Symmetric Space-times from Thermodynamics 7 1 Jinbo Yang,Hongwei Tan, Tangmei He, Jingyi Zhang∗ 0 Centor for Astrophysics,Guangzhou University,510006,Guangzhou,China 2 n January 12, 2017 a J 1 1 Abstract ] c We apply thermodynamics method to generate exact solution with q - maximumsymmetricsurfaceforEinsteinequationwithoutsolvingit. The r g exact solutions are identified with which people have solved before. The [ horizons structure of solutions are discussed in different situation. These 1 resultssuggestthatplane,spherical,pseudo-sphericalsymmetricspacetime v 6 form a family of solutions, and this relationship can also be extended to 7 Kerr spacetime and Taub-NUT spacetime. It sheds light on exploring 8 more general definition of Misner-Sharp mass. 2 0 Key words: Misner-Sharp mass,Thermodynamics, Maximum symmetric . 1 ,Exact solutions, Kerrfamily , Taub-NUTfamily 0 PACS: 04.70.Dy 7 1 : v 1 Introduction i X r In 1970s, the similarity between four laws of black hole mechanics and four a laws of thermodynamics was discovered[1, 2]. Later, the discovery of Hawking radiation made them more closely linked[3, 4]. And it inspired people to ex- ploredeeperconnectionbetweengravityandthermodynamics. Moreconcretely, people found that Einstein equation could be derived through thermodynam- ics consideration[5, 6]. Further,the idea about entropy force was proposed and got development[7, 8]. Their ideas might partially base on Killing vector field, whichmeant that they consider the static situation. But a unified first law was ∗Correspondingauthor,E-mail: [email protected] 1 discover in un-static spacetime base on the concept of Misner-Sharp mass(MS mass)[9]. And it had application in cosmology [10]. Recently,peoplefoundthattheSchwarzchildsolution,RNsolution,Schwarzchild- (Anti)de Sitter solution and their higher dimension generalization could be de- rived from thermodynamics directly rather than solving Einstein equation [11]. ThederivationisalsobaseontheconceptofMisner-Sharpmass(MSmass). The definitionofMSmassisjustforsphericalsymmetricspacetimeingeneraltheory ofrelativityatthebeginning[12]. Latter,thedefinitionwasgeneralizedtoother gravitytheoryin2codimensionmaximumsymmetricspacetime,justlikeGauss- Bounetgravity[13],4dimensionf(R)theory[14],ndimensionf(R)[15],etc. Then someexactsolutionwerederived. Forexample,topologicalSchwarzchildsolution in 4 dimension GR, include spherical symmetric, plane-symmetric and pseudo- sphericalsymmetric. Andexactsolutionsforhighdimensionf(R)theoryinthe case of f(R)=Rd can also be gain in this way[16]. Actually, the same method can also be applied to generate solutions with U(1) charge and cosmological constant Λ. Some solution which have been derived by solving Einstein equa- tion would be gain simply by this method. Inversely speaking,thermodynamics method is so powerful such that the corresponding spherical symmetric, plane- symmetricandpseudo-sphericalsymmetricspacetimecanformafamilyofexact solutions. It shows the deep connection between gravity and thermodynamics. Thispaperwillbeorganizedinthisway. Section2willintroducehowtoget2 codimension maximum symmetric solution with U(1) charge and cosmological constant Λ through the thermodynamics method. Section 3 will discuss the plane-symmetric case. We identify the solution with plane-symmetric which have been got by solving Einstein equation before. Section 4 will discuss the pseudo-sphericalsymmetriccase. Finally,wewillmakeadiscussioninsection5. 