (9.42) R < a r9 8 8 3 1 R2 < a = = 9 9 8πGρ 3πGρ 9.4 An exact solution for incompressible stars - Schwarzschild’s interior solution133

Stellar mass: 4 4 1 4R M = πρR3 < πρR = 3 3 3πGρ 9G (9.43) 4 1 M < 9G √3πGρ

For a neutron star we can use ρ 1017 g /cm3. An upper limit on the mass is ≈ then M < 2.5 M Substitution for p in the expression for eα gives

α 3 Rs 1 Rs 2 e = 1 1 3 r (9.44) 2r − R − 2r − R The line element for the interior Schwarzschild solution is 2 3 R 1 R dr2 ds2 = 1 s 1 s r2 dt2 + + r2dΩ, r R 3 Rs 2 − 2 − R − 2 − R ! 1 3 r ≤ r r − R (9.45) Chapter 10

Cosmology

10.1 Comoving coordinate system

We will consider expanding homogenous and isotropic models of the universe. We introduce an expanding frame of reference with the galactic clusters as reference particles. Then we introduce a ’comoving coordinate system’ in this frame of reference with spatial coordinates χ, θ, φ. We use time measured on standard clocks carried by the galactic clusters as coordinate time (cosmic time). The line element can then be written in the form:

ds2 = dt2 + a(t)2[dχ2 + r(χ)2dΩ2] (10.1) − (For standard clocks at rest in the expanding system, dχ = dΩ = 0 and ds2 = dτ 2 = dt2). The function a(t) is called the expansion factor, and t is called − − cosmic time. The physical distance to a galaxy with coordinate distance dχ from an ob- server at the origin, is:

dlx = √gχχdχ = a(t)dχ (10.2)

Even if the galactic clusters have no coordinate velocity, they do have a radial velocity expressed by the expansion factor. The value χ determines which cluster we are observing and a(t) how it is moving. 4-velocity of a reference particle (galactic cluster):

dxµ dxµ uµ = = = (1, 0, 0, 0) (10.3) dτ dt

duµ duµ This applies at an abritrary time, that is dt = 0. Geodesic equation: dt + µ α β µ Γ αβu u = 0 which is reduces to: Γ tt = 0

0 0 0 1 Γµ = gµν (g + g + g ) = 0 (10.4) tt 2 νt,t tν,t tt,ν z}|{ z}|{ z}|{ We have that g = 1. This shows that the reference particles are freely falling. tt −

134 10.2 Curvature isotropy - the Robertson-Walker metric 135

10.2 Curvature isotropy - the Robertson-Walker met- ric

Introduce orthonormal form-basis:

ˆ ωtˆ = dt ωχˆ = a(t)dχ ωθ = a(t)r(χ)dθ ˆ (10.5) ωφ = a(t)r(χ) sin θdφ

Using Cartans 1st equation:

dωµˆ = Ωµˆ ωνˆ (10.6) − νˆ ∧ to find the connection forms. Then using Cartans 2nd structure equation to calculate the curvature forms:

ˆ Rµˆ = dΩµˆ + Ωµˆ Ωλ (10.7) νˆ νˆ λˆ ∧ νˆ d d Calculations give: (notation: · = dt , 0 = dχ )

a¨ ˆ ˆ Rtˆ = ωtˆ ωi, ωi = ωχˆ, ωθ, ωφ ˆi a ∧ 2 a˙ r00 ˆ ˆ ˆ ˆ Rχˆ = ωχˆ ωj, ωj = ωθ, ωφ (10.8) ˆj a2 − ra2 ∧ 2 2 ˆ a˙
1 r ˆ ˆ Rθ = + 0 ωθ ωφ φˆ a2 r2a2 − r2a2 ∧
The curvature of 3-space (dt = 0) can be found by putting a = 1. That is:

r ˆ Rχˆ = 00 ωχˆ ωj 3 ˆj − r ∧ 2 (10.9) ˆ 1 r ˆ ˆ Rθ = 0 ωθ ωφ 3 φˆ r2 − r2 ∧
The 3-space is assumed to be isotropic and homogenous. This demands

2 r00 1 r0 = − = k , (10.10) − r r2 where k represents the constant curvature of the 3-space.

2 r00 + kr = 0 and r0 = 1 kr (10.11) ∴ − p Solutions with r(0) = 0, r0(0) = 1 :

√ kr = sinh(√ kχ) (k < 0) − − r = χ (k = 0) (10.12) √kr = sin(√kχ) (k > 0) 136 Chapter 10. Cosmology

The solutions can be characterized by the following 3 cases:

r = sinh χ, dr = 1 + r2dχ, (k = 1) − r = χ, dr = dχ, p (k = 0) (10.13) r = sin χ, dr = 1 r2dχ, (k = 1) − p In all three cases one may write dr = √1 kr2dχ, which is just the last equation − above. 2 dr2 We now set dχ = 1 kr2 into the line-element : − ds2 = dt2 + a2(t) dχ2 + r2(χ)dΩ2 − dr2 (10.14) = dt2 + a2(t) + r2dΩ
2 − 1 kr2  −  The first expression is known as the standard form of the line-element, the second is called the Robertson-Walker line-element. The 3-space has constant curvature. 3-space is spherical for k = 1, Euclidean for k = 0 and hyperbolic for k = 1. − Universe models with k = 1 are known as ’closed’ and models with k = 1 − are known as ’open’. Models with k = 0 are called ’flat’ even though these models also have curved space-time.

10.3 Cosmic dynamics

10.3.1 Hubbles law

The observer is placed in origo of the coordinate-system; χ0 = 0. The proper distance to a galaxy with radial coordinate χe is D = a(t)χe. The galaxy has a radial velocity:

dD a˙ a˙ v = = aχ˙ = D = HD whereH = (10.15) dt e a a

The expansion velocity v is proportional to the distance D. This is Hubbles law.

10.3.2 Cosmological redshift of light

∆te : the time interval in transmitter-position at transmission-time ∆t0 : the time interval in receiver-position at receiving-time

Light follows curves with ds2 = 0, with dθ = dφ = 0 we have :

dt = a(t)dχ (10.16) − 10.3 Cosmic dynamics 137

t 6

t0 + ∆t0

t0

te + ∆te

te - χ

χ0 = 0 χe Figure 10.1: Schematic representation of cosmological redshift

Integration from transmitter-event to receiver-event :

t0 dt χ0 = dχ = χ a(t) − e Zte Zχe t0+∆t0 dt χ0 = dχ = χ , a(t) − e Zte+∆te Zχe which gives t0+∆t0 dt t0 dt = 0 (10.17) a − a Zte+∆te Zte or

t0+∆t0 dt te+∆te dt = 0 (10.18) a − a Zt0 Zte

Under the integration from te to te + ∆te the expansion factor a(t) can be considered a constant with value a(te) and under the integration from t0 to t0 + ∆t0 with value a(t0), giving:

∆te ∆t0 = (10.19) a(te) a(t0) 138 Chapter 10. Cosmology

∆t0 and ∆te are intervals of the light at the receiving and transmitting time. Since the wavelength of the light is λ = c∆t we have:

λ0 λe = (10.20) a(t0) a(te)

