Introduction to Numerical Relativity

Introduction to Numerical Relativity Given a manifold M describing a spacetime with 4-metric we want to foliate it via space-like, threedimensional hypersurfaces: {Σ i }. We label such hypersurfaces with the time coordinate t, Σ t. We define the 1-form Ω a = a t, (1) with the normalization Ω 2 = g ab a t b t = 1 α 2, (2) Thefunctionαiscalledthelapsefunctionanditisstrictlypositiveforspacelikehypersurfacesα(t,x i ) > 0. We define also the unit normal vector to the hypersurface Σ t n a = αg ab Ω b = αg ab b t. (3) The sign is chosen so that the related unit vector is future directed. i.e. n a n a = 1, n a can be viewed as the four velocity of an observer moving orthogonally to the hypersurfaces of Σ t Such an observer will have a four-acceleration, a b a b = n a a n b, (4) With the above vector we can construct the spatial metric induced on the 3 dimensional hypersurface The inverse can be found by raising the indices with g ab γ ab = g ab +n a n b, (5) γ ab = g ac g bd γ cd = g ab +n a n b, (6) The metric γ ab is purely spatial, lives within the hypersurface. Its contraction with the normal vector is zero n a γ ab = n a g ab +n a n a n b = n b n b = 0. (7) The projector operator projects a 4-dimensional tensor into a spatial slice, γ a b = g a b +n a n b, (8) in particular the projection of an arbitrary4-vector v a, γ a bv b is purely spatial. In order to project tensors of higher rank, each free index has to be contracted with the projection operator. Sometimes the notation is used in the literature. Any tensor can be decomposed into its spatial and timelike parts, T ab = γ a c γ b d T cd, (9) v a = δ a bv b = (γ a b n a n b )v b = v a n a n b v b, (10) T ab = T ab n a n c T cb n b n c T ac +n a n b n c n d T cd. (11) 1

We will need a 3-dimensional covariant derivative D a that maps spatial tensors into spatial tensors. We also require that this covariant derivate must be compatible with the induced metric, D c γ ab = 0. For a scalar function this derivate is defined as D a f = γ b a bf, (12) and for a spacetime tensor: D a T b c = γ d aγ b eγ f c d T e f, (13) then D c γ ab = γ d cγ f aγ g b d γ fg = γ d cγ f aγ g b d (g fg +n f n g ) = γ d cγ f aγ g b(n f d n g +n g d n f ) = 0, where we have used property (7) and the compatibility of g ab with the 4-covariant derivative. Extrinsic curvature The extrinsic curvature, can be found projecting gradients of the normal vector on Σ t. K ab = γ c aγ d b c n d = γ c aγ d b (c n d) (14) also, it can be writen as K ab = γ c aγ d b c n d = (δ c a +n c n a )(δ d b +n d n b ) c n d = (δ c a +n a n c )δ d b c n d = a n b n a a b. Perhaps the most useful expression is K ab = 1 2 L nγ ab = ( a n b +n a n d d n b ). (15) Gauss-Codazzi equations Let us consider the definition of 3d Riemann tensor, acting on a pure spatial vector (D i D j D j D i )ω k = 3 R l ijk ω l. (16) Using the defintion of 3d covariant derivative D i D j ω k = γi a γb j γc k a(γb l γm c lω m ) = γi a γjγ b k c a b ω c +γi a γjγ b k m ( a γb)( l l ω m )+γiγ a jγ l k( c a γc m )( l ω m ) = γi a γb j γc k a b ω c γk m (nl l ω m )K ij K ik Kj c ω c, To prove the last equality, let us consider first c γ b a = c(g b a +n an b ) = c (n a n b ) = n a c n b +n b c n a = n a (K b c +n c D b lnα) n b (K ac +n c D a lnα) 2

