A No Black Hole Theorem
AA No Black Hole Theorem
Gavin S. Hartnett , Gary T. Horowitz , and Kengo Maeda Department of Physics, UCSB, Santa Barbara, CA 93106, USA Faculty of Engineering, Shibaura Institute of Technology, Saitama 330-8570, JAPAN [email protected], [email protected], [email protected]
Abstract
We show that one cannot put a stationary (extended) black hole inside certain gravitatingflux-tubes. This includes an electric flux-tube in five-dimensional Einstein-Maxwell theory,as well as the standard flux-branes of string theory. The flux always causes the black hole togrow indefinitely. One finds a similar restriction in a Kaluza-Klein setting where the higherdimensional spacetime contains no matter. a r X i v : . [ h e p - t h ] O c t Introduction
Our basic understanding of black holes includes a series of “no-hair” theorems which state thatcertain types of matter must vanish outside stationary black holes. Intuitively, this says that thematter either falls into the black hole or radiates out to infinity. These theorems were originallyproven for linear fields, but they were thought to hold more generally. It was later realized thatmany types of hair are indeed possible. One class of examples consists of black holes in anti-deSitter space. Perhaps the simplest case is a charged scalar field which can exist outside certaincharged black holes [1, 2]. A large class of asymptotically flat examples involves theories that admitstationary solitons. One can often put a small stationary black hole inside the soliton withoutdestroying it [3]. This includes solutions of Einstein-Yang-Mills theory [4, 5, 6], and monopoles ofthe Einstein-Yang-Mills-Higgs system [7]. (For a comprehensive review of these solutions, see [8].)One can also put a rotating black hole inside a rotating boson stars [9, 10].There are some cases where one cannot put a static black hole inside a static soliton. Forexample, if one puts a small black hole inside a static perfect fluid star it will grow and slowlyconsume the star. Similarly, one cannot put a static black hole inside a static boson star [11].The matter content of the soliton determines whether or not one can put a static black hole inside.In higher dimensions, there are extended black holes such as black strings and black branes.There are also extended solitons such as flux-branes. These are higher dimensional generalizationsof “Melvin’s magnetic universe” [12] which describes a static, cylindrically symmetric, gravitatingmagnetic flux-tube. A natural question is whether one can put a static black brane inside a staticflux-brane. Since there is an exact four dimensional static solution describing a Schwarzschildblack hole inside Melvin’s magnetic universe [13], one might expect the answer is yes. We willshow that this is incorrect. If one puts a black brane inside a flux-brane, it will necessarily growand consume the brane.The simplest example starts with an electric flux-tube in five dimensions. As this solution doesnot appear to exist in the literature, we will numerically construct it in Sec. 4. That is, we find astatic, cylindrically symmetric solution of the five-dimensional Einstein-Maxwell theory describinga self-gravitating electric flux-tube. The product of four-dimensional Schwarzschild and a line is asimple five-dimensional black string with the same symmetry as the flux-tube. Our result showsthat one cannot put a thin black string inside the flux-tube and keep the solution stationary.A much larger class of examples that have been discussed in the literature are the flux-branesof string theory [14], which are sourced by one of the higher rank forms in supergravity. We willreview these solutions in Section 4. Our result implies that one cannot put a stationary blackbrane inside any of these flux-branes .As a final application of our result, one can consider higher dimensional vacuum solutions witha U (1) symmetry. Under Kaluza-Klein reduction there is a Maxwell field, and one can construct an A subtlety here is that although the metric is static, the scalar field has a time dependent phase and is notstatic itself. It was earlier shown [15] that certain rank forms cannot exist outside static black holes which are asymptoticallyflat in all directions. (cid:96) a of the event horizon (in D spacetime dimensions): dθdλ = − D − θ − σ ab σ ab − T ab (cid:96) a (cid:96) b (1.1)where λ is an affine parameter, θ is the expansion, and σ ab is the shear of the null geodesiccongruence. For a stationary black hole, the left hand side must vanish. Since the theories we areinterested in all satisfy the null energy condition, the right hand side is the sum of three negativeterms. In order for a stationary black hole to exist, each one much vanish. This is indeed possiblefor black holes placed inside some solitons [8], but we will see that for the flux-branes, the lastterm is always nonzero. In the Kaluza-Klein example where the higher dimensional solution hasno matter, we will see that the shear is necessarily nonzero on the horizon causing the horizon togrow.The outline of this paper is as follows. In the next section we prove that a uniform blackbrane cannot remain stationary inside a flux-brane. In section 3, we discuss the generalization tononuniform black branes. We will prove that there cannot exist stationary black branes which arenonuniform in a compact direction and argue that similar results hold for noncompact directionsalso. In section 4 we discuss the flux-branes in detail, and construct some numerically. We concludein section 5 with some open questions. The two appendices contain some technical details. Consider a theory of gravity in D dimensions coupled to a closed ( p + 1)-form field strength F p +1 .We will assume a general (Einstein-frame) action of the form S = (cid:90) d D x √− g (cid:20) R −
