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Citation Penna, Robert F. "Energy extraction from boosted black holes: Penrose process, jets, and the membrane at infinity." Phys. Rev. D 91, 084044 (April 2015). © 2015 American Physical Society

As Published http://dx.doi.org/10.1103/PhysRevD.91.084044

Publisher American Physical Society

Version Final published version

Citable link http://hdl.handle.net/1721.1/96722

Terms of Use Article is made available in accordance with the publisher's policy and may be subject to US copyright law. Please refer to the publisher's site for terms of use. PHYSICAL REVIEW D 91, 084044 (2015) Energy extraction from boosted black holes: Penrose process, jets, and the membrane at infinity

Robert F. Penna* Department of Physics and Kavli Institute for Astrophysics and Space Research, Massachusetts Institute of Technology, Cambridge, Massachusetts 02139, USA (Received 12 March 2015; published 20 April 2015) Numerical simulations indicate that black holes carrying linear momentum and/or orbital momentum can power jets. The jets extract the kinetic energy stored in the ’s motion. This could provide an important electromagnetic counterpart to gravitational wave searches. We develop the theory underlying these jets. In particular, we derive the analogues of the Penrose process and the Blandford-Znajek jet power prediction for boosted black holes. The jet power we find is ðv=2MÞ2Φ2=ð4πÞ, where v is the hole’s velocity, M is its mass, and Φ is the magnetic flux. We show that energy extraction from boosted black holes is conceptually similar to energy extraction from spinning black holes. However, we highlight two key technical differences: in the boosted case, jet power is no longer defined with respect to a Killing vector, and the relevant notion of black hole mass is observer dependent. We derive a new version of the membrane paradigm in which the membrane lives at infinity rather than the horizon and we show that this is useful for interpreting jets from boosted black holes. Our jet power prediction and the assumptions behind it can be tested with future numerical simulations.

DOI: 10.1103/PhysRevD.91.084044 PACS numbers: 04.40.Nr, 04.70.-s

I. INTRODUCTION (where the black hole carries linear momentum), there are negative energy trajectories. We use these trajectories to Recent numerical simulations [1–8] and analytic esti- derive the analogue of the Penrose process. This gives a mates [9–12] suggest black holes carrying linear and orbital simple example of energy extraction from boosted black momentum can power jets. The jets are driven by electro- holes. It may be useful for describing the interactions of magnetic fields tapping the kinetic energy stored in the stars with moving black holes. black hole’s motion. The power of the simulated jets scales 2 Our second goal is to develop the analogue of the approximately as v , where v is the hole’s velocity [5]. Blandford-Znajek (BZ) model [15,16]. In the original Such jets could be an important electromagnetic counter- BZ model, electromagnetic fields tap a spinning black part to gravitational wave signals because v ∼ 1 in the final hole’s rotational energy and drive jets. The BZ jet power stages of black hole- and black hole-black hole prediction is currently being tested against astrophysical mergers. This paper develops the theory underlying observations of spinning black holes [17,18]. these jets. We develop the analogue of the BZ jet power prediction Our first goal is to develop the analogue of the Penrose for boosted black holes in Sec. III. For small v, we find process [13,14] for boosted black holes. The original Penrose process is a simple mechanism for extracting 1 v 2 rotational energy from Kerr black holes. It relies on the P ¼ Φ2; ð1Þ fact that certain geodesics near spinning black holes have jet 4π 2M negative energy (with respect to global time). In the original Penrose process, a particle with positive energy travels where Φ is the magnetic flux at infinity and M is the black toward the black hole and decays into two daughter hole’s rest mass. This is similar to the BZ prediction for particles. One of the daughter particles falls into the black spinning black holes but with v=ð2MÞ in place of the hole with negative energy and the other returns to infinity. horizon angular velocity ΩH, the flux evaluated at infinity The final particle has more energy than the original and the rather than the horizon, and a slightly different normali- ’ 2 black hole s mass decreases. zation constant. The v scaling is consistent with earlier We derive the analogous process for boosted simulations [3–5,8] and estimates [9–11]. Our formula Schwarzschild black holes in Sec. II. In the rest frame predicts jets from boosted black holes and spinning black of a Schwarzschild black hole there are no negative energy holes have comparable strength when v=ð2MÞ ∼ Ω . ’ H trajectories and it is impossible to lower the black hole s Numerical simulations suggest the true power of jets from mass via the Penrose process. However, in a boosted frame boosted black holes is lower by as much as a factor of 100 [5]. We discuss possible reasons for this discrepancy in *[email protected] Sec. III but save a detailed comparison for the future.

1550-7998=2015=91(8)=084044(15) 084044-1 © 2015 American Physical Society ROBERT F. PENNA PHYSICAL REVIEW D 91, 084044 (2015) Our third goal is to develop a new version of the where pvffiffiffiffiffiffiffiffiffiffiffiffiffiis a constant parameter, 0

084044-2 ENERGY EXTRACTION FROM BOOSTED BLACK HOLES: … PHYSICAL REVIEW D 91, 084044 (2015) 10 of understanding black hole jets. It may also be relevant for describing the interactions of stars with moving black holes. Consider a particle with 4-momentum 5

uμ0 ¼ðuτ0 ;ux0 ;uy0 ;uz0 Þð13Þ

0 ¼ − 0 – M 0 and energy E uτ in the boosted frame (3) (6). In this x frame the black hole carries momentum along z.A coordinate transformation gives

