Hyperbolically symmetric static fluids: A general study
aa r X i v : . [ g r- q c ] J a n Hyperbolically symmetric static fluids: A general study
L. Herrera ∗ Instituto Universitario de F´ısica Fundamental y Matem´aticas,Universidad de Salamanca, Salamanca 37007, Spain
A. Di Prisco † Escuela de F´ısica, Facultad de Ciencias, Universidad Central de Venezuela, Caracas 1050, Venezuela
J. Ospino ‡ Departamento de Matem´atica Aplicada and Instituto Universitario de F´ısica Fundamental y Matem´aticas,Universidad de Salamanca, Salamanca 37007, Spain
We carry on a comprehensive study on static fluid distributions endowed with hyperbolical sym-metry. Their physical properties are analyzed in detail. The energy density appears to be necessarilynegative, which suggests that any possible application of this kind of fluids requires extreme physicalconditions where quantum effects are expected to play an important role. Also, it is found that thefluid distribution cannot fill the region close to the center of symmetry. Such a region may be repre-sented by a vacuum cavity around the center. A suitable definition of mass function, as well as theTolman mass are explicitly calculated. While the former is positive defined, the latter is negative inmost cases, revealing the repulsive nature of gravitational interaction. A general approach to obtainexact solutions is presented and some exact analytical solutions are exhibited.
PACS numbers: 04.40.-b, 04.40.Nr, 04.40.DgKeywords: Relativistic Fluids, nonspherical sources, interior solutions.
I. INTRODUCTION
In recent papers [1, 2] an alternative global descrip-tion of the Schwarzschild black hole has been proposed,motivated on the one hand by the well known fact thatany transformation that maintains the static form of theSchwarzschild metric (in the whole space–time) is unableto remove the coordinate singularity in the line element[3].On the other hand, based on the physically reasonablepoint of view that any equilibrium final state of a physicalprocess should be static, it would be desirable to have astatic solution over the whole space–time.However, as is well known, no static observers can bedefined inside the horizon (see [4, 5] for a discussion onthis point), this conclusion becomes intelligible if we re-call that the Schwarzschild horizon is also a Killing hori-zon, implying that the time–like Killing vector existingoutside the horizon, becomes space–like inside it.Thus, outside the horizon (
R > M ) one has the usualSchwarzschild line element corresponding to the spheri-cally symmetric vacuum solution to the Einstein equa-tions, which in polar coordinate reads (with signature+2) ds = − (cid:18) − MR (cid:19) dt + dR (cid:0) − MR (cid:1) + R d Ω ,d Ω = dθ + sin θdφ . (1) ∗ [email protected] † [email protected]; adiprisc@fisica.ciens.ucv.ve ‡ [email protected] This metric is static and spherically symmetric, mean-ing that it admits four Killing vectors: χ ( ) = ∂ t , χ ( ) = − cos φ∂ θ + cot θ sin φ∂ φ χ ( ) = ∂ φ χ ( ) = sin φ∂ θ + cot θ cos φ∂ φ . (2)However, when R < M the signature remains +2 butthe g tt and the g RR terms switch their signs, which ex-plains the fact that the time–like Killing vector outsidethe horizon, becomes space–like inside it. Also, an ap-parent line element singularity appears at R = 2 M . Ofcourse, as is well known, these drawbacks can be removedby coordinate transformations, but at the price that, asmentioned before, the staticity is lost within the horizon.In order to save the staticity inside the horizon, themodel proposed in [1] describes the space time as consist-ing of a complete four dimensional manifold (describedby (1)) on the exterior side of the horizon and a second(different) complete four dimensional solution in the in-terior of it. More specifically, to ensure that the vector χ ( ) = ∂ t be time–like (inside the horizon), a change insignature as well as a change in the symmetry at the hori-zon was required. The θ − φ sub-manifolds have sphericalsymmetry on the exterior and hyperbolic symmetry inthe interior. The two meet only at R = 2 M , θ = 0.Thus the model permits for a change in symmetry,from spherical outside the horizon to hyperbolic insidethe horizon. Doing so, one has a static solution every-where, but the symmetry of the R = 2 M surface is dif-ferent at both sides of it. We have to stress that we donot know if there is any specific mechanism behind sucha change of symmetry and signature. However, the mainpoint is that the change of symmetry (and signature) wasthe way followed in [1] to obtain a globally static solution.Thus, the solution proposed for R < M is (with sig-nature (+ − −− )): ds = (cid:18) MR − (cid:19) dt − dR (cid:0) MR − (cid:1) − R d Ω ,d Ω = dθ + sinh θdφ . (3)This is a static solution with the ( θ, φ ) space describinga positive Gaussian curvature. It admits the four Killingvectors χ ( ) = ∂ t , χ ( ) = − cos φ∂ θ + coth θ sin φ∂ φ χ ( ) = ∂ φ χ ( ) = sin φ∂ θ + coth θ cos φ∂ φ . (4)A solution to the Einstein equations of the form givenby (3), defined by the hyperbolic symmetry (4), was firstconsidered by Harrison [6], and has been more recentlythe subject of research in different contexts (see [7–15]and references therein).In [2], a general study of geodesics in the spacetimedescribed by (3) was presented, leading to some inter-esting conclusions about the behaviour of a test particlein this new picture of the Schwarzschild black hole. Ourpurpose in this work is to carry out a complete studyon the physical properties of a fluid distribution in theregion inner to the horizon, endowed with the symmetrygiven by (4). Such a fluid distribution might serve as thesource of (3).The mass function ( m ) as well as the Tolman mass( m T ) are defined for our fluid distribution. It is shownthat within the region r < m the energy density is neg-ative, a discussion about the physical implications of thisfact is presented. A general approach to obtain any ex-act solution corresponding to a spacetime admitting hy-perbolical symmetry is provided and some examples arefound and analyzed. II. BASIC DEFINITIONS, NOTATION ANDEQUATIONS
In this section we shall present the physical variablesand the relevant equations necessary for describing astatic self–gravitating locally anisotropic fluid admittingthe four Killing vectors (4).