12 ( ∇ φ ) − p + 1)! e aφ F p +1 (cid:21) (2.1)Note that we have included a possible (but not required) scalar field with coupling to F p +1 governedby a constant a . (One can also include additional matter fields or Chern-Simons terms and theywill not affect our argument.) The equations of motion following from (2.1) are R ab = τ ab , d (cid:63) (cid:0) e aφ F p +1 (cid:1) = 0 , ∇ φ − ae aφ p + 1)! F p +1 = 0 , (2.2)where τ ab is the trace-reversed stress tensor τ ab = T ab − TD − g ab given by τ ab = 12 ∇ a φ ∇ b φ + e aφ (cid:20) p ! ( F p +1 ) a... ( F p +1 ) b... − p ( D − p + 1)! g ab F p +1 (cid:21) . (2.3)3n addition, one has the constraint dF p +1 = 0. There are “flux-brane” solutions to this theorywhich are nonsingular solutions with at least p + 1 (commuting) translational symmetries whichinclude time, and F p +1 is nonzero when restricted to this homogeneous subspace. We will notneed the detailed form of these solutions to prove our “no black hole theorem” so we delay ourconstruction of these solutions until Sec. 4. In this section we rule out uniform black branes insidethese flux-branes, and in the next section we will argue that similar results hold for the nonuniformcase. Theorem:
Consider a “flux-brane” solution to (2.1), i.e., a nonsingular solution in which F tx ··· x p is nonzero, and all fields are independent of t, x , · · · , x p . Then one cannot put a stationary,translationally invariant, black p -brane in the center of this flux-brane. Proof:
First note that F tx ··· x p ≡ E must be constant. If it were a function of another coordinate,say r , then dF = 0 would require that F has a component F r ··· that depends on t or one of the x i contradicting the translation invariance. We will first rule out static black branes, and thengeneralize to the stationary case. If there were a solution describing a static black p -brane in thecenter of this flux-brane with horizon at r = r , then F tx ··· x p F tx ··· x p → ∞ as r → r . To obtaina flux which is smooth on the future horizon, one can introduce F rx ··· x p ( r ) so that the t and r components of F p +1 combine to give F vx ··· x p = E , where v = t + h ( r ) is a good coordinate nearthe horizon. The static Killing field which is null on the horizon is now ξ = ∂/∂v (since we havenot changed the radial coordinate), so the flux of energy crossing the horizon is T bc ξ b ξ c ∝ e aφ ( ξ b F b ··· )( ξ c F c ··· ) = e aφ F vi ··· j F vi ··· j (2.4)This cannot vanish since the right hand side is a sum of nonnegative terms with at least onepositive contribution coming from F vx ··· x p . This contradicts the assumption that the black branewas static, since the Raychaudhuri equation (1.1) shows that the horizon must grow when thereis an energy flux across the horizon.This argument is easily extended to rule out stationary black branes as well. If a black braneis stationary but not static, the Killing vector which is null on the horizon takes the form ξ = ∂/∂t + v∂/∂x + Ω ∂/∂φ (2.5)where x denotes some direction along the brane, i.e., a linear combination of the x i , and φ denotesa rotation in the transverse space. One can first perform a boost in the ( t, x ) plane to a co-movingframe (˜ t, ˜ x ) where the black brane is at rest. Since the flux is boost invariant, we have F ˜ t ˜ x ··· = E .This effectively removes the second term on the right hand side of (2.5). One can now repeatthe argument above. Good coordinates near the horizon of a rotating black hole take the form v = t + h ( r ), ˜ φ = φ + h ( r ). The second coordinate transformation does not affect the flux, andthe first is identical to the static case. So one again finds that if F p +1 is regular on the future The metric on the x , · · · , x p subspace must be positive definite since the horizon is a null surface, and thesecoordinates denote directions along a cross section of the horizon, not along a null generator. p -brane can carryelectric charge of a p + 2 form, or magnetic charge of a D − ( p + 2) form. It can also carry smearedcharges of lower rank forms. All these charges produce fields which are smooth on the horizonwith no flux of energy crossing it. So they do not affect the above result. They cannot stop theblack hole from growing.There are various extensions of this theorem. A simple one just uses Hodge duality. Considera solution with a magnetic q -form field ˜ F which is nonzero. If there is a function h such that F = h (cid:63) ˜ F satisfies the conditions of the theorem, then one cannot put a stationary black branein such a solution.A less trivial extension is the one mentioned in the introduction. Even if there are no form fields F in the higher dimensional theory, there can be solutions which, after dimensional reduction, havea Maxwell field satisfying the conditions of the theorem. One cannot add stationary black branesto such a solution. For example, consider a higher dimensional vacuum solution of the form ds = g yy ( dy + Exdt ) + g tt dt + g xx dx + g ij dz i dz j (2.6)where the metric functions are independent of t, x, y . After Kaluza-Klein reduction on y , one hasa Maxwell field F xt = E , so one cannot add a stationary black string. We will see an example ofthis type of solution in Sec. 4.Our result certainly does not rule out the familiar planar black hole in AdS × S , even thoughthat solution is sourced by a (self-dual) five-form. The reason is that in the usual Poincare co-ordinates for AdS , the nonzero component of the flux is F trx x x . Since the radial direction isincluded (in which there is no translational symmetry) and a constant r surface is null at thehorizon, the right hand side of (2.4) vanishes. We now ask what happens if we relax the assumption that the black branes are translationallyinvariant. It is likely that static spherical black holes can exist inside these flux-branes. Indeed,exact solutions have been constructed describing a static Schwarzschild black hole inside a magneticflux-tube in both four [13], and higher [16] dimensions. In D = 4, one can dualize the magneticflux-tube to an electric flux-tube, but this is not possible for D >
