General Relativity Volker Schlue

Volker Schlue

Department of Mathematics, University of Toronto, 40 St George Street, Toronto, Ontario M5S 2E4, Canada E-mail address: [email protected]

Preface

I have prepared this manuscript for a course on general relativity that I taught at the University of Toronto in the winter semester 2013. It is based on notes that I took as a student at ETH Zürich in a lecture course that Demetrios Christodoulou gave in the fall of 2005. (Chapters that are in significant parts transcribed from these notes will be marked by a .) His teaching and understanding of general relativity made a great impression∗ on me as a student, and I hope that these lecture notes can contribute to a wider circulation of his approach. I am very grateful to AndréLisibach, who provided me with most of the exercises, as well as complementary notes from a similar course that Prof. Christodoulou gave last fall, which were useful for the preparation of the present manuscript.

Toronto, January 2014 V. Schlue

iii

Contents

Preface iii Chapter 1. The equivalence principle and its consequences ∗ 1 1. Classical theory of gravitation 2 2. Special Relativity 5 3. Geodesic correspondence 20 Chapter 2. Einstein’s field equations ∗ 47 1. Einstein’s field equations in the presence of matter 50 2. Action Principle 66 3. The material manifold 74 4. Cosmological constant 79 Chapter 3. Spherical Symmetry 81 1. Einstein’s field equations in spherical symmetry 81 2. Schwarzschild solution 86 3. General properties of the area radius and mass functions 90 4. Spherically symmetric spacetimes with a trapped surface 92 Chapter 4. Dynamical Formulation of the Einstein Equations ∗ 97 1. Decomposition relative to the level sets of a time function 97 2. Slow Motion Approximation 115 3. Gravitational Radiation 125 Bibliography 149

CHAPTER 1

The equivalence principle and its consequences ∗

There are two motivations for the General Theory of Relativity. (1) Extension of the principle of Special Relativity (invariance of the physical laws under change from one inertial system to another — one such system relative to another is in a state of uniform motion) to all systems of reference in any state of motion whatsoever. (2) Establish a theory of gravitation which is in accordance with Special Relativity. Einstein’s Equivalence Principle relates the two motivations. Remark. (1) is sometimes referred to as the requirement that the laws of physics should be valid in any system of spacetime coordinates. (General Covari- ance.) In fact, however, (1) refers to the requirement of invariance of the physical laws under change of physical description, from one relative to one set of observers to another relative to another set of such observers. The history of an observer is represented in relativity by a timelike curve. A family of observers is given by a family of timelike curves that do not intersect each other. Each set of curves is a foliation of a given spacetime region. See Fig. 1.

Figure 1. Two families of observers. The history of each observer is a timelike curve.

1 2 1. THE EQUIVALENCE PRINCIPLE

Notes on Special Relativity An inertial system is a system of reference relative to which any mass moves on a straight line if no external forces act upon it. In view of Newton’s Laws set in Euclidean space, any such system is equivalent to (is not distinguished from) any other system of reference in uniform relative motion, which leads to the principle of Galilean relativity. The corresponding transformation laws leave time unchanged (up to translations) and thus entail surfaces of absolute simultaneity (level sets of time) which separate the future from the past. See Fig. 2. Special Relativity is based on the premise that light, which prior to Ein- stein had been assumed to propagate in a medium at rest relative to a distinguished frame of reference, in fact propagates in the absence of any such medium on cocentric spheres with respect to any inertial system when emitted from a point source; in other words the speed of light is finite and the same in any inertial system. Einstein realized this observation can only be reconciled if we give up the notion of absolute simultaneity in favour of a notion of simultaneity relative to the observer such that the propagation of light is independent of the system of reference, c.f. Fig. 2. This is precisely achieved in the theory of Special Relativity according to which space and time are unified in Minkowski space (R3+1, m), a 3 + 1-dimensional linear space endowed with a quadratic form m of index 1. A frame of reference corresponds to the choice of a unit timelike vector e, m(e, e) = 1, − which determines a unique spacelike hyperplane Σe = X : m(e, X) = 0 with positive- definite induced metric m 0, a 3-dimensional Euclidean{ space consisting} of all events |Σe ≥ considered simultaneous by observers on Σe at rest relative to the frame defined by e, namely observers whose “world-lines” are straight lines with tangent vector e. The set of null vectors L at a point p, m(L, L) = 0, form a “light cone” consisting of all events reached by light emitted from and received at p. The light cone takes the role of level sets of absolute time in Galilean physics in the sense that it separates the past from the future: The set of time-like vectors T at a point p, m(T,T ) < 0, the interior of the light cone, has two disconnected components referred to as future- and past-directed time-like vectors pointing to events that can possibly be influenced by an event at p, or have possibly influenced p, respectively. The exterior of the cone at p, the set of spacelike vectors X at p, m(X,X) > 0, is “causally disconnected” from p, given that no speed of propagation of any physical action whatsoever has ever been observed to surpass the speed of light. The “Principle of special relativity” asserts that all physical phenomena occur in one frame of reference as they do with respect to another, and in principle cannot be used to distinguish a certain frame of reference, i.e. a specific unit time-like vector e. The elements of the orthogonal group with respect to the Minkowski metric m, m(OX,OY ) = m(X,Y ), are called Lorentz transformations, and a “proper” (time-orientation preserving) subgroup mediates in particular the change of one frame of reference to another: e e0 = Oe, m(Oe, Oe) = m(e, e) = 1. Thus the validity of the principle of special relativity7→ requires the invariance of all physical− laws under Lorentz transformations.

1. Classical theory of gravitation A postulate of the classical theory (mechanics: motion of bodies in a given force ﬁeld & gravitation: the gravitational force ﬁeld of a given distribution of bodies) is the equality (up to a universal constant depending on the choice of physical units) of the passive gravitational mass and the inertial mass. 1. CLASSICAL THEORY OF GRAVITATION 3

Figure 2. Propagation of light in Galilean and Special Relativity.

ψ: Newtonian gravitational potential. mG ψ(t, x): Gravitational force acting on a test body of gravitational − ∇ mass mG at time t, the instantaneous position of the body at time t R 3 ∈ being x R . mG is thus a concept of the classical theory of gravitation. x = x(t): Motion∈ of a test body. If a body is subjected to a given force F when at position x and time t, then its acceleration is: d2x 1 (1.1) 2 (t) = F (x, t) dt mI

This law (Newton’s Second Law of Motion) deﬁnes the inertial mass mI . The postulate is

(1.2) mG = mI . In the classical framework there is no explanation for this equality. Set

(1.3) F = mG ψ(t, x(t)). − ∇ Then mG = mI implies that these cancel in the equations of motion: d2x (1.4) (t) = ψ(t, x(t)) dt2 −∇ That is, the motion of the test body in a given gravitational ﬁeld is independent of the test body.

1.1. Tidal forces. Consider the gravitational ﬁeld in a small space-time region, the neighborhood of an event (t0, x0), then a0 = ψ(t0, x0) is the common −∇ acceleration of all test bodies in this region. Let x = x0(t) be the history of the test 0 mass which we take as a reference mass. In the new reference system x = x x0 the origin is translated at each time to the instantaneous position of the reference− mass. (See Fig. 3, 4.) Note that this is an accelerated reference system. 4 1. THE EQUIVALENCE PRINCIPLE

2 x0

x2 1 x0

x = x0(t)

Figure 3. History of a reference mass.

x0 x x0(t) = 00

x1 Figure 4. Reference coordinate system with origin at each time at the instantaneous postion of the reference mass.

Now consider the motion of another, arbitrary test body in this new reference system: x0 = x0(t).

2 0 d x 0 (t) = ψ t, x0(t) + x (t) a0(t) dt2 −∇ − (1.5) 0 = ψ t, x0(t) + x (t) + ψ t, x0(t) −∇ ∇ 2 0 0 2 = ψ t, x0(t) x (t) + ( x (t) ) −∇ · O | | In terms of rectangular components, and to linear order in x0(t) this equation reads

2 0i 3 2 d x X ∂ ψ 0j (1.6) (t) = t, x0(t) x (t) dt2 − ∂xi∂xj · j=1 and is called the tidal equation. 2. SPECIAL RELATIVITY 5

P 0: Periphery

Figure 5. Measurement of P 0/R in the rotating system.

Remark. This equation governs the distortion of a dust cloud in time: tidal distortions. Take here the reference mass to be the test particle occupying the center of mass of a small dust cloud. 1.2. Equivalence principle. (1) In a freely falling reference frame the gravitational force itself is eliminated. Therefore such a system is equivalent to an inertial reference system in the absence of gravity. What remains however is the differential of the gravitational field which causes tidal distortions. (2) Similarly, an accelerated reference frame in the absence of gravitational fields is equivalent to a stationary reference frame in a gravitational field. As a consequence of the equivalence principle, a theory extending the theory of special relativity must include the theory of gravitation.

2. Special Relativity We shall see that the equivalence (2) leads to Riemannian geometry.

2.1. Uniformly rotating system. Consider a uniformly rotating reference system. 2.1.1. Preliminary considerations. We can ignore the space dimension perpendicular to the rotating plane. We have a stationary system x, and a reference system x0 obtained from x by rotation with constant angular velocity ω. Consider a circle of radius x0 = R with periphery P 0. What is P 0/R as measured in the rotating system?| Suppose| we use rods of unit length in the rotating system for measurement. As seen in the stationary system, the rods layed along the radius are also of unit length but those along the periphery experience a Lorentz-contraction so their length is only √1 v2, where v = ωR. (See Fig. 5.) However, in the stationary system the periphery− of x = x0 = R is P = 2πR. Since the unit rod of the rotating system, when laid along| | the| | periphery, has less than unit length by the factor √1 v2, it follows that − P P 0 2π (2.1) P 0 = = = > 2π √1 v2 ⇒ R √1 v2 − − 6 1. THE EQUIVALENCE PRINCIPLE

(x0, y0) t = 0

Figure 6. History of a rotating observer with respect to the rotating frame. in the rotating system.1 This means that the laws of Euclidean geometry do not hold in the rotating system! 2.1.2. Uniformly rotating observers in Minkowski space. Consider the situation in Minkowski space with 2 space and 1 time dimensions. In rectangular coordinates (t, x, y) of the stationary inertial reference system the metric takes the form: (2.2) ds2 = dt2 + dx2 + dy2 − It is convenient to use polar coordinates (t, r, ϕ) such that x = r cos ϕ, y = r sin ϕ and thus (2.3) ds2 = dt2 + dr2 + r2dϕ2 . − Let us denote by (x0, y0), and (r0, ϕ0) the rectangular and polar spatial coordinates, respectively, in the uniformly rotating system.2 A uniformly rotating observer has the history

(2.4) t (t, r = r0, ϕ = ϕ0 + ωt) 7−→ with respect to the stationary system. In the rotating system the observer remains at its initial position (see Fig. 6.)

(2.5) x0 = r0 cos ϕ0 y0 = r0 sin ϕ0 , and has the history

(2.6) t (t, x0 = x0, y0 = y0) . 7−→ The arc element ds2 is a geometric invariant and can be expressed in terms of the (r0, ϕ0) coordinates; (in agreement with general covariance, compare remark on page 1). We ﬁnd

(2.7) r = r0 ϕ = ϕ0 + ωt

(2.8) dr = dr0 dϕ = dϕ0 + ωdt

1This is a sign of negative curvature. 2 2 2 The domain of these coordinates is the disc ω r0 < 1, because no observer can go faster than the speed of light. 2. SPECIAL RELATIVITY 7 and thus ds2 = dt2 + dr2 + r2dϕ2 − (2.9) 2 2 2 2 = dt + dr + r (dϕ0 + ωdt) − 0 0 or 2 2 2 ωr dϕ0 2 r dϕ (2.10) ds2 = 1 ω2r2 dt 0 + dr2 + 0 0 . − − 0 − 1 ω2r2 0 1 ω2r2 − 0 − 0 Let us denote by q α = 1 ω2r2 : a real number(2.11a) − 0 2 ωr dϕ0 θ = dt 0 : a 1-form(2.11b) − 1 ω2r2 − 0 r2 dϕ2 dσ2 = dr2 + 0 0 : a Riemannian metric(2.11c) 0 1 ω2r2 − 0 then we can write (2.12) ds2 = α2θ2 + dσ2 . − We can view dσ as an arc length element in space. The rays ϕ0 = const. are geodesics of dσ and r0 is arc length along these geodesics:

(2.13) dσ = dr0 : along ϕ0 = const.

The curves r0 = const. are the geodesic circles orthogonal to these rays and r0 dϕ0 (2.14) dσ = p : along r0 = const. 1 ω2r2 − 0 0 So we find for the perimeter P of a circle of radius r0 = R that Z 2π 0 r0dϕ0 2πr0 P 2π (2.15) P 0 = = = = , p 2 p 2 2 2 0 1 ω2r 1 ω2r ⇒ R √1 ω R − 0 − 0 − which coincides with (2.1), (and shows that dσ2 is a Non-Euclidean metric). We are led to think that the coordinates (r0, ϕ0) have no physical meaning in the rotating system. However, what can be measured is the arc length element dσ in (2.12). Any such measurement of arclength in space is carried out simultaneously as judged by the rotating observers, which brings us to the question of simultaneity in the rotating system. An inertial observer corresponds to a straight timelike line in Minkowski space. The set of events which the observer considers simultaneous with an event p of his own is the spacelike hyperplane Σ through p orthogonal (with respect to (2.2)) to the line. (See Fig. 7.) A family of observers defining an inertial system is a family of parallel timelike straight lines. The simultaneous hyperplanes Σ will be parallel and the same at each point. The family of accelerated observers defining the rotating system is (2.4) as shown in Fig. 8. The hyperplane of locally simultaneous events is different at each point. 8 1. THE EQUIVALENCE PRINCIPLE

p Σ

Figure 7. Set of simultaneous events Σ to p as judged by an inertial observer.

t > 0

Σ p

t = 0

Figure 8. History of a rotating observer.

∂ Consider the vectorﬁeld ∂t . Its integral curves are the histories of the uniformly rotating observers, parametrized by t. In the original (stationary) coordinates (t, r, ϕ) the motion of an observer is (t, r, ϕ) (t + b, r, ϕ + ωb), while in the 7→ coordinates of the rotating system this motion is simply (t, r0, ϕ0) (t+b, r0, ϕ0). Let 7→ ∂ ∂ ∂ (2.16) V = = + ω ∂t r0,ϕ0 ∂t r,ϕ ∂ϕ t,r be the vectorﬁeld generating the above motion (a 1-parameter group of translations), and let ΣV be the hyperplane in the tangent space at a given point consisting of all vectors at that point which are orthogonal to V , namely n o (2.17) ΣV = X : g(V,X) = 0 . Here and in the following we now write g = ds2. Prop 2.1. The 1-form θ is characterized by the following two properties: (1) θ V = 1 · (2) ΣV = ker θ = X : θ X = 0 . · 2. SPECIAL RELATIVITY 9

P0 Σ0

Figure 9. Projection map related to uniformly rotating observers.

∂ Proof. By (2.11b) and (2.16) we simply have: θ V = dt ( )r ,ϕ = 1. Now · · ∂t 0 0 dσ2(V, ) = 0 because dσ2 does not contain dt. Therefore · g(X,V ) = α2θ(X)θ(V ) = α2θ(X) = 0 − − is equivalent to θ(X) = 0, since α > 0.

Both α and ΣV have physical signiﬁcance: α: the rate of a uniformly rotating clock relative to a stationary clock: p ds α = g(V,V ) = . − dt r0,ϕ0

ΣV : local simultaneous space corresponding to an observer with tangent vector V to his history.

While ΣV contains the events locally simultaneous to an event on the history of an observer with velocity V , we shall now address the question of simultaneity globally. In other words, given a set of simultaneous events as judged by the rotating observers, which set of events do they correspond to as seen in the stationary system? We can rephrase this questions as follows. Let C0 be a curve in the hyperplane Σ0 = (t, x, y): t = 0 . In polar coordinates we have the parametric equations { }

(2.18) C0 : r0 = r0(λ) , ϕ0 = ϕ0(λ) .

Let P0 be the point on C0 with λ = 0, P0 = (0, r0(0), ϕ0(0))

Question: Find a curve C in spacetime through the point P0 such that C projects to C0 and such that the curve C is horizontal. Projection: Here the projection is a map π of spacetime onto the hyperplane Σ0 deﬁned as follows: π(p) = p0 if p is an event belonging to a history of one of the rotating observers and p0 is the initial position of that observer. In terms of the coordinates (t, r0, ϕ0) this is simply π(t, r0, ϕ0) = (r0, ϕ0). Horizontal: This means that the tangent vector to C at each point belongs to ΣV at that point, i.e. it is orthogonal to the history of the observer through that point. This means that the curve C is locally simultaneous as judged by each observer. 10 1. THE EQUIVALENCE PRINCIPLE

C V ∆t C˙

C0 Σ0

Figure 10. A horizontal curve C that projects to C0.

Remark. The question addresses the problem of measurement in the rotating system. We can think of C0 as a curve on the rotating disc whose length we wish to measure using a ruler on the rotating disc. The curve C describes the collection of events in spacetime that are considered when the measurement of length of C0 occurs simultaneously as seen in the rotating system. The condition that C projects to C0 ensures that the curve whose length is measured is C0 in the rotating system; the condition that C is horizontal means that this measurement occurs locally simultaneous as judged by each uniformly rotating observer on C0. The conditions on C are thus simply:

(1) π(C) = C0 (2) X = C˙ ΣV ∈ By (1) we can assume that the parametric equations for C in the coordinates of the rotating system are:

(2.19) C : t = t(λ) , r0 = r0(λ) , ϕ0 = ϕ0(λ) Now C˙ is a vector with components

dt ∂ dr0 ∂ dϕ0 ∂ (2.20) C˙ = + + , dλ ∂t r0,ϕ0 dλ ∂r0 dλ ∂ϕ0 and we have shown above

(2.21) C˙ ΣV θ C˙ = 0 . ∈ ⇐⇒ · Therefore, in view of (2.11b), condition (2) implies 2 dt ωr dϕ0 (2.22) (λ) 0 (λ) = 0 , t(0) = 0 . dλ − 1 ω2r2 dλ − 0 We can integrate this equation to ﬁnd Z λ 2 0 ωr (λ ) dϕ0 (2.23) t(λ) = t(0) + 0 (λ0)dλ0 . 1 ω2r2(λ0) dλ 0 − 0 In the case where C0 is a circle, r0(λ) = R, we obtain ωR2 (2.24) t(λ) = t(0) + ϕ(λ) ϕ(0) , 1 ω2R2 − − 2. SPECIAL RELATIVITY 11 so the change of t in going around a full circle is

2πωR2 (2.25) ∆t = . 1 ω2R2 − The curve C does not close! See Fig. 10. Finally, observe that dσ is in fact arc length along C. Remark. This result is contrary to our intuition of rigid space and time. It tells us that it is impossible to measure the length of a curve on the rotating disc in a globally instantaneous manner: When measuring the length of a closed curve C0 on the disc initiating from P0 — in such a way that each observer on C0 (and at rest with respect to the uniformly rotating disc) agrees that the measurement occurs instantaneously according to his local standard of simultaneity — then upon return to P0 the proper time α∆t has elapsed in the course of the measurement. This shows in particular that no global standard of simultaneity exists in the rotating system.

Note on vectorfields, 1-forms and metrics At any point p on a differentiable manifold we have a homeomorphism ϕ from a M neighborhood of p onto Rn, n = dim . M n n Consider the coordinate lines ai : R R , t (0, . . . , t, . . . , 0) in R . The coordinate → 7→ vectorfield ei : i = 1, . . . , n at p is defined to be the tangent vector to the curve Ki = 1 ϕ− a : ◦ i d ei f = f Ki for all f C∞( ) . · dt ◦ t=0 ∈ M ˜ ˜ 1 n Since f Ki = f ai, where f = f ϕ− is a function on R , and ◦ ◦ ◦ ∂f˜ ∂ ei f = 0 , we simply write: ei = . · ∂xi | ∂xi p

These vectors form a basis for the n-dimensional vectorspace Tp : the tangent space to at p. Any tangent vector X T can be expanded in thisM basis: M ∈ pM n X i ∂ X = Xp . ∂xi p i=1

The dual space of T , denoted by T∗ is the set of linear functions on T (each pM pM pM element is also called a covector). While a vectorﬁeld X is an assignment p X(p) of a tangent vector at each point, a 1-form is an assignment p θ(p) of a covector7→ at each point. The tensor product θ2 = θ θ, or more generally θ 7→ξ of two 1-forms, is a bilinear function on T at each p :⊗ ⊗ pM ∈ M θ ξ (X ,Y ) = θ(X )ξ(Y ) . ⊗ p p p p p A special 1-form is given by the diﬀerential of function f : R: M → df X = X f . · · 12 1. THE EQUIVALENCE PRINCIPLE

If we denote by xi = ϕi local coordinates in a neighborhood of p then we ﬁnd that

i ( i ∂ ∂ϕ 0 i = j dx = = δij = 6 . · ∂xj ∂xj 1 i = j

1 n The 1-forms (dx ,..., dx ) form the dual basis in Tp∗ , and any covector ξ Tp∗ can be expressed as M ∈ M n X i ξ = ξi dx p . i=1 P i P j ∂ We have that for any 1-form θ = i θidx and any vector X = j X ∂xj it holds that X θ X = θ Xi . · i i A metric g on is an assignment p g of a non-degenerate symmetric bilinear M 7→ p form gp in Tp at each p . Here symmetric means g(X,Y ) = g(Y,X), and nondegenerate: M ∈ M g (X ,Y ) = 0 for all Y T = X = 0 . p p p p ∈ pM ⇒ p A symmetric bilinear form is often called a quadratic form. Since g is symmetric and bilinear we can expand g at any point p in the basis dxi dxj: ⊗ n X g = g dxi dxj , p ij |p ⊗ |p i,j=1 where it is understood that gij = gji. A Riemannian metric is a positive definite metric. It allows us to assign a magnitude (length) to a vector X, p X = g(X,X) 0 , | | ≥ and an angle α to two non-zero vectors: g(X,Y ) = X Y cos α . | || | A Lorentzian metric is an assignment p g of a non-degenerate quadratic form of 7→ p index 1 in Tp at each point p . Recall that the index of a quadratic form is the largest dimensionM of a subspace where∈ M the quadratic form is negative definite. There is therefore a vector T T such that g (T ,T ) < 0; T is a timelike vector at p. By p ∈ pM p p p p multiplying with a suitable non-zero factor we can assume that Tp is unit: gp(Tp,Tp) = 1. Define Σ to be the the orthogonal complement of T in T . Then g is positive− p p pM p|Σp definite. Choosing an orthonormal basis (E1 p,...,En p) (here n = 3) for Σp any vector X T can be expanded as | | p ∈ pM n X X = X0T + XiE , p p i|p i=1 and we have n 2 X 2 g (X ,X ) = X0 + Xi . p p p − i=1

The set of vectors Xp for which gp(Xp,Xp) > 0 are called spacelike. A Lorentzian metric deﬁnes a cone gp(Xp,Xp) = 0 at each point, the set of null vectors at the point p. 2. SPECIAL RELATIVITY 13

ψt(K0) ψt(p0)

K0 p0

Figure 11. Condition of symmetry on the family of observers.

2.2. General accelerated system. In the general situation of a given accelerated reference system spacetime is invariant under a 1-parameter group ψt whose orbits are timelike curves. An orbit t ψt(p0), with p0 fixed, is precisely 7→ the history of an observer initiating at the event p0. A family of observers defining a reference system satisfies a symmetry condition: Given a spacelike curve K, the arclength of the curve Kt = ψt(K) equals that of K0 = K for all t. We shall see that the metric takes the form 2 2 2 X i (2.26) ds = α θ + γ , θ = dt + Aidx − i and that (2.25) generalises in this setting to Z i (2.27) ∆t = Ai dx . C0 General Situation. Let us now assume we are in the situation where a 1- parameter group of diffeomorphisms ψt acts on the spacetime manifold such that (1) the orbits t ψt(p) are timelike curves, 7→ (2) ψt acts by isometries. The second condition ensures that the arclength of a spacelike curve K remains unchanged under the flow of ψt; cf. Fig. 11. Moreover, (2) implies ∗ (2.28) ψt g = g and thus d ∗ (2.29) V g = ψt g = 0 , −L dt t=0 where V is the generating vectorfield of ψt, a vectorfield tangential to the orbits: the velocity of the observers. Indeed, given a curve K : λ K(λ)(λ [a, b]) with 7→ ∈ K(a) = p0, K(b) = q0 the condition that the arclength of Kt : λ ψt(K0(λ)) is independent of t reads 7→ Z b q (2.30) L[Kt] = g(K˙ t(λ), K˙ t(λ)) dλ = L[K0]; a | | ˙ ˙ ˙ ˙ ∗ ˙ ˙ Since Kt(λ) = dψt K0(λ) and g(dψt K0(λ), dψt K0(λ)) = (ψt g)(K0(λ), K0(λ)) we conclude on (2.28)· for every t, and· thus by the· definition of the Lie derivative on (2.29). (C.f. note on pull-backs on pg. 17.) 14 1. THE EQUIVALENCE PRINCIPLE

ψt(p0)

Figure 12. Transversal hypersurface H0 to family of observers ψt(p0).

Let H0 be a transversal hypersurface for the set of orbits of ψt (each observer intersects H0 exactly once); see Fig. 12. Here H0 is a spacelike hypersurface in 3+1-dimensional Minkowski space. Every point p can be written as p = ψt(p0) for 1 2 2 some value t where p0 H0, and we can assign to p the coordinates (t, x , x , x ) 1 2 3 ∈ where (x , x , x ) are the spatial coordinates of p0 in Minkowski space. (We can think of H0 as the space of observers. A body with ﬁxed coordinates p0 is at rest in the accelerated system.) In these coordinates ∂ (2.31) V = (where x = (x1, x2, x3) is held ﬁxed) ∂t x and the condition (2.29) simply becomes

∂gµν (2.32) = 0 , ∂t µ ν where gµν = g(∂/∂x , ∂/∂x )(µ, ν = 0, 1, 2, 3) are the metric components in these coordinates (here also x0 = t). Then g takes the form (2.33) g = α2θ2 + γ − where α = α(x) is a positive function, θ is a 1-form 3 X i (2.34) θ = dt + A,A(x) = Ai(x)dx , i=1 and γ = γ(x) is a Riemannian metric in space, 3 X i j (2.35) γ(x) = γij(x)dx dx . i,j=1 As above, α signiﬁes the rate of clocks carried by the observers whose world lines are the group orbits (relative to the parameter t), and θ deﬁnes the local standard 2. SPECIAL RELATIVITY 15

p = ψt0 (p0)

p0 C0

Figure 13. A horizontal curve C that projects to the closed curve C0 in H0. of simultaneity for the given system of observers. It is a 1-form characterized as above – c.f. Prop. on pg. 8 – by the properties that (1) θ V = 1 (normalisation conditon) · (2) θ X = 0 if and only if X belongs to the hyperplane ΣV of all vectors orthogonal· to V . Let us now turn to the question of simultaneity as in the uniformly rotating case (c.f. pg. 9): Let C0 be a curve in H0 which starts at p0, parametrized by λ such that λ = 0 corresponds to p0. We want to ﬁnd a curve C through p = ψt0 (p0) which is horizontal and projects to C0. (See Fig. 13.) The parametric equations i i i i for C0 are given by x = x (λ), i = 1, 2, 3, and those for C by t = t(λ), x = x (λ), i = 1, 2, 3 because C projects to C0. The condition that C is horizontal means that 3 dt ∂ X dxi ∂ (2.36) C˙ = + ΣV , dλ ∂t dλ ∂xi ∈ i=1 which implies by the second property of θ that 3 dt X dxi (2.37) 0 = θ C˙ = + Ai(x) . · dλ dλ i=1

Consider now the case that C0 is a closed curve so that for some λ0 > 0: x(λ0) = x(0). The start and end point of C0 are thus both p0. By (2.37) we obtain an increment of 3 Z λ0 Z λ0 i dt X dx (2.38) ∆t = dλ = Ai x(λ) dλ , dλ − dλ 0 0 i=1 or Z (2.39) ∆t = A. − C0 16 1. THE EQUIVALENCE PRINCIPLE

(f(x), x) H00

(0, x) H0

Figure 14. Gauge transformations

Remark. In analogy to electromagnetic theory we may view A as the vector potential, and introduce the magnetic ﬁeld (2.40) B = dA.

(B is a 2-form, the exterior derivative of a 1-form.) Consider now a surface S0 in H0 whose boundary is C0: ∂S0 = C0. Then by Stokes’ theorem Z Z (2.41) B = A. S0 C0 We may thus interpret the integral (2.39) as the magnetic ﬂux through a surface enclosed by C0.

Remark. The result does not depend on the choice of the hypersurface H0. In 0 fact, another transversal hypersurface H0 is described as a graph t = f(x) in the coordinates constructed above (where H0 = t = 0 ); see Fig. 14. The coordinate transformation { } (2.42) t0 = t f(x) − leaves θ invariant provided the vector potential A is transformed according to (2.43) A0 = A + df , or in coordinates

0 ∂f (2.44) A (x) = Ai(x) + (x); i ∂xi then indeed (2.45) θ0 = dt0 + A0 = dt + A = θ . 2. SPECIAL RELATIVITY 17

The maps (2.43) are precisely the gauge transformations, which — as in electromagnetic theory — leave the magnetic ﬁeld unaltered: (2.46) B0 = B.

In particular, in view of (2.41), (2.38) does not depend on the choice of H0.

Finally note that γ measures arclength along C. Since C˙ ΣV we have that ∈ arclength of C w.r.t. g equals arclength of C0 relative to γ, which is

v 3 Z λ0 u i j u X dx dx (2.47) L[C] = t γij x(λ) dλ . dλ dλ 0 i,j=1

Note on pull-backs Let ψ : be a diffeomorphism, and α a 1-form on (in general the target). Recall thatM the→ M differential dψ(p) is a linear mapping of T Minto T , and we can pM ψ(p)M define the pull-back of α, ψ∗α: a 1-form on , by M (ψ∗α) X = α(ψ(p)) (dψ(p) X)(X T ) . · · · ∈ pM The metric g is a quadratic form in the tangent space at each point, and

(ψ∗g)(X,Y ) = g (dψ(p) X, dψ(p) Y )(X,Y T ). ψ(p) · · ∈ pM

Interpretation in Vector Bundles The construction of the curve C – as first discussed on page 9 – has a natural interpretation as a horizontal lift in vector bundles. In the situation (2.26) the spacetime manifold is = R (t, x), where we can think of ( , γ) as the manifold of observers. We haveB a projection× N 3π : which tells N B → N 1 us which observer q passes through a given event p: π(p) = q. The fiber π− (q) is a timelike curve in ∈: the N history of an observer q. V is the tangent vector field to these curves parametrizedB by t, and θ V = 1. Recall that the local simultaneous space at p is given by the null space of θ: · Σ = X T : θ(p) X = 0 p ∈ pB · Let now Xq be a tangent vector to at q. Observe that the projection π is surjective. Moreover dπ(p) is an isomorphism ofN Σ onto T . Therefore given any X T there p qN q ∈ qN exists a unique vector X] Σ such that p ∈ p dπ(p) X] = X ; · p q ] The vector Xp is called the horizontal lift of Xq to p; see Fig. 15. The horizontal lift is determined by the 2 conditions: (1) dπ(p) X] = X · p q (2) θ(p) X] = 0 . · p 18 1. THE EQUIVALENCE PRINCIPLE

1 ] π− (q) C0

p ] Xp

Σp

N Figure 15. Horizontal lift of vectors and curves.

In local coordinates (t, x) we have:

X i ∂ ] ]t ∂ X ]i ∂ Xq = X Tq and X = X + X Tp . q ∂xi q∈ N p p ∂t p p ∂xi p∈ B i i ] i i ] t By (1) we have (Xp) = Xq, and (Xp) is determined from (2): X t X i θ = dt + A (x)dxi : θ X] = X] + A (x)X] = 0 i · p p i p i i Therefore, ] ∂ X i ∂ X = A(q) Xq + X . p − · ∂t p q ∂xi p i With the horizontal lift of tangent vectors known, we can deﬁne the horizontal lift of curves: Let C0 : I a curve in the manifold of observers, I R an interval con- → N 1 ⊂ ] taining 0. Let q = C0(0), and p π− (q), then there exists a unique curve C0 such that ∈ ˙] ] C (λ) = (C˙0(λ)) (λ I) . 0 C0(λ) ∈ ] Then the curve C = C0 is precisely the solution to the problem under consideration; c.f. (2.37). That is to say the horizontal lift C(λ) = (t(λ), x(λ)) is the solution to the equation dt(λ) X dxi(λ) = A (C (λ)) , t(0) = 0 . dλ − i 0 dλ i In (2.39) we have shown that the horizontal lift C of a closed curve C0 in does not close in . This is a manifestation of curvature. N B

2.3. Conclusion. We have studied accelerated systems in comoving coordinates (t, p0) such that the histories of the corresponing observers at rest in the accelerated sytem are given by t (t, p0). We have seen that following simul- 7→ taneous events in the accelerated system starting at (t0, p0) along a closed curve λ p(λ) with p(0) = p(λ0) = p0 will – in general – take us back to (t1, p0) at 7→ 2. SPECIAL RELATIVITY 19 advanced coordinate time t1 with ∆t = t1 t0 > 0. We conclude that coordinates do not have physical meaning beyond labeling− spacetime events. The differential dt is not the actual time increment measured by the clocks of the accelerating system, and the level sets of t are not surfaces of simultaneity. Instead the rate of clocks in the accelerating system is α, and the relevant time differential α dt. Moreover θ determines a set of locally simultaneous events, and we have shown that it is impossible in general to find a global hypersurface of simultaneity. Simi- larly we have seen that dp0 does not measure elements of arclength in the space of an accelerating system, but instead it is γ. (E.g. for the uniformly rotating system dϕ0 does not give the arc element of circles when multiplied with r0, but instead p 2 2 it is r0dϕ0/ 1 ω r0.) We are led to conclude finally: what does have physical meaning (and− can be measured) are not coordinates but elements of arclength, namely the metrical ground form 3 2 X µ ν (2.48) ds = gµνdx dx . µ,ν=0 Remark. A study of the metrical ground form first appeared in 1828 in C.F. Gauss’ “General Investigations of Curved Surfaces” and stood at the be- ginning of the development of Riemannian Geometry. A. Einstein arrived at the above conclusion in 1915, and General Relativity is formulated in the context of Lorentzian Geometry.

