Chapter 5 Proof of the Hawking-Penrose Singularity Theorems

Chapter 5

Proof of the Hawking-Penrose Singularity Theorems

5.1 Proof of the Hawking-Penrose Singularity Theo- rem

5.1.1 Introduction

The singularity theorems are based on very powerful indirect arguments which show that black hole and cosmological singularities are generic in classical general relativity. Gives us conﬁdence in the big bang. The classical theory predicts its own breakdown.

In the singularity theorems one does not define a singularity as a place the curvature diverges, one characterises singularities as the “holes” left behind by the removal of their pesence. These “holes” should be detectable by the fact that their will be geodesics that cannot be extended any further in at least one direction but still only have a finite range of affine parameter. In the singularity theorems one does not directly establish that the curvature diverges, but merely that there is an obstruction of sould beome sort to timelike or null geodesics being extendable within the spacetime to infinte length. However, it is not always true that this obstruction indeed arises because of the presence of diverging curvature, and the theorem does not directly show this.

Of course one can just “artiﬁcially” remove points from spacetimes which would then be consid- ered to be singular. However, we can avoid such possibilities by restricting consideration only to spacetimes which are not part of a larger spacetime.

Spacetime itself (the structure ( , g)) consists entirely of regular points at which g is well M behaved.

Deﬁnition A spacetime is singular if it is timelike of null geodesically incomplete but cannot be embeded in a larger spacetime.

494 [?], [57]

Energy condition implies a tendency for geodesics to converge, that is, it guarantees that gravity is attractive. Causality conditions prevents geodesics from converging through causality violations.

Figure 5.1: Sir Roger Penrose and Stephen Hawking. Initiated by Penrose, Penrose and Hawking, together with Robert Geroch, contributed much of the work on the existence of spacetime singularities with the use of point-set topological methods.

5.1.2 Some Basic Terminology

Terms we’ll encounter

Σ Σ Σ spacelike timelike null

Figure 5.2: .

Properties of the spacetime:

Spacetime time orientable: At every point there are two cones of timelike vectors. The Lorentzian manifold is time orientable if a continuous choice of one of the cones, termed future, can be made so it doesn’t ”turn upon itself” and be inconsistent. Spacetime is Hausdorﬀ: Events can be isolated. Spacetime metric is non-degenerate: The metric is non-degenerate.

495 Figure 5.3: An example of a non-orientable space.

Paths: Curves: Trips:

Causally deﬁned sets:

Chronalogical sets, I+ and J +: I+(x) := y : x y is called the chronological future { ∈ M ≪ } of x; I (x) := y : y x is called the chronological past of x; J +(x) := y : x y − { ∈ M ≪ } { ∈ M } is called the causal future of x; J (x) := y : y x is called the causal past of x. The − { ∈ M } chronological future of a set S is deﬁned by the set I+(S)= y : x y for some x ⊆ M { ∈ M ≪ ∈ S . Similar deﬁnitios hold for I (S), J +(S) and J (S). } − − Achronal sets: A set S is achronal if no two ponts of S are timelike related. ⊆ M Achronal bounderies Edge of a closed achronal set:

Σ Σ (a) (b)

Figure 5.4: (a) A closed achronal set with edge. (b) A closed achronal set without edge.

Special regions: convex normal neighbourhoods: An open set U is said to be a convex normal neighbourhood if for any p,q U there is a unique geodesic lying in U joining p to q. ∈ geodescially convex: simple regions:

Causality conditions:

All spacetimes in general relativity locally have the same stucture qualitative causal structure as in special relativity, globally spacetimes may not be not be “causally well behaved”. Here are some of the more imprtant conditions:

496 Chronology condition: There are no closed timelike curves. Causality condition: Closed null curves can exist even when the chronology condition is satisfied, this motivates the definition of the causality conditin. The causality condition is satisfied when there are no closed causal curves. Strong causality: there are no “almost closed” causal curves. This condition will be of particular importance for the singularity theorem. Future (past) distinguishing: Any two points with the same chronological futurre (past) coincide, i.e., ( , g) is future distinguishing at p if I+(p) = I+(q) for q = p and q . M ∈ M 6 6 ∈ M stable causality: Stable causality is violated when a spacetime is ”on the verge” of having closed timelike curves in the sense that an arbitrary small perturbation of the metric can produce a new spacetime that violates the chronology condition. See fig(5.1.2). Global hyperbolicity: One definition of globally hyperbolic is that for ( , g) is strongly M M causal and J +(p) J (q) is compact for any p,q . ∩ − ∈ M

h-null cone

g-null cone

Figure 5.5: The h null cone contains more timelike vectors than the g null cone so there − − is more likelyhood to ﬁnd closed timelike curves in ( , h) than in ( ,g). M M

Topology of the manifold: topological base: compact: Compact basically mens the space is finite in size. It is easily shown that compact spacetimes necessarily have closed timelike curves and so are ruled out. paracompactnes: A space may be non-compact, but if we have an open cover then we can “re- fine” it in a manner that still retains its finiteness properties locally on the space. If a manifold admits a Lorentz metric, it is paracompact. A manifold admits a positive definite metric if and only if it is paracompact. The availibility of a positive-definite metric is used in the singularity theorems. Alexanderoff topology: The sets I+(a) I (b) : a, b are open and form a basis for a { ∩ − ∈ M} topology on , called the Alexanderoff topology. An important result involved in the proof of M the singularity theorems is that when a spacetime is strongly causal, the Alexanderoff topology coincides with the topology of the spacetime.

Domains of dependence: Let S be an achronal subset of . Future and past domains of M dependence os S and the total domain of dependence of S, respectively are deﬁned as follows: D+(S)= x : every past endless causal curve from x intersects S . { ∈ M } D (S)= x : every future endless causal curve from x intersects S . − { ∈ M } D(S)= x : every endless causal curve containing x intersects S . { ∈ M } Obviously, D(S)= D+(S) D (S) ∪ − 497 Cauchy horizons: The future, past or total Cauchy horizon of an achronal closed set S is deﬁned as (respectively): H+(S)= x : x D+(S) but I+(x) D+(S)= , { ∈ M ∈ ∩ ∅} H (S)= x : x D (S) but I (x) D (S)= , − { ∈ M ∈ − − ∩ − ∅} H(S)= H+(S) H (S) ∪ − Cauchy surface: A Cauchy hypersurface for is a non-empty set S for which D(S) = M M partial Cauchy surface: Global hyperbolicity:

H H ) + ( S ( ) S + ( S S ) + ) ( + H D (S) H D (S) S S S H D−(S) ) H D−(S) − S ( ) ( ( S S S − ) ( ) H H

(a) (b) (c)

Figure 5.6: Domains of dependence. (a) The future domains of dependence of the achronal set S. (b) The past domain of dependence of the set S. (c) The total domain of dependence of S.

Energy conditions:

Weak energy condition: The weak energy condition states that T V aV b 0 for any timelike ab ≥ vector V . Strong energy condition: We say that an energy-momentum tensor T satifies the strong energy condition if, for every null vector V , T V aV b 0. Combined with the field equations ab ≥ the strong energy condition implies that for all timelike V , R V aV b 0. ab ≥ Generic condition: The strong energy condition holds. If every timelike or null geodesic contains a point where there is some curvature that is not specially aligned with the geodesic. The generic condition is not satisfied by a number of known exact solutions, however, they are rather special. The generic condition specifically is that t R t tctd = 0 where t is the [a b]cd[e f] 6 tangent vector. If the generic enery conditiond holds, each geodesic will encounter a region of gavitational focussing. As we will prove later, this will imply that there are pairs of conjugate points if one can extend the geodesic far enough in each direction

Closed trapped surface: At a normal closed surface, the outgoing null rays from the surface diverge, while the ingoing rays converge. On a closed trapped surface, both the ingoing and outgoing null rays converge. Future-trapped sets:

Jacobi fields: We have already encountered Jacobi fields in the derivation of the Einstein’s vacuum field equations. They are the connecting vectors of neighbouring geodesics.

Focal points: A focal point is where neighbouring geodesic intersect. Where neighbouring geodesics normal to a surface S intersect is also called a foal point.

498 A3 A2 < A1 outgoing A < A light cone 3 1 A2

ingoing light cone

Figure 5.7: closedtrapArea. A closed trapped surface is when the outgoing light cone also converges.

p V

V T S (a) (b)

Figure 5.8: Focal points. (b) Focal point to a surface.

Conjugate points:

Space of causal curves:

5.1.3 The Singularity Theorem of Hawking and Penrose

It predicts that gravitational collapse , both at the Big Bang and inside black holes, brings about space-time singularities as at which the theory breaks down.

The Strong Energy Condition is the most important energy condition (Gravity is attractive)

The basic concepts needed to formulate and understand the singularity theorems will be developed as we work through the proof.

In the context of the singularity theorems, the working definition of a singularity is a kind of incompleteness of the space-time under consideration, more precisely, it is an obstruction of some sort to time-like or null geodesics from being indefinitely extendible. If the world line of a particle only exists for a finite proper time then clearly something has gone wrong but the nature of the singularity - whether it is physical reason or just a mathematical pathology is in general unclear.

499 Penrose Singularity Theorem Future max. development of an initial data set containing a (future) trapped surface is incomplete. Tools: Theory of geodesics in Lorentzian (Riemm.) geometry. Causal geometry. Raychadhouri equation plays the decisive role. The full force of the Einstein equations is not used.

The singularity theorem give very little information about the nature of the singularities, in eﬀect they deny the existence of timelike or null geodesically complete spacetimes. The reason for the incompleteness is not predicted.

Many people believe that the resolution of the problem of singularities will come from mod- iﬁcations of the Einstein equations due to Quantum Gravity at the Planck scale. Explicitly demonstrated in loop quantum cosmology - a mini-superspace model which appropriately incor- porates the discrete nature of quantum spacetime.

Theorem (Hawking and Penrose (1970))

Theorem 5.1.1 1. R KaKb 0 for every non-spacelike vector K. ab ≥ 2. contains no closed timelike curves M

a b RabK K < 0 (5.1)

(3) Every non-spacelike geodesic, with tangent vector K, contains a point at which

K R K KcKd = 0. (5.2) [a b]cd[e f] 6

(4) There exists at least one of the following:

(i) a compact achronal set without edge, (ii) a closed trapped surface, (iii) the null geodesics from some point are eventually focussed, or the null geodesics from some closed 2-surface are all converging.

Condition (4) (i) is satisfied for any past-directed null-cone in our universe, because of the focussing affect of the black-body radiation; while (ii) is more or less the definition of a black hole, Condition (3) is a genericity condition is true for physical space-times except for certain highly symmetrical solutions.

An alternative version of the theorem is that the following three conditions cannot hold:

Theorem 5.1.2 The following are mutually inconsistent in any space-time:

(a) There are no closed trips

(b) Every endless causal geodesic contains a pair of conjugate points

500 (c) There is a future (past) - trapped set in . S M

In fact it will be the alternative form of the theorem we will prove. After we have done that we will explain how the other theorem follows.

5.1.4 Basic Deﬁnitions

Deﬁnition A Lorentzian manifold is time-orientable if a continuous designation of future- directed and past=directed for non-spacelike vectors can be made over the manifold.

Deﬁnition A path is a smooth map from a connected set in R to . M

Deﬁnition A curve is the point set image of a path. If t γ(t) is a path, the curve determined → by γ(t) will be denoted γ.

Future timelike curves are identiﬁed with worldlines of material particles in . M A timelike curve is maximally extended in the past if it has no past endpoint (such a curve is also called past-inextendible). The idea behind this is that such a curve is fully extended in the past direction, and not merely a segment of some other curve. Similarly deﬁnitions hold for timlike curves extended into the future.

M t0 t1 Figure 5.9: A path is a connected set in R to the space-time manifold . M

The set of future-directed causal curves with a past endpoint a and a future endpoint b will be denoted (a, b). C We will require that any smooth timelike curve to contain any end points it may have. This requirement eliminates curves that fail to be timelike at an end point, and which is null there instead.

Deﬁnition A causal curve is said to be past (future)-inextendable if it has no past (future) end point in . M

501 The fig. (5.8 (a)) is a curve in Minkowski that can be prolonged while remaining timelike and hence is a timelike curve that is not inextendable. we have extended the curve α with a past directed, timelike curve which continues indefinitely. The new curve α′ is past-inextendable. The new curve α′′ is future-inextendable. In fig. (5.8(d)) a point has been removed we cannot prolong this curve any further into the future and so it is future-inextendable.

The null generators of the event horizon in ﬁg.(3.13) in the sense that the continuation of the geodesic further into the past is no longer in the event horizon.

∞ missing b α′ end γ points α α′′ a

(a) (b) (c)−∞ (d)

Figure 5.10: Examples in Minkowski spacetime. (a) The curve α is taken to contain its own past end point a and future end point b. (b) The curve α′ is now future-inextendable . (c) The curve α′′ is now past-inextendable (d) The curve γ is not future-inextendable because it cannot be prolonged any further.

β α (a) (b) (c)

Figure 5.11: (a) A curve in Minkowski spacetime with point removed. (b) A curve is zig zag that it fails to have a well deﬁned tangent vector at a point. (c) Here we have a time like curve which is null at its end point.

Let us model this by the artiﬁcal example given in ﬁg(??) where we have removed a point from Minkowski spacetime. This provides an example of how the causal future of a point p will not neccessarily coincide with the closure of the chronological future of p.

In fact this situatio is more general than the above example. Say inﬁntesiamlly neighbouring null geodesics from p intersect at q. This means the point q will be conjugate to p along the null geodesic γ joing them. For points on γ beyond the conjugate point q there will be a variation of γ that gives a timelike curve from p. Thus γ cannot lie in the boundary of the future of p

502 r

timelike segment timelike q curve

null geodesic

Figure 5.12: In ﬂat spacetime, when a null geodesic curve joins onto a timlike curve, there exists a timelike curve between p and q.

r r r

z 0 z0 x1 q q q z1 x2 N1 N1 N N2 2

N3 p p p (a) (b) (c)

Figure 5.13: Say there are two points p and q connected by a null curve and a point r which is connected to q by a timelike curve. A timelike curve joining to the null geodesic. (b) Contiuing in this way, (c), we “peel” away a timelike curve that joins r and p. beyond the conjugate point q. So γ will have a future endpoint as a generator of the boundary of the future of p.

What we can say is the following.

Proposition 5.1.3 For all subsets S , ⊂ M (1) int J +(S)= I+(S),

(2) J +(S) I+(S). ⊂

Proof:

Obvious.

503 J +(p)

Figure 5.14: Artiﬁcial example of how the causal future of a point p will not neccessarily coincide with the closure of the chronological future of p.

Figure 5.15: The point q is conjugate to p along null geodesics, so a null geodesic γ that joins p to q will leave the bounadry of the uture of p at q.

5.1.5 Achronal Sets

Intuitively, a neighbourhood of a point is a set containing the point where you can move that point some amount without leaving the set.

Deﬁnition If X is a topological space and p X, a neighbourhood of p is a set V , which ∈ contains an open set U containing p,

p U V. ∈ ⊆

Deﬁnition A subset S is achronal provided no two of its points can be joined by a ⊂ M timelike curve.

Lemma 5.1.4 An achronal boundary B is an achronal set.

504 Proof: Let B = ∂I+[S], for some. If two points x,y B satisify x << y, then y I+(x) ∈ ∈ ⊆ I+[S] which is open. Hence y cannot be on the boundary of I+[S].

Lemma 5.1.5 Let (λ) be a causal curve that intersects I˙+[S] at some point p. In the past the C (λ) reamins in I˙+[S] I+[S]. C ∪

Proof: Consider an arbitrary point x on (λ) such that x p and an arbitaryly small neigh- C ≫ bourhood of x. A small deformation of the curve (λ), between p and x, produces a timelike C curve from some point q U(x). Since p I˙+[S], a slight deformation of , keeping it D ∈ ∈ D timelike, produces a curve from q to to some point in I+[S]. But as q is in some arbitrary E small neighbourhood of x, it must also lie in I+[S] or else in its boundary, I˙+[S].

q x

E C D p I+[S] I˙+[S]

Figure 5.16: .

Lemma 5.1.6 If p ∂I+(S) then I+(p) I+(S), and I (p) /I+(S) . ∈ ⊂ − ⊂ M

Note that if q I+(p) then p I (q), and hence I (q) is a neighbourhood of p. Since p is on ∈ ∈ − − the boundary of I+(S), it follows that I (q) I+(S) = , and hence q I+(S). The second − ∩ 6 ∅ ∈ part of the lemma can be proven similarly.

Proposition 5.1.7 Let B = ∂, where F is a future set. Let x B and suppose there is an ∈ open set Q containing x satisfying:

(a) If y Q F , there is a point z in F/Q, such that z y. ∈ ∩ ≪ or equivalently

505 I˙+[S] q + I [S] I˙+[S]

(a) (b)

Figure 5.17: The future set cannot be bounded by timelike curves.

(b) I (y) F/Q = , for all y Q F − ∩ 6 ∅ ∈ ∩ or equivalently

The proof uses the method of taking a limit of causal curves. A technical diﬃculty arises however in that a limit of smooth causal curves need not be smooth. This leads to the need of the notion of a C0 causal curve.

A C0 causal curve is a continuous curve is a continuous curve that can be approximated with arbitrary precision by piecewise smooth causal curve.

pn I+(S) γn p γ

Figure 5.18: .

Theorem 5.1.8 I˙+[S] is generated by null geodesics which have no past endpoints.

Proof: Take an arbitary point p in I˙+[S]. Construct an arbitrary neighbourhood U(p). In U(p) I+[S] construct a sequence of points p which converges to to the point p. For each n, ∩ { n} construct a causal curve γn extending from pn to S. Let qn be the intersection of γn with U˙ (p), the boundary of U[p]. Since U˙ (p) is a compact set, the sequence qn must have a limit point, q (the proof of this fact of point set topology will be given in section 5.1.7). As there are causal

506 U(p) p I˙+[S] p 1 2 q 1 I+[S] p q2

C q

Figure 5.19: .

curves from points pn, arbitrarily close to p to points arbitrarily near q there must be a causal curve, , from p to q. C Since q is a limit point of a sequence of points in I+[S], q either lies in I+[S] or in its boundary I˙+[S], or both. Suppose q / I˙+[S]. Then there is a small neighbourhood U(q) is contained ∈ entirely in I+[S]. Construct a causal curve from p to S by going from p to q along the causal curve , then from q along a timelike curve to some point r U(q), and then from r to S along C ∈ a causal curve. Since this curve from p to S has a timelike segment, it reaches any desired point s in some small neighbourhood V (p). But this means that V (p) I+[S], hence that p / I˙+[S]. ⊂ ∈ This contradicts the deﬁnition of p, and so it must be that q I˙+[S]. ∈

C q

I˙+[S] r

Figure 5.20: .

Proposition 5.1.9 Once a a generator, being followed into the future, enters I˙+[S], it can never leave I˙+[S].

507 Proof: Follow the generator from p to p and futher. This curve can never leave I˙+[S]. C ′

5.1.6 Strong Causality

The strong causality condition holds on if there are no closed or “almost-closed” timelike M or null curves through any point of . If µ is any timelike curve in a spacetime that obeys the M strong causality condition and x is any point not on µ, then there must be some neighborhood of x that does not intersect µ. (Otherwise, µ would accumulate at x, and thereby give an N almost-closed timelike curve.)

Globally hyperbolic causally simple stably causal strongly distinguishing causal → → → → → chronological. → Chronology: There are no closed timelike curves, collection of points π , such that p < M 1 p2 ...pn

Causality: There are no closed causal curves, collection of points pi M,s.t.p < p < ... < { } 1 2 pn

Future / Past Distinguishing condition: Any two points with the same chronological future (past) coincide.

Strong Causality Condition: There are no almost closed timelike curves.

Stable Causality Condition: (M,g) is not ”on the verge” of having a bad causal structure. (There is a neighbourhood of g in the Ck open topology ...) Deﬁnition: There exists a continuous nonzero tvf ta such that the metric g′ab := gab the chronology and causality conditions a slightly stronger condition when the chronology condition is almost violated. Since an arbitrarily small perturbation of the metric will result closed timelike curves, such space-times also seems physically unreasonable. Techniquile reasons for considering strong causality. It will be of central importance in what follows.

Strong causality is violated at a point p if there are timelike curves starting at x which come arbitrarly close to p after leaving a given convex neighbourhood.

We give a more precise deﬁnition of strong causality at a point.

A collection of open sets in forms a neighbourhood basis of some point x if every open M ∈ M set containing x has in it an open set from the collection with each open set in the collection containing x.

Deﬁnition (Strongly Causal) A space-time is said to be strongly causal at a if and M ∈ M only if a has a neighbourhood base U : α Λ with the property that no U is intersected by { α ∈ } α a trip that then leaves Uα and then intersects Uα again. A space-time is strongly causal if it is strongly causal at each point.

508 ℓ1 p M

ℓ2

Figure 5.21: The lines ℓ1 and ℓ2 have been removed. This is a space-time which is causal but fails to be strongly causal.

a a a a

(a) (b)

Figure 5.22: (a) . (b) Suitable causal basis.

In 2 (and elsewhere) one may construct various types of neighbourhoods at a point a. A M neighbourdhood base at a constructed with sets similar to either of the ﬁrst two in the ﬁgure is not a suitable “causality base”. 2 is strongly causal because there exists at each point a a M neighbourhood base consisting of sets similar to either the third or the fourth.

Deﬁnition (Causally Convex) An open set in a space-time is said to be causally convex U M if no causal curve leaves and then reenters . U U

Deﬁnition (Strongly Causal) A space-time is strongly causal at a point p if every open ∈ M neighbourhood of p contains a causally convex neighbourhood of p.

