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Evaporating dynamical horizon with Hawking effect in Vaidya spacetime ∗ Shintaro Sawayama Department of Physics, Tokyo Institute of Technology, Oh-Okayama 1-12-1, Megro-ku, Tokyo 152-8550, Japan (Dated: February 7, 2008) We consider how the mass of the black hole decreases by the Hawking radiation in the Vaidya spacetime, using the concept of dynamical horizon equation, proposed by Ashtekar and Krishnan. Using the formula for the change of the dynamical horizon, we derive an equation for the mass incorporatingtheHawkingradiation. ItisshownthatfinalstateistheMinkowskispacetimeinour particular model. PACSnumbers: 04.25.Dm,04.70.Bw 6 0 I. INTRODUCTION centlyAshtekarandKrishnanderivedanequationwhich 0 describeshow the horizonchangesintime. Itneeds only 2 information of the horizon surface. In the study of black hole evaporation, there has n In section II, dynamical horizon is reviewed. In section been an important issue how black hole mass decreases a III, the location of the dynamical horizon in the Vaidya J as a back reaction of the Hawking radiation[1]. We have 9 to confront with this issue to resolve the information spacetime is identified. And then in section IV, the dynamical horizon equation is written down in the case of 1 loss paradox. There have been many works concerning Vaidya matter with the Hawking effect being taken into black hole evaporation, either in string theories[2][3][4], 2 or semiclassical theory typically using apparent horizon account. Section V is devoted to conclusion and discus- v sions. [5]. Hiscok studied spherical model of the black hole 8 evaporation using the Vaidya metric, which we also use 4 inpresentwork,tosolvetheblackholeevaporationprob- 0 II. DYNAMICAL HORIZON 9 lem. However,he simply set a model not taking account 0 of the field equation. Hajicek’s work[7] treatedthe black 5 hole mass more generally than our present case. How- Ashtekar and Krishnan considered dynamical hori- 0 ever, he did not use the field equation either. One of zon [13][14], and derived a new equation that dictates / how the dynamical horizon radius changes. Apparent c the morerecentstudies is SorkinandPiran’swork[8]on q charged black holes. And neutral case has been done by horizon is a time slice of the dynamical horizon. The - Hamade and Stewart[9]. Their conclusion is that black definition of dynamical horizon is, r g holemassdecreasesorincreasesdependingoninitialcon- : dition. Theyusedamodelofthedoublenullcoordinates, Definition. A smooth, three-dimensional, spacelike v submanifold H in a space-time is said to be a dynamical i and obtained a numerical result. But they did not con- X horizon if it is foliated by preferred family of 2-spheres sider the Hawking effect directly but they used massless such that, on each leaf S, the expansion Θ of a null r scalarfieldasamatter. BrevikandHalnescalculatedpri- (l) a normal la vanishes and the expansion Θ of the other mordial black hole evaporation[10]. Very recently Hay- (n) null normal na is strictly negative. wardstudiedblackholeevaporationandformationusing the Vaidya metric [11]. It seems no analytical equation The requirement that one of the null expansions is hasbeenproposedfortheblackholemasswiththeHawk- zero comes from the intuition that black hole does not ing effect taken into account. emit even light. And the requirement that other null The dynamics for the black hole mass with the Hawking expansion is strictly negative comes from that null effect is a long standing problem. Page[12] derived the equation of mass intuitively, that is M˙ ∝ −M−2. But matter goes in black holes inwards. In this definition, we can recapitulate the important it does not come from the first principle. We will com- formula which gives a change of the dynamical horizon mentonhisintuitiveresultinthefinalsection. Toderive radius by the matter flow, using 3+1 and then 2+1 the equation of mass from the first principle we should decompositions