0 5 E ¼ γðE þ vuzÞ; ð14Þ

where E ¼ −uτ > 0 and uz are the particle’s energy and momentum in the black hole rest frame. The boosted 10 0 energy E is negative when vuz < 0 and jvuzj >E. The 10 5 0 5 10 first condition means the particle and the black hole travel z M in opposite directions along z. If such a particle is accreted, ’ γ FIG. 1 (color online). Ergosphere for a boosted Schwarzschild then the black hole s energy increases by E (because it black hole moving in the þz direction with velocity v ¼ 0.3 adds the particle’s unboosted frame energy to its own), and (solid blue), 0.6 (long dashed orange), 0.9 (dashed green), and it decreases by −γvuz (because it loses kinetic energy). The ¼ 0 0.99 (dotted red). The of a v black hole is condition jvuzj >Emeans the latter effect wins. This is shown for comparison. The ergosphere lags behind the event impossible in flat spacetime, where horizon. qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi E ¼ m2 þ u2 þ u2 þ u2 ≥ ju j: ð15Þ 2 2 x y z z gτ0τ0 ¼ γ ðgττ þ v gzz − 2vgztÞ > 0: ð10Þ However, in black hole spacetimes, (15) is replaced with Plugging in (2) gives the radius of the ergosphere, μν 2 g uμuν ¼ −m , and jvuzj >Eis possible. Roughly speak- ¼ 2 γ2ð1 − θÞ2 ð Þ ing, the gravitational field can contribute a negative rstatic M v cos : 11 potential energy to E. In the boosted Schwarzschild metric, 0 Observers inside the ergosphere cannot remain at rest with particles at infinity have E ≥ 0 but particles at finite radii 0 0 respect to ∂τ0 . Figure 1 shows the ergosphere for several may have E < . values of v>0. The ergosphere is offset from the black So we are led to consider something like the original 0 hole and extends to Penrose process. A positive energy (E > 0) particle at infinity falls toward a boosted black hole and splits in two. 1 þ v One half follows a negative energy trajectory into the hole r ð0Þ¼2M : ð12Þ static 1 − v and the other escapes to infinity. The negative energy particle must move against the direction of the black hole’s For v ¼ 0, the ergosphere coincides with the horizon. For motion and be gravitationally bound. The outgoing particle v → 1, it extends to infinity. This is in marked contrast will have more energy than the original and the black hole with the situation for Kerr black holes, for which the will lose energy. ergosphere is always centered on the horizon and confined We have found numerical solutions for this process. within r ≤ 2M. Assume the particles move in the xz-plane, so uy ¼ 0. Each trajectory is then fully characterized by three constants: uτ, C. Penrose process 2 μ uz, and rest mass m ¼ −u uμ. The trajectories cannot be A classic example of rotational energy extraction from geodesics because ∂z is not Killing, but they could be Kerr black holes is the Penrose process [13,14]. In this achieved by using rocket engines to adjust a freely falling process, a particle falls into the ergosphere of a Kerr black particle’s momentum along x. hole and splits in two. One of the daughter particles falls Figure 1 shows one of our solutions. Particle A, with 2 into the black hole along a negative energy geodesic and the uτ ¼ −3=2, uz ¼ 0.9867, and m ¼ −1, travels from infin- other returns to infinity. The final particle has more energy ity to the interaction point ðr; θÞ¼ð4M; π=8Þ. There it than the original particle and the black hole loses mass. It is splits into massless daughter particles B and C. Particle C useful to work out the analogous process for boosted black falls into the black hole with uτ ¼ −1=5 þ ϵ and −4 holes. This is a warm-up for the more challenging problem uz ¼ −1=5, where ϵ ¼ 10 . Particle B returns to infinity

084044-3 ROBERT F. PENNA PHYSICAL REVIEW D 91, 084044 (2015) believe this is a problem. The energy extracted to infinity in our solutions exactly matches the negative energy carried into the black hole. So the black hole’s energy decreases by the same amount as the energy gained, and it is fairly clear that the black hole is the source of energy. The use of nongeodesic trajectories was forced upon us by the fact that linear momentum is not conserved along geodesics in the Schwarzschild metric. A simple thought experiment illustrates the difficulty. Suppose a particle is dropped from rest into a nonmoving Schwarzschild black hole along a geodesic. The initial momentum of the system is zero. The particle speeds up as it falls toward the hole and crosses the horizon with nonzero linear momentum. So the final momentum of the black hole appears to be nonzero, violating momentum conservation. One way to avoid this problem would be to do a fully general relativistic calculation incorporating the fact that as the particle falls toward the hole, the hole also falls toward the particle. This goes beyond the scope of this paper. An alternate approach, FIG. 2 (color online). Penrose process for a boosted Schwarzs- which we chose, is to use nongeodesic trajectories that child black hole moving in the þz direction with velocity v. conserve linear momentum. This seems to give the closest Timelike particle A (solid red) travels from infinity toward the ð θ Þ¼ð4 π 8Þ analogue to the usual Penrose process for test particles black hole. At the interaction point r; M; = it splits interacting with boosted astrophysical black holes. into massless daughter particles B (dashed blue) and C (dotted magenta). B returns to infinity and C falls into the black hole. For Boosted Schwarzschild black holes are related to static v near 1, C has negative energy and B returns to infinity with Schwarzschild black holes by a Lorentz boost. The Penrose more energy than A. process for static Schwarzschild black holes is impossible, so it may seem puzzling that the Penrose process exists for boosted black holes. A helpful analogy is the billiards with 4-momentum fixed by momentum conservation at the problem of scattering a cue ball off of an eight ball. In one B ¼ A − C θ interaction point: uμ uμ uμ at (r; ). frame, the eight ball is at rest and gains energy from the cue The boosted frame energies (14) are ball, while in another frame the cue ball is at rest and gains energy from the eight ball. Both descriptions are physically 0 ¼ γð3 2 þ 0 9867 Þ ð Þ EA = . v ; 16 equivalent, the point being that the energy, defined as the time component of four-momentum, is not a Lorentz 0 ¼ 0 − 0 ð Þ EB EA EC; 17 invariant. Similarly, in the black hole rest frame the particles lose 0 ¼ γð1 − − 5ϵÞ 5 ð Þ EC v = : 18 energy to the black hole and there is no Penrose process. However, in the boosted frame the black hole’s energy 0 0 (defined as the time component of its ADM 4-momentum) If v is near 1, then C falls into the black hole with EC < 0 0 is larger than its irreducible mass, and it can transfer energy and B returns to infinity with EB >EA. This is a concrete example of energy extraction from boosted black holes. to the particles. Further details of our method for finding these solutions and a second example are given in Appendix B. D. Boosted black strings There may be situations where this process is astrophysi- cally relevant. One can imagine a binary star A that splits The metric apart and creates a hypervelocity star B.OrA could be a 2 single star that is tidally disrupted into streams B and C.We 2 rþ 2 dr 2 2 2 ds ¼ − 1 − dt þ þ r dΩ2 þ dz ð19Þ leave further discussion of these problems for the future. r 1 − rþ=r One difference between our solutions and the usual Penrose process is that we consider nongeodesic trajecto- is a black string in 4 þ 1 dimensions [29]. It is a solution of ries, while the usual Penrose process describes geodesics. the 5d vacuum Einstein equations. The horizon is at r ¼ rþ External forces are required to keep particles on our and has topology S2 × R. trajectories. This could make it difficult to distinguish A boosted black string may be obtained using the whether the energy extracted to infinity is derived from the Lorentz transformation (3)–(4). Now the string carries black hole or from the external forces. However, we do not momentum along z. The ergosurface is at [30]

084044-4 ENERGY EXTRACTION FROM BOOSTED BLACK HOLES: … PHYSICAL REVIEW D 91, 084044 (2015) 2M dr2 ds2 ¼ − 1 − dt2 þ þ r2dΩ2; ð21Þ r 1 − 2M=r

where dΩ2 ¼ dθ2 þ sin2θdϕ2. The fiducial observer (FIDO) frame is pffiffiffiffiffiffiffiffi tˆ e ¼ −gttdt; ð22Þ pffiffiffiffiffiffi tˆ e ¼ grrdr; ð23Þ

