On time dependent black hole solutions

Transcription

1 On time dependent black hole solutions Jianwei Mei HUST w/ Wei Xu, in progress ICTS, 5 Sep. 014

2 Some known examples Vaidya (51 In-falling null dust Roberts (89 Free scalar Lu & Zhang (14 Minimally coupled scalar (Customary built potential, 4D N = 4 gauged supergravity Lu & Vazquez-Poritz (14 Dynamical C-metrics Xu (14 w/ scalar in 3D

3 Our setup Gravity coupled to a scalar field, S = d D x [ g R 1 ] ( Φ U(Φ. The potential U(Φ will be determined a posteriori. We have a good reason to do this. By a scaling of the metric (ˆΦ = ˆΦ(Φ, S = d D x [ g F (ˆΦR 1 ] ( ˆΦ Û(ˆΦ.

4 Solution Ansatz (4D, ds = f 1 (τ, Φdτ + f (τf 3 (ΦdτdΦ + τf 4 (Φd Ω,k, d Ω,k = dx 1 kx + (1 kx dy. The scalar field Φ is used as a coordinate. The functions f 1,, f 4 are arbitrary. General solution, c [ 3 f 1 = c 4 τ(1 + c 1 τ k c 4 coth( Φ c 3 c 1c 4 τ (Φ Φ 0 csch ( Φ ] c 3, 1 f = 1 + c 1 τ, f 3 = c 3 csch ( Φ, f 4 = c 4 csch ( Φ, U = c { } 1 (Φ Φ 0 [ + cosh(φ] 3 sinh(φ, c 3 where c 1, c 3, c 4 and Φ 0 are arbitrary constants. The potential is customary tailored.

5 AdS 4 Limit For AdS 4 limit, let c 1 = 1, c 3 = r0v0, c 4 = r 0 and τ = tanh ( v v 0, Φ = Φ 0 + sinh 1 [ r 0 r A(r, v where A > 0 and A 1 as r. The metric becomes ds = f v dv + f r drdv + fr d Ω, f v = r { A tanh ( v [ r 0 v 0 v 0 f r = + A rv r 0 r A A tanh( v 1 + r 0 v 0 r A Φ 0 sinh 1 ( r 0 r A ] [ 1 + r 0 r A + v 0 v A A tanh( v v 0 ( 1 + r r A A ], ] 1 r },, f R = r A tanh( v v 0, Approaches AdS 4 in the limit r and v/v 0. Differs from Lu & Zhang through A. Singular at v = 0 unless A is tuned accordingly (As in Lu and Zhang.

6 Curvature R = r 0 coth ( v v 0 [ r 4 A 1(r 0 + r A r r 0 v 0 A 1 (r 0 r A ] + 1[r 0 r 0 csch ( v v 0 + r A] v 0 r 0 + r A r v 0 A r0 + r A [ r0 v ( 0 l + r ] 0 sinh 1 r. A

7 CTC and energy conditions No CTC = A > 0 in regions with external contact. The energy condition Tµν ξ µ ξ ν = ξ r (1 + r 0 r A v 0 r r A A(1 + A { r 0 v 0 l + sinh 1 r r 0 0 coth ( v ( r v [ 0 A r A 1 + r 0 r A v 0 r A 6ξ { r 0 r 0 v r 0 r A r A + 6ξ {[ r 0 v 0 r 0 v 0 l + sinh 1 ( v 0 [ξ x (1 x + ξ y r r A 1 x ](1 + A } + ξ r r3 A (1 + r 0 r A r 3 coth ( v 0 v r r A v0 ( 0 (1 + [ A r + ξ A x (1 x + r 0 r A ] (1 + r } 0 3r, A 1ξ r r4 A(1 + r 0 r A ξ y 1 x ξ ] } where T µν = R µν R g µν. The null (ξ = 0 energy condition is always satisfied, T µν ξ µ ξ ν 0. For ξ = 1, the last line contributes negatively to T µν ξ µ ξ ν. (This appears to be common for AdS spacetime.

8 Detailed example Consider A = 1 + A1 r ds = f v dv + f r drdv + f R d Ω. + A(v r + A3(v r 3 +. As r, we find fv = r + A l 1 r tanh ( v r [sech ( v v 0 A 1 v 0 v 0 v 0 l { r 0 sech ( v [( v v 0 v 0 A A 1 4 tanh ( v ] + 1 v [A 1 tanh( v ] v 0 v 0 tanh( v ] ( 4r + 0 A 3 v 0 3v 0 l tanh ( v }. v 0 Will focus on the flat case l 1 = 0 (i.e. Φ 0 = 0. The scalar field still contributes a negative piece to the weak energy condition.