2 Exact Solution from Thermodynamics Consideringstaticspacetimewith2dimensionmaximumsymmetricsurface, no loss general,we give this metric ansatz. dr2 ds2 = f(r)dt2+ +r2dΩ2 (1) 2 − h(r) 2 Generalized MS mass for spacetime with maximum symmetric 2-surface is de- fined as [16]. r M (r)= (k h(r)) (2) ms 2 − In spherical symmetric case, dΩ2 = dθ2+sin2θdϕ2,k = 1. In plane-symmetric 2 case dΩ2 = dx2 + dy2, k = 0. In pseudo-spherical symmetric case dΩ2 = 2 2 dθ2 +sinh2θdϕ2, k = 1. Here we would apply the thermodynamics method − to generatesolutionof Einsteinequation. Makea variationwith respective to r ′ k h(r) rh(r) δM (r)= − − dr ms 2 Consider the first law in an adiabatic Misner-Sharp system with Λ term and q term 1. Λ q2 δM (r)= r2dr+ dr (3) ms 2 2r2 Then,we have equation q2 k h(r) rh′(r)=Λr2+ (4) − − r2 The solution is C q2 Λr2 h(r)=k + (5) − r r2 − 3 C is the integration constant. Substitute h(r) into the formula of MS mass,we get r C q2 Λr2 C q2 Λr3 M (r)= (k (k + ))= + (6) ms 2 − − r r2 − 3 2 − 2r 6 Ontheonehand,h(r)canbesolvedbyabovemethod. Ontheotherhand,the relationshipbetweenKillingsurfacegravityandgeometrysurfacegravityshould be applied here in order to find f(r).Under the metric ansatz,Killing surface gravity is 1 h ′ κ(r)= f (7) 2sf Base on unified first law [9, 11], define geometry surface gravity ,w is the work term.Here is the definition. 1 w= IabT (8) ab −2 Demand Killing surface gravity equal to geometry surface gravity,then M ms κ(r)= 4πrw (9) r2 − 1 Theoriginaltermaboutcharge qdq can’tderiveRNsolution. Itshouldbecorrectedas r q2 . 2r2 3 Work terms with cosmologicalconstant and charge are given here [11] Λ q2 wΛ = 8π , wq = 8πr4 (10) Then, the solution is C q2 Λr2 f(r)=( k + +D)2 (11) − r r2 − 3 r Set D =0,we have C q2 Λr2 dr2 ds2 = (k + )dt2+ +r2dΩ2 (12) − − r r2 − 3 k C + q2 Λr2 2 − r r2 − 3 Generally, the vector-potential of electric-magnetic field is A = qdt. Putting r it into Einstein equation and Maxwell equation for checking, it is indeed elec- tromagnetic vacuum solution with cosmological constant. If k = 1,it’s just RN- (A)dS black hole. And we can see that not only spherical symmetric solutions, but also plane-symmetric solution and pseudo-spherical symmetric solutions with charge and cosmological constant can be derived in this simple way. 2 It’s worth noting that the sign of functions f(r) and h(r) can be nega- tive. In this method, there are hidden assumptions f(r) > 0,h(r) > 0. How- ever,conditions f(r) < 0 and h(r) < 0 are also satisfied Lorentz signature. It can be checked that it’s still the solution of Einstein’s theory. Further more, generally we have f(r)=h(r), so the requirement of Lorentz signature is satis- fied automatically. Thus, in later parts of our paper, we will extend our result to situations about f(r)<0, h(r)<0. 3 Plane-symmetric Solutions In this section, we would discuss k=0 solutions in detail. These are Plane- symmetric cases. They are correspondingto solutions which are already gotby solving Einstein equation,just like Taub spacetime, AdS4 black hole, etc. Let’s consider the solution without q or Λ first. The metric is C dr2 ds2 = ( )dt2+ +r2(dx2+dy2) (13) − −r C −r 2 Set C = q = Λ = 0 except k = 0 case, these solutions would go back to Minkowski spacetime. 