Exact relativistic viscous fluid solutions in near horizon extremal Kerr background
aa r X i v : . [ a s t r o - ph . S R ] S e p Exact relativistic viscous fluid solutions in near horizon extremal Kerr background
Daniel Grumiller and Ana-Maria Piso Institute for Theoretical Physics, Vienna University of Technology,Wiedner Hauptstr. 8-10/136, A-1040 Vienna, Austria, Europe Kavli Institute for Astrophysics, Massachusetts Institute of Technology,77 Massachusetts Ave., Cambridge, MA 02139, USA (Dated: May 28, 2018)Realistic accretion disk models require a number of ingredients, including viscous fluids, elec-tromagnetic fields and general relativistic corrections. Close to the innermost stable circular orbit(ISCO) the latter can be appreciable and (quasi-)Newtonian approximations become unreliable.This is particularly true for nearly extremal black holes like GRS 1915+105, where the ISCO almostcoincides with the black hole horizon. To describe the physics close to the ISCO adequately in asimplified model we approximate the nearly extremal Kerr geometry by the near-horizon extremalKerr geometry and construct in this background relativistic viscous fluid solutions with electromag-netic fields. We discuss some applications of our solutions and possible relations to the Kerr/CFTcorrespondence.
PACS numbers: 04.20.Jb, 04.40.Nr, 04.70.Bw, 47.75.+f, 97.10.Gz, 97.60.Lf
Our understanding of black hole accretion rests on analyticmodels.
Marek Abramowicz [1]
I. INTRODUCTION
John Wheeler coined the term “Black Hole” more thanfour decades ago [2]. At that time Black Holes (BHs)were considered as rather esoteric objects of purely the-oretical interest and little physical relevance (for a text-book on BHs and a history see Ref. [3]). The currentassessment of the role of BHs in physics has changed dra-matically, for several independent reasons. BHs are nowused in various areas of physics, far beyond the originalrealms of application envisaged in the pioneering firstdecade of BH research, including quantum chromody-namics and condensed matter physics. We focus hereon BHs in astrophysics and, to a lesser extent, on BHs inthe context of the gauge/gravity duality (also known asAdS/CFT correspondence). The case for astrophysicalBHs is nicely summarized in Ref. [4].By their very nature, BHs can be seen only indirectly,e.g. through X-ray spectra of accretion disks surroundingthe BH. Therefore, most BHs that have been identifiedthrough astrophysical observations are not isolated ob-jects, but have a binary partner — see for instance thelist in Ref. [5], which contains 20 confirmed BHs. At thetop of that list is a specific BH, GRS 1915+105 in theconstellation Aquila, which has particularly interestingfeatures.The mass of GRS 1915+105 is about 18 solar masses, M = (18 . ± . M ⊙ . This experimental result can be es-tablished accurately by considering quasi-periodic oscil-lations (QPOs) [6]. (Earlier independent measurementsled to compatible results, M = (14 ± M ⊙ , but withhigher uncertainty [7, 8]). The spin of GRS 1915+105was measured more recently [9, 10]. The extraction ofthe spin from the observational data is very involved. The method of Ref. [9, 10] employs a spectral analysisof the X-ray continuum, using a suitable accretion diskmodel to determine the accretion disk parameters andthereby also the BH parameters. The spin, in particular,can be determined very accurately due to its profoundimpact on the BH properties, such as the innermost sta-ble circular orbit (ISCO). The ISCO is the orbit closestto the BH horizon where matter can remain stationaryon a circular orbit. The faster the BH spins, the closerthe ISCO is to the BH horizon. In the limit of maximalspinning (extremal) BHs the ISCO coincides with the BHhorizon. This means that accreting matter in a nearly ex-tremal BH can move much deeper into the gravity well ascompared to non-rotating BHs, and this effect leads to ahardening of the X-ray spectrum and a higher efficiencyfor conversion of accreted rest mass into radiation. Bothof these features are observed for GRS 1915+105. It wasconcluded in Ref. [9] that the dimensionless spin param-eter is nearly maximal, a ∗ > .
98, close to the Thornelimit a ∗ < .
998 [11] and to the theoretical upper bound a ∗ ≤
1, with a ∗ = 1 corresponding to an extremal KerrBH. Therefore, the BH GRS 1915+105 is a nearly ex-tremal Kerr BH.In view of the importance of accretion disks for thedetermination of the BH parameters it would be niceto have some exact results available for nearly extremalKerr BHs. Since back reactions are negligible (seee.g. [12, 13]), a suitable accretion disk model involvesa fluid on the background of a nearly extremal Kerr ge-ometry. This is still a very hard problem, because viscos-ity cannot be neglected and electromagnetic fields shouldbe considered as well, cf. e.g. [14, 15, 16]. A simplifica-tion that is useful in other contexts is the Newtonianapproximation, which works well if the accretion disk issufficiently far away from the BH horizon [14, 15]. How-ever, for nearly extremal BHs accreting matter can comevery close to the BH horizon and the Newtonian approx-imation is not suitable to describe the dynamics close tothe ISCO. Thus, if we are interested to describe the dy-namics of accreting matter close to the ISCO we haveto employ General Relativity. The near-extremality ofGRS 1915+105 may allow for a perturbative expansionin the parameter (1 − a ∗ ) < .
02. To leading order in suchan expansion the dynamics is described by the extremallimit a ∗ = 1. We are specifically interested to describethe dynamics close to the ISCO, so we can simultane-ously impose a second perturbative expansion, namely anear horizon approximation.Our starting point is therefore a background geome-try that is obtained after a double limit: first, assumethat the BH is extremal and second, assume that we arearbitrarily close to the horizon. The geometry obtainedin this way is known as Near Horizon Extremal Kerr(NHEK) and was constructed by Bardeen and Horowitz[17] (the possibility of viewing the vicinity of the extremalKerr BH horizon as a spacetime on its own right was al-ready suggested much earlier by the findings of Bardeenand Wagoner [18]).In this paper we study exact solutions of perfect andviscous fluids on the NHEK background, and also includea discussion of electromagnetic fields.We consider first the standard scenario of a fluid withtimelike velocity, u a u a = −
1, which can describe mas-sive as well as massless particles, depending on the equa-tion of state. We construct exact perfect fluid solutionson the NHEK background for a polytropic equation ofstate. However, the Ansatz that u a be timelike pre-supposes that there is a local heat bath, which definesa preferred rest frame. Temperature arises here as thezero component of a timelike vector, and transformationto a moving frame introduces a Lorentz factor. If theboost between the heat bath and the reference frame ap-proaches the speed of light the Lorentz factor becomessingular, and one should not consider temperature as thezero component of a timelike vector, but instead considera lightlike vector. The only natural way to do this is bydemanding u a u a = 0.The dynamics of a fluid in the NHEK geometry ismapped to the dynamics of a fluid at the extremal KerrBH horizon. Particles moving on stationary orbits atthe BH horizon have to move at the speed of light. Thesame consideration applies to the “heat bath”, becausethere is no meaningful way to define a local rest frame atthe BH horizon. Everything that remains stationary atthe extremal Kerr BH horizon necessarily moves with thespeed of light, and the caveat in the previous paragraphapplies. Thus, we consider a “lightlike heat bath”, in thesense that u a u a = 0. We call a fluid with the property u a u a = 0 “null fluid” (not to be confused with a timelikefluid, u a u a = −
1, that describes lightlike matter, like aphoton fluid).We construct a family of exact solutions for a viscousnull fluid on the NHEK background in the presence ofelectromagnetic fields. For a given velocity profile u a wepredict uniquely (up to an overall rescaling) the viscos-ity function η . The class of velocity profiles that we en- counter has very special properties, and we describe themin detail. To address stability issues we discuss linearizedperturbations and establish some rigidity results, whichpoint to the linearized stability of our exact solutions.Besides the potential phenomenological interest re-lated to observations of GRS 1915+100, there are variouspurely theoretical motivation to consider the NHEK ge-ometry. It is an interesting geometry on its own right anddisplays several remarkable geometric properties, some ofwhich we review in this work. Uniqueness and stabilityresults were established recently [19, 20, 21]. Moreover,the NHEK geometry was exploited in the context of theKerr/CFT correspondence [22], which predicts that anyextremal Kerr BH is dual to a certain conformal fieldtheory (CFT). This conjecture has engendered a lot ofrecent interest (cf. e.g. [23, 24, 25, 26, 27, 28, 29, 30,31, 32, 33, 34, 35, 36, 37]), and it could be rewarding toapply our results also in this context.This paper is organized as follows. In section IIwe review salient features of the NHEK geometry andgeodesics in its background. In section III we constructtimelike perfect fluid solutions for arbitrary polytropicequations of state. We mention difficulties with the in-clusion of viscosity. In section IV we construct null per-fect fluid solutions. We are able to lift them to viscousnull fluid solutions and to include certain electromagneticfields. In section V we address perturbations around ourexact solutions. In section VI we provide a discussion ofour results and put them into the perspective of two per-tinent branches of literature: BH astrophysics and theKerr/CFT correspondence.Before starting we mention some of our conventions.We use signature − , + , + , +. We define the Riemanntensor as R abcd := ∂ c Γ abd − ∂ d Γ abc + Γ ebd Γ ace − Γ ebc Γ ade and the Ricci tensor as R bd := R abad , where we employthe usual Einstein summation convention. We choose thesign of the ǫ -tensor so that ǫ trθφ >
0. Symmetrizationof indices is defined by ∇ ( a u b ) := ( ∇ a u b + ∇ b u a ). Weemploy natural units c = G = 1. II. NEAR HORIZON EXTREMAL KERR
In this section we review the features of Kerr andNHEK most relevant to our work.