This can be interpreted as a “stretching” of the electromagnetic waves due to the expansion of space. The cosmological redshift is denoted by z and is given by: λ0 λe a(t0) z = − = 1 (10.21) λe a(te) − Using a a(t ) we can write this as: 0 ≡ 0 a0 1 + z(t) = (10.22) a

10.3.3 Cosmic fluids The energy-momentum tensor for a perfect fluid (no viscosity and no thermal conductivity) is Tµν = (ρ + p)uµuν + pgµν (10.23) In an orthonormal basis

Tµˆνˆ = (ρ + p)uµˆuνˆ + pηµˆνˆ (10.24)

where ηµˆνˆ is the Minkowski metric. We consider 3 types of cosmic fluid:

1. dust: p = 0, T µˆνˆ = ρuµˆuνˆ (10.25)

1 2. radiation: p = 3 ρ, 4 T = ρu u + pη µˆνˆ 3 µˆ νˆ µˆνˆ ρ (10.26) = (4u u + η ) 3 µˆ νˆ µˆνˆ The trace ρ T = T µˆ = (4uµˆu + δµ ) = 0 (10.27) µˆ 3 µˆ µ 3. vacuum: p = ρ, − T = ρη (10.28) µˆνˆ − µˆνˆ If vacuum can be described as a perfect fluid we have p = ρ , where v − v ρ is the energy density. It can be related to Einstein’s cosmological constant, Λ = 8πGρv. One has also introduced a more general type of vacuum energy given by the equation of state pφ = wρφ, where φ denotes that the vacuum energy is 10.3 Cosmic dynamics 139 connected to a scalar field φ. In a homogeneous universe the pressure and the density are given by 1 1 p = φ˙2 V (φ), ρ = φ˙2 + V (φ) (10.29) φ 2 − φ 2 where V (φ) is the potential for the scalar field. Then we have

1 φ˙2 V (φ) w = 2 − (10.30) 1 ˙2 2 φ + V (φ)

The special case φ˙ = 0 gives the Lorentz invariant vacuum with w = 1. The − more general vacuum is called “quintessence”.

10.3.4 Isotropic and homogeneous universe models We will discuss isotropic and homogenous universe models with perfect fluid and a non-vanishing cosmological constant Λ. Calculating the components of the Einstein tensor from the line-ement (10.14) we find in an orthonormal basis

3a˙ 2 3k E = + (10.31) tˆtˆ a2 a2 2a¨ a˙ 2 k E = . (10.32) mˆ mˆ − a − a2 − a2 The components of the energy-momentum tensor of a perfect fluid in a comoving orthonormal basis are Ttˆtˆ = ρ, Tmˆ mˆ = p. (10.33) Hence the tˆtˆ component of Einstein’s field equations is

a˙ 2 + k 3 = 8πGρ + Λ (10.34) a2 mˆ mˆ components: a¨ a˙ 2 k 2 = 8πGp Λ (10.35) − a − a2 − a2 − where ρ is the energy density and p is the pressure. The equations with vanishing cosmological constant are called the Friedmann equations. Inserting eq. (10.34) into eq. (10.35) gives: 4πG a¨ = a(ρ + 3p) (10.36) − 3 If we interpret ρ as the mass density and use the speed of light c, we get 4πG a¨ = a(ρ + 3p/c2) (10.37) − 3

Inserting the gravitational mass density ρG from eq.(9.21) this equation takes the form 4πG a¨ = aρ (10.38) − 3 G 140 Chapter 10. Cosmology

Inserting p = wρc2 into (9.21) gives

ρG = (1 + 3w)ρ (10.39)

which is negative for w < 1/3, i.e. for φ˙2 < V (φ). Special cases: − dust: w = 0, ρ = ρ • G radiation: w = 1 , ρ = 2ρ • 3 G Lorentz-invariant vacuum: w = 1, ρ = 2ρ • − G − In a universe dominated by a Lorentz-invariant vacuum the acceleration of the cosmic expansion is 8πG a¨ = aρ > 0, (10.40) v 3 v i.e. accelerated expansion. This means that vacuum acts upon itself with repul- sive gravitation. The field equations can be combined into

a˙ 2 8πG Λ k H2 = ρ + (10.41) ≡ a 3 m 3 − a2  

where ρm is the density of matter, Λ = 8πGρΛ where ρΛ is the vacuum energy with constant density. ρ = ρm + ρΛ is the total mass density. Then we may write 8πG k H2 = ρ (10.42) 3 − a2 The critical density ρcr is the density in a universe with euclidean spacelike geometry, k = 0, which gives 3H2 ρ = (10.43) cr 8πG We introduce the relative densities

ρm ρΛ Ωm = , ΩΛ = (10.44) ρcr ρcr Furthermore we introduce a dimensionless parameter that describes the curva- ture of 3-space k Ω = (10.45) k −a2H2 Eq. (10.42) can now be written

Ωm + ΩΛ + Ωk = 1 (10.46)

From the Bianchi identity and Einstein’s field equations follow that the energy- momentum density tensor is covariant divergence free. The time-component expresses the equation of continiuty and takes the form

tˆ νˆ tˆνˆ [(ρ + p)u u ];νˆ + (pη );νˆ = 0 (10.47) 10.3 Cosmic dynamics 141

Since utˆ = 1, umˆ = 0 and ηtˆtˆ = 1, ηtˆmˆ = 0, we get − (ρ + p). + (ρ + p)uνˆ p˙ = 0 (10.48) ;νˆ − or νˆ νˆ ρ˙ + (ρ + p)(u ,νˆ + Γ tˆνˆ) = 0 (10.49) ˆ Here uν = 0 and Γtˆ = 0. Calculating Γmˆ for dωµˆ = Γµˆ ωαˆ ωβ we get ,ν tˆtˆ tˆmˆ αˆβˆ ∧

ˆ ˆ a˙ Γmˆ = Γrˆ + Γθ + Γφ = 3 (10.50) tˆmˆ tˆrˆ tˆθˆ tˆφˆ a

Hence a˙ ρ˙ + 3(ρ + p) = 0 (10.51) a which may be written (ρa3). + p(a3). = 0 (10.52)

Let V = a3 be a comoving volume in the universe and U = ρV be the energy in the comoving volume. Then we may write

dU + pdV = 0 (10.53)

This is the first law of thermodynamics for an adiabatic expansion. It follows that the universe expands adiabatically. The adiabatic equation can be written

ρ˙ a˙ = 3 (10.54) ρ + p − a

Assuming p = wρ we get

dρ da = 3(1 + w) ρ − a 3(1+w) ρ a − ln = ln ρ a 0  0  It follows that 3(1+w) a − ρ = ρ (10.55) 0 a  0  This equation tells how the density of different types of matter depends on the expansion factor ρa3(1+w) = constant (10.56)

Special cases:

dust: w = 0 gives ρ a3 = constant • d Thus, the mass in a comoving volume is constant. 142 Chapter 10. Cosmology

radiation: w = 1 gives ρ a4 = constant • 3 r Thus, the radiation energy density decreases faster than the case with dust when the universe is expanding. The energy in a comoving volume is decreasing because of the thermo- dynamic work on the surface. In a remote past, the density of radiation must have exceeded the density of dust:

3 3 ρd0a0 =ρda 4 4 ρr0a0 =ρra 4 4 ρra ρr0a0 3 = 3 ρda ρd0a0

The expansion factor when ρr = ρd:

ρr0 a(t1) = a0 ρd0 Lorentz-invariant vacuum: w = 1 gives ρ = constant. • − Λ The vacuum energy in a comoving volume is increasing a3. ∝

10.4 Some cosmological models

10.4.1 Radiation dominated model The energy-momentum tensor for radiation is trace free. According to the Ein- stein field equations the Einstein tensor must then be trace free:

aa¨ + a˙ 2 + k = 0 (10.57) (aa˙ + kt)· = 0

Integration gives aa˙ + kt = B (10.58) Another integration gives 1 1 a2 + kt2 = Bt + C (10.59) 2 2

The initial condition a(0) = 0 gives C = 0. Hence

a = 2Bt kt2 (10.60) − p For k = 0 we have

B a = √2Bt , a˙ = (10.61) r2t

The expansion velocity reaches infinity at t = 0, (limt 0 a˙ = ) → ∞ 10.4 Some cosmological models 143

Figure 10.2: In a radiation dominated universe the expansion velocity reaches infinity at t = 0.

4 ρRa = K , a = √2Bt 2 2 (10.62) 4ρRB t = K According to the Stefan-Boltzmann law we then have ρ = σT 4 4B2σT 4t2 = K R → ⇒ K1 K1 (10.63) t = 2 T = T ⇔ r t where T is the temperature of the background radiation.

10.4.2 Dust dominated model From the first of the Friedmann equations we have 8πG a˙ 2 + k = ρa2 (10.64) 3 We now introduce a time parameter η given by dt d 1 d = a(η) = dη ⇒ dt a dη (10.65) da 1 da So: a˙ = = dt a dη We also introduce A 8πG ρ a 3. The first Friedmann equation then gives ≡ 3 0 0 8πG 8πG aa˙ 2 + ka = ρa3 = ρ a 3 = A (10.66) 3 3 0 0 144 Chapter 10. Cosmology

Using η we get 1 da ( )2 = A ka a dη − 1 da A ( )2 = k a2 dη a − (10.67) 1 da A A a = k = 1 k a dη a − a − A r r r where we chose the positive root. We now introduce u, given by a = Au2, u = a A . We then get p da du = 2Au (10.68) dη dη

which together with the equation above give 1 du 1 2Au = 1 ku2 Au2 dη u − p (10.69) ⇓ du 1 = dη √1 ku2 2 − This equation will first be integrated for k < 0. Then k = k , so that −| | du η = + K (10.70) 1 + k u2 2 Z | | η p or arcsinh(√ ku) = + K. The condition u(0) = 0 gives K = 0. Hence − 2 k η 1 a = sinh2 = (cosh η 1) (10.71) −A 2 2 − or A a = (cosh η 1) (10.72) −2k − From eqs. (10.43), (10.44) and (10.66) we have

8πG 2 ρm0 2 A = ρm0 = H0 = H0 Ωm0 (10.73) 3 ρcr0

From egs. (10.45) and (10.46) we get

k = H2(Ω 1) (10.74) 0 m0 − Hence, the scale factor of the negatively curved, dust dominated universe model is 1 Ωm0 a(η) = (cosh η 1) (10.75) 2 1 Ω − − m0 10.4 Some cosmological models 145

Inserting this into eq. (10.65) and integrating with t(0) = η(0) leads to

Ωm0 t(η) = (sinh η η) (10.76) 2H (1 Ω )3/2 − 0 − m0 Integrating eg. (10.69) for k = 0 leads to an Einstein-deSitter universe

t 2 a(t) = ( ) 3 (10.77) t0

Finally integrating eg. (10.69) for k > 0 gives, in a similar way as for k < 0

1 Ωm0 a(η) = (1 cos η) (10.78) 2 1 Ω − − m0 Ωm0 t(η) = (η sin η) (10.79) 2H (Ω 1)3/2 − 0 m0 − We see that this is a parametric representation of a cycloid.

In the Einstein-deSitter model the Hubble factor is

a˙ 2 1 2 1 2 H = = , t = = t (10.80) a 3 t 3 H 3 H

The critical density in the Einstein-deSitter model is given by the first Fried- mann equation: 8πG H2 = ρ , k = 0 3 cr ⇓ (10.81) 3H2 ρ ρcr = , Ω = 8πG ρcr

Example 10.4.1 (Age-redshift relation for dust dominated universe with k = 0)

a a 1 + z = 0 a = 0 a ⇒ 1 + z a a (10.82) da = 0 dz = dz −(1 + z)2 −1 + z

Eq. (10.34) gives

3 a˙ 2 8πG 8πG ρ0a0 = ρ = 3 a 3 3 a (10.83)
8πG = ρ (1 + z)3 3 0 146 Chapter 10. Cosmology

Figure 10.3: For k = 1 the density is larger than the critical density, and the universe is closed. For k = 0 we have ρ = ρcr and the expansion velocity of the universe will approach zero as t . For k = 1 we have ρ < ρ . The → ∞ − cr universe is then open, and will continue expanding forever.

3 2 8πG a˙ 2 da Using H0 = 3 ρ0 gives a = H0(1 + z) . From a˙ = dt we get:

da da dz dt = = a˙ = 5 (10.84) a˙ a a −H0(1 + z) 2

Integration gives the age of the universe:

0 1 dz 2 1 1 0 t = = (10.85) 0 −H 5 3 H 3 0 Z (1 + z) 2 0 (1 + z) 2 ∞ ∞  

2 1 t0 = tH where the Hubble-time tH is the age of the universe, if the expansion 3 ≡ H0 rate had been constant. ’Look-back-time’ to a source with redshift z is:

z dz 2 1 ∆t = t = t 1 (10.86) H 5 3 H − 3 Z0 (1 + z) 2 (1 + z) 2   10.4 Some cosmological models 147

exp.factor a

tangent

today

t 0 time t t H

Figure 10.4: tH is the age of the universe if the expansion had been constant, BUT:The exp.rate was faster closer to the Big Bang, so the age is lower.