4. So to our knowledge exactsolutions describing static spherical black holes in higher dimensional electric flux-branes have notyet been constructed, but are likely to exist.If the horizon is extended in all p directions, it is likely that it cannot remain stationary evenif we relax the assumption that it is uniform. We will prove this when the nonuniform directionsare compact, and then give an argument which applies to noncompact directions.5 .1 Compact case Suppose one direction x is periodically identified with period L . It is easy to rule out stationaryblack branes which are inhomogeneous in this compact direction. Choose a radial coordinate r such that the horizon is at constant r and x is a coordinate along the horizon. Expanding F tx ··· x p in a Fourier series in x we get F tx ··· x p = F (0) tx ··· x p + (cid:88) n (cid:54) =0 F ( n ) tx ··· x p e πinx L , (3.1)The coefficients, F ( n ) tx ··· x p are independent of x i , but can depend on coordinates off the brane, say r . We can similarly expand F trx ··· x p = F (0) trx ··· x p + (cid:88) n (cid:54) =0 F ( n ) trx ··· x p e πinx L , (3.2)where the coefficients F ( n ) trx ··· x p again are independent of x i , but can depend on r . The condition dF p +1 = 0 can also be expanded in a Fourier series and must hold mode by mode. In particular,if we look at just the zero mode, we get ∂ r F (0) tx ··· x p = 0. But F (0) tx ··· x p must be nonzero at largedistance from the brane since we are considering a flux brane. Since it is constant, we can nowapply the argument in the uniform black brane case to conclude that stationary nonuniform blackbranes cannot exist.This argument immediately generalizes to more than one compact direction along the brane.One can Fourier transform F tx ··· x p in all the compact directions. The overall zero mode must againbe constant and nonzero at large distances from the black brane. When the black brane is inhomogeneous along a noncompact direction, the argument is moresubtle. One cannot just apply a Fourier transform in x and look at the k = 0 contribution. Sincewe have a constant flux at infinity, the individual k = 0 mode diverges. In this section we willshow that inhomogeneous black strings (i.e. p = 1) cannot exist within a flux-tube because thehorizon would thin out along the string direction at a rate so rapid that the string would pinchoff. We leave for future work the general p case which we believe will yield similar results.Let us suppose that the non-uniform black string in a flux-tube solution exists and considerthe features it must possess. At a large radial distance R away from the string, there is a uniformelectric field F tx = E . This means that asymptotically the electrostatic potential grows linearlywith x , A t = Ex . In contrast, the horizon must be an equipotential surface A t = 0 independentlyof x , which means that the equipotential surfaces coming in from large radius cannot hit thehorizon. Instead they must bunch up, producing a radial component of the electric field thatgrows with x (see Fig. 1). The radial electric field on the horizon, F rt , will be positive for large This follows either from our previous argument that if F tx is nonzero on the horizon, there must be energy fluxacross the horizon, or simply from the fact that a nonzero A t on the horizon would have diverging norm. > x <
0. This requires that the black string has a positive charge densityfor x > x <
0, which is similar to what happens if one puts a longconducting rod in an electric field; there is charge separation with positive charge accumulating atone end and negative charge at the other. ++++ ++ ++ + + −−−−− −− −− − − +++++ r + rx Figure 1: Non-uniform black string in an electric flux-tube. The black lines are equipotentialsurfaces. The horizon is required to be an equipotential surface with value A t = 0; this forcesthe field lines to bunch up as x → ±∞ , which means that the black string is locally positivelycharged at large positive values of x , and negatively charged at large negative values of x . Thestring remains neutral overall, however. Inset:
In the neighbourhood of some positive value of x ,the non-uniform black string in a flux-tube should be approximated by a uniform charged blackstring with an electric field perturbation.If the horizon indeed extends to x = ±∞ without pinching off at a finite value of x , the x -dependence of the metric should be negligible near the horizon compared with the r -dependence, asthe radial electric field grows indefinitely. Then, the geometry at large x should be well describedby a perturbation of a translationally invariant charged black string solution. To explore thispossibility, we consider the following ansatz for the D = d + 1 dimensional Einstein-Maxwelltheory (2.1) with no dilaton and p = 1: ds d +1 = − e A H − H + dt + e B (cid:18) H − d − − dx + H − d − d − − H − dr (cid:19) + e C r H d − d − − d Ω d − , (3.3) A µ dx µ = (cid:115) d − d − (cid:18) r − r + (cid:19) d − H + e D dt. Here H ± = 1 − ( r ± /r ) d − and A, B, C, D are functions of r and x . When A = B = C = D = 0,this is the translationally invariant electrically charged black string solution. This solution first7ppeared in [17] for the case d = 4, and the general d solution can be derived by dualisingand uplifting the magnetically charged dilatonic black holes of [18, 19]. The horizon topology is S d − × R and is located at r = r + , while the curvature singularity is at r = r − . The extremal limitcorresponds to r + = r − and has zero horizon area. The temperature is T = ( d − πr + (cid:34) − (cid:18) r − r + (cid:19) d − (cid:35) − d − d − . (3.4)To model the putative non-uniform black string in the neighbourhood of some large positive x value, we will start with this charged solution (with A = B = C = D = 0) and add a small electricflux along the x -direction.A linear perturbation of the uniform charged black string can be described by: A ( r, x ) = (cid:15)A ( r ) x, B ( r, x ) = (cid:15)B ( r ) x, C ( r, x ) = (cid:15)C ( r ) x, D ( r, x ) = (cid:15)D ( r ) x, (3.5)where (cid:15) is the small parameter controlling the perturbation and is proportional to the asymptoticvalue of the electric flux. Note that the perturbed gauge potential A t vanishes on the horizon r = r + as required by regularity. Also note that all perturbations have a linear dependence on x .Typically, a perturbation about a translationally invariant solution would be expanded in modes e ikx . Here we are using the fact that we expect the dominant contribution to come from small k ,and have kept only the linear term .The equations governing the perturbation come from linearizing the background Einstein andform equations (2.2): δR ab = δF ( a | c F b ) c − F ac F bd h cd − d − g ab F · δF − d − h ab F (3.6)+ 12( d − g ab F ce F de h cd , ∇ a ( δF ) ab + 12 ∇ a (cid:0) hF ab (cid:1) − ∇ a (cid:0) h c [ a F cb ] (cid:1) = 0 . (3.7)There are 6 independent components of the Einstein equations and 1 independent form equation.Two equations are first order constraints: the rx -component Einstein equation is the momentumconstraint and a linear combination of the diagonal components yields the Hamiltonian constraint.Using these constraints, the system can be shown to reduce to 4 ODE’s, first order in A , C andsecond order in B , D .The boundary conditions we desire are such that the perturbation is regular at the horizonand asymptotically the perturbed metric functions fall-off to zero and the Maxwell perturbation is We checked that the perturbation equations derived in this section are identical to those derived from a per-turbation with an e ikx − dependence once the limit k → (cid:15) ). (cid:15) : A , B , C → , D → const = 1 . (3.8)Requiring the perturbation to be regular at the horizon translates into the Dirichlet condition A ( r + ) = B ( r + ) as well as additional Robin boundary conditions relating functions and theirderivatives at r = r + . This is a boundary value problem, and to solve it numerically we firstconvert to a compactified coordinate and discretize using a spectral grid. It can then be convertedto a simple linear algebra problem of the form M.v = b , with M a matrix and b, v vectors. InFig. 2 we plot the solutions for the representative case of d = 4, T = (cid:112) / / (4 π ), and r + = 0 . (cid:45) (cid:45)
505 r (cid:144) r (cid:43) (cid:144) r (cid:43) D (cid:72) r (cid:76) Figure 2:
Left Panel:
The numerical solution for A (solid), B (dashed), C (dotted) for d = 4and T = (cid:112) / / (4 π ). At the horizon, A ( r + ) = B ( r + ), as required by regularity. Right Panel:
The numerical solution for D for the same dimension and temperature.It turns out that the effect of the perturbation on the horizon geometry can be understoodvery simply. Starting with the uniform charged solution, Eq. (3.3) with A = B = C = D = 0, theeffect of the electric field can be taken into account by promoting the black hole parameters to befunctions of x : r ± = r ± ( x ). To see this, note that requiring that the temperature be independent of x imposes a relation between dr + /dx and dr − /dx . Using this condition, one can calculate dA H /dx in terms of dr + /dx , where A H is the cross-sectional area of the horizon at fixed x . Returning tothe linearized perturbation, one can also calculate dA H /dx in terms of C ( r + ). Equating these twoexpressions yields dr + dx = (cid:18) d − d + 5( d − d − (cid:19) (cid:15)C ( r + ) r + . (3.9) These boundary conditions correspond to the linearisation (in E ) of the electric flux-brane solutions discussedlater in Sec. 4. Although those solutions are not asymptotically flat, they are when restricted to linear order inthe asymptotic value of the electric field E , which is proportional to (cid:15) in the current perturbative treatment. Thedeviations from flatness arise at O ( (cid:15) ). dg xx /dx (cid:12)(cid:12)(cid:12) r + in two different ways: first in terms of dr + /dx , and then interms of the perturbation B ( r + ). Equating the answers yields dr + dx = − (cid:18) d − d + 5( d − (cid:19) (cid:15)B ( r + ) r + . (3.10)Thus we see that if C ( r + ) + ( d − B ( r + ) = 0 (3.11)holds for our numerical solutions, then the horizon geometry is accurately modelled by the uniformcharged string with the parameters r + , r − promoted to functions of x and subject to the constraintthat dT /dx = 0. And indeed, we find that this condition is satisfied for our solutions, up to a verysmall numerical error. Therefore, the perturbed horizon behaves exactly as the uniform chargedblack string made non-uniform by slowly varying parameters r ± ( x ).Since C < , we see that as x → + ∞ the horizon thins out and the electric field evaluated atthe horizon increases. It seems that there are two possibilities: either the horizon radius goes tozero in finite distance, in which case one cannot place inhomogeneous non-compact black stringsin flux-tubes, or the horizon continues to shrink without pinching off as x increases. This wouldbe a new type of black hole solution, a “spiky black hole”, that would look somewhat like Fig. 1 .To investigate whether such a solution is possible, let us analyse the rate at which the perturbedhorizon radius decreases with x . In Fig. 3, we plot C ( r + ) for d = 4 and fixed temperature, in thiscase arbitrarily chosen to be T = (cid:112) / / π . We find that C ( r + ) is always negative, and appearsto be diverging as r + →
