Geometry of Vaidya spacetimes
GGeometry of Vaidya spacetimes
Armand COUDRAY & Jean-Philippe NICOLAS Abstract.
We investigate the geometrical structure of Vaidya’s spacetime in the case of awhite hole with decreasing mass, stabilising to a black hole in finite or infinite time or evaporatingcompletely. Our approach relies on a detailed analysis of the ordinary differential equationdescribing the incoming principal null geodesics, among which are the generators of the pasthorizon. We devote special attention to the case of a complete evaporation in infinite timeand establish the existence of an asymptotic light-like singularity of the conformal curvature,touching both the past space-like singularity and future time-like infinity. This singularity ispresent independently of the decay rate of the mass. We derive an explicit formula that relatesdirectly the strength of this null singularity to the asymptotic behaviour of the mass function.
Keywords.
Black hole, white hole, evaporation, Vaidya metric, Einstein equations withmatter, null singularity.
Mathematics subject classification.
Contents
In 1959, P.C. Vaidya published a paper [13] in which he was solving a long standing openproblem in general relativity: finding a modification of the Schwarzschild metric in order toallow for a radiating mass. His original derivation of the metric was based on Schwarzschildcoordinates. Ten years later, in [14], he observed that using instead what is now known asEddington-Finkelstein coordinates would simplify the construction a great deal. Vaidya’s metric LMBA, UMR CNRS 6205, Department of Mathematics, University of Brest, 6 avenue Victor Le Gorgeu,29200 Brest, France. Emails : [email protected], [email protected]. a r X i v : . [ g r- q c ] J a n s a solution to the Einstein equations with matter in the form of null dust and it has since beenthe object of numerous studies; see the book by J.B. Griffiths and J. Podolsky [7], section 9.5,for a very clear presentation of the metric and an excellent account of the history of theseinvestigations. Many of these works aimed at gluing some part of Vaidya’s spacetime with otherexact spacetimes like Schwarzschild and Minkowski, in order to construct models for evaporatingor collapsing black holes, see for instance W.A. Hiscock [9]. A different approach consists instudying the maximal extension (not necessarily analytical) and the matching of the Vaidyaexterior with some interior metric. This has been done under explicit assumptions on themanner in which the mass varies with time, for instance in [1]. The general problem was firsttackled by W. Israel [10] and more recently by F. Fayos, M.M. Martin-Prats and M.M. Senovilla[4] and F. Fayos and R. Torres [5], this last work studying the possibility of a singularity-freegravitational collapse.In this paper, we study the geometry of Vaidya’s spacetime itself, without extension, in asmuch generality as possible. We treat the case of a radiating white hole that emits null dust andas a result sees its mass decrease. The case of a black hole whose mass increases due to incomingnull dust is obtained by merely reversing the arrow of time. We make only minimal naturalassumptions on the behaviour of the mass function. In the existing literature, many of theprecise behaviours of null geodesics are studied numerically. In contrast, the goal of the presentpaper is to give a mathematical derivation of the geometrical features of Vaidya’s spacetime,by analysing precisely the ordinary differential equation describing the incoming principal nullgeodesics, among which are the generators of the past horizon. This equation is well-known andappears explicitly in I. Booth and J. Martin [3], in which a notion of distance between the eventhorizon and an apparent horizon is introduced. This is done for an evaporating white hole that,asymptotically or in finite retarded time, stabilises to a black hole or evaporates completely.We devote special attention to the case of a complete evaporation in infinite time (the casewhere the mass function vanishes in finite time has been studied in details in [5], with a massfunction that is possibly non differentiable at its vanishing point), for which the spacetime isalready maximally extended and, we prove the existence of a light-like singularity of the Weyltensor, touching both the past spacelike singularity and future timelike infinity . This classicalevaporation in infinite time may perhaps be seen as a classical analogue of the final pop inblack hole quantum evaporation, with the emission of a gravitational wave propagating along alight-cone.The article is structured as follows. In Section 2, we recall the construction of Vaidya’sspacetime as a simple modification of the Schwarzschild metric expressed in retarded Eddington-Finkelstein coordinates ( u = t − r ∗ , r, θ, ϕ ) and we recall the essential features of the geometry:Petrov-type D, principal null directions, expression of the Christoffel symbols, Weyl and Riccitensors in the Newman-Penrose formalism, scalar curvature and also the curvature scalar (alsoreferred to as the Kretschmann scalar). We also present our natural assumptions on the massfunction, which are essentially that it is smooth, decreases and admits finite limits as the retardedtimes tends to ±∞ . Finally, we derive the ordinary differential equation that allows to locate Note that F. Fayos and R. Torres [5] also exhibit a null singularity but under slightly different assumptions.The evaporation ends in finite time, therefore the singularity is not asymptotic but well present in the spacetime.Moreover the evaporation is allowed to end brutally: the mass vanishes in finite retarded time with a possiblynon-zero slope, in which case it is non-differentiable at its vanishing point. v = t + r ∗ in the Schwarzschild case.In Section 4, we give further properties of the principal null geodesics in the case of a completeevaporation in infinite time and we prove that they all end up in the future at a light-likeconformal curvature singularity. This singularity is present independently of the speed at whichthe mass function approaches zero in the future, however its strength seems to be directly relatedto the decay rate of the mass. We also construct families of timelike curves that end up either atthe null singularity or at timelike infinity depending of their “mass”, i.e. the rate of their propertime as measured with respect to the retarded time. The section ends with Penrose diagrams ofVaidya’s spacetime in the case of a complete evaporation in infinite time, showing the variouscongruences.All formal calculations of connection coefficients and curvature tensors have been done usingSage Manifolds [6]. Notations.
