Kadath: a spectral solver and its application to black hole spacetimes

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1 Kadath: a spectral solver and its application to black hole spacetimes Philippe Grandclément Laboratoire de l Univers et de ses Théories (LUTH) CNRS / Observatoire de Paris F Meudon, France philippe.grandclement@obspm.fr March 31 st 2010

2 What is KADATH? KADATH is a library that implements spectral methods in the context of theoretical physics. It is written in C++, making extensive use of object oriented programming. Versions are maintained via Subversion. Minimal website : luthier/grandclement/kadath.html The library is described in the paper : JCP 220, 3334 (2010). Designed to be very modular in terms of geometry and type of equations. LateX-like user-interface.

3 Why the name?

4 Numerical methods

5 Spectral expansion Given a set of orthogonal functions Φ i on an interval Λ, spectral theory gives a recipe to approximate f by f I N f = N a i Φ i i=0 Properties the Φ i are called the basis functions. the a i are the coefficients. Multi-dimensional generalization is done by direct product of basis.

6 Sturm-Liouville problems It is a family of eigenvalue problems on [ 1, 1]. Definition : (pu ) + qu = λwu p is continuously differentiable, strictly positive and continuous at x = ±1. q is continuous, non-negative and bounded. w is continuous, non-negative and integrable. The solutions are the eigenvalues λ i and the associated functions u i. The u i are orthogonal with respect to the measure w : 1 1 u m u n wdx = 0 for m n

7 Link with spectral methods A Sturm-Liouville problem is said to be singular if p vanishes at x = ±1. When the orthogonal functions of a spectral expansion are solutions of a singular Sturm-Liouville problem, one can show that I N u converges to u faster than any power of N (if u is C ). This is called spectral convergence The usual choices of basis for spectral methods will be solutions of singular Sturm-Liouville problems (Legendre or Chebyshev polynomials for instance). For functions less regular (i.e. not C ) the error decrease as a power-law.

8 Coefficient and configuration spaces There exist N + 1 point x i in Λ such that f (x i ) = I N f (x i ) Two equivalent descriptions Formulas relate the coefficient a i and the values f (x i ) Complete duality between the two descriptions. One works in the coefficient space when the a i are used (for instance for the computation of f ). One works in the configuration space when the f (x i ) are employed (for the computation of exp (f))

9 Multi-domain approach Definition Space is split into several touching (not overlapping) domains. in each domain, the physical coordinates X are mapped to the numerical ones X. Spectral expansion is performed with respect to the X. Appropriate matching must be enforced at the boundaries. Why? To have C functions only. To increase resolution where needed. To use different variables or different basis in different regions.

10 Geometries in KADATH 1D space. Cylindrical-like coordinates. Spherical spaces with time periodicity. Polar and spherical spaces. Bispherical geometries. Additional cases should be relatively easy to include.

11 Choice of basis Important step in setting the solver. All the terms involved in the equations must have consistent basis. Scalars Assume that all the fields are polynomials of the Cartesian coordinates (when defined). Express the Cartesian coordinates in terms of the numerical ones. Deduce an appropriate choice of basis. Higher order tensors With a Cartesian tensorial basis: given by gradient of scalars. For other tensorial basis: make use of the passage formulas that link to the the Cartesian one.

12 Example of spectral basis (1) In a spherical geometry, near the origin : x = r sin θ cos ϕ y = r sin θ sin ϕ z = r cos θ Scalar symmetric wrt z = 0 cos (mϕ) and sin (mϕ). cos (2jθ) for m even and sin ((2j + 1) θ) for m odd. T 2i (r) for m even and T 2i+1 (r) for m odd. Scalar anti-symmetric wrt z = 0 cos (mϕ) and sin (mϕ). cos ((2j + 1) θ) for m even and sin (2jθ) for m odd. T 2i+1 (r) for m even and T 2i (r) for m odd.

