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# Lecture on Numerical Relativity

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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

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.

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

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.

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

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.

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

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