Deﬁnition Let N be a simple region containing a. The local future (past) of a point a, + denoted IL (a) (IL−(a)), is

+ Deﬁnition Let N be a simple region containing I−(x) I (y) neighbourhood base L ∩ L z : p r q, where the trip lies in N. { ≪ ≪

In Minkowski space-time the points x and y

509 x + x + b I (a) b I (a) b I+(a) x y y y I−(b) I−(b) a I−(b) a a

(a) (b) (c)

Figure 5.23: Minkowski space-time: b is joined to a by a null geodesic we consider x + ∈ I (a) and y I−(b). In the ﬁrst in case (a) y x. In case (b) y < x but y x. In case (c) x and y are∈ not causally related, in particular≪ y x. 6≪ 6≪

ℓ1 ℓ1 ℓ1 + I+(a) I (a) I+(a) b x b x b x y a y a a I−(b) y I−(b) I−(b)

ℓ2 ℓ2 ℓ2 (a) (b) (c)

+ Figure 5.24: b is joined to a by a null geodesic we consider x I (a) and y I−(b). In ∈ ∈ the ﬁrst case (a) y x there are no closed timelike curves. In case (c) For some x and y. ≪

Proposition 5.1.10 Let a . Strong causality fails at a if there is a point b > a, b = a, ∈ M 6 such that for all x I+(a) and all y I (b), y <

Proof:

Suppose a and b exist satisfying the hypotheses. Separate a and b by disjoint neighbourhoods of a contained in V (a) and W (b). Let U(a) be any neighbourhood of a contained in V (a). Choose x1 I (a) U(a) and x I+(a) U(a) see ﬁg.5.1.6 (a). ∈ − ∩ ∈ ∩ Since x1 << a, and a < b, x1 << b. Thus there is a trip from x1 to b which must leave U(a), see ﬁg(5.1.6a). Choose y on this trip inside W (b). Then y <

Now we consider the converse. We ﬁrst need to introduce the following concept. Let N be a simple region containing a. Take any x>>a such that x and a are joined by a unique timelike

510 b W (b) b W (b) y y

x V (a) V (a) U(a) U(a) x a a x′ x′

(a) (b)

Figure 5.25: converse. (b) That y x to be true, there must be a timelike curve that renters U(a). ≫

geodesic in N. x is said to be in the local future of a, denoted x I+(a). ∈ L

Proposition 5.1.11 Let a . Strong causality fails at a only if there is a point b > a, b = a, ∈ M 6 such that for all x I+(a) and all y I (b), y <

Proof:

If y is in the local past of a, then

+ I (x) I−(x)= z : y <

If strong causality fails at a point a there must be a nested sequence of these neighbourhoods at a, U : i = 1, 2, 3,... U U ; U = a with the property that each U contains the { i } i ⊃ i+1 ∩i{ i} { } i past end point of a trip γi which leaves N and then comes back to enter Ui.

U : i = 1, 2, 3,... ; U U ; U = a. (5.3) i I ⊃ i+1 ∩i{ i}

Being the limit of timelike curves [cidi] cannot by spacelike, but can be null or timelike?? y <

x a b c y for i large enough (5.4) ≪ i ≪ i ≪ i ≪

511 γi

U i p ai di y q x c ci

Figure 5.26: proof of .

b b a i i0 U N i0

p y

x c

c ′ i0 + Figure 5.27: For i i , U I (x), from which we can conclude x a . For i i′ , ≥ 0 i ∈ ≪ i ≥ 0 c I−(y). If we choose i > i , i′ then we have x a and c y i ∈ 0 0 ≪ i i ≪

Remark If, given the conditions of the 5.1.10, b is found to satisfy a b, then for any y ≪ ∈ I (b) I+(a), we have y y; i.e. there are closed trips in . − ∩ ≪ M

Alexandroﬀ Topology

A topological space is Hausdorﬀ if any two given points can be isolated from each other by disjoint sets. Or more presicely, a topological space X is Hausdorﬀ if any two given points p ,p X, there are two open sets U and U with p U and p U but with the intersection 1 2 ∈ 1 2 1 ∈ 1 2 ∈ 1 of U1 and U2 empty.

The ﬁner the topology, the more open sets it has.

Theorem 5.1.12 The following three conditions on a spacetime are equivalent: M

512 ℓ1 ℓ1 + I−(x) I (y) + ∩ I−(x) I (y) x′ x′ ∩ x x x b b p a a a

y′ y′ + y y y I−(x′) I (y′) ∩ ℓ2 ℓ2 (a) (b) (c)

+ Figure 5.28: b a x I (a) and y I−(b). In the ﬁrst case (a) y x timelike curves. (b) ∈ ∈ + ≪ Points in the open set of b are also in I (x) I−(y) In case (c) . ∩

(a) A spacetime is strongly causal; M (b) the Alexandroﬀ topology is Hausdorﬀ;

Proof: If the alexanddorﬀ topology is weaker than the manifold topology, there is an -open M set U(y) such that any A-nbd. of y gets outside of U, (for example in ﬁg(5.1.6)). Since is M regular we may assume this holds for the U as well. Now let V (y) be an arbitrary -open M neighbourhodd of y. Let b I+(y), a I (y). By hypoth, I (b) I+(a) /U = . Choose z ∈ ∈ − { − ∩ } 6 ∅ in this set. Then z V (y), but a z b, and V (y) disconnects the trip [azb]. Since V (y) is 6∈ ≫ ≫ arbitrary, strong causality fails at y.

ℓ1 ℓ1

+ U(y) I−(b) I (a)/U ∩ b z b y y a a U(y) V (y)

ℓ2 ℓ2 (a) (b)

+ Figure 5.29: a I−(y) V (y) and b I (y) V (y). In the ﬁrst case (a) y x timelike ∈ ∩ ∈ ∩ + ≪ curves. (b) Points in the open set of b are also in I (x) I−(y). ∩

Proposition 5.1.13 For each a deﬁne Q x : x lies on a closed trip through a . ∈ M a { ∈ M } Then Q is open, for every a ; and either Q Q , for all a, b . a ∈ M a ∩ b ∈ M 513 + U(y) I−(b) I (a)/U z ∩ b b y y a a U(y) V (y)

(a) (b)

+ Figure 5.30: a I−(y) V (y) and b I (y) V (y). In the ﬁrst case (a) y x timelike ∈ ∩ ∈ ∩ + ≪ curves. (b) Points in the open set of b are also in I (x) I−(y). ∩

Proof:

Proposition 5.1.14 If strong causality fails at a , then one (at least) of the following is ∈ M true:

(1) There are closed timelike curves through a.

(2) There is a future endless null geodesic through a along which strong causality fails; and there are closed tris near a.

(3) Time reverse of (2).

(4) There is an endless null geodesic through a along which strong causality fails.

In a spacetime , let G be deﬁned as the set of points where strong causallity holds, i.e. G M ⊂ M G = x : is strongly causal at x . { M }

Proposition 5.1.15 Let a . Strong causality fails at a there is a point b > a, b = a, ∈ M ⇔ 6 such that for all x I+(a) and all y I (b), y x. ∈ ∈ − ≪

Deﬁnition The future domain of dependence, D+(Σ), is the set of points p in for which M every past-inextendable causal curve through p intersects Σ, ﬁg(5.1.6).

We note that Penrose [?] and Geroch [?] use timelike curves to deﬁne the domain of depnedence, rather than non-spacelike curves used above, which agrees with Hawking and Ellis [7].

The physical signiﬁcance of the future (past) domain of development D+(Σ) is that it is the region in the future (past) of Σ that can be predicted from knowledge of the data on Σ.

514 missing α γ missing

c a d b missing α missing

(a) (b)

Figure 5.31: (a) There is an endless null geodesic along which strong causality is violated. (b) Strong causality is violated everywhere in R.

F := I+(Q) F := I+(Q)

Q(a)= F P ∩ boundary of F a boundary of P

P := I−(Q) P := I−(Q)

+ Figure 5.32: F := I (Q) and P := I−(Q). Q = F P , ∂Q = F ∂ + ∂P . ∩

Deﬁnition Let S be a closed achronal set. The edge of S is deﬁned as a set of points x S ∈ such that every neihbourhood U(x) of x contains y I+(S) and z I (S) and a timelike curve ∈ ∈ − γ from z to y which does not meet S.

An example of a clsosed achronal surface without edge is given if ﬁg ().

Proposition 5.1.16 Let S be achronal and closed. Then: ⊂ M (1) H+(S) is achronal and closed.

(2) D+(S) is closed.

(3) x D+(S) I (x) I+[s] D+(s). ∈ ⇒ { − ∩ ⊂

515 ℓ1

p ℓ2 M q

ℓ3

Figure 5.33: The lines ℓ1, ℓ2 and ℓ3 have been removed.

+ q D+(Σ) p D (Σ) 6∈ ∈ past light cone of p

Σ γ past- not past- inextendable inextendable

Figure 5.34: The future domain of dependence, D+(Σ), of Σ. p is in D+(Σ), q isn’t because ther are past-inextendable causal curve through q that don’t intersect Σ, e.g. the curve γ

(4) ∂D+(S)= H+(S) S. ∪ (5) I+[H+(S)] = I+[s]/D+(S).

Proof:

Deﬁnition Let S be achronal and closed. The edge of S is deﬁned to be ⊂ M

x S : for ally,z, there is a trip[zy] not meetingS . (5.5) { ∈ }

Proposition 5.1.17 (a) x D+(S) implies I (x) edge(S)= ∈ − ∩ ∅ (b) edge(S)= edge(H+(S)).

Proposition 5.1.18 Let x H+(S)/edge(S). Then there is a null geodesic on H+(S) with ∈ future endpoint x.

516 light cone D+(Σ)

Figure 5.35: The future domain of dependence, D+(Σ), of a clsoed Σ in Minkowski spacetime.

D+(Σ) S D+(Σ)

D−(Σ) S

D+(Σ)

Figure 5.36: Domains of dependence.

Deﬁnition A non-empty, closed achronal set S is said to be future-trapped if E+[S] is compact. Past-trapped sets are deﬁned analogously.

If we can associate with the topology some countable set of open sets and use the desirable properties of countability to learn something about the space. In conjuction with the requirement of strong causality, we establish the following result:

Proposition 5.1.19 For x int D+(S) the set J (x) J +[S] is compact. ∈ − ∩

Proof: Assume K := J (x) J +[S] is not compact. − ∩ As we assume K to be non-compact, it has no cover with a ﬁnite subcover. There is an open cover U : i = 1, 2, 3 . . . which is countable, locally ﬁnite and whose elements are simple { i1 } regions.

517 ) H S + + ( + (S H D (Σ) )

S H+(S) S ∩

Figure 5.37: The future domain of dependence.

pn H+(S)

γn p D+(Σ)

Figure 5.38: There exists a subsequence γm of past inextendble causal curves that do not meet S that converges to a past inextendble C0 null geodesic γ starting at p.

5.1.7 The Space of Causal Curves

In a spacetime , let G be deﬁned as the set of all points in which are strong causality M ⊂ M M holds,

G := x : is strongly causal (5.6) { M }

The set where srong causality fails being /G. M

Proposition 5.1.20 G is open.

Proof We prove this by showing that /G is closed. First note that a set is closed in if and M M only if this set contains its own boundary. If we show that the limit point of every convergent series in /G also belongs to /G, then by implication, every point in the boundary of /G M M M also belongs to /G. To this end, consider any series x /G which converges to some M { i} ∈ M point x . Take a simple region containing x , ﬁnd the point b such that for all ∈ M i i

+ a I−(b ) and b I (x ) a b. (5.7) ∈ i ∈ i ≪

518 y x U(x) Σ γ z

Figure 5.39: The edge of a closed achronal surface Σ.

q p Σ r

Figure 5.40: A simple example of a closed achronal surface without edge can be given by considering the spacetime R S with light cones locally at 45 degrees. For the open × + neighbourhood there is no r I−(p) and q I (p) with a timelike curve between them that doesn’t intersect S. The∈ closed achronal∈ set Σ has no edge.

Take the intersection of the geodesic [xibi] with the (compact) boundary of the simple region. These intersections have a cluster point y. Take two disjoint neighbourhoods of x and y. It is easy to construct a trip leaving this neighbourhood of x, entering the neighbourhood of y and then reentering the neighbourhood of x, showing that strong causality fails at x. Thus x /G. ∈ M

The length of a smooth curve is

1/2 q dαa(t) dαb(t) L(α)= g (α(t)) dt (5.8) ab dt dt Zp

The length of a peicewise smooth curve is deﬁned by adding the lengths of its smooth segments.

It is assumed that A and B are closed achronal sets contained in K, a compact subset of G. We denote by (A, B) the set of causal curves from A to B lying in K, and by (A, B) the set CK TK of causal trips from A to B lying in K.

Lemma 5.1.21 Any compact subspace C of a Hausdorﬀ topological space X is closed in X.

519 K

cluster point

Figure 5.41: We can only get convergence to a cluster point in K if the space wasn’t locally ﬁnite.

D+(S)

z0 x z S

+ Figure 5.42: K = J (S) J −(p) K is compact. ∩

Proof: Let C be a compact subspace of a Hausdorff topological space X and let a X C. We ∈ − shall prove that there exist an open set Ua such that a Ua X C. Then X C = a X C Ua, ∈ ⊂ − − ∪ ∈ − so X C is open and hence C is closed. − Since X is Hausdorff, we can choose open subsets U and V (x) of X such that a U , x V (x), x ∈ x ∈ and U and V (x) are disjoint. Choosing such subsets U and V (x) for every point x C, we x x ∈ see that the sets V (x) form an open cover of C. Since we assume C is a compact topological space, it follows that C is the union of the open sets Vx1 ,...,Vxr for some finite collection of points x ,...,x C. Let U = r U . Then U , as a finite intersection of open sets, is open. 1 r ∈ a ∩i=1 xi a Also, a U since a U for every i = 1, 2,...,r. Finally, if b U then for any i = 1, 2,...,r, ∈ a ∈ xi ∈ a b U and hence b V (x ), so b C, since C r V (x ). Thus U X C. ∈ xi 6∈ i 6∈ ⊂∪i=1 i a ⊂ −

520 D+(S)

y y2 1 x2 z1 S

+ Figure 5.43: K = J (S) J −(p) K is compact. ∩

y y y ′ i0

x x

xi x i0 (a) (b)

Figure 5.44: G is an open set.

, as a topological space, is metrizable. Choose an arbitrary metric d compatible with the M manifold topology. Then in G, d is compatible with the Alexandroﬀ topology.

Recall that for a set E and a point p,

d(p, E) := inf d(p,y) : y E (5.9) { ∈ }

For a curve γ and some ǫ> 0 we deﬁne the ǫ band V (γ) about γ as − ǫ

V (γ) := y : d(y, γ) <ǫ . (5.10) ǫ { ∈ M }

For γ (A, B) we deﬁne a metric on the space (A, B) as the radius of the smallest band ∈CK CK

521 which when put around either curve encloses the other as well,

ρ(γ , γ ) := inf ǫ : V (γ ) γ and V (γ ) γ . (5.11) 1 2 { ǫ 1 ⊃ 2 ǫ 2 ⊃ 1}

ρ is the restriction to (A, B) of the Hausdorﬀ metric. (ρ is a pseudo-metric on the power set CK of and a metric on the closed subsets of . The elements of (A, B) are compact and thus M M CK closed).

The causal curves form the complete set of pointwise limits of the causal trips.

Proposition 5.1.22 (A, B) is dense in (A, B). T C

Proof: Let γ (A, B). ∈C

Figure 5.45: approximate a causal curve by a causal trip.

Mathematical preliminaries

Recall that a Cauchy sequence is not necessarily convergent. For example, consider the subspace X = (0, 1] of the real line. the sequence deﬁned by xn = 1/n is easily seen to be a Cauchy sequence in this space, but it is not convergent, since the point 0 is not a point of the space.

A metric space is sequentially compact if every sequence has a convergent subsequence.

Proposition 5.1.23 Let (X, d) be a metric space. If a set C X is compact then it is sequen- ⊆ tially compact.

522 Proof: Let E denote the set of members of a sequence x in C. First say E is ﬁnite. Then { n} at least one point, say x, of E must be repeated inﬁnitely often in the sequence x , and its { n} occurrences form a subsequence converging to x, which is a point in C.

Now suppose E is infite. Suppose E has no limit point in C. Then for each x C, there exists ∈ ǫ(x) such that S (x) contains no point of E other than x. That is, S (x) E = or x . ǫ(x) ǫ(x) ∩ ∅ { } Now by compactness of C, the cover S (x) : x C has a finite subcover. But the union of { ǫ(x) ∈ } any finite subcollection of the Sǫ(x)(x) doesn’t cover E, let alone C. This contradiction proves that E must have a limit point in C.

A space is totally bounded if the cover

S (x) : x X (5.12) { ǫ ∈ } has a ﬁnite subcover for any ǫ> 0.

Roughly, a metric space is complete if every series in it which tries to converge is successful, in the sense that it ﬁnds a point in the space to converge to.

Theorem 5.1.24 Let (X, d) be a metric space. A set C X is sequentially compact if and ⊆ only if it is both complete and totally bounded.

Proof: Let C be sequentially compact. Every sequence has a convergent subsequence. Since the sequence is Cauchy, it follows it converges. We prove total boundedness by contracdiction. Suppose there is some ǫ > 0 such that C is not totally bounded for ǫ. It is then possible to construct a sequence such that no term in the sequence is within ǫ of any other term in the sequence. This sequence has no Cauchy subsequences, for if it did there would be a subsequence who’s terms would get close to each other. Hence there are no convergent subsequences, and hence C is not sequentially compact. We construct the desired sequence as follows: take any point p1 in C. Choose p2 to be any point not in Sǫ(p1). This is possible since C is not totally bounded. Choose p to be any point not in S (p ) S (p ). Again this is possible since C is not 3 ǫ 1 ∪ ǫ 2 totally bounded. Continuing in this way one constructs the desired sequence.

Now, let C be complete and totally bounded. Let E denote the set of members of a sequence x in C. If E is finite it has a trivial convergent subsequence. Suppose, then, this is not { n} the case. As C is totally bounded we can form a finite cover with a set of balls of radius 1. One of these balls, call it S must contain an infinite subset of x . Take the first term in the 1 { i} subsequence, x , to be any term in the sequence x that lies in S . Now form a new cover i1 { i} 1 of C using a finite set of balls of radius 1/2. One of these balls, label it S2, must contain an infinite subset of S E. Take the second term in the subsequence, x , to be any term in the 1 ∩ i2 sequence x that lies in S S and for i > i . Continuing in this way, one constructs a { i} 1 ∩ 2 2 1 Cauchy susequence x . Since C is complete, x converges to a point in C. Thus C is { ik } { ik } sequentially compact.

523 Deﬁnition Given an open cover of a metric space X, a ﬁxed real number ǫ > 0 is called a U Lebesgue number for if for any x X, there exists a set U(x) such that S (x) U(x). U ∈ ∈U ǫ ⊂

Theorem 5.1.25 Any sequentially compact metric space (X, d) is compact.

Proof: First we show there exists a Lebesgue number for any open cover of a sequentially compact metric space X. Suppose be an open cover for which no Lebesgue number exists. U Then for any integer n, there exists some point of X, say xn, such that S1/n(xn) is not contained in some U of . Consider a sequence of such points x . By sequentially compactness, x U { n} { n} has a subsequence x converging to some point x X. Since covers X, x U for some { nk } ∈ U ∈ 0 U . Since U is open, S (x) U for some integer m. Now S (x) contains x for all 0 ∈U 0 2/m ⊂ 0 1/m nk k M. Choose k M so that n m, and write s = n . Take y to be any point in S (x ), ≥ ≥ k ≥ k 1/s s then by

d(x,y) d(x,x )+ d(x ,y) < 1/m + 1/s 2/m ≤ s s ≤ we have S (x ) S (x), and hence S (x ) U . This contradiction shows that there 1/s s ⊂ 2/m 1/s s ⊂ 0 exists a Lebesgue number for . U Now, let be an arbitrary open cover of a sequentially compact metric space X. By the U preceding lemma there exists Lebesgue number ǫ for . As a sequentially compact metric space U is totally bounded, there exists a finite cover for this Lebesgue number S (x ) : x ,x ,...,x { ǫ i { 1 2 n}} for X. Then each S (x ) is contained in some set, say U , in by definition of Lebesgue number. ǫ i i U Since X n S (x ) n U , we have a finite subcover U , U ,...,U of for X. ⊂∩i=1 ǫ i ⊂∩i=1 i { 1 2 n} U

Proposition 5.1.26 Any closed subspace C of a compact metric space is compact. T

Proof: Let be any cover of C by sets open in . Since C is closed, C is open. C U T T − T − together with the collection forms an open cover of . By the compactness of , there is a U T T ﬁnte subcover, say U , U ,...,U , one these being the set C. The other U’s provide a { 1 2 r} T − ﬁnite subcover of for C. U

We now move onto proving (A, B) is a compact metric space. CK

Proposition 5.1.27 If K is compact and contained in G, (A, B) is totally bounded. CK

Proof: We wish to show the cover S (γ ) : γ (A, B) has a finite subcover for any ǫ> 0. { ǫ i i ∈CK } Let ǫ < 0 be given. Since K is compact it is totally bounded and may therefore be covered by finitely many Alexandroff neighbourhoods of diameter less than ǫ/2, say A , A ,...,A . { 1 2 N }

524 Consider the open sets B formed by chains of A ’s, A A , such that they connect { i} i j1 ∪···∪ jn the sets A and B, i.e.,

A A = , A B = , A A = (5.13) ∩ j1 6 ∅ jn ∩ 6 ∅ jk ∩ jk+1 6 ∅ and

I+[A ] A = . (5.14) j ∩ j+2 6 ∅

The condition A A = ensures there exists causal curves from one neighbourhood to the jk ∩ jk+1 6 ∅ next, the condition I+[A ] A = guarantees the chain B contains causal curves from A to j ∩ j+2 6 ∅ i B, by excluding cases such as in ﬁg (5.1.7). B

Aj+2 Aj

Figure 5.46: Imposition of the condition I+[A ] A = avoids cases such as the above. j ∩ j+2 6 ∅

As there are a finite number of possible unions of the finite collection of sets Ai’s, there are finitely many Bj’s. Now, set

B′ := γ (A, B) : γ B . (5.15) i { ∈CK ⊂ i}

By construction, we have B = for each i; since every γ in (A, B) is contained in one of the i′ 6 ∅ CK B ’s we have B = (A, B); γ B implies B S (γ). i ∪{ i′} CK ∈ i′ i′ ⊂ ǫ ......

