3. Perturbation of Kerr BH

Transcription

1 3. Petubation of Ke BH hoizon at Δ = 0 ( = ± ) Unfotunately, it i technically fomidable to deal with the metic petubation of Ke BH becaue of coupling between and θ Nevethele, thee exit a fomalim (Newman-Penoe fomalim) to deal with petubation of Ke geomety.

2 Complex null-tetad and Weyl teno gν = lnν nlν + mmν + mmν whee flat pace example: l Ψ 0 = Clmlm ll = nn = mm = 0 d = dt + d + ( dθ + in θdφ ) = (,, 00, ), n = (, 00,, ), Clnlm Ψ 3 = Clnmn Ψ = Cnmnm ln = mm = Weyl teno component (0 d.o.f. 5 complex d.o.f.) m Newman & Penoe ( 6) Ψ = Ψ ( ) = Clnln Clnmm i = 0, 0,, in θ

3 Loentz tanfomation of tetad fame (6 d.o.f. = 3 complex d.o.f.) () l l, m m+ al, n n+ am+ am+ aal () n n, m m+ bn, l l+ bm+ bm+ bbn (3) i l l, n n/, m e α m tanfomation of Weyl component () Ψ Ψ0 Ψ a Ψ Ψ = a a 0 0 Ψ 3 Ψ3 a 3a 3a 0 Ψ3 3 a a 6a a Ψ Ψ () 3 Ψ 0 b b 6b b Ψ0 3 Ψ b 3b 3b 0 Ψ Ψ = b b 0 0 Ψ Ψ b Ψ Ψ Ψ

4 In paticula, unde () n n, m m+ bn, l l+ bm+ bm+ bbn we have Ψ Ψ = Pb ( ) Ψ b + Ψ b + 6 Ψ b + Ψ b +Ψ P ( b) b 3b b Ψ Ψ = = Ψ + Ψ + Ψ + Ψ 0 Petov type D: P Ψ b b b b ( ) ( ) Ke i type D One can make Ψ 0 =0 by () with b=b In thi cae, {( )( ) ( ) ( )} Ψ = P ( b) = Ψ b b b b + b b b b Hence we have Ψ =0 imultaneouly.

5 The eult i Ψ0 0 = Ψ 0 Q Ψ Ψ 6 ( b b) = = Ψ3 Ψ ( b b) Q If we futhe apply tanfomation () Ψ =Ψ Ψ 3 = a P ( b ) + P ( b ) = Ψ( b b ) ( + a ( b b ) ) a a ( ) Ψ = P ( b) + P ( b) + P ( b) =Ψ + a( b b) 6 ( ) Ψ 3 =Ψ =0 fo a = Only Ψ b b 0 (l and n ae called epeated pincipal null diection) When we conide the petubation, Ψ 0 and Ψ ae both gauge-invaiant and Loentz-invaiant. (Ψ and Ψ 3 ae not Loentz-invaiant.)

6 Fo Ke geomety, Kinneley null-tetad l (,, 0, ) l =Δ + a Δ a ( ) (,, 0, ) n = Σ + a Δ a m = i = iain θ, 0,, in θ + dx dv ( iacoθ ) affinely paametized outgoing null geodeic Weyl M Ψ = ( iaco θ ) 3 Ψ 0 =Ψ =Ψ 3 =Ψ = 0

7 Teukolky equation ψ 0 = z Ψ Teukolky equation Ψ = = [ ] = Σ ˆ L ψ π T z = + iacoθ Tˆ = τ T ν ν nd ode diff opeato Thi may be olved by epaation of vaiable ( ) (, ) i t ψ R Z θϕ e ω = ( ω, lm, ) = pin-weighted pheoidal hamonic

8 ( ) ( ) + aω imϕ iωt ψ = dω Rlmω Slm θ e e lm, lm (, ) Z ω θ ϕ aω aω θ ( inθ θ) + Slm ( θ) = λ Slm ( θ) inθ eigenvalue eq. λ = (l )(l ++ ) in the limit aω 0 ( ) λ ( ) + Δ Δ + R = T T π gdtd Z Tˆ = Ω adial Teukolky eq.