Throughout the paper, we use the formalisms of abstract indices, Newman-Penrose and 2-component spinors.
The Vaidya metric can be constructed as follows. We start with the Schwarzschild metric g = (cid:18) − Mr (cid:19) d t − (cid:18) − Mr (cid:19) − d r − r d ω , d ω = d θ + sin θ d ϕ . We express it in outgoing Eddington-Finkelstein coordinates ( u, r, θ, ϕ ), where u = t − r ∗ and r ∗ = r + 2 M log( r − M ): g = (cid:18) − Mr (cid:19) d u + 2d u d r − r d ω . Vaidya’s metric is then obtained simply by allowing the mass M to depend on u : g = (cid:18) − M ( u ) r (cid:19) d u + 2d u d r − r d ω . (1)Throughout the paper, we assume that M is a smooth function of the retarded time u (2)and we denote ˙ M ( u ) := d M d u .
00 0 = − M ( u ) r , Γ
02 2 = r , Γ
03 3 = r sin θ , Γ
10 0 = − r ˙ M ( u ) − rM ( u ) + 2 M ( u ) r , Γ
10 1 = M ( u ) r , Γ
12 2 = − r + 2 M ( u ) , Γ
13 3 = − r sin θ + 2 M ( u ) sin θ , Γ
21 2 = 1 r , Γ
23 3 = − sin θ cos θ , Γ
31 3 = 1 r , Γ
32 3 = cos θ sin θ . The Weyl tensor has Petrov type D (see [12] for the Petrov classification of the Weyl tensor interms of the multiplicities of its principal null directions), i.e. it has two double principal nulldirections that are given by V = ∂∂r , W = ∂∂u − F ∂∂r . (3)This is well known (see [7]) and can be checked easily by observing that V and W both satisfythe condition ensuring that they are at least double roots of the Weyl tensor (see R. Penrose,W. Rindler [11] Vol. 2, p. 224) C abc [ d V e ] V b V c = C abc [ d W e ] W b W c = 0 . We consider a null tetrad built using the principal null vectors above l = V ,n = W ,m = 1 r √ (cid:18) ∂∂θ + i sin θ ∂∂ϕ (cid:19) , ¯ m = 1 r √ (cid:18) ∂∂θ − i sin θ ∂∂ϕ (cid:19) . It is a normalised Newman-Penrose tetrad, i.e. l a l a = n a n a = m a m a = ¯ m a ¯ m a = l a m a = n a m a = 0 , l a n a = − m a ¯ m a = 1 . Let { o A , ι A } be the spin-frame (a local basis of the spin-bundle S A that is normalised, i.e. o A ι A = 1) defined uniquely up to an overall sign by l a = o A ¯ o A (cid:48) , n a = ι A ¯ ι A (cid:48) , m a = o A ¯ ι A (cid:48) , ¯ m a = ι A ¯ o A (cid:48) . Since the spacetime has Petrov type D, the Weyl spinor Ψ
ABCD has only one non-zero componentwhich is Ψ = Ψ ABCD o A o B ι C ι D = − M ( u ) r . The Ricci tensor is non-zero Ric ( g ) = − M ( u ) r d u ⊗ d u g = 0 , and the only non-zero Newman-Penrose scalar for the Ricci tensor isΦ = 12 R ab n a n b = − ˙ M ( u ) r . The curvature scalar, or Kretschmann scalar, is the total contraction of the Riemann tensorwith itself. It is related to the analogous invariant for the Weyl tensor by the following formula(see C. Cherubini, D. Bini, S. Capozziello, R. Ruffini [2]) k := R abcd R abcd = C abcd C abcd + 2 R ab R ab −
13 Scal g . For Vaidya’s spacetime, we have k = R abcd R abcd = C abcd C abcd = 48 M ( u ) r . (4)The metric (1) is defined on R u × ]0 , + ∞ [ r × S ω and describes a radiative white hole whose massvaries with time as a result of outgoing radiation carried by null dust. It is therefore natural toassume that M ( u ) is a non-increasing function of u ; this amounts to assuming that the null dusthas positive energy. Another natural assumption is that the mass has finite limits as u tends to ±∞ : lim u →±∞ M ( u ) → M ± with 0 ≤ M + < M − < + ∞ . (5)In the case where M is constant on ] − ∞ , u − ] and on [ u + , + ∞ [ with −∞ < u − < u + < + ∞ ,we have a Schwarzschild white hole of mass M − which emits a burst of null radiation betweenthe retarded times u − and u + and eventually stabilises to a Schwarzschild black hole of mass M + < M − (unless M + = 0, in which case the white hole evaporates completely in finite time).If M + >