1 ∆t = t0[1 ] (10.87) − (1 + z)3/2

1 z = 1 (10.88) (1 ∆t )2/3 − − t0

∆t = 0, 99 z = 20, 5 (10.89) t0 ⇒

10.4.3 Friedmann-Lemaître model The dynamics of galaxies and clusters of galaxies has made it clear that far stronger gravitational fields are needed to explain the observed motions than those produced by visible matter (McGaugh 2001). At the same time it has become clear that the density of this dark matter is only about 30% of the critical 148 Chapter 10. Cosmology

density, although it is a prediction by the usual versions of the inflationary universe models that the density ought to be equal to the critical density (Linde 2001). Also the recent observations of the temperature fluctuations of the cosmic microwave radiation have shown that space is either flat or very close to flat (Bernadis et.al 2001, Stompor et al. 2001, Pryke et al. 2001). The energy that fills up to the critical density must be evenly distributed in order not to affect the dynamics of the galaxies and the clusters. Furthermore, about two years ago observations of supernovae of type Ia with high cosmic red shifts indicated that the expansion of the universe is accelerating (Riess et al. 1998, Perlmutter et al. 1999). This was explained as a result of repulsive gravitation due to some sort of vacuum energy. Thereby the missing energy needed to make space flat, was identified as vacuum energy. Hence, it seems that we live in a flat universe with vacuum energy having a density around 70% of the critical density and with matter having a density around 30% of the critical density. Until the discovery of the accelerated expansion of the universe the standard model of the universe was assumed to be the Einstein-DeSitter model, which is a flat universe model dominated by cold matter. This universe model is thor- oughly presented in nearly every text book on general relativity and cosmology. Now it seems that we must replace this model with a new "standard model" containing both dark matter and vacuum energy. Recently several types of vacuum energy or so called quintessence energy have been discussed (Zlatev, Wang and Steinhardt 1999, Carroll 1998). How- ever, the most simple type of vacuum energy is the Lorentz invariant vacuum energy (LIVE), which has constant energy density during the expansion of the universe (Zeldovich 1968, Grøn 1986). This type of energy can be mathemati- cally represented by including a cosmological constant in Einstein’s gravitational field equations. The flat universe model with cold dark matter and this type of vacuum energy is the Friedmann-Lemaître model. The field equations for the flat Friedmann-Lemaître is found by putting k = p = 0 in equation (10.35). This gives

a¨ a˙ 2 2 + = Λ (10.90) a a2 Integration leads to Λ aa˙ 2 = a3 + K (10.91) 3 where K is a constant of integration. Since the amount of matter in a volume 3 3 comoving with the cosmic expansion is constant, ρM a = ρM0a0, where the index 0 refers to measured values at the present time. Normalizing the expansion factor so that a0 = 1 and comparing eqs.(10.42) and(10.91) then gives K = 3 2 (8πG/3)ρM0. Introducing a new variable x by a = x and integrating once more with the initial condition a(0) = 0 we obtain

3K t 2 a3 = sinh2 , t = (10.92) Λ t Λ √  Λ  3Λ 10.4 Some cosmological models 149

The vacuum energy has a constant density ρΛ given by

Λ = 8πGρΛ (10.93)

The critical density, which is the density making the 3-space of the universe flat, is 3H2 ρ = (10.94) cr 8πG The relative density, i.e. the density measured in units of the critical density, of the matter and the vacuum energy, are respectively

ρ 8πGρM ΩM = = 2 (10.95) ρcr 3H ρΛ Λ ΩΛ = = 2 (10.96) ρcr 3H Since the present universe model has flat space, the total density is equal to the critical density, i.e. ΩM + ΩΛ = 1. Eq. (10.91) with the normalization 2 a(t0) = 1, where t0 is the present age of the universe, gives 3H0 = 3K + Λ. Eq. (10.34) with k = 0 gives 8πGρ = 3H2 Λ. Hence K = 8πGρ /3 and 0 0 − 0 3K = 8πGρ0 = ρ0 = ΩM0 . In terms of the values of the relative densities at the Λ Λ ρΛ ΩΛ0 present time the expression for the expansion factor then takes the form

t ΩM0 1 ΩΛ0 a = A1/3 sinh2/3 , A = = − (10.97) t Ω Ω  Λ  Λ0 Λ0 Using the identity sinh(x/2) = (cosh x 1)/2 this expression may be written − pA 2t a3 = cosh 1 (10.98) 2 t −   Λ   The age t0 of the universe is found from a(t0) = 1, which by use of the formula arc tanh x = arc sinh(x/√1 x2), leads to the expression −

t0 = tΛarc tanh ΩΛ0 (10.99)

Inserting typical values t = 15 109years, Ωp = 0.7 we get A = 0.43, t = 0 · Λ0 Λ 12 109years. With these values the expansion factor is a = 0.75 sinh2/3(1.2t/t ). · 0 This function is plotted in fig. 10.5. The Hubble parameter as a function of time is H = (2/3tΛ) coth(t/tΛ) (10.100)

Inserting t0 = 1.2tΛ we get Ht0 = 0.8 coth(1.2t/t0), which is plotted in fig. 10.6 The Hubble parameter decreases all the time and approaches a constant value H = 2/3tΛ in the infinite future. The present value of the Hubble parameter ∞ is 2 H0 = (10.101) 3tΛ√ΩΛ0

The corresponding Hubble age is tH0 = (3/2)tΛ√ΩΛ0. Inserting our numerical 1 9 values gives H = 64km/secMpc− and t = 15.7 10 years. In this universe 0 H0 · 150 Chapter 10. Cosmology

a 1.5

1.25

1

0.75

0.5

0.25

0.2 0.4 0.6 0.8 1 1.2 1.4 t/t0

Figure 10.5: The expansion factor as function of cosmic time in units of the age of the universe.

Figure 10.6: The Hubble parameter as function of cosmic time. 10.4 Some cosmological models 151 model the age of the universe is nearly as large as the Hubble age, while in the Einstein-DeSitter model the corresponding age is t = (2/3)t = 10.5 0ED H0 · 109years. The reason for this difference is that in the Einstein-DeSitter model the expansion is decelerated all the time, while in the Friedmann-Lemaître model the repulsive gravitation due to the vacuum energy have made the expansion accelerate lately (see below). Hence, for a given value of the Hubble parameter the previous velocity was larger in the Einstein-DeSitter model than in the Friedmann-Lemaître model. The ratio of the age of the universe and its Hubble age depends upon the present relative density of the vacuum energy as follows,

t0 2 arc tanh √ΩΛ0 = H0t0 = (10.102) tH0 3 √ΩΛ0 This function is depicted graphically in fig. 10.7 The age of the universe increases

Figure 10.7: The ratio of the age of the universe and the Hubble age as function of the present relative density of the vacuum energy. with increasing density of vacuum energy. In the limit that the density of the vacuum approaches the critical density, thereis no dark matter, and the universe model approaches the DeSitter model with exponential expansion and no Big Bang. This model behaves in the same way as the Steady State cosmological model and is infinitely old. A dimensioness quantity representing the rate of change of the cosmic ex- pansion velocity is the deceleration parameter, which is defined as q = a/aH¨ 2. − For the present universe model the deceleration parameter as a function of time is 1 q = [1 3 tanh2(t/t )] (10.103) 2 − Λ 152 Chapter 10. Cosmology

which is shown graphically in fig. 10.8 The inflection point of time t1 when

Figure 10.8: The deceleration parameter as function of cosmic time.

deceleration turned into acceleration is given by q = 0. This leads to

t1 = tΛarc tanh(1/√3) (10.104)

or expressed in terms of the age of the universe

arc tanh(1/√3) t1 = t0 (10.105) arc tanh √ΩΛ0 The corresponding cosmic red shift is