θˆ pffiffiffiffiffiffi FIG. 3 (color online). Boosted black string with horizon at e ¼ gθθdθ; ð24Þ rþ ¼ 1 and momentum in the þz direction. Timelike particle A (solid red) travels from infinity toward the black string along a ϕˆ pffiffiffiffiffiffiffi e ¼ gϕϕdϕ: ð25Þ geodesic. Inside the ergosphere, it splits into massless daughter particles B (dashed blue) and C (dotted magenta), which also The relationship with KS coordinates (2) is follow geodesics. C has negative energy. B and C both fall into the horizon. 2M dτ ¼ dt þ dr; ð26Þ r − 2M ¼ γ2 ð Þ rstatic rþ: 20 x ¼ r sin θ cos ϕ; ð27Þ

¼ θ ϕ ð Þ Since ∂z is Killing, one might expect to find Penrose y r sin sin ; 28 process solutions using particles following geodesics. This would make the Penrose process for boosted black strings z ¼ r cos θ: ð29Þ easier to understand than the Penrose process for boosted 0 0 0 0 Schwarzschild black holes. Define boosted Schwarzschild coordinates, ðt ;r; θ ; ϕ Þ,by However, there do not appear to be geodesic Penrose 2M process solutions in this spacetime for an entirely new dt0 ¼ dτ0 − dr0; ð30Þ reason. Recent work [31] has shown that if such solutions r − 2M exist, the interaction point cannot be a turning point of the 0 0 0 0 dr ¼ sin θ cos ϕdx þ sin θ sin ϕdy þ cos θdz ; ð31Þ incoming particle. We numerically searched for solutions for which the interaction point is not a turning point but 0 cos θ cos ϕ 0 cos θ sin ϕ 0 sin θ 0 were unable to find any examples. Figure 3 shows a typical dθ ¼ dx þ dy − dz ; ð32Þ r r r failed solution. Particles A, B, and C all follow geodesics with constant energy and momentum along z. Particle A csc θ sin ϕ csc θ cos ϕ dϕ0 ¼ − dx0 þ dy0; ð33Þ enters the ergosphere of the boosted black string and splits r r in two. Particle C falls into the black string with negative energy. However, particle B also falls into the black string. where ðτ0;x0;y0;z0Þ are defined by (3)–(6). In primed The horizon is infinitely extended and it is impossible for B coordinates the black hole carries momentum along z. to travel around it along a geodesic. The reverse transformation is

0 0 2M 0 III. JETS dτ ¼ dt þ dr ; ð34Þ r − 2M The BZ model [15,16] is the electromagnetic younger 0 ¼ θ ϕ 0 þ θ ϕ θ0 cousin of the Penrose process. It describes how electro- dx sin cos dr r cos cos d magnetic fields can extract the rotational energy of Kerr − r sin θ sin ϕdϕ0; ð35Þ black holes and it is widely believed to describe astro- physical jets. In this section, we develop the analogue of the dy0 ¼ sin θ sin ϕdr0 þ r cos θ sin ϕdθ0 BZ jet power prediction for boosted Schwarzschild þ θ ϕ ϕ0 ð Þ black holes. r sin cos d ; 36

dz0 ¼ cos θdr0 − r sin θdθ0: ð37Þ A. Coordinates The Schwarzschild metric in Schwarzschild coordinates, Useful transformations between these reference frames ðt; r; θ; ϕÞ,is are collected in Appendix C.

084044-5 ROBERT F. PENNA PHYSICAL REVIEW D 91, 084044 (2015) B. Jet power at infinity (see, e.g., [20]). Combining the outgoing Define the jet power to be boundary condition with the force-free constraint implies Erˆ ¼ 0 Brˆ ¼ 0 r Z or at large . The outgoing boundary 0 pffiffiffiffiffiffiffi 2 ¼ 2ð 2 − 2Þ¼ dE 0 condition (43) also implies F B E ≡ ¼ − r − 0 θ0 ϕ0 ð Þ 2 2 Pjet 0 T 0 g d d ; 38 2ðB − E Þ at large r. Astrophysical fluids are magnetically dt 2 t rˆ rˆ S dominated because the electric field vanishes in the rest 0 frame of highly ionized plasma. So we choose Erˆ ¼ 0 at so Pjet > corresponds to energy leaving the black hole. ∂ 0 large r. This is the usual choice in astrophysics and it is the The vector t is not Killing, so Pjet may be a function of radius. We are interested in the jet power at infinity (the jet case that has been simulated (e.g., [5]). power at the horizon is computed Appendix D). In our It is helpful to replace E~ ∥ and B~ ∥ with the field line idealized setup, we assume an isolated black hole with a jet velocity vF, defined by E ¼ −vF × B. In components, the extending to infinity. Astrophysical jets extend far beyond fields at infinity become the horizon, so our idealized setup is a good approximation. r θˆ The FIDO-frame components of T 0 are ˆ ˆ v t Eϕ ¼ −Bθ ¼ F Brˆ; ð45Þ 1 − vrˆ 2M F r ¼ γ α2 þ θ rˆ − γ θ rˆ þ γ α θ rˆ T 0 v cos Tˆ v cos Trˆ v sin Tθˆ ; t r t ϕˆ ˆ ˆ v ð Þ Eθ ¼ Bϕ ¼ − F Brˆ: ð46Þ 39 1 − rˆ vF where α2 ¼ 1 − 2M=r. The advantage of the FIDO frame is Plugging into (39) gives the stress-energy tensor at infinity, that Tμˆ νˆ is simply [32] ∥ 2 1 1 r vF 2 2 tˆ tˆ 2 2 T 0 ¼ −γ B þ γv cos θB T ¼ ðE þ B Þ; ð40Þ t 1 − rˆ rˆ 2 rˆ 2 vF θˆ ˆ ˆ ˆ ˆ ˆ v Tt j ¼ Tj t ¼ðE × BÞj; ð41Þ þ γv sin θ F B2; ð47Þ 1 − rˆ rˆ vF ˆ ˆ ˆ ˆ ˆ ˆ 1 2 2 ˆ ˆ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi j k ¼ − j k − j k þ ðE þ B Þδj k ð Þ ˆ T E E B B 2 ; 42 ∥ ≡ ð θˆ Þ2 þð ϕÞ2 where vF vF vF . Assume small velocities: v =r ∼ v ≪ 1. A slowly mov- where E and B are the FIDO-frame electric and magnetic F ing black hole (v ≪ 1) is assumed for simplicity. Small fields. v =r should be a good assumption in this case because we At infinity, the six components of the electromagnetic F expect v =r ∼ v (just as in the BZ model for spinning black field are not all independent because radiation is always F holes). In this limit, outgoing at I þ. In particular, in the large r limit, we have the boundary condition 1 r ∥ 2 2 2 θˆ 2 T 0 ¼ −ðv Þ B þ v cos θB þ vv sin θB ; ð48Þ t F rˆ 2 rˆ F rˆ E~ ∥ ¼ −n × B~ ∥; ð43Þ and θˆ ϕˆ θˆ ϕˆ where E~ ∥ ¼ðE ;E Þ, B~ ∥ ¼ðB ;B Þ, and n is the outward- Z dE0 pffiffiffiffiffiffi pointing unit normal vector (see Appendix E for a deriva- ¼ − r − θ ϕ ð Þ 0 Tt0 gd d ; 49 tion). In components, dt S2