13 Example of spectral basis (2) Vector V = (V x, V y, V z ) Same as symmetric scalars for V x and V y. Same as anti-symmetric scalars for V z. With an orthonormal spherical tensorial basis, one has V θ = V x cos θ cos ϕ + V y cos θ sin ϕ V z cos θ which gives θ component of a vector cos (mϕ) and sin (mϕ). sin (2jθ) for m even and cos ((2j + 1) θ) for m odd. T 2i+1 (r) for m even and T 2i (r) for m odd.

14 Weighted residual method ; the τ-method Let R = 0 be a field equation (like f S = 0). The weighted residual method provides a discretization of it by demanding that (R, ξ i ) = 0 i N Properties (, ) denotes the same scalar product as the one used for the spectral expansion. the ξ i are called the test functions. For the τ method the ξ i are the basis functions (i.e. one works in the coefficient space). Some of the last residual equations must be relaxed and replaced by appropriate matching and boundary conditions to get an invertible system.

15 Additional regularity The choice of basis does not encompass all the regularity conditions. Example In spherical symmetry let f contain only a term in m = 2. The θ basis is then cos (2jθ). f is regular if and only if f can be divided by sin θ, which is not guaranteed by the spectral basis. Another way to looks at regularity If the division by sin θ is performed in the coefficient space, what is computed is f f (θ = 0) R = sin θ using R instead of the right ratio causes the appearance of additional homogeneous solutions (not regular). The additional HS are used to ensure that the solution f is regular. Regularity is required on axis of spherical and bispherical coordinates, and at the origin of the spherical ones.

16 Galerkin method When needed, regularity is enforced by making use of the Galerkin method. Basic idea : expand the functions onto linear combination of the basis functions that fulfill the regularity conditions. Example For instance if one requires that f (0) = 0 one can expand f on the G i = T 2i+2 + ( 1) i. Properties A Galerkin basis is not orthogonal. Usually less functions must be considered to keep the same degree of approximation (the Galerkin basis contains more information than the regular basis). The residual equations are then (R, G i ) = 0.

17 Newton-Raphson iteration Given a set of field equations with boundary and matching equations, KADATH translates it into a set of algebraic equations F ( u) = 0, where u are the unknown coefficients of the fields. The non-linear system is solved by Newton-Raphson iteration Initial guess u 0. Iteration : Compute s i = F ( u i) If s i if small enough = solution. Otherwise, one computes the Jacobian : J i = F u ( ui) One solves : J i x i = s i. u i+1 = u i x i. Convergence is very fast for good initial guesses.

18 Computation of the Jacobian Explicit derivation of the Jacobian can be difficult for complicated sets of equations. Automatic differentiation Each quantity x is supplemented by its infinitesimal variation δx. The dual number is defined as x, δx. All the arithmetic is redefined on dual numbers. For instance x, δx y, δy = x y, x δy + δx y. Given a set of unknown u, and a its variations δ u. When F is applied to u, δ u, one then gets : F ( u), δf ( u). One can show that δ F ( u) = J ( u) δ u The full Jacobian is generated column by column, by taking all the possible values for δ u, at the price of a computation roughly twice as long.

19 Inversion of the Jacobian Consider N u unknown fields, in N d domains, with d dimensions. If one works with N coefficients in each dimension, the Jacobian is a m m matrix with: m N d N u N d For N d = 5, N u = 5, N = 20 and d = 3, one gets m = , which is about 150 Go for a full matrix. Solution The matrix is computed in a distributed manner. Easy to parallelize because of the manner the Jacobian is computed. The library SCALAPACK is used to invert the distributed matrix. Currently, KADATH has been tested with 80 processors and it is enough for m

20 Applications of KADATH Critical collapse solution. Stationary gauge field solutions (Monopoles, spinning Q-balls, vortons). Time-dependent solution : oscillons. and black holes...

21 Numerical black holes

22 Foliation of space-time Spacetime is foliated by a family of spatial hypersurfaces Σ t Coordinate system of Σ t : (x 1, x 2, x 3 ). Coordinate system of spacetime : (t, x 1, x 2, x 3 ).