A. Kerr geometry and ISCO
The Kerr metric in Boyer-Lindquist coordinates is ds = − (cid:16) − M rρ (cid:17) dt − M ra sin θρ dt dφ + (cid:16) r + a + 2 M ra sin θρ (cid:17) sin θ dφ + ρ ∆ dr + ρ dθ , (1)with ρ := r + a cos θ ∆ := r (1 − Mr ) + a (2)It is determined by two physical parameters: the mass M and the angular momentum J = aM . Often it isconvenient to employ the dimensionless Kerr parameter a ∗ = a/M instead of the angular momentum. With noloss of generality we consider only positive angular mo-mentum, J, a, a ∗ ≥
0. The Kerr BH event horizon r BH = M (1 + p − a ∗ ) (3)exists only if the inequality a ∗ ≤ a ∗ = 1 (4)the Kerr BH becomes extremal. If it is violated, a ∗ > To determine the ISCO we consider geodesics of time-like test particles in the plane θ = π/
2, cf. e.g. [40]. Theradial velocity is determined by˙ r V eff = E N (5)with the effective potential V eff = − Mr + L − a E N r − M ( L − a √ E N + 1) r (6)Circularity of the orbits requires V eff = E N and dV eff /dr = 0. These conditions allow to solve for E and L in terms of r . For the ISCO additionally d V eff /dr =0 must hold because this condition separates stable fromunstable orbits. This condition allows to solve for theradius of the ISCO r = r ISCO : r ± ISCO M = 3 + p x + 3 a ∗ ∓ q (3 + x + 2 p x + 3 a ∗ )(3 − x ) (7) This point of view was recently challenged by Jacobson andSotiriou [39] who found that a BH could over-spin by droppinginto it a test particle with suitable spin and/or angular momen-tum, a procedure which requires a careful finetuning. However,as pointed out in their paper, their analysis does neither takeinto account corrections from gravitational radiation nor fromself-force effects, and such effects could spoil the finetuning re-quired for a violation of the cosmic censorship conjecture. where the upper (lower) sign refers to co-rotation(counter-rotation) and x = 1 + (1 − a ∗ ) / h (1 + a ∗ ) / + (1 − a ∗ ) / i (8)For a ∗ → x = 3 and the Schwarzschild result r ± ISCO = 6 M is recovered. In the extremal limit we get x = 1. The co-rotating ISCO lies at r + ISCO = M and thuscoincides with the BH horizon r BH in (3): r + ISCO (extremal Kerr) = r BH (extremal Kerr) = M (9) B. Near horizon extremal Kerr geometry If a ∗ ≪ a ∗ → a ∗ ≈
1, do not allow a very ac-curate Schwarzschild approximation in the near horizonregion close to the ISCO. In that region, however, theyallow for a different approximation in terms of the theNHEK geometry introduced by Bardeen and Horowitz[17]. The NHEK geometry is obtained from the Kerr ge-ometry (1) as follows. One introduces the dimensionlesscoordinatesˆ t = λt M ˆ r = λMr − M ˆ φ = φ − t M (10)and takes the limit λ → t, ˆ r, ˆ φ, θ fixed. Thisyields the NHEK geometry ds = M (1 + cos θ ) (cid:16) − d ˆ t + d ˆ r ˆ r + 4 sin θ (1 + cos θ ) (cid:0) d ˆ φ + d ˆ t ˆ r (cid:1) + dθ (cid:17) (11)The spacetime is no longer asymptotically flat — for in-stance, at θ = 0 the spacetime along the axis is AdS .Thus, one should not think of the NHEK geometry as anapproximation to Kerr in the same way as one considersSchwarzschild as an approximation to Kerr. Rather, thegeometry (11) describes the (infinite) throat geometry ofan extremal Kerr BH, so anything that happens withinthis geometry is related to physical processes at (or veryclose to) the event horizon.The coordinates used in the line element (11) do notcover the full NHEK spacetime. Global coordinates,which we shall call again t, r, θ, φ (at the minor risk ofconfusion with the Boyer-Lindquist coordinates), are ob- In our conventions the relation J = M holds. tained through the coordinate transformationˆ r = ( p r cos t + r ) − (12)ˆ t = ˆ r p r sin t (13)ˆ φ = φ + ln (cid:12)(cid:12)(cid:12) cos t + r sin t √ r sin t (cid:12)(cid:12)(cid:12) (14)The NHEK metric in these global coordinates is thengiven by ds = M (1 + cos θ ) (cid:16) − (1 + r ) dt + dr r + 4 sin θ (1 + cos θ ) ( dφ + r dt ) + dθ (cid:17) (15)The range of the coordinates is t ∈ ( −∞ , ∞ ), r ∈ ( −∞ , ∞ ), φ ∈ [0 , π ), θ ∈ [0 , π ]. Note that θ maintainsits original role as polar angle since it does not appearin any of the coordinate transformations that led fromKerr (1) to NHEK (15). The radial coordinate r is notrestricted to be positive because no singularity is encoun-tered at r = 0.We conclude this brief review with some geometric fea-tures of the NHEK spacetime (15) and refer to [17] fora more complete discussion. The t = const . hypersur-faces are spacelike globally, so there are no closed timelikecurves. For sufficiently small radii, r < / ∂ τ is timelike. It is also timelike if the inequal-ity 2 sin θ < θ holds, which is saturated for thecritical polar angle θ crit . = 0 .