4 C cannot be 0, then it is Taub solution[17]. If C < 0, redefine the parameter C = 2µ. Here µ>0, it’s a constant with mass dimension.Then − 2µ dr2 ds2 = dt2+ +r2(dx2+dy2) (14) − r 2µ r Introduce coordinate transformation z = r2. Thus, the metric in new coordi- 4µ nate frame is µ ds2 = ( dt2+dz2)+4µz(dx2+dy2) (15) z − r −1 In fact, it’s static Taub spacetime. Make scale transformation t µ 3t,z → → µ−13z x 1µ−13x, y 1µ−13y , a more familiar line-element is found as → 2 → 2 1 ds2 = ( dt2+dz2)+z(dx2+dy2) (16) z − r IfC >0,asamateroffact,itisspatialhomogenousbutun-staticTaubsolution. Similarly, the coordinate frame which make metric have following form can be found. 1 ds2 = (dt2 dz2)+z(dx2+dy2) (17) z − r Then, two situation of Taub spacetime are corresponding to C >0 and C <0. Plane-symmetric solution without cosmological constant but with charge is [18, 19] C q2 dr2 ds2 = ( + )dt2+ +r2(dx2+dy2) (18) − −r r2 C + q2 −r r2 Define dz = dr ,so −C+q2 r r2 q2 q2 q2 2Cz =(r+ )2+2( )2log r (19) − C C | − C| It’s the solution found by Letelier and Tabensky in 1974 [20]. Their work is based on the foundation of Patnaik [21]. With the same method, we can define C = 2µ if C < 0. When r > − 0,2µ+q2 >0,so there is no horizoninthis case. IfC >0,refindC =2M,then r r2 in the range of 0 < r < q2 , there is 2M + q2 > 0. So r = q2 is the Killing 2M − r r2 2M horizon. If charge q =0 but Λ=0, there are more complex structure. 6 C Λr2 dr2 ds2 = ( )dt2+ +r2(dx2+dy2) (20) − −r − 3 C Λr2 −r − 3 5 If C >0,Λ>0, it’s a spatial homogenous spcaetime. If C <0,Λ>0 ,demand C = rh3,Λ = 1 , r >0 then −L2 3 L2 h r3 r2 h(r)= h (21) rL2 − L2 Obviously, r = r is Killing horizon. In the range of 0 < r < r ,this metric h h describe a static part of the spacetime. Region behind the horizon is spatial homogenous. Similarly,if C < 0,Λ < 0, it is a static spacetime without Killing horizon.If C >0,Λ<0,there is a horizon. Set C = rh3,Λ = 1 ,r >0,thus L2 3 −L2 h r2 r3 h (22) L2 − rL2 Obviously,r=r is Killing horizon. In the range of r >r , it’s a static region. h h Actually, this is so called AdS4 Schwarzchild black hole with planar horizon which is applied in holographic superconductor [22]. The most complicated case is that all parameters are not equal to 0. It is also applied in the research of holographic superconductor[23] and also be the general case of plane-symmetric solution given by Ref.[19]. C q2 Λr2 dr2 ds2 = ( + )dt2+ +r2(dx2+dy2) (23) − −r r2 − 3 C + q2 Λr2 −r r2 − 3 If Λ < 0,define Λ = 1 . Then,make a scale transformation r rL and −3 L2 → redefine all parameter, C CL, q qL. The equation f(r)=0 leads to → → Cr+q2+r4 − =0 (24) r2 It’s impossibleto have4positive rootdue to the lackofr3 term. Since justtwo parameterC andq, assumer =a andr =b arerealrootsofthe equation, then C and q can be expressed by them. Write down the formula r4 Cr+q2 =(r a)(r b)(r2+(a+b)r+a2+ab+b2) (25) − − − Compare the term with same order,then C =a3+a2b+ab2+b3 q2 =a3b+a2b2+ab3 (26) 6 Further more, assume r = a and r = b are positive. Under this condition, whetherotherrealrootsexistdependonequationr2+(a+b)r+a2+ab+b2 =0. Obviously,(a+b)2 4(a2+ab+b2)= 3a2 2ab 3b2<0. Soit’sevenimpossible − − − − tohaverealroot. WecanclaimthattherearenomorethentwoKillinghorizons inthis case. No lossgeneral,assumea>b,then the regionr>a and0<r <b are static, while the region b<r <a is spatial homogenous un-static. Through similar method, we found that there is no more then one Killing horizon when Λ >0. Suppose the horizon is located at r =r , In the range of h r > r , this region is un-static but spatial homogenous, while r < r is static h h region. 