A. The metric
We consider hyperbolically symmetric distributions ofstatic fluid, which for the sake of completeness we assumeto be locally anisotropic and which may be (or may benot) bounded from the exterior by a surface Σ e whoseequation is r = r Σ e = constant. On the other hand aswe shall see below, the fluid distribution cannot fill thecentral region, in which case we may assume that such aregion is represented by an empty vacuole, implying thatthe fluid distribution is also bounded from the inside bya surface Σ i whose equation is r = r Σ i = constant. The line element is given in polar coordinates (with thesame signature as (3)) by ds = e ν dt − e λ dr − r (cid:0) dθ + sinh θdφ (cid:1) , (5)where, due to the imposed symmetry ν ( r ) and λ ( r ) areexclusively functions of r . We number the coordinates: x = t ; x = r ; x = θ ; x = φ .The metric (5) has to satisfy Einstein field equations G νµ = 8 πT νµ . (6)Let us next provide a full description of the source. B. The source
We shall first consider the most general source, com-patible with staticity and axial symmetry. Afterward weshall impose the hyperbolical symmetry.Thus we may write for the energy momentum tensor T αβ = ( µ + P ) V α V β − P g αβ + Π αβ . (7)The above is the canonical algebraic decomposition ofa second order symmetric tensor with respect to unittimelike vector, which has the standard physical mean-ing when T αβ is the energy-momentum tensor describingsome energy distribution, and V µ the four-velocity as-signed by certain observer.Then, it is clear that µ is the energy density (theeigenvalue of T αβ for eigenvector V α ), whereas P is theisotropic pressure, and Π αβ is the anisotropic tensor. Weare considering an Eckart frame where fluid elements areat rest.Thus, it is immediate to see that µ = T αβ V α V β , (8) P = − h αβ T αβ , Π αβ = h µα h νβ ( T µν + P h µν ) , (9)with h µν = g µν − V ν V µ .The introduction of pressure anisotropy in the study ofself-gravitating fluids (Newtonian or relativistic) is jus-tified by the fact that it appears in a large number ofphysically meaningful situations (see [16] and referencestherein). Furthermore, as it has been recently shown [17],physical processes of the kind expected in stellar evolu-tion will always tend to produce pressure anisotropy, evenif the system is initially assumed to be isotropic. Theimportant point to stress here is that any equilibriumconfiguration is the final stage of a dynamic regime andthere is no reason to believe that the acquired anisotropyduring this dynamic process, would disappear in the finalequilibrium state, and therefore the resulting configura-tion, even if initially had isotropic pressure, should inprinciple exhibit pressure anisotropy.Since we choose the fluid to be comoving in our coor-dinates, then V α = ( e − ν/ , , , V α = ( e ν/ , , , . (10)Let us now define a canonical orthonormal tetrad (say e ( a ) α ), by adding to the four–velocity vector e (0) α ≡ V α ,three spacelike unitary vectors e (1) α ≡ K α = (0 , − e λ/ , , e (2) α ≡ L α = (0 , , − r, , (11) e (3) α ≡ S α = (0 , , , − r sinh θ ) , (12)with a = 0 , , , e α ( a ) is easily computed from thecondition η ( a )( b ) = g αβ e α ( a ) e β ( b ) , e α ( a ) e ( b ) α = δ ( b )( a ) , (13)where η ( a )( b ) denotes the Minkowski metric.In order to provide physical significance to the compo-nents of the energy momentum tensor, it is instructive toapply the Bondi approach [18].Thus, following Bondi, let us introduce a purely lo-cally Minkowski frame (l.M.f) with coordinates ( τ, x, y, z )(or equivalently, consider a tetrad field attached to suchl.M.f.) by: dτ = e ν/ dt ; dx = e λ/ dr ; dy = rdθ ; dz = r sinh θdφ. (14)Denoting by a hat the components of the energy mo-mentum tensor in such l.M.f., we have that the mattercontent is given by b T αβ = µ P xx P xy P yx P yy
00 0 0 P zz , (15)where µ, P xy , P xx , P yy , P zz denote the energy density anddifferent stresses, respectively, as measured by our locallydefined Minkowskian observer.This is the general expression for the energy–momentum tensor (in the l.M.f.) only assuming axialsymmetry. However, as consequence of the hyperboli-cal symmetry of the system, it follows from the Einsteinequations that the off diagonal term P xy vanishes and, ingeneral, P xx = P yy = P zz .The components of our tetrad field in the Minkowskicoordinates readˆ V α = (1 , , , K α = (0 , − , , L α = (0 , , − , , ˆ S α = (0 , , , − , (16) from which we may write b T αβ = ( µ + P zz ) ˆ V α ˆ V β − P zz η αβ +( P xx − P zz ) ˆ K α ˆ K β , (17)where η αβ denotes the Minkowski metric.Then transforming back to the coordinates of (5), weobtain the components of the energy momentum tensorin terms of the physical variables as defined in the l.M.f. T αβ = ( µ + P zz ) V α V β − P zz g αβ +( P xx − P zz ) K α K β . (18)It would be useful to express the anisotropic tensor inthe form Π αβ = Π (cid:18) K α K β + h αβ (cid:19) , (19)with Π = P xx − P zz , (20)and P = P xx + 2 P zz . (21)Or, inversely P zz = P −
13 Π , (22) P xx = P + 23 Π . (23)Since the Lie derivative and the partial derivative com-mute, then L χ ( R αβ − g αβ R ) = 8 π L χ T αβ = 0 , (24)for any χ defined by (4), implying that all physical vari-ables only depend on r .If the fluid is bounded from the exterior by a hypersur-face Σ e described by the equation r = r Σ e = constant ,then the smooth matching of (3) and (5) on Σ e requiresthe fulfillment of the Darmois conditions [19], imposingthe continuity of the first and the second fundamentalforms, which imply e ν Σ e = 2 Mr Σ e − , e λ Σ e = 1 Mr Σ e − , P xx ( r Σ e ) = 0 , (25)and the continuity of the mass function m ( r ) definedbelow. If we assume that the central region is sur-rounded by an empty cavity whose delimiting surface is r = r Σ i = constant , then the fulfillment of Darmois con-ditions on Σ i implies e ν Σ i = 1 , e λ Σ i = 1 , P xx ( r Σ i ) = 0 , (26)and m ( r Σ i ) = 0.If either of conditions above (or both) are not satisfied,then we have to resort to Israel conditions [20], implyingthat thin shells are present at either boundary surface(or both). C. The Einstein equations