82 (about 47 . ◦ ). If neitherof these inequalities hold ∂ τ may become spacelike, justlike in the ergoregion of the Kerr BH. The plot Fig. 1 de-picts the norm squared of ∂ τ . We discuss Killing vectorsin more detail in subsection II C below. While NHEKdoes not exhibit any BH horizon, there is still a sensein which one can label a locus in the NHEK geometrywith the attribute “horizon”, namely the horizon of thePoincare-like patch described by the line-element (11) atˆ r → ∞ . If λ is finite in (10) the limit ˆ r → ∞ indeedcorresponds to the locus of the horizon. In global coor-dinates this horizon is mapped to arbitrary finite valuesof r (if ˆ t is infinite) or to r = ∞ (if ˆ t is finite). Thus, inglobal coordinates there is not just a single value of theradial coordinate r that would correspond to the originalBH horizon, but rather it spreads out through the wholeNHEK geometry. The local equivalence between the Poincare-like horizon ofNHEK and the event horizon of extremal Kerr has an analogin the simpler case of the extremal Reissner-Nordstr¨om BH andits near horizon limit AdS × S , see Ref. [41] for a recent dis-cussion. We thank Tom Hartman for discussions on these issues. FIG. 1: Norm squared of Killing vector ∂ t . The (“ergo”-)region with spacelike ∂ t is plotted in gray, the region withtimelike ∂ t is plotted in white. The vertical critical line isat r = 1 / √
3. The horizontal critical lines are at θ ≈ . θ ≈ . C. Killing vectors, invariants and geodesics
The Kerr geometry (1) is stationary and axially sym-metric. Its Killing vectors k = ∂ t k = ∂ φ (16)persist in the NHEK geometry (15). The latter has anenhanced isometry group [17], namely SL (2 , R ) × U (1),whose algebra is generated by the Killing vectors k , k and two additional Killing vectors k ± defined by k + + ik − = e it √ r (cid:0) r ∂ t − i (1 + r ) ∂ r + ∂ φ (cid:1) (17)Generic Killing vectors are neither spacelike nor timelikeglobally. The only exception is k , which is spacelikeglobally. Scalar fields that are invariant with respect toall four Killing vectors can only depend on the coordinate θ . Vector fields that are invariant with respect to all fourKilling vectors must be of the form (cid:2) A t = rA φ ( θ ) , A r = 0 , A θ = A θ ( θ ) , A φ = A φ ( θ ) (cid:3) (18)Since both the Kerr and the NHEK geometry solve theEinstein equations all scalar invariants constructed fromthe Ricci tensor vanish. There is an interesting scalarinvariant that is non-vanishing for the Kerr geometry,namely the Chern-Pontryagin density (also known as in-stanton density). It is defined byCP := 12 ǫ cdef R abef R bacd (19)and for the Kerr geometry it is given byCP = 96 aM r cos θ ( r + a cos θ ) (cid:0) r − a cos θ (cid:1) (cid:0) r − a cos θ (cid:1) (20)The NHEK geometry also has a non-vanishing Chern-Pontryagin densityCP = 96 cos θM (1 + cos θ ) (cid:0) − θ (cid:1) (cid:0) − cos θ (cid:1) (21)Consistently, the NHEK result (21) coincides with theKerr result (20) in the near horizon extremal limit a = r = M . In both cases the Chern-Pontryagin density inte-grates to zero so that the instanton number vanishes. In-terestingly, the Chern-Pontryagin density (21) of NHEKchanges sign at θ = arccos(1 / √ ≈ . θ = π/ V eff = (cid:16) M − L (cid:17) r − LE r (22)We have normalized the 4-velocity to u a u a = −
1. Theconstants of motion are given by L = u φ + ru t E = 1 + r u t − Lr (23)The constant E N on the right hand side of (5) is deter-mined by E N = 2 E − L − / (2 M ). Circularity of theorbits at r = r c requires that the constants of motion aretuned as follows: r c = 4 EL / (2 M ) − L L = 112 M (24)Therefore, circular orbits are possible only at | r c | = ∞ .For L = 1 / (12 M ) the circular orbit is only marginallystable because d V eff /dr = 0. An interesting propertyof timelike geodesics is that they are confined to finiteradii if 12 L < /M , but can escape to infinity if 12 L > /M . The marginal case leads again to the condition L = 1 / (12 M ).Lightlike geodesics experience a slightly different effec-tive potential. Instead of (22) one obtains V eff = − L r − LE r (25)The constant E N on the right hand side of (5) is nowdetermined by E N = 2 E − L . There are no circu-lar orbits for lightlike test particles because V eff = E N and dV eff /dr = 0 cannot hold simultaneously. Lightlikeparticles moving in the plane θ = π/ θ and u θ are arbi-trary. The two constants identified in (23) take now theform L = sin θ θ ( u φ + ru t ) (26) E = (1 + r )(1 + cos θ )2 u t − Lr (27)Like for the Kerr spacetime [42] a third constant of mo-tion D can be identified. For lightlike test particles thecorresponding first integral is given by( u θ ) (1 + cos θ ) = D + 4 cos θ (cid:16) L − L sin θ (cid:17) (28)For arbitrary θ the effective potential (to be inserted into(5) with E N = 0) from (25) becomes V eff = 2 L (1 + r ) (cid:16) θ ) + cot θ (cid:17) − E + 2 Lr ) (1 + cos θ ) + D (1 + r )2(1 + cos θ ) =: V eff0 (29)In the limit θ → π/ D → E N = 2 E − L . For time-like test particles the first integral receives one additionalcontribution as compared to the lightlike case (28):( u θ ) (1+cos θ ) = D +4 cos θ (cid:16) L − M − L sin θ (cid:17) (30)The effective potential for timelike geodesics also containsone additional term as compared to the lightlike case(29): V eff = V eff0 + (1 + r ) sin θ M (1 + cos θ ) (31)Qualitatively, the potentials (29) and (31) behave likethe simpler potential (22): they contain a term quadraticin r that can change its sign, a term linear in r whosesign is determined by − LE and an r -independent term.The discussion about confined geodesics above thereforegeneralizes straightforwardly to generic geodesics.We stress that a collection of non-interacting non-rotating ( L = 0) timelike test particles (“dust”) cannotbe lifted to a regular perfect fluid solution. This can beseen directly from the effective potential (22), which be-comes positive at sufficiently large r if 12 L < /M .Consequently, the radial velocity determined from (5)eventually becomes imaginary. This observation is just arephrasing of the fact that timelike geodesics with van-ishing (or sufficiently small) angular momentum L areconfined [17]. In the next sections we circumvent thisproblem by allowing for pressure and find fluid solutionsthat are well-defined everywhere. III. TIMELIKE PERFECT FLUID
In this section we consider perfect fluid solutions onthe NHEK background with a timelike velocity 4-vector u a . A. Equations of motion
Without loss of generality we employ the normalization u a u a = − T ab f = ρ u a u b + P ∆ ab ∆ ab := g ab + u a u b (33)must be covariantly conserved ∇ a T ab f = 0 (34)Here ρ is the density, P is the pressure and both areconnected by the polytropic equation of state P = α ρ n (35)For simplicity we assume for the time being n = 1.Contracting the conservation equation (34) with u b leads to the relativistic continuity equation ∇ a (cid:0) ρ u a (cid:1) + P ∇ a u a = 0 (36)Contracting the conservation equation (34) with ∆ ab leadsto the relativistic momentum equations (cid:0) ρ + P (cid:1) u a ∇ a u b + ∆ ab ∇ a P = 0 (37)These are our equations of motion. The simple structureof the NHEK background (15) considerably reduces theamount of terms contributing to the equations of motion.Moreover, by analogy to the Kerr case [43, 44, 45, 46] weshall exclusively consider velocity profiles that are sta-tionary and axisymmetric. u a = u a ( r, θ ) (38)In appendix A the equations of motion are displayed ex-plicitly for arbitrary velocity profiles of the form above(38). For vanishing polar velocity u θ = 0 the continuityequation (36) evaluated on the NHEK background (15)reduces to u r ∂ r ρ + ( ρ + P ) ∂ r u r = 0 [ u θ = 0] (39) B. Exact solutions
We consider now solutions to the equations of motionwith vanishing polar velocity u θ = 0. A simple class ofsuch solutions can be obtained by assuming u r = 0, whichobviously solves the continuity equation (39). Since the equations of motion are still somewhat lengthy we restrictour attention to the plane θ = π/ u φ = 0 and therefore u t = 1 M p (1 + r ) u φ = − r u t (40)Physically, this velocity profile corresponds to choos-ing vanishing angular momentum of the fluid particles,see the left Eq. (23). The continuity equation holdsper Ansatz, and the momentum equations evaluated at θ = π/ ∂ θ ρ = 0 [ θ = π/
2] (41) ∂ r ρ + (1 + α ) rα (1 + r ) ρ = 0 [ θ = π/
2] (42)The momentum equations are solved by the followingdensity and momentum profiles at θ = π/ ρ ( r, θ = π/
2) = ρ (1 + r ) − α α P = α ρ (43)Our next goal is to lift the solution in the plane θ = π/ u t = 1 M p (1 + r )(1 + cos θ ) (44a) u r = 0 (44b) u θ = 0 (44c) u φ = − rM p (1 + r )(1 + cos θ ) (44d) ρ = ρ (cid:0) (1 + r )(1 + cos θ ) (cid:1) − α α (44e) P = α ρ (44f)The configuration (44) solves the equations of motion(36), (37). It is gratifying that reasonable values of theequation of state parameter, α ∈ [ − , n = 1 (and n > θ = π/ ρ ( r, θ = π/
2) = (cid:16) − α + ρ (1 + r ) /n − (cid:17) n − (45)and P = α ρ n . For negative α the solution is always glob-ally well-defined. For positive α and n > α and0 < n < αρ > x a the acceleration g a between these particles can becalculated from the geodesic deviation equation g a = R abcd u b u c x d (46)where u a is the 4-velocity. For the solution (44) and fora vector x θ = 0, x a = 0 otherwise, we obtain g θ = x θ − θM (1 + cos θ ) (47)Interestingly, there is a zero in the acceleration for a cer-tain polar angle, even without the presence of additionalforces. The critical value is given by θ ≈ .
