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Numerical Relativity
Mark A. ScheelWalter Burke Institute for Theoretical Physics
Caltech
July 7, 2015
Mark A. Scheel Numerical Relativity July 7, 2015 1 / 34
Outline:
Motivation3+1 split and ADM equationsWell-posedness and hyperbolicityEvolution system 1: BSSN:
Equations and gauge choices
Evolution system 2: Generalized Harmonic:Equations and gauge choices
Handling singularities inside black holesCurrent status of NR
Mark A. Scheel Numerical Relativity July 7, 2015 2 / 34
Motivation:
Many gravitational wave sources (binaries, supernovae) involvenonlinear gravity. To learn these from LIGO, must solve Einsteinsequations to connect source properties with measured gravitationalwaves.
S
S
1
2
M
M
2
1
-10000 -8000 -6000 -4000 -2000 0
t/M
-0.4
-0.2
0
0.2
0.4
h(t)
The only way to solve nonlinear, dynamical, strong-field Einsteinequations is numerically.
Numerical Relativity
Mark A. Scheel Numerical Relativity July 7, 2015 3 / 34
Statement of the problem:
Einsteins Eqs. G = 8T have time/space mixed up.Metric g unknown analytically in most cases.
GoalRewrite G = 8T to solve for the metric later given data now.
Mark A. Scheel Numerical Relativity July 7, 2015 4 / 34
Analogy: Maxwells equations
Constraints
E = 4 B = 0
Evolution
tE = B 4JtB = E
Figure: C. Reisswig
t = 0
t = 1
t = 2
Note: Evolution eqs. preserve constraints:
t( B) = tB = ( E) = 0t( E 4) = tE 4t
= (B 4J) 4t= 4(t+ J) = 0
Goal: Do the same for Einsteins Eqs. G = 8T
Mark A. Scheel Numerical Relativity July 7, 2015 5 / 34
Analogy: Maxwells equations
Constraints
E = 4 B = 0
Evolution
tE = B 4JtB = E
Figure: C. Reisswig
t = 0
t = 1
t = 2
Note: Evolution eqs. preserve constraints:
t( B) = tB = ( E) = 0t( E 4) = tE 4t
= (B 4J) 4t= 4(t+ J) = 0
Goal: Do the same for Einsteins Eqs. G = 8T
Mark A. Scheel Numerical Relativity July 7, 2015 5 / 34
Analogy: Maxwells equations
Constraints
E = 4 B = 0
Evolution
tE = B 4JtB = E
Figure: C. Reisswig
t = 0
t = 1
t = 2
Note: Evolution eqs. preserve constraints:
t( B) = tB = ( E) = 0t( E 4) = tE 4t
= (B 4J) 4t= 4(t+ J) = 0
Goal: Do the same for Einsteins Eqs. G = 8T
Mark A. Scheel Numerical Relativity July 7, 2015 5 / 34
Arnowitt-Deser-Misner (1962) 3+1 split
Figure: C. Reisswig
t = 0
t = 1
t = 2
Assume spacetime split into slices.Let n be normal vector to slice.nn = 1.
Define3-metric = g + nn .Lapse function :Proper time / coord time.Shift vector :How coords move.
In coordinates: = (0, i)
n = 1(1,i)ds2 = 2dt2 + ij(dxi + idt)(dxj + jdt) ,,,,. . . go from 0 to 3.
i,j,k,. . . go from 1 to 3.
Mark A. Scheel Numerical Relativity July 7, 2015 6 / 34
Arnowitt-Deser-Misner (1962) 3+1 split
Figure: C. Reisswig
t = 0
t = 1
t = 2
Assume spacetime split into slices.Let n be normal vector to slice.nn = 1.
Define3-metric = g + nn .Lapse function :Proper time / coord time.Shift vector :How coords move.
In coordinates: = (0, i)
n = 1(1,i)ds2 = 2dt2 + ij(dxi + idt)(dxj + jdt) ,,,,. . . go from 0 to 3.
i,j,k,. . . go from 1 to 3.
Mark A. Scheel Numerical Relativity July 7, 2015 6 / 34
Arnowitt-Deser-Misner (1962) 3+1 split
Figure: C. Reisswig
t = 0
t = 1
t = 2
Assume spacetime split into slices.Let n be normal vector to slice.nn = 1.
Define3-metric = g + nn .Lapse function :Proper time / coord time.Shift vector :How coords move.
In coordinates: = (0, i)
n = 1(1,i)ds2 = 2dt2 + ij(dxi + idt)(dxj + jdt) ,,,,. . . go from 0 to 3.
i,j,k,. . . go from 1 to 3.Mark A. Scheel Numerical Relativity July 7, 2015 6 / 34
Arnowitt-Deser-Misner (1962) 3+1 split
Define extrinsic curvature:spatial projection of gradient of normal:
K = (n)
In terms of 3-metric, lapse, shift:
Kij = 1
2(tij j;i i;j)
(i;j is spatial covariant deriv, i.e. cov. deriv wrt gij .)
Mark A. Scheel Numerical Relativity July 7, 2015 7 / 34
Arnowitt-Deser-Misner (1962) equationsEvolution equations
tij = 2Kij + j;i + i;j
tKij = ;ij + [(3)Rij +KKij 2KimKmj 8Sij 4( S)ij
]+ mKij;m + 2
m(;iKj)m
(3)R+K2 KijKij = 16 Hamiltonian constraintKij;i K;j = 8Sj Momentum constraint
Matter terms: nnTS nTS TS T
12 evolved variables: ij , Kij .12 evolution equations4 constraints4 free variables: , i.