The non–vanishing components of the Einstein equationsfor the metric (5) and the energy momentum tensor (18)are 8 πµ = − ( e − λ + 1) r + λ ′ r e − λ , (27)8 πP r = ( e − λ + 1) r + ν ′ r e − λ , (28)8 πP ⊥ = e − λ ν ′′ + ν ′ − λ ′ ν ′ ν ′ r − λ ′ r ! , (29)where we have used the standard notation P xx ≡ P r and P zz = P yy ≡ P ⊥ , and primes denote derivatives withrespect to r .It is worth stressing the differences between these equa-tions and the corresponding to the spherically symmetriccase (see for example eqs.(2–4) in [21]).From the equations above or using the conservationlaws T αβ ; α = 0 we obtain, besides the identity ˙ µ = 0(where dot denotes derivative with respect to t ), the cor-responding hydrostatic equilibrium equation (the gener-alized Tolman–Oppenheimer–Volkoff equation) P ′ r + ( µ + P r ) ν ′ r Π = 0 . (30)Let us now define the mass function m = m ( r ). Fordoing so, let us notice that using (3) we have that outsidethe fluid distribution (but inside the horizon) M = − (cid:18) R (cid:19) R , (31)where the Riemann tensor component R , has beencalculated with (3).Then generalizing the above definition of mass for theinterior of the fluid distribution we may write m ( r ) = − (cid:16) r (cid:17) R = r (1 + e − λ )2 (32)where now the Riemann tensor component is calculatedwith (5).Feeding back (32) into (27) we obtain m ′ ( r ) = − πr µ ⇒ m = − π Z r µr dr. (33)Since m as defined by (32) is a positive quantity, then µ should be negative and therefore the weak energy con-dition is violated, a result already obtained in [15]. How-ever it is important to stress that our definition of massfunction differs from the one introduced in [15]. In par-ticular our m is positive defined whereas the expressionused in [15] is negative (for the hyperbolically symmetricfluid).The following comments are in order at this point. • It is worth noticing that while the total energy(mass) of the bounded distribution is unique ( M ),the definition of the energy localized in a given pieceof the fluid distribution is not. As a matter offact, this ambiguity in the localization of energy,which is present even in classical electrodynamics[22], has been extensively discussed in general rela-tivity, leading to different definitions of energy (seefor example [23–29] and references therein). • Our main motivation to study hyperbolically sym-metric fluids is directly related to the black holepicture described in the Introduction, according towhich the region interior to the horizon is describedby (3), whereas the spacetime outside the horizonis described by the usual Schwarzschild metric (1).Now, the parameter M appearing in (3) is the sameas the M appearing in (1), i.e. the total mass ofthe source, which is positive defined. Then, if wewish that our mass function be continuous at theboundary surface of our fluid distribution, whichin turn is the source of (3), it is natural to assume(32) as the definition of the mass function. Howeverit should be clear that any other scenario differentfrom the one described above, allows for other al-ternative definitions of mass function. • The equation (32), as expected, is at variance withthe definition in the spherically symmetric case( e − λ = 1 − mr ), since we are interested in the region2 m > r . • If the energy density is regular everywhere then themass function must vanish at the center as m ∼ r ,this implies (as it follows from (32)) that the fluidcannot fill the space in the neighborhood of thecenter, i.e. there is a cavity around the centerwhich may be, either empty, or filled with a fluiddistribution non endowed with hyperbolical sym-metry. Thus the hyperbolically symmetric fluidspans from a minimal value of the coordinate r until its external boundary. For the extreme case µ = constant , this minimal value r min. is definedby − π µr min. >
1. Obviously, if the energy den-sity is singular in the neighborhood of the center,then this region must also be excluded by physicalreasons.From the above it follows that, strictly speaking, weshould write instead of (33) m = 4 π Z rr min | µ | r dr, (34)where due to the fact that µ is negative, we have replacedit by −| µ | (as we shall do from now on).The situation described above is fully consistent withthe results obtained in [2] where it was shown that testparticles cannot reach the center for any finite value ofits energy.Next, using (28) and (32) we obtain ν ′ = 2 4 πr P r − mr (2 m − r ) , (35)from which we may write (30) as P ′ r + ( P r − | µ | ) 4 πr P r − mr (2 m − r ) + 2 r Π = 0 . (36)This is the hydrostatic equilibrium equation for ourfluid. Let us analyze in some detail the physical meaningof its different terms. The first term is just the gradient ofpressure, which is usually negative and opposing gravity.The second term describes the gravitational “force” andcontains two different contributions: on the one hand theterm P r − | µ | which we expect to be negative (or zero forthe stiff equation of state) and is usually interpreted asthe “passive gravitational mass density” (p.g.m.d.), andon the other hand the term 4 πr P r − m that is propor-tional to the “active gravitational mass” (a.g.m.), andwhich is negative if 4 πr P r < m . Finally the third termdescribes the effect of the pressure anisotropy, whosesign depends on the difference between principal stresses.Two important remarks are in order at this point: • It is worth stressing that while the self–regenerativepressure effect (described by the 4 πr P r term in(36)) has the same sign as in the spherically sym-metric case, the mass function contribution in thesecond term has the opposite sign with respect tothe latter case. This of course is due to the factthat the energy density is negative. • If, both, the p.g.m.d. and the a.g.m. are nega-tive, the final effect of the gravitational interactionwould be as usual, to oppose the negative pres-sure gradient. However, because of the equivalenceprinciple, a negative p.g.m.d. implies a negativeinertial mass, which in turn implies that the hy-drostatic force term (the pressure gradient and theanisotropic term), and the gravitational force term,switch their roles with respect to the positive en-ergy density case.