955 (about54.74 degrees). Curiously, this value of θ also leads to avanishing Chern-Pontryagin density (21).We do not further dwell on the perfect fluid case sinceour main interest are viscous fluids. C. Shear viscosity
In the presence of shear viscosity the energy momen-tum tensor for a viscous relativistic fluid receives an ad-ditive contribution T ab vf = T ab f + Π ab (48)The viscosity tensor Π ab is given byΠ ab = − η (cid:0) ∆ ac ∆ bd ∇ ( c u d ) −
13 ∆ ab ∇ c u c (cid:1) (49)The function η parameterizes the local strength of viscos-ity. Three important properties of the viscosity tensor aresymmetry, tracelessness and transversality:Π ab = Π ba Π aa = 0 Π ab u b = 0 (50)The tracelessness is a consequence of neglecting bulk vis-cosity. The continuity equation generalizing (36) is nowgiven by ∇ a (cid:0) ρ u a (cid:1) + P ∇ a u a + Π ab ∇ a u b = 0 (51)The momentum equations generalizing (37) are nowgiven by (cid:0) ρ + P (cid:1) u a ∇ a u b + ∆ ab (cid:0) ∇ a P + ∇ c Π ca (cid:1) = 0 (52)In the Newtonian limit only the momentum equations re-ceive corrections from viscosity. It may be checked easily that the momentum equations (52) are fulfilled for oursolution (44), provided the viscosity function is chosen as η = η ( θ ) √ r [solves momentum equations] (53)Here η is an arbitrary function of the polar angle θ .Technically, the reason for the persistence of the solution(44) with (53) at the level of momentum equations (52)is that for u r = u θ = 0 the only non-vanishing compo-nents of the viscosity tensor are Π rφ and Π tr (actually,the latter component also vanishes in our case). How-ever, the continuity equation (51) is violated unless η vanishes. Thus there is no simple way to lift our perfectfluid solution (44) to a viscous fluid solution.It is important to understand why this happens, sothat a more refined analysis can resolve this problem. Aswe have pointed out, the perfect fluid continuity equa-tion (36) holds identically for u r = u θ = 0. Therefore,the contribution from the viscosity tensor to the conti-nuity equation would have to vanish by itself. This isnot the case for the configuration (44) with (53) unless η = 0. We conclude that a configuration for a viscousfluid should either contain a radial velocity component, u r = 0, or a polar velocity component, u θ = 0, or both,unless the identity ( ∇ ( a u b ) )( ∇ a u b ) = 0 holds. We havenot succeeded in finding exact solutions with viscosityfor the timelike case. However, we are able to constructsolutions for the lightlike case discussed below. IV. VISCOUS NULL FLUID
In this section we consider lightlike viscous fluids u a u a = 0 (54)and construct exact solutions. We also add an electro-magnetic field. A. Physical preliminaries
Before writing down any equations we want to includesome input from physics, because this will guide us tofinding appropriate solutions of the equations of motion.First of all, as explained in the introduction, the lightlikecase is relevant for physics at or near the ISCO. Second,it is a reasonable approximation to consider vanishingangular momentum, because frame dragging effects act-ing on particles near the ISCO dominate over the relativemovement on top of the frame dragging effects. From theleft Eq. (23) we may therefore assume that u φ = − r u t .Third, we want to include viscosity and have experienceddifficulties with such an inclusion if u r = u θ = 0 (see pre-vious section). Therefore, we should refrain from settingboth of these velocity components to zero. In additionto these physically motivated conditions we assume forthe time being the polar velocity to be negligible as com-pared to the other velocity components. Consequently,we employ again the simplification u θ = 0. The onlyunknown functions in the 4-velocity are therefore u t and u r , and they are related by the null condition u a u a = 0.Putting all these ingredients together leads to the follow-ing Ansatz for the 4-velocity: u t = u t ( r, θ ) u r = (1 + r ) u t u θ = 0 u φ = − r u t (55)Additional constraints come from our desire to eventu-ally match scalar quantities in the NHEK geometry withcorresponding quantities in the Kerr geometry evaluatedat/near the ISCO. However, we have seen in sectionII C that the ISCO at r = M in the Kerr geometry ismapped to arbitrary values of the radial coordinate r in the NHEK geometry. Thus, a sensible matching ispossible only if the corresponding scalar quantities areindependent from r . Since we also require φ and t inde-pendence for our background solution, all scalars in theNHEK geometry that allow for a sensible matching mustbe invariant with respect to all four Killing vectors (16),(17) — they can only depend on the polar angle θ . Thus,we assume ρ = ρ ( θ ) P = 13 ρ η = η ( θ ) (56)where for consistency we have imposed the relativisticequation of state for a lightlike fluid.The energy momentum tensor of a viscous null fluid isgiven by T ab vnf = P (cid:0) u a u b + g ab (cid:1) + Π ab (57)The shear viscosity tensor is slightly different as com-pared to the timelike case (49):Π ab = − η (cid:0) ∆ ac ∆ bd ∇ ( c u d ) −
14 ∆ ab ∇ c u c (cid:1) (58)Instead of the properties (50) we obtainΠ ab = Π ba Π aa = 0 Π ab u a u b = 0 (59)but not necessarily Π ab u b = 0. The continuity equationin general simplifies to u a ∂ a P + u a ∇ b Π ab = 0 (60)Together with the Ansatz u θ = 0, P = P ( θ ), the continu-ity equation (60) turns into a constraint on the viscositytensor. u a ∇ b Π ab = 0 (61)The momentum equations can be presented as (cid:0) u a u b + g ab (cid:1) ∂ a P + 4 P ∇ a ( u a u b ) + ∇ a Π ab = 0 (62)Of course, only three of them are independent from thecontinuity equation (60). B. Exact solutions
Our task is now to find an appropriate velocity pro-file u t ( r, θ ) and expressions for ρ ( θ ) and η ( θ ). Thederivations are presented in appendix B. Thus, given theAnsatz (55), (56) we can present already the most gen-eral exact relativistic viscous fluid solution in the NHEKbackground: u t = u ( θ )(1 + r )(1 + cos θ ) (63a) u r = u ( θ )1 + cos θ (63b) u θ = 0 (63c) u φ = − r u ( θ )(1 + r )(1 + cos θ ) (63d) ρ = ρ (63e) P = 13 ρ (63f) η = η ( θ ) (63g)The functions η and u are not independent from eachother but related by the differential equation η ′ (cid:16) u ′ + 2 sin θ cos θ θ u (cid:17) + η (cid:16) u ′′ + 2 cot θ + sin θ cos θ θ u ′ (cid:17) = 0 (63h)An interesting property of our solution is transversalitywith respect to velocityΠ ab u b = 0 (64)Thus, even though the viscosity tensor (58) in generaldoes not exhibit the transversality property (64), it hasthis property for the most general solution (63) compat-ible with the Ansatz (55), (56). Therefore, our viscositytensor has all the standard properties (50) like for anordinary fluid in the Landau-Lifshitz frame [47]. Otheruseful properties of the viscosity tensor and the velocityprofile (63) are the following ones: ∇ a Π ab = 0 (65a) ∇ a u a = 0 (65b) u a ∇ a u b = 0 (65c) ∇ a u b − ∇ b u a = 0 ⇔ u = const . (65d)Thus, the viscosity tensor and the velocity aredivergence-free (65a), (65b). The velocity profile isgeodetic (65c). It has vanishing twist (vorticity) if andonly if u is constant (65d). In that case the veloc-ity profile coincides with the velocity profile of lightlikegeodesics with L = 0 in (26)-(29). The shear of the ve-locity profile is non-vanishing in general, ∇ ( a u b ) = 0, butits norm vanishes, ∇ ( a u b ) ∇ ( a u b ) = 0, as a direct conse-quence of the properties (65b), (65c).If the velocity profile is known (or can be guessed onphysical grounds) the differential equation (63h) leadsto a prediction for the viscosity profile. Of course, notall solutions that are allowed mathematically make sensephysically — the velocity and/or viscosity profiles mayhave singularities at certain angles, or viscosity may failto be non-negative in the whole spacetime. The simplestsolution (63) that is physically meaningful in the wholeNHEK spacetime has constant u and constant viscosity η >