Exercises (Receding observers in Minkowski spacetime): Consider a set of observers in uniform motion receding from a center. In other words consider a family of observers whose histories are straight lines in Minkowski space intersecting in a single point. These are rays contained in the future of the origin: 3 + 0 1 2 3 1+3 0 2 X i 2 0 I0 = (x , x , x , x ) ∈ R : −(x ) + (x ) < 0 and x > 0 . i=1 + (1) Depict the family of receding observers and the set I0 in diagram. + (2) At each p ∈ I0 construct the hyperplane of simultaneous events Σp to p as judged by the observer passing through p. + (3) Do there exist global hypersurfaces of simultaneity in I0 ? In other words, is the distribution of planes + ∆ := {Σp : p ∈ I0 } + integrable? Hint: Consider the hyperboloids Hτ given by 3 + 0 1 2 3 + 0 2 X i 2 2 Hτ := (x , x , x , x ) ∈ I0 :(x ) − (x ) = τ i=1 + + and show that for a given event p ∈ Hτ the tangent plane at p to Hτ coincides with Σp. + + (4) Consider the hyperboloid H1 . Assign to each event p ∈ H1 the polar normal coordinates (χ, θ, φ), where θ and φ are the polar and azimuthal angle of the 20 1. THE EQUIVALENCE PRINCIPLE

+ standard coordinates on the unit sphere, and χ is the geodesic distance on H1 from + (1, 0, 0, 0) to the event p. Verify that the induced metric on H1 is given by dσ2 = dχ2 + sinh2 χ(dθ2 + sin2 θ dφ2). + Show that the Gauss curvature of H1 is equal to −1. + (5) Show that the Minkowski metric can be written in I0 as ds2 = −dτ 2 + τ 2dσ2 . + + That is to say I0 can be foliated by the hyperboloids Hτ and the induced metric on + 2 2 + Hτ is precisely τ dσ . Also show that the Gauss curvature of Hτ is equal to K = −τ −2 . (6) Consider two receding observers. Suppose a light signal is sent from one to the other. Let v = tanh χ be the velocity of the observer emitting the light as measured in the rest frame of the observer receiving the signal. Show that a redshift is observed with ω e = eχ , ωr

where ωe and ωr are the frequencies of the emitted and the received light respectively.

3. Geodesic correspondence In view of the equivalence principle our conclusions for accelerated systems of reference in the absence of a gravitational field apply to stationary reference frames in the presence of a gravitational field. In particular we know that the metrical ground form (2.48) in the presence of a gravitational field is non-trivial. Since we may change to a freely falling reference frame where the gravitational force is eliminated locally we are left with the differential of the gravitational field. In the following we shall find the laws that govern the metric (2.48) in the presence of a gravitational field by an analogy of tidal motions in Newtonian gravity and geodesic motion of nearby particles on Lorentzian manifolds (with curvature).

3.1. Tidal equation. Recall our discussion in Section 1.1 of tidal forces in the framework of the Newtonian theory. We took a freely falling particle as a reference particle and translated the origin of a new coordinate system at each time to the instantaneous position of this particle.

x0(t): motion of reference particle x(t): motion of arbitrary nearby particle Then we have seen that the displacement

(3.1) y(t) = x(t) x0(t) − to a nearby freely falling particle satisﬁes to ﬁrst order in y (small displacements) the tidal equation (1.6): 2 d y 2 (3.2) = ψ x0(t) y(t) dt2 −∇ · 3. GEODESIC CORRESPONDENCE 21

In matrix notation this simply reads

2 2 i 3 d y d y X j (3.3) = M y , i.e. (t) = Mij(t) y (t) , dt2 − dt2 − j=1 where M is the Hessian matrix of the gravitational potential: 2 ∂ ψ (3.4) Mij(t) = x0(t) . ∂xi∂xj Suppose that the nearby particle is at rest relative to the reference particle at time t = 0, i.e. dy (3.5) = 0 . dt t=0 Since the tidal equation is linear we have (3.6) y(t) = A(t) y(0) for some matrix A(t) depending on t. Clearly, A(0) = 1 (identity matrix), and the initial condition (3.5) for y translates into the analogous conditon on A: dA (3.7) (0) = 0 dt And inserting (3.6) into the tidal equation (3.3) yields of course an identical equation of motion for the matrix A: d2A (3.8) (t) = M(t) A(t) dt2 − Let us introduce a matrix θ(t) by: dA (3.9) θ = A−1 dt Then, dθ d2A dA d = A−1 + A−1 dt dt2 dt dt d2A dA dA (3.10) = A−1 A−1 A−1 dt2 − dt dt d2A = A−1 θ2 , dt2 − and by the matrix tidal equation (3.8) we obtain the following equation of motion for θ: dθ (3.11) = θ2 M dt − − (3.12) θ(0) = 0 We have not yet exploited the fact that the Hessian is symmetric: (3.13) M˜ = M 22 1. THE EQUIVALENCE PRINCIPLE

(For an arbitrary matrix A we denote by A˜ the transpose of A: A˜ij = Aji.) Taking the transpose of (3.11) we obtain as well dθ˜ (3.14) = θ˜2 M dt − − We now decompose (3.15) θ = σ + ω into its symmetric part σ and its antisymmetric part ω: 1 1 (3.16) σ = (θ + θ˜) ω = (θ θ˜) 2 2 − Substracting (3.14) from (3.11) M cancels and we obtain the following equation for ω: dω 1 1 = (θ2 θ˜2) = (σ + ω)2 (σ ω)2 (3.17) dt −2 − −2 − − = (σω + ωσ) − This is a linear homogeneous equation for ω considering σ as given. It follows that if ω vanishes initially, then it vanishes for all time. But here θ(0) = 0, therefore σ(0) = 0 as well as ω(0) = 0. It follows that with our initial conditions ω vanishes identically for all time, therefore (3.18) θ = σ is symmetric. It follows that 2 X 2 2 (3.19) tr(θ ) = tr(θθ˜ ) = (θij) = θ 0 , | | ≥ i,j and taking the trace of (3.11) we obtain d (3.20) tr θ = θ 2 tr M. dt −| | − Note that 2 X ∂ ψ (3.21) tr M(t) = x0(t) = ∆ψ x0(t) , i2 i ∂x and according to the Newtonian theory of gravitation: (3.22) ∆ψ = 0 in vacuum. The meaning of (3.20) becomes evident by looking at the evolution of volume. Consider the volume of a parallelepiped spanned by linearly independent displacements (y1, y2, y2) of nearby particles: Ω(y1, y2, y3). (See Fig. 16.) Let us assume that three nearby particles are initially at rest and at positions (E1,E2,E3): an orthonormal (positive) basis. Then Ω(E1,E2,E3) = 1. Recall that the displacements at time t are then obtained using the linear map A(t): deformation matrix. 3. GEODESIC CORRESPONDENCE 23

y3(t)

E3 y2(t)

y1(t) E1

Figure 16. Volume of parallelepiped.

In fact, yi(t) = A(t)Ei, i = 1, 2, 3. Therefore, we obtain for the volume of the corresponding parallelepiped at time t:

Ω(t) = Ω(y1(t), y2(t), y3(t)) = Ω(A(t)E1,A(t)E2,A(t)E3) (3.23) = det A(t) Ω(E1,E2,E3) = det A(t) Ω(0) = det A(t) This formula holds for any volume form Ω. The volume of a parallelepiped is in fact the unique volume form on a vector space with inner product; c.f. note on volume form on pg. 25. Now consider the function (3.24) µ(t) = log Ω(t) = log det A(t) . Since, by differentiation of a determinant, d dA (3.25) det A(t) = det A(t) tr A−1 dt dt we have d (3.26) µ(t) = tr θ(t) , µ(0) = 0 , dt or Z t (3.27) µ(t) = tr θ(t0)dt0 . 0 Moreover, by (3.12) θ(0) = 0 we also have dµ (3.28) (0) = 0 , dt and 2 d µ d (3.29) (0) = tr θ = tr M . dt2 dt t=0 − t=0 Remark. We have shown that tidal forces deform shapes in such a way that volumes remain the same up to third order. Note that while equality to first order is simply a consequence of the fact that all particles constituting the shape are initially at rest, the equality for volumes in second order is precisely the equation for the gravitational field in vacuum, namely the Poisson equation (3.22) for the Newtonian potential. 24 1. THE EQUIVALENCE PRINCIPLE

Elevator

7→

Earth

Figure 17. Elevator experiment.

Elevator experiment Tidal forces are easily visualised in the following experiment: Consider a distribution of masses in a freely falling elevator in the gravitational ﬁeld of the Earth. See Fig. 17. From the point of view of a freely falling observer at the center of the elevator the masses accelerate towards her horizontally and away from her vertically. An initially spherical distribution of nearby masses surrounding the observer is deformed to a prolate spheroid of the same volume. The equality of the volumes is precisely the equation of the gravitational ﬁeld.

Note on volume forms in vector spaces In this note we discuss the notion of a volume form Ω in a n-dimensional vector space, and show that (3.17) holds in general. Moreover, we prove that if Vn is endowed with an inner product, then Ω is in fact unique. n n Let X1,...,Xn be vectors in an n-dimensional space V . A volume form Ω in V is a n-linear form in Vn which is totally antisymmetric and nondegenerate. That is to say Ω(X1,...,Xn) depends linearly on Xi if we ﬁx all vectors X1,...,Xn except Xi, and changes sign if we interchange two vectors: Ω(...,X ,...,X ,...) = Ω(...,X ,...,X ,...) . i j − j i n Moreover it being nondegenerate means there is a n-tuplet of vectors (X1,...,Xn) in V such that Ω(X ,...,X ) = 0 . 1 n 6 If Vn is endowed with an inner product , then this automatically deﬁnes a volume n h· ·i n form on V . For, we can then choose a basis (E1,...,En) for V which is orthonormal: E ,E = δ . h i ji ij We then set

Ω(E1,...,En) = 1 . 3. GEODESIC CORRESPONDENCE 25

However, an inner product only determines an orthonormal basis up to rotation. Let (E10 ,...,En0 ) be another such orthonormal basis. We must verify that also Ω(E10 ,...,En0 ) = 1. (Otherwise the definition does not make sense.) To be precise we shall show that a volume form is uniquely determined by the inner product and orientation. Let us first return to Ω(X1,...,Xn). We can expand any vector Xi in the basis (E1,...,En): n X j Xi = (Xi) Ej : i = 1, . . . , n . j=1 j th The coefficients (Xi) : j = 1, . . . , n form the i column of a matrix X. Then Ω(X1,...,Xn) is determined. Therefore a volume form Ω is uniquely determined by the basis (E1,...,En). Let us verify that in the case n = 2: 1 2 1 2 X1 = (X1) E1 + (X1) E2 X2 = (X2) E1 + (X2) E2 . By multilinearity 1 1 1 2 Ω(X1,X2) =(X1) (X2) Ω(E1,E1) + (X1) (X2) Ω(E1,E2) 2 1 2 2 + (X1) (X2) Ω(E2,E1) + (X1) (X2) Ω(E2,E2) =(X )1(X )2 (X )2(X )1 = det X 1 2 − 1 2 This is in fact the definition of the determinant. Given a linear transformation A of Vn there is a matrix A which represents the linear map in the basis (E ,...,E ). Since A E is a vector, it can be expanded in the basis 1 n · i (E1,...,En): n X j AEi = Ai Ej . j=1 j The coefficients Ai constitute the corresponding matrix, and thus

Ω(AE1, . . . , AEn) = (det A)Ω(X1,...,Xn) .

Given a basis (E1,...,En) we obtain another basis (E10 ,...,En0 ) by transformation with an element of the general linear group GL(n, R): n X j E0 = G E : i = 1, . . . , n where det G = 0 . i i j 6 j=1 n The set of all bases in V has two components, one obtained from (E1,...,En) with det G > 0, the other with det G < 0. Selecting one of the two classes, say the one contain- ing (E1,...,En) which one then refers to as positive bases, corresponds to introducing an orientation on Vn. Hence Ω(E10 ,...,En0 ) > 0 on all positive bases. Consider transformations O which leave the inner product invariant: OX,OY = X,Y h i h i We can represent O in a positive orthonormal basis (E1,...,En), then the columns of the matrix n X j Ei0 = Oi Ej : i = 1, . . . , n j=1 26 1. THE EQUIVALENCE PRINCIPLE are another orthonormal basis:

E0,E0 = O E ,O E = E ,E = δ h i ji h · i · ji h i ji ij This condition reads OO˜ = 1 , the defining condition of an orthogonal matrix. (Here O˜ is the transpose of O.) Therefore (det O)2 = 1, or det O = 1, with det O = +1 if the primed basis is positive. ± We defined Ω(E1,...,En) = 1. The definition makes sense because if (E10 ,...,En0 ) is another positive orthonormal basis then

Ω(E10 ,...,En0 ) = (det O)Ω(E1,...,En) = 1 , because det O = 1 . It follows that if Vn is a oriented vector space with an inner product, then Vn has a unique volume form Ω.

Exercises 1. (Ratio of volumes): Consider a freely falling reference particle in a gravitational field in vacuum. Suppose the particle is surrounded by a sphere of volume Ω which is initially at rest relative to the reference particle. Show that in fact up to fourth order the volume of the sphere does not change in time: Ω(t) = 1 + O(t4) . Ω(0) What is the obstruction to the equality of volumes to higher order in time? 2. (Convergence of particles): Consider a freely falling reference particle in vacuum as in Exercise 1. Assume now in addition that the Hessian matrix of the gravitational potential does not vanish initially in the vicinity of the reference particle. Recall that the displacement of a nearby particle y(t) is given in terms of its initial displacement y(0) (we assume that the particle is initially at rest relative to the reference particle) by y(t) = A(t)y(0) , where A is the deformation matrix. Show that there is an initial displacement y(0) 6= 0 and a time t∗ > 0 such that y(t∗) = 0. Give a physical interpretation of this result. Hints. Recall the definition of the strain matrix dA θ = A−1. dt To solve Exercise 2 one may proceed as follows: (1) Show that under the assumption that the Hessian of the gravitational field does not vanish identically initially we have 3 d tr θ < 0 . dt3 t=0 (2) Deduce from the tidal equation the following ordinary differential inequality: d 1 tr θ ≤ − tr θ2 . dt 3 (Hint: Decompose θ into its diagonal and its traceless part.) 3. GEODESIC CORRESPONDENCE 27

(3) Show that there exists a time t∗ > 0 such that tr θ → −∞ as t → t∗. (Hint: Derive an equation for u = −1/ tr θ.) (4) Show that there exists a tangent vector y(0) 6= 0 such that y(t∗) = 0. (Hint: What is the implication of (3) for A(t) as t → t∗?)

3.2. Theory of Geodesics. According to the equivalence principle the gravitational force is eliminated in a freely falling reference frame. A reference system is given by a family of timelike curves in the spacetime. We have seen — in our discussion of accelerated reference frames in Section 2.2 — that in the presence of a gravitational field the spacetime is curved; i.e. it is not Minkowski space as in Special Relativity, but a Lorentzian manifold with curvature (as we shall discuss in some detail in the next Chapter). Which timelike curves in the spacetime are described by freely falling bodies? Hypothesis: (Einstein) A freely falling test particle in a (non-trivial) gravitational field corresponds to a timelike geodesic in a spacetime with curvature. Remark. It is an important feature of the theory that this hypothesis can be recovered from the equations of motion. We shall return to this question in our discussion of the Einstein equations in the presence of matter. It is thus a consequence of the equivalence principle that the gravitational force is an aspect of the spacetime itself. The aim of this section is to relate Newton’s law for the gravitational potential to an equation for the spacetime curvature. Since in a freely falling reference frame only tidal forces can be measured it can only be the differential of the gravitational field that contributes to spacetime curvature. To derive the law — the Einstein vacuum equations — we shall look at a family of nearby timelike geodesics, corresponding to the trajectories of nearby freely falling particles in the Newtonian theory. 3.2.1. Jacobi equation. Consider a timelike reference geodesic Γ0 (a freely falling reference test particle) in a spacetime ( , g). Let Γ0 be parametrized by arc length t (proper time), and let T be theM unit future directed (parallel) tangent vector field along Γ0:

(3.30) T T = 0 . ∇ Let O be an arbitrarily chosen origin on Γ0 where we set t = 0. Consider a spacelike vector X at O orthogonal to T , and let K0 be the spacelike geodesic through O with initial tangent vector X at O. Then X extends along K0 as the p tangent vector ﬁeld of K0, and the magnitude of X, X = g(X,X), is constant | | along K0. Here K0 is parametrized by λ = X s, where s is the arclength along | | K0. Let us now extend T along K0 by parallel transport; we obtain a unit timelike future directed vectorﬁeld along K0, which is orthogonal to X:

(3.31) g(T,X) = 0 along K0 . 28 1. THE EQUIVALENCE PRINCIPLE

Γλ

Γ0(t)

Γ0

T T

O X K0 K0(λ)

Figure 18. Construction of a timelike geodesic congruence.

At each point K0(λ) along K0, we draw the timelike geodesic towards the future with initial tangent vector T . We call this geodesic Γλ, and the family of curves a timelike geodesic congruence; see Fig. 18. Extend T along Γλ to be the tangent vector field to Γλ; since the tangent vector is parallel transported along a geodsic, we have (3.32) g(T,T ) = 1 . − Remark. The timelike curves Γλ correspond to the histories of nearby freely falling bodies in the Newtonian theory. Remark. In principle, we may observe three different kinds of behaviour: the distance to the nearby geodesic remains constant; the distance decreases; or the distance increases. As we shall describe in detail these cases are related to zero, positive, or negative (sectional) curvature, respectively; see Fig. 19. We define a timelike 2-dimensional (ruled) surface [ (3.33) S = Γλ , λ and for each t 0 a curve Kt S by: ≥ ⊂ (3.34) Kt(λ) = Γλ(t) . Then extend the vectorfield X to the surface S to be the tangent field of the curves Kt. The parameters t, λ can be thought of as coordinates on S; with respect to these coordinates: ∂ ∂ (3.35) T = ,X = , ∂t ∂λ and thus (3.36) [T,X] = 0 . 3. GEODESIC CORRESPONDENCE 29

Plane Sphere Hyperboloid

Figure 19. Three characteristic cases for geodesic congruences: Constancy of distance to nearby geodesics (zero curvature); De- crease of distance (positive curvature); Increase of the distance to a nearby geodesic (negative curvature).

∂ φ ∂t

Tp ∂ (t, λ) ∂λ Xp p

Γ0

Figure 20. Push-forward of coordinate vectorﬁelds to S.

Remark. More precisely, we have a mapping

φ :(t, λ) φ(t, λ) = Γλ(t) = Kt(λ); 7→ see Fig. 20. If p = φ(t, λ) then ∂ ∂ Tp = dφ(t, λ) ,Xp = dφ(t, λ) . · ∂t (t,λ) · ∂λ (t,λ) In other words the vectorﬁelds T and X on S are the push-forward of coordinate vectorﬁelds in the (t, λ) plane, ∂ ∂ T = φ∗ ,X = φ∗ . ∂t ∂λ Therefore, h ∂ ∂ i h ∂ ∂ i [T,X] = φ∗ , φ∗ = φ∗ , = 0 . ∂t ∂λ ∂t ∂λ 30 1. THE EQUIVALENCE PRINCIPLE

Prop 3.1. We have that T is orthogonal to X everywhere on S: g(T,X) = 0 on S.

Proof. We know this is true along K0. Diﬀerentiating the function g(T,X) along Γλ we obtain T g(T,X) = g( T T,X) + g(T, T X) . ∇ ∇ Since the integral curves of T are geodesics the ﬁrst term vanishes by (3.30). Now, by (3.36): T X X T = [T,X] = 0 ; ∇ − ∇ therefore 1 g(T, T X) = g(T, X T ) = X g(T,T ) = 0 , ∇ ∇ 2 since g(T,T ) = 1 is a constant function on S. Hence − T g(T,X) = 0 , which says that g(T,X) is constant along Γλ. Since the integral curves of T initiate on K0 where g(T,X) = 0, it follows that

g(T,X) = 0 along Γλ . Remark. Here we have used that the connection in general relativity is sym- metric3 and compatible with the metric. While having a symmetric connection means X Y Y X = [X,Y ] , ∇ − ∇ for any vectorﬁelds X, Y , metric compatibility is equivalent to Z g(X,Y ) = g( Z X,Y ) + g(X, Z Y ) , ∇ ∇ for any vectorﬁelds X, Y , Z.

The curves Kt for t > 0 are not in general geodesics. The arclength of Kt between the parameter values 0 and λ is: Z λ 0 0 L Kt[0, λ] = X Kt(λ )dλ . 0 | |

The magnitude X Γ0(t) thus measures the deviation of nearby geodesics. Therefore we have to study| | (3.37) X(t) = X : the displacement at time t . |Γ0(t) Remark. The vectorﬁeld (3.37) corresponds precisely to the displacement (3.1) in the tidal equation.

T X: First rate of change of the displacement. ∇2 X: Second rate of change of the displacement. ∇T 3this is the diﬀerence to gauge theories, not the metric. 3. GEODESIC CORRESPONDENCE 31

Since T X = X T and X T = 0 along K0 (T was parallel transported along ∇ ∇ ∇ K0), we have

(3.38) T X t=0 = 0 , ∇ | which corresponds in the Newtonian theory to the condition that the nearby test particles are initially at rest relative to the reference particle. As regards the second rate of change we have 2 X = T T X = T X T. ∇T ∇ ∇ ∇ ∇ Now recall that the deﬁnition of curvature is equivalent to:4

(3.39) [ X , Y ]Z Z = R(X,Y )Z ∇ ∇ − ∇[X,Y ] Since 2 X = T X T = X T T + [ T , X ]T = [ T , X ]T, ∇T ∇ ∇ ∇ ∇ ∇ ∇ ∇ ∇ we obtain, by (3.39), in view also of the antisymmetry of the curvature,

[ T , X ]T = T + R(T,X) T = R(X,T ) T, ∇ ∇ ∇[T,X] · − · the so-called Jacobi equation:

(3.40) 2 X = R(X,T ) T ∇T − · This is a second order equation for the vectorfield X along Γ0. The displacement field X to nearby geodesics is called a Jacobi field 5. We have derived the Jacobi equation, or Geodesic deviation equation, by variation through geodesics. We express (3.40) in components relative to a comoving orthonormal frame along Γ0. First choose an orthonormal basis (E1,E2,E3) for Σ0, the hyperplane in TO consisting of all vectors orthogonal to T . In particular X(0) Σ0. In general,M for every t 0 we denote by ∈ ≥ n o (3.41) Σt = Y T : g(Y,T ) = 0 ∈ Γ0(t)M the hyperplane in the tangent space to at Γ0(t) defined by all vectors orthogonal M to T . Now recall from Prop. 3.1 that X is orthogonal to T along Γ0, and thus

(3.42) X(t) Σt (t 0) . ∈ ≥ We can propagate the vectors Ei : i = 1, 2, 3 along Γ0 by parallel transport, i.e. we deﬁne vectorﬁelds Ei : i = 1, 2, 3 along Γ0 such that

(3.43) T Ei = 0 : i = 1, 2, 3. ∇ Since the inner products remain unchanged we have an orthonormal basis for Σt at each t 0 which we denote by (E1(t),E2(t),E3(t)). Complemented with ≥ E0 = T we obtain an orthonormal basis (Eµ : µ = 0 ..., 3) for T at each Γ0(t)M

4C.f. notes on connections and the curvature tensor on pg. 39. 5 In fact, we have restricted ourselves to the case where K0 is orthogonal to Γ0 at O; X is then called a normal Jacobi ﬁeld. 32 1. THE EQUIVALENCE PRINCIPLE t 0, i.e. along Γ0. But since X(t) Σt we can expand X(t) in the basis ≥ ∈ (Ei(t): i = 1, 2, 3): 3 X i (3.44) X(t) = X (t) Ei . i=1 Remark. The quadruple (t, X1,X2,X3) deﬁnes cylindrical normal coordinates in a neighborhood of Γ0. We say that p = expΓ0(t)(X(t)) if p lies on the spacelike geodesic from Γ0(t) with initial tangent vector X(t) Σt at arclength X(t) . We then assign to p the coordinates (t, X1,X2,X3),∈ and p lies in the |geodesic| hypersurface n o Ht = exp X : X Σt = exp Σt . ∈ We have that τ (t, λ1τ, λ2τ, λ3τ) are spacelike geodesics for each (λ1, λ2, λ3) 3 7→ ∈ R , while τ (τ, 0, 0, 0) is a timelike geodesic; c.f. note on pg. 37 on Riemannian 7→ normal coordinates at a point. Finally note that the curves Kt constructed above do not lie in Ht in general.

Take the covariant derivative (now everything is evaluated on Γ0): 3 XdXi (3.45) T X = Ei , ∇ dt i=1 because T Ei = 0 (parallel transport), and ∇ 3 2 i 2 Xd X (3.46) X = Ei . ∇T dt2 i=1 On the other side, 3 X i (3.47) R(X,T ) T = X Ri0 E0 . · · i=1 Consider now the vector

(3.48) Yi = Ri0 E0 = R(Ei,T ) T, · · which can be expanded in the basis (Eµ : µ = 1,..., 3): 3 0 X j (3.49) Yi = Yi E0 + Yi Ej . j=1 We find the coefficients by taking inner products with the orthonormal basis vectors, in fact 0 j (3.50) Y = g(Yi,E0) , and Y = g(Yi,Ej) . i − i Recall the definition of the curvature tensor: (3.51) R(W, Z, X, Y ) = g(W, R(X,Y ) Z) . · 3. GEODESIC CORRESPONDENCE 33

Thus, 0 (3.52) Y = R(T,T,Ei,T ) = 0 . i − Remark. Here we have used the symmetry property (3.53) R(X,Y ) Z = R(Y,X) Z, · − · which is clear from the deﬁnition (3.39). More properties of the curvature tensor are discussed in the note on page 39. Furthermore, j (3.54) Yi = R(Ej,T,Ei,T ) . Remark. Note that the expression (3.54) is symmetric in (i, j) by the pair symmetry of the curvature tensor; c.f. Notes on page 39. The term on the right hand side of the Jacobi equation is therefore 3 3 X jX R(X,T ) T = X Ri0j0Ei . · j=1 i=1 Thus equating the two sides of (3.40) yields

2 i 3 d X (t) X j (3.55) = Ri0j0 X (t) . dt2 − j=1 3.2.2. Einstein vacuum equation. In conlusion we have shown that in a paral- lelly propagated orthonormal frame (E1,E2,E3) along the reference geodesic Γ0, P i i.e. T Ei = 0, the components of the displacement vector X(t) = i X (t)Ei(t) to a∇ nearby timelike geodesic satisfy the equation

2 i 3 d X (t) X j (3.56) = Mij(t) X (t) , dt2 − j=1 where M is the symmetric matrix

(3.57) Mij = Ri0j0 , or indeed the matrix valued function M(t) along Γ0:

(3.58) Mij(t) = R(Ei,T,Ej,T ) . Γ0(t) Thus the Jacobi equation (3.56) is identical in form to the Newtonian tidal equation (3.3). In the Newtonian theory we have (3.59) tr M = ∆ψ = 0 : in vacuum . The vacuum equations of the relativistic theory (of general relativity) should therefore be (3.60) tr M = 0 . 34 1. THE EQUIVALENCE PRINCIPLE

3.2.3. First variation equation. We conclude our comparison of the tidal equation to the Jacobi equation with a discussion of velocities. Recall that we have extended T from O along K0 by parallel transport. This corresponds to vanishing initial velocity, 3 X dXi dXi (3.61) 0 = X T O = T X O = (0) Ei = = 0 : i = 1, 2, 3 . ∇ | ∇ | dt ⇒ dt t=0 i=1

The curve K0 here lies in the geodesic spacelike hypersurface H0. This is a special case of a orthogonal hypersurface to Γ0. Let now more generally Γ0 be a timelike geodesic normal to a spacelike hypersurface M (not necessarily geodesic). The natural way to extend T from O to M in this case is to define T at every point p M to be the unit future-directed (timelike) normal to M at p. Then let Γp be the∈ timelike geodesic normal to M at p. We define the second fundamental form k of a spacelike hypersurface M with unit (future-directed timelike) normal T to be the bilinear form in TpM at each p M defined by: ∈ (3.62) k(X,Y ) = g( X T,Y ): X,Y TpM ∇ ∀ ∈ Remark. In fact, k is symmetric. So k is a quadratic form in TpM at each p M. To show this, extend X, Y to vectorfields on M tangential to M, then: ∈ k(X,Y ) = g(T, X Y ) , since g(T,Y ) = 0 . − ∇ Similarly with X and Y interchanged, so taking the difference,

k(X,Y ) k(Y,X) = g(T, X Y Y X) = g(T, [X,Y ]) = 0 , − − ∇ − ∇ − since [X,Y ] is also tangential to M.

] We deﬁne a linear transformation k in TpM at each p M by: ∈ ] (3.63) g(k X,Y ) = k(X,Y ): X,Y TpM, · ∀ ∈ in other words ] (3.64) k X = X T. · ∇ In particular at O M: ∈ ] (3.65) T X O = X T O = k X O ∇ | ∇ | · | is the initial velocity at t = 0. With respect to the orthonormal basis (Ei : i = 1, 2, 3) we deﬁne

(3.66) kij = k(Ei,Ej) . Then 3 ] X (3.67) k Ei = kijEj , · j=1 3. GEODESIC CORRESPONDENCE 35

] since g(k Ei,Ej) = kij, and we obtain

3 ] X i (3.68) k X O = kijX Ej O . · | | i,j=1 Therefore

i 3 dX X j (3.69) = kij O X t=0 . dt t=0 | | j=1

We now deﬁne – similarly to Kt above – for each t (at least in a neighborhood of O) a spacelike hypersurface Mt by n o (3.70) Mt = Γp(t): p M ∈

The vectorfield T extends to spacetime (at least a neighborhood of Γ0) to be the tangent vector field to the timelike geodesic congruence Γp. Then T remains orthogonal to each spacelike hypersurface Mt. Remark. We have already made this observation in the proposition on pg. 30, for its proof only relies on the fact that the geodesic tangent field T is orthogonal to M initially. Indeed, given any vectorfield X on M tangential to M, we denote by φt the flow generated by T (i.e. φt(p) = Γp(t) for all p M), and define ∈ (3.71) Xq = dφt Xp , where q Mt is given by q = φt(p), p M. · ∈ ∈ The vectorfield X extended this way to spacetime is tangential to each Mt and satisfies [T,X] = 0. Then 1 (3.72) T g(T,X) = g( T T,X) + g(T, T X) = X g(T,T ) = 0 , ∇ ∇ 2 which yields that if g(T,X) = 0 initially then it vanishes on each Mt, thus in spacetime.

Since T is the unit future-directed timelike normal to each Mt we have that at each t 0: ≥ i 3 dX X j (3.73) (t) = kij(t)X (t) , dt j=1 where kij(t) are the components of the second fundamental form of Mt at the point Γ0(t) with respect to the parallel transported frame (E1,E2,E3). Recall now from our discussion of the tidal equation, the dependence on the initial data being linear, we can write

3 i X j (3.74) X (t) = Aij(t) X (0) , j=1 36 1. THE EQUIVALENCE PRINCIPLE r q

T pM Figure 21. Polar normal coordinates. and the deformation matrix A satisﬁes the same equation as in the Newtonian theory; c.f. (3.8): d2A (3.75) = MA. dt2 − Recall the deﬁnition of the rate of strain matrix dA (3.76) θ = A−1 . dt We have, from (3.74) and (3.75):

i 3 3 dX X dAij j X j (3.77) (t) = (t)X (0) = θij(t) X (t) . dt dt j=1 j=1 The comparison to (3.73) yields

(3.78) θij = kij which is also called the ﬁrst variation equation.

Introductory remarks on normal coordinates and curvature Consider a (n-dimensional) Riemannian manifold ( , γ). Let us fix a point p and consider all tangent vectors N of unit length N 2M= γ(N ,N ) = 1 at p: ∈ M p | p| p p n 1 2 S − = N T : N = 1 p p ∈ pM | p| n 1 Any vector X T can written as X = r N , for some r 0 and N S − . Each p ∈ pM p p ≥ p ∈ p Xp defines a geodesic, Xp being the initial velocity at p. Recall the velocity of a geodesic is constant. We assign to a point q in a neighborhood of p the polar normal coordinates (r, Np) with origin at p if q is the point at parameter value 1 along the geodesic from p with initial velocity Xp = rNp. The point q is therefore defined only by Xp. (See also figure 21.) To obtain rectangular normal coordinates we choose an orthonormal basis (E ,...,E ) for T . Then any X T has a unique expansion 1 n pM p ∈ pM n X i Xp = x Ei . i=1 3. GEODESIC CORRESPONDENCE 37

Then we assign to q the rectangular coordinates (x1, . . . , xn). Note that q sX sX r = X = γ(X ,X ) = δ xixj = (xi)2 . | p| p p ij i,j i Hence the distance from the origin is the same expression as in Euclidean geometry. We now consider the case where is 2-dimensional and introduce the concept of M Gauss curvature. In two dimensions Np is an element of the unit circle and characterized by a single angle ϕ; the polar normal coordinates are thus simply (r, ϕ). The Lemma of Gauss says that the geodesic rays corresponding to the lines of constant ϕ are orthogonal to the geodesic circles Sr: the set of all points which are at distance r from a the point p. The metric takes the form: ds2 = dr2 + R2(r)dϕ2 .

We see that R(r)dϕ is the element of arc length along the geodesic circles Sr. The condition that the metric is locally Euclidean at p is simply R(r) ∂R 1 as r 0 or (0, ϕ) = 1 . r → → ∂r The Gauss curvature K(p) is then defined in polar normal coordinates by ∂2R + KR = 0 . ∂r2 Alternatively consider the circumference C(r) of the geodesic circle Sr: Z 2π C(r) = R(r, ϕ)dϕ . 0 Then another definition of the Gauss curvature is given by d Z Z r Z 2π 2π = K = KR drdϕ , − dr Dr 0 0 where Dr is the geodesic disc of radius r: the set of all points at distance less or equal r from p. We can expand the right hand side to leading order, Z 2 K = Kp πr + higher order terms, Dr and on the left hand side similary, a C(r) = 2πr + 3 r3 + higher order terms. 6 Therefore the Gauss curvature appears as the coefficient to the third order correction to the circumference, K = a . p − 3 Finally let us mention Gauss’ theorem for a geodesic triangle: If α, β, γ are the inner angles of a geodesic triangle enclosing the area T (see figure 22) then ⊂Z M α + β + γ π = K. − T The generalisation thereof is the Gauss-Bonnet theorem for closed surfaces S: Z K = 2πχ , S where χ = 2(1 g) is the Euler characteristic, and g the genus of S (g = 0 for a sphere, g = 1 for a torus).− 38 1. THE EQUIVALENCE PRINCIPLE

T γ α

Figure 22. A geodesic triangle.