Consequently γ c iγ a j c γ b a = n b K ij, (17) and the contraction in second term of the right hand side γ a i γb j ( aγ l b ) = nl K ij. Secondly, since ω a is purely spatial n a ω a = 0 we rewrote the third term in the second line as Finally (D i D j D j D i )ω k gives γ l j nm l ω m = γ l j ω m l n m = ω m K m j. (18) 3 R ijk l ω l = γ a iγ b jγ c kr abc l ω l ω l K l j K ik +ω l K il K jk. (19) Then we have the Gauss Codazzi equations 3 R ijkl = γ a iγ b jγ c kγ d lr abcd +K il K jk K ik K jl, (20) Hamiltonian constraint if we contract Eq. (20) with γ ik 3 R jl = γ b j γd l (R bd +n a n c R abcd )+K jk K k l KK jl, (21) where we have used γ ac = g ac +n a n c. Multipliying again with γ jl 3 R = (R+2R ad n a n d )+K il K il K 2, (22) Notice that if we use Einstein s equations R+2R ad n a n d = 2G ad n a n d = 16πT ad n a n d, (23) in the previous formula we get ρ = T ad n a n d 3 R+K 2 K il K il = 16πρ, (24) which is known as the Hamiltonian constraint. Momentum constraint Let us consider other projections D i D j n k = γi a γb j γk c ad b n c = γi a γb j γk c a(γb d γc e dn e ) = γi a γjγ b c k a b n c +γiγ a jγ b c( k a γb) d d n c +γi a γjγ b c( k a γe) c b n e = γi a γb j γk c a b n c γc k K ijn d d n c γj b Kk i n e b n e = γi a γjγ b c k a b n c (D k lnα)k ij where we have used D c lnα = n d d n c and n e b n e = 0. Then if we rename the index k i D i D j n k = γ a i γb j γk c a b n c (D k lnα)k ij (25) D i D j n i = γ a cγ b j a b n c (D i lnα)k ij, (26) 3

or k j and i j to keep the index i on n i D j D i n i = γj a γb c a b n c (D i lnα)k ij, (27) we can construct the Riemann tensor using the nonconmutativity of ( a b b a ) as D i D j n i D j D i n i = γcγ a j( b a b b a )n c = γcγ a jr b c abd n d = R bd γjn b d (28) Taking into account that D i n i = K and D j n i = K i j we get D i Kj i D jk = R bd γj b nd, (29) Using the projected Eintein s equation R bd γj bnd = 8πT bd γj bnd. We get the Momentum constraint. D i Kj i D jk = 8πS j (30) where we have defined S j = T bd γj b. Evolution of the extrinsic curvature From the contaction of the Gauss-Codazzi equations (21) 3 R jl = γjγ b l d (R bd +n a n c R abcd )+K il Kj i KK jl, (31) the term that involves n a n c R abcd contains second time derivatives of the metric. Let us consider R abcd n d = ( a b b a )n c (32) and a b n c = a ( K bc n b D c lnα) (33) = a K bc +(K ab +n a D b lnα)d c lnα n b a D c lnα (34) where we have used b n c = K bc n b D c lnα. Thus the projection of R abcd n d = ( a b b a )n c γjγ b l d n a n c R abcd = γjγ a ln c b n d R abcd = γjγ a ln c b ( a K bc + b K ac )+(D j lnα)d l lnα+d j D l lnα. (35) The first term of the right hand side of the last expression can be rewritten as γjγ a ln c b a K bc = γjk a bl a n b = γjk a bl (Ka b +n a D b lnα) = K kl Kj k, (36) and the second term γj a γc l nb b K ac = γj a γc l (L nk bc K ab c n b K bc a n b ) = L n K jl +2K jk Kl k, (37) hence we can write the contraction as γj a γc l R abcdn b n d = L n K jl +K jk Kl k + 1 α D jd l α. (38) Let us go back to the term γi bγd j R bd. Using the Einstein s equations γ b iγ d jr bd = 8πγ b iγ d j(t bd 1 2 g bdt). (39) 4

Let us consider the second term g bd g ef T ef γiγ b j d = (γ ef n e n f )γ di γjt d ef (40) = (γ ef n e n f )γ ij T ef (41) = γ ef γ ij T ef γ ij ρ (42) = γ ij (S ρ), (43) with S = γ ij S ij = γ ij γi e γf j T ef = γ ef T ef, S ij = γi c γd j T cd, (44) From the Gauss Codazzi equation (21) 3 R jl = γjγ b l d R bd + 1 α D jd l α+l n K jl +2K kl Kj k KK jl (45) and using γ b jγ d l R bd = 8π[S jl 1 2 γ jl(s ρ)], (46) we have the evolution for K jl L n K jl = 3 R jl 1 α D jd l α+2k kl K k j KK jl 8π[S jl 1 2 γ jl(s ρ)]. (47) 5