0. The divergence is well fitted by a power law, C ( r + ) ∼ − αr − β + + γ, (3.12)with α, β, γ positive fit parameters that are in principle functions of d and T . Surprisingly, we findthat γ = 2 independent of d or T , and that β is independent of T and takes on the values: d β β → d → ∞ . Lastly, α depends on both d and T , but importantly is always positive. We can now use the fact that the perturbed black string iswell modelled as the uniform black string with x -dependent parameters by combining Eq.’s (3.9),(3.12) to find that for small r + r + ( x ) β ∼ c − c (cid:15)x, (3.13)with c a constant of integration which we take to be positive, and c a positive constant. Clearly r + ( x ) pinches off at finite x , which is very strong evidence against the existence of these “spikyblack holes”. This result, together with the proofs of Sec.’s 2 and 3.1, rules out any sort of extendedblack string in a flux-tube. Of course, even if this solution were found to exist, quantum effects would become important since the curvatureis growing large as the horizon shrinks and comes closer to the singularity at r = r − . p = 1 for the non-uniform, non-compact case, it seemslikely that the result holds for higher p as well. To fully answer this question one should repeatthe analysis done here. The relevant uniform charged black brane solutions can be constructed ina very similar manner to (3.3), although an additional uplift would be required. (cid:45) (cid:45) (cid:45) r (cid:43) C (cid:72) r (cid:43) (cid:76) Figure 3: C ( r + ) for d = 4 and T = (cid:112) / / (4 π ). This data, along with Eq. (3.9), shows that theblack string is tapering off as x increases. The apparent divergence as r + → x distance. In this section we construct various flux-brane solutions to (2.2). They can be described by thegeneral ansatz ds = e α ds p +1 + e β dr + e γ ds D − p − , F p +1 = E vol( ds p +1 ) , (4.1)where α, β, γ and the dilaton φ are all functions of the radial coordinate only, and ds p +1 , ds D − p − are Einstein metrics obeying ( p +1) R µν = λ p +1 δ µν , ( D − p − R ij = λ D − p − δ ij , (4.2)where λ p +1 and λ D − p − are constants, indices belonging to the ( p +1)-dimensional space are labelledwith Greek letters, and the indices belonging to the ( D − p − i, j, .. Note that the form equations are already satisfied: F p +1 is trivially closed, and although its Hodge dual in general possess a non-trivial radial dependence,it also acquires a dr -leg, and is therefore also closed. Also note that the gauge freedom associatedwith the choice of radial coordinate has not been fixed yet. We will find it convenient to work indifferent gauges, therefore we leave it un-fixed for now. The equations of motion for this ansatzare presented in Appendix A. 11 .1 Melvin Flux-branes The above ansatz includes both the original four-dimensional Melvin fluxtube [12], as well as itsgeneralization to higher dimensions and non-trivial dilaton [19] . The geometry of the originalMelvin solution consists of the two-dimensional Lorentz invariant worldvolume of the flux-tube,the radial direction, and a transverse circle. We will call flux-branes Melvin-like if the flux-branehas the same cohomogeneity as the original four-dimensional solution. Thus we set p + 1 = D − p = ( D − ds = Λ a D − D − (cid:0) η µν dx µ dx ν + dr (cid:1) + Λ − D ) a D − D − r dϕ , (4.3) e φ = Λ − a ( D − a D − D − , F p +1 = E vol( η µν dx µ dx ν ) , where Λ is given by Λ = (cid:18) a ( D −
2) + 2( D − D − E r (cid:19) . (4.4)Here we have imposed the gauge α = β which is common for these solutions. For these flux-branesof cohomogeneity-2, exact solutions can be found because the ( D − p − S , which has no curvature. For flux-branes of higher cohomogeneity the circle ispromoted to a sphere, the curvature of which induces a coupling between the metric functions andmakes the equations much harder to solve analytically [14].Note that the magnetic dual of (4.3) involves a two-form Maxwell field. For p = 1 and thespecial value of the dilaton coupling a = − (cid:112) D − / ( D − D + 1 dimensions (see Appendix B). So for this value of a (in the dual magneticframe), the magnetic dual of the above solution comes from dimensional reduction of a vacuumsolution. In fact, it can be obtained from flat Minkowski spacetime by dimensionally reducingalong a combination of a translation and a rotation in a perpendicular plane [20].Our theorem is of limited applicability for this class of flux-branes for the following reason.Near r = 0, the backreaction of the flux-brane can be neglected and the spacetime looks flat.It is known that there are no vacuum black holes with horizon topology R D − × S , and so thestationary black holes excluded by the above theorem do not exist in the absence of flux. In orderto obtain flux-brane solutions that could in principle admit small black branes at their centre, the S must be replaced with a higher dimensional sphere. We now generalize the Melvin flux-branes to branes of higher cohomogeneity, so that the transversespace includes a sphere of dimension two or greater, while keeping the p + 1-dimensional worldvol-ume to be Minkowski space. These solutions have been studied before [14, 21, 22] and include the Our conventions differ from those of [19], in particular we are choosing to work with an electric D − p +1-formas opposed to the dual magnetic 2-form. It is possible that the non-standard asymptotics of the flux-brane geometries allow for these horizon topologies,i.e. that black branes of a large enough radius do exist in the flux-brane. Our theorem then implies that these blackbranes must not be stationary.