1/3 a0 2ΩΛ0 z(t ) = 1 = 1 (10.106) 1 a(t ) − 1 Ω − 1  − Λ0 

Inserting ΩΛ0 = 0.7 gives t1 = 0.54t0 and z(t1 = 0.67. The results of analysing the observations of supernova SN 1997 at z = 1.7, corresponding to an emission time t = 0.30t = 4.5 109years, have provided e 0 · evidence that the universe was decelerated at that time (Riess n.d.). M.Turner and A.G.Riess (Turner and Riess 2001) have recently argued that the other supernova data favour a transition from deceleration to acceleration for a red shift around z = 0.5. Note that the expansion velocity given by Hubble’s law, v = Hd, always decreases as seen from fig. 10.6. This is the velocity away from the Earth of the cosmic fluid at a fixed physical distance d from the Earth. The quantity a˙ on the other hand, is the velocity of a fixed fluid particle comoving with the expansion 10.4 Some cosmological models 153 of the universe. If such a particle accelerates, the expansion of the universe is said to accelerate. While H˙ tells how fast the expansion velocity changes at a fixed distance from the Earth, the quantity a¨ represents the acceleration of a free particle comoving with the expanding universe. The connection between these two quantities are a¨ = a(H˙ + H2). The ratio of the inflection point of time and the age of the universe, as given in eq.(10.105), is depicted graphically as function of the present relative density of vacuum energy in fig. 10.9 The turnover point of time happens earlier the

Figure 10.9: The ratio of the point of time when cosmic decelerations turn over to acceleration to the age of the universe. greater the vacuum density is. The change from deceleration to acceleration would happen at the present time if ΩΛ0 = 1/3. The red shift of the inflection point given in eq.(10.106) as a function of vacuum energy density, is plotted in fig. 10.10 Note that the red shift of future points of time is negative, since then a > a0. If ΩΛ0 < 1/3 the transition to acceleration will happen in the future. The critical density is

2 ρcr = ρΛ tanh− (t/tΛ) (10.107)

This is plotted in fig. 10.11. The critical density decreases with time. Eq. (10.106) shows that the relative density of the vacuum energy is

2 ΩΛ = tanh (t/tΛ) (10.108) which is plotted in fig. 10.12. The density of the vacuum energy approaches the critical density. Since the density of the vacuum energy is constant, this is 154 Chapter 10. Cosmology

Figure 10.10: The cosmic red shift of light emitted at the turnover time from deceleration to acceleration as function of the present relative density of vacuum energy.

Figure 10.11: The critical density in units of the constant density of the vacuum energy as function of time. 10.4 Some cosmological models 155

Figure 10.12: The relative density of the vacuum energy density as function of time. better expressed by saying that the critical density approaches the density of the vacuum energy. Furthermore, since the total energy density is equal to the critical density all the time, this also means that the density of matter decreases faster than the critical density. The density of matter as function of time is

2 ρM = ρΛ sinh− (t/tΛ) (10.109) which is shown graphically in fig. 10.13 The relative density of matter as func- tion of time is 2 ΩM = cosh− (t/tΛ) (10.110) which is depicted in fig. 10.14 Adding the relative densities of fig. 10.13 and fig. 10.14 or the expressions (10.107) and (10.109) we get the total relative density ΩT OT = ΩM + ΩΛ = 1. The universe became vacuum dominated at a point of time t2 when ρΛ(t2) = ρM (t2). From eq.(10.109) follows that this point of time is given by sinh(t2/tΛ) = 1. According to eq.(10.99) we get arc sinh(1) t2 = t0 (10.111) arc tanh(√ΩΛ0) From eq.(10.97) follows that the corresponding red shift is

1/3 z(t ) = A− 1 (10.112) 2 −

Inserting ΩΛ0 = 0.7 gives t2 = 0.73t0 and z(t2) = 0.32. The transition to accelerated expansion happens before the universe becomes vacuum dominated. 156 Chapter 10. Cosmology

Figure 10.13: The density of matter in units of the density of vacuum energy as function of time.

Figure 10.14: The relative density of matter as function of time. 10.5 Inflationary Cosmology 157

Note from eqs.(10.103) and (10.108) that in the case of the flat Friedmann- Lemaître universe model, the deceleration parameter may be expressed in terms of the relative density of vacuum only, q = (1/2)(1 3Ω ). The supernova Ia − Λ observations have shown that the expansion is now accelerating. Hence if the universe is flat, this alone means that ΩΛ0 > 1/3. As mentioned above, many different observations indicate that we live in a universe with critical density, where cold matter contributes with about 30% of the density and vacuum energy with about 70%. Such a universe is well described by the Friedmann-Lemaître universe model that have been presented above. However, this model is not quite without problems in explaining the observed properties of the universe. In particular there is now much research directed at solving the so called coincidence problem. As we have seen, the density of the vacuum energy is constant during the expansion, while the density of the mat- ter decreases inversely proportional to a volume comoving with the expanding matter. Yet, one observes that the density of matter and the density of the vacuum energy are of the same order of magnitude at the present time. This seems to be a strange and unexplained coincidence in the model. Also just at the present time the critical density is approaching the density of the vacuum energy. At earlier times the relative density was close to zero, and now it changes approaching the constant value 1 in the future. S. M. Carroll (Carroll 2001) has illustrated this aspect of the coincidence problem by plotting Ω˙ Λ as a function of ln(t/t0). Differentiating the expression (10.108) we get

tΛ dΩΛ sinh(t/tΛ) = 3 (10.113) 2 dt cosh (t/tΛ) which is plotted in fig. 10.15 Putting Ω¨ Λ = 0 we find that the rate of change of ΩΛ was maximal at the point of time t1 when the deceleration of the cosmic expansion turned into acceleration. There is now a great activity in order to try to explain these coinci- dences by introducing more general forms of vacuum energy called quintessence, and with a density determined dynamically bythe evolution of a scalar field (Turner 2001). However, the simplest type of vacuum energy is the LIVE. One may hope that a future theory of quantum gravity may settle the matter and let us un- derstand the vacuum energy. In the meantime we can learn much about the dynamics of a vacuum dominated universe by studying simple and beautiful universe models such as the Friedmann-Lemaître model.

10.5 Inflationary Cosmology

10.5.1 Problems with the Big Bang Models The Horizon Problem The Cosmic Microwave Background (CMB) radiation from two points A and B in opposite directions has the same temperature. This means that it has been 158 Chapter 10. Cosmology

t t Figure 10.15: Rate of change of ΩΛ as function of ln( ). The value ln( ) = 40 t0 t0 − corresponds to the cosmic point of time t 1s. 0 ∼

radiated by sources of the same temperature in these points. Thus, the universe must have been in thermic equilibrium at the decoupling time, t = 3 105years. d · This implies that points A and B, “at opposite sides of the universe”, had been in causal contact already at that time. I.e., a light signal must have had time to move from A to B during the time from t = 0 to t = 3 105 years. The points · A and B must have been within each other’s horizons at the decoupling. Consider a photon moving radially in space descibed by the Robertson- Walker metric (10.14) with k = 0. Light follows a null geodesic curve, i.e. the curve is defined by ds2 = 0. We get dt dr = . (10.114) a(t) The coordinate distance the photon has moved during the time t is t dt ∆r = . (10.115) a(t) Z0 The physical distance the light has moved at the time t is called the horizon distance, and is t dt l = a(t)∆t = a(t) . (10.116) h a(t) Z0 To find a quantitative expression for the “horizon problem”, we may consider a model with critical mass density (Euclidian spacelike geometry.) Using p = wρ and Ω = 1, integration of equation (10.36) gives