Notes on connections and the curvature tensor in general relativity Recall that the definition of curvature is equivalent to R(X,Y ) Z = [ , ]Z Z. · ∇X ∇Y − ∇[X,Y ] We shall see in the next Chapter that R(X,Y ) has the cyclic property R(X,Y ) Z + R(Y,Z) X + R(Z,X) Y = 0 , · · · and satisfies the Bianchi identity ( R)(Y,Z) + ( R)(Z,X) + ( R)(X,Y ) = 0 . ∇X ∇Y ∇Z These identities come from the Jacobi identity for linear operators. The cyclic property follows by considering vectorfields as linear operators on functions, X : f X f , 7−→ · while the Bianchi identity follows by considering vectorfields as linear operators on vectorfields: X : Y Y. 7−→ ∇X Here denotes the covariant derivative, a linear operator on vectorfields with the properties:∇

(1) X (fY ) = X(f)Y + f X Y . (2) ∇ Y = f Y . ∇ ∇fX ∇X In general relativity it comes from a connection which is symmetric and compatible with the metric. The two conditions are:

(1) X Y Y X = [X,Y ] (symmetry) (2) Zg∇ (X,Y− ∇) = g( X,Y ) + g(X, Y ) (metric compatibility) ∇Z ∇Z The cyclic property and the Bianchi identity only depend on the connection being symmetric, and do not rely on the metric property. However, metric compatibility implies the antisymmetry of the curvature tensor in the first two slots: R(W, Z, X, Y ) = R(Z, W, X, Y ) − (Note that the antisymmetry in the last two slots is an immediate consequence of the definition above.) Recall here the definition of the Riemann curvature tensor, a fully covariant tensor of fourth rank: R(W, Z, X, Y ) = g(W, R(X,Y ) Z) . · We can interpret this antisymmetry of the curvature tensor as follows: 3. GEODESIC CORRESPONDENCE 39

The curvature Rp at a point p in the spacetime is an antisymmetric bilinear form in the tangent space T to at p with values in theM space of linear transformations of pM M Tp . Let use denote in general by (U, V ) the space of linear maps from a vector space U toM a vector space V . So if X , Y L T then p p ∈ pM R (X ,Y ) (T , T ) p p p ∈ L pM pM depends linearly on Xp, Yp, and Rp(Xp,Yp) = Rp(Yp,Xp). With Xp, Yp both fixed R (X ,Y ) Z is an element of T depending− linearly on Z . The curvature R of the p p p · p pM p spacetime ( , g) is an assignment p Rp at each p . If X, Y , Z are vectorfields on , then R(MX,Y )Z is the vectorfield7→ whose value at ∈p M is R (X ,Y ) Z . M ∈ M p p p · p Consider, by comparison, a 2-form ω on , that is an assignment p ωp of an antisymmetric bilinear real valued form in T M. Then if Π is a 2-dimensional7→ plane in pM Tp , and (E1,E2) an orthonormal basis for Π, and (X1,X2) any pair of vectors in Π, weM have 1 2 Xi = Xi E1 + Xi E2 : i = 1, 2 , and hence 1 1 X1 X2 ω(X1,X2) = det[X] ω(E1,E2) , where [X] = 2 2 . X1 X2

So if (E10 ,E20 ) is another orthonormal basis with the same orientation, then 1 2 Ei0 = Oi E1 + Oi E2 i = 1, 2 , where O SO(2), and it follows that ∈ ω(E10 ,E20 ) = ω(E1,E2) . Thus ω can be thought of as assigning a real number to an oriented plane, which we may denote by ωΠ. Similarly Rp assigns a linear transformation of Tp to each oriented plane Πp (a 2- dimensional subspace of T ). This linear transformationM can be thought of as follows: pM Consider an infinitesimal closed curve C on the plane Πp, (see Fig. 23); given any vector V T we parallel transport V along the curve C, and let us denote by V 0 the vector ∈ pM obtained upon completing the contour. If the curvature at p is not zero, then V 0 does not coincide with V , and there is a difference ∆V = V 0 V ; in fact − ∆V = R V, ∆A Πp · where ∆A is the area of Πp enclosed by the curve C. The connection being metric means that magnitudes of vectors are preserved by parallel transport. So upon completing the contour C we obtain a vector V 0 which is different from V but of the same magnitude V 0 = V . | | | | Here the metric g is Lorentzian, so V 0 = g(V 0,V 0) = g(V,V ) = V means that V 0 is | | | | related to V by a Lorentz transformation; i.e. V 0 = ΛV for some Λ SO(3, 1). Moreover, ∈ since V 0 = V +(∆A) RΠ V we see that RΠp is an element of the Lie algebra of the Lorentz group. · We express this condition in an orthonormal basis (E0,E1,E2,E3) at a point p, such that gµν = g(Eµ,Eν ) is the Minkowski metric in rectangular coordinates [g ] = [η ] = diag( 1, 1, 1, 1) . µν µν − 40 1. THE EQUIVALENCE PRINCIPLE

∆V V

Πp

Figure 23. Illustration of the linear transformation RΠp .

Consider ﬁrst the condition g(V,V ) = g(V 0,V 0). Expanding V , V 0 in the above basis, and ν denoting by [Λµ] the matrix representing Λ with respect to the above basis,

X µ X µ X β V = V Eµ ,V 0 = V 0 Eµ , ΛEα = ΛαEβ , µ µ β it is straight-forward to show that the condition reads:

X µ ν gµν ΛαΛβ = gαβ . µ,ν For an infinitesimal curve C the transformation Λ differs from the identity by an infinitesimal ϑ: µ µ µ Λα = δα + ϑα . Inserting into the condition above implies

X µ ν ν µ X µ X ν gµν ϑαδβ + ϑβδα = 0 , or equivalently gµβϑα + gαν ϑβ = 0 , µ,ν µ ν P µ which is the statement that ϑαβ = µ gαµϑβ is antisymmetric in α, β:

ϑαβ + ϑβα = 0 .

µ The matrix [(RΠp ) ν ] representing RΠp with respect to the basis (Eµ : µ = 0,..., 3) therefore has the property that

X µ gλµ(RΠp ) ν = (RΠp )λν µ

µ is antisymmetric in λ, ν. [In fact, as we have sketched above, given any matric [Λ ν ] P µ such that [Λκν ] is skew symmetric, where Λκν = µ gκµΛ ν , then exp Λ is an element of the Lorentz group SO(3, 1), (and conversely).] This is equivalent to R(W, Z, X, Y ) = R(Z, W, X, Y ). For, in terms of the basis basis (E : µ = 0,..., 3), the linear transfor- − µ mation Rµν = R(Eµ,Eν ) is represented as follows: X (R ) E = (R )α E , µν · β µν β α α 3. GEODESIC CORRESPONDENCE 41 and X R = R(E ,E ,E ,E ) = g(E ,R E ) = g(E , (R )λ E ) αβµν α β µ ν α µν · β α µν β λ λ X λ X λ = (Rµν ) βg(Eα,Eλ) = gαλ(Rµν ) β . λ λ

The condition that Rµν is an element of the Lie algebra of the Lorentz group thus reads: R = R . βαµν − αβµν

3.3. Ricci curvature. We conclude this chapter with a discussion of the geometric meaning of the equations

tr M = 0 , where Mij = Ri0j0 . 3.3.1. Sectional curvature. The curvature tensor (3.79) R(W, Z, X, Y ) = g(W, R(X,Y ) Z) · is antisymmetric in (W, Z) as well as in (X,Y ). This property is a manifestation of the fact that the connection is compatible with the metric, (c.f. Note on pg. 39). Moreover we have the pair symmetry (3.80) R(X, Y, W, Z) = R(W, Z, X, Y ) . This is a consequence of the cyclic identity6: (3.81) R(X,Y ) Z + R(Y,Z) X + R(Z,X) Y = 0 . · · · Let Π be an arbitrary 2-dimensional plane in Tp at a point p , and (E1,E2) an orthonormal basis for this plane. Then M ∈ M

(3.82) KΠ = R(E1,E2,E1,E2) is independent of the choice of an orthonormal basis for Π7 and also independent 0 0 of the orientation of Π. Indeed if (E1,E2) is another orthonormal basis for Π, 0 1 2 E1 = O1E1 + O1E2 (3.83) 0 1 2 E2 = O2E1 + O2E2 , then 0 0 0 0 2 (3.84) R(E1,E2,E1,E2) = (det O) R(E1,E2,E1,E2) .

KΠ is called the sectional curvature of the plane Π.

6We will return to the cyclic identity when we discuss the Bianchi identities in the next chapter. 7 This also means that KΠ is invariant under Lorentz transformations of Π. 42 1. THE EQUIVALENCE PRINCIPLE

3.3.2. Jacobi equation. Recall the orthonormal frame (T,E1,E2,E3) that we constructed in our discussion of the Jacobi equation. If we set

X t=0 = E1 t=0 , then we obtain 2 1 d X = M11 = KΠ01 dt2 t=0 − t=0 t=0 where Π01 is the plane spanned by E0 = T , and E1. In general, v 3 u 3 X i p uX i 2 (3.85) X = X Ei , X = g(X,X) = t (X ) , | | i=1 i=1 also d2 X (3.86) | | = KΠ X , dt2 t=0 − | | where Π is the plane spanned by T , X. Remark. Review here Fig. 19 on page 29. We see that the distance to a nearby geodesic remains constant, decreases, or increases according as to whether the sectional curvature of the plane spanned by the tangent vector of the reference geodesic and the displacement vector is zero, positive or negative respectively.

Remark. Given a plane Π Tp , then consider the surface HΠ = exp Π. ⊂ M Then KΠ is in fact the Gauss curvature of the surface HΠ at the origin p. 3.3.3. Ricci curvature. Consider then 3 3 3 X X X (3.87) tr M = Mii = Ri0i0 = Ki , i=1 i=1 i=1 where Ki = KΠi0 is the sectional curvature of the plane spanned by E0 and Ei. We define the (E0,E0) component of the Ricci curvature by 3 X (3.88) Ric(E0,E0) = Ki . i=1 ˆ V ˆ ˆ Given any vector V , if V is not unit length we define V = |V | . Then Ric(V, V ) is simply defined as above taking E0 = Vˆ , and (3.89) Ric(V,V ) = V 2 Ric(V,ˆ Vˆ ) . | | This defines a symmetric bilinear form in Tp at each p by polarisation: M ∈ M 1n o (3.90) Ric(U, V ) = Ric(U + V,U + V ) Ric(U V,U V ) 4 − − − We shall from now on use the notation: (3.91) S = Ric . 3. GEODESIC CORRESPONDENCE 43

Alternatively we can express the Ricci curvature in terms of the curvature transformation (3.39). In an arbitrary basis we have 3 X α (3.92) Sµν = Ric(Eµ,Eν) = (Rαν) µ . α=0 We usually denote α α (3.93) (Rµν) β by R βµν . In terms of the inverse of g we have α X −1 αβ (3.94) (Rαν) µ = (g ) Rαµβν . β Since X α X −1 αβ (3.95) Sµν = R µαν = (g ) Rαµβν , α α,β the pair symmetry Rµναβ = Rαµβν implies the symmetry

(3.96) Sνµ = Sµν . The vacuum Einstein equations are therefore

(3.97) Sµν = 0 .

Exercises In the following exercises we study cylindrical normal coordinates which correspond to a freely falling laboratory. Recall our discussion of the Jacobi equation in cylindrical normal coordinates: Let Γ0 be a timelike reference geodesic with unit tangent vector field T , 0 parametrized by arclength x . We have ∇T T = 0. In the local simultaneous space Σx0 at each point along Γ0, Σ 0 = X ∈ T 0 M : g(T,X) = 0 , x Γ0(x ) we have an orthonormal basis (E1,E2,E3) which is parallel transported along Γ0: ∇T Ei = 0. 0 1 2 3 To each point p in a neighborhood of Γ0 we assign cylindrical normal coordinates (x , x , x , x ) if 1 2 2 p = exp 0 X ,X = x E + x E + x E , Γ0(x ) 1 2 3 0 namely if p lies on the spacelike geodesic with initial tangent vector X at Γ0(x ) at arclength p 1 2 2 2 3 2 distance r from Γ0: r = |X| = (x ) + (x ) + (x ) . Then on Γ0: ∂ ∂ T = ,E = : i = 1, 2, 3 . ∂x0 i ∂xi 1. (Expansion of the metric to first order): Show that to first order in r, the metric is equal to the Minkowski metric, i.e. 0 1 2 3 2 gµν (x , x , x , x ) = ηµν + O(r ). Hints: Show that as a consequence of metric compatibility ν ν ∂µgαβ = gνβ Γµα + gαν Γµβ , where Γ are the connection coefficients. The problem thus reduces to showing α Γµν = 0 along Γ0, which can be done as outlined in the following steps: 44 1. THE EQUIVALENCE PRINCIPLE

(1) Given any curve γ(τ) = (x0(τ), x1(τ), x2(τ), x3(τ)) in local coordinates, show that the geodesic equation ∇γ˙ γ˙ = 0 is equivalent to

2 µ 3 α β d x X µ dx dx + Γ = 0 : µ = 0, 1, 2, 3 . dτ 2 αβ dτ dτ α,β=0

(2) Use the fact that the curve τ 7→ (τ, 0, 0, 0) is a geodesic (namely Γ0) to deduce µ 0 0 that Γ00(x , 0, 0, 0) = 0 for all x . 0 (3) Use the fact that for any vector k = (k1, k2, k3) and any x the curve τ 7→ (x0, k1τ, k2τ, k3τ) is a geodesic, and deduce that for all x0 and τ: i j µ 0 1 2 3 k k Γij (x , k τ, k τ, k τ) = 0 . 0 (4) Use the parallel transport equation for Ei to deduce that for all x : µ 0 Γ0i(x , 0, 0, 0) = 0. 2. (Expansion of the metric to second order): Find the second order terms in the µ expansion of Exercise 1 in terms of the curvature components R νλσ. In particular, prove: 3 1 X k l 3 g = δ − R x x + O(r ): i, j = 1, 2, 3 . ij ij 3 ikjl k,l=1 Hints: Similarly to Exercise 1, show that: σ σ ∂α∂µgνλ = (∂αΓµν )gσλ + (∂αΓµλ)gνσ : along Γ0. σ µ The problem therefore reduces to expressing ∂αΓµν in terms of R νσλ, which can be done as follows: (1) Note that we have µ µ µ R νκλ = ∂κΓλν − ∂λΓκν along Γ0. Now deﬁne µ µ µ µ S κλν = ∂κΓλν + ∂λΓνκ + ∂ν Γκλ and derive µ µ µ µ 3 ∂κΓνλ = R νκλ + R λκν + S κλν : along Γ0. µ (2) Use hint (3) from Exercise 1 to show that S ijk = 0: i, j, k = 1, 2, 3 along Γ0. µ µ µ (3) Use the hints from Exercise 1 to show that ∂0Γ00 = ∂0Γ0i = ∂0Γij = 0 along Γ0. µ µ (4) Use (1) and (3) to ﬁnd ∂iΓ0j and ∂iΓ00 along Γ0. 3. (Riemannian sectional curvature and Gauss curvature): Consider a Riemannian manifold M of dimension n > 2. Let Π ⊂ TpM be a 2-dimensional plane in the tangent space at a point p ∈ M. Let HΠ be the geodesic surface

HΠ = expp(Π) = {expp(X): X ∈ Π} , consisting of all geodesics starting at p with an initial tangent vector X in Π. Show that the Gauss curvature of the surface HΠ at p is equal to the Riemannian sectional curvature RΠ = R(E1,E2,E1,E2), where (E1,E2) is an orthonormal basis for Π. Hints: Introduce normal coordinates (x1, . . . , xn) with origin at p, such that 1 2 3 n 3 n HΠ = {(x , x , x , . . . , x ): x = ... = x = 0} .

Use the result of Exercise 1 to calculate the circumference Cr of a geodesic circle 1 2 1 2 2 2 2 Sr = {(x , x ):(x ) + (x ) = r } in HΠ. Finally recall from the note on pg. 37 that the Gauss curvature K appears in the formula dC r = 2π − πKr2 + O(r3) . dr CHAPTER 2

Einstein’s ﬁeld equations ∗

In the previous chapter we have arrived at a uniﬁed theory of space, time and gravitation according to which spacetime is a 3 + 1 dimensional manifold endowed with a metric g of index 1 whose curvature represents tidal gravitationalM forces. The Einstein vacuum equations

(0.1) Sµν = 0 , were found in analogy to Poisson’s equation for the gravitational potential in vacuum: (0.2) ∆ψ = 0 . In this chapter we discuss Einstein’s field equations in the presence of matter, which can be thought of as the analogue of (0.3) ∆ψ = 4πµ , where µ is the mass density.1 Remark. A comparison to the theory of electromagnetism provided some guidance for Einstein from early on. The situation can be compared to electrostatics, where the electric potential φ satisfies (0.4) ∆φ = 4πρ , − and ρ is the electric charge density: the source of the electric field E = φ. It is known however that moving charges create a magnetic field B, according−∇ to Ampère’slaw of magnetostatics: (0.5) B = 4πJ ∇ × Only Maxwell’s theory of electrodynamics unites the sources (ρ, J), the fields (E,B), and the potentials (φ, A) as time and space components of differential forms on Minkowski space and constitutes a theory in agreement with special relativity. In the Newtonian theory the mass density µ is the source of the gravitational field. If general relativity relates to Newtonian gravitation as Maxwell’s theory does to electrostatics, then the question is what is the source in general relativity? We shall see that the energy density ε will play the role of the source in general relativity. In special relativity ε is a component of the energy-momentum stress tensor. While this analogy is described further in the following note, we next discuss the energy-momentum tensor in general relativity.

1in units where Newton’s consant G = 1.

45 46 2. EINSTEIN’S FIELD EQUATIONS ∗

Notes on Maxwell’s theory In electrostatics the electric field E is determined from the electric charge density ρ by the system of equations (0.6) E = φ −∇ (0.7) E = 4πρ , ∆φ = 4πρ . ∇ · − These are complemented by the equations of magnetostatics for the magnetic field B, given an electric current density J (in this note we keep explicitly the speed of light c in the equations): J (0.8) B = 4π ∇ × c (0.9) B = 0 , ∇ · The last equation of course states the absence of a magnetic charge density. In a dynamical situation Gauss’ law E = 0 is replaced by Faraday’s induction law: ∇ × 1 ∂B (0.10) E + = 0 , ∇ × c ∂t and Maxwell corrected Ampèreslaw to be: 1 ∂E J (0.11) B = 4π . ∇ × − c ∂t c Taking the divergence of this equation we obtain the continuity equation ∂ρ (0.12) + J = 0 , ∂t ∇ · which expresses the conservation of charge. This is the integrability condition of the Maxwell equations. Remark. This logic reappears in general relativity: If we introduce a source it has to satisfy an integrability condition for the field equations. We can introduce a vector potential A so that (0.13) B = A, ∇ × as a general solution to (0.9). Then in view of (0.10) we have 1 ∂A (0.14) E = φ . −∇ − c ∂t Now it turns out that Maxwell’s equations can be unified in the spacetime point of view of special relativity: Introduce the potential 1-form 3 X µ (0.15) A = Aµdx , µ=0 and consider the electromagnetic field 2-form (0.16) F = dA, where d is the exterior derivative. Then (0.17) dF = 0 . 2. EINSTEIN’S FIELD EQUATIONS ∗ 47

These are precisely the homogeneous Maxwell equations. Here, in rectangular coordinates we set x0 = ct so time intervals become length, and the metric is 3 X (0.18) g = dx0 dx0 + dxi dxi ; − ⊗ ⊗ i=1 moreover µ 1 µν (0.19) A = (g− ) Aν (0.20) A0 = φ , Ai = Ai : the components of the vector potential. Remark. Given a 1-form θ then dθ is a 2-form deﬁned by (dθ)(X,Y ) = X(θ(Y )) Y (θ(X)) θ([X,Y ]) . − − Similarly, if ω is a 2-form then dω is a 3-form given by dω(X,Y,Z) = X(ω(Y,Z)) + Y (ω(Z,X)) + Z(ω(X,Y )) ω([X,Y ],Z) ω([Y,Z],X) ω([Z,X],Y ) − − − Note if ω = dθ then dω = 0. We have 3 X (0.21) F = F dxµ dxν µν ⊗ µ,ν=0 (0.22) F = ∂ A ∂ A , µν µ ν − ν µ ∂ ∂ (to see this simply take X = ∂xµ and Y = ∂xν in F (X,Y ) = dA(X,Y ) and note that [X,Y ] = 0 in rectangular coordinates); and moreover ∂φ 1 ∂Ai (0.23) F = ∂ A ∂ A = = E i0 i 0 − 0 i −∂xi − c ∂t i ∂Aj ∂Ai (0.24) F = ∂ A ∂ A = = ( A) = B . ij i j − j i ∂xi − ∂xj ijk ∇ × k ijk k Now, the inhomogeneous Maxwell equations are J µ (0.25) F µν = 4π ; where J 0 = cρ , J i = J i : electric charge density. ∇ν c F being antisymmetric the divergence of the left hand side vanishes identically: (0.26) F µν = 0 . ∇µ∇ν The integrability condition for Maxwell’s equations is therefore: J µ = 0(0.27) ∇µ 3 ∂J 0 X ∂J i ∂ρ (0.28) + = + J = 0 : in rectangular coordinates. ∂x0 ∂xi ∂t ∇ · i=1 In this way space and time components are united in this theory: F unites (E,B), A unites (φ, A), and J unites (ρ, J). The spacetime current J µ acts as a source for Maxwell’s equations. We compare the sources: Electrostatics: ρ Newtonian gravitation: µ Maxwell’s theory: J µ General relativity: ? 48 2. EINSTEIN’S FIELD EQUATIONS ∗

We shall see that the energy density ε will play the role of the source in general relativity; (in a quasi-Newtonian situation the leading term is µc2). In special relativity ε is the component T 00 of a contravariant tensorﬁeld T µν , then energy momentum stress tensor: T 00 = ε : energy density T 0i = pi : momentum density T ij : stress. In fact, in Maxwell’s theory: 1 1 ε = E 2 + B 2 4π 2 | | | | 1 (0.29) pi = (E B)i 4π × 1 1 1 T ij = E E B B + δ E 2 + B 2 . 4π − i j − i j 4π ij 2 | | | |

1. Einstein’s field equations in the presence of matter 1.1. Energy-momentum-stress tensor. The energy-momentum-stress ten- ∗ sor T is a 2-contravariant symmetric tensorfield on (a quadratic form in Tx , the cotangent space at each x ) that vanishesM in vacuum. In componentsM relative to an arbitrary frame ∈ M 3 X µν µν νµ (1.1) T = T Eµ Eν ,T = T . ⊗ µ,ν=0 There is a corresponding 1-covariant-1-contravariant tensorfield with components µ X µν (1.2) T λ = T gνλ , ν X µ (1.3) so ( T )Eλ = T λEµ . − − µ + Consider the hyperboloid Hx of future-directed unit timelike vectors u at x. Such a vector u is the 4-velocity of an observer at x. Then

(1.4) T u Tx − · ∈ M is the energy-momentum density of matter relative to the observer with 4-velocity u at x. We have the following physical requirement: This energy-momentum density is a non-spacelike future-directed vector. Dominant energy condition: The linear transformation T maps the + − future hyperboloid Hx into the closure of the open future cone at x, for each x . (See Fig. 1.) ∈ M We decompose T u into a component colinear to u, εu, and a component − · orthogonal to u, p, lying in Σu, the local simultaneous space of the observer with 4-velocity u:

(1.5) T u = εu + p , p Σu = X : g(u, X) = 0 . − · ∈ { } 1. EINSTEIN’S FIELD EQUATIONS IN THE PRESENCE OF MATTER 49

T u − ·

+ Hx

Figure 1. Dominant energy condition.

ε is called the energy density of matter at x relative to the observer with 4-velocity u at x, and p is called the momentum density of matter at x relative to the same observer. We have X µ λ κ X κ λ (1.6) ε = g(T u, u) = gµκT u u = Tκλu u · λ µ,κ,λ κ,λ where X µ X µν (1.7) Tκλ = gµκT λ = gµκgνλT µ µ,ν are the components of a quadratic form in Tx at each x , a 2-covariant symmetric tensorfield on . We then write: M ∈ M M (1.8) ε = T (u, u) . The dominant energy condition reads: (1.9) p ε for every u H+ . | | ≤ ∈ x In particular, ε 0. 1.1.1. Proper≥ energy density. We minimize ε with respect to u, defining the proper energy density (1.10) ρ(x) = inf T (u, u) . + u∈Hx + Note that Hx is non-compact, and consequently the infimum is not necessarily achieved. Remark. This is in contrast to its analogue in Euclidean space, the unit sphere which is compact. Indeed, consider a quadratic form S on the vector space EO of all displacements from an origin O in Euclidean space. We then consider e = inf S(u, u) , u∈SO where SO is the unit sphere in EO: SO = u EO : u = 1 . This is the lowest eigenvalue of S. Here, since the unit sphere{ is∈ compact,| | the} infimum is achieved at some u1 SO, and also at u1, i.e. at some pair of antipodal points in SO; however, uniqueness∈ may not hold.− The infimum may be achieved on a great 50 2. EINSTEIN’S FIELD EQUATIONS ∗ circle, when the first two eigenvalues coincide, or even on the whole sphere, if all three eigenvalues coincide. In the last case 2 S(u, u) = e u : for all u EO , | | ∈ i.e. S is then proportional to the Euclidean inner product. + In other words, a minimizing sequence, i.e. a sequence (un) Hx such that + ⊂ T (un, un) ρ as n , may run off to infinity in Hx . Then the vectors un would approach→ a generator→ ∞ of the future-directed null cone at x. Moreover, even if the infimum is achieved at some u H+, it may not be unique. ∈ x Example: Let K be a future-directed null vector and T = K K, ⊗ i.e. T µν = KµKν . Then ρ = 0 but the limit is not achieved. To see this + we construct a sequence of timelike vectors un H as follows: Let E0 ∈ x be a timelike vector at x, and take un to lie in the plane spanned by E0 and K. There is another null vector K such that g(K,K) = 2; then − un = an K + anK ,

1 = g(un, un) = 2 anan g(K,K) = 4 anan . − − Consider a sequence such that an . So → ∞ 2 2 2 1 T (un, un) = (g(K, un)) = ( 2an) = 4an = 2 0 , − 4an → and thus ρ(x) = 0, but it is not achieved. A material continuum2 is matter whose energy-momentum-stress tensor T has at each point x where it does not vanish the property that (i) the infimum is achieved, + (ii) it is achieved at a unique ux H . ∈ x Then the vector ux is called the material velocity. (And u is a vectorfield defined on the support of the matter.) It is in particular a critical point of T (u, u) under + the constraint that u H . So ux is an eigenvector of T : ∈ x ν ν (1.11) Tµνu = ρ gµνu , x − x where ρ is the corresponding eigenvalue. Remark. We may view this as a constraint variational problem: Minimize µ ν µ ν Tµνu u under the constraint that gµνu u = 1 (and u future-directed). By the method of Lagrange multipliers, the condition− for u to be a critical point is ν ν Tµνu = λgµνu , where λ is the Lagrange multiplier, or eigenvalue. Multiplying by uµ we obtain λ = T (u, u), and so λ = ρ. − − 2also “ponderable matter”. 1. EINSTEIN’S FIELD EQUATIONS IN THE PRESENCE OF MATTER 51

Example: The electromagnetic ﬁeld is not a material continuum. Since the Faraday tensor is antisymmetric,

Fµν = Fνµ , − there are two conjugate null eigenvectors L, L such that ν ν FµνL = α gµνL ν ν FµνL = α gµνL − g(L, L) = 2 . − Then F = α + β⊥ , where is the area 2-form of the timelike plane Π spanned by L, and L, and ⊥ is the area 2-form of the spacelike plane which is orthogonal to the timelike plane Π. In rectangular coordinates (t, x, y, z) at a given point, F = α dt dx + β dy dz , ∧ ∧ that is the electric and magnetic fields E, B only have x-components: ∂ ∂ E = α ,B = β . − ∂x ∂x As we shall see (c.f. note on page 2, in particular (0.29)), the energy- momentum-stress tensor in this situation takes the form: α2 + β2 0 0 0  t 1  0 α2 β2 0 0  x (1.12) T =  − −  8π  0 0 α2 + β2 0  y 0 0 0 α2 + β2 z + The infimum is achieved on the curve Hx Π, and thus in particular not unique. (C.f. exercises below for the details∩ of this proof.) 1.1.2. Proper stress tensor. Given a material continuum let us now define the proper stress tensor S (not to be confused with Ricci curvature) by (1.13) S = T ρ u u . − ⊗ Then (1.14a) Sµν = T µν ρ uµuν µ µν µ −µ ν (1.14b) S λ = S gνλ = T ν ρ u uλ , uλ = u gνλ , µ− (1.14c) Sκλ = gκµS = Tκλ ρ uκuλ , λ − and we have λ λ (1.15) Sκλu = Tκλu + ρ uκ = ρuκ + ρuκ = 0 . − S is a 2-covariant symmetric tensor vanishing on u, so S is in fact a tensor on ⊥ Σx = ux : the local simultaneous space at x. The metric gx restricted to Σx is positive definite. So we have the standard eigenvalue problem for S on Σx. There 52 2. EINSTEIN’S FIELD EQUATIONS ∗ are 3 eigenvalues p1, p2, p3 called the principal pressures, and 3 corresponding eigenvalues E1, E2, E3, which form an orthonormal basis for Σx. Prop 1.1. For a material continuum, the dominant energy condition is equivalent to:

pi ρ for i = 1, 2, 3 . | | ≤ Proof. We show that this inequality implies the positivity condition, and we leave the converse as an exercise. Choose a basis (Eµ : µ = 0,..., 3) such that E0 = u is the material velocity and Ei : i = 1, 2, 3 are the eigenvectors of the proper stress S. Let v be a timelike future-directed vector, s 0 X i 0 X i 2 v = v E0 + v Ei , v > (v ) ; i i then we need to show that T v is a future directed causal vector. We decompose also T v with respect to− the· above basis: − · T v = εu + p , p u⊥ . − · ∈ Then ε = g(T v, u) = ρ g(u, v) = ρv0 , · − and 3 3 X i X i p = S v = v S Ei = v piEi , − · − · − i=1 i=1 and therefore 3 3 2 X i 2 2 X i2 2 02 2 p = v pi ρ v < ρ v = ε , | | ≤ i=1 i=1 because pi ρ. | | ≤ Remark. The proper stress tensor S has the following physical meaning: We know S lives on Σu, the local simultaneous space of the material (it vanishes on u). Let Π be a plane in Σu through the origin. There is a covector N of unit length whose null space is Π, i.e. Π = X Σ: N X = 0 . Π is oriented, so it + { ∈ · − } divides Σu into a postive side Σu , and a negative side Σu :

+ n o − n o Σ = X Σu : N X > 0 , Σ = X Σu : N X < 0 . u ∈ · u ∈ · The vector

S N Σu · ∈ represents the force per unit area exerted by the material across the surface with tangent plane Π. 1. EINSTEIN’S FIELD EQUATIONS IN THE PRESENCE OF MATTER 53

1.1.3. Perfect ﬂuids. A special case of particular interest is if all principal pressures are equal:

(1.16) p1 = p2 = p3 = p .

Then the material is called a perfect ﬂuid, and the common value of pi : i = 1, 2, 3 is called the pressure p. Moreover, in that case (1.17) S = p g . Σx Σx If X, Y are arbitrary vectors at x , then ∈ M (1.18) S(X,Y ) = S(ΠX, ΠY ) = p g(ΠX, ΠY ) , where Π is the orthogonal projection to Σx. In terms of components in an arbitrary frame (Eµ : µ = 0, 1, 2, 3) we have X µ X µ (1.19a) X = X Eµ , ΠX = (ΠX) Eµ , µ µ (1.19b) ΠX = X + g(u, X)u , g(u, u) = 1 , µ µ ν µ µ µ− (1.19c) (ΠX) = Π νX , Π ν = δ ν + u uν , and it follows that (1.20) Sµν = p gµν + uµuν . Thus

(1.21) Tµν = ρ uµuν + Sµν = p gµν + (ρ + p) uµuν is the energy-momentum-stress tensor of a perfect ﬂuid.

Exercises We study the energy-momentum-stress tensor of the electromagnetic field: 1 1 T = F F α − g F F αβ . µν 4π µα ν 4 µν αβ Recall that in a given orthonormal frame its components in terms of the electric field E and magnetic field B are given by (0.29). 1. (Proper energy density of an electromagnetic plane wave) Consider an electromagnetic plane wave in Minkowski space. We assume that in stationary coordinates (t, x, y, z) the wave propagates in the positve x-direction, and that the components of the electric field E and the magnetic field B are only a function of u = t − x (here c = 1): ∂ ∂ E = E (t − x) + E (t − x) y ∂y z ∂z ∂ ∂ B = B (t − x) + B (t − x) . y ∂y z ∂z Moreover, we assume E and B are compactly supported in u. (1) Use Maxwell’s equations to show that

By = −Ez,Bz = Ey. Sketch E, B in the y-z-plane, and their support on the t-x-plane. Calculate the magnitude of the momentum vector E × B. 54 2. EINSTEIN’S FIELD EQUATIONS ∗

(2) Recall the deﬁnition of the proper energy density (1.10). Show that ρ vanishes but that the inﬁmum is not attained. Give a physical interpretation. + Hint: Consider a sequence (un) in Π ∩ Hp , where Π the t-x-plane (and p a point in the support of E and B).