D, p, a ):NS flux-branes correspond to (
D, p, a ) = (10 , , −
1) and (10 , , D, a ) = (cid:0) , − p (cid:1) ; and the two flux-branes of M-theory have ( D, p, a ) = (11 , ,
0) or (11 , , D = 5Perhaps the conceptually simplest generalization of the above flux-brane solutions is the case of a(non-dilatonic) electric flux-tube in D = 5. A flux-tube corresponds to p = 1, therefore our ansatzis ds = e α (cid:0) − dt + dx (cid:1) + e β dr + e γ d Ω , F = Edx ∧ dt, (4.5)A convenient gauge choice comes from examination of the constraint equation (A.4). No derivativesof β appear, therefore if we impose a gauge by directly fixing e γ to be a β -independent function(here we will use e γ = r ), then the constraint can be used to algebraically solve for β and thesystem will reduce to a single ODE for α (it seems that the convenience of this gauge has notbeen appreciated before in the literature). One can scale E out of the equation since rescaling thecoordinates t, x, r by λ rescales the metric by λ . Einstein’s equation remains unchanged if we alsorescale F by λ . The resulting ODE for α is:3 z (4 e α + z ) α (cid:48)(cid:48) + 3(8 e α − z ) α (cid:48) − z ( − e α + z )( α (cid:48) ) + 2 z (6 e α − z )( α (cid:48) ) − z = 0 , (4.6)where z ≡ Er .We were unable to determine the general solution analytically, but a simple closed form solutiondoes exist, and is e α = (cid:114) Er, e β = 52 , e γ = r . (4.7)This solution is clearly singular at r = 0 and hence is not of physical interest. Nevertheless, it isquite useful since it can be shown that this solution is an attractor solution for the large- r behaviourof general solutions [14]. To see this, linearise around the exact singular solution, α = α singular + δα .The solution to the linearised perturbation takes the form δα = r − [ c sin(2 ln r ) + c cos(2 ln r )] , (4.8)where c , are constants of integration. For large r , the perturbation oscillates with a decayingenvelope and asymptotes to the above exact singular solution, indicating that it is indeed anattractor.The solution we are interested in is regular along the axis r = 0. We did not succeed infinding it analytically, but the equations can be solved numerically once the boundary conditionsare supplied. The large- r behaviour is that of the attractor, and the small r behaviour is simplyregularity at the origin. There will be a curvature singularity unless e α → const, and e β → r = 0 yields e α = 1 + z z
405 + O (cid:0) z (cid:1) , e β = 1 + 7 z
36 + 61 z O (cid:0) z (cid:1) (4.9)13e have used the scaling symmetry of the Minkowski directions to set an overall constant to 1.The boundary condition required by regularity is then A (0) = 1, A (cid:48) (0) = 0, where A = e α . InFig. 4 we plot the numerical solution for this flux-tube. From the plot, the non-singular solutioncan be seen to asymptote to the attractor solution at large r .This electric flux-tube is much better behaved asymptotically than the Melvin flux-branes. Onecan see from (4.3) that in the Melvin case, the S in the transverse space shrinks to zero size as r → ∞ . In the current solution, the effect of the flux-tube on the asymptotic geometry is weaker,and the S in the transverse space continues to grow asymptotically. However, the solution is stillnot asymptotically flat in the transverse directions. It is asymptotically a cone. E x p (cid:64) Α (cid:68) E x p (cid:64) Β (cid:68) Figure 4: The metric functions e α (left panel) and e β (right panel) for a non-dilatonic electricflux-tube in five dimensions ( D = 5, p = 1, a = 0). The solid curves correspond to the numerical,non-singular solution, and the dotted curve corresponds to the singular attractor solution.Applying the results of Sections 2 and 3 to this solution, we see that we cannot put a stationaryblack string inside the electric flux-tube. Stated intuitively, there is no way to prevent stress energyfrom flowing into the black string, which would cause it to grow.There is a simple physical interpretation of this result in terms of a stretched horizon . Thestretched horizon has finite electrical conductivity, so if one puts a black string along an electricfield, a current will flow. Since the resistance is nonzero, the current will generate heat and causethe black string to grow. So a static black string is impossible. This is very similar to the discussionin [23] where an exact solution was found showing a black hole growing when an electric field isapplied. D = 5An interesting extension of this five-dimensional example is to include a dilaton, and consider itin the context of Kaluza-Klein theory. As before, we choose to impose the gauge e γ = r , and We thank Juan Maldacena for suggesting this. β using the constraint. Now the system has been reduced to two coupled ODE’s in α and φ . As noted in Ref. [14], the dilaton equation of motion becomes a copy of the α equation ofmotion if we set φ = − a α. (4.10)Therefore the system has again been reduced to a single ODE, and the inclusion of the dilatoncomes at no cost in complexity. We omit the presentation of the equation as it is rather lengthy.Once again, we find that the general solution must be solved for numerically, and that a simple,singular solution exists.For the case a = a KK = − (cid:112) /