Then the collection S (γ ) so determined is a ﬁnite cover of (A, B). { ǫ i } CK

Proposition 5.1.28 (A, B) is complete. CK

525 Proof: Let γ (A, B) be a Cauchy sequence. Cover K by finitely many Alexandroff { i}⊆CK neighbourhoods each of which is contained in a simple region whose closure lies in G. Let x { i} be a sequence on A formed by intersections of the γ . As x is Cauchy and A is compact i { i} (being a closed subset of the compact set K) x converges to a point x A. Let A be an i 0 ∈ 1 Alexandroff neighbourhood in the cover which contains x0. Introduce Minkowskian coordinates in A centered at x . Since x A0 there is some interval [0,t ) such that for each t satisfying 1 0 0 ∈ 1 1 0 t

γ2 γ1

x0(t) t

Figure 5.47: The geodesic λ.

......

Repeat the process, taking the Minkowskian coordinates in A2 to have the value (t1, 0, 0, 0) at x0(t1). The process is is ﬁnite since the curve x0(t) so constructed is a causal curve and cannot re-enter any Alexandroﬀ neighbourhood it has left. Therefore x0(t) must strike B, since B is closed.

Theorem 5.1.29 (A, B) is a compact metric space. CK

Proof: (A, B) is totally bounded and complete and hence compact. CK

Corollary 5.1.30 Let S be closed, achronal, and strongly causal. Let x intD(S). Then ∈ (A, B) is compact CK

526 Proof: (x,S) K J (x) J +[S], which is compact and contained in a strongly causal CK ⊂ ≡ − ∩ region.

Corollary 5.1.31 Let S be closed, achronal, and strongly causal. Let a, b intD(S). Then ∈ (A, B) is compact CK

S b

Figure 5.48: The sequence is a Cauchy sequence in (a, b), but it is not convergent, since there is a point missing. C

Deﬁnition A spacetime is said to be globally hyperbolic if and only if is strongly causal M M and (a, b) is comapct, for all a, b . C ∈ M

Deﬁnition A Cauchy hypersurface for is a closed achronal set S for which D(S)= . M M

Proposition 5.1.32 has a Cauchy hypersurface implies the spacetime is globally hyper- M M bolic.

Proof: Corollary 5.1.31.

Deﬁnition A partial Cauchy surface S is deﬁned as an acausal set without edge.

A partial Cauchy surface is called a Cauchy surface or a global Cauchy surface if D(S)= . M + Let S be a partial Cauchy suurface. Then let = D (S) D−(S). Even though may not be N ∪ + M globally hyperbolic and S is not a Cauchy surface, the region intD (S) or intD−(S) is globally hyperbolic in its own right as the surface S serves as a Cauchy surface for the manifold int . N 527 An important property of globally hyperbolic spacetimes, or any globally hyperbolic subset of , M which is relevant for the singularity theorems is the existence of maximum length non-spacelike geodesics between pairs of causally related points. This we will see in the next section.

If one removes a point from Minkowski spacetime, the resulting spacetime admits no Cauchy M surface and is therefore not gloabally hyperbolic. q

H+(S) λ H+(S)

p D+(S)

t =0 S

Figure 5.49: Minkowski space time with a point removed is not globally hyperbolic. The point q is not in D+(S) as there are non-spacelike curves like λ which do not meet S in the past.

The requirement that a spacetime posses a Cauchy hypersurface is very strong as we will see, such spacetimes have very regular global behaviour and all the pathological features in previous sections are excluded. It has been shown that a globally hyperblic spacetime is homeomorphic to R S where S is a three-dimensional submanifold and for each t R, the spacelike hypersurface × ∈ St of constant time is a Cauchy surface for the spacetime. A globally hyperblic spacetime has a unique topological structure and admits no topology change.

There are spacetimes homeomorphic to R S which do not have Cauchy hypersurfaces. × In quantum gravity, however, diﬀerent kinds of topologies and, in particular, topology changes are conceivable. Perhaps it is possible to construct the quantum theory of the gravitational ﬁeld based on the calssical assumption that = R σ and then lift this restriction in the quantum M × theory.

5.1.8 Conjugate Points

Mathematical preliminaries

Proposition 5.1.33 If f : is a continuous map of topological spaces and is compact T1 →T2 T1 then f( ) is compact. T1

528 Proof: Suppose that is an open cover of f( ) by sets open in . By continuity of f, U T1 T2 f 1(U) is open in for every U in . By compactness of , there is a ﬁnite subcover − T1 U T1 f 1(U ),f 1(U ),...,f 1(U ) , and it is easily seen that U , U ,...,U is then a ﬁnite sub- { − 1 − 2 − r } { 1 2 r} cover of for f( ). U T1

Corollary 5.1.34 Compactness is a toplogical property: if and are homeomorphic then T T2 is compact if and only if is compact. T1 T2

Corollary 5.1.35 Any continuous map from a compact space to a metric space M is bounded.

Proof: We show that any compact subset C of a metric space is bounded, the corollary then follows from proposition 5.1.33. Let x be any point of M. Then C B (x ), since for 0 ⊂ ∪n∞=1 n 0 any x C, d(x,x )

Deﬁnition If f is a real-valued function on a metric space E and p E we say that f attains a 0 ∈ maximum at p if f(p ) f(p) for all p E, and that f attains a minimum at p if f(p ) f(p) 0 0 ≥ ∈ 0 0 ≤ for all p E. ∈

Corollary 5.1.36 Any continuous real-valued function on a nonempty compact metric space attains a maximum at some point, and also attains a minimum at some point.

x0 x0 (a) (b)

Figure 5.50: (a) An upper semi-continuous function. (b) A lower semi-continuous function.

529 Deﬁnition A real-valued function f deﬁned on a topological space is said to be upper T semicontinuous at a point x if, given any ǫ> 0, there exists a neighbourhood of x in 0 ∈T 0 which f(x)

lim sup f(x) f(x0). x x0 ≤ →

The function f is called upper semicontinuous if it is upper semicontinuous at every point of its domain. Similarly, f is said to be lower semicontinuous at x0 if, given any ǫ> 0, there exists a neighbourhood of x in which f(x) >f(x ) ǫ. 0 0 −

Lemma 5.1.37 Any compact subspace C of a Hausdorﬀ topological space X is closed in X.

Theorem 5.1.38 Let f : M R + be a lower semicontinuous function on the Hausdorﬀ → ∪{ ∞} space M. Suppose that there is a real number r such that a) [f r]= x M : f(x) r = and ≤ { ∈ ≤ } 6 ∅ b) [f r] is sequentially compact. ≤

Then there is a minimizing point x0 for f on M:

f(x0) = inf f(x).

Proof: We begin by showing indirectly that f is lower bounded. If f is not bounded from below there is a sequence xn n N such that f(xn) < n for all n N. For sufficiently large n the { } ∈ − ∈ elements of the sequence belong to the set [f r], hence there is a subsequence y = x which ≤ j n(j) converges to a point y M. Since f is lower semicontinuous we know f(y) lim infj f(yj), ∈ ≤ →∞ a contradition since f(y ) < n(j) . We conclude that f bounded from below and thus j − → −∞ has a finite infimum:

530

Threfore there is a minimizing sequence xn n N whose elements belong to [f r] for suﬃciently { } ∈ ≤ large n. Since [f r] is sequentially compact there is again a subsequence y = x which ≤ j n(j) converges to a point x [f r]. Since f is lower semmicontinuous we conclude 0 ∈ ≤

I f(x0) lim inf f(yj)= I. ≤ ≤ j →∞

Theorem 5.1.39 A ﬁnite upper semicontinuous function f deﬁned on a compact topological space is bounded from above. T

Proof:

Theorem 5.1.40 A ﬁnite upper semicontinuous function f deﬁned on a compact topological space achieves its least upper bound on . T T

Proof:

Conjugate points

We consider a congruence of null geodesics with aﬃne parameter v and tangent V a. The a expansion of the geodesics may be deﬁned as θ = DaV , where Da is the covariant derivative. The propagation equation for θ leads to this inequality:

dθ 1 θ2 R V aV b. (5.16) dv ≤−2 − ab

Suppose that (i) R V aV b 0 for all null vectors V a (this is called the null convergence con- ab ≥ dition), (ii) the expansion, θ, is negative at some point v = v0 on a geodesic γ, and (iii) γ is complete in the direction of increasing v (i.e., γ is defined for all v v ). Then θ along ≥ 0 → −∞ γ a finite affine parameter distance from v0. maximum and minimum of functions of one variable. If f ′(a) = 0, then we know that f(x) is stationary at x = a. We then go on to study the sign of f ′′(a). If it is positive, then f(x) has a minimum value at x = a, if it is negative, then f(x) has a maximum value at x = a, and if f ′′(a)=0.

531 τ(α) := b f(α, t)dt where f = ( tat )1/2. recall a − a R dτ = 0. (5.17) dα α=0

for independent of the variation function η.

Interested in the sign of the second variation of the length function,

d2τ (5.18) dα2 α=0

For it to be a maximum the sign of must be negative independently of the choice of the deviation function η(x). conjugate point if

d2τ = 0. (5.19) dα2 α=0

Proposition 5.1.41 Let A and B be achronal topological manifolds in a strongly causal spacetime . Then the length function is upper semi-continuous on (A, B). M T

Proof: Let γ γ in (A, B). We wish to show that lim sup L(γ ) L(γ ) Choose a k → 0 T { k } ≤ 0 subsequence of γ and relabel so that L(γ ) lim sup L(γ ) . It is suﬃcient to prove the { k} k → { k } proposition in the case that γ is a geodesic and (A, B) N, a simple region. 0 T ⊂

Let xk,yk be intersections of γk with A and B respectively; and let µk be the unique causal geodesic [x y ] lying in N. Since γ γ , we have x x and y y . k k k → 0 k → 0 k → 0 Also, since L is a continuous function on geodesics joinig two continuous surfacse, we have L(µ ) L(γ ). By local maximality of geodesics, for every k we have L(γ ) L(µ ). Thus k → 0 k ≤ k lim L(γ ) = a lim L(µ ) , and so the superior limit of the original sequence is less than or { k } ≤ { k } equal to L(γ0).

Two points joined by a timelike curve can be connected by a broken null geodesic, hence, the length function is not a lower continous function.

Theorem 5.1.42 There is a curve γ (A, B) with maximum length. ∈CK

An upper semi-continuous function on a compact set attains its maximum.

Note that γ needn’t be unique and needn’t be a geodesic.

532 Figure 5.51: Two points joined by a timelike curve can be connected by a broken null geodesic.

bb b b

aa a a

Figure 5.52: A sequence of null curves may converge to a timelike curve.

5.1.9 Timelike Congruences

We are considering a smooth congruence of timelike geodesics. The geodesics are parameterized by proper tiem τ...

....so that the vector ﬁelds V a of tangents is normalised to unit length: V aV = 1. a − We deﬁne the transverse metric or “spatial metric” by

hab = gab + VaVb. (5.20)

It is easily seen that the four tensor

a ac a a h b = g hcb = δ b + V Vb (5.21) is the projection operator onto the subspace of the tangent space perpendicular to V a as

a b a a b a a b a h bV = (δ b + V Vb)V =0 andif W Va then h bW = W .

The expansion, twist and shear of the ﬁeld

The expansion, shear and twist of the ﬁeld are deﬁned by

533 B B B

γ 1 γ2 K γ

A A A (a) (b) (c)

Figure 5.53: If A and B are parallel in Minkowskia spacetime then γ1 and γ2 are maximal. (b) Here there is only one element in K (A, B) which is necessarily maximal. It is not a geoesic. (c) Here the maximal eleementC is a trip.

θ = h ch d V (5.22) ab a b ∇(d c)

θ = habh ch d V a b ∇(d c) = hcd η (5.23) ∇(d c)

ω = h ch d V (5.24) ab a b ∇[d c]

1 σ = h ch d V θh (5.25) ab a b ∇(d c) − 3 ab

θ = habh ch d V a b ∇(d c) = hcd V (5.26) ∇d c

The Raychaudhuri equations

Note that V can be decomposed as ∇b a

1 V = θh + σ + ω (5.27) ∇b a 3 ab ab ab

V c ( V ) = V c V + V c( )V ∇c ∇b a ∇b∇c d ∇c∇b −∇b∇c d = V c V + R dV cV ∇b∇c d cba d = (V c V ) ( V c)( V )+ R dV cV ∇b ∇c d − ∇b ∇c a cba d = ( cV )( V )+ R dV cV (5.28) − ∇ b ∇a c cba d 534 where we used V c V = 0. In the following it should be kept in mind that θ and σ are ∇c d ab ab symmetric while ωab is anitsymmetric. Taking the trace over a and b we obtain,

1 1 V c θ = θhca + σca + ωca θh + σ + ω R V cV d ∇c − 3 3 ac ac ac − cd 1 = θ2 σ σab + ωabω R V cV d (5.29) −3 − ab ab − cd and so

dθ 1 = θ2 σ σab ωabω R V cV d (5.30) ds −3 − ab − ab − cd where ω ωab 0 and σ σab 0. This is the Raychaudhuri equation and is of great importance ab ≥ ab ≥ for the singularity theorem. Note that vorticity induces expansion which is in analogy with centrifugal force while shear induces cotraction.

We turn to the corresponding equation for the shear,

The trace-free, symmetric part

The trace-free part of a tensor Xab is

1 T = X h hcdX ab ab − 3 ab cd

The symmetric part of a tensor Tab is

1 1 (T + T ) T = X h hcdX . 2 ab ba ≡ (ab) (ab) − 3 ab cd

1 T = δc δd h hcd X . (ab) (a b) − 3 ab cd 

The trace-free, symmetric part of ( V c)( V )+ R V cV d is − ∇b ∇c a cbad

535 1 1 V h hcd V = ( V )( V c) h hef ( V )( V c) + R V cV d ∇(b a) − 3 ab ∇d c − ∇c (a ∇b) − 3 ab ∇c e ∇f cbad 1 h hef R V cV d −3 ab c(fe)d 1 = + R V cV d h (gef + V eV f )R V cV d · · · cbad − 3 ab c(fe)d 1 = + R V cV d h gef R V cV d · · · cbad − 3 ab cfed 1 = + R V cV d + h gef R V cV d · · · cbad 3 ab fced 1 = + R V cV d + h R V cV d (5.31) · · · cbad 3 ab cd

1 1 h V c θ = h V c (hef V ) −3 ab ∇c −3 ab ∇c ∇f e 1 = h hef V c ( V ) −3 ab ∇c ∇f e 1 1 = h hef V c( V V ) h hef [V c V ] −3 ab ∇c∇f e −∇f ∇c e − 3 ab ∇f ∇c e 1 1 = h hef R V cV d h hef [ (V c V ) ( V c)( V )] −3 ab edcf − 3 ab ∇f ∇c e − ∇f ∇c e 1 1 = h hef R V cV d + h hef ( V )( V c) (5.32) −3 ab c(fe)d 3 ab ∇c e ∇f

Therefore

1 1 V h hcd V = ( V )( V c)+ R V cV d h V c θ (5.33) ∇(b a) − 3 ab ∇d c − ∇c (a ∇b) cbad − 3 ab ∇c

536 Derivative of the shear

1 V c σ = V c V θh ∇c ab ∇c ∇(b a) − 3 ab 1 = V c ( V ) h V c θ ∇c ∇(b a) − 3 ab ∇c 1 = ( V )( V c)+ R V cV d h V c θ − ∇c (a ∇b) c(ba)d − 3 ab ∇c 1 1 1 = θh + σ + ω θhc + σc + ωc + a b −2 3 ac ac ac 3 b b b ↔ 1 +R V cV d h V c θ cbad − 3 ab ∇c 1 1 2 2 = θ2h + θσ + θω + σ σc + σ cω σc ω + ω ωc + a b −2 9 ab 3 ab 3 ab ac b a cb − b ca ac b ↔ 1 +R V cV d h V c θ cbad − 3 ab ∇c 1 2 = θ2h + θσ + σ σc + ω ωc − 9 ab 3 ab ac b ac b 1 1 +R V cV d h θ2 σ σcd + ωcdω R V cV d cbad − 3 ab −3 − cd cd − cd 2 1 1 = θσ σ σc ω ωc + h (σ σcd ωcdω )+ R V cV d + h R V cV d −3 ab − ac b − ac b 3 ab cd − cd cbad 3 ab cd 2 1 = θσ σ σc ω ωc + h (σ σcd ωcdω ) −3 ab − ac b − ac b 3 ab cd − cd 1 +R V cV d h hef R V cV d (5.34) cbad − 3 ab c(fe)d where we used (5.31).

c d Symmetric trace-free part of RcbadV V

c d The symmetric trace-free of RcbadV V is

1 1 R V cV d h hef R V cV d = δe δf h hef R V cV d. (5.35) cbad − 3 ab cefd (a b) − 3 ab cefd In the following we will need the following

c d c d RcabdV V = RdabcV V c d = RbcdaV V c d = RcbadV V (5.36)

c d c d c d This implies Rc(ba)dV V = RcbadV V and Rc[ba]dV V = 0. So that the RHS of (5.35) can be written,

537 1 δeδf h hef R V cV d. (5.37) a b − 3 ab cefd We use the decomposition of the Riemann tensor:

1 R = C g R g R Rg g . (5.38) cbad cbad − c[d a]b − b[a d]c − 3 c[a d]b

Notice it is anti-symmetric in a and d. Using that Ccbad has the same symmetries of Rcbad (in particular R = R ), means that the diﬀerence R C in the above equation is also cbad adcb cbad − cbad anti-symmetric in b and c.

We take (5.37) and decompose it as follows

1 δeδf h hef (C + [C R ])V cV d. (5.39) a b − 3 ab cefd cefd − cefd First note

1 1 δeδf h hef C V cV d = δeδf h (gef + V eV f ) C V cV d a b − 3 ab cefd a b − 3 ab cefd c d = CcbadV V (5.40) where we have used C V cV dV eV f 0 and that the Weyl tensor is trace-free in any two cefd ≡ indices.

Contracting both sides of (5.38) with V cV d we obtain

1 R V cV d = C V cV d g R V cV d g R V cV d Rg g V cV d. cbad cbad − c[d a]b − b[a d]c − 3 c[a d]b

b b a Using gabV = habV + V

1 g R V cV d = [g R g R ]V cV d − c[d a]b −2 cd ab − ca db 1 = [R + V R V d] 2 ab a bd 1 g R V cV d = [g R g R ]V cV d − b[a d]c −2 ab cd − bd ac 1 = [ g R V cV d + V R V c] 2 − ab cd b ac 1 1 1 Rg g V cV d = R[g g g g ]V cV d −3 c[a d]b −2 · 3 ca db − cd ab 1 1 = R[V V + g ] −2 · 3 a b ab 538 Bringing these results together we obtain

1 (R C )V cV d = g R V cV d g R V cV d Rg g V cV d cbad − cbad − c[d a]b − b[a d]c − 3 c[a d]b 1 1 1 1 = R g R V cV d + V R V d Rh (5.41) 2 ab − 2 ab cd (a b)d − 2 · 3 ab

Using δb = h b V V b and the anti-symmetry in d and f, and the anti-symmetry in c and e of a a − a R C , cefd − cefd

1 1 δf δe h hef R V cV d = C V cV d + (h f V V f )(h e V V e) h hef a b − 3 ab cefd cbad a − a b − b − 3 ab (R C ) V cV d cefd − cefd 1 = C V cV d + h f h e h hef (R C ) V cV d cbad a b − 3 ab cefd − cefd (5.42)

Using (5.41) in the above gives

1 1 δf δe h hef R V cV d = C V cV d + h f h e h hef a b − 3 ab cefd cbad a b − 3 ab 1 1 1 1 R g R V cV d + V R V d Rh 2 fe − 2 fe cd (f e)d − 2 · 3 fe 1 1 1 = C V cV d + h eh f R h hef R + cbad 2 a b ef − 2 · 3 ab ef 1 1 h f h eg h hef g R V cV d − 2 a b fe − 3 ab fe cd 1 + h f h eV R V d h hef V R V d a b (f e)d −3 ab (f e)d =0 =0 1 1 1 1 1 | h f{zh eRh }+ | h{zhef Rh} . − 2 · 3 a b fe 2 · 3 · 3 ab fe (5.43)

It is easy to check

h f h = h h f h eh = h hef h = 3 hef g = hef (h V V ) = 3. a bf ab a b fe ab fe fe fe − f e Substituting this into (5.43) gives

1 1 1 δf δe h hef R V cV d = C V cV d + h ch dR h hcdR . (5.44) (a b) − 3 ab cefd cbad 2 a b cd − 3 ab cd 539 And so (5.34) becomes

dσ 2 1 ab = θσ σ σc ω ωc + h (σ σcd ωcdω ) ds −3 ab − ac b − ac b 3 ab cd − cd 1 1 +C V cV d + h h Rcd h h Rcd . (5.45) cbad 2 ac bd − 3 ab cd 

Since the Wely tensor is trace-free is does not appear in the equation in the expansion equation(5.30). however as the terms 2σ2 (= 2σabσ ) occurs on the right-hand side of the expansion equa- − − ab tion, the Weyl tensor induces convergence indirectly by inducing shear.

The Weyl tensor is the part of the Reiemann tensor not depending on the Ricci tensor. From the Einstein equations,

1 R g R + Λg = 8πT . ab − 2 ab ab ab we see that it is the Rici tensor, Rab, is determined locally by the matter distributionis. Thus the Weyl tensor is the part of the curvature which is not determined locally by the matter distribution.