4 Pseudo-spherical Symmetric Spacetime If the 2-dimension maximum symmetric surface is pseudo-sphere,then we have this solution C q2 Λr2 dr2 ds2 = ( 1 + )dt2+ +r2(dθ2+sinh2θdϕ2) − − − r r2 − 3 1 C + q2 Λr2 − − r r2 − 3 (27) The spacetime described by this metric is the general situation with k = 1. − And the case of q = Λ = 0 is called topological Schwarzchild spacetime in Ref.[16], which has already been got in Ref.[24]. Pseudo-spherical symmetric solutionwith negative cosmologicalconstantis presentedin literatures [25, 26]. The simplest case is C dr2 ds2 = ( 1 )dt2+ +r2(dθ2+sinh2θdϕ2) (28) − − − r 1 C − − r If C < 0, there is one horizon, region inside the horizon is static. Instead, outsidethehorizonisun-static. Ithasbeengivenin[24,27].In[27],thek = 1 − tranvalable wormhole was found. Solution with charge but without cosmologicalconstant is C q2 dr2 ds2 = ( 1 + )dt2+ +r2(dθ2+sinh2θdϕ2) (29) − − − r r2 1 C + q2 − − r r2 Then we just discuss its horizons structure, roots of equation 1 C + q2 =0 − − r r2 are C C C C r+ = + ( )2+q2 , r− = ( )2+q2 (30) −2 2 −2 − 2 r r 7 No matter the sign of C, it’s just one positive root r+. Thus, the region 0 < r <r+ is static, while r >r+ is un-static but spatial homogenous. General solution with charge and cosmologicalconstantmay be a new solu- tion. C q2 Λr2 dr2 ds2 = ( 1 + )dt2+ +r2(dθ2+sinh2θdϕ2) − − − r r2 − 3 1 C + q2 Λr2 − − r r2 − 3 (31) When Λ < 0, things are quite different. Similar to what we do for plane- symmetric solution,assume r =a and r =b are real roots, thus r4 r2 Cr+q2 =(r a)(r b)(r2+(a+b)r+a2+ab+b2 1) (32) − − − − − And C =a3+a2b+ab2+b3 (a+b) − q2 =a3b+a2b2+ab3 ab (33) − a and b cannot both be positive unless condition a2+b2+ab > 1 is satisfied. However,this conditionalso leadto that equationr2+(a+b)r+a2+ab+b2 − 1 = 0 has no positive roots. Then we can claim that there are no more than two horizons in this case. The results is similar to plane-symmetric solutions. Assumea>b,thenregionr >aand0<r <barestatic,whileregionb<r<a isspatialhomogenousun-static. Butoneinterestingthingisthatwhenq =0,it maystillhavetwopositiveroots. Stillassumetheyareaandb,thena2+b2+ab should equal to 1. If Λ > 0, we can define Λ = 1 , and make the same scale transformation. 3 L2 Now, there are r4+r2+Cr q2 =(r a)(r b)(r2+(a+b)r+a2+ab+b2 k) − − − − C = a3 a2b ab2 b3 a b − − − − − − q2 = a3b a2b2 ab3 ab (34) − − − − There is no more then one positive root due to the positivity of q2. The results is also similar to plane-symmetric solutions. It is un-static outside, but static inside. 