2. (Proper energy density of the electromagnetic field) Here we shall show that the electromagnetic field is not a material continuum in general. To find a convenient frame, first consider the eigenvalue problem for the Faraday tensor F : ν ν Fµν v = λ gµν v

Here λ ∈ C is a root of the characteristic polynomial: det A(λ) = 0, where A = (Aµν ) is the matrix with components,

Aµν (λ) := Fµν − λ gµν , formed from the electromagnetic ﬁeld tensor Fµν and the metric gµν at a point. Recall that 4 Fµν = −Fνµ. The vector v is the corresponding eigenvector, an element v ∈ C . (1) Show that if λ is an eigenvalue of F so is −λ and λ. There are 4 complex roots of the characteristic polynomial. Show that the following generic cases are possible: (a) Four real eigenvalues: 0 < λ1 < λ2, λ3 = −λ1, λ4 = −λ2. (b) Four imaginary eigenvalues: iµ1, iµ2, iµ3 = −iµ1, iµ4 = −iµ2, where µ1, µ2 ∈ R, 0 < µ1 < µ2. (c) Four complex eigenvalues: λ + iµ, λ − iµ, −λ − iµ, −λ + iµ, where λ, µ ∈ R. (d) Two real and two imaginary eigenvalues: λ, −λ, iµ, −iµ, where λ, µ ∈ R. (2) Show that only the case (d) is possible in the physical case of 3 + 1 dimensions. Hints: (a) Given a real eigenvalue λ the corresponding eigenvector L is also real. Show that in the case (a) of two real eigenvalues 0 < λ1 < λ2 the corresponding eigenvectors L1, L2 satisfy

g(Li,Li) = 0, g(Li,Lj ) = 0 , simply by using the antisymmetry of F :

F (Li,Li) = 0,F (Li,Lj ) = −F (Lj ,Li) . Then deduce a contradiction. (b) To a complex eigenvalue λ corresponds a complex eigenvectors M. In (b), let M1 = R1 + iS1 and M2 = R2 + iS2 be the complex eigenvectors corresponding to the imaginary eigenvalues iµ1, iµ2 (Ri, Si real vectors). Similarly to (i) show that the anti-symmetry of F implies

g(Ri,Si) = 0, g(Ri,Ri) = g(Si,Si) as well as

g(Ri,Rj ) = g(Si,Sj ) = 0, g(Ri,Sj ) = 0. Show that this is impossible in 3 + 1 dimensions. (c) For (c) use the fact that if M is a complex eigenvector corresponding to the complex eigenvalue λ it follows that M is a complex eigenvector corresponding to the complex eigenvalue λ. Then proceed similary to (a) and (b) to rule out case (c). (3) Let F be given by F = α dt ∧ dx + β dy ∧ dz, Show that this form of F corresponds to the physical case (d) above and ﬁnd the eigenvalues in terms of α, β. 1. EINSTEIN’S FIELD EQUATIONS IN THE PRESENCE OF MATTER 55

Verify that ∂ ∂ ∂ ∂ L = + ,L = − , ∂t ∂x ∂t ∂x ∂ ∂ ∂ ∂ M = + i , M = − i , ∂y ∂z ∂y ∂z are the corresponding eigenvectors. Conversly, show that in the physical case (d) generically we can choose the suitable Lorentz frame such that F is of the above form. (4) Finally, verify that in the above frame (ii) the energy-momentum stress tensor takes the form (1.12). Show that the inﬁmum of T (u, u) is assumed on the curve u ∈ Π ∩ H+ where Π is the t-x-plane, and is thus in particular not uniquely achieved.

1.2. Local energy-momentum conservation laws. We now view T as a symmetric 2-covariant tensorfield. The energy-momentum-stress tensor satisfies the conservation laws: (1.22) T = 0 . ∇ · In terms of components relative to an arbitrary frame (Eµ : µ = 0, 1, 2, 3) we have αβ (1.23) E T = ( µT )Eα Eβ ; ∇ µ · ∇ ⊗ and T is a vectorfield ∇ · µν (1.24) T = ( νT )Eµ . ∇ · ∇ The energy-momentum conservation laws are therefore µν (1.25) νT = 0 , ∇ expressed in components with respect to an arbitrary frame. Remark. For a material continuum these are the equations of motion of the continuum. In particular, this encompasses Newton’s law of motion. Remark. Note that in the case of the perfect fluid (1.21) the equations of motion (1.25) are underdetermined: All in all there are 4 equations for 5 unknowns (ρ, p, uµ: 3 independent components). One possibility is to study a barotropic fluid where ρ = ρ(p) . In general, let v be the specific volume per particle, s the specific entropy, and e the energy per particle (this contains mc2 where m is the rest mass per particle). The mechanical properties of a fluid are specified once we stipulate the caloric equation of state: (1.26) e = e(v, s) According to the first law of thermodynamics we have (1.27) de = p dv + θ ds , − 56 2. EINSTEIN’S FIELD EQUATIONS ∗ where p is the pressure and θ the temperature. Let now be ρ = e/v the energy per unit volume, and n = 1/v the number of particles per unit volume, and define the particle current: (1.28) Iµ = nuµ . Then the additional equation is the differential conservation law of particle number: µ (1.29) µI = 0 . ∇ 1.2.1. Quasi-Newtonian hierarchy. We analyze the components of the energy- momentum-stress tensor in a quasi-Newtonian situation. Let (E0,E1,E2,E3) be an orthonormal frame field, E0 being the 4-velocity of a background system of observers in Minkowski space. We express the components of the material velocity in conventional units, (where c denotes the speed of light): 3 3 0 X i 0 2 X i 2 (1.30) u = u E0 + u Ei , g(u, u) = (u ) + (u ) = 1 , − − i=1 i=1 i 1 v (1.31) u0 = , ui = c q 2 q 2 1 |v| 1 |v| − c2 − c2 Recall (1.14) (1.32) T µν = ρuµuν + Sµν , where the proper stress tensor S has the property (1.15) that µν (1.33) S uν = 0 . In the quasi-Newtonian limit this reads: 00 0i 00 0-component: S u0 + S ui = 0 S = 0 as c , −→ → ∞ (1.34) i0 ij i0 i-component: S u0 + S uj = 0 S = 0 as c , −→ → ∞ because

(1.35) u0 1 , and ui 0 , as c . −→ − −→ → ∞ Thus in the non-relativistic limit Sµν reduces to Sij, and the condition (1.15) reduces to (1.36) Sµ0 = 0 . Here T has the units of energy density, and ρ is dominated by the contribution of the rest mass density µ, i.e. (1.37) ρ µc2 as c . ∼ → ∞ We thus obtain the quasi-Newtonian hierarchy:  T 00 = ρ(u0)2 + S00 µc2 : T 00 = (c2)  ∼ O (1.38) T 0i = ρu0ui + S0i µcvi = cpi : T 0i = (c) ∼ O T ij = ρuiuj + Sij µvivj + Sij : T ij = (1) ∼ O 1. EINSTEIN’S FIELD EQUATIONS IN THE PRESENCE OF MATTER 57

Here pi = µvi is the spatial momentum density, and µvivj is also referred to as kinematic stress.

Notes on the Newtonian limit of the conservation laws It might be instructive here to also write out the conservation laws (1.25) in the quasi- Newtonian situation (1.38). On Minkowski space (0.18), the components of (1.25) are µν ∂µT = 0 : 3 ∂T 00 X ∂T 0i + = 0 : µ = 0 , ∂x0 ∂xi i=1 3 ∂T i0 X ∂T ij + = 0 : µ = i . ∂x0 ∂xj j=1 However, in view of (1.38) – and noting that x0 = ct – in the limit c these become: → ∞ 3 ∂µ X ∂pi + = 0 , ∂t ∂xi i=1 3 ∂pi X ∂T ij + = 0 , where T ij = µvivj + Sij . ∂t ∂xj j=1 Consider the flow lines of the non-relativistic material velocity, dxi = vi(t, x(t)) ; dt here the velocity vectorfield v is assumed to be known. This defines a map, Ft(y) = x(t, y), where x(t, y) is the flow line with initial condition xi(0) = yi; here x = (x1, x2, x3), 1 2 3 y = (y , y , y ). Given a “portion” Ω of the material, let us denote by Ωt the same collection of particles in the material under time evolution, namely

Ωt = Ft(Ω) . Let us then consider the mass M and momentum P of the material portion: Z M(t) = µ dV Ωt Z P i(t) = pi dV Ωt In general, for any function f on spacetime, we have: d Z Z ∂f Z f dV = dV + f v dS, dt Ωt Ωt ∂t ∂Ωt · where dS is the oriented area element of the boundary ∂Ωt, i.e. dS = NdA, where N is the unit outward normal, and dA is the scalar area element. Thus by Gauss’ divergence theorem d Z Z ∂f f dV = + (fv) dV dt Ωt Ωt ∂t ∇ · 58 2. EINSTEIN’S FIELD EQUATIONS ∗

In rectangular coordinates dV = dx1dx2dx3 and (fv) = ∂ (fvi). Therefore: ∇ · i dM Z n∂µ ∂(µvi)o = + i dV = 0 , dt Ωt ∂t ∂x i.e. conservation of mass is equivalent to the ﬁrst conservation law. Furthermore, dP i Z n∂pi ∂(pivj)o = + j dV, dt Ωt ∂t ∂x where pivi = µvivj is the kinematic stress. Now, by Newton’s second law, the rate of change of the momentum equals the force F which is exerted on the material by its surroundings: dP i = F i . dt Newton’s third law asserts in turn that this is equal and opposite the force exerted by the portion of the material on its surroudnings. So Z Z Z ij i ij ij ∂S F = S Nj dA = S dSj = j dV, − ∂Ωt ∂Ωt Ωt ∂x ij where S Nj is i-th component of the force per unit area across ∂Ωt, and dSj = NjdA. We conclude that the conservation of momentum corresponds to the diﬀerential conservation law, ∂pi ∂ + µvivj + Sij = 0 . ∂t ∂xj

Exercises In these exercises we study the equations of motion of a perfect ﬂuid. These are the conservation laws (1.25) and (1.29) for the energy-momentum-stress tensor (1.21) and the particle current (1.28). 1. (Non-relativistic limit of the conservation laws.) We consider the equations of motion in Minkowski space and take the material velocity u to be of the form (1.31) in rectangular coordinates. Moreover, we assume that ρ = µc2 + h where µ is the mass density and h is the internal energy density; c.f. (1.37). (1) Derive in the non-relativistic limit c → ∞ the conservation of mass law, ∂µ + ∇ · (µv) = 0 . ∂t Hints: Express the components of the energy-momentum-stress tensor explicitly in orders of c, and derive the conservation law from the 0-component of (1.25). (2) Derive in the nonrelativistic limit c → ∞ the energy conservation law ∂ε + ∇ · f = 0 , ∂t 1 2 where ε is the total energy density given by ε = 2 µ|v| + h and f is the total energy ﬂux given by f i = (ε + p) vi. Hints: Consider the current J µ = T 0µ − µc2uµ , and proceed similarly to (1). 1. EINSTEIN’S FIELD EQUATIONS IN THE PRESENCE OF MATTER 59

2. (Adiabatic condition.) In addition to the equations of motion (1.25) and (1.28) we have the equation of state (1.26) which expresses the energy per particle e as a function of the volume per particle v and the entropy per particle s. We have ρ = e/v, n = 1/v, and according to the ﬁrst law of thermodynamics the pressure p and the temperature θ are then given by (1.27). Prove the adiabatic condition

∇us = 0 , namely that the entropy is constant along the flow lines of the fluid. µν Hints: Consider the u component of (1.25), i.e. the equation uν ∇µT = 0, and use (1.29) in µ conjuction with the above relations to show that u ∇µs = 0. 3. (Pressureless fluid.) Show that if p = 0 then the integral curves of u are timelike geodesics.

1.2.2. Considerations for the ﬁeld equations. Let us now return to Einstein’s vacuum equations

(1.39) Sµν = 0 , where from now on S denotes again the Ricci curvature: α (1.40) Sµν = R µαν .

The component S00 = S(E0,E0) is taken with respect to a unit future-directed 2 tangent ﬁeld to a reference geodesic Γ0. Recall that Ri0j0 corresponds to ( ψ)ij ∇ in the Newtonian theory, whence we get the correspondence3:

(1.41) S00 4π T00 ∼ S00 = tr M = 4πµ : in the Newtonian theory. However, in view of the quasi-Newtonian hierarchy (1.38) there is a certain ambiguity as to what the right hand side of (1.39) in the presence of matter should be. In fact, if we set

(1.42) Sµν = a Tµν + b gµν tr T, in the presence of matter, then we obtain the Newtonian theory in the limit c for all a, b such that → ∞ (1.43) a + b = 4π . For, with respect to an orthonormal frame 3 −1 µν X (1.44) tr T = (g ) Tµν = T00 + Tii = T00 + (1) , − − O i=1 and so in the non-relativisitic limit:

(1.45) S00 = a T00 b tr T = (a + b) T00 . − This ambiguity will be broken by the requirement that the conservation laws (1.22) are a consequence of the ﬁeld equations. We shall illustrate that logic ﬁrst with the example of Maxwell’s theory; (c.f. note on pg. 48).

3in units where c = 1, and G = 1. 60 2. EINSTEIN’S FIELD EQUATIONS ∗

1.2.3. Conservation laws in electromagnetic theory. In Maxwell’s theory the electromagnetic ﬁeld F is a 2-form on Minkowski space . The homogeneous equations are M

dF = 0(1.46) (1.47) F = dA, where A is determined from F only up to a diﬀerential of a function f, i.e. A0 is equivalent to A if

(1.48) A0 = A + df (gauge transformations).

In an arbitrary system of local coordinates (1.47) reads

(1.49) Fµν = ∂µAν ∂νAµ , − and (1.46) is an equation for the cyclic sum:

(1.50) ∂µFνλ + ∂νFλµ + ∂λFµν = 0 .

The inhomogeneous equations are

µν µ (1.51) νF = 4πJ ∇ µν −1 µκ −1 νλ (1.52) F = (g ) (g ) Fκλ , where J µ denotes the electric current density.

Remark. In macroscopic media the inhomogeneous equations are replaced by

µν µ (1.53) νG = 4πJ , ∇ where Gµν is the electromagnetic displacement. Similary to F , which decomposes into the electric, and magnetic ﬁeld E, B,

k (1.54) F0i = Ei ,Fij = εijkB ,

G decomposes into the electric and magnetic displacement ﬁelds D,H:

k (1.55) G0i = Di ,Gij = εijkH .

Since

µν (1.56) µ νF = 0 : identically, ∇ ∇ the inhomogeneous equations (1.51) imply the conservation of the electric current:

µ (1.57) µJ = 0 ; ∇ namely local conservation of electric charge. 1. EINSTEIN’S FIELD EQUATIONS IN THE PRESENCE OF MATTER 61

1.2.4. Conservation laws and the field equations. In general relativity the equations of motion are: µν (1.58) νT = 0 . ∇ We require that these should be the consequence of the field equations of gravitation. In (1.42) we have seen that (1.59) Sµν = a T µν + κ (g−1)µν tr T has the correct Newtonian limit provided b (1.60) κ = , a(1 + κ) = 4π . a Taking the trace, we obtain (1.61) tr S = a(1 + 4κ) tr T, and so κ (1.62) aκ tr T = tr S. 1 + 4κ Inserting this back into (1.59) we have equally: (1.63) Sµν λ (g−1)µν tr S = a T µν , − κ where λ = 1+4κ . The requirement that (1.58) are a consequence of the field equations is therefore satisfied if and only if µν −1 µν (1.64) ν S λ(g ) tr S = 0 : identically. ∇ − In the next section we prove that this is satisfied if and only if 1 1 (1.65) λ = , i.e. κ = . 2 −2 So the Einstein equations in the presence of matter are: 1 (1.66) Sµν (g−1)µν tr S = a T µν − 2 In units where c = 1 and G = 1: a = 8π. The equations are often simply written as (1.67) Eµν = a T µν , where E is the Einstein tensor: 1 (1.68) Eµν = Sµν (g−1)µν tr S. − 2 In rationalized gravitational units, (1.69) 4πG = 1 , a = 2 . 1.3. Bianchi identities. A this Section we prove certain identities for the curvature tensor, known as the Bianchi identities, which are central to Einstein’s equations. 62 2. EINSTEIN’S FIELD EQUATIONS ∗

1.3.1. Jacobi identity. Let A, B, C be linear operators acting on some (linear) space X. Then we have the Jacobi identity: (1.70) [A, [B,C]] + [B, [C,A]] + [C, [A, B]] = 0 If X is vectorfield on it defines two kinds of linear operators: M (1) X acts on functions f by f X f. 7→ · (2) X acts on vectorfields Y by X X Y . 7→ ∇ The first kind (together with the symmetry of the connection) yields the cyclic identity of the curvature: (1.71) R(X,Y ) Z + R(Y,Z) X + R(Z,X) Y = 0 . · · · The second kind yields the Bianchi identity:

(1.72) ( X R)(Y,Z) + ( Y R)(Z,X) + ( Z R)(X,Y ) = 0 . ∇ ∇ ∇ Let us ﬁrst return to (1). We have (1.73) [X, [Y,Z]]f + [Y, [Z,X]]f + [Z, [X,Y ]]f = 0 . Using the symmetry of the connection we have

(1.74) [X, [Y,Z]] = [X, Y Z Z Y ] = X Y Z X Z Y + X. ∇ − ∇ ∇ ∇ − ∇ ∇ ∇[Z,Y ] And thus,

0 = X Y Z X Z Y + X ∇ ∇ − ∇ ∇ ∇[Z,Y ] (1.75) + Y Z X Y X Z + [X,Z]Y ∇ ∇ −∇ ∇ ∇ + Z X Y Z Y X + Z ∇ ∇ − ∇ ∇ ∇[Y,X] Now recalling the deﬁnition of curvature (3.39) from Chapter 1 we see that group- ing the terms as indicated this implies (1.71). Setting X = Eµ, Y = Eν, and Z = Eβ, where (Eµ : µ = 0, 1, 2, 3) is an arbitrary frame, we can also write α α α α α (1.76) R βµν + R µνβ + R νβµ = 0 , where R βµν = (Rµν) β ; this together with the symmetries α α (1.77) R = R ,Rβαµν = Rαβµν , βνµ − βµν − implies the pair symmetry:

(1.78) Rµναβ = Rαβµν . Now consider (2). We have

(1.79) [ X , [ Y , Z ]]W = X [ Y , Z ]W [ Y , Z ] X W ∇ ∇ ∇ ∇ ∇ ∇ − ∇ ∇ ∇ = X R(Y,Z) W + W R(Y,Z) X W X W ∇ · ∇[Y,Z] − · ∇ − ∇[Y,Z]∇ = ( X R)(Y,Z) W + R( X Y,Z) W + R(Y, X Z) W ∇ · ∇ · ∇ · + R(X, [Y,Z]) W + W. · ∇[X,[Y,Z]] 1. EINSTEIN’S FIELD EQUATIONS IN THE PRESENCE OF MATTER 63

Therefore,

0 = ( X R)(W, Z) W + ( Y R)(Z,X) W + ( Z R)(X,Y ) W ∇ · ∇ · ∇ · +R( X Y,Z) W + R(Y, X Z) W ∇ · ∇ · + R( Y Z,X) W +R(Z, Y X) W (1.80) ∇ · ∇ · + R( Z X,Y ) W + R(X, Z Y ) W ∇ · ∇ · + R(X, [Y,Z]) W + R(Y, [Z,X]) W +R(Z, [X,Y ]) W · · · + W + W + W ; ∇[X,[Y,Z]] ∇[Y,[Z,X]] ∇[Z,[X,Y ]] the last line vanishes identically by the Jacobi identity (1.70), while the remaining terms cancel as indicated. This proves the Bianchi identity (1.72). With respect to arbitrary frame Eµ : µ = 0,..., 3, setting X = Eµ, Y = Eν, Z = Eλ, W = Eβ, α α (1.81) ( X R)(Y,Z) W = ( E R)νλ Eβ = (( µR)νλ) Eα = ( µR )Eα , ∇ · ∇ µ · ∇ β ∇ βνλ the Bianchi identity reads: α α α (1.82) µR + νR + λR = 0 . ∇ βνλ ∇ βλµ ∇ βµν 1.3.2. Contracted Bianchi identities. Setting α = ν and noting that X ν (1.83) R βνλ = Sβλ , ν we obtain after summation α (1.84) µSβλ + αR λSβµ = 0 , ∇ ∇ βλµ − ∇ or α (1.85) αR = λSµβ µSλβ ∇ βλµ ∇ − ∇ the ﬁrst contracted Bianchi identity. Remark. Note that the right hand side is antisymmetric in (λ, µ) and the equation can expressed schematically as div R = curl S. Now multiply the contracted Bianchi identity α −1 µλ (1.86) αR = νSλµ λSνµ by (g ) . ∇ µνλ ∇ − ∇ Note on the left hand side −1 µλ α −1 αβ −1 µλ −1 αβ α (1.87) (g ) R µνλ = (g ) (g ) Rβµνλ = (g ) Sβν = S ν , while on the right hand side: −1 µλ (1.88) (g ) Sλµ = tr S −1 λµ α (1.89) (g ) λSνµ = αS . ∇ ∇ ν 64 2. EINSTEIN’S FIELD EQUATIONS ∗

We obtain the second contracted Bianchi identity:

µ 1 (1.90) µS ν tr S = 0 ∇ ν − 2∇ 1.3.3. Relation to Einstein equations. We have derived that

µν 1 −1 µν (1.91) ν S (g ) tr S = 0 : identically. ∇ − 2 1 As already discussed, we must therefore set λ = 2 in (1.64) to obtain the conservation laws (1.25) as a consequence of Einstein equations. The latter therefore read 1 (1.92) Sµν (g−1)µν tr S = a T µν , − 2 where a = 2 in rationalized gravitational units. Remark. A ﬁnal remark on units: The Einstein tensor has the dimension of curvature, namely L−2, and the energy momentum tensor that of energy density, or mass density times c2, i.e. ML−3[c2], where c denotes the speed of light. Newton’s constant G has dimensions T −2L3M −1 (from Newton’s law for two point masses). Hence G L T µν M [ ] = , [ ] = , c2 M c2 L3 and the Einstein equations in standard units are 8π G (1.93) Eµν = T µν . c4 2. Action Principle In this section we discuss the insight that the equations of the general theory can be derived from an action principle. 2.1. General framework. We consider a functional of the metric g and the matter ﬁelds m, on a bounded region in space time: A Z U (2.1) [g, m; ] = L[g, m] dµg A U U

The form L[g, m]dµg is called a Lagrangian, where L[g, m] depends only locally on g and m. 2.1.1. Matter equations. We consider a variation of the action with respect A to the matter ﬁelds mt, and denote d (2.2)m ˙ = mt . dt t=0 In other words we consider a variation of while keeping the metric g ﬁxed: A (2.3) (t) = [g, mt; ] A A U Assume mt is independent of t in , som ˙ = 0 in . Here is a bounded M\U M\U U region in spacetime. Assume in fact that there is an open set 0 such that 0 U U ⊂ U 2. ACTION PRINCIPLE 65

0 0 so that the above holds with replaced by . So mt is independent of t in , m˙ = 0 in 0. Thusm ˙ =U 0 in a neighborhoodU of the boundary of . M\U The Euler-LagrangeM\U equations of matter are: U d (2.4) ˙ = (t) = 0 A dtA t=0 For all such variationsm ˙ , we can write Z ˙ (2.5) g = M m˙ dµg A U · so the equations of matter are (2.6) M = 0 . 2.1.2. Gravitational equations. Let us now consider a variation of the action with respect to the metric g while keeping the matter ﬁelds m ﬁxed. In other 0 words we consider a 1-parameter family of metrics gt agreeing on , and set M\U (2.7) (t) = [gt, m; ] . A A U Here d (2.8)g ˙ = gt dt t=0 vanishes on 0. We can write M\U Z ˙ (2.9) m = G g˙ dµg A U · and the Euler-Lagrange equations of gravitation are

˙ = 0(2.10) A m (2.11) G = 0 . These should be the Einstein equations. 2.2. Gravitational action. We shall show that the Einstein equations are the Euler-Lagrange equations of the following action. Take

(2.12) [g, m; ] = G[g; ] + M [g, m; ] , A U A U A U where G is the gravitational action given by A Z 1 (2.13) G[g; ] = LG[g] dµg ,LG = tr S. A U U −4 Here S is the Ricci curvature of g. This observation is due to Hilbert. Consider the variation of (2.13) through a family of metrics gt, and denote as above by d (2.14) g˙ = gt , dt t=0 d (2.15) ˙G = G(t) , A dtA t=0 66 2. EINSTEIN’S FIELD EQUATIONS ∗ where here Z

(2.16) G(t) = LG[gt] dµgt . A U Let us ﬁrst calculate d (2.17) LG[gt] . dt t=0 We have −1 µν (2.18) tr St = (gt ) (St)µν where (St)µν is the Ricci curvature of gt. In local coordinates: α α (2.19) R βµν = (Rµν) β

(2.20) Rµν = ∂µΓν ∂νΓµ + ΓµΓν ΓνΓµ − − α α (2.21) (Γµ) β = Γµβ

α α α α γ α γ (2.22) R = ∂µΓ ∂νΓ + Γ Γ Γ Γ βµν νβ − µβ µγ νβ − νγ µβ α α α α γ α γ (2.23) Sµν = R = ∂αΓ ∂νΓ + Γ Γ Γ Γ µαν νµ − αµ αγ νµ − νγ αµ ˙ ˙ α ˙ We can now find S. Γµβ are the components of a tensorfield Γ. α α α β α β α β α β S˙µν = ∂αΓ˙ ∂νΓ˙ + Γ˙ Γ + Γ Γ˙ Γ˙ Γ Γ Γ˙ νµ − αµ αβ µν αβ µν − νβ αµ − νβ µα (2.24) α α = αΓ˙ νΓ˙ ∇ µν − ∇ αµ Therefore, −1 µν −1 µν −1 µν (2.25) (tr S)˙= (g )˙ Sµν + (g ) S˙µν = (g )˙ Sµν + tr S,˙ where −1 µν α −1 µν α (2.26) tr S˙ = α (g ) Γ˙ ν (g ) Γ˙ . ∇ µν − ∇ αµ Define the vectorfield Iα by (2.27) Iα = (g−1)µνΓ˙ α (g−1)µαΓ˙ ν , µν − νµ then α (2.28) tr S˙ = αI . ∇ Since also (2.29) (g−1)˙= g−1gg˙ −1 − −1 µν −1 µα −1 νβ (2.30) (g )˙ = (g ) (g ) g˙αβ , − we obtain α αβ (2.31) (tr S)˙ = αI S g˙αβ . ∇ − Next, we calculate

d (2.32) dµgt . dt t=0 2. ACTION PRINCIPLE 67

Here dµg is the volume form corresponding to a metric g; c.f. note on pg. 69. Since (det g)˙ = (det g) tr(g−1g˙)(2.33) −1 µν (2.34) (det g)˙ = (det g)(g ) g˙µν , and p p 1 −1 µν (2.35) ( det g)˙ = det g (g ) g˙µν , − − 2 we obtain 1 −1 µν (2.36) (dµg)˙ = (g ) g˙µν dµg . 2 Thus in view of (2.31) and (2.36) these calculations yield: Z 1 n α αβ o (2.37) ˙G = αI E g˙αβ dµg A −4 U ∇ − Here 1 (2.38) Eαβ = Sαβ (g−1)αβ tr S, − 2 and (2.39) Iα = (g−1)µνΓ˙ α (g−1)µαΓ˙ ν . µν − µν Finally, define the dual 3-form I∗ to the vectorfield I by ∗ α (2.40) Iβγδ = I (dµg)αβγδ . Then by Stokes theorem, Z Z α ∗ (2.41) αI dµg = I = 0 , U ∇ ∂U since I∗ vanishes in a neighborhood of ∂ . Hence we are left with Z U (2.42) ˙G = G g˙ dµg , A U · where G is the 2-contravariant symmetric tensorfield 1 (2.43) Gαβ = Eαβ . 4

Note on the volume form corresponding to a metric

At each x (dµg)x denotes the volume form in Tx determined by the orientation of . If (E∈ M,E ,E ,E ) is a positive orthonormal basisM of T then M 0 1 2 3 xM dµg(E0,E1,E2,E3) = 1 . ∂ ∂ ∂ ∂ Suppose now that the coordinate basis ( ∂x0 , ∂x1 , ∂x2 , ∂x3 ) is positive. We can expand ∂ X = Aα E . ∂xµ µ α α 68 2. EINSTEIN’S FIELD EQUATIONS ∗

Then ∂ ∂ X g = g( , ) = Aα η Aβ , µν ∂xµ ∂xν µ αβ ν α,β since g(Eα,Eβ) = ηαβ. In terms of matrices, we have g = AηA˜ , so taking determinants det g = det A˜ det η det A = (det A)2 , − and det A > 0 since we assumed that we have a positive basis. Thus p det A = det g , − and ∂ ∂ ∂ ∂ p dµ , , , = det A dµ (E ,E ,E ,E ) = det A = det g . g ∂x0 ∂x1 ∂x2 ∂x3 g 0 1 2 3 − By total antisymmetry, ∂ ∂ ∂ ∂ p dµ , , , = det g [µνκλ] , g ∂xµ ∂xν ∂xκ ∂xλ − where [µνκλ] is the fully antisymmetric 4-dimensional symbol.

2.3. Energy-stress tensor. In the previous section we have discussed the gravitational action: Z 1 (2.44) G[g; ] = tr S dµg , A U U −4 We have shown that for variations of the metric g, Z 1 µν 1 −1 µν (2.45) ˙G = S (g ) tr S g˙µν dµg . A U 4 − 2 Let us now return to (2.12). The action M contains the matter fields m: AZ (2.46) M [g, m; ] = L[g, m] dµg A U U In general, for variations of the matter fields m while keeping the metric g fixed, we have Z ˙ (2.47) M g = M m˙ dµg . A U · The equations of matter are simply

(2.48) ˙M = 0 : M = 0 . A g The equations of gravitation should be the variation of the action (2.12) with respect to the metric, while keeping m ﬁxed:

(2.49) ˙ = ˙G + ˙M = 0 A m A A m These are the Einstein equations 1 (2.50) Sµν (g−1)µν tr S = 2 T µν . − 2 2. ACTION PRINCIPLE 69

2.3.1. Deﬁnition. We are thus led to the deﬁnition of the energy-stress tensor T µν in the context of the action principle: Z ˙ 1 µν (2.51) M m = T g˙µν dµg A −2 U Remark. This corresponds to the principle of virtual work in classical mechanics.

If LM [g, m](x) depends on g only through g(x) (so does not depend on the derivatives of g), then

˙ ∂LM (2.52) LM m = g˙µν . ∂gµν Recalling that

1 −1 µν (2.53) (dµg)˙ = (g ) g˙µν dµg 2 we see that Z ˙ ∂LM 1 −1 µν (2.54) m = + (g ) LM g˙µν dµg . A U ∂gµν 2 Comparing to (2.51) we conclude that the deﬁnition of T µν in this case reduces to

µν ∂LM −1 µν (2.55) T = 2 (g ) LM . − ∂gµν − 2.3.2. Conservation Laws. The question that we address next is: What is the connection between the local energy-momentum conservation laws (1.25) and the equations of matter (2.48). We observe that M should be a geometric invariant. Let us make this more precise: Let Abe a spacetime domain and let φ be a diffeomorphism of onto itself.U In ⊂ other M words, φ is a diffeomorphism of the spacetime manifold U onto itself such that φ( ) = . Now ( , φ∗g) is geometrically equivalent to M( , g). But there should alsoU beU a notionM of “pullback” for the matter fields m soM that ( , φ∗g, φ∗m) is physically equivalent to ( , g, m). M M Remark. If the matter is an electromagnetic field then m = A is the electromagnetic potential 1-form and φ∗m = φ∗A is the standard pullback. Physical requirement: ∗ ∗ (2.56) M [φ g, φ m; ] = M [g, m; ] . A U A U Consider ∗ ∗ (2.57) M (t) = M [φ g, φ m; ] , A A t t U where φt is a 1-parameter group of diffeomorphisms such that φt( ) = . Then U U φt is generated by a vectorfield X, X(p) being the tangent vector to the orbit 0 0 φt(p) through p. Suppose in fact that φt is the identity outside , where is a U U 70 2. EINSTEIN’S FIELD EQUATIONS ∗

0 0 subdomain with ; so if p then φt(p) = p, so X(p) = 0. Thus X vanishes in a neighborhoodU ⊂ U of the∈ boundary U \ U ∂ of . We have by the chain rule: U U d M (2.58) A = ˙M + ˙M , dt t=0 A m A g where in the ﬁrst termm ˙ = 0, and

d ∗ (2.59)g ˙ = φ g = X g , dt t t=0 −L and in the second termg ˙ = 0, and

d ∗ (2.60)m ˙ = φ m = X m . dt t t=0 −L Then the physical requirement

d M (2.61) A = 0 dt t=0 reads Z n 1 µν o (2.62) T ( X g)µν + M X m dµg = 0 , U −2 −L · −L for any vectorﬁeld X with support in 0. Now by Leibniz rule, U (2.63) ( X g)(Y,Z) = X g(Y,Z) g( X Y,Z) g(Y, X Z) , −L − −L − −L and the connection being metric is equivalent to: (2.64) X g(Y,Z) = g( X Y,Z) + g(Y, X Z) . ∇ ∇ Moreover, for any two vectorﬁelds X, Y , we have X Y = [X,Y ], and since the connection is symmetric −L

(2.65) X Y Y X = [X,Y ] . ∇ − ∇ Substituting these above we obtain:

( X g)(Y,Z) = g( X Y,Z) + g(Y, X Z) g([X,Y ],Z) g(Y, [X,Z]) (2.66) −L ∇ ∇ − − = g( Y X,Z) + g( Z X,Y ) . ∇ ∇ Given an arbitrary frame Eα : α = 0,..., 3 set Y = Eµ, Z = Eν, then

(2.67) ( X g)µν = g( E X,Eν) + g( E X,Eµ) . −L ∇ µ ∇ ν α Expanding E X = ( µX )Eα we obtain ∇ µ ∇ α α ( X g)µν = gαν µX + gαµ νX (2.68) −L ∇ ∇ = µXν + νXµ , ∇ ∇ α where Xµ = gµαX . So

1 µν 1 µν µν (2.69) T ( X g)µν = T µXν + νXµ) = T νXµ , −2 −L −2 ∇ ∇ − ∇ because T µν is symmetric. 2. ACTION PRINCIPLE 71

Suppose we are considering a solution of the matter equations M = 0. Then the term M X m in (2.62) vanishes and we obtain · −L Z µν (2.70) T νXµ dµg = 0 , U − ∇ for all vectorfields X with compact support in . Integrating by parts this is equivalent to: U Z µν (2.71) νT Xµ dµg = 0 , U ∇ for all such vectorfields X. This is finally equivalent to the local energy-momentum conservation laws µν (2.72) νT = 0 . ∇ Remark. This is a version of Noether’s theorem. An advanced treatment of the action principle in partial differential equations can be found in [Chr00]. Example: The action of the electromagnetic field is given by Z 1 µν (2.73) E[g, A; ] = F Fµν dµg A U 16π U where

(2.74) F = dA,Fµν = ∂µAν ∂νAµ . − For variations with respect to the electromagnetic potential A be obtain

(2.75) F˙µν = ∂µA˙ ν ∂νA˙ µ Z − ˙ 1 µν ˙ E g = F Fµν dµg A 8π U (2.76) Z Z 1 µν 1 µν = F µA˙ ν dµg = µF A˙ νdµg . 4π U ∇ −4π U ∇ So we obtain Maxwell’s equations (1.51) as the matter equations of the matter action

(2.77) EM = E + M , A A A if we deﬁne the electric current J by Z ˙ µ ˙ (2.78) M g = J Aµdµg . A − U Finally, with the energy-stress tensor deﬁned by (2.51) we obtain from (2.73) explicitly in terms of F :

µν 1 µ να 1 −1 µν αβ (2.79) T = F F (g ) F Fαβ . 4π α − 4 The energy-momentum conservation laws (1.25) then automatically hold as discussed above. 72 2. EINSTEIN’S FIELD EQUATIONS ∗

1 f − (y)

ux f x

y M N

Figure 2. Mapping f from spacetime to material manifold . M N 3. The material manifold The material manifold is a 3-dimensional manifold each point of which represents a material particle. is endowed with thoseN structures which are independent of the embedding ofN the material in space. The dynamics is then described by a mapping f : :4 M → N y = f(x): the particle occupying the event x. f −1(y): the world line in of a material point y in . M N We require the inverse image f −1(y) of a material point y to be a timelike curve. There is then a unique future-directed timelike unit vector ux tangent to f −1(y) at x, y = f(x). This is the material velocity. (See Fig. 2.) Consider the diﬀerential (df)x of the mapping f at x. The null space (kernel) −1 of (df)x is the tangent line at x to the world line f (y). This is the linear span of ux. The local simultaneous space Σx of the material particle at x is the orthogonal complement (relative to gx) of ux in Tx . Then (df)x Σ (df restricted to Σx) M | x is a (linear) isomorphism of Σx onto Ty , y = f(x). We also require that this isomorphismN is orientation preserving. ( is oriented and time oriented; is oriented.) The material manifold is endowedM with a volume form ω. In fact,N in ﬂuid mechanics there is no otherN structure on : R N R ω: represents the number of particles contained in the material domain . R ⊂ N Consider the pull-back f ∗ω. This is a 3-form on . In fact, consider the exterior derivative of the 3-form f ∗ω: df ∗ω. This is a 4-formM on , equivalent to ∗ M a function φ, df ω = φ dµg. However, exterior derivatives commute with pullbacks so df ∗ω = f ∗dω . But dω = 0 because there is no non-trivial 4-form on a 3-dimensional manifold. So we obtain: (3.1) df ∗ ω = 0 .