0, the future event horizon is at r = 2 M + but the location of the past horizon isnot so clear. For u < u − , it is located at r = 2 M − and it is a null hypersurface with sphericalsymmetry. Therefore, it is the hypersurface generated by the family of curves indexed by ω ∈ S : γ ( u ) = ( u, r = r ( u ) , ω ) , u ∈ R , (6)that are such that r ( u ) = 2 M − for u ≤ u − and have the property of being null, i.e. g ( ˙ γ ( u ) , ˙ γ ( u )) = 1 − M ( u ) r ( u ) + 2 ˙ r ( u ) = 0 . (7)Hence, the function r ( u ) satisfies the following ordinary differential equation˙ r ( u ) = − (cid:18) − M ( u ) r ( u ) (cid:19) , (8)with r > r ( u ) = 2 M − for u ≤ u − . 5 emark 2.1. We shall often (starting immediately below) identify the solutions to (8) with thecurves (6) satisfying (7) and simply refer to the integral lines of (8) . If we no longer assume that M ( u ) = M − in a neighbourhood of −∞ , the past horizon willbe spanned by solutions to (8) such that r > u →−∞ r ( u ) = 2 M − .The ODE (8) is in fact the general equation for a null curve that is transverse to the levelhypersurfaces of u (i.e. to ∇ u , that is a normal and tangent vector field to these hypersurfaces)and orthogonal to the orbits of the rotation Killing vectors. Vaidya’s spacetime comes equippedwith a null congruence, given by the lines of constant u and ω , which are the integral lines of ∇ u = g − (d u ) = ∂∂r = V , (9)where g − is the inverse Vaidya metric given by g ab ∂ a ∂ b = 2 ∂ u ∂ r − (cid:18) − M ( u ) r (cid:19) ∂ r − r − ∂ ω , (10) ∂ ω denoting the euclidean inverse metric on S . The integral lines of (8) provide us with asecond null congruence that is transverse to the first one, corresponding to the lines of constant v and ω in the case of the Schwarzschild metric. Remark 2.2.
Note that the tangent vector to the integral curves of (8) is exactly ∂∂u − F ∂∂r = W .
The integral curves of (8) are therefore the integral lines of the principal null vector field W .Since the spacetime is not vacuum, we do not have the Goldberg-Sachs Theorem that wouldensure that these are geodesics. However we shall see in Subsection 3.2 Proposition 3.3 thatthese curves are indeed geodesics; they are the family of incoming principal null geodesics andform the second natural null congruence of Vaidya’s spacetime. Similarly the integral lines of V are also geodesics (see Proposition 3.2), they are the outgoing principal null geodesics of Vaidya’sspacetime. In this section, we analyse the qualitative behaviour of solutions to Equation (8), with specialemphasis on the solutions generating the past horizon. Our main results are proved under theassumption that˙ M ( u ) < u − , u + [ , −∞ ≤ u − < u + ≤ + ∞ , ˙ M ≡ u with constant mass alternate with intervals on which the mass decreases.We also ignore, for similar reasons, cases where ˙ M vanishes at isolated points.6 .1 General properties We start with an obvious observation.
Lemma 3.1.
On an interval ] u , u [ on which ˙ M ( u ) does not vanish everywhere, r ( u ) cannotbe identically equal to M ( u ) . Proof. If r ( u ) = 2 M ( u ) satisfies (8) on ] u , u [, then2 ˙ M ( u ) = − (cid:18) − M ( u )2 M ( u ) (cid:19) = 0 on ] u , u [ , which contradicts the assumption.Then, we give an important estimate that is a consequence of the local uniqueness of solutionsto the Cauchy problem for (8). Lemma 3.2.
Let (] u , u [ , r ) be a solution to (8) such that, for a given u ∈ ] u , u [ , we have r ( u ) ≥ M ( u ) . Let us assume that ˙ M ( u ) < for all u ∈ ] u , u [ , then r ( u ) > M ( u ) on ] u , u [ . Proof.