2 a t 3+3w . (10.117) ∝ 10.5 Inflationary Cosmology 159

Inserting this into the expression for lh and integrating gives 3w + 3 l = t. (10.118) h 3w + 1 Let us call the volume inside the horizon the “horizon volume” and denote it by V . From equation (10.118) follows that V t3. At the decoupling time, the H H ∝ horizon volume may therefore be written 3 td (V ) = V , (10.119) H d t 0  0  where V0 is the size of the present horizon volume. Events within this volume are causally connected, and a volume of this size may be in thermal equilibrium at the decoupling time. Let (V0)d be the size, at the decoupling, of the part of the universe that corresponds to the present horizon volume, i.e. the observable universe. For our Euclidean universe, the equation (10.117) holds, giving

2 3 a (td) td w+1 (V ) = V = V . (10.120) 0 d a3(t ) 0 t 0 0  0  From equations (10.119) and (10.120), we get

3w+1 (V0)d td w+1 = . (10.121) (V ) t H d  0 

4 V0)d 4 Using that td = 10− t0 and inserting w = 0 for dust, we find = 10 . (VH )d Thus, there was room for 104 causally connected areas at the decoupling time within what presently represents our observable universe. Points at opposite sides of our observable universe were therefore not causally connected at the decoupling, according to the Friedmann models of the universe. These models can therefore not explain that the temperature of the radiation from such points is the same.

The Flatness Problem Ω = ρ According to eq. (10.42), the total mass parameter ρcr is given by k Ω 1 = . (10.122) − a˙ 2 By using the expansion factor (10.117) for a universe near critical mass density, we get 2 3w+1 Ω 1 t ( 3w+3 ) − = . (10.123) Ω 1 t 0 −  0  For a radiation dominated universe, we get Ω 1 t − = . (10.124) Ω 1 t 0 − 0 160 Chapter 10. Cosmology

Measurements indicate that Ω0 1 is of order of magnitude 1. The age of 17 − the universe is about t0 = 10 s. When we stipulate initial conditions for the 43 universe, it is natural to consider the Planck time, tP = 10− s, since this is the limit to the validity of general relativity. At earlier time, quantum effects will be important, and one can not give a reliable description without using quantum gravitation. The stipulated initial condition for the mass parameter then becomes that Ω 1 is of order 10 60 at the Planck time. Such an extreme − − fine tuning of the initial value of the universe’s mass density can not be explained within standard Big Bang cosmology.

Other Problems The Friedman models can not explain questions about why the universe is nearly homogeneous and has an isotropic expansion, nor say anything about why the universe is expanding.

10.5.2 Cosmic Inflation Spontaneous Symmetry Breaking and the Higgs Mechanism 0 The particles responsible for the electroweak force, W  and Z are massive (causing the weak force to only have short distance effects). This was originally a problem for the quantum field theory describing this force, since it made it difficult to create a renormalisable theory1. This was solved by Higgs and Kibble in 1964 by introducing the so-called Higgs mechanism. 0 The main idea is that the massive bosons W  and Z are given a mass by interacting with a Higgs field φ. The effect causes the mass of the particles to be proportional to the value of the Higgs field in vacuum. It is therefore necessary for the mechanism that the Higgs field has a value different from zero in the vacuum (the vacuum expectation value must be non-zero). Let us see how the Higgs field can get a non-zero vacuum expectation value. The important thing for our purpose is that the potential for the Higgs field may be temperature dependent. Let us assume that the potential for the Higgs field is described by the function 1 1 V (φ) = µ2φ2 + λφ4, (10.125) 2 4

where the sign of µ2 depends on whether the temperature is above or below a critical temperature Tc. This sign has an important consequence for the shape of the potential V . The potential is shown in figure 10.16 for two different 2 temperatures. For T > Tc, µ > 0, and the shape is like in fig. 10.16(a), and 2 there is a stable minimum for φ = 0. However, for T < Tc, µ < 0, and the shape is like in fig. 10.16(b). In this case the potential has stable minima for µ φ = φ = | | and an unstable maximum at φ = 0. For both cases, the  0  √λ

1 2 2 The problem is that the Lagrangian for the gauge bosons can not include terms like m Wµ , which are not gauge invariant 10.5 Inflationary Cosmology 161 potential V (φ) is invariant under the symmetry transformation φ φ (i.e. 7→ − V (φ) = V ( φ)). − The “real” vacuum state of the system is at a stable minimum of the poten- tial. For T > Tc, the minimum is in the “symmetric” state φ = 0. On the other hand, for T < Tc this state is unstable. It is therefore called a “false vacuum”. The system will move into one of the stable minimas at φ = φ . When the  0 system is in one of these states, it is no longer symmetric under the change of sign of φ. Such a symmetry, which is not reflected in the vacuum state, is called spontaneously broken. Note that from figure 10.16(b) we see that the energy of the false vacuum is larger than for the real vacuum.

Figure 10.16: The shape of the potential depends on the sign of µ2. (a): Higher temperature than the critical, with µ2 > 0. (b): Lower temperature than the critical, with µ2 < 0.

The central idea, which originated the “inflationary cosmology”, was to take into consideration the consequences of the unified quantum field theories, the gauge theories, at the construction of relativistic models for the early universe. According to the Friedmann models, the temperature was extremely high in the early history of the universe. If one considers Higgs fields associated with GUT models (grand unified theories), one finds a critical temperature Tc correspond- 14 ing to the energy kTc = 10 GeV , where k is Boltzmann’s constant. Before 35 the universe was about t1 = 10− s old, the temperature was larger than this. Thus, the Higgs field was in the symmetric ground state. According to most of the inflation models, the universe was dominated by radiation at this time.

When the temperature decreases, the Higgs potential changes. This could happen as shown in figure 10.17. Here, there is a potential barrier at the critcal temperature, which means that there can not be a classical phase transition. The transition to the stable minimum must happen by quantum tunneling. This is called a first order phase transition. 162 Chapter 10. Cosmology

Figure 10.17: The temperature dependence of a Higgs potential with a first order phase transition.

Guth’s Inflation Model Alan Guth’s original inflation model (Guth 1981) was based on a first order phase transition. According to most of the inflationary models, the universe was dominated 35 by radiation during the time before 10− s. The universe was then expanding so fast that there was no causal contact between the different parts of the universe that became our observable universe. Probably, the universe was rather homo- geneous, with considerable spacelike variations in temperature. There was also areas of false vacuum, with energy densities characteristic of the GUT energy scale, which also controls it’s critical temperature. While the energy density 4 of the radiation decreased quickly, as a− , the energy density of vacuum was 35 constant. At the time t = 10− s, the energy density of the radiation became less than that of the vacuum. At the same time, the potential started to change, such that the vacuum went from being stable to being an unstable false vacuum. Thus, there was a first order phase transition to the real vacuum. Because of the inhomogeneouty of the universe’s initial condition, this happened with different speed at differing places. The potential barrier slowed down the process, which happened by tunneling, and the universe was at several places considerably undercooled, and there appeared “bubbles” dominated by the energy of the false vacuum. These areas acted on themselves with repulsive gravity. By integrating the equation of motion for the expansion factor in such a vacuum dominated bubble, one gets