Derivative of vorticity

And from the anti-symmetric part of (5.28) yields

V c ω = V c V ) ∇c ab ∇c ∇[b a] = ( V )( V c)+ R V cV d − ∇c [a ∇b] c[ba]d = ( V )( V c) − ∇c [a ∇b] 1 = [( V )( V c) a b] −2 ∇c a ∇b − ↔ 1 1 2 2 = θ2h + θσ + θω + σ σc + 2σc ω + ω ωc a b −2 9 ab 3 ab 3 ab ac b b ac ac b − ↔ 2 = θω 2σc ω (5.46) −3 ab − [b a]c and so

dω 2 ab = θω 2σc ω . (5.47) ds −3 ab − [b a]c

540 Conjugate points

A pair of points p,q γ if there is a jacobi ﬁeld ηa which is not identically zero but vanishes at ∈ both p and q. Or p and q are conjugate if an an an inﬁnitesimally nearby geodesic intersects γ at both p and q.

Equation of geodesic deviation

Let γ be a timelike geodesic with tangent ξa. Consider a congruence of timelike geodesics a a passing through p. It is convenient to introduce an orthonormal basis of spatial vectors e1, e2, a a µ a µ e3 orthogonal to ξ and parallely propagated along γ. The components, η := η ea , of the vector ηa which represents the separation between a geodesic γ(s) and a neighbouring geodesic then satisﬁes the Jacobi equation

d2ηµ = R µξαηβξν (5.48) dτ 2 − αβν α,β,νX

The value of η at time s depends linearily on the initial data η(0) and dηµ/ds(0) at p. Since, by construction, ηµ(0) = 0 for this congruence, we must have

3 dην η (s)= A (s) (0) (5.49) µ µν ds Xν=1

The matrix Aαβ

By (5.49) we have

3 d2ηµ d2 dην = Aµ (s) (0) ds2 ds2 ν ds ν=1 X = R µξαηβξσ − αβσ α,β,σX 3 dην = R µξαξσ Aβ (s) (0) (5.50) − αβσ ν ds ν=1 α,β,σX X 

µ so that A ν(s) satisﬁes the equation

d2Aµ ν (s)= R µξαξσAβ (τ) (5.51) ds2 − αβσ ν α,β,νX

541 With

3 dηµ dAµ dην (τ)= ν (τ) (0) (5.52) dτ dτ dτ ν=1 X clearly, we also have

dA A (0) = 0, µν (0) = δ . (5.53) µν dτ νµ

d By deﬁnition of dτ

d η = V η (5.54) dτ α α;β β or in terms of Aαβ

3 3 dA dην dην µν (τ) (0) = V A (τ) (0) (5.55) dτ dτ α;β βν dτ Xν=1 Xν=1 or

dA αβ (τ)= V A (τ) (5.56) dτ α;γ γβ

Assumming Aαβ is invertible we can write

1 d V (τ)= A− A (5.57) α;β γβ ds αγ

This allows the vorticity and the expansion tensor as well as the expansion scalar to be expressed 1 in terms of the matrix Aαβ and it’s inverse A−αβ:

1 d ω = A− A (5.58) αβ − γ[α ds β]γ

1 d θ = A− A (5.59) αβ γ(α ds β)γ

dA 1 θ = trB = tr A− (5.60) dτ 

542 (so that (det A)3 measures a spacelike 3-volume element orthogonal to γ and transported along the curves of the congruence)? proof? consider

d dA ∂ det A = ij det A (5.61) dτ dτ ∂Aij

Recall from (C.5.9)

∂ 1 det A = A˜ij = det A(A− )ji (5.62) ∂Aij where A˜ij are the cofactors of A. Together we get

d dAij 1 det A = det A (A− ) (5.63) dτ dτ ji 

So (5.91) can be written

d θ = (ln detA ) (5.64) dτ | |

1 σ = h ch d η θh (5.65) ab a b ∇(d c) − 3 ab

Consider the quantity

1 d d A ω A = ( A )A A A (5.66) αδ αγ αδ 2 dt αβ αδ − αβ dt αδ 

It is easy to verify that this quantity is constant along γ(t) by diﬀerating both sides and using (5.51),

d 1 d d d (A ω A ) = A A A A ds αδ αγ αδ 2 ds γα ds γβ − γβ ds γα 1 d2 d2 = A A A A 2 γα ds2 γβ − γβ ds2 γα 1 = (A R A A R A ) 2 γα γ4δ4 δβ − γβ γ4δ4 δα 1 = (A R A A R A ) 2 γα γ4δ4 δβ − γβ δ4γ4 δα = 0 (5.67)

543 where we used Rγ4δ4 = Rδ4γ4. The RHS of (5.66) vanishes where Aαβ is zero and so must be zero all along γ(t),

AαδωαγAαδ = 0. (5.68)

When Aαβ is invertable, we can apply the inverse matrix of A to both sides and so conclude that ωαβ = 0 when Aαβ is non-singular.

d2 Zα = R Zβ (5.69) ds2 − α4γ4

d2 A = R A (5.70) ds2 αβ − α4γ4 γβ

1 The expansion θ is defined as (detA)− d(detA)/ds. Since Aαβ obeys the ordinary differential equation (), d(detA)/ds will be finite.

Proposition 5.1.43 (The focussing theorem) Take any inextendable geodesic γ(λ). If at a b some point γ(λ1) (λ1 > 0), the expansion has a negative value θ1 < 0 and if RabV V everwhere positive then there will be a point conjugate to q along γ(λ) between γ(λ ) and γ(λ + (3/ θ )). 1 1 − 1

The Raychaudhuir equation for non-rotating ﬂow implies

1 θ˙ θ2. (5.71) ≤−3

Writing it in the form θ 2θ˙ (1/3) − ≤−

λ2 θ(λ2) 2 2 1 θ(λ2) 1 dλθ− θ˙ = dθθ− = θ− (λ2 λ1) (5.72) − θ(λ1) ≤−3 − Zλ1 Zθ(λ1) rearranging gives:

1 1 (λ2 λ1) − θ(λ ) θ(λ )− − (5.73) − 2 ≥ − 1 − 3 From this inequality we see that if the congruence is initially converging (θ(0) < 0), then θ(λ) → within an aﬃne parameter λ 3/ θ(0) . This demonstrates the irreversible tendency −∞ ≤ | | of neighbouring geodesics to converge. Of course, this represents merely a singularity in the congruence, not a singularity in the structure of spacetime itself. However, this result is key to the singularity theorems, as the above may not be possible if spacetime is geodesically incomplete, and as already stated, this incompleteness is interpreted as evidence of a singularity.

544 The time-reversed version of the argument holds (s s, V a V a, and ω ω = 0) of this →− →− →− theorem where if θ(0) > 0 at a point then there was must have been a point an affine parameter λ 3/ θ(0) distance away in the past where θ( λ) ≤ | | − → −∞ The Weyl tensr expresses the tidal force that a body experiences when moving along a geodesic. The Weyl tensor differs from the Riemann curvature tensor in that it does not contains information on how the volume of the body changes, but rather only on how the shape of the body is distorted by the tidal force. The Ricci curvature contains precisely the information about how the volume changes in the presence of tidal forces. The Weyl curvature is the only part of the curvature that exists in free space. The effect of the Ricci tensor are due to a matter distribution and has a positive focussing effect.. attractive instability of gravity. proposition (5.1.43), all that is required for the existence of conjugate points γ is that R ξaξb 0 ab ≥ everywhere on γ and R ξbξd = 0 at least one point of γ. We prove this in section 5.1.11 on abcd 6 the equivalence of the two versions of the singularity theorem.

∂τ b ∂ (α) = f(α) ∂α a ∂α Z b = Xa ( T aT )1/2 ∇a − a Za b 1 = T Xa T bdt − f b ∇a Za b 1 a b = TbT aX dt − a f ∇ Z b b a b b a = T a(X Tb/f)dt + X T a(Tb/f)dt − a ∇ a ∇ Zb Z = XbT a (T /f)dt (5.74) ∇a b Za

d2τ b = Xc [XbT a (T /f)]dt (5.75) dα2 ∇c ∇a b Za

545 d2τ b = Xc [XbT a (T /f)]dt dα2 ∇c ∇a b α=0 Za b b c a c b = X (X cT ) a(Tb/f)dt + weused X cX = 0, a ∇ ∇ ∇ Z b a c + X T X c a(Tb/f)dt b ∇ ∇ = Xb(T c Xa) (T /f)dt + weused T a Xa = Xa T a ∇c ∇a b ∇a ∇a Za b b b a c b a c d + X T X a c(Tb/f)dt + X T X Rcab Td/fdt a ∇ ∇ a Zb Z b = XbT c [Xa (T /f)]dt + XbT aXcR dT /fdt (5.76) ∇c ∇a b cab d Za Za where we used

(T /f)= (T /f)+ R d(T /f) (5.77) ∇c∇a b ∇a∇c b cab b choosing the curve parameterization so that f = 1 along the geodesic γ0 and also choosing the a deviation vector to be orthogonal to T along γ0.

d2τ b = Xc T c (T a X )+ R dT aT Xc dt dα2 { ∇c ∇a b cab d } α=0 Za b b = X ( )bdt (5.78) O Za

(5.78)

µ 1 µ ν the matrix A is non-singular between p and q, Y = ν(A− ) νX . P Conjugate points for Timelike geodesics normal to spacelike three-surface

A point p on a geodesic γ of the geodesic congruence orthogonal to Σ is said to be conjugate to Σ along γ if there exists an orthogonal deviation vector ηa of the congruence which is nonzero on Σ but vanishes at p. Intuitively, p is conjugate to Σ if two “inﬁntesimally nearby” geodesics orthogonal to Σ cross p as in ﬁg ().

The three dimensional manifold Σ is deﬁned locally by

f = 0

546 p

T Σ

Figure 5.54: Displacement vectors for a hypersurface.

ab where f is second diﬀerentiable and g ∂af∂bf < 0. The unit normal vector N to the surface Σ is

gad f N a = ∇d ( gbc f f)1/2 − ∇b ∇c We deﬁne the second fundamental form χ of Σ as

χ = h ch d N (5.79) ab a b ∇d c where as before

hab = gab + NaNb is the induced metric tensor on Σ (also called the ﬁrst fundamental form). It follows that χab is symmetric,

χ = h ch d N ab a b ∇d c ∂ f 1 = h ch d ∇d c ∂ f a b f − c ∇d f k k k k h ch d(∂ ∂ f Γe ∂ f) 1 = a b d c − dc e (h c∂ f)h d f − a c b ∇d f k k k k = h ch d N a b ∇c d = χba (5.80)

e e c where we used Γdc =Γcd and ha ∂cf = 0. The congruence of timelike geodesics orthonormal to Σ will consist of the timelike geodesics whose unit tangent vector V is the same as the unit normal N at Σ. Then

547 V = χ at Σ. (5.81) ∇b a ab

When the congruence of geodesics are normal to the hypersurface the vorticity vanishes. Proof: The unit normal vector to the hypersurface is

f V = ∇a a f k k We calculate the vorticity,

ω = h ch d V ab a b ∇[d c] h ch d f f = a b ∇c ∇d 2 ∇d f −∇c f k k k k h ch d f f 1 1 = a b ∇d∇c −∇c∇d + f f 2 f ∇c ∇d f −∇d ∇c f k k k k k k = 0 (5.82) since f f = 0 and h c f = 0. ∇c∇d −∇d∇c a ∇c For the vector η which represents the separtion of a neighbouring geodesic normal to Σ from a geodesic γ(s) normal to Σ, we again have the equation of geodesic deviation for the separating vector,

d2 η (s)= R η (s). ds2 α − α4γ4 γ

Also

d η (s)= V η . ds α ∇β α β

At a point on γ(s) at Σ we have the initial condition

d η = χ η (5.83) ds α αβ β

We express the Jacobi ﬁelds along γ(s) which satisfy the above condition as

3 µ µ ν η (s)= A ν(s)η (0) (5.84) Xν=1

548 µ This A ν(s) satisﬁes the same diﬀerential equation

So that

3 dηµ dAµ (τ)= ν (τ)ην (0) (5.85) dτ dτ ν=1 X by

dηα (τ)= V ηβ (5.86) dτ ν;β or in terms of Aαβ(τ)

3 dA 3 αβ (s)ηβ(0) = V Aγ (s)ηβ(0) (5.87) ds α;γ β ν=1 ν=1 X X or

dA αβ (s)= V A (s) (5.88) ds α;γ γβ which is of the same form as (5.56). This allows the vorticity and the expansion tensor as well 1 as the expansion scalar to be expressed in terms of the matrix Aαβ and it’s inverse A−αβ in the same way as before where there were diﬀerent boundary conditions for Aαβ:

1 d ω = A− A (5.89) αβ − γ[α ds β]γ

1 d θ = A− A (5.90) αβ γ(α ds β)γ

dA 1 θ = trB = tr A− (5.91) dτ Again the quantity

1 d d A ω A = ( A )A A A (5.92) αδ αγ αδ 2 dt αβ αδ − αβ dt αδ is constant.

Thus on the hypersurface we have the boundary conditions

549 dA dA A (0) = δ , µν (0) = νµ (0). (5.93) µν νµ dτ dτ

Proposition 5.1.44 Let be a spacetime satisfying R ξaξb 0 for all timelike ξa. Let Σ be M ab ≥ a spacelike hypersurface with K = θ < 0 at a point q Σ. Then within a proper time τ 3/ K ∈ ≤ | | there exists a point p conjugate to Σ along the geodesic γ orthogonal to Σ and passing through q, assuming that γ can be extended that far.

This is proved using the Raychaudhuri equation (5.30) as in proposition 5.1.43. If the congruence is hypersurface orthogonal, we have ωab = 0, so that the third term vanishes. The second term, σ σab, is manifestly nonpositive. Thus we have − ab

dθ 1 + θ3 0, (5.94) ds 3 ≤ as before, which leads to

1 1 1 − θ(s) θ− (0) s . (5.95) − ≥ − − 3 We are supposing that θ(0) < 0, i.e. the congruence is initialy converging.

Again the time reverse version holds, important in proof of simple singularity theorem on co- mological singulairty presented in the previous appendix. p s q r

Figure 5.55: .

Conversely

ωab = 0

550 implies

V = f a ∇a

This is guaranteed by the Frobenius theorm.

The boundary conditions (5.93) will also be encountered in proposition 5.1.53.

5.1.10 The Singularity Theorems

I+[S]

E+[S]

Figure 5.56: Future horozmos .

I+[S]

E+[S]

Figure 5.57: “doubling” future horozmos. E+[S] is compact

Corollary 1: Let F be a future set, γ a null geodesic on ∂F . Then γ contains no proper segment (??a segment not γ itself or an empty set??) which has a pair of conjugate points.

Proof: If x

Corollary 2: Let be a spacetime satisfying: M (1) There are no closed trips

(2) Every endless null geodesic in contains a pair of conjugate points. M

551 Then is strongly causal. M Proof:

If not, and there are no closed trips, strong causality fails on an endless geodesic γ. By hypothesis, there are conjugate points A and B on γ with a < A, b>B. Then strong causality fails at a and b, and by the theorem, I+(a) I (b) = and there are closed trips (see remark following ∩ − 6 ∅ 5.1.10), which is false.

Proposition 5.1.45 Let S be future trapped and suppose strong causality holds in I+[S]. Then there is a future endless trip γ such that γ intD+(E+[S]). ⊂

Proof: First we nedd to show H+[E+[S]], if it exists, must be non-compact. The basic idea of the argument is given in ﬁg (5.1.10)

Figure 5.58: From the fact that H is a Cauchy horizon it follows that through every point of H there passes a maximally extened past-directed null geodesic that remains in H. Since H is compact, such a curve would have to come back arbitrarily close to itself - reentering some Alexandoﬀ neighbourhood and so violating strong causality.

Let H = H+[E+[S]]. Any trip which leaves intD+(E+[S]) crosses H. If H = , then the ∅ proposition is proven. If not, then H must be non-compact. For, if H is compact, there are ﬁnitely many Alexandoﬀ neighbourhoods covering H, say B1,...,Bk. There exists a point I+[S]/D+(E+[S]); suppose p B . Then there is a point q I+[S]/D+(E+[S]) such that ∈ i i ∈ q << p and q B B for some i . The process continues by induction to yeild p >> i1 i1 ∈ i1 − i 1 q >> q >> >> q >> . However, when the process has been repeated k times two i1 i2 · · · ik · · · (at least) of the q’s must lie in the same Bi1 . By consrtuction, the trip between these two must + leave and then reenter Bi1 contradicting the fact that I [S] is strongly causal.

So H is non-compact. Let ξ be any nowhere vanishing timelike vector field on . Given M a non-zero vector field ξ defined over the manifold, this can be used to define a congruence of curves over the manifold (a congruence of curves is defined such that only one curve goes through each point of the manifold). We use ξ to construct the family of trajectories from E+[S]. If one of these trajectories leaves intD+[E+[S]], it intersects H. This mapping will have to be one-to-one as no point in this boundary can procede itself. If all these trajectories leave, then a homeomorphism is established between E+[S] and H, which is impossible since E+[S] is compact, and H is not. So one trajectory remains inside.

552 + + H (E [S]) H+(E+[S])

ξ(x)

E+[S]

Figure 5.59: E+[S] is compact by deﬁntion, however its Cauchy horizon is non-compact. As the two subsets can not homeomorphic there must be at least one trajectory γ which remains in intD+(E+[S]).

Theorem 5.1.46 The following are mutually inconsistent in any spacetime:

(a) There are no closed trips

(b) Every endless causal geodesic contains a pair of conjugate points

(c) There is a future (past) - trapped set in . S M

Proof: Under the assumptions that all three conditions are satisfied, (a) and (b) imply that M is strongly causal. This implies the existence of a future endless geodesic in intD+(E+(S)) say + γ. Define the compact set T = I−(γ) E (S). We argue that T is past-trapped. Since γ was + + ∩ contained in intI (E (S))), E−(T ) would consist of T and a portion of J˙−(γ), see fig (5.1.10). Since γ was future inextendible, the null geodesic segments generating J˙−(γ) could have no future endpoints. But by (b) every inextendible non-spacelike geodesic contains a pair of conjugate + + points. Thus each generating segment of J˙−(γ) would enter I−(γ). However, as intD (E [S]) is globally hyperbolic, within this region on the null geodesic segments generating J˙−(γ) there cannot occur a pair of conjugate points and so they intersect T . E−(T ) is be compact being the intersection of the closed set J˙−(γ) with a compact set generated by null geodesic segments from T of some bounded affine length. Then, by the time-reversed version of proposition 5.1.45 there exists α, a past endless causal geodesic in intD−(E−(T )).

Choose a sequence ai receding into the past on α without limit point and a sequence ci on γ going to the future. The sets J (c ) J +(a ) are compact, for all i, and so there is a maximal − i ∩ i geodesic µ in (a , c ) for each i. The intersection with T (as it is compact) have a limit point i C i i q and a limiting direction. Construct the causal geodesic µ which has this direction at q. some nearby geodesic must meet γ more than once. But as γ is the limit curve of the sequence γ , so the nearby geodesic will meet at least one member γ of the sequence γ . This would { i} k { i} 553 H H

H H

T E−[T ] past end E−[T ] past end point point

Figure 5.60: γ is a future endless causal geodesic in intD+(E+(S)). H = H+(E+[S]). As the intersection of the closed set J˙−(γ) with a compact set generated by null geodesic segments from T of some bounded aﬃne length, E−[T ] is compact

mean we could lengthen that γk by rounding oﬀ corners, which contradicts that γk is the curve of maximum length from bk to ak.

5.1.11 Implication of the “Displayed” Singularity Theorem from the Established Version

We now establish theorem 5.1.1 from theorem 5.1.2. move on to establish between

Theorem 5.1.47 if were timelike and null geodesically complete, (1) and (2) would imply M (a) by 5.1.48. (3) is the same as (b). (1) and (4) would imply (c): in case (i) would be the compact achronal set without edge and L

+ E ( )= E−( )= (5.96) L L L

Refering to the two versions given in 5.1.1

Proof: To be given.

554 λ

Figure 5.61: The geodesic λ.

Theorem 5.1.48 If for every non-spacelike vector K we have

a b RabK K < 0 (5.97) and

(2) Every non-spacelike geodesic, with tangent vector K, contains a point at which

K R K KcKd = 0. (5.98) [a b]cd[e f] 6 then every inextendible non-spacelike geodesic will contain a pair of conjugate points.

Proof: To be given.

We say that a spacetime satisfying the condition that every timelike or null geodesic contains a c d point at which K[aRb]cd[eKf]K K is not zero satisﬁes the generic condition.

Before we give the proof of this theorem we discuss whether it is reasonable to assume the generic condition in physically realistic solutions. First note that for the timelike case the condition is V R V V cV d = 0 equivalent to R V bV c = 0: [a b]cd[e f] 6 abcd 6

V R V V cV d = 0 (5.99) [a b]cd[e f] 6

Contracting this with V aV f , using R = R = R and V 2 = 1 this becomes abcd − bacd − abdc −

555 ak

a3 γk a Limiting 2 Limiting geodesicγ geodesicγ γ a1 2 conjugate γ1 points ofγ

b2 b nearby 3 geodesic

bk (a) (b)

Figure 5.62: The limit geodesic γ contains conjugate points.

R V bV c = 0. (5.100) abcd 6

Multiplying R V cV d =0 by V V and antisymmetrizing between a and b, and between e and bcde 6 a f f gives us back Eq.(5.99), so the two conditions Eq.(5.99) and Eq.(5.100) are equivalent 3.

One would expect every timelike geodesic to encounter some mater or some gravitational radia- b d tion and so contain some point where the tidal force is non-zero, i.e. a point where RabcdV V is non-zero. Thus one expect every timelike geodesic to contain pairs of conjugate points, provided that it could be extended suﬃciently far in both directions.

