Chapter 1 General Relativity In this chapter, we review the basic formalism of the standard gravity theory. We do not focus on details, but on the basic facts necessary for cosmological applications to be developed in following chapters. 1.1 Special Relativity Before we start discussing General Relativity (GR), let us review some concepts in Special Relativity (SR) and covariant equations. It all starts with problems faced with Electromagnetism (E&M) at the end of the 19th century, which required a redefinition of our classical notions of space and time. 1.1.1 Electromagnetic Equations We start with a brief review of the electromagnetic equations. Maxwell Equations The Maxwell equations describe the production and propagation of the Electromagnetic (E&M) fields. They are written in differential form as ρ E = (Gauss Law) (1.1) ∇· ǫ0 B = 0 (Nonexistence of Magnetic Monopoles) (1.2) ∇· ∂B E = (Faraday induction Law) (1.3) ∇× − ∂t ∂E B = µ j + µ ǫ (Ampere Law) (1.4) ∇× 0 0 0 ∂t where ρ is the charge density and j is electric current density. 11 12 CHAPTER 1. GENERAL RELATIVITY Charge Conservation Taking the divergence of Ampere Law: ∂( E) B = µ j + µ ǫ ∇· ∇·∇× 0∇· 0 0 ∂t ∂(ρ/ǫ ) = µ j + µ ǫ 0 0∇· 0 0 ∂t ∂ρ = µ ( j + ) 0 ∇· ∂t Therefore, charges are conserved through the continuity equation ∂ρ + j = 0 (1.5) ∂t ∇· Electromagnetic Potentials It is convenient to define electromagnetic potentials through the source-free Maxwell Eqs. Firstly, since B = 0 B = A (1.6) ∇· → ∇× Using this expression in Faraday Law, we have ∂ A ∂A ∂A E = ∇× = E + = 0 (1.7) ∇× − ∂t ∇× − ∂t → ∇× ∂t and the term in parenthesis must be the gradient of a scalar field φ. The electric potential φ and the magnetic vector potential A are thus defined as ∂A E = φ (1.8) −∇ − ∂t B = A (1.9) ∇× Gauge Transformation: Invariance If φ e A are solutions of the Maxwell Eqs., then the potentials φ′ and A′ defined by ′ ∂f φ = φ (1.10) − ∂t ′ A = A + f (1.11) ∇ for a general function f(x,t), are also solutions, because ′ ′ ′ ∂A ∂f ∂A ∂( f) ∂A E = φ = φ + ∇ = φ = E (1.12) −∇ − ∂t −∇ ∇ ∂t − ∂t − ∂t −∇ − ∂t ′ ′ B = A = A + ( f)= B (1.13) ∇× ∇× ∇× ∇ Therefore, we have the freedom to choose the function f conveniently without changing the fields. The choice of f implies determining a gauge. The gauge used for the solutions to electro- magnetic waves is the Lorenz gauge ∂φ A + µ ǫ = 0 (Lorenz Gauge) (1.14) ∇· 0 0 ∂t 1.1. SPECIAL RELATIVITY 13 Electromagnetic Waves Inserting the potentials into the Maxwell Eqs, we have ∂A ∂ A ρ E = ( φ )= 2φ ∇· = (1.15) ∇· ∇· −∇ − ∂t −∇ − ∂t ǫ0 and B = ( A)= ( A) 2A ∇× ∇× ∇× ∇ ∇· −∇ ∂E ∂ ∂A = µ j + µ ǫ = µ j + µ ǫ φ 0 0 0 ∂t 0 0 0 ∂t −∇ − ∂t ∂φ ∂2A = µ j µ ǫ µ ǫ (1.16) 0 −∇ 0 0 ∂t − 0 0 ∂t2 These two equations imply: ∂ A ρ 2φ + ∇· = (1.17) ∇ ∂t −ǫ0 ∂2A ∂φ 2A µ ǫ = µ j + A + µ ǫ (1.18) ∇ − 0 0 ∂t2 − 0 ∇ ∇· 0 0 ∂t Choosing the Lorenz gauge ∂φ A + µ ǫ = 0 (1.19) ∇· 0 0 ∂t these equations become 2 2 1 ∂ φ 2 ρ φ = 2 2 + φ = (1.20) −c ∂t ∇ −ǫ0 1 ∂2A 2A = + 2A = µ j (1.21) −c2 ∂t2 ∇ − 0 i.e., the potentials propagate according to the classical non-homogenous wave equation with con- stant speed equal to the speed of light c2 = 1/µ ǫ . Unification: E&M Optics. 0 0 ↔ Lorentz Force Given the E&M fields, corresponding E&M forces F act on particles of charge q and are given by: F = q(E + v B) (1.22) × 1.1.2 Special Relativity Postulates The E&M wave equations have a constant speed of propagation. Questions: 1: With respect to what reference frame should c be measured? 2: How can we explain effects such as the disappearance of magnetic forces in a reference frame that moves with the charge? 14 CHAPTER 1. GENERAL RELATIVITY These issues motivated the development of special relativity, which solves these problems and change our classical concepts of space and time, requiring only two postulates (in fact, only one): Postulate 1: The laws of physics are the same in all inertial frames. Postulate 2: The speed of light is the same in all inertial frames. The constancy of c follows from postulate 1, since E&M is a set of physical laws in which c=const. Therefore Postulate 2 is redundant given Postulate 1. 