4This may not be deﬁned on the whole spacetime M because there could be vacuum regions. 3. THE MATERIAL MANIFOLD 73

∗ Consider f ω Σ . Since (df)x Σ is an orientation preserving isomorphism this | x | x must be a volume form on Σx. But Σx is already endowed with a volume form, namely dµΣx from the spacetime metric gx; this is the volume form of the positive deﬁnite metric gx Σ . Thus there is a function n(x) > 0 such that | x

∗ (3.2) f ω = n(x) dµΣx . Σx n(x): is the number of particles per unit volume in the local rest frame of the material. v = 1/n: is the specific volume, or volume per particle. 3.1. Continuum Mechanics. In fluid mechanics the volume per particle v and the entropy per particle s are an element of the thermodynamic state space: + + (3.3) (v, s) R R . ∈ × Thermodynamic state space: the space of local thermodynamic equi- + + librium states. For a fluid this is simply R R . State function: a function κ(v, s) on the thermodynamic× state space. We shall first discuss the isentropic case, where the entropy per particle is constant throughout the fluid. This is followed by a more general discussion, which includes changes in temperature. 3.1.1. Isentropic case. The Lagrangian in fluid mechanics is LM [g, f] dµg where

(3.4) LM [g, f](x) = κ(v(x)) ; 1 here v(x) = n(x) and s is a constant. We shall analyze the dependence of n(x) on gx. Let us first extend the relation (3.2) defining n(x) to the whole of Tx . Note that M ∗ (3.5) (f ω)(ux,Xx,Yx) = 0 , since ux belongs to the null space of (df)x. Let us extend dµΣx to a 3-form dµΣx defined on all of Tx by M

(3.6) dµΣx (Xx,Yx,Zx) = dµΣx (ΠXx, ΠYx, ΠZx) , where Π denotes the orthogonal projection to Σx: (3.7) ΠX = X + g(u, X)u . Then we have in fact:

(3.8) dµΣx (Xx,Yx,Zx) = dµg(ux,Xx,Yx,Zx) . The extended relation is therefore: ∗ (3.9) (f ω)x = n(x) dµΣx . The left hand side is independent of the metric. The problem thus reduces to analyzing the dependence of dµΣx on gx. Let us work in a coordinate frame ∂ ( ∂xµ : µ = 0, 1, 2, 3); recall that ∂ ∂ ∂ ∂ p (3.10) dµg( , , , ) = det g [αβγδ] . ∂xα ∂xβ ∂xγ ∂xδ − 74 2. EINSTEIN’S FIELD EQUATIONS ∗

Then we have ∂ ∂ ∂ p (3.11) dµ ( , , ) = uα [αβγδ] det g . Σx ∂xβ ∂xγ ∂xδ − We know the dependence of the volume element on gx, namely

p 1p −1 µν (3.12) ( det g)˙= det g (g ) g˙µν , − 2 − −1 but how does ux depend on gx? The tangent line to f (y) at x is independent of gx. So ux depends on gx only through the normalization condition:

(3.13) gx(ux, ux) = 1 − It follows thatu ˙ x is colinear to ux: (3.14)u ˙ α = λuα To determine λ we vary the normalization condition (3.13): µ ν µ ν (3.15) 2 gµνu˙ u +g ˙µνu u = 0 . Then substitute (3.14) to obtain µ ν µ ν (3.16) 2λ gµνu u +g ˙µνu u = 0 . or 1 µ ν (3.17) λ = g˙µνu u . 2 Hence: 1 −1 µν (dµΣ)˙βγδ = λ(dµΣ)βγδ + (g ) g˙µν(dµΣ)βγδ (3.18) 2 1 −1 µν µ ν = (g ) + u u g˙µν(dµ ) . 2 Σ βγδ Now varying the relation (3.9), namely ∗ (3.19) (f ω)βγδ = n (dµΣ)βγδ , with respect to g, we have: ∗ (f ω)˙βγδ = 0(3.20)

(3.21) 0 =n ˙ (dµΣ)βγδ + n (dµΣ)˙βγδ .

Substituting the expression (3.18) for (dµΣ)˙βγδ we conclude that:

1 −1 µν µ ν (3.22)n ˙ + n (g ) + u u g˙µν . 2 Now recall our earlier expression (2.55) for the energy stress tensor:

µν ∂LM −1 µν (3.23) T = 2 (g ) LM . − ∂gµν −

Here LM = κ(v), and ∂L ∂κ ∂n 1 ∂κ (3.24) M = = n (g−1)µν + uµuν . ∂gµν ∂n ∂gµν −2 ∂n 3. THE MATERIAL MANIFOLD 75

Therefore ∂κ ∂κ T µν = n κ (g−1)µν + n uµuν (3.25) ∂n − ∂n ∂κ = κ uµuν + n κ (g−1)µν + uµuν . ∂n − We recognize the energy-momentum-stress tensor (1.21) of the perfect fluid (3.26) T µν = ρ uµuν + p (g−1)µν + uµuν , with ∂κ (3.27) ρ = κ , p = n κ . ∂n − Note, ρ: is energy per unit volume, e = ρ/n: energy per particle. In terms of e and v the relation ∂ρ (3.28) p = n ρ ∂n − becomes simply ∂e (3.29) p = . −∂v This is the first law of thermodynamics (1.27), (3.30) de = p dv , when ds = 0 . − 3.1.2. General case. In the more general case the state function κ is a function of n and the entropy per particle s: (3.31) κ = κ(n, s) . Here s is considered to be independent of the metric g. The temperature θ is defined by ∂κ (3.32) = n θ . ∂s Let us recall (3.9). In an arbitrary frame, ∗ α (3.33) (f ω)βγδ = n (dµΣ)βγδ = n u (dµg)αβγδ . Therefore the 3-form f ∗ω is dual to the vectorfield I, (3.34) Iα = n uα : the particle current, and the equation (3.1) reads in terms of I: ∗ µ (3.35) df ω = 0 µI = 0 . ⇐⇒ ∇ The equations of motion of a fluid are the two conservation laws: µ µI = 0 : local particle conservation law,(3.36) ∇µν νT = 0 : local energy-momentum conservation laws.(3.37) ∇ 76 2. EINSTEIN’S FIELD EQUATIONS ∗

The action functional for the ﬂuid is: Z (3.38) M [g, f, s; ] = κ(n, s)dµg . A U U The matter equations are obtained as above from variations with repect to f, Z ˙ ˙ (3.39) M = M f dµg , A g,s U · while the temperature θ on is deﬁned according to (3.32) by the variation of the action with respect to s:M Z ˙ (3.40) M = nθ s˙ dµg . A g,f U We can proceed as in Section 2.3.2 to derive from the invariance of the action the condition that

0 = ˙M = ˙M + ˙M + ˙M A A f,s A g,s A g,f Z n o 1 µν ˙ (3.41) = T g˙µν + M f + nθ s˙ dµg U −2 · Z n ν a o µ = νTµ + Ma ∂µf + nθ ∂µs X dµg U ∇ where we have used as before that µ µ (3.42) f˙ = X f = X f = X ∂µf , s˙ = X s = X s = X ∂µs , −L · −L · (3.43) g˙µν = ( X g)µν = νXν + νXµ . −L ∇ ∇ Therefore in view of the conservation laws (3.37), a (3.44) Ma ∂µf + nθ ∂µs = 0 . Since df u = 0, multiplying by uµ this reduces to · µ (3.45) u ∂µs = 0 , or us = 0 . ∇ This is the adiabatic condition, which states that the entropy is constant along the flow lines of the fluid. 3.1.3. Dust model. Let us finally consider the trivial case p = 0. Recall that according to the first law of thermodynamics the energy per particle e, the volume per particle v, and the pressure are related by ∂e (3.46) e = e(v, s) , p = . −∂v Thus p = 0 is equivalent to e being just a function of s, e = m(s), m the rest mass of the particle. Since the entropy per particle s is constant along the flow lines by (3.45) (as long as the flow is smooth), the particle conservation law (3.36), µ µ µ (3.47) µI = 0 ,I = n u , ∇ implies µ µ µ µ (3.48) µ(ρ u ) = µ(ne u ) = n u µe + e µ(nu ) = 0 , ∇ ∇ ∇ ∇ 4. COSMOLOGICAL CONSTANT 77 since we have e ρ = v : energy density in the local rest frame of the fluid, 1 n = v : number of particles per unit volume, where v is volume per particle. Moreover, the energy-momentum-stress tensor reduces to: (3.49) T µν = ρ uµ uν , so the energy-momentum conservation laws immediately imply, µν µ ν µ ν ν µ (3.50) νT = ( νu )ρ u + u ν(ρ u ) = ρ u νu = 0 . ∇ ∇ ∇ ∇ This is equivalent to the geodesic equation:

(3.51) uu = 0 . ∇ The ﬂow lines of a pressureless ﬂuid are timelike geodesics. Remark. This proves in a trivial case the geodesic hypothesis of Chapter 1, Section 3.2.

4. Cosmological constant An important modiﬁcation of the Einstein equations was proposed in 1917 which amounts to adding a cosmological term: 1 (4.1) Sµν gµν tr S + Λ gµν = 2Tµν , − 2 where Λ is the cosmological constant. Remark. Observe that the conservation laws (1.25) still hold by virtue of the Bianchi identities and metric compatibility. Remark. Note that in the absence of matter (4.1) reduces to

(4.2) Sµν = Λ gµν .

Exercises 3+1 On a domain of Minkowski space (R , η) consider the conformally ﬂat metric

2 1 0 2 1 2 2 3 3 2 gµν = Ω ηµν , where Ω(x) = c , hx, xi = −(x ) + (x ) + (x ) + (x ) , 1 + 4 hx, xi and c is a constant. (1) Show that the Riemann curvature tensor takes the form Rαµβν = c gαβ gµν − gαν gβµ .

(2) Show that every spacelike plane Π has sectional curvature KΠ = c. (3) Show that every timelike plane Π has sectional curvature KΠ = −c. 3+1 (4) Consider the case c = 1. Find and sketch the domain in R where the metric g is deﬁned. Show that the x0-axis is a geodesic and prove that the point (2, 0, 0, 0) is not reached in ﬁnite proper time. 78 2. EINSTEIN’S FIELD EQUATIONS ∗

3+1 (5) Consider the case c = −1. Again, ﬁnd and sketch the domain in R where the metric is deﬁned. Verify that the x0-axis is a geodesic. Show that the past of the point with coordinates (2, 0, 0, 0) is the whole hyperplane x0 = 0. Now consider the points p with coordinates (2 + ε, 0, 0, 0) and q with coordinates (−2 − ε, 0, 0, 0), for any ε > 0. Show that q and p can be connected by a curve with arbitrarily large arc length. Remark. The spacetime discussed in (4) is called de Sitter space and the one in (5) anti de Sitter (AdS). CHAPTER 3

Spherical Symmetry

In Section 1 we impose spherical symmetry and thus reduce the Einstein equations, 1 (0.1) Sµν gµν tr S = 2 Tµν , − 2 to a system of partial diﬀerential equations on a 1 + 1-dimensional quotient manifold; our exposition here follows [Chr95]. In Section 2 we discuss the Schwarzschild spacetime as the only solution to the vacuum equations. In Sections 3, 4 we study (in the presence of matter) the concepts and questions that emerge in Section 2 — in particular the notion of a black hole and the issue of singularities — from an evolutionary point of view. This being only a brief introduction to the study of spherically symmetric self-gravitating systems in general relativity, I encourage the reader to consult the original references [Chr95, Chr99b, Daf05].

1. Einstein’s field equations in spherical symmetry 1.1. Spacetime manifold . A spacetime is spherically symmetric if the rotation group SO(3) acts on theM spacetime manifold ( , g) by spacelike isometries. M The group orbits are then metric 2-spheres (each group orbit is diffeomorphic 2 to S ). Moreover there is an orthogonal family of timelike hypersurfaces which intersect each sphere exactly once and which are totally geodesic (that is a geodesic of the hypersurface is a geodesic of the spacetime). Consider on one hand the map determined by the group orbits from one of these hypersurfaces H onto another H0: a point p H is mapped to a point p0 H0 if and only if p and p0 lie on the same group∈ orbit S; c.f. Fig. 1. This mapping∈ is an isometry. On the other hand consider the map of one of the group orbits S onto another S0 determined by the orthogonal hypersurfaces: q S is mapped to q0 S0 if and only if q, q0 lie on the same timelike hypersurface H∈. This mapping is a∈conformal isometry, in fact the scale factor is constant. The quotient space is the space of group orbits: (1.1) = /SO(3) . Q M In general is a 1+1-dimensional manifold with boundary. The boundary of , which we denoteQ by Γ, corresponds to the set of fixed points of the group action.Q It is a timelike geodesic in . M 79 80 3. SPHERICAL SYMMETRY

p0 H0

S p q0 S0

q S H H

Figure 1. Mapping determined by group orbits, and mapping determined by orthogonal hypersurfaces, respectively.

Q Q The Lorentzian manifold is then endowed with a metric g, and ( , g) is represented by any one of theQ isometric timelike hypersurfaces H. The spacetimeQ M metric g then takes the form M Q S (1.2) g =g + g , and S ◦ (1.3) g= r2(x) γ , (x ) , ∈ Q ◦ 2 where r is a function on , and γ is the metric of the standard unit sphere S in 3 Qa A R . In local coordinates (x : a = 0, 1) on , and local coordinates (y : A = 2, 3) 2 Q on S we have M Q a b 2 ◦ A B (1.4) g =gab (x)dx dx + r (x) γAB (y)dy dy . Then 2 2 2 (1.5) Area (S) = r Area (S ) = 4π r . Note that r vanishes on Γ, the center of symmetry.1 a A The non-vanishing connection coeﬃcients are Γbc,ΓBC , a −1 ab ◦ (1.6a) Γ = r (g ) ∂br γBC , BC − B 1 (1.6b) Γ = ∂ar δBC . aC r ◦ ◦ 2 Since the Ricci curvature of (S , γ) is simply γ, ◦ ◦ (1.7) S=γ ,

1 We assume r : Q → R is a smooth function with r > 0 in the interior of Q, and g(∇r, ∇r) > 0 near Γ. 1. EINSTEIN’S FIELD EQUATIONS IN SPHERICAL SYMMETRY 81 and the Ricci curvature of ( , g) is Kg, where K is the Gauss curvature of ( , g), Q Q Q (1.8) S= Kg , we readily verify that the components of the Ricci curvature of are M M 2 (1.9a) S ab = K gab a br − r ∇ ∇ M S aA = 0(1.9b) M a ◦ (1.9c) S AB = 1 (r ar) γAB . − ∇ ∇ In view of the Einstein equations (0.1) the energy-momentum-stress tensor is also spherically symmetric, and necessarily takes has the form: a b 2 ◦ A B (1.10) Tab(x)dx dx + r (x)S(x) γAB (y)dy dy , where S is a function on . Thus the Einstein equations read: Q 2 1 (1.11a) Kgab a br = 2 Tab gab(tr T + 2S) − r ∇ ∇ − 2 a 2 (1.11b) 1 (r ar) = r tr T. − ∇ ∇ − This is a non-linear wave equation for r on . The trace of (1.11a) is Q 1 a (1.12) K r = 2S. ( = a) − r − ∇ ∇ Substituting for r from (1.11b) we obtain 1 a (1.13) K = 1 ∂ r ∂ar + tr T 2S. r2 − − We substitute this in the ﬁrst equation to obtain the Hessian equations:

1 a (1.14) a br = 1 (∂ r)(∂ar) gab r Tab gab tr T . ∇ ∇ 2r − − − The system (1.13), (1.14) is equivalent to (1.11a), (1.11b). In spherical symmetry the energy-momentum conservation laws reduce to: 2 ab a (1.15) b r T = 2r ∂ r S . ∇ This follows easily from aν ab aA ab a µA A aµ (1.16) νT = bT + AT = bT + Γ T + Γ T = 0 , ∇ ∇ ∇ ∇ Aµ Aµ using (1.6) and (1.10). Remark. In fact, one can show that the Hessian equations (1.14) in conjuction with the energy-momentum conservation laws (1.15) imply the equation for the Gauss curvature (1.13). Also the trace of the Hessian equations is (1.11b). We conclude that the Hessian equations (1.14) together with the energy- momentum conservation laws (1.15) contain all the information in the Einstein equations (0.1) in spherical symmetry. 82 3. SPHERICAL SYMMETRY

Figure 2. Mass content of a sphere.

1.2. Mass function. We define the mass function m on by: Q 2m −1 ab (1.17) 1 = (g ) ∂ar ∂br . − r Remark. m(x) represents the mass-energy content of the sphere (a group orbit) corresponding to the point x . ∈ Q Differentiate m and use the Hessian equations to obtain the mass equations: 2 b (1.18) ∂am = r Tab gab tr T ∂ r . − Note that S does not enter. Remark. It turns out that the mass equations together with the trace of the Hessian equations form a system equivalent to the original Hessian equations. Also, in view of (1.17) we can rewrite (1.13) as 2m (1.19) K = + tr T 2S. r3 − 1.3. Null coordinates on . We choose a function u whose level curves are outgoing null curves, and whichQ is increasing towards the future. The function u is then determined up to a transformation of the form (1.20) u f(u) , 7→ where f is an increasing function. We also choose a function v whose level curves are incoming null curves, and increasing towards the future; then similarly v is defined up to a transformation (1.21) v g(v) , 7→ where g is an increasing function. We then define: ∂ ∂ (1.22) L = 2 ,L = 2 . ∂v ∂u Thus we are using v as a parameter along the outgoing null curves, u as a parameter along the incoming null curves. L, L are future-directed null vectorfields, outgoing and incoming, respectively. Then set (1.23) g(L, L) = 2 Ω2 ; − this defines a function Ω > 0 on . The metric on in u, v coordinates takes the form:2 Q Q Q (1.24) g= Ω2 du dv . − 2We assume here and in the following that the coordinates u, v exist in fact globally on Q. This rules out cases where Q is e.g. a cylinder. 1. EINSTEIN’S FIELD EQUATIONS IN SPHERICAL SYMMETRY 83

The transformations (1.20), (1.21) thus correspond to conformal transformations. Remark. We can thus assume that the image of (u, v) is a bounded subset 1+1 of R (by applying another conformal transformation). Note 1 2 (1.25) guv = Ω . −2 Q The only non-vanishing connection coeﬃcients of g are: 2 ∂Ω 2 ∂Ω (1.26) Γu = , Γv = . uu Ω ∂u vv Ω ∂v In null coordinates the Gauss curvature of is given by: Q 4 ∂2 log Ω (1.27) K = Ω2 ∂u ∂v The Hessian equations (1.14) in null coordinates read: ∂2r 2 ∂Ω ∂r (1.28a) = r Tuu ∂u2 − Ω ∂u ∂u − ∂2r 1 ∂r ∂r Ω2 (1.28b) + = + r Tuv ∂u ∂v r ∂u ∂v − 4r ∂2r 2 ∂Ω ∂r (1.28c) = r Tvv . ∂v2 − Ω ∂v ∂v − The deﬁning equation of the mass function (1.17) becomes 2m 4 ∂r ∂r (1.29) 1 = , − r −Ω2 ∂u ∂v and the mass equations are 2 ∂m 2r ∂r ∂r (1.30a) = Tuv Tuu ∂u Ω2 ∂u − ∂v 2 ∂m 2r ∂r ∂r (1.30b) = Tuv Tvv . ∂v Ω2 ∂v − ∂u Also note the positivity condition on the energy-momentum-stress tensor implies

(1.31) Tuu 0 ,Tuv 0 ,Tvv 0 . ≥ ≥ ≥ Indeed, since the positivity condition is equivalent to T (X,Y ) 0 for any pair of future directed non-spacelike vectors X, Y at a point, we take (≥X,Y ) to be (L, L), (L,L), (L,L) and the conditions (1.31) immediately follow. Remark. Note that (1.28a) (which is the (u, u) component of the Einstein equations (1.11a)) can also be written as: 1 ∂r r (1.32) ∂u = Tuu . Ω2 ∂u −Ω2 Since by (1.31) the right hand side is negative this implies the convexity of r. The equation (1.32) and its analogue for the (v, v) component are also known as Raychaudhuri equations. 84 3. SPHERICAL SYMMETRY

2. Schwarzschild solution The Schwarzschild solution is a spherically symmetric solution of the vacuum equations.3 Let us now assume that Tab = 0. Note that the conservation laws (1.15) imply S = 0. Since the right hand side of the mass equations (1.18) vanishes, m is a constant. We take m > 0. 4 The (u, v) component (or trace) of the Hessian equations becomes ∂2r 1 ∂r ∂r Ω2 (2.1) + = . ∂u ∂v r ∂u ∂v − 4r Solving (1.29) for Ω2 we substitute in (2.1) to obtain: ∂2r 2m ∂r ∂r (2.2) r = . ∂u ∂v r 2m ∂u ∂v − We now deﬁne a new unknown r∗ by: ∗ r (2.3) r = r + 2m log 1 , 2m − so that dr∗ 1 (2.4) = . dr 1 2m − r Then the equation satisﬁed by r becomes in terms of the new unknown r∗: ∂2r∗ (2.5) = 0 . ∂u ∂v The general solution is given by: (2.6) r∗ = f(u) + g(v) . Given r∗, if r∗ > 0 there is a unique value of r corresponding to r∗. However, if r∗ < 0 there are two values of r corresponding to r∗, one greater and one less than 2m; see Fig. 3. We require that the representation in terms of null coordinates be such that r∗ = — which corresponds to r = 2m — is contained in the uv plane. To satisfy−∞ this requirement we take (2.7a) f(u) = 2m log u | | (2.7b) g(v) = 2m log v | | so that (2.8) f(u) + g(v) = r∗ = 2m log( u v ) . | || | Then r∗ r r (2.9) uv = e 2m = e 2m 1 . | | 2m −

3In fact it is the spherically symmetric solution of the vacuum equations; the precise statement of this uniqueness result is the content of Birkhoﬀ’s theorem. 4If m = 0, then Q × SO(3) is Minkowski space. 2. SCHWARZSCHILD SOLUTION 85

r∗

Figure 3. “Tortoise coordinate” r∗ and area radius r.

The condition that u, v are increasing towards the future selects the sign, and we obtain, ﬁnally:

r r (2.10) uv = e 2m 1 . − 2m − Thus we have  < 0 : r > 2m  (2.11) uv = 0 : r = 2m > 0 : r < 2m . We distinguish four regions, as depicted in Fig. 4: I: u < 0 , v > 0 II: u > 0 , v < 0 III: u > 0 , v > 0 IV: u < 0 , v < 0 . The level curves of r are hyperbolas in the uv-plane, timelike for r > 2m, spacelike for r < 2m. r = 0 is the spacelike hyperbola uv = 1; there are 2 branches, one in region III and one in region IV. Similarly a timelike hyperbola r > 2m has 2 branches, one in region I and one in region II. Recall (1.19), which reduces to 2m (2.12) K = . r3 Since m > 0 is constant, the curvature blows up at r = 0. Q Next let us ﬁnd g explicitly. We have ∂r∗ 2m ∂r∗ 2m (2.13) = , = . ∂u u ∂v v From these we obtain ∂r 2m2m (2.14a) = 1 ∂u − r u ∂r 2m2m (2.14b) = 1 ∂v − r v 86 3. SPHERICAL SYMMETRY u v r = 0

∂r ∂r r < 2m III: ∂u < 0 , ∂v < 0 .

III r > 2m + + Hin Hout H0 ∂r ∂r ∂r ∂r II: ∂u > 0 , ∂v < 0 . II I I: ∂u < 0 , ∂v > 0 .

Hin− Hout− IV

∂r ∂r IV : ∂u > 0 , ∂v > 0 .

r = 0

Figure 4. Global causal geometry of the Schwarzschild solution. and thus ( ∂r > 0 : in II, IV (2.15a) ∂u < 0 : in I, III ( ∂r > 0 : in I, IV (2.15b) ∂v < 0 : in II, III . In other words, I: r increases outward, and decreases inward. This is the exterior region. II: r decreases outward. III: r decreases towards the future. This is the trapped region. IV: r increases toward the future. This is the anti-trapped region. We substitute the solution (2.10) for r in the formula (1.29) for Ω to obtain: Ω2 2m∂r∗ ∂r∗ (2.16) = 1 4 − − r ∂u ∂v

3 2 32m − r (2.17) Ω = e 2m . r We can use (2.17) to show that the singularity at r = 0 is reached by any observer in the trapped region in ﬁnite proper time. Consider the timelike geodesics γα(λ) which are the rays from the origin in region III: u = αλ, v = α−1λ (α a positive constant). Then

2 r r (2.18) uv = λ = e 2m 1 . − 2m 2. SCHWARZSCHILD SOLUTION 87

As λ 1, r 0, and in fact → → r 2 (2.19) λ2 = 1 + (r3) , − 2m O so 1 r (2.20) p 1 as r 0 or λ 1 . 2(1 λ) 2m −→ → → − The arc length of these geodesics from the origin (λ = 0) to r = 0 (λ = 1) is simply r Z 1 p Z 1 du dv Z 1 (2.21) g(γ ˙α(λ), γ˙α(λ))dλ = Ω dλ = Ω dλ < , 0 − 0 dλ dλ 0 ∞ because 4m (2.22) Ω 1 as λ 1 . −→ (2(1 λ)) 4 → − While the timelike hyperbolas of constant r > 2m in regions I and II are actually timelike metric cylinders in , the hyperbolas of constant r < 2m in regions III and IV are spacelike cylindersM in . The pair of null lines u = 0, v = 0 are theM level curves r = 2m. We have the branches, (c.f. Fig. 4): + Hout: u = 0 , v > 0 is an outgoing future null cylinder. + Hin: v = 0 , u > 0 is an incoming future null cylinder. H0: u = 0 , v = 0 is a single sphere. − Hout: v = 0 , u < 0 is an outgoing past null cylinder. − Hin: u = 0 , v < 0 is an incoming past null cylinder. + Hout is the outer component of the past boundary of III or the future boundary of I. It is called the outer future event horizon. Regions I and II are asymptotically + ﬂat. Hout is the boundary of the region accessible to observations from the inﬁnity of region I. I is also called the domain of outer communications. The spacelike rays through the origin in the uv plane do not intersect regions III, IV, but go from II to region I, as we move outward. They correspond to spacelike hypersurfaces in which are isometric to each other. In each such M hypersurface H0 is a minimal sphere (the origin in the uv plane is the critical point of the function r on ). It is also called the bifurcation sphere; see Fig. 5. In the domain of outerQ communications we can introduce a coordinate which is constant on the rays through the origin:

u + v (2.23) t = 4m artanh u v − 88 3. SPHERICAL SYMMETRY

Figure 5. The bifurcation sphere H0.

Then with r viewed as a function of (u, v) implicitly deﬁned by (2.10) we readily verify: Ω2 h i Ω2dudv = (v du u dv)2 (v du + u dv)2 − 4uv − − (2.24) 2m 1 = 1 dt2 + dr2 , − − r 1 2m − r and the Schwarzschild metric in the exterior region takes the classical form:

2m 2m−1 ◦ (2.25) g = 1 dt2 + 1 dr2 + r2 γ − − r − r Remark. I recommend Chapter 2 of [DR08] for further reading on the Schwarzschild solution and black holes more generally.

3. General properties of the area radius and mass functions We have discussed in Section 2 the Schwarzschild solution. The question arises whether the characteristic properties of this vacuum solution — in particular the existence of a future event horizon — can form in evolution. Physically speaking the question is: can black holes form from the collapse of stars? We are thus interested in solutions to Einstein’s equations which arise from initial data. In other words we consider spacetimes which admit a Cauchy hypersurface.

Evolutionary Hypothesis: There is a spacelike curve Σ0 such that each past-directed causal curve from any point in terminates⊂ Q at a single Q point on Σ0. 3. GENERAL PROPERTIES OF THE AREA RADIUS AND MASS FUNCTIONS 89

Γ0 (u, v)

(u0, v) Σ0

Figure 6. Cauchy hypersurface Σ0.

Note that Σ0 is the past boundary of . (See ﬁgure 6.) We have seen that in region I of theQ Schwarzschild solution — i.e. the domain of outer communications — the area radius is increasing outwards and decreasing inwards. Let us assume that this holds along Σ0. Hypothesis on initial data: Along Σ0, ∂r ∂r < 0 , > 0 . ∂u ∂v Under these assumptions we can now show that in general — i.e. for any matter model satisfying the dominant energy condition — the area radius is decreasing inwards everywhere on . Q Prop 3.1. Under the above hypotheses we have ∂r < 0 : on . ∂u Q Proof. Recall that we can rewrite (1.28a) as

∂ 1 ∂r −2 (3.1) = r Ω Tuu . ∂u Ω2 ∂u − −2 By the positivity condition (1.31) we thus have that Ω ∂ur is a nonincreasing function of u. In view of the evolution hypothesis the past directed (ingoing) null line from any point (u, v) intersects Σ0 at a point (u0, v) with u u0; see Fig. 6. Therefore ∈ Q ≥ 1 ∂r 1 ∂r (3.2) (u, v) (u0, v) < 0 , Ω2 ∂u ≤ Ω2 ∂u by assumption on the initial data. We can now see from (1.29) that  > 0 if r > 2m ∂r  (3.3) = 0 if r = 2m ∂v < 0 if r < 2m , which partitions (by Prop. 3.1) the manifold into three regions: Q Regular region : Here ∂vr > 0, and r > 2m, R Apparent horizon : Here ∂vr = 0, and r = 2m, A Trapped region : Here ∂vr < 0, and r < 2m. T 90 3. SPHERICAL SYMMETRY

Next we can show the property that lends its name to the trapped region: Prop 3.2. The future-directed outgoing null curve from any point in the trapped region is contained in the trapped region. Furthermore, a future-directed outgoing null curve from any point on the apparent horizon cannot enter the regular region.

Proof. Exercise. With these properties of the radius at hand, let us turn to the mass function. Remarkably, we have the following monotonicity properties: Prop 3.3. In the regular region, ∂m ∂m 0 , 0 . ∂u ≤ ∂v ≥ Proof. In view of Prop. 3.1 these inequalities follow immediately from the mass equations (1.30). Finally, let us prove that under the above hypotheses the mass is a nonnegative function on . Q Prop 3.4. Under the above hypotheses, m 0 : on . ≥ Q Proof. Let (u1, v1) . We can assume without loss of generality that ∈ Q (u1, v1) lies in the exterior region, for otherwise 2m r > 0. By Prop. 3.2 the ≥ past-directed (outgoing) null curve (u1, v), lies in the exterior region as well for all v < v1, until by the evolutionary hypothesis it terminates at a point (u1, v0), v0 < v1, which either lies on Σ0 or Γ. In any case we have by Prop. 3.3 that

(3.4) m(u1, v1) m(u1, v0) . ≥ 5 Since m vanishes on Γ, and m is nondecreasing outwards along Σ0 by Prop. 3.3, we have m(u1, v0) 0 in either case. ≥ Remark. Moreover, one can show that equality m = 0 holds at a point x if and only if the energy momentum tensor and the curvature K of vanish∈in Q the entire domain of dependence of the sphere x ; see [Chr95]Q for further discussion. ∈ Q

4. Spherically symmetric spacetimes with a trapped surface The general properties of Section 3 allow us to introduce unambiguously the notions of future null infinity and Bondi mass. Recall that the coordinate transformations u f(u), v g(v) correspond to conformal transformations which we can use to7→ map into7→ a bounded subset 1+1 Q of R . The depiction of in this manner is commonly referred to as the Penrose Q 5by regularity of Γ, which we assume. 4. SPHERICALLY SYMMETRIC SPACETIMES WITH A TRAPPED SURFACE 91 diagram of . The boundary of in the topology of the plane is called the Penrose boundaryM which we denoteQ by ∂ = .6 Q Q\Q The unique limit point on ∂ of Σ0 in the outgoing direction is called spacelike infinity, and denoted by i0. LetQ us then denote by + ∂ the set of all points on the boundary which are the limiting points of outgoingI ⊂ nullQ curves along which the area radius tends to infinity; in other words (u∗, v∗) + if ∈ I (4.1) lim r(u, v) = , v :(u=u∗,v)∈Q ∞ and (u∗, v∗) ∂ . This set is called future null infinity. Note that + may be empty. ∈ Q I Prop 4.1. If + is non-empty, then + is a connected incoming null curve emanating from i0I. I Proof. To see that + is a segment of an incoming null curve, we show that I + 0 no incoming null curve emanating from Σ0 intersects . Let i have coordi- I nates (u0, v0) and consider v1 < v0, u1 > u0 such that (u1, v1) Σ0; then by ∈ Prop. 3.1 we have r(u, v1) < r(u1, v1) < for all u > u1 with (u, v1) , so + ∞ ∈ Q+ does not intersect the incoming null line v = v1. For completeness of let I ∗ + ∗ ∗ I (u , v0) and u0 < u < u ; again by Prop. 3.1 we have r(u, v) > r(u , v) while ∈ I∗ + limv→v r(u , v) = . Thus also limv→v r(u, v) = , and (u, v0) . 0 ∞ 0 ∞ ∈ I In the following we shall assume that future null infinity is not empty. In fact, we shall include it in our notion of “asymptotic flatness”:

Definition 4.2. We say that Σ0 is asymptotically flat if (i) ∂vr > 0 in a neighborhood of i0, and (ii) the limit of r = at i0. Note that then m is nondecreasing in a neighborhood of i0, and we require∞ that (iii) it has a finite limit at i0. Finally we require that (iv) + is not empty. I Remark. The condition (iv) is strictly speaking not an assumption on the inital data, but it is reasonable to include it in the definition. For if we consider the case where the energy-momentum tensor is compactly supported on Σ0 we obtain with our argument of Section 2 that a neighborhood of i0 in is isometric to the Schwarzschild solution. Similary we expect to infer that +Qis non-empty I from sufficiently fast decaying matter fields along Σ0. The fundamental question is however if future null infinity is complete, which is conjectured to be true in the context of gravitational collapse (Penrose). Weak cosmic censorship conjecture: For “generic” asymptotically flat initial data on Σ0 the maximal future development of solutions to Ein- stein’s field equations in the presence of “appropriate” matter possess a complete future null infinity.