As in the timelike case, this condition will be satisﬁed for a null geodesic which passes through some matter provided that the matter is not pure radiation () and moving in the direction of the geodesic tangent vector K. It will be satisﬁed in empty space if the null geodesic contains some point where the Weyl tensor is non-zero and where K does not lie in one of the directions c d at the point for which K[aRb]cd[eKf]K K = 0.

We break it down proof of 5.1.48 down into timelike then null case. First timelike case, in which it reads: If R V aV b 0 and if at some point p = γ(s ) the tidal force R V bV d is non-zero, ab ≥ 1 abcd there will be values s0 and s2 such that q = γ(s0) and r = γ(s2) will be conjugate along γ, providing that γ(s) can be extended to these values.

3How do we know that the righthand side of Eq.(5.99), when contracted with V aV f is not zero? Well, b c c d if RabcdV V = 0 this would mean that that V[aRb]cd[eVf]V V = 0 which is false.

556 Mathematical preliminaries

Some mathematical preliminaries are needed in order to make lemma 5.1.52 more rigorous, although it is written so that it can be ’understood’ without going into all such details.

Deﬁnition A set B is bounded if there is a point p in S and a radius r > 0 such that B is a subset of the neigbourhood Nr(p).

Deﬁnition A function f from S to T is said to be bounded if the image set f(S)= f(p) : p { ∈ S is a bounded subset of T . }

Proposition 5.1.49 Suppose that A is a compact subset of S and that f is a continuous function from A to T . Then the image f(A)= f(p) : p A is a compact subset of T . In particular, { ∈ } f is bounded.

Proof:

Suppose is an open cover of f(A). Let be the collection of subsets of S consisting of the V U inverse images of the sets in the collection . As f is a continuous function the inverse image of V an open set is open in A. Given a point p in A, f(p) belongs to some V , so p belongs to the ∈V set f 1(V ) from the collection . Thus is an open cover of A. Let U ,...,U be a ﬁnite − U U { 1 n} subcover. Then each U is f 1(V ) for some V in the collection , and the collection V is a j − j j V { j} ﬁnite subcover of f(A).

The properties of a norm are:

f 0, f = 0 f 0 k k ≥ k k ⇐⇒ ≡ cf = c f , c R k k | | · k k ∈ f + g f + g . (5.101) k | ≤ k k k k

Any norm deﬁnes a metric via

d(f, g)= f g 0, d(f, g)= f g = 0 f g k − k ≥ k − k ⇐⇒ ≡ d(f, g) = f g = g f = d(g,f) k − k k − k d(f, h)= f h = (f g) + (g h) k − k k − − | (f g) + (g h) ≤ k − k k − k = d(f, g)+ d(g, h). (5.102)

Thus everything proved for metric spaces carries over to the case of normed linear spaces.

557 Deﬁnition Supremum (sup) or least upper bound. If S is a nonempty set of real numbers, then a real number u is an upper bound of S if x u or every x in S. A number v is aleast ≤ upper bound of S if v is an upper bound and no number less than v is an upper bound of S. Thus, the least upper bound is the smallest real number that is greater than or equal to every number in S.

The result concerning uniform continuity pertains to metric spaces. the standard metric in the space of continuous functions is obtained by starting with the norm

f = sup f(x) : x X (5.103) k k {| | ∈ }

We check the that the norm f has the properties that deﬁne a norm: k k

sup f(x) : x X 0, sup f(x) : x X = 0 f 0 {| | ∈ } ≥ {| | ∈ } ⇐⇒ ≡ sup cf(x) : x X = c sup f(x) : x X , c R {| | ∈ } | | · {| | ∈ } ∈ sup f(x)+ g(x) : x X sup f(x) : x X + sup g(x) : x X . (5.104) {| | ∈ } ≤ {| | ∈ } {| | ∈ }

We can deﬁne a distance

d(f, g)= f g = sup f(x) g(x) : x X (5.105) k − k {| − | ∈ } for arbitrary f, g N. The distance between f and g is the maximum distance between their ∈ graphs. The fact that (5.105) it is a metric follows at once from the properties (5.101):

We can deﬁne diﬀerent norms for matrices.

This is well deﬁned and ﬁnite because the image of a compact space under a continuous function is compact, in particular bounded.

Lemma 5.1.50 Let X and Y be metric spaces and f : X Y a continuous function. If X is → compact, then f is uniformally continuous.

Proof:

To prove the lemma we start with a number ǫ> 0 and ﬁnd a number δ > 0 such that if p,q X ∈ are any points such that dX (p,q) < δ then dY (f(p),f(q)) <ǫ.

First for each p X we ﬁnd a number δ(p) such that if q X and ∈ ∈

dX (p,q) < δ(p) then dY (f(p),f(q)) < ǫ/2; (5.106)

558 this is always possible since f is continuous at p. Let B(p) be the open ball in X with center p and radius δ(p)/2. The open balls B(p) with p ranging over all points of X provides an open cover for X. Since X is compact, there is a finite sub-cover. Thus there exists a finite number of points of X, say p1,p2,...,pn, such that X is the union of their open balls. Now define

δ = min δ(p )/2, δ(p )/2,...,δ(p )/2 { 1 2 n }

This will be the δ that satisﬁes our original demand. For suppose that p,q X, with ∈

dX (p,q) < δ. (5.107)

For some i we have p B(p ), so that ∈ i

dX (pi,p) < δ(pi)/2.

Also

d (p ,q) d (p ,p)+ d (p,q) X i ≤ X i X < δ(pi)/2+ δ δ(p ). ≤ i

Thus

d (p ,p), d (p ,q) δ(p ) X i X i ≤ i

Now from (5.106) this means

dY f(pi),f(p) < ǫ/2 and dY f(pi),f(q) < ǫ/2. Therefore

dY (f(p),f(q)) dY f(p),f(pi) + fY f(pi),f(q) ≤ ǫ ǫ < + = ǫ (5.108) 2 2 whenever (5.107) is satisﬁed.

559 We prove some facts about the differentiability, smoothness and continuity of Taylor series. Functions that have derivatives of all orders are called smooth. If a function is differentiable at a point, then it must also be continuous there (the converse isn’t always true, for example the function y = x is continuous at x = 0 but not differentiable there). | | Before we move on to this question, let us remind the ourselves of one method used t determined if a series expansion to be absolutely convergent, namely the ratio test for absolute convergence: let an be a positive-term series, and suppose P a lim n+1 = L< 1. n →∞ an

If L< 1 then the series is convergent. The proof of this is as follows: suppose limn (an+1/an)= →∞ L< 1. Let r be any number such that 0 L

an+1 < r or an+1 < anr. an

Substituting N, N + 1, N + 2, . . . for n, we obtain

aN+1 < aN r 2 aN+2 < aN+1r < aN r 3 aN+3 < aN+2r < aN r (5.109) and in general

m aN+m < aN r whenever m> 0

Recall the basic comparison test; given two positive-terms series an and bn, if bn converges and a < b for every positive integer n greater than some N, then obviously the series n n P P P an converges. It follows from the comparison test that the seies P a + a + a + . . . N+1 N+2 · · · N+m converges, sinse its terms are less than the correspondig terms of the convergent geometric series

a r + a r2 + a r3 + .... N N · · · N

Since the convergence or divergence is unaﬀected by discarding a ﬁnite number of terms, the series n∞=0 an. P

560 Lemma 5.1.51 If the series

∞ n anx (5.110) n=0 X is absolutely convergent and has radius of convergence R > 0, then the associated function n f(x)= n∞=0 anx is diﬀerentiable at every point of the range of convergence and P

∞ n 1 2 f ′(x)= nanx − = a1 + 2a2x + 3a3x + . . . (5.111) nX=0 Moreover, it will be smooth.

Proof:

First we prove that if the series (5.110) has a radius of convergence R then so does

∞ n ∞ n (n + 1)an+1x and (n + 2)(n + 1)an+2x (5.112) nX=0 Xn=0 The second of these have the same relation to the first as the first does to (5.110), so we need only n consider the first. Multiplicatin by x gives the series n∞=0 nanx , which therefore converges for n the same values of x and has the same radius of convergence ∞ (n + 1)a x itself. P n=0 n+1 Note P

n 1 n 2 n 1 n n (y x)(y − + y − x + + x − )= y x − · · · −

n n y x n 1 n 1 n 2 n 1 n 1 a − nx − = a (y − + y − x + + x − nx − ) n y x − n · · · − − n 1 n 1 n 2 n 2 n 2 = an[(y − x − ) + (y − x − )x + + (y x)x − ] − n 1 n 1 − n 2 n ·2 · · − y − x − y − x − y x n 2 = a (y x) − + − x + + − x − n − y x y x · · · y x − − − (5.113)

Since x < r and y < r, | | | |

k k y x k 1 k 2 k 1 k 1 − = y − + y − x + + x − < kr − . (5.114) y x | · · · |

−

Using (5.114) in (5.113) we get

561 n n y x n 1 n 2 n 3 n 3 n 2 a − nx − < x y a [(n 1)r − + (n 2)r − r + + 2rr − + r − ] n y x − | − | | n| − − · · · − n 2 = x y an r − [(n 1) + (n 2) + + 1] | − | | | − − · · · 1 n 2 = n(n 1) x y a r − (5.115) 2 − | − | | n|

The power series with coeﬃcients a has the same radius of convergence as that with a . | n| n

∞ n 2 ∞ n n(n 1) a r − = (n + 2)(n + 1) a r − | n| | n+2| Xn=2 nX=0 = K < + (5.116) ∞ we get

f(y) f(x) ∞ n 1 1 − nanx − < K y x (5.117) y x − 2 | − | − n=1 X proving (5.111).

Theorem for timelike geodesics from a point

Lemma 5.1.52 For any s∗ > 0, there exists C > 0 such that if max A′ (0) C, A(0) = I3 3, | αβ |≥ × T r(A (0)) 0, A (0) symmetric, ′ ≤ ′

d2A αβ + A R = 0 (5.118) ds2 αγ γ4β4 then det(A)(s ) = 0 for some s [0,s ]. 1 1 ∈ ∗

Proof:

As (5.118) is a linear ODE, the above condition can be restated as follows: for any s∗ > 0, there exists σ > 0 such that if max A = 1, A(0) = σ I where 0 < σ < σ, tr(A (0)) 0, A (0) | αβ′ | A A ′ ≤ ′ symmetric,

d2A αβ + A R = 0, ds2 αγ γ4β4

562 then det(A)(s ) = 0 for some s [0,s ]. The equivalence of the two statements is established 1 1 ∈ ∗ as follows: we have

max A′ (0) = C˜ where C˜ C. | αβ | ≥

Set

˜ ˜ Aαβ′ (0) := Aαβ′ (0)/C then

A˜αβ(0) = δαβ/C˜

Given

1 1 C˜ ≤ C

Set

1 1 σ ˜ = and σ = , σ ˜ σ. A C˜ C A ≤

We then have max A˜ (0) = 1, A˜(0) = σ where 0 < σ < σ and obviously, trA˜ (0) 0 and | αβ′ | A˜ A˜ ′ ≤ A˜′(0) is symmetric. Given that (5.118) is a linear ODE we have

d2A˜ αβ + A˜ R = 0 ds2 αγ γ4β4 and the equivalence is established.

Now we turn to the proof of the second form of the lemma. Obviously, at s = 0 we have that the eigenvalues of Aαβ are all equal to σA > 0. To prove the lemma we shall show that there is a negative eigenvalue of A for some s [0,s ], as this implies an eigenvalue must become αβ 0 ∈ ∗ zero at some point s [0,s ] and hence det(A)(s ) = 0 for some s [0,s ]. Let us proceed. 1 ∈ 0 1 1 ∈ ∗

Write Aαβ(s) as a ﬁrst order Taylor polynomial plus ’remainder’,

Aαβ(s)= Aαβ(0) + Aαβ′ (0)s + rαβ(s)

We can ﬁnd an expression for the remainder in terms of the value of Aαβ′′ for some value of s, sαβ, in the interval [0,s]. To this end, if t is any number in the interval [0,s], let Hαβ be the function deﬁned by

563 (s t)2 H (t) := A (s) A (t)+ A′ (t)(s t) r (s) − αβ αβ − αβ αβ − − αβ s2 If we diﬀerentiate both sides of this equation with respect to t, regarding s as a constant, we obtain

(s t) H′ (t)= A′′ (t)(s t)+ r (s) 2 − αβ − αβ − αβ · s2

By the deﬁntion of Hαβ we easily see that Hαβ(s) = 0. We also note that

s2 H (0) = A (s) [A (0) + A′ (0)s] r (s) αβ αβ − αβ αβ − αβ s2 = A (s) [A (0) + A′ (0)s + r (s)] αβ − αβ αβ αβ = 0.

Hence there is a number sαβ between 0 and s such that Hαβ′ (sαβ) = 0; that is

(s sαβ) A′′ (s )(s s )+ r (s) 2 − = 0. − αβ αβ − αβ αβ · s2 giving

A′′ (sαβ) r (s)= αβ s2. αβ 2! where s is some function of s and where 0 s s. Using this we obtain αβ ≤ αβ ≤

s2 A (s)= σ δ + sA′ (0) + A′′ (s (s)), (5.119) αβ A αβ αβ 2 αβ αβ which is what we wished to prove. Next we want to show that A (s) on [0,s ] is bounded by | αβ | ∗ a constant which is independent of the boundary conditions specified in the lemma. To this end we note that the differential equation (5.118) is equivalent to a system of first order differential equations which can be written in matrix form as

d Aαβ(s) 0 δαγ Aγβ(s) dAαβ = dAγβ (5.120) ds (s) R4α4γ (s) 0 (s) ds ! − ds ! the solution of which is

564 Aαβ(s) 0 sδαγ Aγβ(0) dAαβ = exp s dAγβ . (5.121) (s) R4α4γ (t)dt 0 (0) ds ! − 0 ds ! R We see from this is that any solution for Aαβ(s) will be of the general form

d A (s)= h (s)A (0) + g (s) A (0) (5.122) αβ αγ γβ αγ ds γβ where the functions hαγ (s) and gαγ (s) are solely determined by the matrix

0 sδ exp αγ (5.123) s R (t)dt 0 − 0 4α4γ R and as such are independent of the intial conditions placed on A . Now are h (s) and g (s) αβ | αγ | | αγ | bounded?

Can we use the same kind proof involved in the ratio test but for a series where the ‘coeﬃcients’ are matrices instead of numbers? Put

s Q (s) := R (t)/s αγ − 4α4γ Z0 and let us expand (5.123):

0 δ s exp αγ Q s 0 αγ δ 0 0 δ s2 Q 0 s3 0 Q = αγ + s αγ + αγ + αγ 0 δ Q 0 2! 0 Q 3! Q2 0 αγ αγ αγ αγ s4 Q2 0 s5 0 Q2 s6 Q3 0 + αγ + αγ + αγ + . . . 4! 0 Q2 5! Q3 0 6! 0 Q3 αγ αγ αγ hαβ corresponds to the top left hand quadrant:

s2 s4 s6 s2r h (s)= δ + Q + (Q2) + (Q3) + + (Qr) + . . . (5.124) αγ αγ 2! αγ 4! αγ 6! αγ · · · (2r)! αγ

3 where we deﬁne (Q )αγ := δαγ . And gαβ corresponds to the top right hand quadrant:

s3 s5 s7 s2r+1 g (s)= sδ + Q + (Q2) + (Q3) + + (Qr) + . . . (5.125) αγ αγ 3! αγ 5! αγ 7! αγ · · · (2r + 1)! αγ

Consider

565 s2 s4 s6 h (s) δ + Q + (Q2) + (Q3) + (5.126) | αγ |≤ αγ 2! | αγ | 4! | αγ | 6! | αγ |

Deﬁne q as

q := max Q for all α, γ { | αγ | } and deﬁne the matrix D to be the 3 3 matrix with all entries equal to 1. Then obviously ×

n n n 1 n (Q ) (qD) = 3 − q D | αγ |≤ we then have

2 4 6 2r s s 2 s 2 3 s r 1 r h (s) δ + qD + 3q D + 3 q D + + 3 − q D + . . . (5.127) | αγ |≤ αγ 2! αγ 4! αγ 6! αγ · · · (2r)! αγ

We are now in a position to apply the ratio test

s2(n+1)3nqn+1 (2n)! s2 3q lim 2n n 1 n = lim · = 0. (5.128) n (2(n + 2))! · s 3 − q n (2n + 2)(2n + 1) →∞ →∞

Therefore the series on the right side of the inequality (5.1 27) converges and hence the series expansion for h (s) converges absolutely and is h (s) is bounded. Similarly for g (s) we | αβ | | αβ | | αβ | write

3 5 7 2r+1 s s 2 s 2 3 s r 1 r g (s) sδ + qD + 3q D + 3 q D + + 3 − q D + . . . (5.129) | αγ |≤ αγ 3! αγ 5! αγ 7! αγ · · · (2r + 1)! αγ

We apply the ratio test

s2(n+1)+13nqn+1 (2n + 1)! s2 3q lim 2n+1 n 1 n = lim · = 0. (5.130) n (2(n + 1) + 1)! · s 3 − q n (2n + 3)(2n + 2) →∞ →∞

Therefore the functions g (s) and h (s) are bounded. Now applying the conditions of the | αβ | | αβ | lemma, namely that A (0) = ǫ δ and max A (0) = 1, to (5.122) we deduce: αβ A αβ | αβ′ |

d A (s) ǫ h (s) + g A (0) | αβ | ≤ A| αβ | | αγ ds γβ | d ǫ h (s) + g A (0) ≤ A| αβ | | αγ ||ds γβ | ǫ h (s) + g (s) (5.131) ≤ | αβ | | αγ | 566 Thus we see that Aαβ(s) on [0,s∗] is bounded by a constant which is independent of A. Now, as R4α4γ is bounded on [0,s∗], by (5.118), Aαβ′′ is also bounded on [0,s∗]. Hence,

A′′ (s ) C ∗ , (5.132) | αβ αβ |≤ s where Cs∗ is independent of the choice of A(0) and A′(0).

Next, A (0) is symmetric. Let λ λ , and λ be the eigenvalues of A with λ λ λ . αβ′ 1 2 3 αβ′ 1 ≤ 2 ≤ 3 We claim that λ 1 . The proof of this is based on a result of constrained minima: the 1 ≤ − 2 eigenvectors can always be normalised to unity, therefore we need only consider xα such that 2 2 2 x1 + x2 + x3 = 1 (this is the constraint); the quantity Aαβ′ xαxβ will obtain a minimum if and only if

∂ ∂ 2 2 2 Aαβ′ xαxβ = λ (x1 + x2 + x3) ∂xα ∂xα or, taking into account that Aαβ′ is symmetric, if and only if

Aαβ′ xβ = λxα i.e., minimisation occurs when xβ an eigenvector of Aαβ′ . The coresponding value Aαβ′ xαxβ is equal to the eigenvalue λ1, due to normalisation. Therefore

λ1 = min x2+x2+x2 Aαβ′ (0)xαxβ. { 1 2 3}

Case 1. A (0) = 1 for some α (no summation over al). Without loss of generality, we can | αα′ | assume that α = 1. For the case where A (0) = +1, A (0) + A (0) + A (0) 0 implies 11′ 11′ 22′ 33′ ≤ A (0) + A (0) 1. Which means that either A (0) or A (0) is less than or equal to 1 . 22′ 33′ ≤ − 22′ 33′ − 2 By taking x = (1, 0, 0) we have λ A (0), by taking x = (0, 1, 0) we have λ A (0), and 1 ≤ 11′ 1 ≤ 22′ by taking x = (0, 0, 1) we have λ A (0). As a result we have λ 1 . For A (0) = 1, we 1 ≤ 33′ 1 ≤− 2 11′ − have A (0) + A (0) +1, implying λ A (0) = 1. 22′ 33′ ≤ 1 ≤ 11′ − Case 2. A (0) = 1 for some α = β. Without loss of generality, we assume α = 1 and β = 2. | αβ′ | 6 For A (0) = +1 we let x = (1/√2, 1/√2, 0). We have 12′ −

567 1 A11′ (0) A12′ (0) √2 1 1 · 1 λ1 ( , , 0) A′ (0) A′ (0) ≤ √2 −√2  12 22 ·   − √2  0 · · ·     A (0) A (0)   1 11′ − 12′ = (1, 1, 0) A (0) A (0) 2 −  12′ − 22′  · 1   = (A′ (0) 2A′ (0) + A′ (0)) 2 11 − 12 22 1 = (A′ (0) 2+ A′ (0)). (5.133) 2 11 − 22

If A (0)+A (0) 1, then λ 1 . If A (0)+A (0) 1, then A (0) 1 (by T r(A (0)) 11′ 22′ ≤ 1 ≤− 2 11′ 22′ ≥ 33′ ≤− ′ ≤ 0). So λ is still less than or equal to -1. For the case A (0) = 1, we take x = (1/√2, 1/√2, 0) 1 12′ − in which case

1 A11′ (0) A12′ (0) √2 1 1 · 1 λ1 ( , , 0) A′ (0) A′ (0) ≤ √2 √2  12 22 ·   √2  0 · · ·     A (0) + A (0)   1 11′ 12′ = (1, 1, 0) A (0) + A (0) 2  12′ 22′  · 1   = (A′ (0) + 2A′ (0) + A′ (0)) 2 11 12 22 1 = (A′ (0) 2+ A′ (0)). (5.134) 2 11 − 22

Proceeding as before, we also ﬁnd λ 1. The most deﬁnitive statement we can conclude 1 ≤ − from all of the above is that

1 λ (5.135) min ≤−2 which is what we set out to prove.