ˆ ˆ ˆ ˆ Eθ ¼ Bϕ;Eϕ ¼ −Bθ: ð44Þ so integrating (48) over the sphere at infinity gives the jet ∥ rˆ power. It depends on the unknown functions vF and B . This eliminates two components of the fields at infinity. Unlike the original BZ model for spinning black holes, We further enforce the force-free constraint E · B ¼ 0, there are no exact force-free solutions to be our guide. which is a good approximation for astrophysical black hole There are less symmetries than in the BZ model, so it is magnetospheres [15,16]. In astrophysical jets, the force- unclear whether exact solutions are possible. ∥ rˆ free condition breaks down far from the black hole, in the For the moment, the best guide to vF and B are so-called load region, where gas kinetic energy becomes numerical simulations. Numerical simulations of BZ jets comparable to the magnetic energy of the jet. The force-free tend to relax to field geometries with ΩF=ΩH ≈ 1=2 (where condition is a good approximation between the horizon and ΩF and ΩH are the field line and horizon angular veloc- the load region. The load is believed to be sufficiently far ities), and Brˆ is roughly uniform on the horizon (at least for from the black hole so that for our purposes we may place it low black hole spins) [20,33]. The field is approximately a

084044-6 ENERGY EXTRACTION FROM BOOSTED BLACK HOLES: … PHYSICAL REVIEW D 91, 084044 (2015) split monopole. The split monopole is in some sense the We begin by reviewing the standard black hole mem- simplest solution and it acts like a ground state, while brane paradigm. A modern derivation is based on an action higher order multipoles are radiated away. principle [19]. Consider an observer who remains forever in We assume jets from boosted black holes are similar and the black hole exterior. Such an observer cannot receive guess signals from the black hole interior, so the interior can be eliminated from their calculations. In particular, given a ∥ v =r 1 Lagrangian, L, they can use the action F ¼ ; ð50Þ v=ð2MÞ 2 Z pffiffiffiffiffiffi ¼ 4 − L ð Þ rˆ S d x g ; 52 and that B is a function of r only. In this case, only the first exterior term on the rhs of (48) contributes to the integral (49) and the jet power is with domain of integration restricted to the black hole δ exterior. The variation of this action, S, gives boundary 1 v 2 terms supported on the horizon. To obtain the correct P ¼ Φ2; ð51Þ jet 4π 2M equations of motion, the boundary terms need to be eliminated by adding surface terms to the action. The Φ ¼ð2π 2 rˆÞ surface terms encode the properties of the membrane on the where r B r→∞ is the flux through a hemisphere at infinity. This is similar to the BZ prediction, horizon. In particular, they fix the membrane’s current BZ ¼ Ω2 Φ2 ð6πÞ ð2 Þ density and stress-energy tensor. Further imposing the Pjet H H= , but with v= M playing the role of ΩH, the flux measured at infinity rather than the horizon, boundary condition that all waves are ingoing at the and a slightly different normalization constant. The upshot horizon fixes the resistivity and viscosity of the membrane. is that boosted black holes and spinning black holes have The true horizon is a null surface. It is convenient to jets of comparable strength when v=ð2MÞ ∼ ΩH (for fixed define the membrane on a stretched horizon, a timelike magnetic flux). surface some small distance above the true horizon, and The jet power observed in numerical simulations of then take the true horizon limit. boosted black holes appears to be smaller than (51) by as This section is based on the observation that the same þ much as a factor of 100 [5]. It may be that (50) is an recipe works at I . Consider an observer who remains overestimate of the field line velocity. It may also be forever in the black hole exterior. They cannot receive þ relevant that the simulated jets do not extend over a full 4π signals from beyond I . Define “stretched infinity” to be a steradians. It will be interesting to understand this differ- timelike surface some large but finite distance from the 0 ence better but we save a more detailed comparison for the black hole (see Fig. 4). Let M be a truncated spacetime future. ending at stretched infinity. Given a Lagrangian L, use the The membrane paradigm gives a dual description of action black holes as conductive membranes [16,20]. The power Z pffiffiffiffiffiffi radiated by a conductor moving through a magnetic field ¼ 4 − L ð Þ 2 2 S d x g ; 53 scales with velocity and field strength as P ∼ v B [34,35]. M0 So the black hole jet power may also be expected to scale as 0 v2B2 [5]. The jet power formula (51) confirms this expect- with domain of integration M . Varying this action gives ation. The power radiated by a conductor scales with the boundary terms supported on stretched infinity, which must size of the conductor as L2 [34,35]. This shows up in our be canceled by adding surface terms to the action. These formula as a factor of M2. surface terms fix the current and stress-energy tensor of the

IV. THE MEMBRANE AT INFINITY The BZ model has an elegant formulation in the black hole membrane paradigm [16,20]. In this picture, the black hole is represented by a fluid membrane at the horizon. The black hole’s mass and angular momentum are stored in the membrane’s stress-energy tensor and jets are powered by electromagnetic torques acting on the membrane. In our model of jets from boosted black holes, the energy flux at the horizon need not match the energy flux at infinity because ∂t0 is not Killing. So in this section we will reformulate the membrane paradigm such that the mem- FIG. 4. Black hole Penrose diagram. Stretched infinity (dotted) brane lives at infinity (where the jet power is evaluated) is a timelike surface some large but finite distance from the rather than the horizon. black hole.

084044-7 ROBERT F. PENNA PHYSICAL REVIEW D 91, 084044 (2015) membrane at infinity. The boundary condition that all E~ ∥ ¼ ρ~j; ð60Þ waves are outgoing at I þ fixes the resistivity and viscosity of the membrane. on the membrane at infinity. Equations (43) and (57) at stretched infinity differ from the black hole horizon A. Membrane current versions by relative minus signs, but these signs cancel in (60). So the resistivity of the membrane at infinity has To derive the electromagnetic properties of the mem- the same value as in the usual black hole membrane brane at infinity, consider the Maxwell action paradigm, ρ ¼ 1 ¼ 377 Ω. Z pffiffiffiffiffiffi 1 ¼ 4 − − 2 þ ð Þ S d x g 4 F J · A : 54 B. Membrane stress-energy tensor Now consider the Einstein-Hilbert action. Varying the Varying this action gives a term which is a total derivative action on M0 gives a surface term supported on stretched Z infinity. Eliminating this surface term endows stretched pffiffiffiffiffiffi ab 4 infinity with a stress-energy tensor − ∂að −gF δAbÞd x; ð55Þ 1 ¼ − ð − Þ ð Þ and integrating by parts gives a surface term supported on tab 8π Khab Kab ; 61 stretched infinity, where h is the induced metric on stretched infinity, Z pffiffiffiffiffiffi ab − 3 − ab δ ð Þ d x hF na Ab; 56 a ¼ a ð Þ Kb n jb; 62