First, note that if r ( u ) = 2 M ( u ), then ˙ r ( u ) = 0, while ˙ M ( u ) <
0, hence thereexists ε > u , u + ε [ we have r ( u ) > M ( u ). If r ( u ) > M ( u ), then we havethe same conclusion by continuity.Now, let u be the lowest value of u in ] u , u [ such that r ( u ) = 2 M ( u ). Then (8) impliesthat ˙ r ( u ) = 0 > M ( u ) and therefore, there exists δ > r ( u ) < M ( u ) in ] u − δ, u [.By continuity of r and M , there exists u ∈ ] u , u [ such that r ( u ) = 2 M ( u ). This contradictsthe assumptions on u . It follows that r ( u ) > M ( u ) on ] u , u [.The asymptotic behaviour of maximal solutions to (8) in the past is unstable. One solutionhas a finite limit 2 M − ; it corresponds to the past event horizon. All other solutions either endat past null infinity or reach the past singularity in finite retarded time. The following theoremgives a complete classification of the solutions to (8) in terms of their behaviour in the past andalso describes precisely their behaviour in the future. Theorem 3.1.
Under Assumptions (2) , (5) and (11) , there exists a unique maximal solution r h to (8) such that lim u →−∞ r h ( u ) = 2 M − . • If either M + > or u + = + ∞ , r h exists on the whole real line, r h ( u ) → M + as u → + ∞ and any other maximal solution r to (8) belongs to either of the following two categories:1. r exists on the whole real line, r ( u ) > r h ( u ) for all u ∈ R , lim u →−∞ r ( u ) = + ∞ and lim u → + ∞ r ( u ) = 2 M + ;2. r exists on ] u , + ∞ [ with u ∈ R and satisfies: r ( u ) < r h ( u ) for all u ∈ ] u , + ∞ [ , lim u → u r ( u ) = 0 and lim u → + ∞ r ( u ) = 2 M + . • If M + = 0 and u + < + ∞ , r h exists on an interval ] − ∞ , u [ with u + ≤ u < + ∞ and lim u → u r h ( u ) = 0 . The other maximal solutions are of two types: . r exists on ] −∞ , u [ with u ≤ u < + ∞ , r ( u ) > r h ( u ) on ] −∞ , u [ , lim u → u r ( u ) = 0 and lim u →−∞ r ( u ) = + ∞ ;2. r exists on ] u , u [ with −∞ < u < u ≤ u , r ( u ) → as u tends to either u or u and r ( u ) < r h ( u ) on ] u , u [ . Proof.
Step 1: uniqueness of a maximal solution with finite limit as u → −∞ . First, if a solution r exists on an interval of the form ] − ∞ , u [ and has a finite limit at −∞ , then this limitmust be 2 M − . Indeed let us denote this limit by l , using (8),lim u →−∞ ˙ r ( u ) = − (cid:18) − M − l (cid:19) . So ˙ r also has a finite limit at −∞ and this limit must be zero in order not to contradictthe finite limit of r ( u ), i.e. l = 2 M − .Then let us show that there is at most one solution to (8) defined on an interval of theform ] − ∞ , u [ such that lim u →−∞ r ( u ) = 2 M − . Let us assume that there are two such solutions r and r . Then ψ = r − r satisfies˙ ψ ( u ) = Mr − Mr = − Mr r ψ (12)and lim u →−∞ ψ ( u ) = 0 . However, since lim u →−∞ − Mr r = − M − < , if follows that unless ψ is identically zero,(log( | ψ | )) (cid:48) −→ − M − as u → −∞ and ψ blows up exponentially fast at −∞ . Since we know that ψ tends to zero at −∞ , weconclude that ψ is identically zero, i.e. r = r . Step2: construction of the past horizon.
Now we construct a solution to (8) that tends to 2 M − at −∞ . Let us first consider the case where u − = −∞ . For each n ∈ N , we define r n to bethe maximal solution to (8) such that r n ( − n ) = 2 M ( − n ). It exists on an interval of theform ] u n , u n [, u n < − n < u n . Let u n = min { u n , u + } . By Lemma 3.2, r n ( u ) > M ( u ) on] − n, u n [, hence ˙ r n < r n ( u ) < M ( − n ). These a priori boundsimply that u n ≥ u + . Therefore, u n = + ∞ in the case where u + = + ∞ . If u + < + ∞ and M + >
0, then we could have r n ( u + ) = 2 M + , in which case u n = + ∞ and r n ( u ) = 2 M + for u ≥ u + , or r n ( u + ) > M + and then r n ( u ) > M + on [ u + , u n [ since two solutions cannot8ross, whence ˙ r n is negative there and we infer u n = + ∞ . If u + < + ∞ and M + = 0 thenon ] u + , u n [, r n satisfies the simple ODE ˙ r n = − , and r n reaches 0 in finite retarded time. Hence in this case u n < + ∞ .Using again the fact that, by uniqueness of solutions to the Cauchy problem for (8), twosolutions cannot cross, we infer that the sequence u n is increasing. Let u = lim n → + ∞ u n . For any compact interval I of ] −∞ , u [, there