3, the flux-tube can be uplifted to a vacuum solution of Einsteingravity (see Appendix B). The flux-tube has been geometrized, and the six-dimensional line elementis now ds = e − α ( dy + Exdt ) + e α (cid:0) − dt + dx (cid:1) + e α +2 β dr + e α +2 γ d Ω . (4.11)where we have chosen the gauge A = Exdt . While the metric (4.11) has three translational sym-metries, they do not all commute. The Killing fields include ∂/∂t, ∂/∂y , and ∂/∂x − Et∂/∂y . Thisis a timelike version of the Bianchi Type II symmetries of a homogeneous anisotropic cosmology.To gain some insight into this Ricci-flat solution, we can examine its large- r behaviour, which isgoverned by the singular attractor solution: e α = (cid:18) (cid:19) / ( Er ) / , e β = 2716 , e γ = r . (4.12)Note that the KK circle is pinching off at infinity.Since one cannot put a stationary black string in the five-dimensional flux-tube, one cannot puta stationary black two-brane in this vacuum solution. One can see this directly in six-dimensions, byderiving the following contradiction. Adding a stationary uniform black two-brane would producea metric of the following form: ds = g yy ( dy + Exdv ) + h (cid:2) − f dv + 2 dvdr (cid:3) + g xx dx + r d Ω (4.13)where v is an ingoing null coordinate, and f vanishes at some radius r denoting the horizon.The metric functions g xx , g yy , h depend only on r . The vector (cid:96) = ∂/∂v − Ex∂/∂y is null on thehorizon but it is not a Killing vector. So the six-dimensional spacetime does not have a Killinghorizon. One can show that the vector (cid:96) is tangent to a null geodesic, and has zero expansion,but nonzero shear. This violates the Raychaudhuri equation (1.1) and shows that the assumptionof a stationary solution is inconsistent with the field equations. To see why the shear is nonzero,consider a small bundle of light rays extended in the x and y directions with thickness ∆ x, ∆ y .Starting at v = v , this bundle has a rectangular cross-section. Under evolution by (cid:96) , the rectanglegets distorted with geodesics pushed forward in the y direction by an amount proportional to x .This produces shear.This argument is a Lorentzian version of the one given by Iizuka et al [24] who consider fivedimensional black holes having Bianchi symmetry on the horizon. They show that for Bianchitypes II, VI , or VII , if ∂/∂v − (cid:96) is not a Killing field, then the shear will be nonzero and thesolution cannot be stationary. 15 .2.3 General Case Having studied five-dimensional flux-tubes we now discuss the general case. Much of the aboveanalysis carries over in higher dimensions. The gauge e γ = r is imposed, and the constraintequation is used to algebraically solve for β . The system now only involves two coupled andundetermined functions, α and φ . As in the five-dimensional dilatonic flux-brane, the φ equationof motion becomes identical to the α equation of motion after a rescaling [14] φ = − a ( D − D − p − α. (4.14)Therefore the equations of motion for this general ansatz have been reduced to a single ODE, thepresentation of which we omit as it is quite long. As before a simple analytic family of singularsolutions exists: e α = (cid:20) (2 + a )( D −
2) + 2 p ( D − − p D − D − p −
3) ( Er ) (cid:21) D − p − a D − p )( D − p − , (4.15) e β = ( a ( D − D − p −
3) + 2( p + 1)( D − p − ) ( a ( D −
2) + 2 ( D ( p + 1) − ( p + 1) − D − p −
3) ( a ( D −
2) + 2( p + 1)( D − p − . (4.16)These solutions have been noticed by numerous authors for various values of the parameters( D, p, a ) [14, 21, 22]. Although clearly singular at r = 0, these exact solutions are still quiteuseful as they are often attractors for the large- r behaviour of more general solutions [14]. Asbefore, this can be determined by linearising around the exact solution, α = α singular + δα . Thesolution to the linearised perturbation takes the form δα = r − q [ c sin( ν ln r ) + c cos( ν ln r )] , (4.17)where c , are constants of integration, and q, ν are constants depending on ( D, p, a ). If both q, ν >