D. The Riemann and the Weyl tensor
As is well known, the Riemann tensor may be ex-pressed through the Weyl tensor C ραβµ , the Ricci tensor R αβ and the scalar curvature R , as R ραβµ = C ραβµ + 12 R ρβ g αµ − R αβ δ ρµ + 12 R αµ δ ρβ − R ρµ g αβ − R ( δ ρβ g αµ − g αβ δ ρµ ) . (37) In our case, the magnetic part of the Weyl tensor van-ishes and we can express the Weyl tensor in terms of itselectric part ( E αβ = C αγβδ V γ V δ ) as C µνκλ = ( g µναβ g κλγδ − η µναβ η κλγδ ) V α V γ E βδ , (38)with g µναβ = g µα g νβ − g µβ g να , and η µναβ denoting theLevi–Civita tensor.The electric part of the Weyl tensor for our metric (5)may be written as E αβ = E (cid:18) K α K β + 13 h αβ (cid:19) , (39)satisfying the following properties: E αα = 0 , E αγ = E ( αγ ) , E αγ V γ = 0 , (40)where E = − e − λ ν ′′ + ν ′ − λ ′ ν ′ − ν ′ r + λ ′ r ! − r (cid:0) e − λ (cid:1) . (41)Using the field equations (27)–(29), (32) and (41) thefollowing relationship may be obtained3 mr = 4 π | µ | + 4 π Π − E . (42)Taking the r -derivative of the expression above and using(33) we find E = 4 πr Z r ˜ r | µ | ′ d ˜ r + 4 π Π . (43)Finally, inserting (43) into (42) we obtain m ( r ) = 4 π r | µ | − π Z r ˜ r | µ | ′ d ˜ r. (44)Equation (43) relates the Weyl tensor to two fun-damental physical properties of the fluid distribu-tion, namely: energy density inhomogeneity and localanisotropy of pressure, whereas (44) expresses the massfunction in terms of its value in the case of a homogeneousenergy density distribution, plus the change induced byenergy density inhomogeneity. In the expressions (43),(44) above it must be kept in mind that the center of thedistribution should be excluded. E. Tolman mass
An alternative definition to describe the energy contentof a fluid sphere was proposed by Tolman many years ago[23] . The Tolman mass generalized for any fluid elementof our static fluid distribution inside Σ e reads m T = Z π Z π Z r √− g ( T − T − T ) d ˜ rdθdφ (45)= 2 π ( coshπ − Z r e ( ν + λ ) / ˜ r ( −| µ | + P r + 2 P ⊥ ) d ˜ r. Using the field equations (27)–(29), the integration of(45) produces m T = ( coshπ − e ( ν − λ ) / r ν ′ , (46)or combining (35) with (46) m T = ( coshπ − e ( ν + λ ) / (4 πP r r − m ) . (47)In the light of equation (47) (or (46)) and (36)(or (30)),the usual physical interpretation of m T as a measure ofthe active gravitational mass becomes evident. It must bestressed the fact that this quantity is negative provided4 πP r r < m , which would imply the repulsive characterof the gravitational interaction in the spacetime underconsideration. Indeed, let us consider the four–acceleration a α , de-fined as usually by a α = V α ; β V β , (48)that in our case may be written as a α = aK α , (49)where a = ν ′ e − λ/ , (50)which allows to write a = 2 m T r e − ν/ ( coshπ − . (51)Thus the four-acceleration is directed inwardly (if4 πP r r < m ). Now, let us recall that a µ represents theinertial radial acceleration which is necessary in order tomaintain static the frame by canceling the gravitationalacceleration exerted on the frame. Therefore the fact thatthe four–acceleration is directed radially inward, revealsthe repulsive nature of the gravitational force.Next, taking the r -derivative of (45) and using (47) wefind m ′ T − r m T = − ( coshπ − e ( ν + λ ) / r (4 π Π + E ) , (52)whose integration produces m T = ( m T ) Σ e (cid:18) rr Σ e (cid:19) + ( coshπ − r Z r Σ e r e ( ν + λ ) / ˜ r ( E + 4 π Π) d ˜ r, (53)or using (43) m T = ( m T ) Σ e (cid:18) rr Σ e (cid:19) + ( coshπ − r Z r Σ e r e ( ν + λ ) / ˜ r π ˜ r Z ˜ r | µ | ′ s ds + 8 π Π ! d ˜ r. (54)The above expressions are equivalent to the ones obtainedfor the spherically symmetric case [30].We shall next present the orthogonal splitting of theRiemann tensor, and express it in terms of the variablesconsidered so far. Doing so we shall be able to define thestructure scalars for our fluid distribution. III. THE ORTHOGONAL SPLITTING OF THERIEMANN TENSOR AND THE STRUCTURESCALARS
Following the orthogonal splitting scheme of the Rie-mann tensor considered by Bel [31], let us introducethe following tensors (we shall follow closely, with somechanges, the notation of [32]), Y αβ = R αγβδ V γ V δ , (55) Z αβ = ∗ R αγβδ V γ V δ = 12 η αγǫµ R ǫµ βδ V γ V δ , (56) X αβ = ∗ R ∗ αγβδ V γ V δ = 12 η ǫµαγ R ∗ ǫµβδ V γ V δ , (57)where ∗ denotes the dual tensor, i.e. R ∗ αβγδ = η ǫµγδ R ǫµαβ .It can be shown that the Riemann tensor can be ex-pressed through these tensors in what is called the or-thogonal splitting of the Riemann tensor (see [32] for de-tails). However, instead of using the explicit form of thesplitting of Riemann tensor (eq.