Mark A. Scheel Numerical Relativity July 7, 2015 8 / 34
Free evolution:
Figure: C. Reisswig
Use evolution equations toevolve forward in time.Check that constraints arestill satisfied.
Solve constraints at t = 0.
Mark A. Scheel Numerical Relativity July 7, 2015 9 / 34
Just put equations on a computer and solve them?
ADM equations developed in 1960s (ADM), 1970s (York).First attempt at binary BH simulation 1964 (Hahn, Lindquist)In summary, the numerical solution of the Einstein field equationspresents no insurmountable difficulties.First successful binary BH simulation 2005 (Pretorius).
Why is it so difficult?
ADM equations are not well-posed.ADM evolution equations amplify small errors in constraints.Black holes have physical singularities inside.Coordinates dont mean anything.
You need to choose them.Almost all choices are bad.
Mark A. Scheel Numerical Relativity July 7, 2015 10 / 34
Just put equations on a computer and solve them?
ADM equations developed in 1960s (ADM), 1970s (York).First attempt at binary BH simulation 1964 (Hahn, Lindquist)In summary, the numerical solution of the Einstein field equationspresents no insurmountable difficulties.First successful binary BH simulation 2005 (Pretorius).
Why is it so difficult?
ADM equations are not well-posed.ADM evolution equations amplify small errors in constraints.Black holes have physical singularities inside.Coordinates dont mean anything.
You need to choose them.Almost all choices are bad.
Mark A. Scheel Numerical Relativity July 7, 2015 10 / 34
Well-posedness
Hadamard 1902:
A problem is well-posed if and only if:A solution exists.The solution is unique.The solution depends continuously on initial and boundary data.
Example 1 (Hadamard 1923):
2t u 2xu = 0, x [0, 1]
Initial data: u = 0, tu =sin(2nx)
(2n)P, P 1.
Bdry conditions: u = 0 at x = 0, 1.
Solution: u(x, t) =sin(2nx) sin(2nt)
(2n)P+1.
For n, initial data 0 and u(x, t) 0. = well-posed
Mark A. Scheel Numerical Relativity July 7, 2015 11 / 34
Well-posedness
Hadamard 1902:
A problem is well-posed if and only if:A solution exists.The solution is unique.The solution depends continuously on initial and boundary data.
Example 1 (Hadamard 1923):
2t u 2xu = 0, x [0, 1]
Initial data: u = 0, tu =sin(2nx)
(2n)P, P 1.
Bdry conditions: u = 0 at x = 0, 1.
Solution: u(x, t) =sin(2nx) sin(2nt)
(2n)P+1.
For n, initial data 0 and u(x, t) 0. = well-posed
Mark A. Scheel Numerical Relativity July 7, 2015 11 / 34
Well-posedness
Hadamard 1902:
A problem is well-posed if and only if:A solution exists.The solution is unique.The solution depends continuously on initial and boundary data.
Example 1 (Hadamard 1923):
2t u 2xu = 0, x [0, 1]
Initial data: u = 0, tu =sin(2nx)
(2n)P, P 1.
Bdry conditions: u = 0 at x = 0, 1.
Solution: u(x, t) =sin(2nx) sin(2nt)
(2n)P+1.
For n, initial data 0 and u(x, t) 0. = well-posed
Mark A. Scheel Numerical Relativity July 7, 2015 11 / 34
Well-posedness
Example 2:
2t u+2xu = 0, x [0, 1] only change is sign
Initial data: u = 0, tu =sin(2nx)
(2n)P, P 1.
Bdry conditions: u = 0 at x = 0, 1.
Solution: u(x, t) =sin(2nx)sinh(2nt)
(2n)P+1.
For n, initial data 0 but u(x, t). = ill-posed
In other words, small perturbation at t = 0 produces arbitrarilylarge solution at given finite time.
Mark A. Scheel Numerical Relativity July 7, 2015 12 / 34
Well-posedness
Example 2:
2t u+2xu = 0, x [0, 1] only change is sign
Initial data: u = 0, tu =sin(2nx)
(2n)P, P 1.
Bdry conditions: u = 0 at x = 0, 1.
Solution: u(x, t) =sin(2nx)sinh(2nt)
(2n)P+1.
For n, initial data 0 but u(x, t). = ill-posed
In other words, small perturbation at t = 0 produces arbitrarilylarge solution at given finite time.
Mark A. Scheel Numerical Relativity July 7, 2015 12 / 34
Hyperbolicity of PDEsCan write any PDE system as set of 1st order PDEs:
t~u+ Ai i~u = ~B.
~u is a vector of variables, Ai are matrices, ~B is a vector of RHS terms.Ai and ~B can depend on ~u but not its derivatives.
Example: scalar wave equation in 1D: 2t 2x = 0
Define x t. then
t = ,t + x = 0,
t + x = 0,
~u =
~B = 0
0
Ax = 0 0 00 0 1
0 1 0
Mark A. Scheel Numerical Relativity July 7, 2015 13 / 34
Hyperbolicity of PDEsCan write any PDE system as set of 1st order PDEs:
t~u+ Ai i~u = ~B.
~u is a vector of variables, Ai are matrices, ~B is a vector of RHS terms.Ai and ~B can depend on ~u but not its derivatives.
Example: scalar