6 Q is a manifold with boundary Γ ∪ Σ0 in the sense of “manifold with boundary”. The 1+1 “Penrose boundary” refers the boundary of Q as a set in R which by convention does not include Γ ∪ Σ0. 92 3. SPHERICAL SYMMETRY

Remark. This conjecture captures the heuristic expectation that disregarding exceptional initial conditions a self-gravitating system cannot develop singularities which are observable from infinity (“naked” singularities), even though the observation continues for all time. Consider the causal past of +, denoted by J −( +), namely the set of points in which can be reached fromI + with a past-directedI causal curve. As a consequenceQ of Prop. 3.2 the causalI past of + in is a subset of the regular region. This domain is also called the domainI of outerQ communcations. An important observation is that for the completeness of future null infinity it suffices — in cases where the following extension principle holds — that the spacetime contains a single trapped surface, i.e. that the trapped region is not empty [Daf05]. Extension Principle: A matter model is said to satisfy the extension principle if, informally, a “first singularity” emanating from the regular region can only occur on the centre; or more precisely, if p Γ is a point on the boundary of the regular region which is not∈ atR\ the centre, and q R I−(p) a point in its chronological past such that J −(p) J +(q) ∈ p ∩ then in fact p . ∩ \{ } ⊂ R ∪ A ∈ R ∪ A Prop 4.3. If is non-empty, and the matter model satisfies the extension principle, then +A∪Tis future complete. I Proof. See [Daf05]. Remark. The weak cosmic censorship conjecture has been proven for the collapse of a self-gravitating scalar field by Christodoulou in a series of papers culminating in [Chr99a]; (see also [Chr99b] for an overview). It is shown in particular that generically singularities are always “preceded” by trapped surfaces7, and the completeness of future null infinity can then be inferred from Prop. 4.3. Given the completeness of + we define the black hole region to be I (4.2) = J −( +) , B Q\ I and say, if non-empty, that the spacetime contains a black hole. The future boundary ofB the causal past of + in is called the future event horizon, I Q (4.3) + = ∂J −( +) . H I ∩ Q Clearly, + is a subset of ; however only if the above extension criterion holds thenH the event horizonR ∪ terminates A at the future end point i+ of +, and moreover the following picture of emerges: Fig. 7, (see [Daf05] forI further discussion). Q Finally, let us mention that the mass function extends to future null infinity as follows: Assuming that the mass function is uniformly bounded along Σ0,

7It is shown that generic perturbations of initial data leading to “naked singularities” give rise to solutions with the stated property. 4. SPHERICALLY SYMMETRIC SPACETIMES WITH A TRAPPED SURFACE 93

+ + B H I Γ

Σ0

Figure 7. Penrose diagram of a spherically symmetric black hole spacetime.

− + sup m M0, we have by Prop. 3.3 that m M0 on J ( ) and m extends to Σ0 ≤ ≤ I a function M on +. Again, in view of Prop. 3.3 the Bondi mass M, I (4.4) M(u) = lim m(u, v) , v→v∗ is a (not necessarily diﬀerentiable) monotone nonincreasing function of retarded time u, and we call Mf = lim inf M the ﬁnal Bondi mass. Note that by Prop. 3.4 in particular

(4.5) Mf 0 . ≥ In the proof of Prop. 4.3 a lower bound on Mf is established in terms of the area radius of the event horizon; a bound of this form is referred to as a Penrose inequality. Prop 4.4. Under the assumptions of Prop. 4.3 we have + r 2Mf : along . ≤ H In particular, i+ / +. ∈ I Proof. See [Daf05]. A broader discussion of the weak and strong cosmic censorship conjectures along with an introduction to the above mentioned scalar ﬁeld model can be found in [Chr99b].

CHAPTER 4

Dynamical Formulation of the Einstein Equations ∗

In this chapter we discuss the dynamical formulation of general relativity be- ginning with the decomposition of the Einstein equations with respect to the level sets of a time function. We then discuss a few exciting consequences of the theory, namely the non-linear laws of post-Newtonian physics in Section 2, and gravitational waves in Section 3.

1. Decomposition relative to the level sets of a time function 1.1. Choice of a Time Function. A time function on a spacetime manifold ( , g) is a differentiable function t on such that for every future directed timelikeM vector V at any point p weM have V t > 0. This implies that if q lies in the chronological future of p —∈ that M is there is· a future-directed timelike curve from p to q — then we have (1.1) t(q) > t(p) , so a time function increases toward the future. We consider the vectorfield M defined by (1.2) g(M,V ) = V t , − · for any vector V Tp , at any point p . In terms of components in an arbitrary coordinate∈ frame:M ∈ M µ ν µ (1.3) gµνM V = V ∂µt , − so the definition says: µ −1 µν (1.4) M = (g ) ∂νt . − Then M is timelike future-directed, (1.5) g(M,M) < 0 . This is equivalent to t being a time function. Let us then define a positive function φ by: 1 1 (1.6) φ = = . p p −1 µν g(M,M) − (g ) ∂µt∂νt − − This is called the lapse function associated to t. Consider the vectorfields colinear to M: N: the unit future-directed vectorfield colinear to M, T : the vectorfield colinear to M satisfying T t = 1. · 95 96 4. DYNAMICAL FORMULATION OF THE EINSTEIN EQUATIONS ∗

We have, 1 (1.7) N = p M = φM , g(M,M) − and (1.8) T = φ N ; for µν M t g ∂νt∂µt (1.9) φ N t = φ2 M t = · = − = 1 , · · g(M,M) g(M,M) − − so (1.10) T = φN = φ2M. Since the vectorﬁelds M, N, T are colinear, they have the same integral curves. The integral curves of T are parametrized by t. The integral curves of N are parametrized by arclength s (proper time). Along a given integral curve, ∂ ∂ (1.11) T = ,N = . ∂t ∂s So the relation T = φN reads: (1.12) ds = φ dt , along the given integral curve. Therefore the proper time that elapses along an integral curve C between the parameter values t1 < t2 is

Z t2 (1.13) L[C] = φ C dt ; t1 | this is the meaning of the “lapse” function. Consider now the hypersurfaces Σt, the level sets of the time function t: n o (1.14) Σc = p : t(p) = c ∈ M

The tangent space TpΣc is the subspace of Tp consisting of those vectors X M ∈ Tp such that X t = 0 , M · n o (1.15) TpΣc = X Tp : X t = 0 . ∈ M · By the deﬁnition of M, N, T , this is equivalent to (1.16) g(M,X) = 0 , g(N,X) = 0 , g(T,X) = 0 .

So TpΣc is the orthogonal complement of Np in Tp . Thus N is the future M directed unit normal to Σc. An arbitrary level set of t will be denoted by Σt. The integral curves of T are orthogonal to the Σt. 1. DECOMPOSITION RELATIVE TO THE LEVEL SETS OF A TIME FUNCTION 97

X Σt p

Figure 1. Level sets of a time function t and integral curves of T .

1.2. First and second fundamental form of Σt. The ﬁrst fundamental form is the induced metric gt on Σt. This is the restriction

(1.17) gt p = gp TpΣt of gp to TpΣt.TpΣt is a spacelike subspace, being the orthogonal complement of N. Therefore gt is positive deﬁnite. We then say that the hypersurface Σt is spacelike. The second fundamental form of Σt is a bilinear form kt in TpΣt at each p Σt ∈ (a 2-covariant tensorﬁeld on Σt). Given a pair of vectors X, Y TpΣt, ∈ (1.18) kt(X,Y ) = g( X N,Y ) . ∇ Note that since N is unit, we have 1 g( X N,N) = X(g(N,N)) = 0 , ∇ 2 so the vector X N is tangential to Σt, X N TpΣt. ∇ ∇ ∈ Prop 1.1. The second fundamental form is symmetric

(1.19) kt(Y,X) = kt(X,Y ) , so kt is a quadratic form in TpΣt at each p Σt. ∈ Proof. Extend X, Y to vectorﬁelds tangential to Σt. Then

kt(X,Y ) = g( X N,Y ) = g(N, X Y ) , ∇ − ∇ since g(N,Y ) = 0. Hence

kt(X,Y ) kt(Y,X) = g(N, X Y Y X) = g(N, [X,Y ]) = 0 , − − ∇ − ∇ − because [X,Y ] is tangential to Σt, both X, Y being tangential vectorﬁelds to Σt. 98 4. DYNAMICAL FORMULATION OF THE EINSTEIN EQUATIONS ∗

Kτ Σt+τ

ψτ

Σt

Figure 2. Illustration of curve Kτ in Στ .

Let ψt be the 1-parameter group of diffeomorphisms generated by T . Then if p Σt, ψτ (p) is a point on Σt+τ at parameter value τ along the integral curve ∈ of T initiating at p. Given a curve K0 on Σt with end points a ; b, we define the curve Kτ on Σt+τ by Kτ = ψτ (K0). Let then L(τ) be the arc length of Kτ . We shall show that d Z (1.20) L(τ) = kt(X,X)ds , τ=0 dτ K0 if K0 is parametrized by arclength and X is its unit tangent vectorfield. If K0 is parametrized with an arbitrary parameter λ rather than arclength then Z d kt(X,X) (1.21) L(τ) = dλ , dτ τ=0 X K0 | | p where X = g (X,X) is the magnitude of X = K˙ 0, the tangent field of K0. | | t Let K0 be parametrized by an arbitrary parameter λ, and let K˙ 0 be the tangent vectorfield to K0, K˙ 0(λ) being the tangent vector at K0(λ). Then K˙ τ , the tangent vectorfield to Kτ = ψτ K0 is given by: ◦ (1.22) K˙ τ (λ) = dψτ K˙ 0(λ) . · The arc length L(τ) of Kτ is given by Z r ˙ ˙ (1.23) L(τ) = gt+τ Kτ (λ), Kτ (λ) dλ ; I (here I is the parameter interval). On the surface S = τ Kτ we may define the vectorfield X by: ∪ ˙ (1.24) XKτ (λ) = Kτ (λ) We see that

(1.25) (ψτ )∗X = X, 1. DECOMPOSITION RELATIVE TO THE LEVEL SETS OF A TIME FUNCTION 99 by deﬁnition of X; (c.f. notes on push-forward on pg. 101). Hence

(1.26) [T,X] = 0 .

We want to ﬁnd

d Z g K˙ , K˙ dL dτ t+τ τ τ τ=0 (1.27) (τ) τ=0 = q dλ . dτ I ˙ ˙ 2 gt(K0, K0)

Now d (1.28) g K˙ τ , K˙ τ = T g(X,X) , dτ t+τ τ=0 · and X is tangential to the surfaces Σt+τ . Therefore (1.29) T g(X,X) = T g(X,X) = 2g(X, T X) = 2g(X, X T ) , · · ∇ ∇ because T X X T = [T,X] = 0, the connection being symmetric. Since ∇ − ∇ g(X, X N) = k(X,X) by deﬁnition of the second fundamental form, while g(X,N) = 0 we∇ obtain:

(1.30) T g(X,X) = 2φ k(X,X) .

We conclude Z ˙ ˙ dL(τ) φ k(K0, K0) (1.31) = dλ dτ τ=0 I K˙ 0 | | More generally we have the inﬁnitesimal version:

Prop 1.2 (First variational formula). If X, Y are vectorﬁelds tangential to Σt which are Lie transported by T , that is [T,X] = [T,Y ] = 0, then

T g(X,Y ) = 2 φ k(X,Y ) .

Push-forward of a vectorfield X by a diffeomorphism ψ Let ψ be a diffeomorphism of onto itself. Let also X be a vectorfield on . The push- M M 1 forward of X by ψ, ψ X is another vectorfield on , defined as follows: Let q = ψ− (p), ∗ M then (ψ X)p = dψ Xq, a vector at ψ(q) = p. Let Y be another vectorfield, generating ∗ · the 1-parameter group ψs. Then

X (ψs) X Y X = lim − ∗ . s 0 −L → s A basic result in diﬀerential geometry is that X = [Y,X]. −LY 100 4. DYNAMICAL FORMULATION OF THE EINSTEIN EQUATIONS ∗

1.3. Codazzi equations. A level set Σt together with the first fundamental form gt is a Riemannian manifold. It has a Riemannian connection and associated covariant derivative, denoted by . The covariant derivative of the ambient space ( , g) is denoted by . ∇ M ∇ Prop 1.3. The covariant derivative of (Σt, g ) is given by ∇ t X Y = Π X Y, ∇ ∇ (X, Y tangential to Σt), where Π is the orthogonal projection to Σt. Proof. Define by the projection of and show that this defines a connection which is compatible∇ with the metric and∇ symmetric. Since there is a unique Riemannian connection corresponding to a given metric, must coincide with ∇ the connection of the induced metric gt. Let X, Y be vectorfields tangential to Σt then X Y Y X = Π( X Y Y X) = Π[X,Y ] = [X,Y ] ∇ − ∇ ∇ − ∇ since the ambient connection is symmetric, and since the commutator of two vectorfields tangential to Σt is also tangential to Σt. Next, let also Z be a vectorfield tangential to Σt; then

Z(g(X,Y )) = Z(g(X,Y )) = g( Z X,Y ) + g(X, Z Y ) = ∇ ∇ = g( Z X,Y ) + g(X, Z Y ) . ∇ ∇ Let θ be a 1-form on Σt. Then X θ, X being a vectorﬁeld tangential to Σt, is deﬁned by ∇

(1.32) ( X θ) Y = X(θ(Y )) θ X Y ∇ · − · ∇ for any vectorfield Y tangential to Σt. Similarly for covariant tensorfields, in particular for the second fundamental form kt of Σt. Let X, Y , Z be vectors tangential to Σt at some point, and extend them to vectorfields tangential to Σt. Then

(1.33) ( X k)(Y,Z) = X(k(Y,Z)) k( X Y,Z) k(Y, X Z) . ∇ − ∇ − ∇ Now k(Y,Z) = g( Y N,Z) and hence ∇ (1.34) X(k(Y,Z)) = g( X Y N,Z) + g( Y N, X Z) . ∇ ∇ ∇ ∇ We substitute, interchange X, Y and substract:

(1.35) ( X k)(Y,Z) ( Y k)(X,Z) = ∇ − ∇ = g([ X , Y ]N,Z) k([X,Y ],Z) k(Y, X Z) + k(X, Y Z) ∇ ∇ − − ∇ ∇ + g( Y N, X Z) g( X N, Y Z) ∇ ∇ − ∇ ∇ The ﬁrst two terms on the right hand side are

(1.36) g([ X , Y ]N N,Z) = g(R(X,Y )N,Z) = R(Z,N,X,Y ) . ∇ ∇ − ∇[X,Y ] The remaining four terms cancel because

(1.37) g( Y N, X Z) = g( Y N, X Z) = k(Y, X Z) , ∇ ∇ ∇ ∇ ∇ 1. DECOMPOSITION RELATIVE TO THE LEVEL SETS OF A TIME FUNCTION 101 and similarly with X, Y interchanged. For, let W be the vectorﬁeld W = X Z, ∇ then ΠW = X Z; but for any vector W , ∇ (1.38) ΠW = W + g(W, N)N so g( Y N,W ) = g( Y N, ΠW ) because ∇ ∇ 1 (1.39) g( Y N,N) = Y (g(N,N)) = 0 . ∇ 2 In conclusion, we have arrived at the Codazzi equations:

(1.40) ( X k)(Y,Z) ( Y k)(X,Z) = R(Z,N,X,Y ) ∇ − ∇ for any three vectors X, Y , Z tangent to Σt at a point.

1.4. Convenient frame. Let E0 = N. We supplement E0 to a frame with vectorﬁelds (Ei : i = 1, 2, 3) tangential to the Σt and let Ei be Lie transported by T :

(1.41) [T,Ei] = 0 i = 1, 2, 3 .

In fact we may deﬁne the (Ei : i = 1, 2, 3) ﬁrst just on Σ0 and then extend them to each Σt by the push-forward of ψt. That is to say with q Σ0, p = ψt(q) Σt, we set ∈ ∈

(1.42) Ei(p) = dψt Ei(q) , · then Ei thus extended to the spacetime satisﬁes (ψt)∗Ei = Ei, hence [T,Ei] = 0. In such a frame

g00 = 1 (= g(N,N))(1.43) − g0i = 0 (= g(N,Ei))(1.44)

(1.45) gij = gij (= g(Ei,Ej)) , and the ﬁrst fundamental formula becomes: ∂gij (1.46) = 2φ kij ∂t Similarly, the Codazzi equations become:

(1.47) ikjm jkim = Rm0ij ∇ − ∇ Consider the contraction of these equations with g. While on the left hand side −1 im (1.48) (g ) kim = tr k , we ﬁnd on the right hand side: −1 im (1.49) (g ) Rm0ij = S0j = 2T0j by the Einstein equations in this frame. The contracted Codazzi equations are thus the “perp-parallel” components of the Einstein equations: i (1.50) kji j tr k = 2T0j ∇ − ∇ Remark. We refer to the components of the Einstein equations G = 0 in the decomposition relative to the level sets of a time function Σt as follows: 102 4. DYNAMICAL FORMULATION OF THE EINSTEIN EQUATIONS ∗

: This is simply the equation G00 = 0. ⊥⊥ : The “perpendicular-parallel” components are the equations G0j = 0, ⊥k E0 = N being perpendicular, and Ej being parallel to Σt. : These are the equations Gij = 0. kk 1.5. Gauss equations. In this section we derive the ( )-components of the Einstein equations. ⊥⊥ We denote by R(W, Z, X, Y ) the curvature tensor of the induced metric g. In this Section, W, Z, X, Y are vector ﬁelds tangential to Σt, and (1.51) R(W, Z, X, Y ) = g(W, R(X,Y ) Z) , · where R(X,Y ) denotes the intrinsic curvature transformation to Σt,

(1.52) R(X,Y ) Z = X Y Z Y X Z Z. · ∇ ∇ − ∇ ∇ − ∇[X,Y ] What is the relation between the intrinsic curvature tensor R of (Σt, g) and the curvature tensor R of the ambient spacetime restricted to directions tangential to the spacelike hypersurface Σt? Consider any vectorﬁeld U (not necessarily tangential to Σt) on . Then M (1.53) g(W, ΠU) = g(W, U) where Π is the orthogonal projection to Σt: (1.54) ΠU = U + g(N,U)N

Recalling that X Y = Π X Y we thus have ∇ ∇ (1.55) g(W, R(X,Y ) Z) = g(W, X Y Z Y X Z Z) . · ∇ ∇ − ∇ ∇ − ∇[X,Y ] Now,

(1.56) X Y Z = X (Π Y Z) = Π X Y Z + ( X Π) Y Z, ∇ ∇ ∇ ∇ ∇ ∇ ∇ · ∇ where (1.57) ( X Π) U = X (ΠU) Π X U = X U + g(U, N)N Π X U ∇ · ∇ − ∇ ∇ − · ∇ = X U + g( X U, N)N + g(U, X N)N + g(U, N) X N Π X U ∇ ∇ ∇ ∇ − · ∇ = g(U, X N)N + g(U, N) X N. ∇ ∇ We apply this to the case where U = Y Z. We have ∇ 1 (1.58) g( X N,N) = X g(N,N) = 0 , ∇ 2 · so X N is tangential to Σt. We have ∇ (1.59) g( X N,Y ) = k(X,Y ) ∇ for any vector Y tangential to Σt. At a given point p Σt, we consider X as a ∈ vector at p. Moreover X N is considered a vector at p Σt tangent to Σt, that depends linearly on X:∇ ∈

X TpΣt X N TpΣt ∈ 7−→ ∇ ∈ 1. DECOMPOSITION RELATIVE TO THE LEVEL SETS OF A TIME FUNCTION 103

] This is a linear transformation of TpΣt which we shall call k : ] (1.60) X N = k X ∇ · The linear transformation k] corresponds to the symmetric bilinear form k: (1.61) g(k] X,Y ) = k(X,Y ) · ] The components of k with respect to the frame (Ei : i = 1, 2, 3) are ] m −1 mi (1.62) (k ) j = (g ) kij . We were considering

(1.63) g(W, X Y Z) = g(W, Π X Y Z + ( X Π) U) , ∇ ∇ ∇ ∇ ∇ · where U = Y Z. The ﬁrst term of ∇ (1.64) ( X Π) U = g(U, X N)N + g(U, N) X N ∇ · ∇ ∇ does not contribute as g(W, N) = 0. With the above we ﬁnd

(1.65) g(U, N)g(W, X N) = g(U, N)k(X,W ) , ∇ and,

(1.66) g(U, N) = g( Y Z,N) = g(Z, Y N) = k(Y,Z) . ∇ − ∇ − Therefore we obtain R(W, Z, X, Y ) = g(W, Π X Y Z Π Y X Z Π Z) ∇ ∇ − ∇ ∇ − ∇[X,Y ] (1.67) k(X,W )k(Y,Z) + k(Y,W )k(X,Z) − = R(W, Z, X, Y ) k(X,W )k(W, Z) + k(Y,W )k(X,Z) . − Setting W = Em, Z = Ei, X = En, Y = Ej these equations read:

(1.68) Rminj + kmnkij kmjkni = Rminj − The ﬁrst contracted Gauss equations are obtained by contracting these equations with g, yielding the intrinsic Ricci curvature to Σt on the left hand side: −1 mn (1.69) (g ) Rminj = Sij On the right hand side: −1 µν −1 mn (1.70) (g ) Rµiνj = Sij = R0i0j + (g ) Rminj . − Also −1 mn (1.71) (g ) kmnkij = tr k kij −1 mn ] n (1.72) (g ) kmiknj = (k ) iknj . ] n n We now denote (k ) i simply by k i. The contracted Gauss equations therefore read: n (1.73) Sij + tr k kij k knj = R0i0j + Sij − i The second contaction with g yields the intrinsic scalar curvature of Σt on the left hand side: −1 ij (1.74) tr S = (g ) Sij 104 4. DYNAMICAL FORMULATION OF THE EINSTEIN EQUATIONS ∗

Also, −1 ij n n i −1 ij −1 mn 2 (1.75) (g ) k knj = k k = (g ) (g ) kimkjn = k , i i n | | is the sum of the squares of the entries in an orthonormal frame. On the right hand side −1 ij (1.76) (g ) R0i0j = S00 −1 ij −1 µν (1.77) S00 + (g ) Sij = (g ) Sµν = tr S. − The second contracted Gauss equations are thus the ( )-component of the Ein- stein equations: ⊥⊥

2 2 (1.78) tr S + (tr k) k = 2S00 + tr S = 4 T00 − | | Remark. Recall that T00 is the energy density of matter as seen by the observers whose world lines are the normal lines. 1.6. Acceleration equations. In the dynamical formulation of general relativity we have a formal relation with classical mechanics:

gij: position, kij: velocity. With the ﬁrst fundamental form understood as position, the correspondence of the second fundamental form with velocity is simply expressed in

1 ∂gij (1.79) kij = . 2φ ∂t The ( )-components of the Einstein equations will precisely be the equations for the accelerationkk : 1 ∂kij (1.80) . φ ∂t Let us however first discuss the acceleration in the real sense, namely the acceleration of the observers. 1.6.1. Proper acceleration. Consider a future-directed timelike curve Γ with a unit future-directed tangent field u. The (proper) acceleration is simply uu, a vectorfield along the curve, which is orthogonal to the curve since ∇ 1 (1.81) g( uu, u) = u g(u, u) = 0 . ∇ 2 “Proper acceleration” is thus geodesic curvature. In Section 1.1 we have discussed a 3-parameter family of timelike curves: the normal lines. N is here in the role of the velocity u. The acceleration is therefore N N; a vectorfield tangential to Σt. To calculate N N we use the gradient vectorfield∇ M: ∇ µ −1 µν µ (1.82) M = (g ) ∂νt = ∂ t − − (1.83) N = φM T = φN 1. DECOMPOSITION RELATIVE TO THE LEVEL SETS OF A TIME FUNCTION 105

Consider ﬁrst M M in arbitrary local coordinates: ∇ µ ν µ (1.84) ( M M) = ∂ t ν∂ t ∇ ∇ So µ ν ν gµλ( M M) = ∂ t ν∂λt = ∂ t λ∂νt ∇ ∇ ∇ (1.85) 1 ν −3 = ∂ ∂ t∂νt = φ ∂ φ , 2 λ λ where we have used the fact that the Hessian is symmetric (cf. note on p. 107):

(1.86) µ∂νt = ν∂µt ∇ ∇ We conclude that µ −3 µ (1.87) ( M M) = φ ∂ φ ∇ µ −2 µ (1.88) ( N M) = φ ∂ φ ∇ and in view of

(1.89) N N = φ N M + (Nφ)M ∇ ∇ also µ −1 µ µ (1.90) ( N N) = φ ∂ φ + (Nφ)N . ∇ Therefore

(1.91) N N = Π U, ∇ · where (1.92) U µ = φ−1∂µφ = ∂µ log φ , and Π denotes the orthogonal projection to Σt as above. Since (1.93) U = φ−1 φ = log φ ∇ ∇ (1.94) Π U = φ−1 φ = log φ · ∇ ∇ we ﬁnally obtain that

(1.95) N N = log φ . ∇ ∇ Thus in the formal analogy with mechanics, log φ plays the role of the New- tonian potential.

Notes on the Hessian of a function The Hessian of a function f is deﬁned by ( 2f)(X,Y ) = ( df)(X,Y ) = ( df)(Y ) = X(df Y ) df( Y ) = X(Y f) Y f . ∇ ∇ ∇X · − ∇X −∇X · It is symmetric because ( df)(X,Y ) ( df)(Y,X) = XY f Y Xf ( Y X)f = 0 ∇ − ∇ − − ∇X − ∇Y by the symmetry condition of the connection: Y X = [X,Y ] . ∇X − ∇Y 106 4. DYNAMICAL FORMULATION OF THE EINSTEIN EQUATIONS ∗

In local coordinates this follows immediately from the symmetry properties of the connection coefficients: ∂ f = ∂ ∂ f Γα ∂ f . ∇µ ν µ ν − µν α 1.6.2. Second variational forumula. Let X, Y be vectorfields defined and tangential to Σ0. Extend them to by the condition that they are Lie transported by T : M (1.96) [T,X] = [T,Y ] = 0 .

Then X, Y are tangential to each Σt. Consider now (1.97) T k(X,Y ) . · That is: (1.98) T g( X N,Y ) = φN g( X N,Y ) = ∇ ∇ n o = φ g( N X N,Y ) + g( X N, N Y ) ∇ ∇ ∇ ∇ We have

(1.99) N X N = X N N + R(N,X) N + N ∇ ∇ ∇ ∇ · ∇[N,X] and, Y being tangential to Σt, −1 −1 g( X N N,Y ) = g( X (φ φ),Y ) = g( X (φ φ),Y )(1.100) ∇ ∇ ∇ ∇ ∇ ∇ (1.101) g(R(N,X) N,Y ) = R(Y,N,N,X) = R(X,N,Y,N) . · − Now N = φ−1T and [T,X] = 0, so (1.102) [N,X] = [φ−1T,X] = X(φ−1) T = φX(φ−1) N − − −1 −1 (1.103) N = φX(φ ) N N = X(φ ) φ ∇[N,X] − ∇ − ∇ (1.104) g( N,Y ) = X(φ−1)(Y φ) . ∇[N,X] − Therefore, −1 −1 2 (1.105) g( X (φ φ),Y ) + g( N,Y ) = φ ( φ)(X,Y ) , ∇ ∇ ∇[N,X] ∇ and we conclude −1 2 (1.106) g( N X N,Y ) = φ φ(X,Y ) R(X,N,Y,N) . ∇ ∇ ∇ − This is the ﬁrst term in (1.98). For the second term, we write −1 (1.107) N Y = Y N + [N,Y ] = Y N + φY (φ )N, ∇ ∇ ∇ and so ] ] (1.108) g( X N, N Y ) = g( X N, Y N) = g(k X, k Y ) . ∇ ∇ ∇ ∇ · · The result is the second variational formula, (1.109) ( T k)(X,Y ) = T k(X,Y )) = −L 2 n o = ( φ(X,Y ) + φ R(X,N,Y,N) + g(k] X, k] Y ) , ∇ − · · 1. DECOMPOSITION RELATIVE TO THE LEVEL SETS OF A TIME FUNCTION 107

timelike geodesic N

X Σ0

Figure 3. Geodesic congruence and normal lines to Σ0 . which expressed in the convenient frame, setting X = Ei, Y = Ej, where (1.110) (k] X)m = (k])m = km · i i (1.111) (k] Y )m = (k])m = km · j j ] ] m n m (1.112) g(k X, k Y ) = g k k = k kmj · · mn i j i then reads

∂kij n m o (1.113) = i jφ + φ Ri0j0 + k kmj . ∂t ∇ ∇ − i Remark. In the case φ = 1 the normal lines are timelike geodesics parametrized by arc length: −1 (1.114) N N = φ φ = 0 . ∇ ∇ The Σt are the level sets of the distance function from Σ0. k is the second fun- ] damental form of the hypersurfaces Σt. Here we expressed k in terms of its components in a Lie transported frame. In Section 3.2.3 we found that its components in a parallel orthonormal frame form the rate of strain matrix θ associated to the geodesic congruence of the normal lines, see (3.78). The initial condition θ = 0 for the Jacobi equation corresponds to parallel transport of N along a spacelike geodesic. Note that in the case φ = 1: N = T . Here N is not parallel transported along Σ0 but is simply the unit normal to Σ0. Now ] (1.115) N X = X N = k X, ∇ ∇ · where the left hand side is the velocity V , and X on the right hand side the displacement in the Jacobi equation. The initial condition k = 0 corresponds to Σ0 being totally geodesic for the spacelike geodesic congruence orthogonal to N at a point. Draw all spacelike geodesics orthogonal to a given timelike geodesic through a given point on Σ0, as in Fig. 3. The unit tangent field X to this spacelike 108 4. DYNAMICAL FORMULATION OF THE EINSTEIN EQUATIONS ∗ geodesic congruence – not defined at the origin – satisfies: X X = 0. Hence ∇ (1.116) k(X,X) = g( X N,X) = g(N, X X) = 0 . ∇ − ∇ At the origin X is any unit vector, and it follows that k = 0 at the origin. For this reason the initial condition for the Jacobi equation becomes V = 0, and the two conditions indeed coincide. 1.6.3. The ( )-components of the Einstein equations. Substitute in the second variation formulakk (1.113) for the curvature from the first contracted Gauss equation (1.73) to obtain:

∂kij 2 n m o (1.117) = φ φ S¯ij + tr k kij 2k kmj Sij ∂t ∇ij − − i − The ( )-components of the Einstein equations kk 1 (1.118) Sij = 2 Tij g tr T − 2 ij are thus expressed in the evolution equations

∂kij 2 n m o (1.119) = φ φ S¯ij + tr k kij 2k kmj 2Tij + g tr T . ∂t ∇ij − − i − ij

In the mechanical correspondence, where g plays the role of position and kij that of velocity related by (1.79), we can think of (1.119) as an equation for the acceleration. It captures the dynamics of the gravitational system. The ( ) components of the Einstein equations are complemented by the ( ) and (kk ) equations (1.50) and (1.78) which are constraints on the inital conditions.⊥k They⊥⊥ are called constraint equations. 1.6.4. Trace of second variation formula. We shall also derive an equation for the trace of the second fundamental form k. Since ∂ ∂ −1 ij −1 ij ∂kij ij (1.120) tr k = (g ) kij = (g ) 2φ k kij , ∂t ∂t ∂t − we can substitute from (1.119) to obtain

∂ n 2 o (1.121) tr k = φ φ tr S + (tr k) 3T00 + tr T . ∂t 4 − − Then substitute for tr S from the second contracted Gauss equation (1.78) to obtain

∂ n 2 o (1.122) tr k = φ φ k + T00 + tr T . ∂t 4 − | | 1.7. Summary. We have discussed the decomposition of the Einstein equations (in rationalized gravitational units where 4πG = 1; also at this point c = 1) with respect to the level sets Σt of an arbitrary time function t, and associated lapse function φ: −2 −1 µν (1.123) φ = (g ) ∂µt∂νt . − We have discussed the normal lines, a family of timelike curves normal to Σt (which can be viewed as the histories of a system of observers), parametrized by t. 1. DECOMPOSITION RELATIVE TO THE LEVEL SETS OF A TIME FUNCTION 109

The tangent vectorfield is denoted by T , T t = 1, and related to the unit tangent field N by T = φN. · Moreover we use a frame field (Eµ : µ = 0, 1, 2, 3), where E0 = N and the Ei : i = 1, 2, 3 are tangential to Σt and Lie transported by T :[T,Ei] = 0. In this frame we have:

(1.124) g00 = 1 , gi0 = 0 , gij = g . − ij We denote by S the Ricci curvature of ( , g), and by S the Ricci curvature M of (S, g). Let us also introduce pi = T i0: momentum density relative to the system of observers whose world lines are the normal lines. ρ = T 00: energy density relative to the same system of observers. We have seen that the ( )- and ( )-components of the Einstein equations with respect to this decomposition⊥k are⊥⊥constraint equations, which may referred to as the momentum constraint and energy constraint respectively: j (1.125) kij i tr k = 2T0i = 2pi ∇ − ∇ − 2 2 (1.126) tr S k + (tr k) = 4T00 = 4ρ − | | The ﬁrst variation formula

∂gij (1.127) = 2φ kij ∂t can be viewed as the deﬁnition of the second fundamental form k, and for the second variation formula we have found:

∂kij 2 n m o n 1 o (1.128) = φ φ Sij 2kimk + tr k kij + 2φ Tij g tr T ∂t ∇ij − − j − 2 ij Note also that (1.129) tr T = ρ + tr T, − where T ij = Tij are the components of the stress tensor relative to the system of observers whose histories are the normal lines. The proper acceleration of these observers is given by

(1.130) N N = log φ , ∇ ∇ which suggests that log φ is to be identiﬁed with the Newtonian potential. α Given the frame (Eα : α = 0, 1, 2, 3) we can deﬁne the dual frame (θ : α = α α 0, 1, 2, 3) by δ (Eβ) = δβ , and α β (1.131) g = gαβθ θ , gαβ = g(Eα,Eβ) . ⊗ −1 −1 0 Here E0 = N = φ T = φ ∂t and so θ = φdt, and 0 0 i j g = θ θ + gij θ θ (1.132) − ⊗ ⊗ = φ2dt2 + g θi θj − ij ⊗ 110 4. DYNAMICAL FORMULATION OF THE EINSTEIN EQUATIONS ∗

Σt

Σ0 U0 Figure 4. Variation of a spacelike hypersurface.