and also the Gauss-Bonet theorem. treat the Einstein equation with the back reaction term by the Hawking radiation. However, the Einstein equa- R R ( 2 − 1)= T τâξb d3V tion cannot be analytically solved, because the equation 2G 2G Z ab (R) ∆H contains fourth derivative terms as back reaction. Re- 1 + (|σ|2+2|ζ|2)d3V. (1) 16πGZ ∆H Here R ,R the radius of the dynamical horizon , T 2 1 ab ∗Electronicaddress: [email protected] is the stress-energy tensor, |σ|2 = σabσab,|ζ|2 = ζaζa 2 where σ is the shear, and ζa := qãbrˆc∇ l , with two ab c b dimensional metric q˜ , and ξa := N la (N := |∂R|) ab (R) R R where R is the radius of the dynamical horizon. This is the dynamical horizon equation that tells how râ la the horizon radius changes by the matter flow, shear and expansion. In the spherically symmetric case that na we shall consider in what follows the second term of the right hand side vanishes. Although in the case −tâ of quantum field theory in curved space time, the FIG. 1: For the case that the dynamical horizon decreases, dominant energy condition does not hold[16][17], we we should choose la = −tâ +râ so that la points into the can use the dynamical horizon equation because it is dynamical horizon. valid even when the black hole radius decreases. And the dynamical horizon equation is a consequence of the Einstein equation. We use the dynamical horizon with dynamics equation in place of the Einstein equation. r Because of the negative energy, we can expand the ∗ r =r+2M(v)ln −1 . (6) definition of the dynamical horizon in the case of time- (cid:18)2M(v) (cid:19) like. Thenthedefinitionofthedynamicalhorizonisnow, For later convenience, we write, Definition (modified version). A smooth, three- dr dimensional, spacelike or timelike submanifold H in a= dr∗. (7) a space-time is said to be a dynamical horizon if it is foliated by preferred family of 2-spheres such that, There are two null vectors, on each leaf S, the expansion Θ of a null normal la (l) lt −a−1 vanishesandtheexpansionΘ ofthe othernullnormal na is strictly negative. (n) la =lr∗= a−1 , (8) lθ 0 We can easily calculate the timelike dynamical horizon lφ  0  equation only replacing râ and τâ. And the way of 3+1 corresponding to the null vector va, and the other is and 2+1 decomposition is replaced. The timelike dynamical horizon was defined by Ashtekar and Galloway nt −a−1 and called timelike membrane [15]. na =nr∗=−F−F2Gaa−1. (9) nθ 0     nφ  0      III. VAIDYA SPACETIME Here we multiply a−1 so that la =va. This choice of the null vector la is explained in figure 1. From now on we The Vaidya metric is of the form put, ds2 =−Fdv2+2Gdvdr+r2dΩ2, (2) 2M(v) F = 1− (10) (cid:18) r (cid:19) where F and G are functions of v and r, and va is null vectorandristhearearadius,andM isthemassdefined G = 1, (11) by M = r(1− F ), a function of v and r. This metric is 2 G2 in a similar form to the Schwarzschild metric, assuming spherically symmetric. By substituting the Vaidya met- that M(v) is a function of v only. For a constant M, ric (2) into the Einstein equationso that we can identify the metric coincides with the Schwarzschild metric. We the energy-momentum tensor T as ab calculate the expansions Θ and Θ of the two null (l) (n) vectors la,na, because the definition of the dynamical 2 8πTvv := r2(FM,r+GM,v) (3) horizon requires one of the null expansions to be zero and the other to be minus. The result is, 2G 8πT := − M (4) vr r2 ,r 1 Θ = (2F −a) (12) 2G,r (l) r 8πT := . (5) rr rG 1 −2F2+aF −2a2 Θ = . (13) (n) ar(cid:18) −F +2a (cid:19) We do not need to check that the solution of the dynamical horizonequationsatisfies the Einstein equation. From Θ =0 we get, (l) Because we would like to consider the Schwarzschildlike ∗ ∗ metric, we set v =t+r , where r is tortoise coordinate 2F −a=0. (14) 3 wecancheckthattheothernullexpansionΘ isstrictly (19)(20) with la =va, we see (n) negative. Thereforeinthiscase,wecanapplythedynam- ical horizon equation. In the usual Schwarzschildmetric Ttl = −Tvv+Tvr∗ withdynamics,bothexpansionsbecomezero. Thisisthe 1 = (−FM −M −aM ) one of the reasonswhy we choosethe Vaidya metric. By 4πr2 ,r ,v ,r inserting equation (7) to equation (14), we obtain 1 5 = − M . (22) 4πr22 ,v r a = F(cid:18)1−2M,vln(cid:18)2M −1(cid:19) With tâ being the unit vector in the direction of ta, we obtain r + M . (15) M(r/2M −1) ,v(cid:19) T =− 1 5M F−1. (23) tˆl 4πr22 ,v Notethata isproportionaltoF. NowwesolveΘ =0, (l) For the dynamical horizon integration (1), we get to determine the dynamical horizon radius as r2 5 M2 2F −a = 2F 4πr2 T dr = (1+e−v/2M)dM, (24) r Zr1 D tˆl D 2ZM1 − F 1−2M ln −1 ,v (cid:18) (cid:18)2M (cid:19) where we have used r + M(r/2M −1)M,v(cid:19) e−v/2M F = , (25) 1+e−v/2M = 0. (16) and the fact From this equation we obtain, −1 dM v r =−e−v/2M 2(1+e−v/2M)+ e−v/2M ,(26) 1+ −2M ln D −1 dv (cid:18) M (cid:19) ,v (cid:18) (cid:18)2M (cid:19) +M(r /r2DM −1)M,v(cid:19)=0. (17) cahbaonvgeincaglctuhleatiinotne,grwaetiotrneavtarMia,bvleanfrdomF−r1DwtiothMr.DIfinxtehde, D because these functions are used only in the integration. Inserting equation (24) to the dynamical horizon equa- HereF =0isalsothe solutionofthedynamicalhorizon. tion (1), we obtain Thedynamicalhorizonradiusr isgivenbysolving(17) D as 1 M2 (2M +2Me−v/2M) r =2M +2Me−v/2M. (18) 2 (cid:12) D (cid:12)M1 (cid:12) Note that this dynamical horizon radius is outside the = M2 5(1+e−v/2M)d(cid:12)M. (27) Z 2 r =2M, that is othor solution. M1 Taking the limit M →M =M, we obtain 2 1 IV. DYNAMICAL HORIZON EQUATION WITH − 3(1+e−v/2M)+ v e−v/2M =0. (28) HAWKING RADIATION 2 2M This equation is the dynamical horizon equation in the At first, we should derive the energy-momentum case that only the Vaidya matter is present. There is no tensor T for the integration of the dynamical horizon tˆl solution of this equation, except the trivial one (F = 0 equation. ForthisendwederiveitfromthegivenVaidya or r =2M), so matter. For G = 1, F = 1− 2M(v), the non-vanishing r components of the energy-momentum tensor becomes r =2M(v). (29) D 1 Here M(v) is the arbitrary function only of the v, which T = (FM +M ) (19) vv 4πr2 ,r ,v represent the Vaidya black hole spacetime. 1 Next, we take into account the Hawking radiation. To Tlr∗ = −4πr2M,ra (20) solve this problem, we use two ideas that is to use the Tr∗r∗ = 0. (21) dynamical horizon equation, and to use the Vaidya metric. The reason to use the dynamical horizon equation Here we have made the coordinate transformation from comes from the fact that we need only information of ∗ r to r . Writing Ttl in terms of Tvv and Tvr∗ given by matter near horizon, without solving the full Einstein 4 equation with back reaction being the fourth order dif- namical horizon equation (1), we obtain ferential equations, for a massless scalar field. For the matter on the dynamical horizon, we use the result of 1(2M +2Me−v/2M) M2 Candelas [18], which assumes that spacetime is almost 2 (cid:12) (cid:12)M1 static and is valid near the horizon, r ∼2M. =b M2 22(1+e−v/2M)4e−v(cid:12)(cid:12)/M Z M2 T = −T M1 tl tt 1 ∞ dωω3 × 2(1+e−v/2M)+ v e−v/2M dM = (cid:18) M (cid:19) 2π2(1−2M/r)Z e8πMω −1 0 = 1 , (30) + M2 5(1+e−v/2M)dM. (36) 2cM4π2(1−2M/r) ZM1 2 Taking the limit M →M =M, we finally get 2 1 where we have used a well known result, 3 v − (1+e−v/2M)+ e−v/2M ∞ dωω3 π4 2 2M Z eaω−1 = 15a4, (31) 22(1+e−v/2M)4 0 =b M2 and where c = 61440. This matter energy is negative ×ev/M 2(1+e−v/2M)+ v e−v/2M , (37) near the event horizon. In the dynamical horizon equa- (cid:18) M (cid:19) tion, if black hole absorbs negative energy, black hole or radius decreases. This is one of the motivations to use thenegativeenergytensor. Nextwereplacelengthoftto 8b(1+e−v/2M)4ev/M unit length,because inthe dynamicalhorizonequationtˆ M2 = −3(1+e−v/2M)+ v e−v/2M is used, so 2 2M v × (1+e−v/2M)+ e−v/2M . (38) tˆ0 =F−1/2, l0 =F−1/2, (32) (cid:18) 2M (cid:19) Thisisthemainresultofthepresentworkthatdescribes and therefore, the energy tensor becomes how the mass of black hole decreases. This equation is the transcendental equation, so usually it cannot be solved analytically. However, with the right hand side 1 Ttˆl = 2M4cπ2(1−2M/r)2. (33) depending only on −v/2M, we can easily treat Eq.