a where h is the determinant of the induced metric on is its extrinsic curvature, K ¼ K a, and jb is the three- stretched infinity and na is the outward-pointing spacelike covariant derivative on stretched infinity. This is the same unit normal at stretched infinity. We conclude that stretched stress-energy tensor that appears in the original black hole infinity carries a current membrane paradigm [16,19] but with an overall minus sign. As in the previous section, the sign comes from the a ¼ − ab ð Þ j F nb: 57 fact that tab is obtained from an integration by parts and stretched infinity is an outer boundary of spacetime. Its time component, Equation (61) is the same as the Brown-York stress-energy tensor but with an overall minus sign. We explain the origin σ ¼ − tb ¼ − ð Þ F nb E⊥; 58 of this difference below. ’ ’ Just as the membrane s current terminates electric and is the membrane s charge density and it terminates the magnetic fields, the membrane’s stress-energy tensor cre- normal component of the electric field at stretched infinity. a ates a discontinuity in the extrinsic curvature. The dis- The spatial components of j form a surface current continuity is given by the Israel junction condition [16,19] terminating the tangential components of the magnetic field, 1 ¼ ð½ − ½ Þ ð Þ tab 8π K hab K ab ; 63 B~ ∥ ¼ nˆ × ~j: ð59Þ where ½K¼Kþ − K− is the difference between the The current (57) contains an overall minus sign relative extrinsic curvature of stretched infinity as defined with to the current in the usual black hole membrane paradigm respect to the spacetime outside stretched infinity and as [16,19]. This may be traced to the integration by parts of defined with respect to the spacetime inside. The extrinsic (55) and the fact that stretched infinity is an outer boundary curvature appearing in (61) is K−, so the Israel junction 0 of M whereas the horizon is an inner boundary. The minus implies Kþ ¼ 0. In other words, the membrane stress- sign has a simple physical interpretation: outward-pointing energy tensor (61) terminates the gravitational field outside radial field lines begin at positive charges on the stretched stretched infinity. The Brown-York stress-energy tensor is horizon and terminate at negative charges on stretched defined so as to terminate the gravitational field inside I þ. infinity. The charge density of stretched infinity vanishes in This explains the relative minus sign between (61) and the the true infinity limit. However, the surface area blows up in Brown-York stress-energy tensor. this limit, so the total charge of the membrane at infinity The analogue of the electromagnetic outgoing boundary remains finite. condition (43) is encoded in the relationship between the a At stretched infinity, we have the outgoing boundary extrinsic curvature of stretched infinity, K b, and the condition (43). Combined with (57), it implies Ohm’slaw, extrinsic curvature of true infinity,

084044-8 ENERGY EXTRACTION FROM BOOSTED BLACK HOLES: … PHYSICAL REVIEW D 91, 084044 (2015) a a k b ¼ l jb; ð64Þ and so,

þ where l is the future-directed null generator of I . Null nˆ ¼ ∂r ¼ −∂t ¼ −2∂u ¼ −l: ð73Þ generators are normal to true infinity because it is a null surface, and the future-directed null generator plays the role As stretched infinity approaches true infinity, nˆ → −l,as of the outward-pointing normal in the definition of extrinsic claimed. This explains the minus sign in (66). curvature for null surfaces. Enforcing the boundary condition (66) turns tab into the As stretched infinity approaches true infinity, stress-energy tensor of a viscous fluid. Split spacetime into space and time by fixing a family of fiducial observers with na → −la; ð65Þ four-velocity Ua such that Ua → la at true infinity. (For Schwarzschild, these are the FIDOs.) Define constant-time and so surfaces to be surfaces to which Ua is orthogonal. The metric on a two-dimensional constant-time slice of a a K b → −k b: ð66Þ stretched infinity is Iþ The minus sign reflects the fact that all radiation at is γAB ¼ hAB þ UAUB; ð74Þ outgoing. At a black hole horizon the sign would be positive. where uppercase indices A; B; … indicate tensors living on To summarize, the membrane at infinity differs from the these slices. membrane at the horizon by two extra minus signs. The The time-time component of the extrinsic curvature is first minus sign is the overall sign in (61). This minus sign þ a b appears because I is an outer boundary of spacetime U Ubk a ¼ −κ; ð75Þ rather than an inner boundary. The second extra minus sign þ κ a∇ b ¼ κ b is the sign in (65). This minus sign appears because I where the surface gravity, , is defined by l al l , satisfies an outgoing rather than an ingoing boundary and we have used Eqs. (64) and (65). Decompose the condition. These two minus signs are independent. For space-space components of the extrinsic curvature into a example, at I − only the first extra minus sign would appear. traceless part and a trace, At a horizon only the second extra minus sign 1 would appear. kAB ¼ σAB þ γABθ; ð76Þ To clarify the minus sign in (65), consider the 2 Schwarzschild spacetime (21). Ingoing and outgoing where σAB is the shear and θ the expansion. The time-space Eddington-Finkelstein coordinates are b A ¼ 0 ¼ A ¼ components vanish: U kb . The trace is k k A κ þ θ ¼ þ ð Þ . v t r 67 Plugging into (61) gives the stress tensor of the mem- brane at infinity, u ¼ t − r; ð68Þ 1 1 ¼ −σ þ γ θ þ κ ð Þ where tAB 8π AB AB 2 : 77 dr 2M −1 ¼ 1 − : ð69Þ It is the usual stress tensor of a two-dimensional viscous dr r Newtonian fluid with pressure p ¼ κ=ð8πÞ, shear viscosity η ¼ 1=ð16πÞ, and bulk viscosity ζ ¼ −1=ð16πÞ. Stretched infinity is a timelike surface at some large but Equations (61) and (66) differ from the stretched horizon finite radius. Its outward-pointing unit spacelike normal is versions by relative minus signs but these signs cancel in (77), so the viscosity parameters of the membrane at nˆ ¼ ∂ : ð70Þ r infinity are the same as in the standard membrane paradigm The future-directed null generator of true infinity is at the black hole horizon.

l ¼ 2∂u; ð71Þ C. Jets revisited Consider the momentum flux, where the normalization is a convention that leads to Z simpler formulas. On Iþ, v ¼ const and dv ¼ 0. In this pffiffiffiffiffiffi dP r – ¼ T 0 −gdθdϕ; ð78Þ case, (67) (68) give dt0 z

du ¼ 2dt ¼ −2dr; ð72Þ at stretched infinity for a boosted Schwarzschild black hole.