8πGρc a = eHt, H = . (10.126) r 3 10.5 Inflationary Cosmology 163

By inserting the GUT value above, we get H = 6.6 1034s 1, i.e. H 1 = · − − 1.5 10 35s. With reference to field theoretical works by Sindney Coleman and · − others, Guth argumented that a realistic duration of the nucleation process hap- 33 pening during the phase transition is 10− s. During this time, the expansion factor increases by a factor of 1028. This vacuum dominated epoch is called the inflation era. Let us look closer at what happens with the energy of the universe in the course of it’s development, according to the inflationary models. To understand this we first have to consider what happens at the end of the inflationary era. When the Higgs field reaches the minimum corresponding to the real vacuum, it starts to oscillate. According to the quantum description of the oscillating field, the energy of the false vacuum is converted into radiation and particles. In this way the equation of state for the energy dominating the development of the expansion factor changes from p = ρ, characteristic for vacuum, to p = 1 ρ, − 3 characteristic of radiation. The energy density and the temperature of the radiation is then increased 33 enourmously. Before and after this short period around the time t = 10− s the radiation energy increases adiabatically, such that ρa4 = constant. According to Stefan-Boltzmanns law of radiadion, ρ T 4. Therefore, aT = constant during ∝ adiabatic expansion. This means that during the inflationary era, while the expansion factor increases exponentially, the energy density and temperature of radiation decreases exponentially. At the end of the inflationary era, the radiation is reheated so that it returns to the energy it had when the inflationary era started. It may be interesting to note that the Newtonian theory of gravitation does not allow an inflationary era, since stress has no gravitational effect according to it.

The Inflation Models’ Answers to the Problems of the Friedmann Models

The horizon problem will here be investigated in the light of this model. The problem was that there was room for about 10000 causally connected areas inside the area spanned by our presently observable universe at the time. Let us calculate the horizon radius lh and the radius a of the region presently within the horizon, l = 15 109ly = 1.5 1026cm, at the time t = 10 35s when the h · · 1 − inflation started. From equation (10.118) for the radiation dominated period before the inflatinary era, one gets

25 l = 2t = 6 10− cm. (10.127) h 1 ·

The radius, at time t1, of the region corresponding to our observable universe, Ht 35 is found by using that a e during the inflation era from t1 = 10− s to 33 1 ∝ 11 t2 = 10− s, a t 2 in the radiation dominated period from t2 to t3 = 10 s, 2 ∝ 17 and a t 3 in the matter dominated period from t until now, t = 10 s. This ∝ 3 0 164 Chapter 10. Cosmology

gives 1 2 Ht1 2 3 e t2 t3 28 a1 = lh(t0) = 1.5 10− cm. (10.128) eHt2 t t ·  3   0  We see that at the beginning of the inflationary era the horizon radius, lh, was larger than the radius a of the region corresponding to our observable universe. The whole of this region was then causally connected, and thermic equilibrium was established. This equilibrium has been kept since then, and explains the observed isotropy of the cosmic background radiation. We will now consider the flatness problem. This problem was the necessity, in the Friedmann models, of fine tuning the initial density in order to obtain the closeness of the observed mass density to the critical density. Again, the inflationary models give another result. Inserting the expansion factor (10.126) into equation (10.122), we get

k 2Ht Ω 1 = e− , (10.129) − H2 where H is constant and given in eq. (10.126). The ratio between Ω 1 at the − end of and the beginning of the inflationary era becomes

Ω2 1 2H(t2 t1) 56 − = e− − = 10− . (10.130) Ω 1 1 − Contrary to in the Friedmann models, where the mass density moves away from the critical density as time is increasing, the density approaches the critical density exponentially during the inflationary era. Within a large range of initial conditions, this means that according to the inflation models the universe should still have almost critical mass density. Bibliography

Bernadis et.al, P. d.: 2001, Multiple peaks in the angular power spectrum of the cosmic microwave background: Significance and consequences for cos- mology, astro-ph/0105296. Brill, D. R. and Cohen, J. M.: 1966, Phys. Rev. 143, 1011. Carroll, S. M.: 1998, Quintessence and the rest of the world., Phys. Rev. Lett. 81, 3067–3070. Carroll, S. M.: 2001, Dark energy and the preposterous universe., astro- ph/0107571. Grøn, Ø.: 1986, Repulsive gravitation and inflationary universe models., 54, 46– 52. Grøn, Ø. and Flø, H. K.: 1984, Differensialgeometri, Departmental. Course compendium. Guth, A. H.: 1981, The inflationary universe: a possible solution to the horizon and flatness problems, Phys. Rev. D23, 347–356. Linde, A.: 2001, Inflation and string cosmology, astro-ph/0107176. McGaugh, S.: 2001, Constraints on the radial mass distribution of dark matter halos from rotation curves, astro.ph/0107490. Misner, C. W., Thorne, K. S. and Wheeler, J. A.: 1973, Gravitation, first edn, W. H. Freeman and company, San Francisco. Perlmutter et al., S.: 1999, Measurements of omega ans lambda from 42 high- redshift supernovae., Astrophys. J. 517, 565–586. Pound, R. V. and Rebka, G. A., J.: 1960, Phys. Review Letters 4, 337. Pryke et al., C.: 2001, Cosmological parameter extraction from the first season of observations with dasi., astro-ph/0104490. Ravndal, F.: 1978, Inføring i generell relativitetsteori, Departmental. Course compendium. Riess, A. G.: n.d., The farthest known supernova: Support for an accelerating universe and a glimpse of the epoch of deceleration., astro-ph/0104455.

165 166 BIBLIOGRAPHY

Riess et al., A. G.: 1998, Observational evidence from supernovae for an accel- erating universe and a cosmological constant., Astron. J. 116, 1009–1038.

Stompor et al., R.: 2001, Cosmological implications of the maxima-i high resolution cosmic microwave background anisotropy measurement., astro- ph/0105062.

Turner, M. S.: 2001, Dark energy and the new cosmology., astro-ph/0108103.

Turner, M. S. and Riess, A. G.: 2001, Do Sn Ia provide direct evidence for past deceleration of the universe?, astro-ph/0106051.

Zeldovich, Y.: 1968, The cosmological constant and the theory of elementary particles, Sov. Phys. Usp. 11, 381–393.