9 ouce tem ˆ T = τ T ν ν t = iω, ϕ = im z z τ ν = L D 0 + D0 L z lmν + mlν z z z z ( ) L z L ll D zd zmm z 0 ν 0 0 ν Δ z z τ ν = L D + D L Σ ν + ν z z z z ( nm mn) Δ z + L z L0 zσ nnν + D 0 z D 0 mmν z L D mω mω m θ + a ωinθ + cotθ inθ ( + ) ( ) iω a ima M + + inθ Δ opeation: m m ω ω

10 Teukolky-Staobinky identity Relation between R & - R (Ψ 0 & Ψ ) K Δ ( D ) [ Δ R] =C R D= i, D = + i Δ ( D) [ R] = C R K = ( + a ) ω am Δ = M+ a R ( D) Δ ( D ) [ Δ R] = C K Δ Δ Δ = ( D ) ( D) [ R] C R = ( + ω ω ) ( ) + 36 ω 36 ω C Q a m a Q a m a Im[ C] ( Q )( 96a ω 8aωm) ω ( M a ) + + = Mω Q = λ + ( + ) = Q a 0 limit: Ca ( = 0) = λ( λ+ ) + imω = C 0

11 Chandaekha tanfomation (fo a=0) Fo the Schwazchild cae, we have two independent fomalim: RWZ fomalim & Teukolky fomalim How ae they elated? Radial =- Teukolky eq. 3 [ + ] = 0 = A( ) B( ) Y ; Y R( ) + Chandaekha ( 83) d ± = ± iω d * d ( 3M) A ( ) = ln = d * Δ Δ = ( M) Δ B( ) = ( λ + 6M) 5 λ = ( l )( l+ )

12 Δ Δ Setting Y = ( ) Thu, X [ V( ) ] X = = =Δ+ + = + + one find X atifie RW eq.: Δ R Y X J J X Δ Δ R= J J X + + [ ] ( ) [ ] d J = iω ± d ± Δ Invee tanfomation: J ± = D ( a = 0) D( a = 0) C 0 5 R 3 R X = J J = Δ Δ C0 = λ( λ+ ) imω No uch tanfomation i known fo Ke cae

13 Gavitational adiation fom EMRI G ν [ g ] = 8 π GT BH gν = gν + hν ν M >> v/c can be lage M, J BH v [ ] = 8 ν δ G h π G T ν compact ta ( ~M ) can be appoximated by a point paticle: ( ) ( x z( )) ν dz dz δ τ ν T = dτ d τ d τ g

14 up Geen function method Bounday condition fo homogeneou mode down in out up in R( ) R ( ) d R ( ) T( x ) + W W Wonkian R up R in Teukolky eq. i not eal no ym. unde complex conjugation: in out But it i ymmetic unde c.c.+( ): R R R = R in out At ψ =Ψ ~ h ih ( ) + de = dt πω R 3

15 Mano-Takaugi-Suzuki fomalim up in Ω ( x) gd x' Ω ( x ) T( x ) + W ψ Ω = ( ) ( θ, ϕ ) R Z e ω up up i t R in () up at mall and ( ) R Ω = difficult to obtain by eie expanion. (Mano et al.(996)&(997)) ( ) ( θϕ, ) R Z e ω in in i t at lage ae not To evaluate the Wonkian R in R up ( ) ( ) up in W R R we need to evaluate at lage. (o at mall ) Need to contuct a olution valid in the whole ange of

16 MTS method Expanion nea hoizon (fo implicity, et a = 0) eie in tem of hypegeometic function R in ~ pn+ ν ( x) F( n+ ν iε, n ν iε, iε; x) ν: eigenvalue (to be detemined) x = M ε = Mω ( Newton limit: ε 0, ν l o -( l+ ) ) Teukolky equation lead to 3-tem ecuion elation convegence condition detemine ν

17 Analytic continuation of F ( α, β, γ ; x) to lage x Γ( α ) Γ( β ) Γ ( ) ( α ) Γ( β α ) α F α, β, γ ; x = ( x) F α, γ β, α β + ; Γ( γ ) Γ( γ α ) x Γ( β ) Γ( α β ) β + ( x) F β, γ α, β α + ; Γ( γ β ) x One find with Expanion aound = ζ ν ν ν C n n= iz n z b e z F (,; iz) 0 confluent hypegeometic fcn. (Coulomb wave fcn.) MTS find exactly the ame 3-tem ecuion elation a the one nea hoizon matching at z< & (-x)>> detemine (matching egion: M << < ) z = ε( x) = ω ν ζ C ν ν ζc ζ 0

18 Leading ode wavefom degw deobit Enegy balance agument i ufficient: = dt dt Fo example, wave fom fo quai-cicula obit i detemined by ate of change of GW fequency f: df dt = de dt obit de df obit de obit dt de obit df = 0 + O + ( ) ( O ) ( ) ( ) ( geodeic O O ) = + +

19 Enegy lo ate fo cicula obit on the equatoial plane in Ke backgound to PN ode = v = ( MΩ) / 3 q a/ M Obital feq. a meaued by obeve at infinity