(4.6) in [32]), we shallproceed as follows (for details see [33], where the generalnon–static case has been considered).Using the Einstein equations we may write (37) as R αγβδ = C αγβδ + 28 πT [ α [ β δ γ ] δ ] + 8 πT ( 13 δ α [ β δ γδ ] − δ [ α [ β δ γ ] δ ] ) , (58)then feeding back (7) into (58) we split the Riemanntensor as R αγβδ = R αγ ( I ) βδ + R αγ ( II ) βδ + R αγ ( III ) βδ , (59)where R αγ ( I ) βδ = 16 πµV [ α V [ β δ γ ] δ ] − πP h [ α [ β δ γ ] δ ] +8 π ( µ − P )( 13 δ α [ β δ γδ ] − δ [ α [ β δ γ ] δ ] ) (60) R αγ ( II ) βδ = 16 π Π [ α [ β δ γ ] δ ] (61) R αγ ( III ) βδ = 4 V [ α V [ β E γ ] δ ] − ǫ αγµ ǫ βδν E µν (62)with ǫ αγβ = V µ η µαγβ , ǫ αγβ V β = 0 , (63)and where the vanishing of the magnetic part of the Weyltensor ( H αβ = ∗ C αγβδ V γ V δ ) has been used.Using the results above, we can now find the explicitexpressions for the three tensors Y αβ , Z αβ and X αβ interms of the physical variables, we obtain Y αβ = 4 π µ + 3 P ) h αβ + 4 π Π αβ + E αβ , (64) Z αβ = 0 , (65)and X αβ = − π | µ | h αβ + 4 π Π αβ − E αβ . (66)As shown in [33], the tensors above may be expressedin terms of some scalar functions, referred to as structure scalars, by decomposing them into their trace–free partand their trace, as X αβ = X T h αβ X T F (cid:18) K α K β + h αβ (cid:19) , (67) Y αβ = Y T h αβ Y T F (cid:18) K α K β + h αβ (cid:19) . (68)These scalars in turn may written as: X T = − π | µ | , (69) X T F = 4 π Π − E , (70)or using (43) X T F = − πr Z r ˜ r | µ | ′ d ˜ r, (71)and Y T = 4 π ( −| µ | + 3 P ) , (72) Y T F = 4 π Π + E , (73)or using (43) Y T F = 8 π Π + 4 πr Z r ˜ r | µ | ′ d ˜ r. (74)From the above it follows that local anisotropy of pres-sure is determined by X T F and Y T F by8 π Π = X T F + Y T F . (75)To establish the physical meaning of Y T and Y T F letus get back to equations (45) and (54), using (72) and(74) we get m T = ( m T ) Σ e (cid:18) rr Σ e (cid:19) + ( coshπ − r Z r Σ e r e ( ν + λ ) / ˜ r Y T F d ˜ r, (76)and m T = ( coshπ − Z r ˜ r e ( ν + λ ) / Y T d ˜ r. (77)We see that Y T F encompasses the influence of the localanisotropy of pressure and density inhomogeneity on theTolman mass. Or, in other words, Y T F describes howthese two factors modify the value of the Tolman mass,with respect to its value for the homogeneous isotropicfluid. This fact was at the origin of the definition of com-plexity provided in [34]. Indeed, if we assume that thehomogeneous (in the energy density) fluid with isotropicpressure is endowed with minimal complexity, then thevariable responsible for measuring complexity, which wecall the complexity factor, should vanish for this kind offluid distribution, as it happens for Y T F .Also, it is worth noticing that from (77) Y T appears tobe proportional to the Tolman mass “density”. IV. ALL STATIC SOLUTIONS
We shall next present a general formalism to expressany static hyperbolically symmetric solution in terms oftwo generating functions. Afterward we shall also presentsome explicit solutions and their generating functions.The procedure is similar to the one proposed for thespherically symmetric case (see [21], [35]).Thus, from (28) and (29) we may write8 π ( P r − P ⊥ ) = ( e − λ + 1) r (78) − e − λ ν ′′ + ν ′ − λ ′ ν ′ − ν ′ r − λ ′ r ! , which, by introducing the following auxiliary functions ν ′ z − r ; e − λ = y, (79)becomes y ′ + y (cid:18) z ′ z + 2 z − r + 4 zr (cid:19) = 2 z (cid:18) r − π Π (cid:19) . (80)The integration of (80) produces e λ ( r ) = z e R ( zr +2 z ) dr r (cid:16) R z (1 − π Π r ) r e R ( zr +2 z ) dr dr + C (cid:17) . (81)Therefore, any static solution is fully described by thetwo generating functions Π and z .For the physical variables we may write4 π | µ | = m ′ r , (82)4 πP r = z (2 mr − r ) − m + rr , (83)8 πP ⊥ = (cid:18) mr − (cid:19) (cid:18) z ′ + z − zr + 1 r (cid:19) + z (cid:18) m ′ r − mr (cid:19) . (84)We shall now find some explicit solutions and theirrespective generating functions. A. The conformally flat solutions
Due to the conspicuous role played by the Weyl tensorin the structure of the fluid distribution, as indicated by(43) and (53), it is worth considering the special case E = 0 (conformal flatness).Thus we shall now proceed to integrate the condition E = 0 , (85)which, using (41), may be written as (cid:18) e − λ ν ′ r (cid:19) ′ + e − ( ν + λ ) (cid:18) e ν ν ′ r (cid:19) ′ − (cid:18) e − λ + 1 r (cid:19) ′ = 0 . (86)Introducing the new variables y = e − λ ; ν ′ u ′ u (87)the equation(86) is cast into y ′ + 2 y (cid:0) u ′′ − u ′ /r + u/r (cid:1) ( u ′ − u/r ) + 2 ur ( u ′ − u/r ) = 0 , (88)whose formal solution is y = e − R k ( r ) dr (cid:20)Z e R k ( r ) dr f ( r ) dr + C (cid:21) , (89)where C is a constant of integration, and k ( r ) = 2 ddr h ln (cid:16) u ′ − ur (cid:17)i , (90) f ( r ) = − ur ( u ′ − u/r ) . (91)Changing back to the original variables, eq.(89) becomes ν ′ − r = e λ/ r p αr e − ν − , (92)where α is a constant of integration which using thematching conditions (25) becomes α = M (9 M − r Σ e ) r e . (93)Next, (92) may be formally integrated, to obtain e ν = αr sin (cid:18)Z e λ/ r dr + γ (cid:19) (94)where γ is a constant of integration which using (25)reads, γ = arcsin r Σ e vuut (cid:16) Mr Σ e − (cid:17) M (9 M − r Σ e ) − (cid:18)Z e λ/ r dr (cid:19) Σ e . (95)It is worth noticing the difference between (94) andthe corresponding expression in the spherically symmet-ric case (equation (40) in [36]).Obviously the conformally flat condition only deter-mines one generating function, therefore in order to finda specific model we have to impose an additional restric-tion. As an example we shall consider the extreme case P r = 0. This solution represents the hyperbolically sym-metric analogue of the model I found in [36] for the spher-ically symmetric case.Thus assuming P r = 0, we find from (28) ν ′ = − (1 + e λ ) r . (96)Replacing (96) in (41) and using E = 0 the followingrelationship may be easily obtained8(1 + e λ ) + (1 + e λ ) + 3 rλ ′ − rλ ′ e λ = 0 , (97)or g (9 g − − g ′ r (3 g −