exists n ∈ N such that the sequence ( r n ) n ≥ n is well-defined, increasing and bounded on I . Hence, Lebesgue’s dominated convergencetheorem implies that the sequence ( r n ) converges in L (] − ∞ , u [) towards a positivefunction r h such that 2 M ( u ) ≤ r h ( u ) ≤ M − ∀ u ∈ ] − ∞ , u [ . (13)Moreover, 1 /r n also converges towards 1 /r h in L (] − ∞ , u [), because, for any givencompact interval I , it is a well-defined, decreasing and bounded sequence on I for n largeenough. This implies, by equation (8) for r n , that ˙ r n converges in L (] − ∞ , u [) and byuniqueness of the limit in the sense of distributions, the limit must be ˙ r h . Consequently r h is a solution to (8) in the sense of distributions. An easy bootstrap argument then showsthat r h is a strong solution to (8) and is in fact smooth on ] − ∞ , u [.Besides, by (13) and the fact that M ( u ) → M − as u → −∞ , it follows thatlim u →−∞ r h ( u ) = 2 M − . In the case where u − > −∞ , we simply need to consider the maximal solution to (8) suchthat r ( u − −
1) = 2 M − . This solution exists on an interval of the form ] − ∞ , u [ andsatisfies (13).Let us now turn to the value of u and the behaviour of r h in the future. If either M + > u + = + ∞ , then by (13), we have u = + ∞ . By (13) again, ˙ r h ( u ) < R and itfollows that r h ( u ) has a finite limit l as u → + ∞ . If M + >
0, we have˙ r h ( u ) → − (cid:18) − M + l (cid:19) as u → + ∞ and we must have l = 2 M + or contradict the finite limit of r h . If M + = 0 and u + = + ∞ ,then if l (cid:54) = 0, ˙ r h ( u ) → −
12 as u → + ∞ which is incompatible with u = + ∞ . Finally, if M + = 0 and u + < + ∞ , then (13) impliesthat u ≥ u + . If r h ( u + ) = 0 then the solution terminates at u = u + , u = u + . Otherwise, u > u + and on [ u + , u [ we have ˙ r h ( u ) = − ,
9s long as r h ( u ) remains positive. Therefore, we have r h ( u ) = r h ( u + ) − ( u − u + ) for u + ≤ u ≤ u + + r h ( u + )and the integral curve ends at u = u + + r h ( u + ), i.e. u is finite and is equal to u + + r h ( u + ).Hence for M + = 0 and u + < + ∞ , the past event horizon vanishes in finite retarded time u + + r h ( u + ) and there is no future event horizon. Step 3: classification of the other maximal solutions. • We begin with the case where either u + = + ∞ or M + >
0. Let (] u , u [ , r ) be amaximal solution to (8). Let u ∈ ] u , u [ and assume that 0 < r ( u ) < r h ( u ) (resp. r ( u ) > r h ( u )). By uniqueness of solutions to the Cauchy problem for (8), solutionscannot cross, so for all u ∈ ] u , u [ we have 0 < r ( u ) < r h ( u ) (resp. r ( u ) > r h ( u )).1. Case where 0 < r ( u ) < r h ( u ). Let us first assume that r ( u ) > M ( u ). If r ( u ) > M ( u ) on its interval of existence, then r ( u ) is bounded between 2 M ( u )and r h ( u ) and we must have ] u , u [ = R . However, we then have r ( u ) → M − as u → −∞ and this contradicts the uniqueness of r h . It follows that there exists u ∈ ] u , u [ such that r ( u ) = 2 M ( u ). Therefore r ( u ) < M ( u ) on ] u , u [(the proof is similar to that of Lemma 3.2), r is an increasing function on thisinterval and ˙ r is decreasing. This implies that r ( u ) must reach 0 in finite time inthe past and keep on existing towards the past as long as it has not reached 0.Hence u > −∞ and r ( u ) → u → u . Since r ( u ) = 2 M ( u ), then we haveby Lemma 3.2 that r ( u ) > M ( u ) on ] u , u + [ and since solutions do not cross, r ( u ) ≥ M ( u ) on [ u + , u [. Hence2 M ( u ) ≤ r ( u ) < r h ( u ) on ] u , u [ . This implies that u = + ∞ and lim u → + ∞ r ( u ) = 2 M + .If r ( u ) = 2 M ( u ), then we can repeat the arguments above, replacing u by u ;we infer: u > −∞ and lim u → u r ( u ) = 0, u = + ∞ and lim u → + ∞ r ( u ) = 2 M + .If r ( u ) < M ( u ) then r ( u ) increases as long as r ( u ) < M ( u ). Either r ( u ) < M ( u ) on its whole interval of existence (note that this requires M + > u = + ∞ , or there exists u ∈ ] u , u [ such that r ( u ) = 2 M ( u ) andwe can then use the same reasoning as before on ] u , u [ and infer that u = + ∞ .In the latter case, we have as before lim u → + ∞ r ( u ) = 2 M + . In the former, r ( u )has a finite positive limit l as u → + ∞ and (recall that we must have M + > r ( u ) → − (cid:18) − M + l (cid:19) as u → + ∞ and we must have l = 2 M + in order not to contradict u = + ∞ . In both cases,we have u > −∞ and r ( u ) → u → u .2. If r ( u ) > r h ( u ) then r is a decreasing function on its interval of existence andis bounded below by r h . This implies that u = + ∞ . Moreover, on its wholeinterval of existence, r satisfies − < ˙ r ( u ) < u = −∞ . Since r is a decreasing function on R , it has a limitas u → −∞ and we have seen above that this limit cannot be finite, hencelim u →−∞ r ( u ) = + ∞ . The solution r also has a limit l as u → + ∞ and since r is a decreasing functionand r ( u ) > r h ( u ) > M ( u ) on R , it follows that 2 M + ≤ l < + ∞ . If M + > l > r also has a limit as u → + ∞ given bylim u → + ∞ ˙ r ( u ) = − (cid:18) − M + l (cid:19) . This must be zero in order not to contradict the finite limit of r . Hence l = 2 M + .If M + = 0, then unless l = 0 we have thatlim u → + ∞ ˙ r ( u ) = − r ( u ) must reach 0 in finite retarded time and contradicts u = + ∞ . Hence in this case we have l = 0 = 2 M + . • In the case where M + = 0 and u + < + ∞ , the proof uses exactly the same argumentsas in step 3 and the end of step 2.If the mass decreases only for a finite range of u , we have not been able to rule out cases forwhich we have r h ( u + ) = 2 M + , nor have we managed to find explicit examples of this situation.It is however easy to see that there are cases for which r h ( u ) > M + for all u ∈ R . Proposition 3.1.
In the case where u ± are both finite, assume that u + − u − < M − − M + ) ,then r h ( u + ) > M + and therefore r h ( u ) > M + for all u ∈ R . Proof.
It is a simple observation. We have r h ( u + ) − r h ( u − ) = (cid:90) u + u − ˙ r h ( u )d u = − (cid:90) u + u − (cid:18) − M ( u ) r h ( u ) (cid:19) d u = −
12 ( u + − u − ) + (cid:90) u + u − M ( u ) r h ( u ) d u > −
12 ( u + − u − ) . Since r h ( u − ) = 2 M − , r ( u + ) > M − −
12 ( u + − u − ) > M + . Since on ] u + , + ∞ [, 2 M + is a solution to (8), then by uniqueness of solutions we must have r h ( u ) > M + for all u > u + . This proves the proposition.11 .2 The second optical function The function u is an optical function, which means that its gradient is a null vector field, orequivalently that u satisfies the eikonal equation g ( ∇ u, ∇ u ) = 0 . (14)An important property of optical functions is that the integral lines of their gradient are nullgeodesics with affine parametrisation. This is established in [8]. The more complete Propositions(7.1.60) and (7.1.61) in Penrose and Rindler Vol 2 [11] state that for a null congruence, thefollowing three properties are equivalent :1. it is hypersurface-orthogonal;2. it is hypersurface-forming;3. it is geodetic and twist-free.We recall the proof of the fact that the integral curves of an optical function are null geodesics,as it is a straightforward calculation. Lemma 3.3.
Let ξ be an optical function and denote L = ∇ ξ . The integral curves of L aregeodesics and L corresponds to a choice of affine parameter, i.e. ∇ L L = 0 . Proof.
The proof is direct : ∇ L L b = ∇ ∇ ξ ∇ b ξ , = ∇ a ξ ∇ a ∇ b ξ , = ∇ a ξ ∇ b ∇ a ξ since the connection is torsion-free,= ∇ b ( ∇ a ξ ∇ a ξ ) − (cid:16) ∇ b ∇ a ξ (cid:17) ∇ a ξ , = 0 − ∇ a ξ ∇ a ∇ b ξ since ∇ ξ is null and the connection torsion-free,= −∇ ∇ ξ ∇ b ξ . Since (see (9)) ∇ u = ∂∂r = V and V is a principal null direction of the Weyl tensor, a consequence of Lemma 3.3 and of (14)is the following. Proposition 3.2.
The integral lines of V are affinely parametrised null geodesics; they are theoutgoing principal null geodesics of Vaidya’s spacetime. We now establish the existence of a second optical function.12 roposition 3.3.
There exists a function v defined on R u × ]0 , + ∞ [ r × S ω , depending solelyon u and r , such that ∇ v is everywhere tangent to the integral lines of (8) . This means that g ( ∇ v, ∇ v ) = 0 , i.e. v is an optical function. The integral lines of (8) , which are also the integrallines of ∇ v , are therefore null geodesics and their congruence generates the level hypersurfacesof v . Since the integral lines of (8) are also tangent to W (defined in (3) ) it follows that theyare the incoming principal null geodesics of Vaidya’s spacetime. Proof.
