aa r X i v : . [ g r- q c ] D ec Thin shell model revisited
Sijie Gao ∗ and Xiaobao Wang † Department of Physics, Beijing Normal University,Beijing 100875, ChinaNovember 6, 2018
Abstract
We reconsider some fundamental problems of the thin shell model.First, we point out that the “cut and paste” construction does not guar-antee a well-defined manifold because there is no overlap of coordinatesacross the shell. When one requires that the spacetime metric across thethin shell is continuous, it also provides a way to specify the tangent spaceand the manifold. Other authors have shown that this specification leadsto the conservation laws when shells collide. On the other hand, the well-known areal radius r seems to be a perfect coordinate covering all regionsof a spherically symmetric spacetime. However, we show by simple butrigorous arguments that r fails to be a coordinate covering a neighbor-hood of the thin shell if the metric across the shell is continuous. Whentwo spherical shells collide and merge into one, we show that it is possiblethat r remains to be a good coordinate and the conservation laws hold.To make this happen, different spacetime regions divided by the shellsmust be glued in a specific way such that some constraints are satisfied.We compare our new construction with the old one by solving constraintsnumerically. Since the pioneering work by Israel [1], the thin shell model has been exten-sively studied and has a wide application in gravitational collapse, cosmology,wormhole theory, etc. Such a model is important because it is an idealization ofthe real matter distribution and has given many interesting solutions in generalrelativity and alternative gravity theories.Despite great successes of this model, some fundamental problems still re-main to be answered. It is well known that a thin shell is a three dimensionalhypersurface (in four dimension spacetimes) which can be constructed by the “cut and paste” approach [2]. Let M and M be two distinct spacetimes with ∗ [email protected] † { x µ } and { x µ } . We can assign a metric g ab ( x µi ) to M i , i = 1 , i . If we wish to unifythe two spacetimes, it is natural to glue them by identifying their boundaries,i.e., the new spacetime M = M ∪ M connects the two distinct spacetimes atthe hypersurface Σ ≡ Σ = Σ (see Fig. 1). There is always a discontinuity ofthe extrinsic curvature across Σ which is related to the surface matter distribu-tion on Σ. So the first derivative of the metric is discontinuous at Σ, which leadsto the famous junction conditions [1]. It is also a convention to require that theinduce metric h ab is continuous. Howerve, there is no general requirement onthe continuity of g ab across Σ.Before we go further, we should notice that any metric is defined on a man-ifold. An overlooked question is: have we had a well-defined manifold by theabove construction? If we go through the general properties of manifold, we willfind immediately that the answer is “not yet”. A manifold allows to be coveredby more than one coordinate system. But neither x µ nor x µ covers a neigh-borhood of a point on Σ. Although both coordinate systems give coordinateson Σ, their overlap is only a three dimensional region, not an open set of themanifold [3]. An obvious consequence is that the tangent space of each pointon Σ is not uniquely defined by the construction so far. Before we glue M and M together, we only have two “half tangent spaces” at their boundaries.Identifying the boundaries does not give a unified tangent space for each pointat the boundary. There is an ambiguity for each tangent vector u a ∈ V at theboundary to find its “other half” u a ∈ V satisfying u a + u a = 0, where V and V are the tangent spaces of M and M , respectively. There are two equivalentways to fix the ambiguity. First, we can extend { x µ } to M such that there isa four dimensional overlap of { x µ } and { x µ } . By this way, M is a well-definedmanifold. Second, Note that the three dimensional tangent space of Σ has noambiguity by construction. Thus, we only need to assign one transversal vector u a in V with a negative transversal vector u a in V such that u a + u a = 0 (seeFig. 