2) = 0 , (98)where e − λ = 2 g −
1. The integration of (98) produces C r = 4 g g − , (99)where C is a constant of integration.Next, combining (92) and (96) we obtain e ν = αr (2 g − g (9 g − . (100)For the physical variables we get | µ | = 3 g πr (2 g − g − , (101) P ⊥ = 34 πr g (3 g − . (102)From the above it follows that g > / e ν is positive, and defining the minimum value of r forthe fluid distribution. The model may be completed byassuming an empty vacuole with boundary surface r = r min . Since P r is zero, then both the a.g.m. and thep.g.m.d. are negative. It should be noticed that whilethe matching conditions may be satisfied on Σ e , the massfunction would be discontinuous across Σ i . Thus a thinshell appears at the boundary surface r = r min . The generating functions for this model are easilyfound to be z = g − r (2 g − , (103)and Π( r ) = − πr g (3 g − . (104) B. Two anisotropic solutions from given energydensity profiles
We shall next find two solutions, by extending to thehyperbolically symmetric case the procedure developedin [37] which allows to find an anisotropic solution fromany known isotropic one, in the spherically symmetriccase.The basic ansatz of the method is based on a specificform of the anisotropy, more specifically it is assumedthat P ⊥ − P r = C ( −| µ | + P r ) ν ′ r, (105)where C is a constant measuring the anisotropy of thepressure.Then, using (105) in (30) we obtain P ′ r + ( −| µ | + P r ) ν ′ h = 0 , (106)with h ≡ − C .Obviously h = 1 corresponds to the isotropic pressurecase. Then assuming the energy density distribution ofa given isotropic solution we may find the correspondinganisotropic model satisfying (105).
1. The incompressible fluid
Let us first consider a fluid distribution with con-stant energy density ( µ = constant ). If the fluid hasisotropic pressure then the solution is unique and wouldbe the hyperbolically symmetric version of the interiorSchwarzschild solution, such a solution has been foundin [15]. However if the pressure is anisotropic thereare an infinite number of possible solutions. Here wefind a solution which would be the generalization of theBowers–Liang solution [38], for the hyperbolically sym-metric fluid.Thus, assuming µ = constant we may integrate (106)obtaining P r − | µ | = βe − νh/ , (107)where β is a constant of integration.From the junction condition ( P r ) Σ e = 0, we obtain for β β = −| µ | e ν Σ e h/ , (108)producing P r = | µ | h − e ( ν Σ e − ν ) h/ i . (109)Also, from (33) we have m ( r ) = 4 π | µ | r . (110)0Combining (109) with (28) we obtain ν ′ (cid:18) πr | µ | − (cid:19) − π | µ | r π | µ | re ( ν Σ e − ν ) h/ = 0 , (111)where (32) has been used.The integration of (111) produces e νh = 14 " (cid:18) π | µ | r e − (cid:19) h/ − (cid:18) π | µ | r − (cid:19) h/ , (112) where junction conditions (25) have been used.Combining the above expression with (109) we obtain, P r = | µ | (cid:20)(cid:16) π | µ | r e − (cid:17) h/ − (cid:16) π | µ | r − (cid:17) h/ (cid:21) (cid:16) π | µ | r e − (cid:17) h/ − (cid:16) π | µ | r − (cid:17) h/ , (113)and for the tangential pressure we have P ⊥ = | µ | (cid:20)(cid:16) π | µ | r e − (cid:17) h/ − (cid:16) π | µ | r − (cid:17) h/ (cid:21) (cid:16) π | µ | r e − (cid:17) h/ − (cid:16) π | µ | r − (cid:17) h/ + (1 − h )8 π | µ | r (cid:16) π | µ | r e − (cid:17) h/ (cid:16) π | µ | r − (cid:17) h − (cid:20) (cid:16) π | µ | r e − (cid:17) h/ − (cid:16) π | µ | r − (cid:17) h/ (cid:21) . (114)As mentioned before, the fluid distribution describedabove cannot fill the whole space, but is restricted bya minimal value of the r coordinate, satisfying r min > q π | µ | . For 0 < r < r min we may assume an emptycavity surrounding the center. Also, it is a simple matterto check that both, the a.g.m. and the p.g.m.d., arenegative. As in the previous case, this solution may bematched on Σ e but not on Σ i , due to the discontinuityof the mass function across that hypersurface.For h = 1 we recover the incompressible isotropic fluidsolution found in [15], whereas for h = 1 we obtain the so-lution equivalent to the Bowers–Liang fluid distribution,corresponding to the hyperbolically symmetric case.It is not difficult to see that the two generating functionsof this solution are z = 1 r " π | µ | r e − h/ − ( π | µ | r − h − ( π | µ | r − π | µ | r e − h/ − ( π | µ | r − h/ , (115)andΠ( r ) = − (1 − h )8 π | µ | r (cid:16) π | µ | r e − (cid:17) h/ (cid:16) π | µ | r − (cid:17) h − h π | µ | r e − h/ − ( π | µ | r − h/ i . (116)
2. Tolman VI type solution