0. According to (65d) this is the only solution withvanishing vorticity. A solution with vorticity is given by u ∝ cos θ + ln tan θ and constant viscosity η >
0. Thissolution has a velocity profile with a logarithmic singu-larity at the poles θ = 0 , π , but is regular and physicallyacceptable otherwise. C. Conserved currents
We construct now conserved currents by following astandard procedure. We start with the equations of mo-tion (62) and contract them with the Killing vectors (16),(17). By virtue of the Killing equation ∇ ( a k b ) 0 , , ± = 0we obtain currents J a , , ± = P k a , , ± + 4 u a u b k b , , ± + Π ab k b , , ± (66)that are conserved ∇ a J a , , ± = 0 (67)We discuss now briefly some properties of these currentsevaluated on the solutions (63). The current related toenergy flux is given by J a = P δ at + 4 u a u t + Π at (68)Depending on the velocity profile function u ( θ ) thiscurrent may be spacelike or timelike in certain regionsof spacetime. For large values of | r | the current (68)is spacelike between the two horizontal asymptotes de-picted in Fig. 1, essentially the “ergoregion”, and timelikeelsewhere, regardless of the velocity profile. The currentrelated to angular momentum flux is given by J a = P δ aφ + Π aφ (69)For any velocity profile function u ( θ ) this current isspacelike everywhere, except at the poles θ = 0 , π whereit becomes lightlike. This result can be seen easily byexploiting the propertyΠ ab Π bc = 0 if a = θ, φ or c = θ, φ (70)which implies J a J a ∝ k a k a ∝ g φφ . The remaining twocurrents have no analogue in the Kerr spacetime, and wehave not found a simple physical interpretation for them.Integrating the current conservation equation (67) oversome spacetime volume V and using Gauss’ law leads tointegral identities Z ∂V d x p | γ | J a , , ± n a = 0 (71) Here ∂V is the boundary of the spacetime volume, n a the outward pointing unit normal and γ the determinantof the induced metric at the boundary. Of particularinterest are boundaries at constant radius, r = r , so that n a = √ g rr ( ∂ r ) a and p | γ | = √− g √ g rr . If we consideras volume V an interval r ∈ [ r , r + ε ], with the othercoordinates arbitrary, then the integral identity (71) inthe limit of ε → Z r = r d x √− g ∂ r J r , , ± = 0 (72)Thus, if the local identity ∂ r J r , , ± = 0 (73)holds the integral identity (72) is implied automatically.A particularly simple case is the integral identity (72)for the angular momentum flux J . Dropping the trivial t and φ integrations we obtain π Z dθ sin θ (1 + cos θ ) ∂ r Π rφ = 0 (74)With the formula for Π rφ from the appendix (B1e) andthe solution for u t (63a) we obtainΠ rφ = 4 η u sin θ (1 + cos θ ) ≥ η and the time component of the velocity (andhence u ) must be non-negative.We postpone applications of the integral identities (72)and a comparison to analog identities in the Kerr case tothe conclusions. D. Electromagnetic field
To describe a relativistic viscous plasma we introducethe abelian field strength F ab = ∇ a A b − ∇ b A a = ∂ a A b − ∂ b A a (76)in terms of the 4-vector potential A a . Then the homoge-neous Maxwell equations ǫ abcd ∇ b F cd = 0 (77)hold automatically. The inhomogeneous Maxwell equa-tions ∇ a F ab = 4 π j b (78) Note that the determinant of the metric is independent from r . j a (with some convenientnormalization). The full energy momentum tensor T ab vnp = T ab vnf + T ab M (79)receives a contribution from the Maxwell field: T ab M = 14 π (cid:16) F ac F bc − g ab F cd F cd (cid:17) (80)We consider here exclusively solutions where the Maxwellenergy momentum tensor is conserved by itself. ∇ a T ab M = 0 (81)We have found several solutions with vanishing or non-vanishing current. The latter all turned out to be prob-lematic in the following sense: there is always a partof the spacetime where the current becomes spacelike, j a j a >
0. This need not be a generic feature, but itis a feature present in all the solutions we have found.Therefore, we consider only solutions with j a = 0.A particular solution of this type is given by the gaugefield A t = B ln (1 + r ) − E (cid:0) cos θ + ln tan θ (cid:1) (82a) A φ = 2 B arctan r (82b) A r = 0 (82c) A θ = 0 (82d)The field strength can be decomposed into electric andmagnetic parts with respect to a 4-vector n a , which isusually assumed to be timelike and normalized. Since wedo not have some preferred timelike vector field available,we choose instead n a = u a . The associated magnetic fieldis given by B a = n b ǫ bacd F cd with the non-vanishingcomponents B θ = − B u t M sin θ B φ = E u t (1 + cos θ )2 M sin θ (83)The associated electric field is given by E a = n b F ba , withthe non-vanishing components E θ = E u t M sin θ E φ = B u t (1 + cos θ )2 M sin θ (84)The action invariant − F ab F ab for the solution (82) isgiven by − F ab F ab = E − B M (1 + r ) sin θ (85)The instanton invariant ǫ abcd F ab F cd for the solution (82)is given by ǫ abcd F ab F cd = 8 EBM (1 + r ) sin θ (86)Both invariants exhibit second order poles at the twopoles θ = 0 , π . The action invariant is zero if E = ± B ,the instanton invariant is zero if either E or B vanishes.Our discussion of electromagnetic fields is far frombeing exhaustive, but it clearly demonstrates that non-vanishing electromagnetic fields can be included straight-forwardly. V. PERTURBATIONS