8 5 Discussion In this paper,we apply MS mass and unified first law to derive a family of 2 codimension maximum symmetric solution in electromagnetic vacuum with cosmological constant,include famous solution like Schwarzchild-(anti) de Sit- ter solution, RN-(anti) de Sitter solution, Taub solution, AdS4 black hole,etc. Thereare3subclassesaccordingtok. Concretely,k =1correspondtospherical symmetry, k =0 correspondto plane-symmetry, k = 1 correspondto pseudo- − sphericalsymmetry. In everyclass, C,q andΛ determine the horizonstructure. It’s worth noting that all exact solutions for k = 1, k = 0 and k = 1 form a − family whith pattern like this: Change sinθ to sinhθ while k = 1 to k = 1, − pseudo-sphericalsymmetricsolutioncanbegeneratedfromsphericalsymmetric solution. For plane symmetric solution, change sinθ to θ while k =1 to k =0, and interpretθ as radius coordinate onthe plane. Take x=θcosϕ,y =θsinϕ, we have dθ2+θ2dϕ2 = dx2+dy2. We are inspired by this pattern. Although the thermodynamics method cannot derive Kerr solution, however, it’s well known that Kerr spacetime can be regarded as the solution deformed from the simplest spherical symmetric solution called Schwarchild spacetime. We have learned that there are plane symmetric solution called Taub spacetime and pseudo-spherical symmetric solution called topological Schwarzchild spacetime oranti-Schwarzchildspacetime. Inprinciple,theycanalsogeneratecorrespond- ing spinning solution like Kerr spacetime. We do find them out here. Firstly, Kerr spacetime is given here 2Mr ρ2 ds2 = (1 )dt2+ dr2+ρ2dθ2 − − ρ2 r2+a2 2Mr − (35) 2Mr 2Mr +(r2+a2+ a2sin2θ)sin2θdϕ2 2 asin2θdtdϕ ρ2 − ρ2 Inthisformula,ρ2 =r2+a2cos2θ. Then,inordertoextendtoplanesymmetric case,changesomedetailsbyfollowingrules. Change1infrontofdt2 andr2+a2 under dr2 to 0; change cosθ to 1 in ρ2 and replace every sinθ by θ, we have 2Mr ρ2 ds2 = dt2 dr2+ρ2dθ2 ρ2 − 2Mr (36) 2Mr 2Mr +(r2+a2+ a2θ2)θ2dϕ2 2 aθ2dtdϕ ρ2 − ρ2 Calculating the Ricci tensor and RλσµνR , the result shows that R = 0 λσµν µν and RλσµνR ρ−12. It shows that this is indeed the exact solution of vac- λσµν ∝ 9 uum Einstein equation. This modify-by-hand method seems questionable, but one can find the generalized Newman complex transformation from Exp.(14). First,usepolarcoordinateandreplaceµby M,thenintroducetheEddington − coordinate du=dt dr/(−2M), to rewrite the line element (14). − r 2M ds2 = du2 2dudr+r2dθ2+r2θ2dφ2 (37) r − Select tetrad as follow 1 ∂ ka = ( )a √2 ∂r 1 ∂ f ∂ la = (( )a ( )a) √2 ∂u − 2 ∂r (38) 1 ∂ i ∂ ma = (( )a ( )a) √2r ∂θ − θ ∂ϕ 1 ∂ i ∂ m¯a = (( )a+ ( )a) √2r ∂θ θ ∂ϕ Then, the generalized Newman complex transformationfor plane symmetric is θ2 ′ ′ u =u ia , r =r ia (39) − 2 − ′ ′ So, du =du iaθdθ and dr =dr. Substitute them into the line element, − and introduce du′ =dt 2Mr dr, then one can get Exp.(36) − r2+a2 And there is no singularity for this spacetime since RλσµνR is finite λσµν everywhere due to ρ2 a2. For pseudo symmetrical solution, we apply similar ≥ rules and get 2Mr ρ2 ds2 = ( 1 )dt2+ dr2+ρ2dθ2 − − − ρ2 (r2+a2) 2Mr − − (40) 2Mr 2Mr +(r2+a2+ a2sh2θ)sh2θdϕ2 2 ash2θdtdϕ ρ2 − ρ2 Inthisformula,ρ2 =r2+a2ch2θ. Similarly,thereisR =0andRλσµνR µν λσµν ∝ ρ−12. Itisalsosingularity-freesinceρ2 a2. ThisissocalledS-Kerrsolution[28, ≥ 29] which is one kind of S-brane[30]. The the generalized Newman complex transformation to derive it is as following. Let’s start with pseudo spheri- cal symmetric Schilwarzchild solution under the Eddington coordinate du = dt dr/( 1 2M), − − − r 2M ds2 = ( 1 )du2 2dudr+r2dθ2+r2sinhθ2dφ2 (41) − − − r − 10