Special and General Relativity Lecture Notes: Day 17 (11/04/08) Shawn Mitryk Contents 1 Horizons 1.1 Definitions . . . . . . . 1.2 Null Solutions . . . . . 1.3 Schwarzschild Metric . 1.4 Isotropic Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 2 2 2 4 2 Comparison to Electrodynamics 5 3 Next Class 3.1 Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 6 1 1 Horizons 1.1 Definitions • Event Horizon - global • Apparent Horizon - local (convenient for Numerical Relativity) • Killing Horizon - Not a ”horizon,” rather a geodesically complete null surface 1.2 Null Solutions • Minkowski: X 2 − T 2 = 0 • Schwarzschild: R2 − T 2 = 0 1.3 Schwarzschild Metric ds2 = −[1 − 1 2M 2 ]dt + dr2 + r2 dΩ2 r [1 − 2M ] r (1) r 2M we obtain: Making the transformation T 2 − R2 = [ 2M r − 1]e ds2 = 32M 3 −r e 2M (−dT 2 + dR2 ) + r2 dΩ2 r (2) In Mikoswski coordinates there is a killing vector of the form: K = x∂T +T ∂x This is much like the angular momentum: Pz = x∂y − y∂x Taking the vector K µ = [X, T, 0, 0]: We can show: Kµ K µ = −X 2 + T 2 =⇒ 0 for X = ±T K µ orbits cover the killing horizon X 2 − T 2 = 0. −→ The Killing vector is null on the killing horizon. but K µ ∇µ K ν = −κK ν 6= 0 Taking X µ = (T, X, 0, 0) and U µ = (1, 1, 0, 0) ∝ K µ Then: κ2 = κ2 = κ2 = κ2 = 1 − (∇µ Kν )(∇µ K ν ) 2 1 − (g µρ gνσ ∇µ K ν )(∇ρ K σ ) 2 1 − (−1 − 1) = 1 2 ±1 In schwartzchild: We consider the killing vector K = R∇T + T ∇R 2 (3) (4) (5) (6) Noting: K µ Kµ 32M 3 −r e 2M (−R2 + T 2 ) = 0@R = ±T r = (7) Taking the transformation: V = U = ∂ ∂T = 1 (V − U ) 2 1 T − R =⇒ T = (V + U ) 2 ∂U ∂ ∂V ∂ + ∂T ∂U ∂T ∂V T + R =⇒ R = (8) (9) (10) Thus we can define the killing vector in the form: K K K = R∂T + T ∂R ∂ ∂ 1 ∂ ∂ 1 (V − U )( + ) + (V + U )(− + ) = 2 ∂U ∂V 2 ∂U ∂V ∂ ∂ = −U +V ∂U ∂V (11) (12) (13) Then recalling: V U v = (14) e 4M −e = −u 4M (15) Thus K becomes: K = −U ( ∂t ∂ ∂r ∂ ∂t ∂ ∂r ∂ + )+V( + ) ∂u ∂t ∂u ∂r ∂v ∂t ∂v ∂r (16) We then want to find a function f (r) such that f 0 dr = −U dV − V dU dr −V dr Thus we find: dv = −U f 0 and du = f 0 And finally: = −U V (17) f (r) = v−u (18) f (r) = f (r) Taking the ratio: Differentiating: V U e 4M r r [ − 1]e− 2M 2M (19) t = −e 2M dV V − 2 dU U U 3 t = e 2M dt 2M (20) From which we can read: dt dV dt dU 2M −2M = t V U e 2M −2M 2M V = t 2 U U e 2M = = (21) (22) Thus we can calculate the killing vector K: = −U [ K 2M ∂ ∂ −2M ∂ ]+V[ ] = 4M U ∂t V ∂t ∂t K µ = (4M, 0, 0, 0) (23) (24) From which we calculate: κ2 = κ2 = −1 ∇µ K ν ∇µ Kν 2 1 − (g µρ gνσ ∇µ K ν )(∇ρ K σ ) 2 (25) (26) ∇0 K 0 = Γ000 K 0 (27) ∇0 K 1 = Γ100 K 0 (28) ∇1 K 0 = (29) ∇1 K 1 = Γ001 K 0 Γ101 K 0 (30) Finally: κ2 = κ2 = κ = 1 1 2 ) − [g 00 g00 (∇0 K 0 )2 ] + g 00 g11 (∇0 K 1 )2 + g 11 g00 (∇1 K 0 )2 + g 11 g11 ∇1 K(31) 2 2 ( M2 (1 − 2M 1 2M 2 M 4M r )) − [−(1 − ) ( 2 )2 + − r ] (32) 2M 2M 2 r r (1 − r ) (1 − r )2 16M 4 −→r=2M 1 r4 (33) Defining a ”more useful” form: k = κ2 = κ = 1.4 ∂ ∂t M2 r4 M 1 −→r=2M 2 r 4M (34) (35) (36) Isotropic Coordinates The Schwarzschild metric reads: ds2 = − (1 − (1 + M 2 2R ) dt2 M 2 ) 2R + (1 − 4 M 4 ) (dR2 + R2 dΩ) 2R (37) Figure 1: Kruskal Mapping Conformal Mapping: R → R1 maps the left side of the plot to the right side. For small M R 1: ds2 2 = −(1 − 2M 2 2M )dt + (1 + )(dR2 + R2 dΩ2 ) R R (38) Comparison to Electrodynamics Consider a charge density ρ: Z Q = ρdV Z 1 Q = EdΩ 4πε0 Z p Q = − d2 x γ (2) nµ σν F µν (39) (40) (41) ∂Σ But this has no General Relativity equivalent. There exists no volume density for mass (or energy). For r = constant: r 2M , 0, 0) (42) σν = (0, 1 − r r 2M nµ = (− 1 − , 0, 0, 0) (43) r We note the following representations for the Energy: Z p 1 EKomar = d2 x γ (2) nµ σν ∇µ K µ 4πG ∂Σ (44) (45) 5 Note that K ν is a timelike killing vector →r→∞ 1 Another form: EADM = 1 16πG Z ∂Σ p d2 x γ (2) σi (∂j hji − ∂i hjf ) where gµν = nµν + hµν has a solution if h is of the order enough for integral to converge). 1 rn (46) (falls off fast Finally we also note the angular momentum: Z p 1 d2 x γ (2) nµ σν ∇µ Rµ JKomar = 8πG ∂Σ (47) (48) where Rµ is the rotational killing vector 3 ∂ ∂φ Next Class 3.1 Reading • Ch. 33 in Gravitation (Misner, Thorne, and Wheeler) • Section 6.7 and Chapter 7 in Spacetime and Geometry (Carroll) 6