If we choose (Ei : i = 1, 2, 3) to be a coordinate frame (which in general we do not ∂ need) Ei = ∂xi then (1.133) g = φ2dt2 + g dxidxj . − ij 1.8. Maximal foliations. We shall now make a choice of a time function: (1.134) tr k = 0

Consider a limiting case, where the level sets Σt coincide with Σ0 outside a domain 0 Σ0 with compact closure in Σ0; see figure 4. Then φ vanishes outside U ⊂ 0 on Σ0, and outside the corresponding domain t = Φt( 0) on Σt. Let V (t) be U U U the volume of t. Then U dV Z d Z (1.135) = dµg = φ tr k dµg , t=0 t t=0 Σ 0 dt U0 dt U0 0 by the first variational formula, because Z Z p 1 2 3 (1.136) V (t) = dµgt = det gt θ θ θ U0 U0 ∧ ∧ and ∂ p 1 ∂gij p (1.137) det g = (g−1)ij det g ∂t t 2 ∂t t ∂gij (1.138) = 2φ kij . ∂t The condition that the volume of 0 is stationary means that U d (1.139) V (t) = 0 dt t=0 for any such variation, that is Z (1.140) φ tr k dµg = 0 U0 for any nonnegative function φ vanishing outside 0 in Σ0, and any 0 with com- U U pact closure in Σ0. This is thus equivalent to the conditon (1.134). Remark. The above formulas hold in fact for arbitrary variations vanishing outside 0. A general variation does not correspond to a foliation by the level sets of aU time function, but to a normal differentiable homotopy. In other words, consider a differentiable mapping H : I Σ0 , where I = ( ε, ε) is a × → M − parameter interval, such that H{λ}×Σ0 is a diffeomorphism onto its image Σλ, a spacelike hypersurface in , and H{0}×Σ0 is the identity mapping on Σ0. Such a mapping H is called a differentiableM homotopy, which is called normal if for any 1. DECOMPOSITION RELATIVE TO THE LEVEL SETS OF A TIME FUNCTION 111 point p Σ0, H is a timelike curve in orthogonal to the Σλ. Along each ∈ I×{p} M such timelike curve can define φ = dτ/dλ which generalizes the lapse function of a foliation and is not in general nonnegative. Then as before Z (1.141) φ tr k dµg = 0 tr k = 0 on Σ0 . U0⊂Σ0 ⇒ A spacelike hypersurface Σ in is called maximal if M (1.142) tr k = 0 . It corresponds to a maximum of the volume under compactly supported variations. The motion is analogous to that of a minimal surface in Euclidean space (Plateau’s problem). Here we consider complete spacelike hypersurfaces, in fact Cauchy hypersurfaces. In Minkowski spacetime the only complete maximal spacelike hypersurfaces are the spacelike hyperplanes. In a general spacetime which is asymptotically flat at spacelike infinity there is a unique maximal hypersurface asymptotic to a given hyperplane at spacelike infinity. A time function t shall be called maximal if each of its level sets Σt is a maximal hypersurface and the associated lapse function φ satisfies

(1.143) φ 1 at along each Σt. −→ ∞ Then t becomes proper time at infinity. 1.9. Equations of motion. Let us finally discuss the decomposition of the conservation laws µν (1.144) νT = 0 . γ ∇ The connection coefficients Γαβ are defined by γ (1.145) E Eβ = Γ Eγ , ∇ α αβ where E0 = N, and the tangential Ei : i = 1, 2, 3 are transported by T = φN:

(1.146) [T,Ei] = 0 We have seen that the acceleration of the observers is −1 −1 −1 ij (1.147) E E0 = N N = φ φ = φ (g ) jφ Ei , ∇ 0 ∇ ∇ ∇ and thus 0 i −1 −1 ij (1.148) Γ = 0 Γ = φ (g ) jφ . 00 00 ∇ Moreover, j (1.149) E E0 = E N = k Ej , ∇ i ∇ i i hence 0 j j (1.150) Γi0 = 0 Γi0 = ki ; next, −1 −1 E Ei = N Ei = φ T Ei = φ E T ∇ 0 ∇ ∇ ∇ i (1.151) −1 −1 j = φ E (φN) = φ (Eiφ)N + k Ej ∇ i i 112 4. DYNAMICAL FORMULATION OF THE EINSTEIN EQUATIONS ∗ hence 0 −1 j j (1.152) Γ = φ iφ Γ = k . 0i ∇ 0i i Note that: (1.153) Γ0 = Γ0 . 0i 6 i0 Finally, 0 m (1.154) E Ej = Γ N + Γ Em , ∇ i ij ij where 0 (1.155) Γ = g( E Ej,N) = g(Ej, E N) = kij , ij − ∇ i ∇ i and m ¯m (1.156) Γij = Γij are the connection coeﬃcients of the induced connection on Σt: m (1.157) E Ej = Γ¯ Em ∇ i ij m (1.158) E Ej = Π E Ei = Γ Em , ∇ i ∇ i ij where Π is the orthogonal projection to Σt. Now the equations of motion are: µν µ λν ν µλ (1.159) Eν(T ) + Γνλ T + Γνλ T = 0 However, for a maximal foliation: ν 0 i Γν0 = Γ00 + Γi0 = tr k = 0(1.160) ν 0 i −1 i (1.161) Γ = Γ + Γ = φ jφ + Γ¯ . νj 0j ij ∇ ij With our previous notation (1.162) T 00 = ρ , T 0i = pi ,T ij = T¯ij , the conservation laws henceforth read:

1 ∂ρ i −1 i ij (1.163) + ip + 2φ ( iφ) p + kijT¯ = 0 φ ∂t ∇ ∇

i 1 ∂p ij −1 ij −1 i i j (1.164) + jT¯ + φ ( jφ)T¯ + φ ( φ)ρ + 2k p = 0 φ ∂t ∇ ∇ ∇ j Remark. The last terms can be compared with the Lorentz force density in Maxwell’s theory: i ν i 0 i j i i j k (1.165) F νJ = F 0J + F jJ = E σ + εjkJ B , where J 0 = σ denotes the charge density, and J i the current density.

Exercises 2. SLOW MOTION APPROXIMATION 113

1. (Kasner metric) Let us look at the Kasner metric 3 2 X 2p i 2 g = −dt + t i (dx ) . i=1

(1) Calculate the components of the second fundamental form kij . (2) Calculate the components of the Ricci tensor Sij . (3) Show that the Kasner metric is a solution of the Einstein vaccum ﬁeld equations if and only if 3 3 X X 2 pi = pi = 1. i=1 i=1 2. (Scalar curvature of maximal hypersurface) Suppose we have a maximal foliation. Look at the scalar curvature R of a maximal hypersurface. Show that R ≥ 0 with R = 0 at a point if and only if kij = 0 and ρ = 0 at this point.

2. Slow Motion Approximation The slow motion approximation consists formally in the limit c . −→ ∞ 2.1. Field equations in conventional units. In units where c = 1 conventional time t is related to the spacetime coordinate x0 by 6 (2.1) x0 = c t . We shall keep rationalized units where 4πG = 1, and the Einstein equations are 1 2 (2.2) Sµν (g−1)µν tr S = T µν . − 2 c4 We shall consider the decomposition of the Einstein equations with respect to a maximal time function x0 = ct in these units. Recall also the quasi-Newtonian hierarchy discussed in Section 1.2.1 according to which T 00 = c2 ρ ρ : mass density T 0i = c pi pi : momentum density T ij = T¯ij : stress and in particular T 00 T 0i T ij. In these units the constraint equations read tr k = 0(2.3a) j 2 (2.3b) kij = pi ∇ −c3 4 (2.3c) tr S k 2 = ρ , − | | c2 and (1.122) becomes the lapse equation: 1 1 (2.4) φ = φ k 2 + ρ + tr T . 4 | | c2 c4 114 4. DYNAMICAL FORMULATION OF THE EINSTEIN EQUATIONS ∗

The ﬁrst and second variation formulas are now 1 ∂gij (2.5a) = 2φ kij c ∂t

1 ∂kij m (2.5b) = i jφ φ Sij 2kimk c ∂t ∇ ∇ − − j 1 1 1 1 + 2 φ ρ g + T¯ij g tr T¯ . 2 c2 ij c4 − 2 ij Finally introducing c into the equations of motion we obtain:

∂ρ i i 1 ij + φ i p + 2( iφ)p + φ kijT¯ = 0(2.6a) ∂t ∇ ∇ c i ∂p ij ij 2 i i j + φ jT¯ + ( jφ)T¯ + c ( φ)ρ + 2cφ k p = 0(2.6b) ∂t ∇ ∇ ∇ j −2 −1 We have to determine φ to order c and kij to order c just to obtain the Newtonian theory. And to obtain the equations of motion correct to order c−2 we −4 −3 −2 have to determine φ to the order c , kij to order c , and gij to order c . 2.2. Slow motion limits. From the momentum constraint (2.3b) we see that we may assume that the limit 3 (2.7) lim c kij = lij c→∞ exists, so

1 −3 (2.8) kij = lij + o c . c3 Definition 2.1. We say f = oc−α, with α 0, if f and its derivatives up to certain ﬁxed order, after multiplying by cα, tend≥ to 0 as c . → ∞ We also make the basic assumption that (2.9) lim φ = 1 , c→∞ which is now seen to be consistent with the lapse equation (2.4). From the ﬁrst variation formula (2.5a) we then have

∂gij 2 −2 (2.10) = lij + o c . ∂t c2 We also assume that on Σ0 the ﬁrst fundamental form gij tends to the Eu- clidean metric as c . Choosing rectangular coordinates of the limiting ﬂat metric we have: → ∞

(2.11) lim g = δij : on Σ0 . c→∞ ij Hence by (2.10) also for each t > 0:

(2.12) lim g = δij : on Σt . c→∞ ij We shall denote the covariant derivative with respect to the limiting Euclidean metric simply by , and the Laplacian by . ∇ 4 2. SLOW MOTION APPROXIMATION 115

Consider the coeﬃcients of φ in the lapse equation: 1 1 (2.13) lim c2 k 2 + ρ + tr T¯ = ρ c→∞ | | c2 c4 Therefore (2.14) lim c2φ 1 = ψ c→∞ − exists and satisﬁes (2.15) ψ = ρ . 4 ψ is the Newtonian potential. The equations of motion (2.6) in the limit c then become: → ∞ ∂ρ i + ip = 0(2.16a) ∂t ∇ i ∂p ij i + jT¯ + ( ψ)ρ = 0(2.16b) ∂t ∇ ∇ We now consider the second variation equation (2.5b). Multiply (2.5b) by c2 and take the limit c . In view of (2.8) the left hand side vanishes in the limit, and we conclude that→ ∞ 2 (2.17) lim c Sij = Qij c→∞ exists and is given by

(2.18) Qij = i jψ + ρ δij . ∇ ∇ The trace of this equation is (2.19) tr Q = ψ + 3ρ = 4ρ 4 by (2.15). This coincides with the limit of the energy constraint (2.3c). We can substitute for ρ in terms of ψ in (2.18) to obtain:

(2.20) Qij = i jψ + ψ δij ∇ ∇ 4 We take the trace-free part of that: 1 Qˆij = Qij δij tr Q (2.21) − 3 1 = i jψ δij ψ ∇ ∇ − 3 4 Now we take the symmetric curl of Qˆ with respect to the limiting Euclidean metric. That is deﬁne (c.f. notes on pg. 120):

1 ab ab (2.22) cij = curl Qˆij = aQˆbj + aQˆbi 2 i ∇ j ∇ We have ab ab 1 aj 1 a (2.23) aQˆbj = a b jψ a ψ = a ψ i ∇ i ∇ ∇ ∇ − 3 i ∇ 4 3 ij ∇ 4 116 4. DYNAMICAL FORMULATION OF THE EINSTEIN EQUATIONS ∗ because the covariant derivatives a with respect to the ﬂat metric δ commute. This expression in antisymmetric in∇ i, j, hence

(2.24) cij = 0 . Prop 2.2 (Bach’s Theorem). Let (Σ, g) be a 3-dimensional manifold and S the Ricci curvature of g. Define the Bach tensor B by B = curl S.ˆ Then the vanishing of B is necessary and sufficient for the existence of a positive function χ on Σ such that g = χ4e where e is flat. In other words, then g is conformally flat. Remarks on Bach’s theorem. Consider first the linearized version: We consider a 1-parameter family of metrics gij(λ) such that gij λ=0 = eij is the flat metric, and denote | dgij g˙ = . ij dλ λ=0 Let Sij(λ) be the Ricci curvature of gij(λ), and d S˙ij = Sij . dλ λ=0 Then m m S˙ij = mΓ˙ jΓ˙ , ∇ ij − ∇ im where denotes the covariant derivative with respect to e, and ∇

˙ m d m 1 −1 mn Γij = Γij = (e ) ig˙ jn + jg˙ in ng˙ ij . dλ λ=0 2 ∇ ∇ − ∇ Suppose now that the metric variation is conformal: d g˙ = 4χe ˙ ij , χ˙ = χ λ=0 , ij dλ | then m m m m Γ˙ = 2 iχ˙ δ + jχ˙ δ χ˙ eij ij ∇ j ∇ i − ∇ and m Γ˙ = 6 iχ˙ . im ∇ Substitute in the expression for S˙ij to obtain: 1 S˙ij = 2 i jχ˙ χ˙ eij 3 i jχ˙ = i jχ˙ χe˙ ij 2 ∇ ∇ − 4 − ∇ ∇ −∇ ∇ − 4

Conversely, S˙ij corresponds to a conformal variation if there is a solutionχ ˙ of this equation: 1 tr S˙ = 4 χ˙ 2 − 4 Substitute then for χ˙ in the equation for S˙ij to obtain: 4 1 1 1 i jχ˙ = S˙ij eij tr S˙ = Aij −∇ ∇ 2 − 4 2 2. SLOW MOTION APPROXIMATION 117

The integrability condition is:

kAij iAkj = 0 . ∇ − ∇ In the exact version of Bach’s theorem 1 Aij = Sij g tr S. − 4 ij The integrability condition is equivalent to ki kAij = 0 . m ∇ Let us decompose Sij into its trace-free part Sîj and its trace-part: 1 Sij = Sîj + g tr S 3 ij Then 1 Aij = Sîj + g tr S, 12 ij and the anti-symmetric part of the integrability condition above is equivalent to mj ki kAij = 0 , l m ∇ or in view of the identity mj ki = (g−1)jkδi (g−1)jiδk : l m l − l j Alj l tr A = 0 . ∇ − ∇ Now substituting for Aij in term of the original Sij this condition reads j 1 Slj l tr S = 0 . ∇ − 2∇ This however is an identity, the second contracted Bianchi identity (in 3 dimensions). So the integrability condition above reduces to its symmetric part, which is curl Aij = 0 . Finally we show that (curl A)ij = (curl Sˆ)ij = Bij is the Bach tensor. Indeed the additional term in Aij does not contribute to the curl: ab ab a a a(g tr S) + a(g tr S) = + a tr S = 0 . i ∇ bj j ∇ bi i j j i ∇ We have seen in the discussion of the linearized version above that once the integrability condition is satisfiedχ ˙ is determined by the equation: 1 (2.25) χ˙ = tr S˙ 4 −8 In the exact theory this equation is: −5 (2.26) tr S = 8χ eχ , − 4 4 where Sij is the Ricci curvature of gij = χ eij. 118 4. DYNAMICAL FORMULATION OF THE EINSTEIN EQUATIONS ∗

Now we have shown that the symmetric curl of the trace-free part of the limiting Ricci curvature (2.17) vanishes (2.24). Therefore by virtue of Bach’s theorem, 4 −2 (2.27) gij = χ eij + o c , or, after linearizing, −2 (2.28) gij = 1 + 4χ ˙ eij + o c , where in view of (2.25) and (2.19) the functionχ ˙ satisﬁes 1 (2.29) χ˙ = ρ , 4 −2 and thus in comparison to (2.15) in fact 1 (2.30)χ ˙ = ψ . −2 In conclusion:

−2 −2 (2.31) g = 1 2c ψ eij + o c ij − where e is the ﬂat metric.

Note on symmetric curl operator In general if θ is a symmetric 2-covariant tensorﬁeld in a 3-dimensional manifold (Σ, g) the symmetric curl of θ is deﬁned by: 1 (curl θ) = ab θ + ab θ ij 2 i ∇a bj j ∇a bi where is the covariant derivative of g and is the volume form of g, and indices are raised with∇ respect to g.

3 Since the Σt are diﬀeomorphic to R , e is isometric to the Euclidean metric. Consider in general a 1-parameter family of metrics e(t) depending on the time t which are isometric to the Euclidean metric. Then there are new spatial coordinates Xk : k = 1, 2, 3 such that ∂Xk ∂Xl (2.32) eij = δ . ∂xi ∂xj kl Writing X~ = (X1,X2,X3) this is ∂X~ ∂X~ (2.33) eij = . ∂xi · ∂xj Let us deﬁne now, ∂X~ (2.34) V~ = , (x constant). ∂t 2. SLOW MOTION APPROXIMATION 119

This is the velocity of the normal lines (the observers) with respect to the X~ - constant coordinate lines (that is the conformally Euclidean coordinate lines). Let us also define ∂X~ (2.35) βi = V~ . · ∂xi Then differentiating (2.33) with respect to t (at x constant), 2 ∂eij ∂V~ ∂X~ ∂X~ ∂V~ ∂βj ∂βi ∂ X~ (2.36) = + = + 2V~ . ∂t ∂xi · ∂xj ∂xi · ∂xj ∂xi ∂xj − · ∂xi∂xj e m Now if Γij are the connection coefficients of eij with respect to the x-coordinates then in fact e k ∂2X~ (2.37) Γ βk = V~ . ij · ∂xi∂xj This follows from e k ∂xk ∂2X~ (2.38) Γij= , ∂X~ · ∂xi∂xj which is the general formula for the transformation of the connection coefficients under change of coordinates in view of the fact that the connection coefficients in the X~ coordinates vanish. Therefore (2.36) actually is:

∂eij e e (2.39) = βj+ j βi = ( β] e)ij , ∂t ∇ ∇ −L where β] is the vectorﬁeld corresponding to the 1-form β, ] i −1 ij (2.40) (β ) = (e ) βj .

These formulas hold for a general 1-parameter family of metrics eij(t). Let us now return to the study of the post-Newtonian approximation. We diﬀerentiate (2.31) with respect to t (at x constant):

∂gij −2 ∂ψ ∂eij −2 (2.41) = 2c δij + + o c ∂t − ∂t ∂t The limit 2 (2.42) lim c βi = αi c→∞ exists and 2 ∂eij (2.43) lim c = iαj + jαi . c→∞ ∂t ∇ ∇ We compare this result to the limit of the ﬁrst variation formula (2.10) derived above and obtain: 2 c ∂gij ∂ψ 1 (2.44) lij = lim = δij + iαj + jαi . c→∞ 2 ∂t − ∂t 2 ∇ ∇ The condition (2.3a) implies (2.45) tr l = 0 , 120 4. DYNAMICAL FORMULATION OF THE EINSTEIN EQUATIONS ∗ which is ∂ψ (2.46) α = 3 . ∇ · ∂t Eliminating ∂tψ in (2.44) then yields: 2 (2.47) 2lij = iαj + jαi δij α ∇ ∇ − 3 ∇ · The momentum constraint (2.3b) becomes in the limit j (2.48) lij = 2pi ; ∇ − and from the above,

j 1 (2.49) 2 lij = αi + i( α) . ∇ 4 3∇ ∇ · We thus obtain the following elliptic system for α: 1 (2.50) αi + i( α) = 4pi 4 3∇ ∇ · − Taking the divergence gives (2.51) ( α) = 3 p , 4 ∇ · − ∇ · which in view of (2.46) really reads: ∂ψ (2.52) = p 4 ∂t −∇ · This is the integrability condition of the post-Newtonian system. By virtue of (2.15) the integrability condition (2.52) is equivalent to the continuity equation (2.16a): ∂ρ (2.53) + p = 0 . ∂t ∇ · Prop 2.3. The solution of the elliptic system (2.50) (tending to zero at inﬁn- ity) is: Z ( i 0i j 0j ) 1 7δij (x x )(x x ) j 0 3 0 (2.54) αi(x) = 0 + − 0 3− p (x ) d x 8π 3 x x x x R | − | | − |

Consider now the transformation of coordinates (t, x) to (t, X), Xa = Xa(t, x)(2.55) ∂Xa (2.56) dXa = V adt + dxi . ∂xi Using (2.32) we obtain i j a a b b (2.57) eijdx dx = δab dX V dt dX V dt . − − Let us then ﬁnd the spacetime metric to order c−2 (after setting x0 = ct): (2.58) g = c2φ2(dt)2 + g dxidxj , − ij 2. SLOW MOTION APPROXIMATION 121 where 1 1 (2.59a) φ = 1 + ψ + ω + oc−4 c2 c4 4 −2 (2.59b) gij = χ eij + o c 1 1 (2.59c) χ = 1 ψ + oc−2 . − 2 c2 Thus:

2 2 2 a b (2.60) g = c dt 2ψdt + δabdX dX − − 1 n 2 2 a a bo −2 + ψ + 2ω dt 2αa dtdX 2ψ δabdX dX + o c c2 − − − Here ω is the correction to the Newtonian potential. In fact set: 1 (2.61) ω = lim c4 φ 1 ψ c→∞ − − c2 We revisit the lapse equation (2.4) to ﬁnd the equation for ω: h 1 i φ = φ k 2 + c2ρ + tr T¯ 4 | | c4 Recall that k = oc−6. We have to determine both sides to order c−4. On | | the left hand side, we have to consider (2.59b) because the c−2 contribution to g cancels. However, we may set eij = δij because this is a scalar equation for the 3-dimensional manifold (Σt, g), (and thus does not depend on the choice of coordinates on Σt and we may choose the coordinates X~ w.r.t. which eij = δij). In general, 1 ∂ p ∂φ (2.62) φ = det g (g−1)ij 4 √det g ∂xi ∂xj Here, p 6 −1 ij −4 (2.63) det g = χ , (g ) = χ δij , so

1 2 i 1 2 (2.64) φ = i χ φ = φ + χ φ , 4 χ6 ∇ ∇ χ4 4 χ∇ · ∇ with respect to the coordinates X~ . Substitute (2.59a) and (2.59c) then using (2.15) we obtain: 2 1 1 (2.65) φ + χ φ = ρ + ω ψ 2 + oc−4 4 χ∇ · ∇ c2 c4 4 − |∇ | On the right hand side, 1 1 1 (2.66) k 2 + c2ρ + tr T¯ = ρ + tr T¯ + oc−4 | | c4 c2 c4 122 4. DYNAMICAL FORMULATION OF THE EINSTEIN EQUATIONS ∗

Taking also into account (2.59c) for the factor χ−4 on the left hand side, and (2.59a) for the factor φ on the right hand side, we finally obtain the equation: (2.67) ω = ρ ψ + ψ 2 + tr T¯ 4 − |∇ | The first two terms on the right hand side are quadratic in Newton’s potential and thus display the non-linearity of the Einstein field equations. This leads to 3-body forces.

Exercises Consider an axially symmetric stationary mass distribution of compact support. That is take ρ such that ∂ρ = 0 , supp ρ = Ω ∂t 3 where Ω has compact closure in R , and take p = ρv with the integral curves of v being circular orbits. Show that the leading order term of α at spatial inﬁnity depends only on the angular momentum of the mass distribution. Hints: We may choose rectangular coordinates (x1, x2, x3) such that the total angular momentum vector Z k k i j L = ε ij x p dx Ω q 3 P3 i 2 lies in the x -axis. Moreover set r = i=1(x ) , and show that to leading order in 1/r: 1 1 x · x0 = + + or−2 x0 ∈ Ω . |x − x0| r r3 Now: (1) Use the continuity equation to show that the total momentum vanishes: Z P i = pidx = 0 Ω (2) Deﬁne Z Dij = xipj dx , Ω and show that D is antisymmetric and moreover 1 Dij = εij Lk . 2 k (3) Show that to leading order in 1/r: 1 1 α = ε Lj xk i 2π r3 ijk Note that this implies that upon introducing polar coordinates (r, θ, φ) the cross term of the metric (2.60) is given by: 2 1 |L| − α dtdxi = − sin2 θ dtdφ c2 i πc2 r

3. Gravitational Radiation In this section we discuss gravitational radiation. As a precursor we first derive the radiation field of scalar fields, and then treat electromagnetic waves using a suitable null decomposition. 3. GRAVITATIONAL RADIATION 123

Figure 5. Example of an axially symmetric and rotating mass conﬁguration.

(t, x)

Ct−0 (t, x)

Σt0

− Figure 6. Truncated light cone Ct0 .

3.1. Scalar wave equation in Minkowski space. Consider the scalar wave equation in Minkowski space,

(3.1) φ = ρ , −1 µν ∂2 P3 ∂2 1 2 3 where = (g ) µ ν = 2 + 2 in rectangular coordinates (t, x , x , x ). ∇ ∇ − ∂t i=1 ∂xi Here ρ is a given source, and let φ be a solution with trivial initial data at t = t0: ∂φ (3.2) φ = 0 , = 0 . t=t0 ∂t t=t0 − We denote by Ct0 (t, x) the past light cone of the point (t, x) truncated by Σt0 = 3 (t, x): t = t0, x R , (c.f. ﬁgure 6): { ∈ } − n 0 0 0 2 0 2 0 o C (t, x) = (t , x ):(t t ) x x = 0, t0 t t t0 − − | − | ≤ ≤

3.1.1. Representation formula. Then, by Kirchhoﬀ’s formula, 1 Z ρ(t x x0 , x0) φ(t, x) = − | − | d3x0 x0∈ 3 0 −4π R0 x x (3.3) t−|x−x |≥t0 | − | 1 Z = ρ dµC−(t,x) −4π C− (t,x) t0 124 4. DYNAMICAL FORMULATION OF THE EINSTEIN EQUATIONS ∗

Letting t0 we obtain the retarded solution → −∞ Z Z 3 0 1 1 0 0 d x (3.4) φ(t, x) = ρ dµC−(t,x) = ρ(t x x , x ) 0 . −4π − −4π 0 3 − | − | x x C (t,x) x ∈R | − |

Notes on Kirchhoﬀ’s formula The representation formula (3.3) is easily obtained with the method of spherical means. We can take, without loss of generality the point (t, x) on the time axis by translation. Any point in spacetime, away from the axis, can be assigned spherical coordinates p (t, r, ϑ), where r = (x1)2 + (x2)2 + (x3)2, and ϑ S2. The Minkowski metric in these ∈ coordinates reads g = dt2 + dr2 + r2 γ◦ and − ∂2φ ∂2φ 2 ∂φ 1 φ = + + + ◦/ φ , − ∂t2 ∂r2 r ∂r r2 4

◦ where / denotes the Laplacian on (S2, γ◦). Given any function f in spacetime, we denote 4 2 ◦ 2 by f¯(t, r) the mean value of f over (S , γ). Since γ = r γ, dµ = r dµ ◦ , and the area t,r γ γ 2 of St,r is 4πr we have 1 Z 1 Z f¯(t, r) = f(t, r, ϑ)dµ = f(t, r, ϑ)dµ ◦ , 2 γ γ 4πr 4π 2 St,r S

◦ that is f¯(t, r) is simply the mean value of f(t, r, ϑ) on S2. Note that / φ = 0, because S2 is closed (without boundary). We conclude from (3.1) that φ¯ satisﬁes4 ∂2φ¯ ∂2φ¯ 2 ∂φ¯ + + =ρ ¯ , − ∂t2 ∂r2 r ∂r Now set ψ = rφ¯, σ = rρ¯. These functions of (t, r) satisfy the 1 + 1-dimensional wave 2 2 equation ∂t ψ + ∂r ψ = σ, and ψ vanishes on the time axis (r = 0). Setting v = t + r, and u = t− r this becomes − ∂2ψ 4 = σ , − ∂u∂v where the initial data to be trivial on t = t0 by (3.2). Since, by continuity of φ, limr 0 φ¯(t, r) = φ(t, 0), we obtain φ(t, 0) as an integral of σ over the line v = t: → ∂ψ Z t 2φ(t, 0) = lim 4 = σ du . − r 0 − ∂v 2t t → 0− But since σ(t0, r0) is the mean value of ρ(t0, r0) over the sphere St0,r0 , the integral of σ on the line v = t is really the integral of ρ over the past light cone Ct−0 (t, 0) of (t, 0) truncated at the past hypersurface t = t0:

Z t Z t t0 Z t t0 Z − ¯ 1 − dr σdu v=t = 2 rφ(t r, r)dr = φ dµγ . 2t t 0 − 2π 0 St−r,r r 0− We have thus obtained, after a suitable space translation, the formula (3.3). 3. GRAVITATIONAL RADIATION 125

(u + r, rξ)

ξ Σu

C−(u + r, rξ)

U Figure 7. Plane approximation of C−(u + r, rξ) . ∩ U

3.1.2. Radiation field. Set 2 (3.5) x = r ξ , ξ S , ∈ (3.6) r = x , t = u + r . | | Then with ξ, u fixed, the equations t = u + r, x = rξ are those of a light ray parametrized by r, namely the light ray in the direction ξ which starts at the spatial origin at t = u. Let us suppose that the source ρ has compact support in a neighborhood of the spatial origin. Then for any fixed bounded spacetime region , U −(u + r, rξ) P (u, ξ) , as r , U ∩ C → U ∩ → ∞ where P (u, ξ) is a null plane corresponding to the given ray (u, ξ), c.f. figure 7:

n 0 0 0 0 0 3o (3.7) P (u, ξ) = (t , x ): t = u + ξ x , x R · ∈ Since p p x x0 = x 2 2x x0 + x 2 = r2 2rξ x0 + x0 2 | − | | | − · | | − · | | r (3.8) 2 1 = r 1 ξ x0 + x0 = r ξ x0 + o(r−1) , − r · r2 | | − · we have (3.9) t x x0 = t r + ξ x0 + o(r−1) = u + ξ x0 + o(r−1) , − | − | − · · and it follows that 1 Z 1 Z (3.10) lim (rφ)(u + r, rξ) = ρ = ρ(u + ξ x0, x0)d3x0 . r→∞ −4π P (u,ξ) −4π x0∈R3 · 126 4. DYNAMICAL FORMULATION OF THE EINSTEIN EQUATIONS ∗

L ∂ ∂t L

N St,r

Σt

Figure 8. Null normals L, L to wave fronts St,r.

Remark. The function thus deﬁned, (3.11) Ψ(u, ξ) := lim (rφ)(u + r, rξ) r→∞ is also called the radiation ﬁeld.

3.1.3. Tangential derivatives. Consider the spheres St,r on Σt of radius r, n o (3.12) St,r = (t, x): x = r | | We may think of these spheres as wave fronts. Every St,r has two null normals at each point, the outgoing null normal L and the incoming null normal L, normalized so that their ∂t components are equal to 1, c.f. ﬁgure 8: ∂ ∂ (3.13) L = + N,L = N, ∂t ∂t − where N is the outward unit normal to St,r in Σt: xi ∂ ∂ (3.14) N = = ξi . r ∂xi ∂xi We denote by Π the orthogonal projection to the tangent space to St,r at a point. th Let us also introduce the generators Ωi : i = 1, 2, 3 of the rotations about the i rectangular coordinate axis:

j ∂ (3.15) Ωi = x . ijk ∂xk We also consider 1 j ∂ (3.16) Ωˆ i = Ωi = ξ . r ijk ∂xk Note that 3 3 X a b X j k j k Ωˆ Ωˆ = ijaξ ikbξ = (δjkδab δjbδka)ξ ξ i i − (3.17) i=1 i=1 = δab ξaξb = Πab , − or 3 X (3.18) Ωˆ i Ωˆ i = Π . ⊗ i=1 3. GRAVITATIONAL RADIATION 127

Prop 3.1. Given any function ρ of compact spacelike support, we have Z ˆ µ Ωi (ξ) µρ = 0(3.19) P (u,ξ) ∇ Z µ (3.20) L (ξ) µρ = 0 . P (u,ξ) ∇ Proof. Recall (3.7) for the deﬁnition of the plane P (u, ξ). We have

∂ 0 0 k 0 0 ρ(u + ξ x , x ) = ( 0ρ ξ + kρ)(u + ξ x , x ) ∂x0k · ∇ ∇ · and since ρ has compact spacelike support Z ∂ ρ(u + ξ x0, x0) d3x0 = 0 . 0k x0∈R3 ∂x · Hence Z Z k kρ = ξ 0ρ . P (u,ξ) ∇ − P (u,ξ) ∇ Now recall that ˆ 0 ˆ k j Ωi = 0 , Ωi = ijkξ , and thus Z Z ˆ µ j k Ωi µρ = ijkξ ξ 0ρ = 0 . P (u,ξ) ∇ − P (u,ξ) ∇ Moreover, recall that L0 = 1 ,Lk = ξk , and therefore: Z Z Z µ k 2 L µρ = 0ρ + ξ kρ = 0ρ ξ 0ρ = 0 . P (u,ξ) ∇ P (u,ξ) ∇ ∇ P (u,ξ) ∇ − | | ∇

The Ωi : i = 1, 2, 3, span the tangent space to St,r at each point, and together with L they represent the tangential derivatives to the surfaces of constant retarded time u. A simple consequence of the above general proposition is a decay statement for these tangential derivatives. Cor 3.2. Let φ be a solution to (3.1, 3.2) with a source ρ of compact spatial support. Then (3.21) lim r Lφ = 0 , r→∞

(3.22) lim r Ωˆ iφ = 0 , i = 1, 2, 3. r→∞

Proof. We diﬀerentiate the equation (3.1) to obtain that ∂µφ : µ = 0, 1, 2, 3, satisﬁes the wave equation ∂φ ∂ρ = ; ∂xµ ∂xµ 128 4. DYNAMICAL FORMULATION OF THE EINSTEIN EQUATIONS ∗ in other words the generators of the translations along the coordinate axes commute with the wave operator. Therefore Z µ 1 µ lim r L ∂µφ(u + r, rξ) = L ∂µρ = 0 r→∞ −4π P (u,ξ) Z ˆ µ 1 ˆ µ lim r Ωi ∂µφ(u + r, rξ) = Ωi ∂µρ = 0 r→∞ −4π P (u,ξ) by the previous proposition.