As a second application of the approach sketched abovewe shall now find a solution inspired in the well knownTolman VI model [39]. It is worth recalling that in thespherically symmetric case with isotropic pressure, theequation of state of this model corresponds to the equa-tion of state of a Fermi gas in the limit of very largeenergy density. Thus, following the approach described above, let usassume µ = Kr ; ⇒ m = − πKr (117)where K is a negative constant.Then (106) for this case may be written as P ′ r = αP r r + βrP r + γr , (118)where the constants α, β, γ are defined by α ≡ πKh πK + 1 ; β ≡ πh πK + 1 ; γ ≡ πhK πK + 1 . (119)Equation (118) may be integrated, producing P r = (2 + α + ǫ )(2 + α − ǫ )( r ǫ Σ e − r ǫ )2 βr [ r ǫ (2 + α − ǫ ) − r ǫ Σ e (2 + α + ǫ )] , (120)where ǫ ≡ p α + α − γβ , and the boundary con-dition P r ( r Σ e ) = 0 has been used. Thus the matching onΣ e is assured, while the discontinuity of the mass func-tion (117) across Σ i produces a thin shell on the innerboundary surface.For the metric functions and the tangential pressure thecorresponding expressions are e − λ = − (8 πK + 1) , (121) e ν = ( Mr Σ e − ǫ ) n r n (2+ ǫ )Σ e n r n (2 − ǫ ) [ r ǫ (2 + α − ǫ ) − r ǫ Σ e (2 + α + ǫ )] n o , (122)with n = πβ (8 πK +1) , and1 P ⊥ = (2 + α + ǫ )(2 + α − ǫ )( r ǫ Σ e − r ǫ )2 βr [ r ǫ (2 + α − ǫ ) − r ǫ Σ e (2 + α + ǫ )] + 2 π ( h − πK + 1) (cid:26) r ǫ Σ e (2 + α + ǫ )(2 − ǫ ) − r ǫ (2 + α − ǫ )(2 + ǫ )2 βr [ r ǫ (2 + α − ǫ ) − r ǫ Σ e (2 + α + ǫ )] (cid:27) . (123)As it is obvious this fluid distribution is singular atthe center, and therefore the central region should beexcluded by elementary physical reasons. Furthermorefrom (121) it follows that 8 π | K | >
1, implying α > , β < , γ < , n > z = r ǫ (2 n + nǫ + 2)(2 + α − ǫ ) − r ǫ Σ e (2 + α + ǫ )(2 n − nǫ + 2)2 r [ r ǫ (2 + α − ǫ ) − r ǫ Σ e (2 + α + ǫ )] , (124)and Π( r ) = − π ( h − πK + 1) (cid:26) r ǫ Σ e (2 + α + ǫ )(2 − ǫ ) − r ǫ (2 + α − ǫ )(2 + ǫ )2 βr [ r ǫ (2 + α − ǫ ) − r ǫ Σ e (2 + α + ǫ )] (cid:27) . (125) C. A model with vanishing complexity factor
As mentioned before, the scalar Y T F has been shown tobe a suitable measure of the complexity of the fluid distri-bution (see the discussion on this issue in [34]), thereforeit would be interesting to find a model (besides the homo-geneous and isotropic solution) satisfying the conditionof vanishing complexity ( Y T F = 0). Since there are aninfinite number of such solutions, we have to impose anadditional restriction in order to obtain a specific model.Here we shall assume (besides the vanishing complexityfactor), the condition P r = 0.Thus, assuming P r = 0, we obtain from (28) ν ′ = − g (2 g − r , (126)where g is defined by e − λ = 2 g − . (127)Next, imposing Y T F = 0 in (76) it follows that m T = ( m T ) Σ e r r . (128)The combination of (46),(126),(127) and (128) produces e ν = 4( m T ) Σ e r r ( coshπ − (2 g − g . (129)On the other hand the condition Y T F = 0 may bewritten as g ′ r (1 − g ) + g (5 g −
2) = 0 , (130) whose solution reads C r = g (5 g − , (131)where C is a constant of integration.Then for the physical variables we obtain | µ | = 34 πr g (2 g − g − , (132) P ⊥ = 38 πr g ( g − . (133)In this case, the fluid distribution is restricted by aminimal value of the r coordinate, satisfying g ( r min ) > r min is obtained from (131). For0 < r < r min we may assume, as in precedent mod-els, an empty cavity surrounding the center. Also as inprecedent models, the discontinuity of the mass functionacross Σ i implies that a thin shell appears on it. Finally,since the radial pressure is assumed to be zero, both thea.g.m. and the p.g.m.d. are negative.The generating functions for this model are easilyfound to be z = g − r (2 g − , (134)and Π( r ) = − πr g ( g − . (135) D. The stiff equation of state
Finally we shall consider a couple of solutions satis-fying the so called stiff equation of state, which as far2as we know was proposed for the first time by Zeldovich[40], and is believed to be suitable to describe ultradensematter (in particular for neutral vector mesons ω and φ ). In its original form it assumes that energy densityequals pressure (in relativistic units). In our case we shallassume | µ | = P r . (136)then (36) becomes P ′ r + 2 r Π = 0 . (137)To obtain specific solutions, additional information isrequired. Here, as examples, we shall consider two par-ticular cases. P ⊥ = 0 . Let us first assume that tangential pressure vanishes.Then (137) can be easily integrated, producing P r = Kr ⇒ | µ | = Kr , (138)where K is a positive constant of integration.The above equation, together with (32), (33) and (35)produces m = 4 πKr, e − λ = 8 πK − , ν = constant. (139)In this model, both, the a.g.m. and the p.g.m.d. van-ish.Obviously there are not vanishing pressure surfaces forthis solution, and the corresponding generating functionsare Π = Kr , z = 1 r . (140) Y TF = 0 . Let us next consider the simplest stiff fluid model (i.e.the one satisfying, besides (136), the vanishing complex-ity factor condition).Then, using this latter condition in (74) and feedingback the resulting expression into (137) one obtains P ′′ r + 3 r