1). Then any other vector is uniquely assigned a negative vector by theaddition rule.To be definitive, we assume that Σ is timelike and the induced metric isgiven by h ab = g iab + n ia n ib , i = 1 , , (1)where n ia is the spacelike normal of Σ. Since h ab is the same from both sides,we see immediately that the spacetime metric g ab is continuous if and only if n a = n a (2)By our argument above, Eq. (2) also uniquely fixes the tangent space of anypoint on Σ. In spherically symmetric spacetimes, if a few shells collide, it hasbeen shown by Langlois, Maeda and Wands (LMW) [4] that this identificationleads to the conservation of energy and momentum at the collision point. TheLMW method has been further applied to bubble and brane collisions [6]-[8].In a spherically symmetric spacetime, the radial coordinate r is the arealradius of the sphere formed by the SO(3) isometry [3]. So r is a well-defined2igure 1: M and M are connected at Σ. For each u a , one needs to specify a u a such that u + u a = 0.function and seems to be the only natural coordinate covering different regionsdivided by the shells. However, we show in section 2 that if the metric is con-tinuous across the shell, r is no longer a good coordinate for points on the shell.Therefore, if r is a good coordinate, we must choose other identifications whichbreak down the continuity of metrics and generally violate the conservation lawsas well. In section 3, we consider the simplest collision: two shells merge into oneafter they collide. we derive some constraint equations such that r remains tobe a good coordinate and the conservation laws hold. By imposing appropriateinitial conditions, we find that these equations are solvable at least numerically. r coordinate We consider a spherical shell Σ moving in a spherical spacetime. The coordinateson the two sides of the shell are labeled by ( t , r ) and ( t , r ), where we havedropped the ( θ, φ ) coordinates for simplicity (see Fig. 2).The metrics on both sides are in the form ds i = − f i ( r ) dt + f − i ( r ) dr + r d Ω (3)where i = 1 , f i ( r ) = 1 − M i r (4)Note that r = r by continuity, but t is discontinuous across the shell. We maywrite the four-velocity of the shell as u a = ˙ t i (cid:18) ∂∂t i (cid:19) a + ˙ r (cid:18) ∂∂r i (cid:19) a (5)3igure 2: A spherical shell Σ moving with four-velocity u a .Note that we have used ˙ r instead of ˙ r i because r = r . The normalizationcondition g ab u a u b = − t i = ± s f i + ˙ r f i (6)The normal vector of Σ is of the form n ai = ˙ rf i ( r ) (cid:18) ∂∂t i (cid:19) a + p ˙ r + f i (cid:18) ∂∂r i (cid:19) a (7)Now we have three orthogonal and normal tetrads related by the followingLorentz transformation [4] (cid:18) u a n a (cid:19) = Λ( α i ) q f i (cid:16) ∂∂t i (cid:17) a √ f i (cid:16) ∂∂r i (cid:17) a (8)where Λ( α ) = (cid:18) cosh( α ) sinh( α )sinh( α ) cosh( α ) (cid:19) (9)and α i = sinh − ˙ r √ f i (10)Therefore, q f (cid:16) ∂∂t (cid:17) a √ f (cid:16) ∂∂r (cid:17) a = Λ( α − α ) q f (cid:16) ∂∂t (cid:17) a √ f (cid:16) ∂∂r (cid:17) a (11)4e should emphasize that the continuity of metric is crucial to derive thisformula. However, we show now that this treatment is inconsistent with theassumption that r is a good coordinate.As we have mentioned above, the two sets of coordinates { t , r } and { t , r } do not have a four-dimensional overlap in the neighborhood of Σ. We need firstextend the coordinates smoothly such that they have a four-dimensional overlapregion O p where p ∈ Σ. If r is a good coordinate everywhere, we should have r = r = r (12)in O p .Then we can write down the transformation at p (cid:18) ∂∂t (cid:19) a = ∂t ∂t (cid:18) ∂∂t (cid:19) a + ∂r ∂t (cid:18) ∂∂r (cid:19) a (13)The second term vanishes due to r = r . So (cid:18) ∂∂t (cid:19) a = ∂t ∂t (cid:18) ∂∂t (cid:19) a (14)Similarly, (cid:18) ∂∂r (cid:19) a = ∂t ∂r (cid:18) ∂∂t (cid:19) a + ∂r ∂r (cid:18) ∂∂r (cid:19) a = ∂t ∂r (cid:18) ∂∂t (cid:19) a + (cid:18) ∂∂r (cid:19) a (15)Note that α = α due to f = f . Therefore, Eq. (11) indicates that (cid:16) ∂∂t (cid:17) a isnot parallel to (cid:16) ∂∂t (cid:17) a . This contradicts Eq. (14).Another quick way to see the breakdown of r is to notice that the θθ com-ponent of the extrinsic curvature of Σ is given by [5] K θθ = rn r = r p f + ˙ r (16)which is obviously discontinuous across the shell. This discontinuity leads tothe junction condition. Note that n r = n a ( dr ) a (17)So if n a = n a , the discontinuity of n r indicates that r is not a qualified functionin any neighborhood of p ∈ Σ. In this section, we shall match different sides of shells in a way such that r is agood coordinate across all shells. Then we shall discuss the conservation lawswhen shells collide. For simplicity, we consider the collision of two shells. Afterthe collision, they merge as one shell. 5igure 3: Two shells collide and stick together. As shown in Fig. 3, Σ and Σ represent two shells before the collision and Σ represents the shell after the collision. The spacetime is then divided into threeparts covered by coordinates { t i , r i } , i = 2 , ,
6. Applying Eqs. (14) and (15)to Σ , we have (cid:18) ∂∂t (cid:19) a = T (cid:18) ∂∂t (cid:19) a (18) (cid:18) ∂∂r (cid:19) a = R (cid:18) ∂∂t (cid:19) a + (cid:18) ∂∂r (cid:19) a (19)where T = ∂t ∂t and R = ∂t ∂r .Similarly, on Σ and Σ we have (cid:18) ∂∂t (cid:19) a = T (cid:18) ∂∂t (cid:19) a (20) (cid:18) ∂∂r (cid:19) a = R (cid:18) ∂∂t (cid:19) a + (cid:18) ∂∂r (cid:19) a (21) (cid:18) ∂∂t (cid:19) a = T (cid:18) ∂∂t (cid:19) a (22) (cid:18) ∂∂r (cid:19) a = R (cid:18) ∂∂t (cid:19) a + (cid:18) ∂∂r (cid:19) a (23)Combining Eqs. (22), (20) and (18), we have (cid:18) ∂∂t (cid:19) a = T T T (cid:18) ∂∂t (cid:19) a (24)6hile Eqs. (23), (21) and (19) give (cid:18) ∂∂r (cid:19) a = R T T (cid:18) ∂∂t (cid:19) a + R T (cid:18) ∂∂t (cid:19) a + R (cid:18) ∂∂t (cid:19) a + (cid:18) ∂∂r (cid:19) a (25)Therefore, T T T = 1 (26) R T T + R T + R = 0 (27)On the other hand, the four-velocity of Σ reads u a = ˙ t (cid:18) ∂∂t (cid:19) a + ˙ r (cid:18) ∂∂r (cid:19) a = ˙ t (cid:18) ∂∂t (cid:19) a + ˙ r (cid:18) ∂∂r (cid:19) a (28)where ˙ t = s f + ˙ r f , ˙ t = s f + ˙ r f (29)Then junction condition reads m = r (cid:18)q f + ˙ r − q f + ˙ r (cid:19) (30)Substituting Eqs. (18) and (19) into Eq. (28) yields˙ t (cid:18) ∂∂t (cid:19) a + ˙ r (cid:18) ∂∂r (cid:19) a = ˙ t T (cid:18) ∂∂t (cid:19) a + ˙ r R (cid:18) ∂∂t (cid:19) a + ˙ r (cid:18) ∂∂r (cid:19) a (31)which gives ˙ t = ˙ t T + ˙ r R (32)Similarly, On Σ and Σ , we have˙ t = ˙ t T + ˙ r R (33)˙ t = ˙ t T + ˙ r R (34) The conservation law is m u a + m u a − m u a = 0 (35)Note that u a = ˙ t (cid:18) ∂∂t (cid:19) a + ˙ r (cid:18) ∂∂r (cid:19) a (36)7 a = ˙ t (cid:18) ∂∂t (cid:19) a + ˙ r (cid:18) ∂∂r (cid:19) a (37) u a = ˙ t (cid:18) ∂∂t (cid:19) a + ˙ r (cid:18) ∂∂r (cid:19) a = ˙ t T (cid:18) ∂∂t (cid:19) a + ˙ r (cid:20) R (cid:18) ∂∂t (cid:19) a + (cid:18) ∂∂r (cid:19) a (cid:21) = ( ˙ t T + ˙ r R ) (cid:18) ∂∂t (cid:19) a + ˙ r (cid:18) ∂∂r (cid:19) a (38)where Eqs. (20) and (21) have been used.So the conservation law yields m ˙ r + m ˙ r − m ˙ r = 0 (39) m ˙ t + m ˙ t − m ( ˙ t T + ˙ r R ) = m ˙ t + m ˙ t − m ˙ t = 0 (40)where we have used˙ t ≡ ∂t ∂τ (cid:12)(cid:12) = ∂t ∂t ∂t ∂τ + ∂t ∂r ˙ r = T ˙ t + R ˙ r (41)in the last step.Eq. (39) is the r component of Eq. (35), which is particularly simple. Thecorresponding equations in the LMW method are not equivalent to our Eqs.(39) and (40), although the vector form (35) of the conservation law is the samein both methods. Now we have 15 independent variables: m , ˙ r , m , ˙ r , m , ˙ r f , f , f T , T , T , R , R , R while there are 10 equations: 3 junction conditions, Eqs. (26) and (27), Eqs.