Notes on transversal derivatives and energy ﬂux

Recall the outgoing and incoming (future-directed) null normals L, L to the spheres St,r, (3.13). In terms of retarded and advanced time coordinates, u = t r, and v = t + r, respectively, we have − ∂ ∂ L = 2 ,L = 2 . ∂v ∂u We consider now the limit of the transversal derivative Lφ. On one hand, using the identities in the proofs of Prop. 3.1 and Cor. 3.2, Z Z Z 1 µ 1 k 1 lim r Lφ = L µρ = 0ρ ξ kρ = 2 0ρ . r →∞ −4π P (u,ξ) ∇ −4π P (u,ξ) ∇ − ∇ −4π P (u,ξ) ∇ On the other hand, with the already found above, 2∂uΨ = limr rLφ. Thus →∞ ∂Ψ 1 Z ∂ρ = . ∂u −4π P (u,ξ) ∂t The transversal derivative is related to the energy radiated to infinity, or energy flux. In particular, the power radiated to future null infinity at retarded time u, is given by Z ∂Ψ2 G(u) = 2 (u, ξ) dµ ◦ (ξ) . γ S2 ∂u This can be seen using the method of energy currents; c.f. exercises.

Exercises 1. (Tangential and longitudinal decay) Recall the setting and the decay statements of Corollary 3.2. In this exercise we shall look at the limits to next order in r. Show that for a retarded solution φ to 3.1 with a source ρ of compact spatial support we have: (1) lim rΩiφ = ΩiΨ r→∞ Hints: 2 1 2 (a) Use polar coordinates (θ, ϕ) on S such that ξ = sin θ cos ϕ, ξ = sin θ sin ϕ, ξ3 = cos θ and arrange the coordinate axes in such a way that ∂ ∂ ∂ Ω = = x1 − x2 . 3 ∂ϕ ∂x2 ∂x1 In this setting it suﬃces to show that ∂Ψ lim r Ω3φ = . r→∞ ∂ϕ 3. GRAVITATIONAL RADIATION 129

(b) To prove (1) make use of the commutation relation [Ωi, ] = 0 to infer that 1 Z lim r Ω3φ(u, ξ) = − Ω3ρ. r→∞ 4π P (u,ξ) (2) lim r2 Lφ = −Ψ r→∞ Hints: µ ∂ (a) Consider the scaling vector ﬁeld S := x ∂xµ and compute the commutator [S, ] to conclude that Z lim rSφ = (Sρ + 2ρ). r→∞ P (u,ξ) (b) Next compute that ∂ ∂ρ ∂ρ ∂ρ x0i ρ(u + ξ · x0, x0) = t + xi (u + ξ · x0, x0) − u (u + ξ · x0, x0), ∂x0i ∂t ∂xi ∂t and integrate by parts to show that Z ∂Ψ (Sρ + 2ρ) = −Ψ + u . P (u,ξ) ∂u

1 (c) Finally note that S = 2 (uL + vL) and use ∂Ψ lim rLφ = 0, lim rLφ = 2 r→∞ r→∞ ∂u together with u = t − r, v = t + r to prove (b). 2. (Energy flux) The energy flux through a segment of a null hypersurface of constant advanced time v with respect to the timelike vectorfield ∂t is given by

Z u2 Z 2 F (v)[u1, u2] = T (L, ∂t) r dµ◦du , γ 2 u1 S where T denotes the energy momentum tensor of the scalar ﬁeld, a 2-covariant tensorﬁeld which 1 αβ is given by Tµν [φ] = ∂µφ ∂ν φ − 2 gµν g ∂αφ∂β φ . Show that

Z u2 lim F (v)[u1, u2] = G(u) du , v→∞ u1 where G(u) is defined as in the note on page 130. Hints: 2 2 (1) Express ∂t in terms of L, L, and verify T (L,L) = (Lφ) , and T (L,L) = |∇/φ|γ , where ∇/ denotes the connection on the sphere St,r. P3 2 2 2 (2) Show that i=1(Ωiφ) = r |∇/φ|γ and use Corollary 3.2 to compute the limit. 3. (Incoming and outgoing radiation) Consider the incoming radiation field of the retarded 2 solution of (3.1), namely for fixed advanced time v, and ξ ∈ S , consider the limit of rφ along the past-directed light ray t = v − r, x = rξ, parametrized by r in the direction ξ, as r → ∞. (1) Show that for a spatially compact source, there is no incoming radiation. In other words, show that the incoming energy flux vanishes. Hints: (a) Show that 1 Z lim (rφ)(v, ξ) = − lim ρ r→∞ 4π u→−∞ P (u,ξ) 130 4. DYNAMICAL FORMULATION OF THE EINSTEIN EQUATIONS ∗

(b) The incoming energy ﬂux is given by

Z v2 Z F [v1, v2] = G(v)dv , G(v) = lim T (L, ∂t)dµγ ; r→∞ v1 Sv−r,r c.f. Exercise 2 above. Show that G(v) = 0. (2) Find the conditions on the source so that the total energy radiated to inﬁnity is ﬁnite. Hint: The amount of energy radiated from the past up to retarded time u0 is given by

Z u0 F−(u0) = G(u)du . −∞

Make a suitable assumption on the support of ρ, and ∂tρ that ensures the ﬁniteness of F−(∞).

3.2. Maxwell’s equations. Consider now Maxwell’s equations in Minkowski space,

αFβγ + βFγα + γFαβ = 0(3.23a) ∇ ∇ µν ∇µ (3.23b) νF = J ∇ for the Faraday tensor F in the presence of sources J. 3.2.1. Wave equations. We take the derivative λ of the inhomogoneous equation (3.23b), in covariant form, and antisymmetrize∇ in λ, µ:

ν (3.24) ( λFµν µFλν) = λJµ µJλ ∇ ∇ − ∇ ∇ − ∇ From the homogeneous equations (3.23a) we have

(3.25) λFµν µFλν = λFµν + µFνλ = νFλµ , ∇ − ∇ ∇ ∇ −∇ ν and ν = Fλµ. We thus obtain a scalar wave equation for each rectangular component−∇ ∇ of F−:

(3.26) Fλµ = λJµ + µJλ −∇ ∇ Conversely, a solution F to these wave equations whose initial data satisﬁes Maxwell’s equations is actually a solution of Maxwell’s equations for all time. For, suppose we have a solution to (3.26) then

( αFβγ + βFγα + γFαβ) = α βJγ + α γJβ ∇ ∇ ∇ − ∇ ∇ ∇ ∇ (3.27) β γJα + β αJγ − ∇ ∇ ∇ ∇ γ αJβ + γ βJα = 0 , − ∇ ∇ ∇ ∇ ν and, by charge conservation Jν = 0, ∇ ν ν ν ( Fµν) = µJν + νJµ = (Jµ) , ∇ −∇ν ∇ ∇ ∇ (3.28) ( Fµν Jµ) = 0 . ∇ − 3. GRAVITATIONAL RADIATION 131

3.2.2. Null decomposition of Faraday tensor. We decompose the Maxwell ﬁeld relative to the wave fronts into:

α, α: St,r-1-forms, that is 1-forms in Minkowski spacetime which vanish on ⊥ the orthogonal space (TpSt,r) to St,r, ρ, σ: functions . The null decomposition of the Faraday tensor is then deﬁned by: κ λ κ λ (3.29a) αµ = Πµ L Fκλ αµ = Πµ L Fκλ 1 (3.29b) ρ = LkLλF 2 κλ κ λ (3.29c) σ µν = Πµ Πν Fκλ ,

k where ij = ξ ijk is a St,r-2-form proportional to the area form of St,r. 3.2.3. Radiative properties. We now consider the radiation ﬁelds

(3.30) fµν(u, ξ) = lim (rFµν)(u + r, rξ) , r→∞ which by (3.10), and (3.26), are given by 1 Z (3.31) fµν(u, ξ) = ( µJν νJµ) . 4π P (u,ξ) ∇ − ∇ Recall the decomposition (3.29) of the Faraday tensor F . We denote the corresponding decomposition of the limit f by α[f] , α[f] , ρ[f] , σ[f] . Speciﬁcally,

(3.32a) α[f](Ωˆ i) = f(Ωˆ i,L) , α[f](Ωˆ i) = f(Ωˆ i,L) , 1 (3.32b) ρ[f] = f(L,L) , σ[f](Ωˆ i, Ωˆ j) = f(Ωˆ i, Ωˆ j) . 2 Now, by (3.31) and Prop. 3.1 we have that (3.33) α[f] = 0 , σ[f] = 0 , and Z 1 µ ν (3.34) ρ[f] = L L µJν . 8π P (u,ξ) ∇ Furthermore, since ∂ ∂ ∂ ∂ (3.35) L = + ξi ,L = ξi , ∂x0 ∂xi ∂x0 − ∂xi we have Z 1 i i i j (3.36) ρ[f] = ( 0J0 ξ iJ0 + ξ 0Ji ξ ξ iJj) , 8π P (u,ξ) ∇ − ∇ ∇ − ∇ 132 4. DYNAMICAL FORMULATION OF THE EINSTEIN EQUATIONS ∗ and again by Prop. 3.1, Z Z i (3.37) ξ iJ0 = 0J0 , − P (u,ξ) ∇ P (u,ξ) ∇ Z Z i j j (3.38) ξ ξ iJj = ξ 0Jj . P (u,ξ) ∇ − P (u,ξ) ∇ Therefore Z 1 i j (3.39) ρ[f] = ( 0J0 ξ ξ iJj) . 4π P (u,ξ) ∇ − ∇ Now recall the formula (3.17). By Prop. 3.1 we have 3 Z Z X a b (3.40) 0 = Ωˆ Ωˆ aJb = (δab ξaξb) aJb , i i ∇ − ∇ i=1 P (u,ξ) P (u,ξ) or Z Z i j i (3.41) ξ ξ iJj = Ji . P (u,ξ) ∇ P (u,ξ) ∇ Hence Z 1 i (3.42) ρ[f] = ( 0J0 Ji) = 0 , 4π P (u,ξ) ∇ − ∇ by the current conservation law: µ 0 i i (3.43) 0 = Jµ = J0 + Ji = 0J0 + Ji . ∇ ∇ ∇ −∇ ∇ In conclusion, only the α-component survives in the limit. We denote (3.44) a = α[f] , which by Prop. 3.1 reduces to Z 1 κ λ (3.45) aµ(u, ξ) = Πµ L λJκ . −4π P (u,ξ) ∇ Since Z λ (3.46) L λJκ = 0 , P (u,ξ) ∇ and ∂ (3.47) L + L = 2 , ∂x0 we ﬁnally obtain Z 1 k ∂Jk (3.48) a0 = 0 , ai = Πi . −2π P (u,ξ) ∂t 3. GRAVITATIONAL RADIATION 133

3.2.4. Dipole approximation. If the time scale of variation of the source is much larger than its spatial dimension then we can replace the integral over the null plane P (u, ξ) by an integral over the spacelike plane Σu. We can then express (3.48) in terms of the dipole moments Z (3.49) Di(t) = xiρ(t, x)d3x , Σt where ρ is the charge density. In view of conservation of charge (3.43),

∂ρ k (3.50) + Jk = 0 , ∂t ∇ we have, after integration by parts,

i Z Z dD i ∂ρ 3 i k 3 (3.51) (t) = x (t, x) d x = x kJ d x = dt Σt ∂t − Σt ∇ Z Z i k 3 i 3 = kx J d x = J d x . Σt ∇ Σt Therefore, taking another time-derivative, 2 i Z i d D ∂J 3 (3.52) 2 (t) = d x . dt Σt ∂t Thus, we have in this approximation, 2 1 d k (3.53) a (u, ξ) = Πik(ξ) D (u) . i −2π dt2 3.2.5. Polarisation. Note finally that the Maxwell radiation field can be written as: 1 1 (3.54) fµν = a Lν + a Lµ −2 µ 2 ν Let us denote by e, and b the limits of the electric and magnetic fields respectively,

(3.55) ei = lim rEi , bi = lim rBi . r→∞ r→∞ Then 1 (3.56) ei = fi0 = a , 2 i and

1 jk 1 jk 1 1 1 jk bi = i fjk = i ( ajξk + akξj) = i ξjak (3.57) 2 2 −2 2 2 1 = a k = e k . 2 k i k i That is to say, the electric and magnetic ﬁelds are orthogonal in the limit, c.f. ﬁg- ure 9. 134 4. DYNAMICAL FORMULATION OF THE EINSTEIN EQUATIONS ∗

Figure 9. Electric and magnetic components of the radiation ﬁeld.

3.3. Gravitational waves. We turn to the gravitational case. We recall the Bianchi identities from Section 1.3,

(3.58) αRβγδ + βRγαδ + γRαβδ = 0 , ∇ ∇ ∇ which after the first contraction become α (3.59) Rαβγδ = γSβδ δSβγ . ∇ ∇ − ∇ We now substitute for Sµν from the Einstein equations 1 (3.60) Sµν = 2T˜µν , T˜µν = Tµν gµν tr T, − 2 to obtain the inhomogeneous equations α (3.61) Rαβγδ = 2 γT˜βδ δT˜βγ . ∇ ∇ − ∇ The are analogous to the Maxwell equations (3.23b) and we can proceed as in Section 3.2.1 to derive a wave equation for the curvature tensor. Apply to (3.61) and antisymmetrize in (β, ), ∇ α α (3.62) Rαβγδ β Rαγδ = ∇ ∇ − ∇ ∇ = 2 γT˜βδ β γT˜δ δT˜βγ + β δT˜γ . ∇ ∇ − ∇ ∇ − ∇ ∇ ∇ ∇ Next we wish to commute the covariant derivatives on the left hand side, which produces quadratic terms in the curvature, by its very definition. After commutation, however, the l.h.s. becomes by (3.58): α α α Rαβγδ βRαγδ = Rαβγδ + βRαγδ (3.63) ∇ ∇ − ∇ ∇ ∇ α∇ ∇ = αRβγδ = Rβγδ −∇ ∇ − Therefore we obtain the following wave equation for the curvature tensor: (3.64) Rαβγδ = 2 α γT˜βδ β γT˜αδ α δT˜βγ + β δT˜αγ + Qαβγδ ∇ ∇ − ∇ ∇ − ∇ ∇ ∇ ∇ Here Qαβγδ are expressions that are quadratic in the curvature. We shall now impose of a major simplification which is often referred to as the weak field approximation. Weak field approximation: We break the connection between the metric and the curvature, and consider the wave equations (3.64) and the Bianchi identities (3.58), (3.61) as equations for a tensorfield Rαβγδ in the Minkowski spacetime. Here Rαβγδ is to posses the algebraic symmetries of the curvature tensor. Moreover we drop the quadratic terms Qαβγδ in (3.64). 3. GRAVITATIONAL RADIATION 135

Set −1 αγ (3.65) Sβδ = (g ) Rαβγδ .

Let us verify that a solution Rαβγδ of (3.64) whose initial data satisﬁes the Einstein equations is actually a solution of the Einstein equations (3.60). Take the trace of (3.64), then Sβδ satisﬁes: α α (3.66) Sβδ 2T˜βδ = 2 β T˜αδ 2 δ T˜αβ + 2 β δ tr T˜ − − ∇ ∇ − ∇ ∇ ∇ ∇ Now taking into account the conservation laws, µν (3.67) νT = 0 , ∇ we have

µν 1 µ (3.68) νT˜ = tr T, ∇ −2∇ and the right hand side above vanishes:

α α (3.69) 2 β T˜αδ 2 δ T˜αβ + 2 β δ tr T˜ = − ∇ ∇ − ∇ ∇ ∇ ∇ = β δ tr T + δ β tr T 2 β δT = 0 . ∇ ∇ ∇ ∇ − ∇ ∇ Therefore, in the weak ﬁeld approximation, any solution of (3.64) satisﬁes (3.70) Sµν 2T˜µν = 0 . − 3.3.1. Asymptotics of linearized gravitational waves. The retarded solution of the linearized wave equation (3.71) Rαβγδ = 2 α γT˜βδ β γT˜αδ α δT˜βγ + β δT˜αγ ∇ ∇ − ∇ ∇ − ∇ ∇ ∇ ∇ in rectangular coordinates relative to the Minkowski metric g is given by:

(3.72) lim rRαβγδ(u + r, rξ) = qαβγδ(u, ξ) = r→∞ Z 1 = α γT˜βδ β γT˜αδ α δT˜βγ + β δT˜αγ . −2π P (u,ξ) ∇ ∇ − ∇ ∇ − ∇ ∇ ∇ ∇

But, by the equations (3.60) we have that Sµν = 2T˜µν = 0 outside the support of the source, in particular at infinity −1 αγ (3.73) (g ) qαβγδ = 0 . 3.3.2. Null decomposition of the curvature tensor. Consider the curvature tensor of a solution of the Einstein equations in a vacuum region. Let S be a spacelike 2 surface diffeomorphic to S (a wave front). Let L, L be respectively the outgoing, incoming future directed null normals to S, normalized according to g(L, L) = 2. L, L are determined up to the transformation L aL, L a−1L where a −is a positive function on S. (These correspond to Lorentz→ transformations→ in the orthogonal timelike plane, c.f. figure 10.) The curvature then decomposes into: 136 4. DYNAMICAL FORMULATION OF THE EINSTEIN EQUATIONS ∗

L L S L p L

(T S)⊥ M p Figure 10. Null normals to S.

α, α: symmetric trace-free 2-covariant-tensorﬁelds on S deﬁned by α(X,Y ) = R(X,L,Y,L)(3.74a)

(3.74b) α(X,Y ) = R(X, L, Y, L) X,Y TpS. ∀ ∈ β, β: 1-forms on S deﬁned by R(X,Y,Z,L) = (X,Y )β(Z)(3.75a)

(3.75b) R(X,Y,Z,L) = (X,Y )β(Z) X,Y,Z TpS, ∀ ∈ where denotes the area 2-form of S. ρ, σ: functions deﬁned by (3.76a) R(X,Y,Z,W ) = (X,Y )(Z,W )ρ 1 (3.76b) R(X,Y,L,L) = (X,Y ) σ X,Y,Z,W TpS. 2 ∀ ∈ Let us introduce a frame for the spacetime along S by setting M (3.77) E1,E2 TS,E3 = L ,E4 = L. ∈ In this frame,

g33 = g44 =0 , g34 = 2 , gAB = δAB , A, B = 1, 2(3.78a) − 1 (3.78b) (g−1)33 = (g−1)44 =0 , (g−1)34 = , (g−1)AB = δAB . −2 Moreover, we can write

(3.79a) αAB = α(EA,EB) , βA = β(EA) , (3.79b) αAB = α(EA,EB) , βA = β(EA) , and

(3.80a) αAB = RA3B3 , αAB = RA4B4

(3.80b) RABC3 = ABβC ,RABC4 = ABβC 1 (3.80c) RABCD = ABCDρ , RAB34 = ABσ , 2 and the metric can also be written as −1 1 1 X (3.81) g = E3 E4 E4 E3 + EA EA . −2 ⊗ − 2 ⊗ ⊗ A=1,2 3. GRAVITATIONAL RADIATION 137

To see that α, α are traceless note that, in vacuum,

−1 αγ 1 1 X (3.82a) 0 = S33 = (g ) Rα3γ3 = R3343 R4333 + RA3A3 = tr α , −2 − 2 A=1,2

(3.82b) 0 = S44 = tr α . Also, in vacuum,

−1 αγ 1 X (3.83) 0 = SAB = (g ) RαAγB = R3A4B + R4A3B + RCACB , −2 C=1,2 and X X (3.84) RABAD = ABADρ = δBD ρ ; A=1,2 A=1,2 that is 1 (3.85) R3A4B + R4A3B = ρδAB 2 Since, by the cyclic property of the curvature,

(3.86) RA3B4 RB3A4 = RA3B4 + R3BA4 = RBA34 = RAB34 , − − we have 1 (3.87) RA3B4 RB3A4 = ABσ 2 − and we conclude that

(3.88) RA3B4 = ρ δAB + σ AB . Furthermore, in vacuum,

−1 αγ 1 1 X (3.89) 0 = SA3 = (g ) RαAγ3 = R3A43 R4A33 + RBAB3 , −2 − 2 B=1,2 and X X ∗ (3.90) RABA3 = βAAB = βB A=1,2 A=1,2 so that

1 ∗ (3.91) RA334 = β . 2 A Similarly,

1 ∗ (3.92) RA434 = β . 2 − A

Let us return to the discussion of Section 3.3.1. In view of Proposition 3.1, and that α, β, ρ do not contain L in their deﬁnition, it follows that the corresponding 138 4. DYNAMICAL FORMULATION OF THE EINSTEIN EQUATIONS ∗ components of q vanish. Consider now the σ component of q: Z 1 ˆ ˆ 1 ˆ α ˆ β γ δ ˜ ˜ (3.93) q(Ωi, Ωj,L,L) = Ωi Ωj L L α γTβδ β γTαδ 2 −2π P (u,ξ) ∇ ∇ − ∇ ∇ α δT˜βγ + β δT˜αγ = 0 . − ∇ ∇ ∇ ∇ So σ[q] vanishes as well. Thus only the α component remains: (3.94) a = α[q]

Z ˆ ˆ 1 ˆ α β ˆ γ δ ˜ ˜ (3.95) a(Ωi, Ωj) = Ωi L Ωj L α γTβδ β γTαδ −2π P (u,ξ) ∇ ∇ − ∇ ∇ Z ˜ ˜ 1 ˆ α β ˆ γ δ ˜ α δTβγ + β δTαγ = Ωi L Ωj L β δTαγ . − ∇ ∇ ∇ ∇ −2π P (u,ξ) ∇ ∇ Again by Prop. 3.1, Z Z Z µ 1 µ µ (3.96) L µρ = 2 L + L µρ = 2 0ρ . P (u,ξ) ∇ 2 P (u,ξ) ∇ P (u,ξ) ∇ Therefore Z Z 2 ˜ ˆ ˆ 2 ˆ α ˆ γ 2 ˜ 2 ˆ m ˆ n ∂ Tmn (3.97) a(Ωi, Ωj) = Ωi Ωj 0Tαγ = Ωi Ωj 2 . −π P (u,ξ) ∇ −π P (u,ξ) ∂t Since a is trace-free, we conclude that Z 2 2 m n 1 mn ∂ Tmn (3.98) aij = Πi Πj Π Πij 2 . −π − 2 P (u,ξ) ∂t 3.3.3. Quadropole formula. In the case of a slowly varying source we may replace the intgral over P (u, ξ) by the integral over Σu. Now consider the quadropole moment Z i j 3 (3.99) Qij(t) = x x ρ(t, x) d x . R3 In the context of the slow motion approximation of Section 2, we have in particular the limit of the continuity equation (2.16a), which then implies: Z dQij ∂ρ (3.100) = xixj (t, x)d3x = dt R3 ∂t Z Z i j k 3 j i i j 3 ij = x x kp d x = x p + x p d x = D (t) − R3 ∇ R3 Furthermore, we substitute from the equations of motion (disregarding non-linear terms), i ∂p ik (3.101) = kT , ∂t −∇ 3. GRAVITATIONAL RADIATION 139 to obtain that ij Z dD i jk j ik 3 (3.102) = x kT + x kT d x = dt − R3 ∇ ∇ Z Z i jk j ik ij 3 = kx T + kx T = 2 T d x . R3 ∇ ∇ R3 Therefore Z 2 4 ∂ Tmn d (3.103) 2 2 = 4 Qmn(t) R3 ∂t dt and we conclude that, for a slowly varying source,

4 1 m n 1 mn d (3.104) a (u, ξ) = Π Π Π Πij (ξ) Qmn(u) ij −π i j − 2 dt4 m m i m where Πi (ξ) = δi ξ ξ . Finally, we note that− the radiation field of linearized gravitational waves takes the form: 1 (3.105) qαβγδ = a LβLδ a LβLδ a LαLδ + a LαLγ . 4 αγ − αδ − βγ βδ Remark. In [CK93] D. Christodoulou and S. Klainerman published a rig- orous analysis of the exact asymptotic behaviour of gravitational waves. In particular, it turns out that the nonlinearity of the Einstein equations accounts for an effect on gravitational radiation that is on the same level of magnitude as the linear effects discussed above [Chr91]. Thus the weak field approximation does not reflect correctly the nature of gravitational waves. 3.4. Gravitational wave experiments. In this section we explain the ex- perimental implications of the theory of gravitational waves. 3.4.1. Gravitational wave detectors. A gravitational wave detector consists of three test masses, a reference mass m0, and two masses m1, m2, arranged in the classical interferometer configuration depicted in figure 11. Each of the masses m1, m2 is positioned at a distance d0 from the reference mass m0, and any changes in distance are measured by comparing light travel times from the reference mass m0 to m1 and m2, respectively, as depicted in the spacetime diagram of figure 12 for the light beam that is split at m0, and reflected at m1, m2. The experiment is either performed on earth, in which case the test bodies are suspended from pen- dulums, or in outer space. On earth, for periods much smaller than the pendulum period, the motion can be considered to be free in the horizontal plane, (defined by the three masses). If the experiment is performed in space, then the motion of the test masses is free anyway. The reference mass m0 describes a geodesic Γ0 in spacetime, and m1, m2 are freely falling on nearby geodesics Γ1,Γ2. Let us pick an orthonormal basis (E1,E2,E3) for the spacelike hyperplane Σ0, n o Σ0 = X T : g(X, Γ˙ 0(0)) = 0 . ∈ Γ0(0)M 140 4. DYNAMICAL FORMULATION OF THE EINSTEIN EQUATIONS ∗

m1 mirror horizontal plane

laser beam m2 m0 beam splitter

Figure 11. Gravitational wave detector.

spacetime same phase

m2 m1

hypersurface of constant phase

m0 Figure 12. Spacetime diagram of a light beam used to measure distances in a gravitational wave detector.

Ei Γ0(t)

Σt

Γ0

Figure 13. Timelike geodesic Γ0 described by m0.

In general, see figure 13, we define, n o Σt = X T : g(X,T ) = 0 , ∈ Γ0(t)M where T = Γ˙ 0(t) denotes the unit tangent vector field along Γ0. We now consider 3. GRAVITATIONAL RADIATION 141

T p

Γ0(t) X

Γ0

Figure 14. Exponential map with base points on Γ0. for each t the spacelike geodesic hyperplane

Ht = expΓ0(t)(Σt); here p = expΓ0(t)(X) is the point at parameter value 1 along the spacelike geodesic issuing at Γ0(t) with initial tangent vector X, c.f. ﬁgure 14. That is to say

(3.106) dist(p, Γ0(t)) := d = X , p = exp (X) . | | Γ0(t) We parallel propagate the basis (E1,E2,E3) along Γ0 to obtain an orthonormal basis for each Σt. Any event p in a (cylindrical) spacetime neighborhood of Γ0 is then assigned the coordinates 1 2 3 1 2 3 (t, x , x , x ) if p = expΓ0(t)(X) with X = x E1 + x E2 + x E3 .

Since (E1,E2,E3) is an orthogonal basis, we have p (3.107) X = (x1)2 + (x2)2 + (x3)2 = d . | | In this coordinate system, (as shown in the exercises on page 44 on cylindrical normal coordinates), 2 (3.108) gµν ηµν = Rd , − O where R denotes the magnitude of the curvature components along Γ0. As we shall see, the displacements of the test masses due to the passage of a gravitational wave train are (Rτ 2), where τ is the period (time scale) of variation of the curvature. O Assumption: We assume

d0 1 . τ This is satisfied for typical sources and detectors: For example for a −3 binary neutron star system τ 10 s = 300 km, and d0 = 3 km for −2 ∼ Virgo, so d0/τ = 10 . Under this assumption (1) the speed of light can be taken to be 1 in the above coordinates, so differences in distance accurately reflect differences in light travel time. 142 4. DYNAMICAL FORMULATION OF THE EINSTEIN EQUATIONS ∗

wave fronts

space E3

horizontal plane

Figure 15. Vertical incoming radiation.

(2) we can replace the geodesic equation by the Jacobi equation, (recall here the discussion in Ch. 1, Section 3): 2 i d X j (3.109) = RiT jT X , where RiT jT = R(Ei,T,Ej,T ) . dt2 − Let us now suppose that the incoming gravitational waves are coming from the vertical direction in space, with respect to the horizontal plane deﬁned by the masses, c.f. ﬁgure 15. That means the normal N to the wave fronts in Ht is E3. − Since we have previously chosen (E1,E2,N) to be a postive basis, (E1,E2,E3) is a negative basis, and thus (E2,E1,E3) again a positive basis. Moreover,

(3.110) L = T E3 ,L = T + E3 . − Recall, from Section 3.3, aAB −2 (3.111) α = R(EA,L,EB,L) = + o r , AB r and all other curvature components are o(r−2) at least. i th Let us denote by x(A) the i coordinate of the test mass mA, A = 1, 2. Then, the Jacobi equation (3.109) reduces to, (see exercises below), 2 3 d x(A) (3.112a) = or−2 dt2 2 B d x(A) 1 (3.112b) = a xC + or−2 , dt2 −4r BC (A) with initial conditions 3 B B (3.113) lim x (t) = 0 , lim x (t) = d0δA , t→−∞ (A) t→−∞ (A) (3.114) lim x˙ i (t) = 0 . t→−∞ (A) −1 3 To order o(r ) we then obtain x(A) = 0: the masses remain on the horizontal plane. Since the fractional displacements are small, we can substitute on the right B the initial values for x(A), to obtain: 2 A d x(B) d0 (3.115) = a . dt2 −4r AB 3. GRAVITATIONAL RADIATION 143

m0 m2

Figure 16. Instantaneous acceleration of the test masses.

Remark. Recall a is a symmetric trace-free 2 2-matrix. This means that AB × the instantaneous acceleration of m2 is obtained from that of m1, by counter- clockwise rotation by 90◦, c.f. figure 16, (and exercises below). Integrating once, we obtain the velocities, Z t A d0 (3.116)x ˙ (B) = aAB(u)du . −4r −∞ Integrating again, we obtain the displacement of the masses: Z t Z t0 A A A d0 n o 0 (3.117) ∆x(B)(t) = x(B)(t) x(B)( ) = aAB(u)du dt . − −∞ −4r −∞ −∞ Set 1 Z t (3.118) ΞAB(t) = aAB(u)du −2 −∞ then we have Z t A d0 (3.119) ∆x(B)(t) = ΞAB(u)du . 2r −∞ A quantity of interest, both theoretically and experimentally, is the overall displacement of the test masses: (3.120) lim ∆xA (t) = xA ( ) xA ( ) t→∞ (B) (B) ∞ − (B) −∞ In the framework of the linear theory this vanishes provided that: (1) in the initial and final states the astronomical bodies are at rest in their respective center of mass frames, and (2) the final center of mass frame is at rest relative to the initial center of mass frame. Indeed, in the context of the weak field approximation discussed in Section 3.3 we have established the quadrupole formula according to which 4 d Qmn (3.121) a . AB ∼ dt4 144 4. DYNAMICAL FORMULATION OF THE EINSTEIN EQUATIONS ∗

Therefore 3 d Qmn (3.122) ΞAB , ∼ dt3 and the instantaneous displacement results in: 2 2 d Qmn d Qmn (3.123) ∆xA (t) (t) ( ) (B) ∼ dt2 − dt2 −∞ However, for an N-body configuration the quadropole moment is given by N X i j (3.124) Qij = MαXαXα α=1 so that N N ¨ X ˙ i ˙ j X ¨ i j i ¨ j (3.125) Qij = 2 MαXαXα + Mα XαXα + XαXα α=1 α=1 takes the same value initially (at t = ) and finally (t = ) provided the bodies are at rest relative to their center of mass−∞ frame, and the final∞ center of mass frame is at rest relative to the initial frame. Remark. This is not true in the framework of the fully nonlinear theory. In fact, a permanent displacement was proven to occur and is known as the memory 2 effect [Chr91]. Let us denote, for each (u, ξ), ξ S , by ∈ F (u, ξ) = Ξ(u, ξ) 2 | | the radiative power per unit solid angle, in the direction ξ, at retarded time u. Then in the full theory the permanent displacement of the test masses is given by

A d0 (3.126) lim ∆x (t) = ΣAB , t→∞ (B) 2r where Z ∞ ΣAB(ξ) = ΞAB(u, ξ)du −∞ However, we have an explicit formula, Z 0 0 0 ◦ ΣAB(ξ) = GAB(ξ, ξ )E(ξ )dµγ(ξ ) S2 (where GAB is a certain kernel), which relates the total displacement to the total energy radiated per unit solid angle in the direction ξ: Z ∞ F (u, ξ)du = E(ξ) −∞ 0 2 Note that the energy radiated in all directions ξ S contributes to the displacements of masses located in a certain direction ξ from∈ the source. See [Chr91] for more details.

Exercises 3. GRAVITATIONAL RADIATION 145

1. (Equations of motion) Derive the equations of motion (3.112) from the Jacobi equation (3.109) using the results of Section 3.3. Hint: Use 1 1 T = L + L ,E = L − L , 2 3 2 to write all components of the curvature in terms of the null decomposition of Section 3.3.2. For example, 3 1 1 X −2 R(E ,T,E ,T ) = R(L, L, L,L) = − R(L,E , L, E ) = −ρ = o(r ) 3 3 4 2 A A A=1 where we have used the (L,L) component of the vaccuum Einstein equation, and the form (3.81) of the metric. 2. (Polarisation) Show that the speciﬁc form of the (approximate) equations of motion (3.115) implies that the instantaneous acceleration of m2 is obtained from that of m1, by counter-clockwise rotation by 90◦.

Bibliography

[Chr91] Demetrios Christodoulou, Nonlinear nature of gravitation and gravitational-wave experiments, Physical Review Letters 67 (1991), no. 12, 1486–1489. [Chr95] Demetrios Christodoulou, Self-gravitating relativistic fluids, Arch. Rational Mech. Anal. 130 (1995), 343–400. [Chr99a] , The instability of naked singularities in the gravitational collapse of a scalar field, Annals of Mathematics 149 (1999), 183–217. [Chr99b] , On the global initial value problem and the issue of singularities, Class. Quan- tum Grav. 16 (1999), A23–A35. [Chr00] , The action principle and partial differential equations, Annals of Mathematics Studies, vol. 146, Princeton University Press, 2000. [CK93] Demetrios Christodoulou and Sergiu Klainerman, The global nonlinear stability of the minkowski space, Princeton Mathematical Series, vol. 41, Princeton University Press, 1993. [Daf05] Mihalis Dafermos, Spherically symmetric spacetimes with a trapped surface, Classical and Quantum Gravity 22 (2005), 2221–2232. [DR08] Mihalis Dafermos and Igor Rodnianski, Lectures on black holes and linear waves, Pro- ceedings of the Clay Mathematics Summer School, arXiv:0805.3880, 2008.

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