P ′ r = 0 , (141)whose solution reads P r = br − a (142)where a and b are two positive constants of integration. Then from (32) and (33) it follows at once m = 4 πr (cid:18) b − ar (cid:19) , (143)from which we easily obtain λ . Finally, feeding back theseexpressions into (35) we may obtain ν .Assuming the fluid distribution to be bounded fromthe exterior by the surface Σ e described by r = r Σ e = constant , then we may write P r = b (cid:18) r − r e (cid:19) , (144)and m = 4 πbr r e (cid:0) r e − r (cid:1) . (145)Thus, while matching conditions are satisfied on Σ e , theyare not on Σ i .From the above expressions it follows at once that4 πr P r − m = − πbr r e . (146)Finally, for the tangential pressure we obtain P ⊥ = − br e (147)Thus, for this model the p.g.m.d vanishes, whereas, un-like the previous case, the a.g.m. ( m T ) does not, and isnegative. V. CONCLUSIONS
Motivated by the physical interest of the black holepicture described in the Introduction, and which assumesthat the spacetime inside the horizon is described by (3),we have carried out a general study on the properties ofstatic fluid distributions endowed with hyperbolical sym-metry, which eventually could serve as the source of suchspacetime. Thus we have found that such fluid distri-butions may be anisotropic in the pressure, with onlytwo main stresses unequal and the energy density is nec-essarily negative. Furthermore, the fluid cannot fill thewhole space within the horizon, the central region beingexcluded. This is so, whether the energy density withinthe fluid distribution is regular or not. This last resultimplies that the central region should consist in a vac-uum cavity, or should be described by a different type ofsource. On the other hand the fact that the fluid distri-bution cannot attain the center concurs with the resultobtained in [2] indicating that no test particle with finiteenergy can reach the center.The violation of the weak energy condition ( µ < πP r r < m ), requires some discussion.3Let us start by mentioning that in spite of the fact thatfrom classical physics considerations we expect the en-ergy density to be positive, negative energy densities areoften invoked in extremes cosmological and astrophysicalscenarios, usually in relation with possible quantum ef-fects, of the kind we could expect within the horizon (see[41–45] and references therein).Besides, it is worth recalling that at purely classicallevel, it has been shown that any spherically symmetricdistribution of charged fluid (independently of its equa-tion of state) whose total mass-radius and charge corre-spond to the observed values of the electron, must havenegative energy distribution (at least for some values ofthe radial coordinate) [46–48]. The possible origin ofthis intriguing result may be found in a remark by Pa-papetrou about the finiteness of the total mass of theReissner-Nordstrom solution [49]. Indeed, since the elec-trostatic energy of a point charge is infinite, the onlyway to produce a finite total mass is the presence of aninfinite amount of negative energy at the centre of sym-metry. Without entering into a detailed analysis of thisissue, which is beyond the scope of this manuscript, wespeculate that the violation of the weak energy conditionmight be related to quantum vacuum of the gravitationalfield.Next, we recall that in [2] it has been obtained that anytest particle within the horizon, for the metric (3), wouldexperience a repulsive force. In the case of a fluid distri-bution, this repulsive nature of the gravitational interac-tion was already brought out in Section II as due to thefact that the a.g.m. (if 4 πP r r < m ) is negative. On theother hand, as we have already mentioned, we expect thep.g.m.d. to be negative, or at most zero, which accord-ing to the equivalence principle (stating that the inertialmass equals the passive gravitational mass) would implythat the inertial mass is negative too, therefore a negativepressure gradient even if directed outwardly, would pushany fluid element inwardly. In other words, the forcesacting on any fluid element look as if they have switchedtheir role, with respect the positive energy density case.Finally, we have developed a general formalism to ex-press any static hyperbolically symmetric fluid solutionin terms of two generating functions. Some explicit solu-tions have been found and their physical variables havebeen exhibited. In all of them it appears clearly that the central region cannot be filled with the fluid distri-bution. Further study on the physical properties of thesesolutions, although out of the scope of this work, wouldbe necessary. ACKNOWLEDGMENTS
This work was partially supported by Ministerio deCiencia, Innovacion y Universidades. Grant number:PGC2018–096038–B–I00.
Appendix A: Some basic formulae
In what follows we shall deploy some formulae used inour discussion.
1. Christoffel symbols Γ = ν ′ , Γ = ν ′ e ( ν − λ ) , Γ = λ ′ , Γ = − re − λ , Γ = − re − λ sinh θ, (A1)Γ = 1 r , Γ = −
12 sinh 2 θ, Γ = coth θ, Γ = 1 r .
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