0, then the perturbation oscillates with a decaying envelope and asymptotes to the aboveexact singular solution for large r , indicating that it is an attractor. We have checked the singularsolution is an attractor for the following solutions: the two 5d flux-tubes presented above, and allof the flux-branes of string theory/M-theory with co-dimension greater than two.The solutions we are interested in are, of course, nonsingular at the origin. We did not succeedin finding analytic non-singular solutions, but the equations can be solved numerically, just as theywere in the D = 5 cases considered above. We therefore see that there is a large class of non-singular flux-brane solutions for which our theorem applies, including flux-branes that appear instring/M-theory. These flux-branes cannot be “blackened” without introducing time-dependence . As Ref. [14] worked with the Hodge dual picture, their expression is related to ours, (4.14), via p +1 → − ( p +1). Here a comment is in order: The flux-brane solutions are qualitatively different from the p -brane solutions ofstring theory. The p -branes are singular supergravity solutions that possess degrees of freedom associated withtheir worldvolume, whereas the flux-branes are completely non-singular (the simple family of singular solutions isnot physical), and have no worldvolume degrees of freedom. Therefore, the inability to “blacken” the flux-branescarries no implications for a worldvolume theory. Discussion
We have shown that one cannot put stationary black branes inside flux-branes. This was rigorouslyestablished for uniform black branes or black branes that are nonuniform in compact directions.For the case of black strings inside a flux-tube, we have given numerical evidence that this resultextends to the noncompact case, and we expect that it extends to all cases where the horizon isnoncompact.A translationally invariant black string (or black brane) is subject to Gregory-Laflamme insta-bilities. One can stabilize it by compactifying the direction it is extended along. Our result showsthat even when it is stable, the black string cannot remain stationary inside a flux tube.An interesting open question in this compactified context is the following . In vacuum gravitywith one direction compactified on a circle, there is a well studied black hole - black string transitionin the space of static solutions (see [25] and references therein). One can start with a small sphericalblack hole and increase its size. One obtains a continuous family of solutions in which the sphericalblack hole gets distorted when it approaches the size of the circle. It then makes a transition to anonuniform black string and eventually turns into a uniform black string. Now suppose we startwith an electric flux-tube wrapping the compact direction. As we argued in section 2, a smallstatic spherical black hole should still exist. We can continuously increase the size of this blackhole and it is likely that there will again be a transition to a nonuniform black string. But wehave seen that the nonuniform black string cannot be static. So when does the static solution stopexisting? The intuitive picture of the electric field inducing currents on the stretched horizon givenin section 4.2.1 suggests that it will not be until the nonuniform black string forms.What is the analog of this intuitive picture for higher rank forms? Can one understand theabsence of static black two-branes inside F flux vacua, by postulating that the stretched hori-zon contains strings which move when placed inside this flux? This seems likely since one candimensionally reduce a flux p -brane along p − p − Acknowledgements
It is a pleasure to thank H. Elvang, V. Hubeny, J. Maldacena, and A. Puhm for discussions.This work was supported in part by NSF grant PHY12-05500, by NSF grant PHYS-1066293 andthe hospitality of the Aspen Center for Physics. It was also supported in part by JSPS KAKENHIGrants No. 26400280. We thank Henriette Elvang for raising this question. Equations of Motion
Here we present the equations of motion (2.2) for the ansatz (4.1). The form equations are alreadysatisfied as noted above. The three independent (trace reversed) Einstein equations are R µν = (cid:2) − (cid:2) α (cid:48)(cid:48) − α (cid:48) β (cid:48) + ( p + 1)( α (cid:48) ) + mα (cid:48) γ (cid:48) (cid:3) e − β + λ p +1 e − α (cid:3) δ µν , (A.1)= τ µν = − m D − E e aφ − p +1) α δ µν .R rr = (cid:2) − ( p + 1) α (cid:48)(cid:48) − mγ (cid:48)(cid:48) + ( p + 1) α (cid:48) β (cid:48) + mβ (cid:48) γ (cid:48) − ( p + 1)( α (cid:48) ) − m ( γ (cid:48) ) (cid:3) e − β (A.2)= τ rr = 12 e − β ( φ (cid:48) ) + p D − E e aφ − p +1) α .R ij = (cid:2) − (cid:2) γ (cid:48)(cid:48) − β (cid:48) γ (cid:48) + m ( γ (cid:48) ) + ( p + 1) α (cid:48) γ (cid:48) (cid:3) e − β + λ m e − γ (cid:3) δ ij (A.3)= τ ij = p D − E e aφ − p +1) α δ ij . where m = D − p − r . A very useful linearcombination of these equations is the “Hamiltonian constraint” G rr = e − β (cid:20) m ( m − γ (cid:48) ) + 12 p ( p + 1)( α (cid:48) ) + m ( p + 1) α (cid:48) γ (cid:48) (cid:21) (A.4) − m λ m e − γ −
12 ( p + 1) λ p +1 e − α = T rr = 14 ( φ (cid:48) ) e − β + E e aφ − p +1) α . Lastly, the dilaton equation of motion is e − ( p +1) α − β − mγ ∂ r (cid:0) e ( p +1) α − β + mγ ∂ r φ (cid:1) + aE e aφ − p +1) α = 0 . (A.5) B Uplift of Kaluza-Klein Electric Flux-tubes
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