084044-9 ROBERT F. PENNA PHYSICAL REVIEW D 91, 084044 (2015) For small v, the only contribution is the term includes all the electromagnetic degrees of freedom we need at infinity. r rˆ rˆ θˆ 0 ¼ − θ ˆ ¼ θ ð Þ Tz sin T θ sin B B : 79 In these variables, the momentum flux (80) is Z pffiffiffiffiffiffi In membrane variables, the momentum flux is dP z ¼ σB −gdθdϕ; ð88Þ Z dt0 dP pffiffiffiffiffiffi ¼ ð~j × B~Þz −gdθdϕ; ð80Þ dt0 which is the magnetic monopole equivalent of a qE Lorentz force. The torques driving standard BZ jets are ϕ where we have used (59). This is the usual expression for a σB Lorentz forces. The energy flux is the same as (82) Lorentz force acting on the membrane at infinity. 2 2 but with j~jj in place of j~jj . The advantage of the dual For small v, the energy flux at infinity is current formulation is that all the variables at infinity live in 0 Z 2 þ 1 dE ˆ ˆ ˆ ˆ pffiffiffiffiffiffi dimensions. ¼ ðEθBϕ − EϕBθÞ −gdθdϕ: ð81Þ dt0 V. CONCLUSIONS ’ Using the outgoing boundary condition (43) and Ohm s We have developed the theory underlying kinetic energy law (60) gives extraction from moving black holes. We derived the 0 Z analogues of the Penrose process and the BZ jet power dE 2pffiffiffiffiffiffi 0 ¼ ρj~jj −gdθdϕ; ð82Þ prediction for boosted black holes. We also derived a new dt version of the membrane paradigm in which the membrane lives at infinity, and we showed that this formalism is useful the usual expression for Joule heating in a resistor. for interpreting energy extraction from boosted black holes. The Penrose processes for boosted black holes and D. Dual current formulation spinning black holes have a similar conceptual basis. In The membrane current, ja, encodes all components of both cases, energy extraction is related to the existence of the electromagnetic field at infinity except Brˆ. There is an negative energy trajectories. BZ jets are a generalized alternate formulation of membrane electrodynamics in version of the Penrose process, with force-free electromag- which all the variables we need at infinity are components netic fields replacing point particles. So jets from boosted of the membrane current. Start not from the usual Maxwell black holes and spinning black holes are also qualitatively action (54), but rather similar. The same language that describes jets from spin- Z ning black holes (e.g., negative energy fluxes inside the 1 pffiffiffiffiffiffi S ¼ − ðFÞ2 −gd4x; ð83Þ ergosphere, torques acting on a membrane) can be applied 4 to jets from boosted black holes. We have highlighted two important technical differences where F is the dual field strength. Then the membrane’s between boosted black holes and spinning black holes. One current density is is that the relevant notion of energy in the boosted case is defined with respect to a vector ∂t0 which is not Killing. As a ¼ − ab ð Þ j F nb; 84 a result, the energy flux at the horizon need not match the energy flux at infinity even for ∂t0 -invariant solutions. One instead of (57). It is a magnetic monopole current. The can construct solutions in which the energy fluxes at the magnetic monopole charge density is horizon and infinity are the same (as we showed in Sec. II), but astrophysically relevant solutions (such as the jets in σ ≡ tˆ ¼ − rˆ ð Þ j B ; 85 Sec. III), are unlikely to have this property. So it is important to compute fluxes at infinity. and it terminates the normal component of the magnetic A second difference between energy extraction from field. The idea of terminating the magnetic field at the boosted black holes and spinning black holes is that the horizon with monopole charges has been suggested by former is an observer-dependent process, while the latter is [36]. The other components of the monopole current are observer independent. This can be traced to the fact that the

θˆ ϕˆ θˆ relevant notion of black hole energy in the boosted case is ¼ ¼ − ð Þ ADM j E B ; 86 the time component of Pμ , which is not a Lorentz

ˆ invariant. The relevant notion of black hole energy in the ϕ ¼ − θˆ ¼ − ϕˆ ð Þ − 2 j E B : 87 spinning case is the norm PADM, which is Lorentz invariant. iˆ rˆ The only component of the field not packaged in j is E , Our discussion of the Penrose process for boosted black rˆ iˆ but force-free jets have E ¼ 0 at stretched infinity. So j holes in Sec. II relied on numerical solutions for trajectories

084044-10 ENERGY EXTRACTION FROM BOOSTED BLACK HOLES: … PHYSICAL REVIEW D 91, 084044 (2015) with constant linear momentum. It may be possible to find area of an event horizon with vanishing expansion is slicing and classify these trajectories analytically. This would invariant. This is a well-known statement (see e.g. [39])but allow one to answer a number of interesting questions. we record a proof here for completeness. For example, what is the maximum energy that can be Consider a foliation of spacetime into spacelike slices, Σ, extracted using the boosted black hole Penrose process as a with future-pointing unit normal ξa. The horizon is a a function of the interaction point ðr; θÞ? The answers are 2-sphere in Σ with outward-pointing unit normal n . The somewhat coordinate dependent, but understanding the induced metric on the horizon is answers in a natural coordinate system would give insight into general features of the process. mab ¼ gab þ ξaξb − nanb: ðA1Þ We have described the analogue of BZ jets and computed the jet power (51) to be The presence of ξa in this formula suggests the area computed using mab might be slicing dependent. Let 1 v 2 P ¼ Φ2; ð89Þ jet 4π 2M ξ ξ − ¼ 2 þ − ð Þ a b nanb kðakbÞ; A2 pffiffiffi at least for small v. This can be tested with numerical where ka ¼ðξa naÞ= 2 are null vectors. We can rescale simulations [3,5]. It will be interesting to use simulations to þ þ a k such that k ¼ l and k k− ¼ −1. understand the distributions of Ω and Brˆ and to compare a F The area of the horizon is the energy and momentum fluxes at the horizon and infinity. On the analytical side, our computations can be Z generalized away from the small v limit and they can be A ¼ m1=2d2x; ðA3Þ generalized from boosted Schwarzschild black holes to S2 boosted Kerr black holes. 2 where m ¼ detðmabÞ. The choice of S depends on the We have shown that it is possible to reformulate the 2 standard membrane paradigm such that the membrane lives slicing, but one S can be carried into another by trans- at infinity rather than the black hole horizon. The mem- lations along l. For the area to be slicing invariant, we brane at infinity has the same resistivity and viscosity require coefficients as in the standard membrane paradigm. The L 1=2 ¼ 0 ð Þ membrane at infinity is useful for understanding jets from lm ; A4 boosted black holes because the energy and momentum fluxes at infinity can be described using the familiar which is equivalent to vanishing expansion: language of dissipation and Lorentz forces acting on a 1 conductor. a 1=2 θ ¼ ∇al ¼ Llm ¼ 0: ðA5Þ The stress-energy tensor of the membrane at infinity is m1=2 the same as the Brown-York stress-energy tensor [37] up to a minus sign. The Brown-York stress-energy tensor is not The second equality in (A5) follows from the Jacobi finite for general asymptotically flat spacetimes but formula for Lie derivatives. Let aij be a nonsingular matrix ¼ requires the addition of Mann-Marolf counterterms [38]. and let a det aij. Then the Jacobi formula is Similar counterterms should be incorporated into the L ¼ jiL ð Þ definition of the membrane at infinity. We hope to explore Xa aa Xaij: A6 the membrane interpretation of these counterterms in the future. To prove this formula, note that the determinant is a polynomial in the aij such that there is the chain rule ACKNOWLEDGMENTS ∂a I thank Richard Brito, Vitor Cardoso, Carlos Palenzuela, LXa ¼ LXaij: ðA7Þ ∂a Paolo Pani, Maria Rodriguez and especially Luis Lehner ij for helpful comments and suggestions. This work was ji supported by a Pappalardo Fellowship in Physics at MIT. Replacing the partial derivatives with aa gives the Jacobi formula. An application of the Jacobi formula gives

APPENDIX A: BOOST INVARIANCE OF 1 1=2 ba c c c a L m ¼ m ðl m ; þ m l; þ m l Þ¼∇ l ; HORIZON AREA l log 2 ab c cb a ac ;b a The discussion of Sec. II A relied on the fact that the area ðA8Þ of a Schwarzschild black hole’s event horizon is boost invariant. This follows from the more general fact that the which is (A5).