By the above,

s s A′ (0) + A′′ (s ) A′ (0) C ∗ . (5.136) | αβ 2 αβ αβ − αβ |≤ 2 s

This implies the deviation from Aαβ′ (0) can be made arbtrarily small by suitable choice of s. Now, the eigenvalues of a 3 by 3 matrix are roots of a cubic equation which has an explicit formula to solve, the minimum eigenvalue is a continuous function of its entries. Thus it is always possible, by choising s small enough, so that the new λ is less than or equal to say 1 . 0 1 − 4 568 In actual fact this choice can be made independent of Aαβ′ (0). We now turn to the technical part of the proof.

We denote by A the space given in the lemma, namely

A := A (0) = σ δ ; A′ (0) = A′ (0); max A′ (0) = 1; tr(A′ (0)) 0) . (5.137) { αβ A αβ αβ βα | αβ | αβ ≤ }

This is compact by the Heine-Borel theorem (a subset of R that is closed and bounded is compact). Then deﬁne the space A˜ by

A˜ := [0,s] A (5.138) × which is also compact. Deﬁne the function h by

s h(s; A (0), A′ (0)) := A′ (0) + A′′ (s (s)) (5.139) αβ αβ αβ 2 αβ αβ

h : A˜ B. →

And by (5.122) we have

h(s; A (0), A′ (0)) := σ h (s)+ g (s)A′ (0) σ δ (5.140) αβ αβ A αβ αγ αβ − A αβ from which we see that h is a continuous function over A˜ implying that B is compact and bounded. Now the minimum eigenvalue λ is a continuous function over B

λ : B C R. → ⊂ implying it is bounded. Moreover, as it is a continuous function over a compact space it is also uniformally continuous, that is, given any ǫ> 0 there exists a δ > 0 such that if p,q B and ∈

dB(p,q) < δ then

λ(p) λ(q) < ǫ. | − |

We pick p to correspond to h(s = 0; A (0)) B and pick ǫ to be equal to 1/4, then we can αβ′ ∈ ﬁnd a δ

569 h(s = 0; Aαβ′ (0)) h(s; Aαβ′ (0)) | s − | = A′ (0) + A′′ (s ) A′ (0) < δ, (5.141) | αβ 2 αβ αβ − αβ | such that

s λ A′ (0) + A′′ (s ) λ A′ (0) < 1/4. (5.142) αβ 2 αβ αβ − αβ

From reference to (5.136), we see this δ can be achieved by taking s to satisfy

sC ∗ s < δ 2 or

δ s< 2Cs∗

Let s0 denote the value of s that corresponds to ǫ = 1/4. The choice s0 is independent of Aαβ(0) and Aαβ′ (0) and follows from the fact that the minimum eigenvalue is a continuous function over a compact space and as such is uniformally continuous.

Now we turn to the full solution with s = s0,

s0 A (s )= σ δ + s (A′ (0) + A )). (5.143) αβ 0 A αβ 0 αβ 2 αβ

Note that what we have established is that the minimum eigenvalue of the matrix Aαβ′ (0) + s0 2 Aαβ(sαβ(s)) is (1/4). Actually what we really need is the corresponding reuslt for this ≤ − s0 matrix multiplied by s0 > 0, namely the matrix s0(Aαβ′ (0) + 2 aαβ)). It is easy to see then for this matrix the minnimum eigenvalue is (s /4). ≤− 0 Now consider (5.143). Say we wanted to ﬁnd the eigenvalues of a matrix M, we would solve the equation

det M λ = 0. (5.144) | − |

What if we then added σAI to M, what would be the new eigenvalues? We would have to solve

det (M + σ I) λ I = 0 | A − new |

570 which is the same as solving

det M (λ σ I) = 0 (5.145) | − new − A |

Comparison of (5.144) with (5.145) implies

λnew = λ + σA.

Obviously, by adding the constant σA times the unit matrix all this does is shift the value of the eigenvalues of M by the positive number σA. By choosing σ small enough we can ensure that the minimum eigenvalue is less than or equal to say s0 (namely, with the choice σ = s /8). − 8 0 We have now established that the full solution for s = s has a minimum eigenvalue (s /8) 0 ≤− 0 provided we take σ small enough. Now, all the einganvalues of Aαβ(0) are σA > 0, but we have shown that there is a negative eigenvalue of Aαβ(s0). By the mean value theorem, there exists a zero eigenvalue of A (s ) for some s [0,s ]. So det(A)(s ) = 0. αβ 1 1 ∈ 0 1

Proposition 5.1.53 If the strong energy condition holds and the generality condition holds at a point then there is a pair of conjugate points along any causal geodesic provided that the geodesic can be extended far enouigh.

Proof: We break the proof into four parts a) Prove there is a conjugate point to the future in the case of non-positive expansion, θ 0. ≤ b) The map η that takes you from the inital condition, using the Jacobi equation, to the ﬁrst focal point is a continuous function. c) For S = b is a 3 by 3 symmetric matrix withtr(b) 0 . We prove that η(S) is bounded. { ≤ } d) Take a future focal pint and prove the existence of a past focal point.

Let γ be a time-like geodesic.

A solution for a Jacobi ﬁeld Aαβ(s) (satisfying (5.70)) along γ(s) is uniquely determined by the value of Aαβ and dAαβ/ds at p. Consider the space of solutions for which

d A = δ and A is a symmetric matrix with trace θ 0. (5.146) αβ|p αβ ds αβ|p |p ≤

We specify

Jα(s)= Aαβ(s)eβ(0). (5.147) alX=1 571 This implies

d2 A (s)e (0) = R A (s)e (0) (5.148) ds2 αβ β − α4γ4 αβ β βX=1 βX=1 or

d2 A = R A (5.149) ds2 αβ − α4γ4 αβ along γ.

We let S = b is a 3 by 3 symmetric matrix withtr(b) 0 . we claim for ony b S, if A (0) = { ≤ } ∈ αβ I3 3, A′ (0) = b × αβ Proof a):

First we prove that for all such solutions there is a ﬁnite interval on γ(s) from the point γ(s1) where the matrix Aαβ(s) ﬁrst becomes singular.

First, for each solution in P there will be some s3 >s1 for which Aαβ(s3) is singular, since either

1 1 d 1 d ω (s)= (A− (s) A (s) A− (s) A (s)) αβ −2 γα ds βγ − γβ ds αγ

So for ωαβ(0) we ﬁnd

1 1 1 ω (0) = (A− (0)A′ (0) A− (0)A′ (0)) αβ −2 γα βγ − γβ αγ 1 = (δ A′ (0) δ A′ (0)) −2 γα βγ − γβ αγ 1 = (A′ (0) A′ (0)) −2 βα − αβ = 0.

d (A (s)ω (s)A (s)) = 0 ds γα γδ δβ

ωαβ(s) will be zero everywhere along γ(s) where Aαβ(s) is non-singular.

(i) θ < 0, in which case it follows from the focussing theorem, or |p For tr(b) < 0, we have at the origin

572 1 θ(0) = tr(Aαβ′ (0)Aαβ− (0)) = tr(b) < 0.

So by the focussing theorem (proporsition 5.1.43) there will be a point where det(A(s)) =0 for some s> 0. For tr(b) = 0 we have θ(0) = 0.

(ii) tr(b)=0 or θ = 0: First suppose that |p

a b ab Rabγ′ (0)γ′ (0) + σabσ = 0

ab as σabσ is nonnegative this implies,

a b ab Rabγ′ (0)γ′ (0) = 0 and σabσ = 0

By the Raychaudhuri equation for shear (5.34) reduces to

dσ ab = R V cV d (5.150) ds cbad at γ(0). Since we have a non-zero tidal force at γ(0) this implies

dσ ab = 0 ds 6 at γ(0), which implies

ab σabσ > 0 on an open segment of γ whose closure includes γ(0). By the Raychaudhuri equation

d 1 θ = R V aV b σ σcd θ2 (5.151) ds − ab − cd − 3 positive σ σcd, (R V aV b 0 for s> 0), will cause θ to become negative for s> 0. Again we cd ab ≥ will have det(A(s)) = 0 for some s> 0. So the Weyl tensor produces convergence indirectly by inducing shear.

Now suppose that

a b ab R γ′ (0)γ′ (0) + σ σ = 0 ab ab 6 then by the Raychaudhuri equation we have det(A(s)) = 0 for some s> 0.

573 Proof b):

Let S = b is a 3 by 3 symmetric matrix with tr(b) 0 . Let η : S [0, + ) such that { ≤ } → ∞ d2 η(b) = min s [0, + ] : det(A(s)) = 0 with Aαβ(0) = I3 3, A′ (0) = b and 2 Aαβ + { ∈ ∞ × αβ ds A R = 0 . We claim η is continuous. αγ γ44β } Suppose the map is not continuous at some point b S. Then there exists a ǫ> 0 such that for ∈ all n N, there exists some b S with max (b ) b < 1 but ∈ n ∈ | n αβ − αβ| n

η(b ) η(b) < ǫ. (5.152) | n − | 6

Note condition (5.152) is equivalent to

η(b ) (η(b) ǫ, η(b)+ ǫ). (5.153) n 6∈ −

Case i) tr(b) < 0 at γ(0).

We have that bn is assumed to converge to b. Now even though we may have tr(bn) = 0 for some n, the serious tr(b ) cannot converge to zero as tr(b) is assumed to be less than zero. { n } Then for some n > N N we will have that ∈

tr(b) tr(b ) < . n 2

tr(bn)

tr(b)/2 tr(b)

Figure 5.63: Since b b, tr(b ) < tr(b)/2 for all n > N. n → n

(see ﬁg 5.63). Recall that the focussing theorem maintains if θ(s) < 0 then θ will tend to −∞ within ∆s 3/ θ(s) . Hence η(b ) [0, 6/tr(b)]. As [0, 6/tr(b)] is a compact set, η(b ) ≤ | | n ∈ − − n converges to ξ [0, 6/tr(b)], where, by assumption of discontinuity, ξ = η(b). ∈ − 6 As

(Abn )αβ(s)= hαβ(s)+ gαγ (s)(bn)γβ

574 is obviously continuous with respect to the components of bn, we have

A (s) A (s). (5.154) bn → b as b b. Indeed, η(b ) converging to ξ means that n → n

By the continuity of Abn (s) with respect to s and (5.154)

A (η(b )) A (ξ) bn n → b

As the determinant is a continuous function of the matrix elements and

det(Abn (η(bn)) = 0 for all n, we have

det(Ab(ξ)) = 0. (5.156)

If ξ < η(b), then by (5.156) there is a contradiction since η(b) is the first point where det(Ab(s)) = 0. If ξ > η(b), then there is a sequence of expansions θ(hn) associated to bn such that θ(hn) tends to when h tends to η(b) from the ‘left’; h is a sequence of values of s tending −∞ n { n} to η(b) and θ(hn) is the expansion corresponding to the solution with intial condition bn with the sequence h chosen such that θ(h ) is finite for finite n. Since h converges to η(b), for { n} n n n > N N we have ∈

h (η(b) ǫ, η(b)). n ∈ −

Again by the focussing theorem if θ(s) < 0 then θ will tend to within ∆s 3/ θ(s) ; for −∞ ≤ | | large n (and with n > N) the expansion θ corresponding to bn is so negative that the ﬁrst point where det(Abn ) = 0, i.e. η(bn), is within the distance

∆s ǫ. ≤

This is achieved for θ(h ) satisfying | n |

θ(h ) 3/ǫ. | n |≥

575 Therefore η(b ) lies inside the open interval (η(b) ǫ, η(b)+ǫ) in contradiction to the assumption n − (5.153).

Case ii) tr(b) = 0 at γ(0). By the above argument, we have tr(b) < 0 on an open interval whose closure contains γ(0). We apply the same arguent as in case i) and ﬁnd a contradiction.

Proof c):

We prove that η(S) is bounded, i.e. η(S) [0,s ]. First let C > 0 be a constant. Deﬁne ⊆ 5

S := b S : max b >C . C { ∈ | αβ| }

By lemma 5.1.52, η(S ) is bounded. The set S S corresponds to the subspace of S consisting C − C of all matrices whose component’s absolute values are less than or equal to C, i.e.,

S S = b S : max b C . − C { ∈ | αβ|≤ }

Being closed and bounded it is compact. As η is continuous η(S S ) will be bouded by − C proposition 5.1.49.

This means that there is some s5 >s1 such that η(S) is contained in the segment from γ(s1) to γ(s5).

γ(s5)

θp < 0

Figure 5.64: η(S) is contained in the bounded segment from γ(s1) to γ(s5).

Proof d):

Then we consider a point r beyond the this interval. Suppose that there is no conjugate to γ(s5) along γ[0,s5]. Otherwise, the propositin is done. We will see that it follows that there must be a point say q before γ(s1) conjugate to r. First we need to prove there is a Jacobi ﬁeld J along γ such that Jα(0) = eα and Jα(s5) = 0. Let us write the general solution of the equation (5.118) in terms of the boundary conditions given at the point r = γ(s5),

d A (s)= h (s s )A (s )+ g (s s ) A (s ) (5.157) αβ αγ − 5 γβ 5 αγ − 5 ds γβ 5

At s = 0, this is

576 d A (0) = h ( s )A (s )+ g ( s ) A (s ) (5.158) αβ αγ − 5 γβ 5 αγ − 5 ds γβ 5

We choose

Aγβ(s5) = 0, (5.159) to make Jα(s5) = 0. Then we have

d A (0) = g ( s ) A (s ). (5.160) αβ αγ − 5 ds γβ 5

d We consider the solutions for which ds Aγβ(s5) is non-singular. As we are taking Aαβ(0) to be non-singular, by detA = detg det d A (s ), g ( s ) is also non-singular. Let us choose ds γβ 5 αγ − 5

d 1 A (s ) = (g− ( s )) (5.161) ds γβ 5 − 5 γβ so that

Aαβ(0) = δαβ. (5.162)

d By this chioce ds Aγβ(s5) is symmetric. The corresponding Jacobi ﬁeld will have zero vorticity: in general we have

1 d d A ω A = A A A A γα γδ δβ 2 γα ds γβ − γβ ds γα the Jacobi ﬁeld vanishes at γ(s5) so

AγαωγδAδβ = 0 at s = s5. As Aαβ has an inverse on γ[0,s5). Therefore ωαβ =0 on γ[0,s5). From this we see that

1 1 1 0= ω (0) = (A− (0)A′ (0) A− (0)A′ (0)) (5.163) αβ −2 γα βγ − γβ αγ this implies

577 1 1 0 = A− (0)A′ (0) A− (0)A′ (0) γα βγ − γβ αγ = δ A′ (0) δ A′ (0) γα βγ − γβ αγ = A′ (0) A′ (0) (5.164) βα − αβ so Aαβ′ (0) is symmetric. We have now established a family of Jacobi ﬁelds that zanish at γ(s5) and correspond to Aαβ′ (0) symmetric, Aαβ(0) = δαβ and zero vorticity.

From this we see that if there is no point conjugate to γ(s5) between γ(s5) and γ(0) the Jacobi ﬁelds which are zero at γ(s5) must have expansion which is positive at γ(0) otherwise they would correspond to families of Jacobi ﬁelds with zero vorticity which have non-posiitve expasion at γ(0). It follows from the (time-reversed) focussing theorem that there must be a point say q before γ(s1) conjugate to r. r θp > 0

γ(s5)

Figure 5.65: The Jacobi ﬁelds which are zero at r must have expansion θ which is positive at p otherwise r would lie in the bounded interval from γ(s1) to γ(s5).

r θp > 0

Figure 5.66: With R V bV c = 0. Non-positive expansion at p (θ > 0) implies there is abcd 6 a point q conjugate to r, in the past of p. This is just the time reversed version of the focussing theorem.

578 Theorem for null geodesics from a point

In the appendix J on horizons we introduced adapted basis to describe the geometry of congruences of null geodesics.

The equation

d2 Zm = R Zn (m,n = 1, 2) (5.165) dv2 − m4n4 along a null geodesic γ(v), a Jacobi ﬁeld along γ(v). The components Zm are taken to be the components with respect to the basis E1 and E2, of a vector in the space Sq at each point q. The point p is conjugate to q along the null geodesic γ(v) if there is a Jocobi ﬁeld along γ(v), not identically zero, which vanishes at q and p.

d Zm(v)= Aˆ Zn . (5.166) mn dv q

The null expansion scalar of S with respect to K is the scalar ﬁeld θˆ on S has a natural geometric interpretation. Let Σ be the intersection of S with hypersurface in which is tranverse to K M near p S; Σ will be a co-dimensional two spacelike submanifold of , along which K is ∈ M orthogonal.

p Σ

Figure 5.67: The null expansion scalar θˆ.

One has as before: AˆlmωˆlkAˆkn = 0, so the vorticity of the Jacobi ﬁelds which are zero at p vanishes so long as Aˆmn is invertible. Also p will be conjugate to q along γ(v) if and only if

1 d θˆ = (det Aˆ)− (det Aˆ) (5.167) dv becomes inﬁnite at p. We have the null version of the focussing theorem:

Proposition 5.1.54 If R KaKb 0 and if at some point γ(v ) the expansion θˆ has negative ab ≥ 1 value θˆ < 0, then there will be a point conjugate to q along γ(v) between γ(v ) and γ(v + (2/ 1 1 1 − θˆ1)) provided that γ(v) can be extended that far.

579 The expansion θˆ of the matrix Aˆmn obeys (??):

d 1 θˆ = R KaKb 2ˆσ2 θˆ2, (5.168) dv − ab − − 2 and so the proof proceeds as before.

Proposition 5.1.55 If R V aV b 0 and if at some point p = γ(v ) KcKdK R V is ab ≥ 1 [a b]cd[e f] non-zero, there will be values v0 and v2 such that q = γ(v0) and r = γ(v2) will be conjugate along γ, providing that γ(v) can be extended to these values.

If K R K KcKd = 0 is non-zero then so is R . The proof is then similar to that of [a b]cd[e f] 6 m4n4 proposition ??.

Theorem for null geodesics orthogonal to a spacelike two-surface

Proposition 5.1.56

5.2 Black Holes

It predicts that gravitational collapse , both at the Big Bang and inside black holes, brings about spacetime singularities as at which the theory breaks down.

Even if one perturbs around the Schwarzschild solution, speciﬁc values of expansions of null congruences may change, but for suﬃciently small perturbations they will remain negative if they are negative for the unperturbed solution.

The Strong Energy Condition

T + T 0 0ˆ0ˆ 0ˆ0ˆ ≥ Xi is the most important energy condition (Gravity is attractive)

580 Focussing Theorem Existance weak strong generic Maximal Raycherdhuri’s Eq’s geodesic space of p conjugate points Energy conditions causal s curves + J [S] J −(p) is compact∪

Singularity Theorems Global Strong Hyperbolicity Chronology Domains of condition Causality Singularities Causal dependence condition non- cosmological Stable Curvature physical Singularities Black holes Causality Singularities

Figure 5.68: Singularity theorems.

5.2.1 Collapse of a Star

In Newtonian gravity the force between particles goes as 1/r2. While the matter of the star is widely spaced the gravitational force between particles is weak. The gravitaional force increases as the particles come closer together. This increases the accelaration of the collapse and if nothing intervenes matter will collapse to a point. A star which has a high temperature has high pressure which balances the gravitational force and stops the collapse.

The gravity compresses the mater against the degeneracy pressure. If the neutron star is too large, the gravitational forces overwhelm the pressure gradients and collapse cannot be halted. The neutron star continues to shrink until it ﬁnally becomes a black hole.

Using the Newtonian order of magnitude argument, we note that for a star with mass M and radius r , the gravitational unit volume is of the order M/r2nm , where nm M/r3 is the 0 0 n n ≃ 0 mass density.

The gravitational force is balanced by a pressure gradient of order P/r0, P being the average pressure in the star. Thus the pressure P can be expressed as:

2 M 2/3 4/3 4/3 P = 4 M n mN (5.169) r0 ≈

When th density is suﬃcienlt low the main contribution to the pressure is from the degneracy

581 Singularity

Empty space

Fluid

Figure 5.69: Diagram of collapse of a star.

of non-relativistic electrons, hence using (4.3.4) we have

~2 5/3 1 P = n me− (5.170)

Equationg the two expressions for the pressure we obtain

2/3 4/3 4/3 ~2 5/3 1 M n mN = n me− (5.171) which gives the value of the number density n as

2 4 3~ 6 n = M mnme − . (5.172)

The above value of n is based on the assumption that the self-gravity of the star is coming into play, this will be valid as long as this n is greater than the value of n given by() where self-gravity 3~ 3 has no inﬂuence on the star as it is too small; and also this n must be less than me − for the correctness of () In terms of pressures the relationship between small and large stasrs can be stated as

2 4/3 2/3 4/3 4/3 e n < M n mn (5.173) or equivalently as

582 Closed trapped surface Singularity

Event horizon

+ I

Star interior Empty space

− I

Figure 5.70: Penrose diagram of collapse of a star.