23 Metric quantities Various functions Lapse N, shift N and spatial metric γ ij. The metric reads ds 2 = ( N 2 N i ) N i dt 2 + 2N i dtdx i + γ ij dx i dx j

24 Projection of Einstein equations Type Einstein Maxwell Constraints Evolution Hamiltonian R + K 2 K ij K ij = 0 E = 0 Momentum : D j K ij D i K = 0 B = 0 γ ij E L N γ ij = 2NK ij t t = 1 ( ) B ε 0 µ 0 K ij B L N K ij = D i D j N+ t t = E N ( R ij 2K ik Kj k + KK ) ij R ij is the Ricci tensor of γ ij and D i the covariant derivative associated with γ ij. K ij is called the extrinsic curvature.

25 Digging the holes The puncture method Solve the equations in the whole space but at two points. Those points are the so-called punctures and the properties of the black holes are encoded in the way the various fields diverge there. Excision method Solve in the region outside the horizon. The properties of the black holes are encoded in the boundary conditions of the fields on the horizon.

26 The apparent horizon Definition It is the outermost trapped surface. A surface is trapped if the expansion of the outgoing light cones is negative (or zero). Properties It is a local notion. It tracks the strong gravitational fields. It is always inside the event horizon. In the stationary case the two types of horizons coincide. Inner BC for the fields can be derived by demanding that the surface of excision are apparent horizons.

27 Conformally flat binary black holes

28 Hypothesis Circular orbits Circular orbits imply that t = Ω ϕ. Working in the corotating frame gives a stationary problem. Translates on the BC for the shift vector. Conformal flatness Assume that γ ij = Ψ 4 f ij. For simplicity only. Prevents the appearance of gravitational waves.

29 Equations Three unknown fields Ψ, N and N. N = NΨ 4  ij  ij 2 D j ln Ψ D j N N i D i Dj N j = ( 2Âij Dj N 6N D j ln Ψ ) Ψ = Ψ5 8 ÂijÂij with Âij = 1 ( D i N j + 2N D j N i 23 ) D k N k f ij One does not solve for the traceless part of the evolution equations.

30 Boundary conditions at infinity With the conformal flatness approximation, the spacetime does not contain gravitational waves = one can enforce asymptotic flatness, hence : N 1 Ψ 1 N Ω ϕ The last equation indicates that one is working in the corotating frame and Ω is the orbital velocity.

31 Inner boundary conditions Inner BC for the fields can be derived by demanding that the surface of excision are apparent horizons: N = N 0, arbitrary choice. r Ψ + Ψ 2a + Ψ3 Ã ij s i s j = 0, the surface is an apparent horizon. N r = N 0, the surface remains an apparent horizon. Ψ2 N θ = 0. N ϕ = (Ω Ω loc ) ϕ loc encodes the rotation state of the holes.

32 Integral equations Virial theorem Ψ 1 + M ADM 2r N 1 M Komar r One demands that, at the first order in 1/r the metric coincides with the Schwarzschild metric : M Komar = M ADM It determines Ω uniquely. Rotation state of the holes Corotation : Ω loc = Ω Irrotation : the local spin vanishes (given by a surface integral on the horizons).

33 Lapse

34 Conformal factor

35 Shift

36 Kerr black hole

37 Problem Remove the conformal flatness : γ ij = Ψ 4 γ ij, where γ ij f ij and det ( γ) = 1. Solve the full set of equations for a single rotating black hole. Equations in the general case for N : trace of the evolution equation. for Ψ : Hamiltonian constraint. for N : momentum constraints. Boundary conditions Inner BC : apparent horizon in the general case Outer BC : asymptotic flatness + corotation.