084044-11 ROBERT F. PENNA PHYSICAL REVIEW D 91, 084044 (2015) APPENDIX B: PENROSE PROCESS SOLUTIONS In this section we detail the numerical method used to find the Penrose process solutions discussed in Sec. II and we give another example of such a solution. Our task is to find three trajectories, A, B, and C, which meet at an interaction point ðr; θÞ such that four- momentum is conserved,

A B C uμ ¼ uμ þ uμ at ðr; θÞ: ðB1Þ

We further require that A and B extend to infinity and C falls into the black hole with negative energy in the boosted KS frame. We assume A is timelike and B and C are null. Each trajectory is then fully characterized by two constants, uτ and uz (we set uy ¼ 0). Given these constants, a trajectory ðrðτÞ; θðτÞÞ is fixed by the differential equations 2 θ θ M cos cos _ FIG. 5 (color online). Penrose process for a boosted Schwarzs- u ¼ − uτ þ r_ − r sin θθ; ðB2Þ z α2r α2 child black hole moving in the þz direction with velocity v. This is similar to Fig. 2, except the momentum of particle A is 2 _2 2 ¼ − uτ þ r þ 2θ_2 ð Þ primarily along z (rather than x) and a larger amount of energy is m α2 α2 r ; B3 extracted (for fixed v). pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi where α ¼ 1 − 2M=r. Four-momenta in KS and uB ¼ uA − uC; ðB9Þ Schwarzschild coordinates are related by t t t

¼ ð Þ B A C uτ ut; B4 uz ¼ uz − uz : ðB10Þ 2 θ θ M sin cos The trajectories of A, B, and C are now fully determined. u ¼ − u þ sin θu þ uθ; ðB5Þ x α2 t r C C A r r We used trial and error to find uτ , uz , and uτ such that the trajectories of A and B extend to infinity and particle C falls 2 θ θ ¼ − M cos þ θ − sin ð Þ into the black hole. uz 2 ut cos ur uθ: B6 α r r Figure 5 shows one such solution. The interaction point is the same as in Sec. II, ðr; θÞ¼ð4M; π=8Þ,but We begin by fixing the interaction point ðr; θÞ and the C C the momentum of A is primarily along z rather than x. two constants uτ and u that define particle C. In Sec. II, A A z Particle A has ðuτ ;u Þ¼ð−10; −9.9376Þ, particle C has ð θ Þ¼ð4 π 8Þ C ¼ −1 5 þ ϵ z we picked r; M; = , uτ = , and ð C CÞ¼ð−6 99 −7Þ C −4 uτ ;uz . ; , and the momentum of particle B uz ¼ −1=5, where ϵ ¼ 10 . A A is fixed by energy conservation. The energy of particle C in Next, we choose ut . In our example, ut ¼ −3=2 . The A the boosted frame is remaining components of particle A’s four-momentum, ur A and uθ , are fixed by 0 ¼ −γð0 01 − 7ð1 − ÞÞ ð Þ EC . v ; B11 A 2 A 2 μ ðu Þ ðu Þ −1 ¼ A ¼ − t þ α2ð AÞ2 þ θ ð Þ u uμ 2 ur 2 ; B7 which is negative for v near 1. A α r For this process to make sense, it is important that uz is A C A C finite at the horizon. A coordinate transformation gives μ C ut ut 2 A C uθ uθ −1=2 ¼ u uμ ¼ − þ α u u þ ; ðB8Þ A α2 r r r2 2M cos θ sin θ u ¼ − u þ cos θu − uθ: ðB12Þ at the interaction point. Equation (B8) follows from z α2r t r r 0 ¼ μ B ¼ −1 − 2 μ C energy conservation: uBuμ uAuμ , and so μ C ¼ −1 2 uAuμ = . The four-momentum of particle A in KS The first two terms on the rhs are infinite at the horizon. We coordinates is given by (B4)–(B6). need to check that these infinities cancel. Let us check this Finally, we fix the four-momentum of particle B using for a radial null geodesic (the general case is not much energy conservation (B1). In particular, harder). In this case,

084044-12 ENERGY EXTRACTION FROM BOOSTED BLACK HOLES: … PHYSICAL REVIEW D 91, 084044 (2015) μ tt 2 rr 2 0 ¼ uμu ¼ −g u þ g ðu Þ : ðB13Þ 2Mv cos θ t r dt ¼ γ 1 þ dt0 α2r It follows that u ¼ g u ¼ u =α2. Plugging into (B12) and r rr t t 2α2ðγ − 1ÞMsin2θ þ γv cos θð4M − rÞ setting r ¼ 2M gives þ dr0 α4r θ 2 sin sin θð2ðγ − 1ÞM cos θ þ α γrvÞ 0 u ¼ uθ; ðB14Þ þ dθ ; ðC9Þ z r α2 which is finite. 2γMv cos θ dr ¼ −γv cos θdt0 þ γcos2θ þ sin2θ − dr0 α2r APPENDIX C: REFERENCE FRAMES − ðγ − 1Þr sin θ cos θdθ0; ðC10Þ The discussion in Sec. III relied on several different γ θ θð2γ − α2ðγ − 1Þ θÞ reference frames. Here we collect some of the relevant θ ¼ v sin 0 þ sin Mv r cos 0 d dt 2 2 dr transformations. r α r The FIDO-frame components of the boosted þðγsin2θ þ cos2θÞdθ0; ðC11Þ Schwarzschild basis vectors are dϕ ¼ dϕ0: ðC12Þ 2Mv cos θ γv cos θ ∂ 0 ¼ γ α þ eˆ − eˆ t α t α r ≪ 1 r The equivalence of (38) and (49) forR v pffiffiffiffiffiffiffifollows from 0 r0 0 0 0 0 þ γ θ ð Þ (C9). To see this, write dE ¼ − T 0 −g dθ dt dϕ ¼ v sin eθˆ ; C1 R pffiffiffiffiffiffi t − r − θ ϕ0 ¼ 0 þ ð Þ Tt0 gdtd d and then note dt dt O v . 2Mα2ðγ − 1Þsin2θ − γðr − 4MÞv cos θ The Schwarzschild components of the KS basis vectors ∂ 0 ¼ eˆ r α3 t r are 1 2 2 2Mγv cos θ þ sin θ þ γcos θ − eˆ ∂ ¼ ∂ ð Þ α α2r r τ t; C13 2 γ M v 2M sin θ cos θ þ sin θ cos θð1 − γÞþ eθˆ ; ðC2Þ α2 ∂x ¼ − ∂t þ sin θ∂r þ ∂θ; ðC14Þ r α2r r sin θð2Mðγ − 1Þ cos θ þ α2γrvÞ 1 ∂θ0 ¼ eˆ ∂ ¼ ∂θ; ðC15Þ α t y r sin θ ðγ − 1Þr sin θ cos θ − eˆ 2 θ θ α r ∂ ¼ − M cos ∂ þ θ∂ − sin ∂ ð Þ z 2 t cos r θ: C16 2 2 α r r þ rðγsin θ þ cos θÞeθˆ ; ðC3Þ Equations (B4)–(B6) follow from these relations. ∂ϕ0 ¼ r sin θeϕˆ : ðC4Þ

