Lecture note on 3+1 formalism of numerical relativity Masaru Shibata (Yukawa Institute, Kyoto U) 2010/06/09 1 Basis equations for Numerical Relativity G GT 8 4 Einstein equation c T 0 u 0 u Yll Q Matter fields: Fj 4 Next time if any []F 0 f p f p S p 2 Others in Numerical Relativity • Imposing gauge conditions • Extracting gravitational waves • Finding black holes (finding apparent horizon) • Adaptive mesh refinement 3 Contents 1. Structure of Einstein‟s equation (briefly) 2. 3+1 formalism of Einstein‟s equation 3. BSSN formalism 4. Gauge conditions 5. Initial value problem 6. Implementation of finite differencing 7. Extracting gravitational waves 8. Adaptive mesh refinement 9. Apparent horizon finder 4 Solution of Einstein’s equation for dynamical phenomena • We have to solve Einstein‟s equations as an initial value problem 4 • Einstein‟s equations, G = 8Gc T , are equations for space and time Space and time coordinates appear in a mixed way; “time coordinate” does not always have the property of time. E.g., Schwarzschild coordinates; t is time for r > 2M, but is not for r < 2M Special formalism is necessary to follow dynamics of a variety of spacetimes 5 Schwarzschild spacetime 1 2 222GM 2 GM 2 2 2 2 2 ds c dt 1 2 dr r d sin d r rc t is a radial t = const coordinate Coordinate t is the time singularity coordinate. Time has to be always “time” in numerical relativity 6 Several formulations 1. 3+1 (N+1) formalism 2. Formulation based on a special (harmonic) coordinates (e.g., Pretorius formalism; also often used in Post-Newtonian theory) 3. Hyperbolic formalism (e.g., Kidder-Scheel-Teukolsky formalism) • Others ….. In this lecture, I focus on the first one 7 Einstein’s equation = hyperbolic equations 1 G G : Einstein tensor G R g R 8 T 4 R : Ricci tensor 2 c R 1 g g g g 2 g G g g g g g g t 2 t : Pseudo tensor of Landau-Lifshitz Og de-Donder Gauge: gg 0 Wave equation 2 G g g g g O g 8 Einstein’s equation is similar to scalar wave equation 2 ① t Post-Newtonian way t ② 3+1 way t or , Kij , gij t i i Hyperbolic way ③ i ti Similar: However, spacetime is not a priori given in general relativity 9 Section I: 3+1 (ADM) formalism Concept 1. Foliate spacetime by spacelike surfaces 2. “Choose” time coordinates for a direction of time at each location 3. Follow dynamics of spacelike surfaces Spacelike Time axes hypersurfaces 10 3+1 decomposition (N+1) Time 4 Evolve gij & Kij 3 Choose time axes at each point Space i t + t nn = -1 2 Draw timelike t unit normal wrt n t spacelike surface, and choose lapse t g ij, K ij 1 Choose a spacelike surface Kij~g ij : Talk later & give space-metric 11 Definition of variables g (g , , k) gij = space metric ij Kij= extrinsic curvature Dynamics = lapse function i = shift vector Gauge Lie derivative g g n n : g i n 0 1 Kn g g L g ij i j 2 n ij 1 tg ij DD i j j i 2 n t ; n 0 12 Line element i Time t t i Space dx t + t t n t 2 2 i i j j ds dt gij dx dt dx dt 13 Structure of variables i 1 n, , n , 0 ; cf. t 1, 0 ij : lapse function, :shift vector; i g ij 2 k , 1/22 , i / ggki, ij i j 2 * ,g ij * ,g / k , 0 ,0 Cf. g ki, g ij * ,g ij * ,g j j j KKij , ij 0 ,0 KK, ij * ,KKij * , 14 Geometric meaning of Kij If space-like hyper-surface is curved, n is not parallel-transported. un0 n n n u ii u