(32),(33),(34), plus two equations of conservation of energy and momentum(Eq. (39) and (40)). We may set initial data: f , f , f , m , m , then the restvariables can be solved. We take f = 0 . , f = 0 . , f = 0 . , ˙ r = 1 , ˙ r = 2 , r = 10 . (42)8hen the ten equations mentioned above give rise to the following numericalsolutions: m = 0 . , m = 0 . , m = 0 . , ˙ r = 1 . T = 0 . , T = 0 . , T = 1 . R = − . , R = − . , R = 0 . T i and R i tell us how the manifold is constructed.As we have discussed, this match of manifold differs from the LMW treatment.However, the initial conditions of Eq. (42) are exactly needed for the LMWmethod (see Appendix A). It is not surprising that the two methods give riseto different solutions for m and ˙ r (see Eq. (43) and Eq. (66)) because thematching conditions are different. In this paper, some fundamental problems of the thin shell model have beenreconsidered and clarified. To make the thin shell spacetime a well-definedmanifold, some extra conditions need to be imposed. Some authors have proventhat the continuity of the metric across the shell leads to the conservation ofenergy and momentum in spherically symmetric spacetimes. However, we showthat in this treatment, the areal radius r is no longer a coordinate covering aneighborhood of the shell. We have then proposed a new matching techniquesuch that the conservation law and the coordinate r are both preserved. In thecase that two shells collide and merge into one shell, we have shown that theinitial conditions that needed to solve all the equations are exactly the sameas in the LMW method. Our work suggests that spacetimes containing thinshells can be matched in different ways and the conservation laws can still bepreserved. Acknowledgements
This research was supported by NSFC Grants No. 11235003, 11375026 andNCET-12-0054.
A Review of LMW mechanism and numericalcalculation
In this appendix, we review LMW’s treatment and apply it to the case in section3. With the same initial conditions, the numerical computation shows that thetwo methods gives different results. 9 .1 One shell
The junction condition of one shell is p f + ˙ r − p f + ˙ r = ρr (45)where r increases from region 1 to region 2 and ρ is the surface density of theshell. Let sinh α i = ˙ r √ f i (46)we have p f i + ˙ r = p f i q α i = p f i cosh α i (47)Let ˜ ρ = ρr (48)Then p f cosh α − p f cosh α = ˜ ρ (49)i.e., 12 p f e α + 12 p f e − α − p f e α − p f e − α = ˜ ρ (50)Eq. (46) leads to e α − e − α e α − e − α = √ f √ f (51)So Eq. (50) may be written as p f e α − p f e α = ˜ ρ (52) p f e − α − p f e − α = ˜ ρ (53) A.2 Three shells
We still consider the collision of two shells as shown in Fig. 3. Applying Eq.(46) to each shell, we havesinh α = ˙ r √ f , sinh α = ˙ r √ f (54)sinh α = ˙ r √ f , sinh α = ˙ r √ f (55)sinh α = ˙ r √ f , sinh α = ˙ r √ f (56)10pplying Eq. (52) to each shell yields˜ ρ = p f e α − p f e α (57)˜ ρ = p f e α − p f e α (58)˜ ρ = p f e α − p f e α (59)Note that ˜ ρ given above has a sign difference from what we gave previouslybecause it corresponds to the shell after the collision.The consistency condition is given by [4] α − α + α − α + α − α = 0 (60)If we define α ij = − α ji (61)The condition becomes α + α + α + α + α + α = 0 (62)For f , ˜ ρ e α e α e α e α e α + ˜ ρ e α + ˜ ρ e α e α e α = p f e α e α e α e α e α e α − p f e α e α e α e α + ... = 0 (63)where we have used Eq. (62). Such relations can be written as˜ ρ e ± α + ˜ ρ e ± α + ˜ ρ e ± α = 0 (64)where α = α + α + α . The other two relations can be obtained byreplacing 2 by 4 and 6. It is also easy to get˜ ρ γ + ˜ ρ γ + ˜ ρ γ = 0 (65)where γ ij = cosh α ij . This is the energy conservation law [4]. A.3 Numerical results
We use the same initial data as given in Eq. (42). Then Eqs. (54) and (55)can be solved directly. Combining Eq. (56) and Eq. (60), one can obtain α = 1 . α = 1 . m = 0 . , ˙ r = 1 . eferences [1] W.Israel, Nuovo Cimento B General Relativity (The University of Chicago Press, Chicago,1984).[4] D.Langlois, K.Maeda and D.Wands, Phys. Rev. Lett.661,