3 2 e mn− < M. (5.174)

On the other hand since

3 2 e mn− < M (5.175) other stuﬀ

dP 2 = ρM(r)r− (5.176) dr − where

r M(r) = 4π ρr2dr. Z0

4 Multiply (5.176) by r and integrate the LHS by parts from 0 to r, and since p = 0 at r0 we obtain:

r0 (M(r ))2 4 pr3dr = 0 . (5.177) 8π Z0

583 M/M ⊙ 1.5 D neutron stars 1.0 B white dwarfs

0.5 C A

log10 R (km) 1 2 3 4

Figure 5.71: starCollaps.

dP On the other hand, since dr is never positive,

r 3/4 d 3 3√2 3/4 2 pr′ < p r . (5.178) dr 4 Z0 

Also p is never greater than ~n3/4, hence,

r0 3 2 4/3 4/3 4/3 pr′ dr′ < ~ nr dr = ~(M(r0)) (4πmn)− . (5.179) Z0 From (5.177), after simpliﬁcation,

3/2 1/2 2 ~3/2 2 M(r0) < (8π) (4π)− mn− < 8 mn− (5.180)

When the body is static, spherically symmetric and is composed of a perfect ﬂuid, the Einstein ﬁeld equations can be reduced to

dp (µ + p)(Mˆ (r) + 4πr3p) = (5.181) dr − r r 2Mˆ (r) − where the radial coordinate is such that the area of the 2-surface (r =constant, t =constant.) is 4πr2. Similar to the Newtonian case the function Mˆ (r), represents the mass deﬁned by the integral:

r Mˆ (r)= 4πr2µdr (5.182) Z0

584 main stream

initial mass expands H layer << 1.3 solar masses Red Giant He

White Dwarf cools

H He Black Dwarf

initial mass < 2−3 solar masses initial mass > 2−3 solar masses Neutron star supported by neutron degeneracy pressure Black Hole

Figure 5.72: .

where µ = ρ(1 + ǫ) is the total energy density, ρ = nmm (n times the mass of the nucleon) and ǫ is the relativistic increase of mass associated with the momentum of the fermions. ?? The function Mˆ (r0) equals the Schwarzschild mass of the exterior Schwarzschild solution for r > r0. For a bounded star Mˆ (r0) will be less than the conserved mass:

r0 4πρr2dr M˜ = (5.183) (1 2M/r)1/2 Z0 − where N is the total number of nucleons in the star. The difference M˜ Mˆ represents the − amount of energy (binding energy) radiated off to infinity since the formation of the star from dispered matter initially at rest.

5.2.2 The Cauchy Problem - Existence and Uniqueness

Local existence and uniqueness (Bruhat, Hugh- Kato-Marsden, Christ.-K) Tools: Classical theory of nonl. hyperb. eqts. Energy estimates, Sobolev inequalities, canonical bootstrap strategy as developed by Sobolev, Friedrichs, Leray, Kato, John etc. 1930-1970. Results: Need (Σ,g,k)

s s 1 g H (Σ), k H − (Σ), s> 5/2 ∈ loc ∈ loc

585 Strictly we can only say a closed trapped surface has been formed if there exists a solid theoretical understanding of the relation between exact solutions of Einstein’s equations and those which approximate them.

Second Order Hyperbolic Equations

The energy inequality relates the size of an integral involving the derivatives of the data and a similar integral of the solution.

The Cauchy problem for the Einstein equations Alan D. Rendall, [8]. the evolution equations are (essentially) hyperbolic but we also have to solve the constraints to produce initial data for the evolution equations. We need to work with function spaces which are adapted tot he hyperbolic equations and are also compatible with elliptic equations.

These spaces can be deﬁned as follows. Choose a Riemannian metric on the manifold S (we are still assuming that S is compact). If f is a C∞ function deﬁne its Sobolev norm of order s to be

s 1/2 k 2 f Hs = f (5.184) k k " S |∇ | # Xk=0 Z where kf denotes the covariant derivative of f of order k, ... The Sobolev space Hs(S) is ∇ defined to be the completion of the space of C∞ functions with respect to this norm. Now the norm depends on the choice of auxiliary Riemannian metric but the norms obtained for different metrics are equivalent, what this means is the following. Say we have two different metrics with respective norms f s and f s , there exists positive finite numbers C and C such that k kH k kH′ 1 2

C f s f ′ s C f s . (5.185) 1k kH ≤ k kH ≤ 2k kH

If a sequence is a Cauchy series in one norm it is a Cauchy series in the other. The two norm spaces are topologically equivalent, the completions are . If a series of functions converges for one norm space it does in the other. In this sense Hs(S) is independent of the auxiliary metric chosen. The relation to the more usual maximum norm is as follows. Suppose m > dim( )/2. Then N

ψ < const. ψ, ˜ . (5.186) k k × k N km

5.2.3 Non-Linear Hyperbolic Diﬀerential Equations

Quasi-Linear First Order PDEs in 2D

∂u ∂u a(u,x,y) + b(u,x,y) = c(u,x,y) (5.187) ∂x ∂y

586 Write

G(x,y,u)= F (x,y,u) u = 0. (5.188) − The normal to the solution surface is in the direction

∂G ∂G ∂G ∂F ∂F G = , , = , , 1 (5.189) ∇ ∂x ∂y ∂u ∂x ∂y − But in the surface, F = u, implying

∂u ∂u G = , , 1 (5.190) ∇ ∂x ∂y − Now let a = (a, b, c)

∂u ∂u a G = a + b c = 0 ·∇ ∂x ∂y −

So that a is perpendicular to the normal to the solution surface. Implying that a lies in the solution surface and we can distinguish in the surface lines everywhere parallel to a.

If we parameterize these lines using the variable τ, then they satisfy

dx dy du = a, = b, = c (5.191) dτ dτ dτ and are called characteristics or characteristic curves.

If initial conditions are given as values of u on some line in the (x,y) plane, this is equivalent − to specifying a line, Γ , in the (x,y) plane. 0 − u

τ increasing

y Γ0

x Figure 5.73: characteristic.

Let Γ0 be parameterized by s, so that Γ0 is x(s),y(s), u(s), we solve the PDE by starting at the initial line and integrating forward and backward along the characteristics.

587 Well-Posed-Problems

1) Existence - the problem must have a solution.

2) Uniqueness - the problem must have just one solution.

3) Continuous dependence on the boundary conditions - a small change in the boundary data must lead to a small change in the solution.

The third requirement is essential for physical reasons. If in a model of some physical situation, a small change in the boundary values, e.g noise or measurement error, leads to a large change in the global solution then there is likely to be something wrong with the model (note this does not exclude the exponential divergences of solutions associated with chaotic behaviour.

The third requirement is of particular importance as we wish to show that we still have a closed- trapped surface when we make a small perturbation away from spherically symmetric black hole.

Examples

Quasi-linear ﬁrst order systems.

Initial value problem (Cauchy problem) for the 1D wave equation

∂2y 1 ∂2y = ∂x2 c2 ∂t2 subject to ∂y y(x, 0) = y (x), (x, 0) = v (x) 0 ∂t 0

The solution exists and is given by uniquely by D’Alambert’s solution:

1 x+ct y(x,t)= y0(x + ct)+ y0(x ct)+ v0(s)ds . 2 − x ct Z − To demonstrate continuous dependence on boundary data consider the two solutions

y = y1(x,t), y = y2(x,t) for initial conditions ∂y y (x, 0) = y (x), 1 (x, 0) = v (x) 1 0 ∂t 0 ∂y y (x, 0) = y (x)+ y(x), 2 (x, 0) = v (x)+ v(x). 2 0 ∂t 0

588 The diﬀerence between the two solutions is

Y (x,t)= y (x,t) y (x,t) 2 − 1 and D’Alambert’s solution gives

1 x+ct Y (x,t)= y(x + ct)+ y(x ct)+ v(s)ds . 2 − x ct Z − This implies

1 x+ct Y y(x + ct)+ y(x ct) + v(s)ds | | ≤ 2| − | | x ct | Z − max y + 2ct max v (5.192) ≤ | | | |

This implies

Second Order Semi-Linear Diﬀerential Equations in 2D

The general equation is

∂2φ ∂2φ ∂2φ ∂φ ∂φ a(x,y) + 2b(x,y) + c(x,y) + f(x, y, φ, , ) = 0. (5.193) ∂x2 ∂x∂y ∂y2 ∂x ∂y

We will ﬁnd that the ﬁrst three terms, involving the 2nd derivatives allow us to clarify the equation into one of three distinct types.

We do a change of coordinates

(x,y) ξ(x,y), η(x,y) 7→ with

∂(ξ, η) = 0 ∂(x,y) 6 so that the transformation is invertible. Let φ(x,y)= ψ(ξ, η). This implies

φx = ξxψξ + ξxψη

2 2 φxx = ξxψξξ + 2ξxηxψξη + ξxψηη + (ξxxψξ + ηxxψη)

Similarly for φxy and φyy. Substituting these into the PDE equation gives

589 Aψξξ + 2Bψξη + Cψηη + F (ξ,η,ψ,ψξ , ψη) = 0, (5.194)

A,B,C being functions of ξ and η:

2 2 A = aξx + 2bξxξy + cξy 1 B = aξ η + 2b(η ξ + ξ η )+ cξ η 2 x x x y x y y y 2 2 C = aηx + 2bηxηy + cηy

This can be expressed as

A B ξ η a b ξ ξ = x x x y (5.195) B C ξ η b c η η y y x y 

5.2.4 Existence and Uniqueness

Banach Fixed Point Theorem. Suppose that X is a nonempty complete met5ric space with metric d and suppose that the function S from X to itself is a strict contraction, that is, for some positive constant ρ< 1, and any x and y in S,

d S(x),S(y) ρd(x,y). (5.196) ≤ Then S has a unique ﬁxed point in X.

Proof: Choose any point x X and deﬁne a sequence by iteration: x = S(x ). Condition 1 ∈ k+1 k (5.196) implies that

2 k d(xk+2,xk+1) ρd(xk+1,xk) ρ d(xk,xk 1) ρ d(x2,x1). ≤ ≤ − ≤···≤

Therefore

d(xk+m,xk+1) d(xk+m,xk+m 1)+ d(xk+m 1,xk+m 2)+ + d(xk+2,xk+1) − − − ≤ k+m 2 k+m 3 k · · · [ρ − + ρ − + + ρ ]d(x ,x ) ≤ · · · 2 1 ρk (x ,x ). (5.197) ≤ 1 ρ 2 1 −

It follows that x is a Cauchy sequence in X, so it converges to a point x X. Now S is { k}k∞=1 ∈ continuous, so

590 d(x,S(x)) = lim d(xk,S(xk)) = lim d(xk,xk+1) = 0. k k →∞ →∞

If x′ is also a ﬁxed point, then

d(x,x′)= d(S(x),S(x′)) ρd(x,x′), ≤ so d(x,x′)=0 and x′ = x

Higer order differential equations are equivalent to systems of first order differential equations. For example, setting

2 n 1 dy d y d − y y1 = y, ,y2 = , ,y3 = 2 ,..., ,yn = n 1 , dx dx dx − the n-th order diﬀerential equation

dny dy dn 1y = F x,y, ,..., − dxn dx dxn 1 − is equivalent to the systems of ﬁrst order diﬀerential equations

dy 1 = y dx 2 dy 2 = y dx 3 . . . dyn 1 − = y dx n dy n = F (x,y ,y ,...,y ). (5.198) dx 1 2 n

5.2.5 Cauchy-Kowalewski Theorem

1 n 1 n Theorem 5.2.1 (Cauchy-Kowalewski theorem): Let t,x ,...,x − be coordinates of R . n Consider a system of partial diﬀerential equations for n unknown functions φ1,...,φn in R , having the form

∂2φ ∂φ ∂φ ∂2φ ∂2φ i = F t,xa; φ ; i , i , i , i (5.199) ∂t2 i j ∂t ∂xa ∂t∂xa ∂xa∂xb

591 a a where Fi is an analytic function of its variables. Let fi(x ) and gi(x ) be analytic functions. Then there is an open neighbourhood of the hypersurface t = t such that within there exists O 0 O a unique analytic solution to the PDE such that

∂φ φ (t ,xa)= f (xa) and i (t ,xa)= g (xa). i 0 i ∂t 0 i

Non-Linear Hyperbolic Diﬀerential Equations the metric deﬁnes the space-time over which it propagates

To obtain a solution of the non-linear equations one employs an iterative scheme on approximate linear equations whose solutions are shown to converge in a certain neighbourhood of the initial surface.

The quasilinear system

∂ψ = Ci(ψ)∂ ψ + D(ψ), (5.200) ∂t i is replaced by a sequence of linear problems

∂ψ n+1 = Ci(ψ )∂ ψ + F (ψ ), n> 0, (5.201) ∂t n i n+1 n all with data ψ0. we solve a succession of linear problems in which the coeﬃcients are determined by the solutions of the previous problem.

ψn+1 = T (ψn). (5.202)

If T has a ﬁxed point ψ, i.e., ψ = T (ψ), then this is a solution of the Cauchy problem.

Provided that these smoothness conditions are satisﬁed, uniqueness, stability and local existence of solutions of the Cauchy problem is guaranteed.

5.2.6 Reduced Einstein Equations

Two metrics g1 and g2 on a manifold are physically equivalent if there is a diffeommorphism which takes g1 into g2, and g1 satisfies the field equations if and only if g2 does. Thus the solutions of the field equations can be unique only up to a diffeomorphism. In order to obtain a definite member of the equivalence class of metrics which represents a spacetime, one introduces a fixed background metric and impose four gauge conditions on the covariant derivatives of the physical metric with respect to the background metric.

Positive deﬁnite metric eab

592 the Einstein-vacuum equations take the reduced form;

ab g ∂a∂b = Nab(g,∂g) (5.203) with N quadratic in the ﬁrst derivatives ∂g of the metric.

Expressed relative to the wave coordinates xa the spacetime metric g takes the form:

g = N 2dt2g (dxi + vidt)(dxj + vjdt) (5.204) − ij where gij is a Riemannian metric on the slices Σt, given by the level hypersurfaces of the time function t = x0, N is the lapse function of the time foliation, and v is a vector-valued shift function.

Harmonic Coordinates these coordinates are functions that solve the wave equation they can be constructed by solving a Cauchy problem for the wave equation.

Energy Estimates energy estimates provide a general tool for deriving a priori estimates for hyperbolic equations. Together with Sobolev inequalities, which were developed for this reason, they allow us to prove local in time existence, uniqueness and continuous dependence on initial data for general classes of nonlinear equations

Klein-Gordon theory:

1 T = ∂ φ∂φ η (∂ φ∂cφ + m2φ2) ab a − 2 ab c

a ∂ (Tab) = 0

a b ∂ (Tabξ ) = 0. (5.205) i) Let S0 be a closed ball on the initial hypersurface Σ0; ii) let Σ1 denote the hypersurface t = t1; iii) let K = D+(S ) J (Σ ); 0 ∩ − 1 iv) let S = D+(S ) Σ and 1 0 ∩ 1

593 Σ S1 1

Σ S2 0 S0

Figure 5.74: EnergyIneq1.

v) let S2 denote the null portion of the boundary of K (see ﬁg. 5.74).

We integrate (5.205)

a b a b a b Tabξ ξ + Tabℓ ξ = Tabξ ξ , (5.206) S 1 S2 S0 Z − Z Z

a b strong energy condition T bv is a future directed timelike or null vector. Consequently we a b − a b have Tabℓ ξ 0. Therefore Tabℓ ξ is nonnegative. we obtain ≥ S2 R ∂φ 2 ∂φ 2 + φ 2 + m2φ2 + φ 2 + m2φ2 . (5.207) ∂t |∇ | ≤ ∂t |∇ | ZS1 " # ZS0 " # it shows that if φ and φ are both C2. Write ψ(x)= φ (x) φ (x) so that ψ =0 on S , then 1 2 1 − 2 0

∂ψ 2 + ψ 2 + m2ψ2 0. (5.208) ∂t |∇ | ≤ ZS1 " # for m = 0 this implies ψ =0 on S and hance ψ vanishes throughout D+(S )as Σ is arbitrary. 6 1 0 1

ab ab ab (S t,a),b)dv = S ,bt,adv + S t,abdv ZUt ZUt ZUt ab = S t,anbdS E(0) + E(t). (5.209) ∂Ut ∂U − Z ∩

Sija b 0. (5.210) i j ≥

Applying this to the ﬁrst term in (5.209)

594 Sab t dv + Sabt dv E(0) + E(t). (5.211) ,b ,a ,ab ≥− ZUt ZUt

Surfaces (t = constant)

(∂ ) U 2

U t=t’ (t′) H (∂ ) U 2 (t′) U

(∂ ) U 1 Figure 5.75: EnergyCond.

5.2.7 Sobolev Inequalities

There is no simple relation between the maximum size of the derivatives of the data at points on the initial surface, and the maximum size of the derivatives of the solution in the domain of dependence. But there is a relation between the size of an integral involing the derivatives of the data and a similar integral for the solution. This integral is referred to as an energy, because in certain special cases it does have this physical interpretation.

By the classical Sobolev inequality,

E exp Ct sup ∂φ(τ) Hs−1 dτ (5.212) ≤ 0 τ t k k ≤ ≤ provided that s> 5/2.

Products

KI LP , P K, L, . (5.213) k J Q N k0 ≤ 2k N kmk N km if n 4 and 2m>n, then for any two ﬁelds KI ,LP W m( ), the product KI LP is also ≤ J Q ∈ N J Q in W m( ). N

595 5.2.8 Developments for the Empty Space Einstein Equations

The reduced Einstein equations

1 Eab (φcd) = 8πT ab (Rˆab Rˆgˆab) (5.214) cd − − 2 are quasi-linear second order hyperbolic equations. i.e

L(K) AabKI + BaP I KQ + CPI KQ = F I (5.215) ≡ J,ab QJ P,a QJ P J where A is a metric of Lorentz signiture.

ab Take suitable trial field φ′ and use this to determine the values of the coefficients A, B and C in the operator E. Using these values we then solve (5.214) as a linear equation with the ab presecribed inital data and obtain a new field φ′′ . They show that under sutiable conditions the map α has a fixed point. This fixed point is the desired solution of the quasi-linear equation.

ˆ M

(t′) U+ H

(t′) U (0) H ∂ U U Figure 5.76: .

The map α : W 4+a( ) W 4+a( ) will take the closed ball W (r) of some radius r into itself U+ → U+ provided φ, (0) ˜ and φ, (0) ˜ satisfy k 0 H ∩ Uk4+a k 1 H ∩ Uk3+a

1 φ, (0) ˜ rP k 0 H ∩ Uk4+a ≤ 2 7,a 1 φ, (0) ˜ rP (5.216) k 1 H ∩ Uk3+a ≤ 2 7,a

They show that α has a ﬁxed point if the above inequalities hold and if r is suﬃciently small.

596 Suppose φ′1 and φ′2 are in W (r). For the ﬁelds φ′′1 = α(φ′1) and φ′′2 = α(φ′2) one establishess the inequality

φ′′1 φ′′2, rQ φ′1 φ′2, (5.217) k − U+k1 ≤ 5k − U+k1 where Q5 is some constant independent of r. Thus for suﬃciently small r, the map α will be contracting in the norm (i.e α(φ1) α(φ2) < φ1 φ2 ). k k1 k − k1 k − k1 Now

n+1 n α(φ) α (φ′1), rQ φ α (φ′1), k − U+k1 ≤ 5k − U+k1

As n the right-hand side tends to zero. This implies that α(φ) φ, = 0 and so that →∞ k − U+k1 α(φ) = φ. Since the map α is contracting the ﬁxed point is unique in W (r). They therefore proved:

Proposition 5.2.2 If ˆg is a solution of the empty space Einstein equations, the reduced empty space Einstein equations have a solution φ W 4+a( ) if φ, (0) ˜ and φ, (0) ∈ U+ k 0 H ∩ Uk4+a k 1 H ∩ ˜ are suﬃciently small. φ, (0) ˜ will be bounded and so φ will be at least C(2+a) . Uk3+a k H ∩U +k4+a −

This shows that if one makes a suﬃciently small perturbation in the intial data of an empty space solution of Einstein equations one obtains a solution in a region . U+

5.2.9 Stability of Closed Trapped Surfaces

5.2.10 Apparent Horizons

The singularities that arise from localized gravitational collapse are associated with black holes. Intuitively, the Cosmic Censorship Hypothesis postulates that all singularities are hidden inside the event horizon, i.e. inside I˙ ( +), the boundary of the past of future null inﬁnity, +, the − I I latter is usually assume to be complete.

5.3 The Big Bang

Cosmic background radiation

597 5.4 Biblioliographical notes

In this chapter I have relied on the following reﬀerences: [?], [?], [57]

The details of th proof of the existence of pair of conjugate points on every causal geodesic follows the proof given in “Causality, Conjugate Points and Singularity Theorems in Space-time” by TONG, Pun Wai.

5.5 Sobolev Spaces

5.5.1 Review of LP Spaces

Given a measurable subset Ω Rn the space Lp(Ω), 1 p < , consists of all measurable ⊂ ≤ ∞ functions f :Ω C with ﬁnite LP norm, →

1/p p f p = f(x) dx < . k kL | | ∞ ZΩ For all values of 1 p the spaces Lp(Ω) are Banach spaces. ≤ ≤∞ Recal the inequality

1/p 1/q 1/r

figi fi gi | |! ≤ | |! | |! Xi Xi Xi where

1 1 1 = + p q r

The following is the integral version of the Holder inequality

fg p f q g r k kL ≤ k kL k kL whenever

1 1 1 = + p q r

γ − ⋆f(x) q Rn C f(x) p Rn (5.218) k|·| kL ( ) ≤ k kL ( )

598 The Holder inequality is

f ⋆ g q Rn f p Rn g ′ k kL ( ) ≤ k kL ( ) k kLp (Rn)

A multi-index is an n tuple −

α = (α1,...,αm) of natural numbers, associated with which is a diﬀerential operator,

α α ∂| |f α = α1 + α2 + + αn ∂ f = α1 αm . | | · · · ∂x1 ...∂xm

Also

1/2 n f = (∂ f,...,∂ ), f = ∂ f 2 ∇ 1 n |∇ |  | j |  Xj=1   Rn Deﬁnition Let Ω be a domain in . Denote by D(Ω) or C0∞(Ω) the set of C∞(Ω) functions with compact support in Ω.

Deﬁnition Weak deriative:

We say that a given function f L1 (Ω) has weak derivative Dα f, provided there exists a ∈ loc w function g L1 (Ω) such that ∈ loc

α (α) g(x)ϕ(x)dx = ( 1)| | f(x)ϕ (x)dx (5.219) − ZΩ ZΩ for all ϕ C (Ω), where α = m α is the order of the derivative. If such a g exists, we ∈ 0∞ | | i=1 i deﬁne P

α Dwf = g.