The metric g can be written as g = F d u (cid:0) d u + 2 F − d r (cid:1) − r d ω . Following the construction of v for the Schwarzschild metric, it is tempting to putd v = d u + 2 F − d r = 2 F − g − ( W ) , however this 1-form is not closed sinced (cid:0) d u + 2 F − d r (cid:1) = − F − ∂F∂u d u ∧ d r = 4 ˙ M ( u ) rF d u ∧ d r which vanishes identically only if the mass M is constant, i.e. in the Schwarzschild case. Weintroduce an auxiliary function ψ > g = Fψ d u (cid:0) ψ d u + 2 ψF − d r (cid:1) − r d ω . Our purpose is to find conditions on ψ that ensure that the 1-form α := ψ d u + 2 ψF − d r isexact. Since we work in the variables ( u, r ) on the simply connected domain R u × ]0 , + ∞ [ r , allthat is required is that α be closed, i.e. thatd α = 2 ∂∂u (cid:18) ψF (cid:19) − ∂ψ∂r = 0 . This equation has the more explicit form ∂ψ∂u − F ∂ψ∂r + 2 F ˙ Mr ψ = 0 . (15)This is an ordinary differential equation along the integral lines of the second principal nulldirection (defined in (3)) W = ∂∂u − F ∂∂r parametrised by u . Let γ ( u ) = ( u, r ( u ) , ω ) be an integral line of W (which is equivalent to r ( u )being a solution to (8)), Equation (15) along γ readsdd u ( ψ ◦ γ ) = (cid:32) − MrF ψ (cid:33) ◦ γ ,
13r equivalently dd u (log | ψ ◦ γ | ) = (cid:32) − MrF (cid:33) ◦ γ . (16)Equation (15) can therefore be integrated as follows. First, we take a hypersurface transverseto all the integral lines of (8), for instance S = { u = 0 } and, we fix the value of ψ on S , say ψ = 1 on S . Then we evaluate ψ on each integral line of (8) by solving the ODE (16). Since theintegral lines of (8) are a congruence of R u × ]0 , + ∞ [ r × S ω , this allows to define ψ on this wholedomain as a smooth (by smooth dependence on initial data) and nowhere vanishing function.The 1-form α is then closed on R u × ]0 , + ∞ [ r × S ω . Since α depends only on u and r , we may seeit as a closed 1-form on R u × ]0 , + ∞ [ r which is simply connected. It follows that α is exact on R u × ]0 , + ∞ [ r × S ω and modulo a choice of hypersurface Σ generated by the integral lines of (8),we can define a function v on R u × ]0 , + ∞ [ r × S ω such that v = 0 on Σ and α = d v . In particular,d v = 2 ψF − g − ( W ) and ∇ v = 2 ψF − W .
We now devote particular attention to the case where M + = 0 and u + = + ∞ . As before,we assume that ˙ M < u − , + ∞ [. As we have established in Theorem 3.1, the past event horizon ends up at r = 0 as u → + ∞ and so do all the integral curves of (8), i.e. all the incoming principal null geodesics. From this,we infer the following theorem. Theorem 4.1.
Whatever the speed at which M ( u ) → as u → + ∞ , we have a null singularityof the conformal structure in the future of our spacetime. More precisely, the Kretschmannscalar k does not remain bounded as u → + ∞ along any integral line of (8) . Proof.
Consider (] u , + ∞ [ , r ) a maximal solution to (8), with u ∈ R ∪ {−∞} . Assumethat k remains bounded along the integral line as u → + ∞ . Then, using (4), so does M/r andit follows that M/r tends to 0 as u → + ∞ along the integral line. This implies in turn that˙ r ( u ) → − / u → + ∞ , which contradicts the fact that r ( u ) → u → + ∞ . Remark 4.1.
If we assume that along the integral lines of (8) , ˙ r ( u ) has a limit as u → + ∞ ,this limit is necessarily zero in order not to contradict the fact that r ( u ) → as u → + ∞ . Thisimplies in turn that along the integral line, M ( u ) r ( u ) → as u → + ∞ , i.e. r ( u ) (cid:39) M ( u ) as u → + ∞ (17) and k (cid:39) M ( u ) as u → + ∞ . (18)14 .2 A family of uniformly timelike congruences Some uniformly timelike curves also end up at r = 0 as u → + ∞ . Let us consider a curve γ ( u ) = ( u, r ( u ) , θ, ϕ ) , such that g ( ˙ γ ( u ) , ˙ γ ( u )) = ε , ε > , (19)then r satisfies the differential equation˙ r ( u ) = ε − (cid:18) − M ( u ) r ( u ) (cid:19) . (20)The tangent vector is τ = ∂ u + (cid:18) ε − (cid:18) − M ( u ) r (cid:19)(cid:19) ∂ r and ∇ τ τ = − M ( u ) r (cid:0) τ − ε ∂ r (cid:1) , so the integral curves of (20) are not geodesics, except for ε = 0. The behaviour of the integralcurves of (20) changes radically according to the value of ε . This is detailed in the next twopropositions. The first one deals with the case where 0 < ε < Proposition 4.1.