1.1.3 Coordinates and Metric Defining contravariant coordinates xµ =(x0,x1,x2,x3)=(ct, x, y, z) , (1.23) the line element ds ds2 = dx2 + dy2 + dz2 c2dt2 = (dx0)2 +(dx1)2 +(dx2)2 +(dx3)2 µ ν − − = ηµνdx dx (1.24) defines the metric ηµν 1 0 0 0 − 0 1 0 0 ηµν = . (1.25) 0 0 1 0 0 0 0 1 Covariant coordinates xµ are defined as x = n xν =( ct, x, y, z) (1.26) µ µν − Similarly, µ µν x = η xν, (1.27) µν µ µν where ηµν is the inverse metric, i.e. η ηνα = δ α. In flat space it turns out that η = ηµν. Throughout we use the Einstein sum convention: crossed repeated indices are summed over: 3 ηµνx ηµνx (1.28) ν ≡ ν ν=0 X 1.1.4 Invariance of the Line Element: Spatial Rotations Let us first consider a simple rotation in 3 dimensions. Consider a rod such that one of its ends is at the origin (0, 0, 0) and the other is at coordinates (x, y, z) in frame K. The rod has length l and therefore l2 = x2 + y2 + z2 (1.29) 1.1. SPECIAL RELATIVITY 15 Now suppose we consider this same rod as seen by a frame K′ that is rotated by angle θ with respect to the z axis of K. In K′, the rod still has one end at the origin (0, 0, 0), but the other one is at new coordinates (x′,y′, z′) given by ′ ct = ct ′ x = x cos θ + y sin θ ′ y = x sin θ + y cos θ ′ − z = z as measured by system K′. Therefore: ct′ 1 0 0 0 ct x′ 0 cos θ sin θ 0 x ′ = y 0 sin θ cos θ 0 y ′ − z 0 0 0 1 z Obviously, if we compute the length l′ measured by K′ we obtain ′ ′ ′ ′ l 2 =(x )2 +(y )2 +(z )2 = x2 + y2 + z2 = l2 Therefore the rotation transformation left the length Figure 1.1: Rotation of frame K into K′ around 2 l of the rod invariant. A similar idea applies in 4- the z axis by angle θ in the xy plane. dimensional space-time, except that instead of rotat- ing purely in space, we introduce an operation that also involves the time coordinate: boosts. 1.1.5 Invariance of the Line Element: Lorentz Boosts Consider an inertial frame K and another K′ that moves relative to K with speed v in the x direction. For simplicity we may assume that the two frames coincide initially, i.e. at t = t′ = 0 we have x = x′ = 0. For both frames c is the same, so considering the trajectory of a light ray, we have that the line element s is null in both frames ′ ′ ′ ′ ′ s2 = x2 + y2 + z2 c2t2 =0= x 2 + y 2 + z 2 c2t 2 = s 2 (1.30) − − It can be shown (see Schutz) that this implies that the interval s2 is the same in all inertial frames, even when s2 = 0. The invariance of s2 is similar to the invariance of the length l2 of a rod 6 in 2 dimensions. But differently from l2, which is always positive, the space time s2 can be either positive (space-like events) or negative (time-like events). The Lorentz transformations relates coordinates xµ and xµ′, keeping s2 invariant (and null in the case of light). Considering e.g. a particle at rest in K′, and therefore at x = vt in K, one can show that the transformation is given by (Exercise: Prove it) ′ x0 = γ(x0 βx1) (1.31) ′ − x1 = γ(x1 βx0) (1.32) ′ ′ − x1 = x2 (1.33) ′ ′ x3 = x3 (1.34) 16 CHAPTER 1. GENERAL RELATIVITY where v β = < 1 (1.35) c 1 γ = > 1 (1.36) 1 β2 − or p µ′ ′ ∂x xµ = xν =Λµ xν (1.37) ∂xν ν with γ βγ 0 0 − βγ γ 0 0 Λµ = − (1.38) ν 0 0 1 0 0 0 0 1 µ 2 2 2 2 2 Notice that detΛ ν = γ β γ = (1 β )γ = 1. A similar (inverse) transformation relates the − ′− µ −1 µ coordinates in the inverse direction (x x), with β β, i.e. (Λ ν) (β)=Λ ν( β). → →− − Similarly, for covariant coordinates the transformation follows ′ ′ν ν α ν γ α ν γβ xµ = ηµνx = ηµνΛ αx = ηµνΛ γ δ α x = ηµνΛ γη xβ (1.39) γβ η ηβα β Λµ |{z} or | {z } ′ ν xµ =Λµ xν (1.40) with Λ ν η Λα ηβν and also given by (as easily shown by matrix multiplication) µ ≡ µα β γ +βγ 0 0 − +βγ γ 0 0 Λ ν = (Λµ ) 1 = (1.41) µ ν 0 0 1 0 0 0 0 1 Boost as space-time rotation with complex time and complex angle Finally, we may check that by identifying γ = cosh φ (1.42) βγ = sinh φ (1.43) or equivalently β = tanh φ, we have cosh φ sinh φ 0 0 − sinh φ cosh φ 0 0 Λµ = − (1.44) ν 0 0 1 0 0 0 0 1 1.1.