599 5.5.2 Hardy-Littlewood-Sobolev Inequality

We split the convolution with singular kernel into two parts:

γ f(x y) f(x y) − ⋆f(x)= −γ dy + −γ dy | · | y R y y

We estimate the ﬁrst term simply by Holder’s inequality,

1/p′ f(x y) γp′ p −γ dy f L y − dy y R y ≤ k k y R | | ! Z| |≥ | | Z| |≥ 1/p′ n ′ ′ γ p γp = R − f Lp y − dy k k y 1 | | Z| |≥ !

The integral converges if γp′ >n,

The second part written as

f(x y) ∞ f(x y) −γ dy | −γ |dy y

Therefore we have found that for every x Rn and every R> 0, ∈

n γ ′ γ n γ − ⋆f(x) C R p − f p + C R − f(x) | · | ≤ 1 k kL 2 M

where the C1,C2 are constants independent of R and x.

Set the two terms on the right hand side equal

n ′ γ n γ C R p − f p = C R − f(x) 1 k kL 2 M

600 i.e.,

p/n C f p R(x)= 1k kL C f(x) 2M where we have used p /(p 1) = p. Since (n γ)p/n = 1 p/q, we have ′ ′ − − −

γ n γ − ⋆f(x) 2C R − f(x) | · | ≤ 2 M 1 p/q C1 f Lp − = 2C k k f(x) 2 C f(x) M 2M 1 p p 1 p p − q q − q q = (C C ) f p f(x) 1 2 k kL M

Taking the Lp norm of both sides

1 p p γ − q q − ⋆f(x) p C f p f p . k| · | kL ≤ k kL kM kL

Corollary 5.5.1 The Sobolev inequality

f q C ∂f p k kL ≤ k kL for n/q = n/p 1, in the non-sharp regime p> 1. −

Proof: Proof of the Sobolev inequality. First we prove

1 f(y) (x y) f(x)= ∇ · n− dy, a.e.(Ω). nΩn 1 Ω x y − Z | − | we then use f (x y) f (x y) to obtain ∇ · − ≤ |∇ | | − |

1 f(y) f(x) |∇ n | 1 dy | |≤ nΩn 1 Ω x y − − Z | − | As we will see, the Sobolev inequality the follows by applying the Hardy-Littlewood-Sobolev inequality.

By the fundamental theorem of calculus, for each w Sn 1 we have ∈ −

601 d f(x)= ∞ f(x + ωr)dr ∞ w ∂ f(x + ωr)dr − dr ≡− i i Z0 Z0

d/2 where Ωd 1 = 2π /(d/2 1)!. − − n 1 n wn is the volume enclosed by the unit sphere S − in R .

Integrating with respect to w,

f(x)dw = ∞ w f(x + rw)drdw − ·∇ ZSn−1 ZSn−1 Z0 implies

1 w f(y) f(x) = ·∇ n 1 dy −nw Rn x y n Z | − | − 1 (x y) f(y) = − ∇ n 1 −nw Rn x y · x y n Z | − | | − | − 1 f(y) (x y) = ∇ · n− dy nw Rn x y n Z | − |

Therefore

1 ∂f(y) 1 1 n f(x) | n| 1 dy = ( − ⋆ ∂f )(x) | |≤ Ωd 1 x y − Ωd 1 | · | | | − Z | − | − we take the Lp norm and use (5.218) to get

1 1 n C q p f L − ⋆ ∂f Lq ∂f L k k ≤ Ωd 1 | · | | | ≤ Ωd 1 k k − − whenever p> 1 and

n 1 1 1 1 − = . − n p − q

602 5.5.3 Sobolev Inequalities

Lemma 5.5.2 By Cauchy’s inequality for non-negative a ,...,a R, we have 1 m ∈

1 m m m 1 ai ai. (5.220) ! ≤ m Yi=1 Xi=1

Proof:

For example

1 2 2 2 1 ai ai ! ≤ 2 Yi=1 Xi=1 as follows from (a a )2 0, which implies 4a a a2 + a2 + 2a a = (a + a )2. 1 − 2 ≥ 2 2 ≤ 1 2 2 2 1 2 Take case m = 3. Consider

3 2 (a1 + a2 + a3) = (a1 + a2 + a3)(a1 + a2 + a3) 2 2 2 = (a1 + a2 + a3)(a1 + a2 + a3 + 2a1a2 + 2a1a3 + 2a2a3)

We make use of 2ab a2 + b2. ≤

1 (a2 + a2 + a2) = [(a2 + a2) + (a2 + a2) + (a2 + a2)] 1 2 3 2 1 2 1 3 1 3 a a + a a + a a ≥ 1 2 1 3 2 3

Hence

3 2 (a1 + a2 + a3) = (a1 + a2 + a3)(a1 + a2 + a3)

= 3(a1 + a2 + a3)(a1a2 + a1a3 + a2a3) 2 2 2 2 2 2 = 3[3a1a2a3 + a1(a2 + a3)+ a2(a1 + a3)+ a3(a1 + a2)] 27a a a ≥ 1 2 3

So we have proved

1 1 (a a a ) 3 (a + a + a ) 1 2 3 ≤ 3 1 2 3

603 Cases m> 3 are proved similarly.

Lemma 5.5.3 The Gagliardo-Nirenberg-Sobolev inequality.

f C (q) f . (5.221) k kp ≤ n k∇ kq

qn p = . n q − In particular for q = 1, p = n(n 1). −

Proof:

n 1/n f n C ∂ f . (5.222) k kL n−1 (Rn) ≤ k j kL1(Rn) jY=1

When n = 2,

f(x ,x ) 2dx dx ∂ f(y ,x ) dy ∂ f(x ,y ) dy dx dx | 1 2 | 1 2 ≤ | y1 1 2 | 1 | y2 1 2 | 2 1 2 ZZ ZZZ Z = ∂ f 1 ∂ f 1 . k y1 kL k y2 kL

Now n = 3,

1/2 1/2 1/2 f(x) 3/2 ∂ f(y ,x ,x ) dy ∂ f(x ,y ,x ) dy ∂ f(x ,x ,y ) dy . | | ≤ | y1 1 2 3 | 1 | y2 1 2 3 | 2 | y3 1 2 3 | 3 Z Z Z 

We integrate with respect to x1

1/2 f(x ,x ,x ) 3/2dx ∂ f(y ,x ,x ) dy | 1 2 3 | 1 ≤ | y1 1 2 3 | 1 Z Z 1/2 1/2 ∂ f(x ,y ,x ) dy ∂ f(x ,x ,y ) dy dx | y2 1 2 3 | 2 | y3 1 2 3 | 3 1 Z Z Z 

By the Holder inequality fg 1 f 2 g 2 , k kL ≤ k kL k kL

604 1/2 1/2 ∂ f(x ,y ,x ) dy ∂ f(x ,x ,y ) dy dx | y2 1 2 3 | 2 | y3 1 2 3 | 3 1 Z Z Z 1/2 1/2 ∂ f(x ,y ,x ) dy dx ∂ f(x ,x ,y ) dy dx ≤ | y2 1 2 3 | 2 1 | y3 1 2 3 | 3 1 Z Z 

1/2 f(x ,x ,x ) 3/2dx ∂ f(y ,x ,x ) dy | 1 2 3 | 1 ≤ | y1 1 2 3 | 1 Z Z 1/2 1/2 ∂ f(x ,y ,x ) dy dx ∂ f(x ,x ,y ) dy dx | y2 1 2 3 | 2 1 | y3 1 2 3 | 3 1 ZZ ZZ 

Similarly integration over x2 leads to

1/2 f(x ,x ,x ) 3/2dx dx ∂ f(y ,x ,x ) dy dx | 1 2 3 | 1 2 ≤ | y1 1 2 3 | 1 2 Z ZZ 1/2 1/2 ∂ f(x ,y ,x ) dy dx ∂ f(x ,x ,y ) dy dx dx | y2 1 2 3 | 2 1 | y3 1 2 3 | 3 1 2 ZZ ZZZ and ﬁnally integration over x3 gives

1/2 f(x ,x ,x ) 3/2dx dx dx ∂ f(y ,x ,x ) dy dx dx | 1 2 3 | 1 2 3 ≤ | y1 1 2 3 | 1 2 3 Z ZZZ 1/2 1/2 ∂ f(x ,y ,x ) dy dx dx ∂ f(x ,x ,y ) dy dx dx | y2 1 2 3 | 2 1 3 | y3 1 2 3 | 3 1 2 ZZZ ZZZ

By Cauchy’s inequality

3 3 1/3 1 ∂ f dx ∂ f dx | j | ≤ 3 | j | jY=1 Z Xj=1 Z

Hence

605 3 1 f 3/2 R3 ∂ f dx k kL ( ) ≤ 3 | j | Xj=1 Z 3 1 2 ∂jf dx ≤ √3 v | | uj=1 Z uX 1 t = f L1(R3) √3k∇ k

Thus we have established that

1 f 3/2 f 1. (5.223) k k ≤ √3k∇ k

It is easy to generalize and obtain

1 f n f 1 Rn . (5.224) k kL n−1 (Rn) ≤ √nk∇ kL ( )

So the inequality of the lemma holds for p = 1.

Now we prove the case p 1, we can use inequality (5.224) to obtain estimates for γ f n , where γ > 1 k| | k n−1

γ 1 γ f n f L1 k| | k n−1 ≤ √nk∇| | k

1 γ 1 = γ f − f √n | | ∇| | Z

By Holder’s inequality

n−1 1 1 n n q p γ − n γ (γ 1)q n p n f · n 1 d x f − d x f d x | | ≤ √n | | · |∇ | Z Z Z where q = p/(p 1). Let γ satify −

γn (γ 1)p = (γ 1)q = − n 1 − p 1 − − that is,

(n 1)p γ = − > 1. n p − 606 Then

(n−1)p (p−1)n − − n p n−p n p n−p 1 np np γ np np p f n−p dnx f n−p dnx f pdnx . | | ≤ √n | | · |∇ | "Z # "Z # Z 

Which implies

(n 1)p f np − f p k k n−p ≤ (n p)√nk∇ k − as required.

Theorem 5.5.4 m n f q Rn C ∂ f p Rn , f C∞(R ) (5.225) k kL ( ) ≤ k kL ( ) ∈ 0 holds for

1 1 m = > 0, m N, (1 p

While for q = , we have ∞

m k n f ∞ Rn C ∂ f p Rn , f C∞(R ) (5.227) k kL ( ) ≤ k kL ( ) ∈ 0 Xk=0 where m > n/p. lemma

Proof: We obtain the cases withh m> 1 by repeated iterations of the case m = 1. For example m = 2:

′ f q Rn C f q n k kL ( ) ≤ k kL (R )

f q Rn C ∂f ′ k kL ( ) ≤ 1k kLq (Rn) for

607 1 1 1 = n q′ − q

′ 2 ∂f q n C ∂ f p Rn k kL (R ) ≤ 2k kL ( ) for

1 1 1 = n p − q′

2 f q Rn C C ∂ f p Rn k kL ( ) ≤ 1 2k kL ( ) where

2 1 1 = . n p − q

Now

m 1 ′ m ∂ − f q n C ∂ f p Rn k kL (R ) ≤ mk kL ( ) for

1 1 1 = . n p − q′ and

m 1 1 1 m 1 ′ f Lq (Rn) C1C2 ...Cm 1 ∂ − f Lq (Rn), for − = . k k ≤ − k k n q′ − q

So that

m f q Rn C C ...C ∂ f p Rn k kL ( ) ≤ 1 2 mk kL ( ) for

m 1 1 = . n p − q

608 Hence we assume m = 1

It only remains to prove the case m = 1, p = 1, q = n/(n 1), −

5.5.4 Classical Sobolev Spaces

Sobolev norm

1/p p f k,p(Ω) := ∂αf Lp(Ω) , for 1 p< . k kW  k k  ≤ ∞ α k |X|≤   Diﬀerentiation in the sense of distributions.

For example

1/p n p p ∂f f 1,p(Ω) := f Lp(Ω) + k kW k k ∂x p  j=1 j L (Ω) X  

Let f, g k,p(Ω) ∈W

1/p p f + g k,p(Ω) = ∂αf + ∂αg Lp(Ω) k kW  k k  α k |X|≤   1/p p ∂ f p + ∂ g p ≤  k α kL (Ω) k α kL (Ω)  α k |X|≤  1/p  1/p ∂ f p + ∂ g p ≤  k α kLp(Ω)  k α kLp(Ω) α k α k |X|≤ |X|≤ = f k,p(Ω) + g k,p(Ω)   k kW k kW

Theorem 5.5.5 The Sobolev k(Ω) space is a Banach space. Wp

Proof: Let v be a Cauchy sequence with respect to the norm k . Since this norm is j p (Ω) { } k · kW α just a combination of p norms of weak derivatives, it follows, that for all α k, D v k·kL (Ω) | |≤ { w j} is a Cauchy sequence with respect to the norm p . k · kL (Ω) First note that if w w in Lp(Ω), then for all ϕ D(Ω) j → ∈ 609 w (x)ϕ(x)dx w(x)ϕ(x)dx, j → ZΩ ZΩ as follows from Holder’s inequality,

w ϕ wϕ p w w p ϕ ∞ 0 k j − kL (Ω) ≤ k j − kL (Ω)k kL (Ω) → as j . To verify Dα v = vα we must show that →∞ w

α α (α) v ϕ(x)dx = ( 1)| | vϕ (x)dx − ZΩ ZΩ for all ϕ D(Ω). ∈

α α v ϕ(x)dx = lim Dwvj ϕ(x)dx j ZΩ →∞ ZΩ α (α) = ( 1)| | v ϕ (x)dx − j ZΩ α (α) = ( 1)| | vϕ (x)dx − ZΩ for all ϕ D(Ω). ∈

For the derivatives ∂αf(x), we have

α α α α ∂˜f(k) = (ik) f˜(k)= i| |k f˜(k)

Then

α 2 α 2 ˜ 2 f Hm(Rn) = ∂ f dx = k f(k) dk k k Rn | | Rn  | |  | | α m Z Z α m | X|≤ | X|≤  

2 2 n f(x) dx = f˜(k) dk, f L2(R ) Rn | | Rn | | ∈ Z Z Obviously

1 1 (1 + k 2)2 = (1 + 2 k 2 + k 4) 1+ k 2 + k 4 (1 + 2 k 2 + k 4)=(1+ k 2)2 2 | | 2 | | | | ≤ | | | | ≤ | | | | | |

610 Since

c (1 + k 2)m kα 2 c (1 + k 2)m 1 | | ≤ | | ≤ 2 | | α m | X|≤

Thus

For example

Thus the norm

1/2 f(k) 2(1 + k 2)mdnk Rn | | | | Z is equivalent to the standard norm in m(Rn). W2 Product theorem

˜ 2 2 s n fg s,2(Ω) = fg (1 + k ) d k k kW | | | | ZΩ

fg˜ 2 = f˜(k + l )˜g(l )f˜(k + l )g˜(l )dnl dnl | | 1 1 2 2 1 2 ZΩ ZΩ

f = f˜(x )(1 + x 2)u/2 1 1 | 1| 2 u/2 f2 = f˜(x2)(1 + x2 ) | |2 t/2 f3 =g ˜(x3)(1 + x3 ) | |2 t/2 f4 = g˜(x4)(1 + x4 ) 2 s| | 2 u/2 2 u/2 2 t/2 2 t/2 φ = (1+ k ) (1 + l )− (1 + k + l )− (1 + l )− (1 + k + l )− | | | 1| | 1| | 2| | 2|

n n n A = f1f2f3f4φd l1d l2d k Z Use Cauchy inequality

A [P (k)P (k)Q(k)]1/2dnk ≤ 1 2 Z 611 where

n P1(k) = f1f3d l1 Z n P2(k) = f2f4d l2 Z n n Q(k) = φd l1d l2 (5.228) Z

2 P 2dnk = f˜(k + l )˜g(l )(1 + k + l 2)t/2(1 + l 2)u/2dnl dnk | 1| 1 1 | 1| | 1| 1 Z Z Z =

Q(k) is bounded

5.5.5 Fractional Hs - Sobolev Spaces

1/2 α α 2 α 2 ∂ f(x) ∂ f(y) f Hs(Rn) = ∂ f(x) dx + | − n+2s | dxdy k k  Rn | | Rn Rn x y  α k Z α =k Z Z |X|≤ |X| | − |   5.6 Worked Exercies and Details

Details Hamiltonian.

N N N a N R = ′′ ′ ′ + ′ + ′ (5.229) 00 2a − 4a N a ra ′ N N N a a R = ′′ ′ ′ + ′ + ′ (5.230) 11 −2N − 4N N a ra ′ r N a 1 R = 1 ′ + ′ (5.231) 22 − 2a N a − a 2 R33 = R22 sin θ. (5.232)

Thus the Ricci scalar

612 2 k + (a/N)(a/N¨ )/c2 + (a/N˙ ) /c2 R = gµν R = 6 (5.233) µν − (a/N)2

2 2 2 G00 = 3R/R˙ c + 3k/R (5.234) k + 2RR/c¨ 2 + R˙ 2/c2 G = (5.235) 11 − 1 kσ2 −

Details Hamiltonian.

(a) Find the action for isotropic spacetimes with scalar matter ﬁeld φ(x).

(b) Check that the equations of motion of this action gives the Friedmann equation

a˙ 2 16πG = ρ(a) (5.236) a 3 

2 3 3 1 pφ 3 ρ(a)= a− (a)= a− 3 + a V (φ) (5.237) H 2 a !

(a)

1 1 S = d4x√ gR + d4x√ g (∂ φ)2 V (φ) , (5.238) 16πG − − 2 µ − Z Z where R is the Ricci scalar curvature and V is a matter potential.

ds2 = N 2(t) dt2 a2(t) dσ2 (5.239) −

1 dσ2 = dr2 + r2( dθ2 + sin2 θ + dφ2) (5.240) 1 Kr2 −

2 1 N 0 0 0 N 2 0 0 0 a2 1+Kr2 0 1 −Kr2 0 0 µν 0 − a2 0 0 (gµν )=  − 2 2  , (g )=  1  . 0 0 a r 0 0 0 a−2r2 0 − 2 2 2 1  0 0 0 a r sin θ   0 0 0 − 2   −   a2r2 sin θ     (5.241)

613 1 R2 g = g00 = 1 g = = − 00 11 g11 1 kσ2 − 1 1 g = = R2σ2 g = = R2σ2 sin2 θ. (5.242) 22 g22 − 33 g33 −

N 2a6r2 sin2 θ det(g )= (5.243) − µν 1 Kr2 −

1 r2 1/2 1 φ˙2 d4x√ g (∂ φ)2 V (φ) = dt dr dθ sin θ dφ Na3 V (φ) − 2 µ − 1 Kr2 2 N 2 − Z Z Z − " # 1 φ˙2 = dtN(t) a3 a3V (φ) K 2 N 2(t) − Z ( ) (5.244) where K

r = dr (5.245) K (1 Kr2)1/2 Z − φ = φ(r)

1 ∂ ∂µφ = φ˙2 + 0 (5.246) µ N 2

3 a˙ 2 1 φ˙2 S = dtN(t) a K + a3 a3V (φ) , (5.247) 8πG − N 2(t) 2 N 2(t) − Z ( )

(b)

∂ d ∂ L L = 0 (5.248) ∂a − dt ∂a˙

∂ 3 a˙ 2 3 φ˙2 L = N(t) K + a2 3a2V (φ) (5.249) ∂a 8πG − N 2(t) 2 N 2(t) − ( )

˙ ˙ ¨ d ∂ 6 d φ 6 ˙ φ φ L = = N(t) 2 (5.250) dt ∂a˙ −8πG dt N(t)! 8πG N (t) − N(t)

614 3 a˙ 2 6 φ¨ φ˙ 3 φ˙2 N(t) K = N˙ (t) + N(t)3a2V (φ) a2 (5.251) 8πG − N 2(t) 8πG N(t) − N 2(t) − 2 N(t) !

Singularity Theorems

Details Hamiltonian. there would not be any null geodesic γ which is directed along a princpal null direction at six or more of its points.

Details Hamiltonian.

may be topologically embedded in a finite dimensional Euclidean space, so is Lindelof. If M M K is non-compact, there is an open cover with nonfinite subcover; by paracompactness, there is a locally finite open refinement of this; since the space is Lindelof, there is a countable subcover of this. Since simple regions form a basis, this final cover may be taken to consist of simple regions.

Not sure if I want this included?

Details Hamiltonian.

µ 1 µ ν the matrix A is non-singular between p and q, Y = ν(A− ) νX . P 1 d 1 d 2ω = (A− ) A (A− ) A (5.252) αβ γβ dt αγ − γα dt βγ

(5.78)

b d2 L = A Y ν (A Y γ)+ R dT aT A Y γ dt (5.253) µν dt2 µγ µab d µγ Za (5.63)

b d d d2 L = lim ǫ 0 A Y β 2 A Y γ + A Y β (5.254) → − αβ dt µγ dt µγ dt2 Za+ǫ 615 (We take the limit because the second derivative of Y β may not be well deﬁned at q.) Integrating the second term by parts

b d2 b d d A Y βA Y γ = (A Y βA ) Y γ dt αβ µγ dt2 − dt αβ µγ dt Za Za b d d d d = A Y β A Y γ + (A Y β)(A Y γ ) − αβ dt µγ dt αβ dt µγ dt Za d d + Y βA A Y β dt (5.255) µγ dt αβ dt substituting this into (5.254)

sp d d d d L = Σ (A Y β)(A Y δ)+ Y β ( A )A A A (5.256) − αβ dt αδ dt dt αβ αδ − αβ dt αδ Z0 

d d ( A )A A A = 2A ω A = 0. (5.257) dt αβ αδ − αβ dt αδ − αδ αγ αδ 

L 0. (5.258) ≤

empty satisfy E1′ (φ′′1) = 0, E2′ (φ′′2) = 0 where E1′ is the Einstein operator with the coeﬃcients A′1, B′1 and C′1 determined by φ′1.

Since the coeﬃecients

616