9 Horizon Consider a simplest case A = 1 + a f (v/r, f v = 1 + r ( 1 + a f r 0 v 0 r [ v 0 r 1 + r 0 + a f r ( sinh 1 r 0 tanh ( v r + a f v 0 + a tanh( v v 0 v 0 v f r 1 + r 0 +a f r In the limit a 0, f v (r h = 0 leads to r h = r 1 a r r 1 sinh 1 ( r / 0 r0 ( = 1 + r 0 r 1 r 1 r1 ]. v 0 r 1 coth ( v v 0, r1 r = f tanh( v v 0 v 0 v f r0 v 0 r 0 + r1 + r 1 sech ( v v 0. + O(a 4, with For large v/v 0, there is always a solution for r 1, and hence r. r oscillates when f (v is periodic.

10 Hawking s area theorem Raychaudhuri equation: for any null geodesic u, (u θ u = 1 θ u ˆσ + ŵ R µν u µ u ν. ŵ = 0 for hypersurface orthogonal vectors; So θ u always decrease along the path, assuming weak energy condition. θ u < 0 will lead to θ u = at finite affine distance. (= One can never pass from θ u < 0 to θ u > 0. θ u must be non-negative on and outside an event horizon. (= Area of horizon will never decrease, assuming the energy condition. The naive formula f v = 0 must fail to locate the horizon in our case.

11 Two seemingly contradicting scenarios If the area of horizon changes with f (v, then it contradicts Hawking s area theorem; If not, why does f (v not have any effect on determining the location of horizon?

12 Apparent horizon : Example Schwarzschild ds = fdt + 1 f dr + r d Ω. Advanced and retarded time: du = dt 1 f dr and dv = dt + 1 f dr = ds = fdudv + r d Ω. Kruskal coordinates, ũ = e u/4m and ṽ = e v/4m = ds = 16m f ũṽ dũdṽ + r d Ω. u = ũṽ f ũ, v = ũṽ f ṽ θ u = 4m r ṽ, θ v = 4m r ũ. Apparent horizon is when θ u = θ v = 0. Working with Kruskal coordinates is important in the process. (Necessary?

13 Figure: The extended spacetime of Schwarzschild. Abstracted from POPE s lecture notes.

14 Apparent horizon for our solution The retarded time ( du = f 0 dv f r dr, r f 0 + v ( f 0f r = 0. f v f v A solution near f v = 0 and in the limit v/v 0 and a 0, f 0 = 1 + f 01 ζ v + a (f 0 + f 03 ζ v, f 01 = f 0 = f 03 = c 01n where some detail is in the next page. + c c 011 (r r 10, r r [ 10 c0n ] + c 00 + c 01 (r r 10 v f, r r 10 f 03n (r r 10 + f 03n + f f 031 (r r 10, r r 10 [ u = v + a c0n ] f (v + c 00 + c 01 (r r 10 ( f r 0 r r 10 r f v dr r r 10.

15 Large v/v 0 and small a fv = (r + a f (1 ζv sinh 1 ( r 0 r 0 v 0 r r + r 0 v a ( ζ v (f + v 0 v f v 0 r + r 0, fr = r( ζv r + r 0 [ 1 a f (r + r 0 ], ζv = sech( v v 0, r h = r 10 + r 1 ζv + a (r 0 + r ζv, sinh 1 ( r 0 / ( r 0 r 10 r 1 = r = = r r 0 r 10 v 0 r 10, r 10 r 0 r + r 10 ( 0 + r 10 v 0 f r 10 v 0 v f (r 0 v 0 r 0, r 0 = + r 10 r 10 r 0 v 0 r 0, + r 10 r 0 + r 10 v 0 [ r 10 (r 0 v 0 r 0 f r + r r 10 + r 10 4 v 0 v f ] (r 0 v 0 r 0. + r 10

16 Apparent horizon: leading order In terms of u and v, ds = f v f 0 (u v, vdudv + f R (u v, vd Ω. Using Kruskal coordinates ũ = e u and ṽ = e v, Expansion of null geodesics, For future null infinity, ds = f v (ũṽ, ṽ dũdṽ + fr (ũṽ, ṽd Ω. f 0 ũṽ u = f 0 f v ũṽ ũ, θ u = (u ln f R, v = f 0 f v ũṽ ṽ, θ v = (v ln f R. θ u ṽ( f 0 f r 0 (r 10 a f r 10 (r 10 + a f ṽ r 0 + r10. r 10

17 Discussion θ u ṽ( f 0 (r10 a f r 0 f r r 10 (r10 + a f ṽ 0 + r10 r10. Location of the apparent horizon is determined by where θ u changes its sign, which is largely controlled by the sign of ṽ, 1 which is unlikely to be affected by small corrections from ζ v or a. As such, the future apparent horizon is at ṽ = 0, so v. In the Schwarzschild case u = t r and v = t + r, r = r + m ln(r m. In our case, one can similarly define t = 1 (u + v, t = v + a f (v [ c0n ] + c 00 + c 01 (r r 10 ( f r 0 r r 10 r f v The apparent horizon does NOT change in time. 1 All functions depends on u through u v = ln( ũṽ. dr r r 10.

18 Thanks!