Let < ε < be given. There exists a unique maximal solution r ε to (20) such that lim u →−∞ r ε ( u ) = 2 M − − ε . This solution exists on the whole real line and r ε ( u ) → as u → + ∞ . Any other maximalsolution r to (8) belongs to either of the two categories :1. r exists on the whole real line, r ( u ) > r ε ( u ) for all u ∈ R , lim u →−∞ r ( u ) = + ∞ and lim u → + ∞ r ( u ) = 0 ;2. r exists on ] u , + ∞ [ with u ∈ R and satisfies : r ( u ) < r ε ( u ) for all u ∈ ] u , + ∞ [ and r ( u ) tends to as u → u and as u → + ∞ .Moreover, the Kretschmann scalar k fails to be bounded on the integral lines of (20) as u → + ∞ .If we assume moreover that ˙ r has a limit as u → + ∞ along the integral lines of (20) , then wehave a similar behaviour for k to that described in Remark 4.1 for the integral lines of (8) ,namely k (cid:39) − ε ) M ( u ) as u → + ∞ . (21)The second proposition treats the cases where ε ≥ Proposition 4.2. If ε ≥ , then all maximal solutions to (20) exist on a interval ] u , + ∞ [ with u > −∞ , r is strictly increasing on ] u , + ∞ [ and r ( u ) → as u → u . Moreover: if ε = 1 , then the limit of r ( u ) as u → + ∞ is finite if and only if M ( u ) is integrable inthe neighbourhood of + ∞ ; • if ε > , then r ( u ) → + ∞ as u → + ∞ . Remark 4.2.
In view of (19) , the proper time along an integral curve of (20) is exactly given(after an adequate choice of origin) by τ = εu . Therefore, the behaviour of the integral lines of (20) as u → + ∞ described in Propositions 4.1 and 4.2 corresponds to their behaviour as propertime tends to + ∞ . Proof of Proposition 4.1.
We observe that Equation (20) can be transformed to (8). Weput ˜ r ( u ) = r ( u )1 − ε , ˜ M ( u ) = M ( u )(1 − ε ) , then Equation (20) becomes ˙˜ r ( u ) = − (cid:32) − M ( u )˜ r ( u ) (cid:33) . The classification of maximal solutions therefore follows directly from Theorem 3.1. The deriva-tion of the behaviour of the Kretschmann scalar along an integral line is also similar to the nullcase. The proof of the lack of boundedness is the same as that of Theorem 4.1 and assumingthat ˙ r has a limit as u → + ∞ along an integral line of (20), this limit must be zero and we infer r ( u ) (cid:39) M ( u )1 − ε . Then (21) follows from (4).
Proof of Proposition 4.2.
Let (] a, b [ , r ) be a maximal solution to (20). • Case where ε = 1 . The differential equation (20) becomes˙ r ( u ) = M ( u ) r ( u ) , or equivalently dd u (( r ( u )) ) = 2 M ( u ) . The function r is strictly increasing and given u ∈ ] a, b [, we have for all u ∈ ] a, b [ r ( u ) = r ( u ) + 2 (cid:90) uu M ( s )d s . Then a is finite, strictly lower than u and is precisely such that (cid:90) u a M ( s )d s = r ( u ) . Also b = + ∞ and lim u → + ∞ r ( u ) = r ( u ) + 2 (cid:90) + ∞ u M ( s )d s . Case where ε > . Now for all u ∈ ] a, b [,˙ r ( u ) > ε − > . (22)It follows that a is finite and lim u → a r ( u ) = 0 . Moreover, let u ∈ ] a, b [, then using the fact that r is strictly increasing on ] a, b [, we havefor all u ∈ ] u , b [, ˙ r ( u ) < ε −
12 + M + r ( u ) , whence b = + ∞ and (22) implies lim u → + ∞ r ( u ) = + ∞ . The two congruences of null geodesics that we have considered (the curves of constant ( u, ω )and of constant ( v, ω )) are inextendible. The spacetime is therefore maximally extended andwe have two global charts R u × ]0 , + ∞ [ r × S ω and R v × ]0 , + ∞ [ r × S ω . Figures 1 to 4 display thePenrose diagram of Vaidya’s spacetime for a mass function that decreases strictly on the wholereal line, tends to 0 as u → + ∞ and to a finite positive limit M + as u → −∞ , with the generalforms of various congruences: the incoming principal null geodesics (lines of constant ( v, ω )) inFigure 1, the outgoing principal null geodesics (lines of constant ( u, ω )) in Figure 2, the timelikecurves given by the integral lines of (20) for 0 < ε < ε ≥ v, ω ). Figure 2: Outgoing principal null congru-ence: lines of constant ( u, ω ).17igure 3: The timelike congruence for ε <
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