In this section we consider small deviations from sta-tionarity and axisymmetry. We decompose all fields intoa background contribution and a fluctuation. Our mainassumption is that the fluctuations in the viscosity func-tion η are negligible as compared to fluctuations in den-sity or fluctuations in velocity, concurrent with previousapproaches (cf. e.g. [48]). We also assume that fluctu-ations do not change the equation of state, P = ρ/ u af = u a + ǫ δu a (87a) ρ f = ρ + ǫ δρ (87b) P f = P + ǫ δP = 13 ρ f = 13 ρ + ǫ δρ (87c) η f = η (87d)Quantities with an index f contain both the backgroundsolution (denoted without index) and the fluctuation (de-noted by δ and scaled by the small parameter ǫ ). Thefluctuation functions δu a and δρ depend on all coordi-nates t , r , θ , φ . We demand that the total velocity belightlike. u af u fa = 2 ǫ u a δu a + O ( ǫ ) = 0 (88)The inner product between the background velocity u a and the velocity fluctuation vector δu a must thereforevanish to leading order in ǫ . In this work we shall ex-clusively consider first order expressions in ǫ , and manyof the equalities below are valid not exactly, but only upto higher order terms. Another assumption concerns thetransversality property (64): we demand that this prop-erty is maintained to leading order in the fluctuations, u b δ Π ab + Π ab δu b = 0 (89)The property (89) guarantees that also at linearized or-der in the fluctuations our viscosity tensor has all thestandard properties (50). For similar reasons as in sec-tion IV A we require that scalar quantities like pressureexhibit no radial dependence, ∂ r δP = 0 (90)We assume further that the fluctuations do not intro-duce angular momentum, δu φ = 0 (91)The conditions (88) and (91) imply that the velocity fluc-tuation vector has only two independent components andconveniently can be decomposed as δu a = u a δu + δ aθ δu θ (92)The first contribution to the velocity fluctuation vector(92) leads to a rescaling of the velocity. We call these1fluctuations “scaling fluctuations”. The second contri-bution gives the velocity profile a θ -component and thusprovides a vertical component to the velocity vector incylindrical coordinates. We call these fluctuations “ver-tical fluctuations”. We shall discuss both contributionsseparately.Before such a discussion we address an important tech-nical issue. Below we shall encounter various partial dif-ferential equations of the form u a ∂ a f = 0 ⇔ ( ∂ t + (1 + r ) ∂ r − r∂ φ ) f = 0 (93)where f is some physical quantity and u a is the back-ground velocity profile given by (63). The general so-lution of the differential equation (93) consistent withperiodicity in the azimuthal angle φ is given by f = ∞ X n =0 f cn ( t − arctan r, θ ) cos (cid:0) n ( φ + 12 ln (1 + r )) (cid:1) + ∞ X n =1 f sn ( t − arctan r, θ ) sin (cid:0) n ( φ + 12 ln (1 + r )) (cid:1) (94)The free functions f cn , f sn depend on two arguments: thefirst one is the combination ( t − arctan r ) and the secondthe polar angle θ . A. Vertical fluctuations
We set δu = 0 and keep only δu θ in the velocity fluc-tuation vector (92).The θ -component of the transversality condition (89)implies η u c ∂ c δu θ = 0 (95)Therefore, δu θ must have the same Fourier decomposi-tion as the function f in (94). The φ -component of thetransversality condition (89) holds identically. The t - and r -components of the transversality condition (89) are re-dundant with each other and yield another condition. u a δ Π ra + Π rθ δu θ = 0 (96)With the definition (58) for the viscosity tensor the con-dition (96) establishes a first order differential equationfor the fluctuation δu θ . ∂ θ δu θ = (cid:16) u ′ u + (1 − θ ) cot θ θ (cid:17) δu θ (97)Its general solution is given by δu θ = u ( θ )sin θ (1 + cos θ ) f ( t, r, φ ) (98)Of course, the integration function f ( t, r, φ ) is not ar-bitrary, but must have the Fourier decomposition (94). For any background profile that has non-vanishing u at the poles θ = 0 , π , the solution (98) necessarily di-verges there. If δu θ diverges at some points it should notbe regarded as a small (first order) fluctuation. There-fore, without having to solve any of the equations of mo-tion we are led to a rigidity result: there are no verticalfluctuations as long as the background velocity does notvanish at the poles θ = 0 , π . This applies to the cases u = const . and u ∝ cos θ +ln tan θ discussed in sectionIV B. Vertical fluctuations with a regular profile for δu θ can exist only if u ( θ ) vanishes at the poles θ = 0 , π . B. Scaling fluctuations
We set δu θ = 0 and keep only δu in the velocity fluc-tuation vector (92).The continuity equation (60) simplifies to u a ∂ a δP + u a ∇ b δ Π ab = 0 (99)Transversality (89) implies u a δ Π ab = − η u b u c ∂ c δu = 0 (100)It can be shown that transversality also implies u a ∇ b δ Π ab = 0. Therefore, the continuity equation fur-ther reduces to u a ∂ a δP = 0 (101)Together with the requirement (90) we obtain the follow-ing result from the general solution (94): δP = δP ( θ ) (102)Similarly, from the condition (100) we obtain δu = ∞ X n =0 δu cn ( t − arctan r, θ ) cos (cid:0) n ( φ + 12 ln (1 + r )) (cid:1) + ∞ X n =1 δu sn ( t − arctan r, θ ) sin (cid:0) n ( φ + 12 ln (1 + r )) (cid:1) (103)The momentum equation (62) with free index φ sim-plifies to ∇ a δ Π aφ = 0 (104)and is fulfilled identically. The momentum equation (62)with free index θ then simplifies to ∂ θ δP = 0 (105)which implies that δP = const . A constant shift of theconstant background pressure can be absorbed by a re-definition of units and therefore no physical pressure fluc-tuations are encountered in scaling fluctuations. The ve-locity fluctuations, however, can be non-trivial.2The remaining two momentum equations with free in-dices t, r are redundant with each other. We consider the r equation. ∇ a δ Π ar = 0 (106)It leads to a second order differential equation in the co-ordinate θ . Since we can linearly superpose fluctuationsit is sufficient to consider one of the Fourier modes. Wetake δu = f cn ( t − arctan r, θ ) cos (cid:0) n ( φ + 12 ln (1 + r )) (cid:1) (107)for some fixed integer n . The second order differen-tial equation descending from the r -momentum equation(106) reads explicitly (cid:16) ∂ θ + (cid:0) η ′ η + 2 u ′ u + (3 − cos θ ) cot θ θ (cid:1) ∂ θ − n (1 + cos θ ) θ (cid:17) f cn = 0 (108)With η and u given this equation can be solved.We consider first the simplest case, u = const . (andtherefore η = const . as well). Then the general solutionfor the velocity fluctuation is given by δu = ∞ X n =0 cos (cid:0) n ( φ + 12 ln (1 + r )) (cid:1)(cid:0) f ccn ( t − arctan r ) cosh Θ n + f csn ( t − arctan r ) sinh Θ n (cid:1) + ∞ X n =1 sin (cid:0) n ( φ + 12 ln (1 + r )) (cid:1)(cid:0) f scn ( t − arctan r ) cosh Θ n + f ssn ( t − arctan r ) sinh Θ n (cid:1) (109)with Θ n := n (cid:0) cos θ + ln tan θ (cid:1) (110)where the functions f ccn , f csn , f scn and f ssn are arbitrary.At the poles θ = 0 , π the Fourier modes diverge becauseof the term ln tan θ appearing in (110). We should dis-card fluctuations that diverge at some points, becauseclearly the linearized Ansatz is no longer valid for suchcontributions. The only admissible fluctuation thereforecomes from the zero mode, n = 0, and the regular partof the fluctuation δu becomes independent of the polarand azimuthal angles, δu reg = f ( t − arctan r ) (111)The arbitrary function f is not fixed by any require-ment so far. It is remarkable that there can be nooscillations in the angular coordinates. If we requirethat the fluctuations vanish at some initial time t = t then f ( t − arctan r ) must vanish for all values of r , which implies that f must vanish