The FIDO-frame components of the Schwarzschild one- APPENDIX D: JET POWER AT THE HORIZON forms are Recall that the jet power (38) is 1 Z ¼ tˆ ð Þ 0 pffiffiffiffiffiffiffi dt e ; C5 dE 0 α ≡ ¼ − r − 0 θ0 ϕ0 ð Þ Pjet 0 Tt0 g d d : D1 dt S2 dr ¼ αerˆ; ðC6Þ ∂ 0 The vector t is not Killing, so Pjet may be a function of 1 ˆ radius. In this section we evaluate the jet power at the dθ ¼ eθ; ðC7Þ r horizon. The astrophysically more interesting observable is the jet power at infinity, which we computed in Sec. III B. 1 r ϕˆ As before, the FIDO-frame components of T 0 are dϕ ¼ e : ðC8Þ t r sin θ 2 r 2 M rˆ rˆ rˆ Equation (39) follows from (C1) and (C6). T 0 ¼ γ α þ v cos θ Tˆ − γv cos θT ˆ þ γvα sin θT ˆ ; t r t r θ The boosted Schwarzschild components of the Schwarzschild one-forms are ðD2Þ

084044-13 ROBERT F. PENNA PHYSICAL REVIEW D 91, 084044 (2015) α2 ¼ 1 − 2 ¼ð 2 Þ2Φ2 ð12πÞ Φ ¼ 2π where M=r. In the FIDO frame, the stress- Pjet v= M H= , where H rBrˆ is the mag- energy tensor has its usual form (40)–(42). netic flux at the horizon. As noted earlier, this need not At the horizon, the six components of the electromag- match the jet power at infinity computed in Sec. III B netic field are not all independent because radiation is because ∂t0 is not Killing. always ingoing at the horizon. In particular, we have the horizon boundary condition [19] APPENDIX E: OUTGOING BOUNDARY E~ ∥ ¼ n × B~ ∥; ðD3Þ CONDITION

θˆ ϕˆ θˆ ϕˆ In Sec. III B, we imposed the outgoing boundary where E~ ∥ ¼ðE ;E Þ, B~ ∥ ¼ðB ;B Þ, and n is the outward- condition (43) pointing unit normal vector. In components,

θˆ ϕˆ ϕˆ θˆ E~ ∥ ¼ −n × B~ ∥ ðE1Þ E ¼ −B ;E¼ B : ðD4Þ Iþ This eliminates two components of the fields at the horizon. at . This boundary condition has appeared before (see, We also have the force-free constraint E · B ¼ 0. e.g., [40]). It differs from the ingoing boundary condition Combined with the horizon boundary condition, it implies imposed at black hole horizons by an overall minus sign. A Erˆ ¼ 0 or Brˆ ¼ 0 at the horizon. We choose Erˆ ¼ 0. simple derivation of the horizon boundary condition has As before, we replace E~ ∥ and B~ ∥ with the field line been given by [16,19]. In this section we adapt their argument to Iþ and derive (E1). velocity vF, defined by E ¼ −vF × B. In components, the fields at the horizon are Equation (E1) is expressed in the FIDO frame. The FIDO frame is singular at Iþ: all of Iþ is mapped to t ¼ r ¼ ∞. θˆ Outgoing Eddington-Finkelstein coordinates, ˆ ˆ v Eϕ ¼ Bθ ¼ F Brˆ; ðD5Þ 1 þ rˆ vF 2M ds2 ¼ − 1 − du2 − 2dudr þ r2dΩ2; ðE2Þ ϕˆ r ˆ ˆ v Eθ ¼ −Bϕ ¼ − F Brˆ: ðD6Þ 1 þ rˆ vF are nonsingular there. Lines of constant u are null, but we can perturb them slightly so that they become timelike near Plugging into (D2) gives the stress-energy tensor at the Iþ. Let E~ and B~ be the electric and magnetic fields horizon, measured by local observers in this frame. E~ and B~ and the

∥ 2 ˆ FIDO-frame fields are related by a Lorentz boost. At α 1 α θ rˆ r vF 2 2 vF 2 stretched infinity, FIDOs move with velocity v ≈−1 with T 0 ¼ γ B þ γv cos θB − γv sin θ B ; t 1 þ rˆ rˆ 2 rˆ 1 þ rˆ rˆ vF vF respect to perturbed Eddington-Finkelstein observers, so ðD7Þ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Eθˆ ≈ γðEθ~ þ Bϕ~ Þ; ðE3Þ ˆ ∥ ¼ ð θˆ Þ2 þð ϕÞ2 where vF vF vF . For small velocities, E ˆ ≈ γðE ~ − B~ Þ; ðE4Þ 1 ϕ ϕ θ r ∥ 2 2 2 θˆ 2 T 0 ≈ ðαv Þ B þ γv cos θB − αv v sin θB : ðD8Þ t F rˆ 2 rˆ F rˆ Bθˆ ≈ γðBθ~ − Eϕ~ Þ; ðE5Þ This is the same as the expression at infinity (48), except ≈ γð þ Þ ð Þ the first and third terms on the rhs differ by relative Bϕˆ Bϕ~ Eθ~ : E6 minus signs and by extra factors of α. At infinity, only ~ ~ the first term on the rhs contributed to the jet power, but If E and B are finite, then it follows from (E3)–(E6) that at the horizon this term has the wrong sign to describe Eθˆ ≈ Bϕˆ and Eϕˆ ≈−Bθˆ on stretched infinity, with equality energy extraction. It describes dissipation on the stretched in the true infinity limit. This proves (E1). The derivation of horizon. At the horizon, energy extraction is provided the ingoing boundary condition at the horizon is similar, by the third term on the rhs of (D8). If we make the except freely falling observers play the role of the perturbed rˆ same assumptions as earlier for B and αvF, we find Eddington-Finkelstein observers [16,19].

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