We present rotating solutions of Einsteins gravity coupled to an effective BornInfeld theory that describes the end of openstring tachyon condensation after the decay of an unstable Dbrane or a braneantibrane system. The geometry of these solutions is that of the rotating frozen star. The solutions are stationary ultracompact can be made nonsingular via a regularization procedure and their exterior geometry is identical for all practical purposes to that of the Kerr solution. The BornInfeld matter consists of radial electricflux tubes that emanate from or end at the ellipsoidal core of the star. Each end of the flux tubes carries an electric charge so that the electric field can be viewed as being sourced by an ellipsoidal charge distribution of positive and negative charges near the center of the star. Meanwhile the stars outer layer is equal and oppositely charged resulting in a vanishing electric field in the external spacetime. We show that these rotating solutions are ultrastable against radial perturbations just like their static frozen star counterparts. They are also effectively immune to ergoregion instabilities as we discuss.
['Gravastars', 'modified gravity', 'GR black holes', 'quantum black holes', 'High Energy Physics - Theory']
Kerr black hole mimickers sourced by a string fluid
{'Kerr Black hole mimickers sourced by a string fluid': 'Ram Brustein (1) , A.J.M. Medved (2 , 3) , Tamar Simhon (1)  \n- (1) Department of Physics, Ben-Gurion University, Beer-Sheva 84105, Israel\n- (2) Department of Physics & Electronics, Rhodes University, Grahamstown 6140, South Africa\n- (3) National Institute for Theoretical Physics (NITheP), Western Cape 7602, South Africa  \nramyb@bgu.ac.il, j.medved@ru.ac.za, simhot@post.bgu.ac.il', 'Abstract': "We present rotating solutions of Einstein's gravity coupled to an effective Born-Infeld theory that describes the end of open-string tachyon condensation after the decay of an unstable D -brane or a braneantibrane system. The geometry of these solutions is that of the rotating frozen star. The solutions are stationary, non-singular and ultracompact, and their exterior geometry is identical, for all practical purposes, to that of the Kerr solution. The Born-Infeld matter consists of radial electric-flux tubes that emanate from, or end at, the ellipsoidal core of the star. Each end of the flux tubes carries an electric charge, so that the electric field can be viewed as being sourced by an ellipsoidal charge distribution of positive and negative charges near the center of the star. Meanwhile, the star's outer layer is equal and oppositely charged, resulting in a vanishing electric field in the external spacetime. We show that these rotating solutions are ultrastable against radial perturbations, just like their static frozen star counterparts. They are also effectively immune to ergoregion instabilities, as we discuss.", '1 Introduction': "The frozen star model [1, 2, 3, 4, 5] describes a type of ultracompact object [6] that is horizonless and regular but, yet, able to mimic a black hole (BH) when viewed by an external observer. Initially, we used it to describe a spherically symmetric and static solution of Einstein's equations and thus a Schwarzschild mimicker. In addition to mimicking the classical aspects of a Schwarzschild BH, the frozen star model was also able to reproduce the thermodynamic properties of an equal-mass Schwarzschild BH [7]. The frozen star is sourced, effectively, by a fluid of maximally negative radial pressure and vanishing transverse pressure throughout the bulk of the interior. 1 This is, of course, rather exotic matter, but its motivation follows from an earlier proposal for describing the interior of a BH mimicker - the polymer model [8, 9] - in which case, the object consists of a collection of long, closed, interacting, fundamental strings that are excited to extremely hot temperatures. The idea is that the frozen star model describes the effective classical geometry and semiclassical aspects that an external observer would attribute to such a BH mimicker, even if she is ignorant about its microscopic description.  \nThe polymer model is premised on the notion that a BH mimicker needs to contain matter that is in a strongly non-classical state so as to account for the maximally entropic state of the object. A consequence of such highly quantum matter is that the interior does not have a classical geometrical description. The frozen star geometry is then meant to capture the main  \nfeatures of the polymer interior but from a classical and semiclassical perspective. This leads to some unorthodox aspects of the frozen star solution, including the highly anisotropic state of matter as described above, as well as the property that every radial surface in the bulk of the interior is a surface of very large (but finite) redshift, same as for the would-be horizon. As unconventional as these properties may be, they do conspire to circumvent singularity theorems [10, 11], constraints on the compactness of matter [12, 13, 14, 15] and issues with energy conservation when accounting for the effects of evaporation [16, 17].  \nImportantly, another microscopic description of the polymer model has recently been identified and presented in [18]. The same frozen star geometry was shown to be sourced by a fluid of cold, open strings resulting from the decay of an unstable D -brane or a brane-antibrane system at the end of open-string tachyon condensation. The effective Lagrangian for this process was originally formulated and developed by Sen [19, 20, 21, 22, 23] (and see [24] for a review with more references). Later on, Sen's Lagrangian was recast by Gibbons, Hori and Yi (GHY) [25] (also see [26]) into a specific Born-Infeld form that is describing a fluid of rigid electric-flux tubes (equivalently, open strings). A similar form of Lagrangian with a two-form field had already appeared in earlier literature in the context of describing the physics of flux tubes and strings: firstly in [27] and then later with regard to the cloud-of-strings model [28], the string-dust model [29] and 'hedgehog' compactifications [30, 31].  \nWhen gravity is neglected, the Born-Infeld theory of Dp -branes leads to BIons, which are spherically symmetric, static and solitonic solutions of finite  \nenergy [32, 33]. These BIons can be viewed as flux tubes that extend from a central point-like source all the way out to infinity. It follows from this point of view that frozen stars can be viewed as gravitationally back-reacted BIons or alternatively, as an example of the gravitational confinement of open strings. What we have found in [18] is that the outermost surface of the star contains an equal and opposite charge that cancels out the point-like source in the center; meaning that the flux tubes terminate at the outer surface, so that both the energy and spatial extent of these BIons are finite. In spite of the attractive force between the opposite charges, stability is maintained because of an effective force that is associated with the constraint of constant mass.  \nUnlike the polymer model, the Born-Infeld framework provides us with a well-defined matter Lagrangian to couple to that of Einstein's theory. This completes the Einstein equations and allows us to describe the dynamics of the frozen star in a generic state that is not necessarily static nor spherically symmetric. Therefore, we can now claim that the frozen star is a consistent and complete model which describes an ultracompact object whose equilibrium state is, for all practical purposes, indistinguishable from that of a Schwarzschild BH for an external observer.  \nThe Born-Infeld framework should eventually enable us to study extensions of our model to situations in which the departures from equilibrium physics are macroscopically large, as would be the case for an analysis on the dynamics of astrophysical BH mergers. As discussed in [34, 35], this is just what is needed if one is to observationally distinguish frozen stars from Schwarzschild BHs, as well as from other BH mimickers.  \nIn this paper, we have a more modest extension in mind, namely, to the case of a rotating (but stationary) frozen star solution; that is, to a Kerr BH mimicker. This can be viewed as a critical first step in a program for extending the frozen star solution beyond the static, spherically symmetric case. The inclusion of rotation is a vitally important step in our program as it allows us to make contact with observational physics. We have, in fact, already managed to incorporate rotation into the original frozen star framework [36]. 2 This task was accomplished by extending the solution to the case of non-vanishing spin parameter a , which is the analogue of the standard Kerr spin parameter, in a way that preserves the basic properties of the frozen star while maintaining the symmetries of the Kerr solution such as the constancy of the Carter constant [62] (which is not a constant a priori ). There was still some degree of arbitrariness in the procedure, but we negated this by verifying that the volume integrals of appropriate elements of the stress tensor are able to reproduce the correct mass and angular momentum of a Kerr BH with the same parametrizations.  \nThe main purpose of the current paper is to show explicitly that the Born-Infeld framework includes the rotating solution as discussed above (see Sec. 3.2). In addition, we will establish that the rotating frozen star is ultrastable (Sec. 4.1), just like its static counterpart, and is essentially immune to the type of ergoregion instabilities that typically afflict rotating, horizonless, compact objects (Sec. 4.2). We will also discuss how the internal geometry is matched to the external Kerr geometry at the star's outer surface (Sec. 5.1)  \nand how it is regularized at the star's center (Sec. 5.2). Some details about the matching procedure will be deferred to an appendix. Section 2 provides a review of [18], whereas Section 3.1 is a review of [36].", '2 Lagrangian and equations of motion': "In this section, we recall formalism and results from [18] for the sake of selfcompleteness and to establish our notations. Many of the expressions below first appeared in [25]. Our conventions are that an index of 0 denotes time, one of a , b , · · · denotes an arbitrary spacetime dimension and one of i , j , · · · denotes an arbitrary spatial dimension. Three large spatial dimensions are assumed for concreteness.  \nOur model describes Einstein's gravity coupled to a Born-Infeld-like theory for the antisymmetric tensor K ab . The action is, up to possible surface terms (including a constant-mass constraint),  \nS GBI = ∫ d 4 x { 1 16 πG √ -gR + 1 2 πα ' √ -1 2 K ab K ab + λ 1 ε abcd K ab K cd + √ -gJ a A a } (1)  \nThe term with the Lagrange multiplier λ 1 and the source J a will be discussed later. To arrive at this action, we followed the analysis of GHY [25] (also, [26]), whose own starting point was Sen's effective action for the final state of the decay of unstable D -branes at the end of tachyon condensation and near the minimum of the tachyon potential where it vanishes [19, 20, 21, 22, 23].  \nThe connection with flux tubes or open strings follows from the observation in [25] that K ab can be identified as a surface-forming bivector, which  \n.  \ncan always be regarded as the cross-sectional slice of the world sheet of an open string.", '2.1 Lagrangian': "Sen's effective Lagrangian 3 takes the following form:  \nL = -V ( T ) √ -Det ( g +2 πα ' F ) + √ -gA a J a , (2)  \nwhere V ( T ) is both the D -brane tension and the tachyon potential, g ab is the background metric, 2 πα ' is the inverse of the fundamental string tension, F ab = ∂ a A b -∂ b A a is the field-strength tensor for the gauge field A a and J a , a 4-vector current density, is the source.  \nAlthough the Lagrangian (2) vanishes when V ( T ) vanishes, the corresponding Hamiltonian  \nH = δ L δ ( ∂ 0 A i ) -L = 1 2 πα ' √ D i D i + P i P i , (3)  \nremains well defined. Here, P i is the conserved momentum (see below), E i = F 0 i is the electric field and D i is the electric displacement, which is defined as the canonical conjugate of the gauge field and thus as D i = δ L δ ( ∂ 0 A i ) . As is commonly the case in Born-Infeld theories, it is D i and not E i which satisfies the Gauss'-law constraint ( ρ e is the volume charge density),  \n∇ i D i = J 0 = ρ e . (4)  \nAlso, in the flux-tube picture, it is D i that determines the direction of the electric flux.  \nThe conserved momentum ( i.e. , the analogue of H for spatial translations) is expressible as  \n1 2 πα ' P i = -F ij D j = ( ⃗ D × ⃗ B ) i , (5)  \nwhere B i is the magnetic induction and defined by  \nB i = 1 2 ε ijk F jk . (6)  \nAlong with the Gauss'-law constraint (4), the fields are required to satisfy Ampere's law ∇ i F i j -∇ 0 E j = J j and the Bianchi identities, which include Faraday's law ⃗ ∇× ⃗ E + ∂ 0 ⃗ B = 0 and, given that there are no sources for magnetic monopoles,  \n∇ i B i = 0 . (7)  \nTo obtain a Lagrangian that can be used near the minimum of the potential, GHY changed variables from the elements F ij to their duals K ij = 2 δ H δ F ij . The fields F ab and, as defined in Eq. (10), K ab should be regarded as independent variables. This is important when the equations of motion (EOM) for K ab are derived.  \nA new Lagrangian can be defined by using a Legendre transformation of the form  \nL ' = H1 2 F ij K ij = 1 2 πα ' √ D i D i -1 2 K ij K ij , (8)  \nwith the latter equality following by way of the identity  \n1 2 K ij K ij = 1 (2 πα ' H ) 2 P i P i D j D j . (9)  \nIt is useful to introduce an anti-symmetric two-form field K ab that acts as an effective field strength,  \nK 0 i = D i , K ij = K ij , (10)  \nso that the new Lagrangian (8) can be written as one that is manifestly of the Born-Infeld form  \nL ' = 1 2 πα ' √ -1 2 K ab K ab . (11)  \nHere, it has been assumed that the electric-field term dominates over the magnetic one and the negative sign is a consequence of g tt < 0 appearing in the contraction when a = 0 and b = i (and vice versa ). In the absence of magnetic sources, the new Lagrangian is equal to the original Hamiltonian and simplifies to L ' = H = E i D i .  \nThere is a technical issue: A canonical analysis of L ' does not lead back to the same Hamiltonian H . However, as explained in [25], this can be corrected by using a Lagrange multiplier (here, denoted by λ 1 ) to impose the constraint K∧K = 0 , which follows directly from Eq. (10). This accounts for the Lagrange-multiplier term in Eq. (1).  \nAnother subtlety concerns the source term. The source term in Eq. (2)  \nis unaffected by the Legendre transformation; however, A a is not the gauge field for the field strength K ab . But, since F and K are to be regarded as independent, one can vary the Lagrangian L ' with respect to either field. The variation with respect to A a leads to the actual Gauss'-law constraint as in Eq. (4). This is quite obvious in the case of no magnetic sources because, then, L ' = E i D i .", '2.2 Equations of motion': "The EOM for K are particularly transparent when expressed in terms of the original field-strength tensor, as these are the same as the original Bianchi identities, d F = 0 . Meanwhile, the Bianchi identities for the new fieldstrength tensor d K = 0 happen to be the same as the EOM for the original Lagrangian.  \nThe gravitational EOM are, as always,  \n1 8 πG G a b = T a b . (12)  \nFor the geometry of interest, the sources are restricted to the core and the outer surface of the star. In this case, as justified in [28], the stress tensor works out as follows:  \nT ab = 2 δ L ' δg ab = 1 2 πα ' K c a K bc √ -1 2 K ab K ab . (13)  \nAlternatively, in terms of the effective electric field and the magnetic field  \nvia the conserved momentum, the stress tensor is given by  \nT 00 = 1 2 πα ' D i D i √ D i D i -1 2 K ij K ij = H , (14)  \nT 0 i = 1 2 πα ' D j K ij √ D i D i -1 2 K ij K ij = -1 2 πα ' P i , (15)  \nT ij = 1 2 πα ' -D i D j + K k i K jk √ D i D i -1 2 K ij K ij = 1 (2 πα ' ) 2 -D i D j + P i P j H , (16)  \nwhere the right-most equalities can be obtained by way of Eq. (9).", '3 Rotating solutions': 'The rotating frozen star has the same exterior geometry as a Kerr BH but the interior geometry is much different, as we will now proceed to describe.', '3.1 The metric and Einstein tensor': "Let us begin here by recalling from [36] the interior (bulk) metric for the rotating frozen star in corotating or zero-angular-momentum-observer (ZAMO) coordinates [63, 64]. These are obtained from Boyer-Lindquist coordinates by the transformation t → χ and ϕ → ϕ + ωχ , where ω = a r 2 + a 2 is the local angular velocity, and a has the same meaning as the spin parameter for the Kerr solution and a dimensionality of length. This transformation ensures that each radial slice in the bulk of the interior is a nearly null surface for which redshift is large but finite. This is the same geometry but with a different radial size as that of the star's outermost surface. The area of  \nany such slice is A ( r ) = 4 πR 2 = 4 π ( r 2 + a 2 ) , with R being the Kerr areal coordinate. The would-be Kerr horizon is located at R + = √ r 2 + + a 2 , with r + = M (1 + √ 1 -a 2 /M 2 ) , where M is the mass of the star.  \nThe interior metric is quite simple,  \nds 2 FS/ZAMO = -ε 2 dχ 2 + dr 2 ε 2 +Σ dθ 2 + R 4 Σ sin 2 θdϕ 2 , (17)  \nwhere ε 2 ≪ 1 is a dimensionless parameter and Σ = r 2 + a 2 cos 2 θ . It should be noted that g ϕϕ differs from that of Boyer-Lindquist Kerr coordinates. This is because, in the spirit of the original frozen star model, the mass M of the star is redefined by a scaling relation: 4 M = R 2 + (1 -ε 2 ) 2 r + → M ( r ) = R 2 (1 -ε 2 ) 2 r . Note that M , like a , has a dimensionality of length and the factor of 1 -ε 2 accounts for the fact that the outermost surface extends slightly past the location of the would-be Kerr horizon (we are neglecting orderε 4 corrections).  \nOn the basis of the antecedent polymer model, we expect that the small parameter ε 2 -which has been called ϵ or ε in some earlier papers - scales as the ratio of the Planck length to the star's radius. This ratio is 10 -38 for a solar-mass ultracompact object, so that ε 2 is extremely small.  \nThe metric (17) is valid for all values of a such that a M ≤ 1 and limits to that of the static frozen star as a → 0 . That the limiting value of a is equal to its limiting value for the Kerr solution can be understood as follows: For our solution, a ( r ≲ r + ) ≤ a ( r ≳ r +) ; otherwise, the star will shed mass until a stationary state is attained. It then follows that a interior ≤ a exterior ≤ M .  \nZAMO coordinates are often convenient to use but hide the fact that the star is rotating. This aspect, however, becomes apparent with the restoration of the original t and ϕ coordinates. This leads to the frozen star metric in its analogue of Boyer-Lindquist coordinates,  \nds 2 FS = ( a 2 sin 2 θ Σ -ε 2 ) dt 2 -2 a sin 2 θ R 2 Σ dtdϕ + 1 ε 2 dr 2 + Σ dθ 2 + R 4 Σ sin 2 θdϕ 2 . (18)  \nIt is useful to compare the frozen star metric in Eq. (17) to the nearhorizon limit of the Kerr metric in ZAMO coordinates. To leading order in ∆ = r 2 -2 Mr + a 2 , the latter metric can be expressed as  \nds 2 Kerr/ZAMO = -∆ Σ dχ 2 + Σ ∆ dr 2 +Σ dθ 2 + R 4 Σ sin 2 θdϕ 2 . (19)  \nSince ∆ vanishes at the horizon where r = r + , it is clear that the frozen star model essentially trades ∆ Σ for ε 2 and then extends this near-horizon form throughout the bulk of the interior. It is also clear from Eqs. (17) and (19) that the metric is continuous at the outer surface; however, its derivatives are not. We correct this later on by introducing a thin transitional layer which interpolates smoothly between the interior and exterior geometries.  \nThe frozen star geometry is regular throughout the interior except for a removable, mild singularity at the center of the star. As discussed later on, this can be regularized by smoothing the metric in a small region near the core of the star. This regularization, as well as the smoothing procedure at the outer surface, does not change any of our results in a significant way. It  \nonly changes physical quantities, such as the mass, angular momentum and charge, by perturbatively small amounts.  \nThe Einstein tensor G µν = R µν -1 2 g µν R has the following diagonal components to leading order in ε 2 :  \nG t t = G r r = -R 2 ( r 2 -3 a 2 cos 2 θ ) ( r 2 + a 2 cos 2 θ ) 3 , (20)  \nG θ θ , G ϕ ϕ = 0 . (21)  \nThere is also a pair of non-vanishing off-diagonal elements, G rθ and its inverse. But these can be removed by diagonalizing the r -θ sector, leaving the diagonal components unchanged up to ε 2 corrections.  \nOne can see that ρ is negative in a region around the axis of rotation where r 2 -3 a 2 cos 2 θ is negative. This does not lead to issues related to causality as discussed at length in [36].", '3.2 Born-Infeld fields and sources': "̸  \nWhen a = 0 , the simplest solution of the EOM is an axially symmetric and apparently static configuration. The solution looks like a static solution because of our choice to use ZAMO coordinates, but it 'knows' about the rotation and so is 'secretly' stationary. For this solution, K tr = -K rt and all other elements of K vanish. One then finds that K ab K ab = -2 D r D r . Then, just as in [18] for the spherically symmetric case, the only non-vanishing elements of T ab are  \n-T t t = T r r = 1 2 πα ' √ D r D r . (22)  \nAs expected, p r = -ρ and p ⊥ = 0 , which corresponds to the so-called string fluid as described in [25].  \nThe Einstein equations (12) along with Eqs. (20) and (22) can be used to obtain  \n1 8 πG R 2 ( r 2 -3 a 2 cos 2 θ ) ( r 2 + a 2 cos 2 θ ) 3 = 1 2 πα ' √ D r D r . (23)  \nIf we adopt the conventions that D r = D r 5 and that D r > 0 in the a → 0 limit, then the magnitude of the electric displacement D r can be written as  \nD r = α ' 4 G R 2 ( r 2 -3 a 2 cos 2 θ ) ( r 2 + a 2 cos 2 θ ) 3 , (24)  \nwhereas D θ , D ϕ are vanishing. 6 In the region where D r < 0 , the electric flux is ingoing towards the center, while in regions of positive D r , the flux is radially outgoing. The configuration of the flux is depicted in Fig. 1 for several values of a .  \nThe conservation of T ab and the antisymmetry properties of K ab imply that, in the bulk of the frozen star, ∇ i D i = 0 , as discussed in Section III of [28] and as can be verified by a direct calculation. It follows that there can be no charges located anywhere in the bulk. This observation supports the fluxtube picture, which helps in understanding the location of the charges. The end of each flux tube carries a charge that is equal in magnitude and opposite in sign to the charge on the opposite end, so that the total charge of the flux  \ntube vanishes. This charge distribution can be substantiated by calling on the Gauss'-law constraint (4) and then choosing a narrow, radially directed Gaussian surface that caps off at each end of the flux tube. The rotating frozen star can therefore be viewed as being sourced by a fluid of electric-flux lines. Some of the flux lines are emanating radially from a source in the central core and extending all the way to the surface of the star. Another group of flux lines are emanating from the special surface where D r vanishes, r 2 -3 a 2 cos 2 θ = 0 , and ending at the center. Yet another group of flux lines emerge from the surface of vanishing D r , extending from there to the outer surface. This is depicted in Fig. 1, where the arrows are pointing at the start and end points of the flux lines in the different regions.  \nAccording to Eq. (24), the flux tubes are radial, but the coordinate r is part of an oblate spheroidal coordinate system and not a standard radial coordinate [65]. The oblate spheroidal coordinates are related to standard Cartesian coordinates as follows:  \nx = R sin θ cos ϕ , y = R sin θ sin ϕ , (25) z = r cos θ .  \nThe radial direction for fixed θ and ϕ coordinates is a hyperbola,  \nx 2 + y 2 a 2 sin 2 θ -z 2 a 2 cos 2 θ = 1 , (26)  \nFigure 1: Shown are cross-sections of fixed ϕ of the frozen star for the cases a = 0 . 3 (left), a = √ 3 / 2 (middle) and a = 1 (right). The star is sourced by a fluid of electric-flux lines that either emanate from or end at an ellipsoidal charge distribution in its core or, alternatively, emanate from the blue curve and end either at the outer surface or the center of the star. The flux is directed along the constantθ hyperbolae. The flux inside the regions bounded by the blue curves is negative, as indicated by the blue, inward-pointing arrows, while the flux outside this region is positive, as indicated by the red, outward-pointing arrows. The outer-surface charge distribution has the same angular distribution as that in the core, but is oppositely charged. The flux lines are rigid and, consequently, the star is ultrastable.  \n<!-- image -->  \nwhile cross sections of surfaces of fixed R are confocal ellipses,  \nx 2 + y 2 R 2 + z 2 r 2 = 1 . (27)  \nThe surface at r = 0 is a degenerate ellipsoid whose height (range of the z coordinate) goes to zero with r but whose surface area, A = 4 πa 2 , nevertheless remains finite.  \nWe will next discuss the charge at the ends of the flux tubes, starting with the one at the outer surface of the frozen star. Let us choose a narrow Gaussian surface whose sides are radially directed with one end just inside the  \nouter surface and the other end just outside the outer surface where D i = 0 for any i . Then using the Gauss'-law constraint (4), we find a surface charge density of σ ( r + ) = -D r . Hence, an outgoing flux leads to a negative surface charge density, while an ingoing flux leads to a positive charge density. The total charge at the surface q ( r + ) is then obtained by integrating D r ,  \nq ( r + ) = 2 π ∫ 1 -1 d [cos θ ] D r ( r + , θ ) = -πα ' G . (28)  \nIn choosing the right-hand side of Eq. (24) to be positive for small a , we have implicitly chosen the signs of the charges to correspond to the case of a positive total charge at the star's center and, therefore, a negative total charge at its outer surface. What is more interesting is the magnitude: The total charge on the surface of a rotating star of any mass is precisely equal to that of a static star of any mass [18].  \nIn the core, the calculation of the charge is more subtle because the geometry degenerates in the limit r → 0 . Therefore, we assume that the charge resides at an infinitesimal but non-zero radius r = δ . Then choosing a radially directed, narrow Gaussian surface with one end at r = δ and the other close to the outer surface of the star at r = r + , we find that the surface densities are exactly equal in magnitude, σ (core) = -σ ( r + ) , as previously argued. Hence, σ (core) = + D r . Notice that the sign of the charge at any given point is determined by the sign of r 2 -3 a 2 cos 2 θ . Most of the flattened ellipsoid at r = δ is negatively charged, while the positive charge is pushed to a narrow ring for which θ → π/ 2 with increasing a .  \nRegardless, the total charge remains constant,  \nq (core) = πα ' G . (29)  \nThe charge distribution on the outer surface can be expanded in multipoles. For small a , the charge distribution is dominated by the monopole. The first correction is a negative quadrupole with a relative strength that scales as a 2 /R 2 + . This can be clearly seen in Fig. 1.  \nThe emerging picture is compelling: When the star is at rest, the flux lines are straight, starting from a point-like charge at the star's center and ending at the outer surface. Then, as the star begins to rotate, the flux lines bend and are pushed away from the center to the surface where r 2 -3 a 2 cos 2 θ = 0 due to the influence of a centrifugal force, as depicted in Fig. 1. Correspondingly, the positive, point-like charge spreads from the center to a ring of radius a on the equator, while the void near the axis of rotation is filled by negative charge - just enough to ensure that the total charge remains the same. The ring of positive charge is at the same position as that of the would-be ring-singularity of the Kerr geometry. For small a , the resulting distribution is mostly quadrupolar. When the rotation is fast enough, the ring of positive charge is pushed further out, whereas the region of negative charge reaches the outer surface and continues to spread over it as the rotation approaches its maximal value of a = M .", '4 Stability': 'In this section, we discuss the stability of the rotating frozen star, beginning with its stability against the radial class of linear perturbations. As expected, the rotating frozen star is shown to be ultrastable; meaning that all linear perturbations of both the matter and metric vanish identically. So that, when in equilibrium, a rotating frozen star is just as stable as a Kerr BH; neither can be excited, although their mass and/or angular momentum can change. Then we discuss the possibility of ergoregion instabilities - those associated with negative-energy regions inside either the star or its external ergosphere - and argue that any such instability could only develop over a time scale that is much larger than the age of the Universe.', '4.1 Ultrastability against radial perturbations': "The property that ensures the ultrastability of the static frozen star, G t t = G r r , remains intact, and so we anticipate that the interior of the rotating frozen star is similarly ultrastable.  \nTo see this explicitly, let us start by linearly perturbing G µν → G µν + δG µν and T µν → T µν + δT µν . Then  \n1 8 πG δG µν = δT µν . (30)  \nLet us next consider the various contributions to the right-hand side of the previous equation. For i , j = θ , ϕ (but not r ), it follows from Eq. (15) that  \nδT 0 i = -1 2 πα ' δ P i (31)  \nand from Eq. (16) that  \n(2 πα ' ) 2 H δT ij = -δ ( D i D j ) + δ ( P i P j ) + (2 πα ' H ) 2 T ij δ ( 1 H ) = 0 , (32)  \nwith the right-most equality due to the vanishing of the background values of the D i 's, P i 's and T ij 's. Notice that this last result indicates that fluctuations in the transverse pressure vanish, δp ⊥ = 0 .  \nAs for δT rj and δT r 0 , let us start with  \n(2 πα ' ) 2 δ ( H T rj ) = -δ ( D r D j ) + δ ( P r P j ) . (33)  \nThen, since the background values of the T rj 's, D j 's and any of the P 's vanish,  \nδT rj = -1 2 πα ' δD j , (34)  \nwhere we have also used that, on the solution, 2 πα ' H = √ D r D r = D r , with the latter equality because of our choice of convention. Next,  \nδT r 0 = -1 2 πα ' δ P r = δ ( D × B ) r = 0 , (35)  \nwhere the final equality comes about because, on the solution, D r is the only non-vanishing component of the electric displacement and all components of the magnetic induction vanish.  \nMoving on to the diagonal components of the t, r sector and again using that, on the solution, D r is the only non-vanishing component of the  \ndisplacement and 2 πα ' H = √ D r D r = D r , we have  \nδT 0 0 = δ H = 1 2 πα ' δD r (36)  \nand  \nδT r r = -1 2 πα ' [ 2 δD r -1 2 δ ( D a D a ) √ D a D a ] = -1 2 πα ' [2 δD r -δD r ] = -1 2 πα ' δD r , (37)  \nwhere, in the second line, the diagonality of the on-solution metric has also been used and the vanishing P 2 contribution has now been discarded from the outset. In other words,  \nδT 0 0 + δT r r = 0 , (38)  \nmeaning that the equation of state is preserved for linear perturbations.  \nWe now pivot to the left-hand side of the linearized Einstein equations; that is, the calculation of δG µν . This can be done by perturbing the metric by four functions of t and r , as is the standard practice for even-parity perturbations [66]. Denoting the four functions by H 0 , H 1 , H 2 and K , we can then write  \nds 2 = -ε 2 (1+ H 0 ) dt 2 +2 ∂ t H 1 dtdr + 1 ε 2 (1+ H 2 ) dr 2 +(1+ K ) [ Σ dθ 2 + R 2 Σ sin 2 θdϕ 2 ] . (39)  \nLet us parameterize the time dependence of all perturbations as e iωt and further assume that ω = 0 . The stability for the case that the perturbations  \n̸  \ndo not depend on t , but only on r , is simpler because K can be set to zero by an r -dependent coordinate transformation.  \nWe proceed by expanding G µν to linear order in the perturbations and to leading order in ε 2 . The calculation is quite involved and Mathematica has been used to perform the expansion. The results for the diagonal components are as follows (where a subscript of 0 on a density denotes a background quantity):  \nG 0 0 = 8 πGρ 0 K , (40)  \nδG r r = -8 πG ( p r ) 0 K + 2 r R 2 ∂ 2 t H 1 -1 ε 2 ∂ 2 t K , (41)  \nδG θ θ = δG ϕ ϕ = -1 2 1 ε 2 ( ∂ 2 t H 2 + ∂ 2 t K ) . (42)  \nWe also find that, for i = θ, ϕ ,  \nδG 0 i = δG ri = 0 (43)  \nand that  \nδG 0 r = 8 πGρ 0 ∂ t H 1 + 1 R 2 ∂ t ( rH 2 -rK + R 2 ∂ r K ) . (44)  \nTo summarize, we obtain, along with δD i = 0 and δ P i = δ P r = 0 , the  \nfollowing set of linearized perturbation equations:  \nρ 0 K = 1 2 πα ' δD r , (45)  \nρ 0 K + 1 8 πG [ 2 r R 2 ∂ 2 t H 1 -1 ε 2 ∂ 2 t K ] = -1 2 πα ' δD r , (46)  \nρ 0 ∂ t H 1 + 1 8 πG [ 1 R 2 ∂ t ( rH 2 -rK + R 2 ∂ r K ) ] = 0 , (47)  \n-1 2 1 ε 2 ( ∂ 2 t H 2 + ∂ 2 t K ) = 0 . (48)  \nFrom the second equation, it follows that K vanishes, otherwise the third term on the left blows up with nothing to compensate for it. Then, from the first and fourth equation, it must be that δD r and H 2 respectively vanish. From the vanishing of K and H 2 , it follows from the third equation that H 1 = 0 . Since δD r vanishes, so must δρ and δp r ( cf , Eqs. (36,37)), as did δp ⊥ .  \nOur conclusion is that all matter perturbations and metric perturbations vanish identically, to leading order in ε 2 , except for H 0 , which is left undetermined. 7 The reason that H 0 evades determination can be traced back to the possibility of eliminating it by a transformation of the time coordinate, which does not change the other components of the metric to leading order in ε 2 . A similar phenomenon was encountered in the analysis of radial stability in the static case. To establish the vanishing of H 0 , one needs to allow deviations from p r = -ρ and then take the limit p r →-ρ a posteriori .", '4.2 Absence of ergoregion instabilities': 'As previously observed in Section 3, the rotating frozen star contains regions of negative energy, or ergoregions, in its interior. These ergoregions are different than the ergoregion of the exterior Kerr geometry, which is also shared by the frozen star.  \nThe exterior ergoregion of horizonless, rotating, ultracompact objects leads to well-known instabilities [67]. The physical origin of such an instability is that a wave or particle that is scattered by the ultracompact object can extract rotational energy from it. A region of negative energy can then lead to superradiant scattering, meaning that a scattered wave can end up with a larger amplitude than that of the original incident wave (see, for example, [68]). This effect will lead to instabilities when the scattering process happens repeatedly; enough times so that the amplitude of the multi-scattered wave can grow exponentially large.  \nAs a horizonless, rotating, ultracompact object, a frozen star might well be suffering from these types of instabilities. However, as argued next, the timescale of any such instability would be extremely large for an astrophysically relevant frozen star, and so the effects of the instability would be irrelevant for all practical purposes.  \nFor an exterior instability to be relevant, the wave has to traverse it many times. However, as explained below, the absorption coefficient of a macroscopic frozen star is effectively unity, up to a correction of order ε 2 , so that an infalling wave effectively crosses the external ergosphere only once.  \nLet us recall from [4] that the light-crossing time for a solar-sized frozen star was estimated to be 10 32 s, which is longer than the lifetime of the Uni-  \nverse! This estimate can be attributed to waves or particles having interior velocities on the order of the extremely small parameter ε 2 , whose magnitude is of the order of the ratio of the Planck length to the (would be) horizon length scale, ε 2 ∼ l P r + . This is because waves are restricted to travelling along radial, null trajectories, dr dt = ε 2 . Since the requisite trip time has to be much, much larger than the light-crossing time, this all but rules out the possibility of this type of instability for any frozen star whose size is of astrophysical relevance.  \nThe argument is valid for both the exterior and interior ergoregions of the rotating frozen star. To understand why the interior ergoregion is not an issue, consider that the scattering surface in this case can either be at the outer layer or in the central core of the star. Furthermore, such an instability could only happen after a given wave has transversed the interior of the star many times. Then, since the crossing time is extremely large, the huge time scale of the instability once again renders it irrelevant for astrophysical objects.', '5.1 Matching the frozen star and Kerr geometries at the outer surface': "To match continuously the interior geometry of the frozen star to the nearhorizon, external Kerr geometry, what is required is a narrow transitional  \nlayer of width λ ≪ r + (but λ ≫ ε 2 r + ), as this allows for the interior metric and its first two derivatives - what is needed to describe the Einstein tensor - to be smoothly connected to the exterior. Such a procedure has already been carried out for the spherically symmetric, static star, as documented in [3].  \nIn practice, we need to find four interpolating polynomials P tt , P rr , P tϕ and P ϕϕ for each of the four metric components g tt , g rr , g tϕ and g ϕϕ . Note that g θθ is already continuous and does not require further consideration. The polynomials need to be of degree 5 so as to ensure the continuity of the metric and its first two derivatives at both ends of the transition layer.  \nIt is useful to recall Eq. (18) for the frozen star metric in its BoyerLindquist-like form, as this will be the choice used in the matching procedure. It can be rewritten as  \nds 2 FS = ( a 2 sin 2 θ Σ -ε 2 ) dt 2 -( 4 Mra Σ +2 a ∆ Σ ) sin 2 θdtdϕ + 1 ε 2 dr 2 + Σ dθ 2 + R 4 Σ sin 2 θdϕ 2 . (49)  \nLet us, for convenience, start the transitional layer at r = r + , where ∆ = R 2 -2 Mr = 0 or R 2 = 2 Mr . We can use this last result to simplify the metric components and their first and second derivatives at r + .  \nThe idea is then to smoothly connect the frozen star metric over a distance λ ≪ r + to the exterior Kerr metric at r = r + + λ . 8 In Boyer-Lindquist  \ncoordinates, as is appropriate for the matching, the latter is  \nds 2 Kerr = -( 1 -2 Mr Σ ) dt 2 -4 Mra Σ sin 2 θdtdϕ + Σ ∆ dr 2 + Σ dθ 2 + ( R 2 + 2 Mra 2 sin 2 θ Σ ) sin 2 θdϕ 2 . (50)  \nUsing Σ = r 2 + a 2 cos θ 2 and ∆ = R 2 -2 Mr , we can simplify the first and last components of the above metric. Respectively,  \n1 -2 Mr Σ = ∆ Σ -a 2 sin 2 θ Σ , (51)  \nand  \nR 2 + 2 Mra 2 sin 2 θ Σ = R 4 Σ -a 2 sin 2 θ ∆ Σ . (52)  \nTo summarize,  \nds 2 FS = ( a 2 sin 2 θ Σ -ε 2 ) dt 2 -( 4 Mra Σ +2 a ∆ Σ ) sin 2 θdtdϕ + 1 ε 2 dr 2 + Σ dθ 2 + R 4 Σ sin 2 θdϕ 2 , (53)  \nds 2 Kerr = ( a 2 sin 2 θ Σ -∆ Σ ) dt 2 -4 Mra Σ sin 2 θdtdϕ + Σ ∆ dr 2 + Σ dθ 2 + ( R 4 Σ -a 2 sin 2 θ ∆ Σ ) sin 2 θdϕ 2 . (54)  \nThe conclusion is that, to perform the smoothing, we need the polynomial expansion of ∆ Σ (and Σ ∆ ) to second order in r -derivatives. We denote this polynomial by Q ds ( r, θ ) and use it to smoothly connect g tt , g rr , g tϕ and g ϕϕ , with g θθ smooth by construction. Then, wherever there is a difference proportional to ∆ Σ between the frozen star and Kerr metrics, we replace it  \nby a polynomial of the form ε 2 + A ( r -r + ) 3 + B ( r -r + ) 4 + C ( r -r + ) 5 , with A, B, C determined by the metric component and its first two derivatives at r = r + + λ .  \nFor example, the polynomial P rr for g rr has to satisfy the following six relations,  \n( P rr , P ' rr , P '' rr ) | ( r + ) = ( 1 ε 2 , 0 , 0 ) , (55)  \n( P rr , P ' rr , P '' rr ) | ( r + + λ ) = ( Σ ∆ , ( Σ ∆ ) ' , ( Σ ∆ ) '' ) | ( r + + λ ) . (56)  \nThe end result of this process are four polynomials of degree 5 which smoothly interpolate the metric and Einstein tensor from the interior of the frozen star to the exterior Kerr geometry, while maintaining the exact angular dependence.  \nThe resulting components of the Einstein tensor, and therefore the stress tensor, scale in a similar way to their counterparts in the static case. The transverse pressures p θ , p ϕ are equal and vanish at both ends of the transitional layer, while scaling in the middle of the layer as 1 /λ . The energy density ρ and magnitude of the radial pressure | p r | decrease from their common frozen star value at r = r + to zero at r = r + + λ . Both are independent of λ at leading order.  \nThe full analysis for the matching procedure is presented in an appendix. The calculations are complicated by the fact that ∆ is a small quantity near the outer surface at r + , but its first and second derivatives are not, so that the orders of the perturbative expansions mix for the first and second derivatives of the metric components.", '5.2 Regularizing the core': "As can be seen in Section 3, the curvature and all non-vanishing components of the Einstein tensor remain regular throughout the interior, except for a mild singularity near the center of the star, similarly to the spherically symmetric, static case. In spite of the singular behavior, the amounts of mass and angular momentum in the core region are perturbatively small, as revealed by direct integration of the energy and angular-momentum densities. (An explicit expression for the latter can be found in [36].) In fact, the situation is even better for the rotating case because, for a non-vanishing a , the singularity is restricted to the equator. Still, it is useful to confirm that a suitable regularization procedure can indeed be enforced.  \nIt turns out that it is enough to regularize the metric by changing the angular function Σ = r 2 + a 2 cos 2 θ in a small core region near r = 0 . For example, Σ is modified for r ≤ η with η ≪ r + and η ≫ ε 2 r + , while Σ is unchanged for r ≥ η . Then, in this core region, we redefine Σ → Σ+ f ( r ) such that f (0) = const. and f ( r = η ) = 0 , f ' ( r = η ) = 0 and f '' ( r = η ) = 0 , so that the metric matches smoothly to the bulk metric of the frozen star. A simple example for such a regulator is  \nf ( r ) =      f ( r ) = ( η -r ) 3 η r ≤ η , 0 r ≥ η . (57)  \nThe resulting Einstein tensor G a b , to leading order in ε 2 , is almost identical to the unregularized tensor, with the only order-unity components (after diagonalization) being G t t and G r r = -G t t . Via Einstein's equations, this  \nequality maintains the equation of state p r = -ρ , but now with  \n8 πGρ = ( a 2 + r 2 + f ( r )) ( r 2 -3 a 2 cos 2 θ + f ( r )) ( a 2 cos 2 θ + r 2 + f ( r )) 3 . (58)  \nThe only difference between the regularized and unregularized cases is the appearances of f ( r ) in the former. Notice that derivatives of f ( r ) do not appear. On the equator, the previously singular energy density on the ring at r = 0 and cos θ = 0 now scales as  \nρ ∼ a 2 + f (0) f (0) 2 = a 2 + η 2 η 4 ∼ a 2 η 4 . (59)  \nIn the limit a → 0 , the energy density scales as 1 η 2 , as it did in the static case.  \nAs for the flux, since D r ∝ ρ , the result of the regularization is to modify what looked like a singular ring of charge to an ellipsoid of charge with regions of positive and negative charge distributions, as discussed in Section 3.2.  \nFrom [36], we know that the mass in the regularized core region scales as M ( r ≤ η ) = 1 2 η , so that M ( r ≤ η ) M = η r + ≪ 1 , and similarly for the ratio of the angular momentum in the core to the total angular momentum, J ( r ≤ η ) J = η r + ≪ 1 .", '6 Discussion': "Wehave shown how the Born-Infeld framework, which was introduced in [18], can be extended from the case of the spherically symmetric, static frozen star  \nto its axially symmetric, rotating counterpart, which was first proposed in [36]. The importance of having an explicit form of matter Lagrangian cannot be overstated. It allows us to study frozen star geometries that have no such symmetries and consider the evolution of these solutions and their stability in fully dynamical scenarios. This will be particularly useful for modeling the early phases of BH mergers, which are expected to be observationally relevant for the purpose of discriminating between frozen stars, Kerr BHs and also between frozen stars and other BH mimickers [34, 35].  \nAn important next step in the just-discussed program would be to extend this framework to the case of the defrosted star [5], for which ρ + p and p ⊥ are no longer vanishing but perturbatively small, yet much larger than ε 2 . Deviating from the ρ + p = 0 equation of state is necessary if we are to study the star's ringdown modes and resulting gravitational-wave emissions because of the ultrastability of the solution for the case ρ + p = 0 . This was made clear in [5], where it was shown how the defrosted star can support excitations whereas the fully frozen star cannot. To extend the framework in this way will, however, entail another level of complexity. In addition to a weak, radially directed electric field, a radially directed magnetic field will also be required. Such a magnetic field can be sourced by a magnetic monopole, meaning that the complete solution is sourced by a dyon. Work along this line is already in progress.", 'Acknowledgments': "We thank Yoav Zigdon for discussions, Suvendu Giri for pointing to us the similarity of the frozen star model to the clouds of string model and to the Born-Infeld BIons and Frans Pretorius for insisting that we find the source matter Lagrangian of the frozen star. We extend our special thanks to Piljin Yi, for helping us understand his work and its implications to the frozen star model. The research is supported by the German Research Foundation through a German-Israeli Project Cooperation (DIP) grant 'Holography and the Swampland' and by VATAT (Israel planning and budgeting committee) grant for supporting theoretical high energy physics. The research of AJMM received support from an NRF Evaluation and Rating Grant 119411 and a Rhodes Discretionary Grant SD07/2022. AJMM thanks Ben Gurion University for their hospitality during past visits.", 'A Smoothing of the outer transitional layer': "Here, we provide more details about the smoothing procedure that made use of an outer transitional layer to connect the frozen star interior to the Kerr exterior.  \nIn the outer transition layer, we use the Kerr metric with ∆ Σ replaced by the degree-5 polynomial Q ds ( r, θ ). The resulting line element is given by  \nds 2 TL = ( a 2 sin 2 θ Σ -Q ds ( r, θ ) ) dt 2 +2 a sin 2 θ ( Q ds ( r, θ ) -R 2 Σ ) dtdϕ + 1 Q ds ( r, θ ) dr 2 +Σ dθ 2 + ( R 4 Σ -a 2 sin 2 θ Q ds ( r, θ ) ) sin 2 θdϕ 2 .  \nThe interpolating polynomial Q ds and it first and second radial derivatives at the inner boundary of the transitional layer, which is located at r = r + , have to agree with the frozen star geometry, leading to the three conditions ( Q ds ( r + , θ ) = ε 2 , Q ' ds ( r + , θ ) = 0 , Q '' ds ( r + , θ ) = 0) . Meanwhile, at the outer boundary of the layer, where r = r + + λ , the polynomial and its derivatives must satisfy conditions that are based on the external Kerr solution,  \n( Q ds ( r + + λ, θ ) = ∆ Σ | r = r + + λ , Q ' ds ( r + + λ, θ ) = ( ∆ Σ ) ' | r = r + + λ , Q '' ds ( r + + λ, θ ) = ( ∆ Σ ) '' | r = r + + λ ) .  \nThe resulting polynomial can be expressed in terms of x = r -r + , Σ + λ = ( r + + λ ) 2 + a 2 cos 2 θ and Q ds ( x, θ ) = ε 2 + A x 3 λ 3 + B x 4 λ 4 + C x 5 λ 5 , where the coefficients A , B , C work out to be 9  \nA = 10 ( a 2 +( r + + λ )( r + + λ -2 M ) Σ 2 + λ -ε 2 ) -8 λ a 2 cos 2 θ ( r + + λ -M ) + ( r + + λ ) ( M ( r + + λ ) -a 2 ) Σ 2 + λ + λ 2 a 4 cos 4 θ -a 2 cos 2 θ ( a 2 +3( r + + λ )( r + + λ -2 M )) + ( r + + λ ) 2 (3 a 2 -2 M ( r + + λ )) Σ 3 + λ , B = 16 λ ( M -r + ) -3 λ 2 Σ + λ -2 λ a 2 (7 r + +6 λ ) + λ ( r + + λ ) ((9 r + -8 M ) + 7 r + ( r + -2 M ) + 2 λ 2 ) Σ 2 + λ -8 λ 2 ( r + + λ ) 2 ( a 2 +( r + + λ )( r + + λ -2 M )) Σ 3 + λ +15 ε 2 , (60) C = 6 λ ( r + -M ) + λ 2 Σ + λ +  \n+ λ a 2 (6 r + +5 λ ) + λ ( r + + λ ) ((7 r + -6 M ) + 6 r + ( r + -2 M ) + λ 2 ) Σ 2 + λ + 4 λ 2 ( r + + λ ) 2 ( a 2 +( r + + λ )( r + + λ -2 M )) Σ 3 + λ -6 ε 2 .  \nWe have verified explicitly that the resulting components of the Einstein tensor - equivalently, the stress tensor - at r = r + are indeed equal to their frozen star counterparts and that they vanish at r = r + + λ as required.  \nDue to the complicated expression for the interpolating polynomial Q ds ( r, θ ), it is not practical nor particularly useful to present the expressions for the Einstein tensor components as a function of r and θ throughout the transitional region. We do, however, present their values at the midpoint of the layer, r = r + + λ 2 , to leading order in λ r + . 10 To arrive at these values, we calculated the Einstein tensor resulting from the metric in Eq. (60), then substituted the leading-order values of the polynomial and its first and second derivatives with respect to r and θ at the midpoint. We then expanded the result in λ/r + and noted the leading-order value. The resulting set of  \nnon-vanishing energy-momentum components at the midpoint are  \nG tt = 3 4 λr + a 2 ( r 2 + -a 2 ) sin 2 θ Σ 2 + , (61)  \nG tϕ = -3 4 λr + a ( r 4 + -a 4 ) sin 2 θ Σ 2 + , (62)  \nG ϕϕ = 3 4 λr + ( a 2 + r 2 + ) ( r 4 + -a 4 ) sin 2 θ Σ 2 + , (63)  \nG rr = 14 11 λ r + Σ + , (64)  \nG rθ = -34 11 a 2 r + sin(2 θ ) Σ 2 + , (65)  \nG θθ = 3 4 λr + ( r 2 + -a 2 ) . 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