in the interval( t − π/ , t + π/ u = const . We consider now generic functions u , η . In that caseit is quite difficult to solve the second order differentialequation (108). Moreover, it may well be that again allFourier modes have to vanish because of regularity re-quirements. We focus therefore on the zero mode n = 0. δu = A ( θ ) f ( t − arctan r ) (112)Then (108) simplifies to A ′′ + A ′ (cid:16) u ′ u + η ′ η + (3 − cos θ ) cot θ θ (cid:17) = 0 (113)where prime denotes differentiation with respect to θ .One solution is given by A = const . The other solutionis obtained by defining y := u η A ′ , which leads to thefirst order differential equation y ′ + y (3 − cos θ ) cot θ θ = 0 (114)The general solution for the function A is given by A ( θ ) = A + A Z θ xu ( x ) η ( x ) sin x dx (115)If either the velocity profile or the viscosity has a zero forsome value of θ the second term in (115) has a singularityand for consistency we must set A = 0. Even if u and η are non-vanishing and regular throughout the wholespacetime we may have to set A = 0 for consistency. Anexample is the simple solution η = const . , u = const . ,which leads to an integral in the second term of (115)that diverges at the poles θ = 0 , π . Typically, the onlyallowed fluctuations have A = 0, A = 0. Without lossof generality we may set A = 1. Consistent linearizedperturbations for the zero mode n = 0 are therefore givenby δu a = u a f ( t − arctan r ) (116)By the same token as before we obtain a rigidity result:if δu a is supposed to vanish at an initial time t = t thefunction f must vanish in the open interval ( t − π/ , t + π/ VI. ASTROPHYSICS AND KERR/CFT
In this section we summarize and interpret our results,and put them into the perspective of the diverse litera-ture.3In section I we considered the black holeGRS1915+105, whose spin is nearly maximal. Thedynamics close to the innermost stable circular orbit(ISCO) could therefore be well described by consideringviscous fluid solutions on an extremal Kerr backgroundin the near horizon region. A non-standard feature ofthe fluid is that on the ISCO its velocity vector shouldnot be timelike but lightlike, since the ISCO coincideswith the extremal horizon. In section II we introducedthe near horizon extremal Kerr (NHEK) geometry(15). The NHEK spacetime is obtained by consideringa double limit of the Kerr spacetime: near horizonapproximation and near extremality of the Kerr blackhole. We reviewed geodesics of timelike and lightlike testparticles in the NHEK background for fixed polar angle θ = π/
2. We generalized these results to arbitrary polarangles θ and polar velocities u θ and identified a thirdconstant of motion D (28), analogous to the Kerr case[42]. In section III we constructed exact solutions for atimelike perfect fluid (44) on the NHEK background,but we had difficulties in including viscosity. Since theNHEK geometry describes the physics on the ISCO,physically relevant results can be obtained if we demandthat the velocity vector be lightlike instead of timelike.Consequently, our next step consisted in finding exactsolutions for a viscous null fluid in section IV. Westarted from the conservation equation of the energymomentum tensor for a viscous null fluid, presentedin (57). The results that we obtained for the velocityprofile, density, pressure and viscosity are presented in(63). For any given velocity distribution we predictuniquely the viscosity function, up to rescaling (63h).We added an electromagnetic field (76). Consequently,the energy momentum tensor receives an additionalcontribution given by (80). For the velocity profile (63),one particular solution for the gauge field that doesnot induce any current takes the form (82). In sectionV we considered first order perturbations around thebackground solution (63). Our assumptions (89) – (91)led us to separate the perturbations into scaling andvertical fluctuations (92). We obtained some rigidityresults from which we concluded that there are nophysically significant first order perturbations.In summary, it is generally not possible to find exactsolutions to the viscous fluid equations in the presenceof electromagnetic fields. We were able to succeed dueto the simplifications implied by the highly symmetricNHEK background (15).We address now potential applications in astrophysics.A quantity of interest in numerical simulations of accre-tion disks is the viscous torque (cf. e.g. [12, 13]; elec-tromagnetic torque [49] does not arise for our solution).Namely, some boundary condition should be imposed onthe viscous torque, and it is debatable what is the ap-propriate boundary condition at the ISCO. It may seemnatural to demand vanishing viscous torque at the ISCO,see e.g. [1] for a review, but this is not necessarily a goodcondition (see for instance [50], where it is also shown that viscous torque can change its sign). Viscous torquein our case is determined by one component of the viscos-ity tensor (B1), namely by Π rφ , integrated over a surfaceof constant radius. This result can be derived by fullanalogy to the Kerr case [50]. The integrated conserva-tion equation for the angular momentum flux J consistsonly of one term, namely the viscous torque integratedalong the ISCO boundary, see (74). Our result (75) showsthat for the NHEK background the viscous torque can-not change its sign. The result (75) also provides theappropriate boundary condition for the viscous torqueas a function of the velocity profile u ( θ ). It could beinteresting to implement (75) as boundary condition forthe viscous torque in numerical simulations of viscousfluids on (nearly) extremal Kerr backgrounds.It is not necessarily straightforward to translate phys-ical results obtained in the NHEK geometry to corre-sponding results in the Kerr geometry. In this contextit is worthwhile mentioning that the NHEK coordinates(15) were obtained from the Boyer-Lindquist coordinates(1) by virtue of two coordinate transformations, none ofwhich involved the polar angle θ . Thus, the θ -dependenceof physical quantities like velocity, viscosity or electro-magnetic fields is susceptible to a direct translation intothe Kerr geometry. This is not true for radial depen-dences. Since the whole NHEK geometry corresponds tothe ISCO in the original Kerr geometry it is not evenclear what a radial dependence in NHEK coordinatesmeans in terms of the original geometry. Therefore, weshould trust only statements that involve polar angles,like the result for the viscous torque discussed in the pre-vious paragraph or the relation (63h) between velocityand viscosity.In our discussion we have encountered some specialvalues for the polar angle θ : the values θ ≈ .
82 (and θ ≈ .
32) separate the “ergo-region” where the Killingvector ∂ t becomes spacelike from the “normal region”,see Fig. 1. The current related to the energy flux J (68)has the same asymptotic behavior as the Killing vector ∂ t , and it is lightlike for the two critical angles mentionedabove. The value θ ≈ .