38 Dirac gauge The spatial metric is not uniquely defined by the evolution equations and must be supplied by a choice of gauge. Dirac gauge Definition D j γ ij = 0 where D stands for the flat covariant derivative. Property : simplifies the expression of R ij which contains only one second order term : R ij = 1 2 γkl D k D l γ ij + first order

39 Equation for the spatial metric γ Given by the evolution equation (with t = 0) : D i Dj N 2 D i N D j Ψ Ψ +2N D i Dj Ψ ( Ψ Ψ 4 2 D j N D i Ψ Dk N + 2 γ D k Ψ ij Ψ Ψ + 2N γ D Dk k Ψ ij Ψ 4A ij N k D k Ψ Ψ + 2N γ ij Dk Ψ D k Ψ Ψ 2 N R ij 6N D i Ψ D j Ψ Ψ 2 + 2NΨ 4 A ik A k j ) + N k Dk A ij + A ik Dj N k + A jk Di N k = 0

40 Behavior on the horizon One can rewrite the evolution equation to make appear explicitly the second order operator in terms of γ : ] H ( γ ij ) = [ γ kl Ψ4 N 2 N k N l D k D l γ ij = S ij The inner boundary condition for the radial component of the shift is N r S = N Ψ 2 sr where s r is the unit normal to the sphere. It follows that H does not contain terms like 2 on the horizon. r2 The evolution equation are degenerated on the horizon.

41 Inner BC for γ ij Possible choice Evolution equation : H ( γ ij ) = S ij Dirac gauge : D j γ ij = 0 Set of equations used (empirical) H ( γ ij ) = S ij for all but (r, θ). θ component of Dirac gauge : D j γ θj = 0 The (ϕ, ϕ) component is not determined by the evolution equation but by demanding that det ( γ) = 1. A posteriori check : all the forgotten equations must be fulfilled (in particular Dirac gauge in whole space).

42 Error on the whole set of eqs

43 Lapse a/m = 0.91

44 Shift a/m = 0.91

45 Spatial metric a/m = 0.91

46 Non-conformally flat binary black holes

47 Hypothesis Try to impose exact circularity without conformal flatness. Difficulties Behavior of γ on the horizons? How to deal with gravitational waves?

48 Behavior on the horizons With exact circular orbit, the only possible state of rotation is (probably) corotation It means that N i S = N Ψ 2 si. As for Kerr, the operator H ( γ ij ) is degenerate on the horizons. The no-boundary treatment seems to work. Quid of the Dirac gauge?

49 Solution with the no-boundary treatment

50 Behavior at infinity Exact circularity is inconsistent with conformal flatness. Spacetime would be filled with gravitational waves. Mathematically γ does not decrease fast enough. Global quantities (like the total energy) are infinite. Proposition Solve the evolution equations only up to r < R At r = R match γ with an appropriate solution. To avoid divergence set γ ij = f ij for r > R in all the other equations. Check that the results do not change for R large enough. This technique has been successfully applied in gauge field theory (in the oscillon case).

51 Asymptotic behavior of H ( γ ij ) H ( γ ij ) = ] [ γ kl Ψ4 N 2 N k N l D k D l γ ij Given the behaviors of N, Ψ and N at infinity, one gets : ] H ( γ ij ) = [ Ω 2 2 ϕ 2 γ ij For each harmonic m the operator becomes an Helmholtz like operator : H m = [ + m 2 Ω 2]

52 Link to a wave operator Far from the source, the metric should obey an homogeneous wave equation : [ ] 2 t 2 γ ij = 0 Circularity implies that t = Ω ϕ so that one gets : [ Ω 2 2 ϕ 2 ] γ ij The same Helmholtz operator as before is recovered.

53 Matching of the metric For r > R one assumes that γ obeys an homogeneous Helmholtz solution. Each spherical harmonic should be matched with : h lm = C lm [(cos α lm ) j l (mωr) + (sin α lm ) y l (mωr)] The matching of the solution and its derivative at r = R should give the solution at r < R AND the amplitudes C lm. However, the phases α lm are a priori unconstrained. A posteriori, one would need to check that the C lm do not depend on R, for R sufficiently large.

54 What is next? Get a stable algorithm solving the evolution equation, with the no-boundary treatment. Test the validity of the matching procedure. Each run is difficult and long because of the many unknowns, complex geometry and complicated equations. The procedure should provide a natural way of extracting the waves and share some light on both post-newtonian and recent numerical results.

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