(38) numerically. Figure 2 is a graph of M as a function of v If the dynamical horizon were inside the event horizon, Calculating the integration on the right hand side of (1) the dynamical horizon radius would be for this matter, r =2M −2Me−v/2M. D r2 b D dr In this case, the dynamical horizon equation would be- Z M4(1−2M/r )2 D D come 4M2(1+e−v/2M)4dr =b DdM −8b(1−e−v/2M)4ev/M Z M4e−v/M dM M2 = 7(1−e−v/2M)− v e−v/2M R2 4(1+e−v/2M)4 2 2M =b e−v/M v Z M2 × (1−e−v/2M)− e−v/2M . R1 (cid:18) 2M (cid:19) v × 2(1+e−v/2M)+ e−v/2M dM. (34) (cid:18) M (cid:19) The singular behavior of this expression excludes its physical relevance. Now we show an approximation of the Eq.(38) in par- Here we insert the expression for r (18)in the first line, and the expression for dr /dM =D2(1+e−v/2M)+ ticular limiting case. Taking the limit M → 0, and D v e−v/2M is used. Here b is a constant calculated in [18] −v/2M =const, we can see that (38) becomes, M bC bvC 1 2 + =0. (39) 1 M2 M3 b= . (35) 30720π Where C ,C are positive constants. or 1 2 If we also take account of the contributionof the Vaidya C v 2 M =− . (40) matter, and inserting this into the integrationto the dy- C 1 5 M Fromtheanalysisofthetranscendentalequation(38),we haveshownthat the blackhole mass eventually vanishes 80 andthe spacetime becomes the Minkowskispacetime in- 60 dependent of the initial black hole mass size. 40 The dynamical horizon method in this paper can take into account of the back reaction of the Hawking radi- 20 ation without solving the field equation which contains v -1500-1250-1000 -750 -500 -250 the fourth order differentials. In the limit of the black hole mass going to zero, the FIG. 2: Numerical calculation oftheblack hole mass M as a derivative of the mass becomes small in proportion to function of v from theequation (60) ∗ the null coordinate (v = t + r ). On the other hand as the black hole mass becomes large, the derivative be- haves the minus of the inverse of the logarithm of the So, in the vanishing process the mass is proportional to mass. Our result, which is different from Page’s result, v. For M →large comesfromthefactthatinthelargemasslimit,theblack C holeradiusbehaveslikequadraticoftheblackholemass. M˙ =− 3 , (41) Thisprobablycomesfromwhenlargemasslimitthatthe logM approximationr →2M is broken. where C is a positive constant. It comes from the limit We would like to compare the present work to the pre- 3 M → ∞ and −v/2M → ∞. In this limit the equation ceding works. Sorkin and Piran or Hamade and Stewart (38) becomes v = −2MlogM. This is different from used a massless scalar field instead of the Hawking ra- Page’s result [12]. Because if M goes to large, the dy- diation as the back reaction directly. The conclusion of namical horizon radius increases as M2, so absorbed en- their paper is that black hole starts with the small mass ergy also become large. From this reason derivative of and it evaporates or increases. However, it is shown in M by v changes. If we do not consider next order, the the present work that even if the black hole starts with derivativeofM becomesM˙ =−C ,sothat4πr2 T4 ≈1, a large mass it always vanishes. 4 D contradicting with Page’s intuition. Although we have treated the black hole evaporation semi-classically, we hope this work will give an intuition to quantization of black holes. V. CONCLUSION AND DISCUSSIONS Acknowledgments We have derived an equation which describes how the black hole mass changes taking into account of the Hawkingradiation,inthespecialVaidyaspacetimewhich We would like to acknowledge A.Hosoya, T.Mishima, becomes the Schwarzschild spacetime in the static case. T.Okamura,andM.Shiinoforcommentsanddiscussions. [1] S.W.Hawking Commun. 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