Zlatev, I., Wang, L. and Steinhardt, P. J.: 1999, Quintessence, cosmic coinci- dence, and the cosmological constant, Phys. Rev. Lett. 82, 896–899. Index

Symbols mini ...... 122 , ν1ν2 ...... 50 radiation...... 121 δS = 0 ...... 59 temperature ...... 121 δ(~r) (Dirac delta function) ...... 4 Born-stiff motion...... 44 γij ...... 34 (tensor product) ...... 26 C ⊗ d2α ...... 51 canonical momentum...... 59 dΩ (solid angle) ...... 5 Cartan dtˆ...... 35 formalism...... 100 dl2 ...... 34 structure equations [ ] (antisymmetric combination) . 32 1st ...... 79, 100 α Γ µν ...... 61 2nd ...... 85, 101 3-space...... 34 surface curvature...... 89 4-acceleration ...... 16, 120 centrifugal acceleration ...... 57 4-momentum ...... 15 centrifulgal barrier ...... 112 4-velocity ...... 15 Christoffel symbols ...... 61 identity ...... 16, 59 in coordinate basis ...... 53 physical interpretation . . . . . 103 A rotating refernece frame . . . . . 74 accelerated frame comoving orthonormal basis . . . . . 31 hyperbolically...... 44 connection coefficients acceleration of particle in a Riemannian space ...... 74 in non-rotating frame ...... 56 in coordinate basis ...... 53 in rotating frame...... 57 rotating reference frame . . . . . 73 acceleration scalar ...... 120 connection forms ...... 100 action ...... 58 conservation Archimedean spiral...... 41 energy-momentum ...... 93 B constant of motion ...... 59 basis continuity equation...... 93 1-form ...... 25 contravariant tensor ...... 26 coordinate basis vector . . 18, 21 convective derivative ...... 93 orthonormal...... 18, 22 coordinate Bianchi cyclic ...... 59, 60 1st identity...... 90, 92 Gaussian...... 54 component form ...... 91 transformation ...... 20, 27 2nd identity ...... 91, 95 coordinate basis ...... 18, 21 black hole connection coefficients...... 53 horizon ...... 120 coordinate system ...... 18

167 168 INDEX

coriolis acceleration ...... 57 Einstein cosmological redshift ...... 135 field equations ...... 96 covariance principle...... 11, 51 principle of equivalence...... 9 covariant equations...... 28 synchronization ...... 35 covariant tensor ...... 26 tensor...... 96 curvature ...... 81 Einstein deSitter universe ...... 144 forms ...... 85, 101 embedding of surface ...... 89 of the Schwarzschild metric 108 Gaussian ...... 88, 89 energy-momentum tensor...... 93 geodesic ...... 87 equivalence, principle of ...... 9 normal...... 87 Euler equation of motion ...... 93 principal ...... 88 Euler-Lagrange equations ...... 59 Ricci...... 91 exterior differentiation...... 49 Riemann tensor ...... 83 surface (Cartan formalism). .89 F field equations D Einstein ...... 96 deflection of light ...... 117 vacuum ...... 97 differential forms first order phase transition . . . . . 160 1-form ...... 25 flatness problem...... 158, 163 2-form ...... 32 forms ...... see differential forms components ...... 33 free fall ...... 104 connection ...... 100 covariant differentiation of . . 76 G curvature ...... 85, 101 Gauss of surface ...... 89 coordinates ...... 54 differentiation of ...... 49 curvature...... 88, 89 directional derivatives of . . . . 76 integral theorem ...... 4 null form...... 33 theorema egregium...... 89 p-form ...... 32 general principle of relativity . . . . 10 transformation ...... 27 geodesic differentiation curvature ...... 87 convective ...... 93 curve...... 57, 61 covariant ...... 53, 55 spatial ...... 65 exterior ...... 49 equation ...... 56 of form ...... 49 postulate ...... 98 Dirac delta function ...... 4 gravitational acceleration...... 103 distance Tolman-Whittaker expression128 coordinate ...... 100 gravitational collapse...... 112 physical...... 100 gravitational time dilation...... 40 Doppler effect ...... 23 Grøn, Øyvind...... 3 Doppler effect, gravitational . . . . . 71 Guth, Alan ...... 161 dust dominated universe ...... 142 H E Hafele-Keating experiment . . . . . 114 Eddington - Finkelstein coordinates Hamilton-principle ...... 58 104 Hawking radiation ...... 121 INDEX 169 heat capacity of black hole . . . . . 121 from Schwarzschild solution 103 Higgs mechanism ...... 159 mixed tensor ...... 26 horizon problem...... 156, 162 momentum-energy tensor...... 93 Hubbles law...... 135 hydrostatic equilibrium ...... 127 N hyperbolically accelerated frame. 44 Newton fluid ...... 94 I hydrodynamics...... 93 incompressible stars ...... 130 law of gravitation ...... 1 inertial reference frame ...... 18 continuous distribution. . . . .2 inertial dragging ...... 123 local form...... 2 inertial reference frame ...... 54 Newtonian incompressible star . 126 inflation ...... 161 O K orbit equation...... 113 Kerr metric ...... 122 orthonormal basis ...... 18 Kretschmann’s curvature scalar 104 Kruskal - Szekers...... 104 P parallel transport ...... 55, 81 L particle trajectory Lagrange Schwarzschild 3-space ...... 110 covariant dynamics...... 57 Poincaré’s lemma ...... 51 equations ...... 59 precession free particle ...... 61 of Mercury’s perihelion . . . . 115 function ...... 58 pressure forces ...... 127 free particles ...... 61 principle of equivalence ...... 9, 103 Laplace equation ...... 3 projectile motion ...... 68 Levi-Civita connection ...... 53 proper distance ...... 135 light deceleration of ...... 108 R deflection of ...... 117 radiation dominated universe. . .141 light cones ...... 106 rank ...... 26 line element ...... 31 rapidity ...... 45 free particle ...... 60 redshift in space...... 34 cosmological ...... 135 spacetime ...... 35 Kerr spacetime ...... 123 Lorentz transformation ...... 23 reference frame...... 17 inertial ...... 18, 54 M relativity, general principle of. . . .10 Mach’s principle ...... 12 Ricci metric curvature ...... 91 covariant differentiation . . . . . 77 tensor, divergence...... 96 Kerr (rotating) ...... 122 identity ...... 90 static,stationary ...... 60 component form ...... 90 tensor...... 28 Riemann curvature tensor . . 83, 101 mini black holes...... 122 components ...... 84 Minkowski space-time rotating black hole ...... 122 170 INDEX

rotating reference frame ...... 65 product ...... 26 connection coefficients...... 73 rank ...... 26 Ricci curvature...... 91 S Riemann curvature . . . . . 83, 101 Sagnac effect ...... 42 transformation properties . . . 27 Schwarzschild theorema egregium ...... 89 3-space ...... 110 tidal forces ...... 5 energy ...... 110 time dilation, gravitational ...... 40 momentum ...... 110 Tolman-Oppenheimer-Volkov eq.130 light cones ...... 106 Tolman-Whittaker expression . . 128 metric ...... 104 travelling time ...... 105 embedding...... 108 Twin “paradox” ...... 69 radius ...... 2, 104, 120 solution U interior ...... 130 universe model outer ...... 100 dust dominated ...... 142 spacetime ...... 104, 120 inflationary ...... 161 simultaneity radiation dominated ...... 141 on Einstein synchronized clocks V 35 vacuum field equations ...... 97 space, geometry of ...... 35 vector singularity basis ...... 14 coordinate ...... 104 covariant differentiation of . . 75 physical...... 104 dimension...... 13 solid angle ...... 4 linearly independent ...... 13 spherical geometry ...... 38 spontaneous symmetry breaking160 Z structure coefficients ...... 23 ZAMO...... 123 in a Riemannian space ...... 74 in coordinate basis ...... 25 surface curvature ...... 89 surface gravity ...... 120

T tangent plane ...... 17 tensor...... 25 antisymmetric ...... 31 construction example ...... 27 contraction of ...... 91 contravariant ...... 26 covariant...... 26 covariant differentiation of . . 77 Einstein ...... 96 metric ...... 28 mixed ...... 26, 28 momentum energy ...... 93