Kij g j u n i 0 Kij denotes the “curved” degree of chosen spatial hyper-surfaces 15 Next step is to rewrite Einstein’s equation by gij & Kij As a first step, it is necessary to define covariant derivative associated with gij Dig jk 0 Required property ijk T lmn : A spacetime tensor ijk h''''''' i j k l m n i j k Define DTTh lmng h g i' g j ' g k ' g l g m g n h ' l' m ' n ' Then, projection i'''' j k i jk'' Dig jk g i g j g k i''' g j k g i g j g k i'''''()g j k n j n k =gi''' g j g k n n n n 0 i j k j'''''' i k k i j OK 16 3+1 definition of Kij 1 Kij g i g j n g i g j n n 2 ggi jn g i n i n g j n n n 0 in j n i n n j 1 Kij i n j j n i n n i n j 2 1 gj in g i j n n g ij 2 1 ggj in i j nng ij 2 1 tg ij DD i j j i 2 17 Usuful, often used relations K n n n n n n D ln Note : n n D ln : acceleration We will use it for many times 18 3+1 decomposition of Einstein’s equation G n n 8 T n n : Hamiltonian constraint G ngkk 8 T n g : Momentum constraint GTgi g j 8 g i g j : Evolution equation • First two eqs = constraint eqs ・・ no second derivative of spatial metric • Last one = evolution eqs ・・ hyperbolic eqs of spatial metric • No time derivative for & k 19 Similar to Maxwell’s eqs i ieE 4 i Constraint eqs i B 0 iii t E B 4 j Evolution eqs i i t BE Step 1: Give an initial condition which satisfies constraints. Step 2: Solve evolution equations. 20 All eqs have to be written by (g, K) • Method: Derive Gauss-Codacci equations Start from definition of 3D Riemann tensor (3) l DDDDRDi j j i k ijk l : l 3 vector Using defintion of 3D covariant derivative, a b c l m DDijk g ijkabclm g g g g a b c a b m l gi g j g k a b c g i g j g k a g b l m a l c m gi g j g k a g c l m a b c a m c gi g j g k a b c n a m g k K ij K ik K j c 21 where we used b b b b b b cacaag g n n ca n n n ac n n ca n b b b na K c n c Dln n K ac n c D a ln c a b b ggi j cg a nK ij l m l m m a gjn l m g j m l n m K j n a 0 Then, DiDDD j j ik gives (3)R l = ga g b g cRKKKK l l l ijk l i j k abc l l j ik l i jk Gauss (3) a b c d Codacci RRKKKKijkl g i g j g k g l abcd il jk ik jl eq. ()N a b c d N 1 Note: RRijklg i g j g k g l abcdKKKK il jk ik jl 22 Gauss-Codacci equations I, continued (3) a b c d RRKKKijklg i g j g k g l abcd il jk ikK jl Multiplying g ik (3) b d a c k Rjlg jg l R bd R abcd n n K jk K l KK jl Multiplying g jl again This will be used later (3)ac il 2 R R 2 Rad n n K il K K a c a c a c Here, R 2 Rad n n 2 Gabb n nn 16Tna (3) 2 il ac RKKKil 16Tab n n Hamiltonian constraint 1 component 23 • Derive Gauss-Codacci equations II Start from k abk c abk dc e Dij D ng ijcab g g D n g ijcabed g g g g n a b k c a b k d c gijcab g g nn g ijc g g ab g d a b k c e gi g j g c a g e bn a b k c k d c b k e gi g j g c a bn g c K ij n d n g j K i n e b n a b k c k gi g j g c a bnD ln Kij d c c e where nd n Dln , n e b n 0 24 • Derive Gauss-Codacci equations II k a b k c k Di D j ng i g j g c a b n Dln K ij (i,k) i a b c i Di D j ng c g j a b n Dln K ij D D nig a g b n c D i ln K (j,k) j i j c a b ij i i a b c Di D j n D j D i n gg c j a b b a n a b c d ggc jRn abd iii a b c d b d Dij n K, D n KRj, g c g j abdnn R bd g j i b d b d DKDKi j j R bdgj n 8 T bd g j n Momentum constraint 3 components 25 Derive evolution equation Start from contracted Gauss-Codacci eq.