955 leads to vanishing Chern–Pontryagin density (21) and vanishing geodesic deviationfor a timelike perfect fluid (47). It could be interestingto look for experimental signatures close to these specialpolar angles in real data.An interesting effect in the rich physics of accretiondisks are quasi-periodic oscillations (QPOs). As men-tioned in section V, we have not found any QPOs forperturbations around our viscous null fluid solution. Thisis consistent with the data selection procedure that ledto the spin results for the black hole GRS1915+105 [9]:the first of the three criteria employed by McClintock etal. states explicitly that QPOs must be absent (or veryweak).Of course, all our results are based upon certain as-sumptions. We discuss now briefly which of them couldbe relaxed and how this would influence our analysis.Our viscous null fluid solution (63) describes only the4physics at the ISCO of an extremal Kerr black hole. Ifwe drop both assumptions we are back to viscous flu-ids on a general Kerr background, which is too compli-cated for analytic studies. It may be possible, however,to use perturbation theory, taking our viscous fluid solu-tion (63) as the leading order result. Perhaps the “nearNHEK”-geometry constructed recently [35] can providethe suitable next-to-leading order background geometry.Another important assumption was that the velocity vec-tor is lightlike. We motivated this physically by arguingthat particles at the ISCO of an extremal Kerr black holecan only move with the speed of light, but slightly outsidethe ISCO the fluid should have a timelike velocity vector.Thus, for certain perturbative considerations the timelikefluid may provide a better starting point. However, wehave not succeeded in finding exact viscous fluid solu-tions, unless the velocity is lightlike. The derivation ofour exact solution (63) required some additional assump-tions, all of which were motivated. Some of them couldbe relaxed, and therefore we collect here our assumptionsfor future generalizations: we assumed vanishing angu-lar momentum; we assumed that the polar componentof the velocity vector is small as compared to the othercomponents, and hence set u θ = 0; we assumed that allscalar quantities depend only on the polar angle θ (56) sothat a matching to scalar quantities in the Kerr geometryis possible. Another interesting generalization would bethe construction of solutions with a globally causal cur-rent. We have found only solutions with electromagneticfields where the current is not causal globally or wherethe current vanishes. Finally, our results on first orderperturbations and our conclusion that QPOs are absentare based on the assumptions (89) and (91). Dropping ei-ther or both of these assumptions might lead to differentresults.Finally, we address briefly some (rather superficial,but expandable) relations to the conjectured (extremal)Kerr/CFT correspondence by Guica et al. [22]. If theconjecture holds then the near-extreme black hole GRS1915+105 is approximately dual to (the chiral half of)a conformal field theory (CFT), with left central charge c L ∼ × . The Kerr/CFT correspondence startsfrom the same assumptions as the present work: a dou- ble limit is taken that leads to the NHEK geometry (15).The chirality of the dual CFT implies that all physicalexcitations must be lightlike, which concurs with our as-sumption that the velocity profile is lightlike. It may beworthwhile to recover some of the results above from theCFT perspective, like the prediction of the viscosity func-tion from the velocity distribution, the specific boundarycondition for the viscous torque, the apparent absence ofQPOs or the appearance of some critical angles θ . Acknowledgments
We thank Bruno Coppi, Florian Preis, Paola Rebusco,Paul Romatschke, Dominik Schwarz and Bob Wagonerfor discussions.DG was supported by the project MC-OIF 021421 ofthe European Commission under the Sixth EU Frame-work Programme for Research and Technological Devel-opment (FP6). During the final stage DG was supportedby the START project Y435-N16 of the Austrian ScienceFoundation (FWF).AP was supported by the project AO/1-5582/07/NL/CB, Ariadna ID 07/1301 of the EuropeanSpace Agency.
APPENDIX A: PERFECT FLUID EQUATIONSIN NHEK
Here we collect some useful formulas for timelike per-fect fluids, u a u a = −
1, in the NHEK geometry. Weassume throughout the paper that the velocity 4-vectorcomponents depend only on r and θ . With this assump-tion the equations of motion simplify considerably. Thecontinuity equation (36) reads explicitly u r ∂ r ρ + u θ ∂ θ ρ + ( P + ρ ) (cid:16) u θ (3 cos θ −
1) cot θ θ + ∂ r u r + ∂ θ u θ (cid:17) = 0 (A1)The momentum equations (37) read explicitly u t ( u r ∂ r P + u θ ∂ θ P ) + ( P + ρ ) (cid:16) ru t u r θ + 4 cos θ − r )(1 + cos θ ) − u t u θ θ cos θ θ − u r u φ θ (1 + r )(1 + cos θ ) + u r ∂ r u t + u θ ∂ θ u t (cid:17) = 0 (A2) u r ( u r ∂ r P + u θ ∂ θ P ) + 1 + r M (1 + cos θ ) ∂ r P + ( P + ρ ) (cid:16) ( u t ) r (1 + r ) cos θ + 6 cos θ − θ ) − u t u φ (1 + r ) 4 sin θ (1 + cos θ ) − ( u r ) r r − u r u θ θ cos θ θ + u r ∂ r u r + u θ ∂ θ u r (cid:17) = 0 (A3)5 u θ ( u r ∂ r P + u θ ∂ θ P ) + 1 M (1 + cos θ ) ∂ θ P + ( P + ρ ) (cid:16) − ( u t ) sin θ (cid:0)
11 + cos θ + r cos θ + 2 cos θ + 92(1 + cos θ ) (cid:1) + (cid:0) ( u r )
11 + r − ( u θ ) (cid:1) sin θ cos θ (1 + cos θ ) − u φ (cid:0) ru t + u φ (cid:1) sin θ cos θ (1 + cos θ ) + u r ∂ r u θ + u θ ∂ θ u θ (cid:17) = 0 (A4) u φ ( u r ∂ r P + u θ ∂ θ P ) + ( P + ρ ) (cid:16) u t u r (cid:0)
11 + r − r cos θ + 6 cos θ − r )(1 + cos θ ) (cid:1) − ru t u θ (cos θ − θ −
3) cot θ θ + 4 u r u φ r (1 + r )(1 + cos θ ) + 2 u θ u φ cot θ θ + u r ∂ r u φ + u θ ∂ θ u φ (cid:17) = 0 (A5)Only four of the five equations above are independent. APPENDIX B: VISCOUS NULL FLUIDEQUATIONS IN NHEK
The shear viscosity tensor (58) for the velocity distri-bution (55) takes the following form:Π tθ = M η (1 + cos θ ) (1 + r ) ∂ θ u t (B1a)Π tφ = 2 M η sin θ θ (cid:0) r (1 + r ) ∂ r u t − u t (cid:1) (B1b)Π rr = − M η θ r (cid:0) r ) ∂ r u t + 2 r u t (cid:1) − M η (1 + cos θ ) ( u t ) ∂ r (cid:0) (1 + r ) u t (cid:1) (B1c)Π rθ = − M η (1 + cos θ ) ∂ θ u t (B1d)Π rφ = 4 M η sin θ θ u t (B1e)Π θθ = 12 M η (1 + cos θ ) ∂ r (cid:0) (1 + r ) u t ) (B1f)Π θφ = 0 (B1g)Π φφ = 2 M η sin θ θ ∂ r (cid:0) (1 + r ) u t (cid:1) (B1h) The remaining half of the components follows from thesymmetry, tracelessness and projection properties (59).With the results (B1) the continuity equation (61) sim-plifies to − η (2 ru t + (1 + r ) ∂ r u t ) = 0 (B2)By integrating (B2) we obtain that u t has the form u t = f ( θ )1 + r (B3)For convenience, we define f ( θ ) = u ( θ )1+cos θ ; consequently,(B3) takes the form u t = u ( θ )(1 + r )(1 + cos θ ) (B4)The t -component of the momentum equations will thenreduce to (63h). The θ momentum equation requires thedensity to be constant, ρ = ρ (63). The r and φ mo-mentum equations are redundant. [1] M. A. Abramowicz (2008), 0812.3924.[2] J. A. Wheeler, Our universe: the known and unknown (1967), public lecture.[3] V. Frolov and I. Novikov,
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