Lecture on Numerical Relativity

transcript

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

Motivation:

Many gravitational wave sources (binaries, supernovae) involvenonlinear gravity. To learn these from LIGO, must solve Einstein’sequations to connect source properties with measured gravitationalwaves.

⇐⇒-10000 -8000 -6000 -4000 -2000 0

The only way to solve nonlinear, dynamical, strong-field Einsteinequations is numerically.

⇒ Numerical Relativity

Statement of the problem:

Einstein’s Eqs. Gµν = 8πTµν have time/space mixed up.Metric gµν unknown analytically in most cases.

GoalRewrite Gµν = 8πTµν to solve for the metric later given data now.

Analogy: Maxwell’s equations

Constraints

∇ · E = 4πρ

∇ ·B = 0

Evolution

∂tE = ∇×B − 4πJ

∂tB = −∇× EFigure: C. Reisswig

Note: Evolution eqs. preserve constraints:

∂t(∇ ·B) = ∇ · ∂tB = −∇ · (∇× E) = 0

∂t(∇ · E − 4πρ) = ∇ · ∂tE − 4π∂tρ

= ∇ · (∇×B − 4πJ)− 4π∂tρ

= −4π(∂tρ+∇ · J) = 0

Goal: Do the same for Einstein’s Eqs. Gµν = 8πTµν

Constraints

∇ · E = 4πρ

∇ ·B = 0

Evolution

∂tE = ∇×B − 4πJ

∂t(∇ ·B) = ∇ · ∂tB = −∇ · (∇× E) = 0

∂t(∇ · E − 4πρ) = ∇ · ∂tE − 4π∂tρ

= ∇ · (∇×B − 4πJ)− 4π∂tρ

= −4π(∂tρ+∇ · J) = 0

Constraints

∇ · E = 4πρ

∇ ·B = 0

Evolution

∂tE = ∇×B − 4πJ

∂t(∇ ·B) = ∇ · ∂tB = −∇ · (∇× E) = 0

∂t(∇ · E − 4πρ) = ∇ · ∂tE − 4π∂tρ

= ∇ · (∇×B − 4πJ)− 4π∂tρ

= −4π(∂tρ+∇ · J) = 0

Arnowitt-Deser-Misner (1962) 3+1 split

Figure: C. Reisswig

Assume spacetime split into slices.Let nµ be normal vector to slice.nµnµ = −1.

Define3-metric γµν = gµν + nµnν .Lapse function α:“Proper time” / “coord time”.Shift vector βµ:“How coords move”.

In coordinates:βµ = (0, βi)

nµ = α−1(1,−βi)ds2 = −α2dt2 + γij(dx

i + βidt)(dxj + βjdt) α,β,µ,ν,. . . go from 0 to 3.i,j,k,. . . go from 1 to 3.

Figure: C. Reisswig

Define extrinsic curvature:spatial projection of gradient of normal:

Kµν = −γλµγρν∇(λnρ)

In terms of 3-metric, lapse, shift:

Kij = − 1

2α(∂tγij − βj;i − βi;j)

(βi;j is spatial covariant deriv, i.e. cov. deriv wrt gij .)

Arnowitt-Deser-Misner (1962) equationsEvolution equations

∂tγij =− 2αKij + βj;i + βi;j

∂tKij =− α;ij + α[(3)Rij +KKij − 2KimK

mj − 8πSij − 4π(ρ− S)γij

]+ βmKij;m + 2βm(;iKj)m

(3)R+K2 −KijKij = 16πρ Hamiltonian constraint

Kij;i −K;j = 8πSj Momentum constraint

Matter terms:ρ ≡ nµnνTµνSµ ≡ γµνnλTνλSµν ≡ γµργνλTρλS ≡ γµνTµν

12 evolved variables: γij , Kij .12 evolution equations4 constraints4 free variables: α, βi.

Free evolution:

Figure: C. Reisswig

Use evolution equations toevolve forward in time.Check that constraints arestill satisfied.

Solve constraints at t = 0.

“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 don’t mean anything.

You need to choose them.Almost all choices are bad.

“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 don’t mean anything.

You need to choose them.Almost all choices are bad.

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(2πnx)

(2πn)P, P ≥ 1.

Bdry conditions: u = 0 at x = 0, 1.

Solution: u(x, t) =sin(2πnx) sin(2πnt)

(2πn)P+1.

For n→∞, initial data→ 0 and u(x, t)→ 0. =⇒ well-posed

Well-posedness

Hadamard 1902:

∂2t u− ∂2xu = 0, x ∈ [0, 1]

(2πn)P, P ≥ 1.

(2πn)P+1.

Well-posedness

Hadamard 1902:

∂2t u− ∂2xu = 0, x ∈ [0, 1]

(2πn)P, P ≥ 1.

(2πn)P+1.

Well-posedness

Example 2:

∂2t u+∂2xu = 0, x ∈ [0, 1] only change is sign

(2πn)P, P ≥ 1.

Solution: u(x, t) =sin(2πnx)sinh(2πnt)

(2πn)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.

Well-posedness

Example 2:

∂2t u+∂2xu = 0, x ∈ [0, 1] only change is sign

(2πn)P, P ≥ 1.

Solution: u(x, t) =sin(2πnx)sinh(2πnt)

(2πn)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.

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 ψ − ∂2xψ = 0

DefineΦ ≡ ∂xψΠ ≡ −∂tψ. then

∂tψ = −Π,

∂tΠ + ∂xΦ = 0,

∂tΦ + ∂xΠ = 0,

ψΠΦ

−Π00

0 0 00 0 10 1 0

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 ψ − ∂2xψ = 0

DefineΦ ≡ ∂xψΠ ≡ −∂tψ. then

∂tψ = −Π,

∂tΠ + ∂xΦ = 0,

∂tΦ + ∂xΠ = 0,

ψΠΦ

−Π00

0 0 00 0 10 1 0

Hyperbolicity of PDEs

∂t~u+ Ai · ∂i~u = ~B.

Pick any spatial unit vector ni.Then niAi is the characteristic matrix in direction ni.

This matrix can have eigenvalues and eigenvectors:~e a · (niAi) = v(a)~e

a (no sum on a).

~e a is the ath eigenvector.v(a) is the ath eigenvalue.Eigenvalues va are also called characteristic speeds.From eigenvectors can form ua = ~u · ~e a, which are calledcharacteristic fields.

∂t~u+ Ai · ∂i~u = ~B.

System isWeakly hyperbolic if all eigenvalues of niAi are real.

Usually ill-posed.Well-posedness depends on details ~B.

Strongly hyperbolic if it is weakly hyperbolic and there is acomplete set of eigenvectors, independent of the solution and ni.

⇒Well-posed.⇒ Only characteristic fields corresponding to negative eigenvaluesneed bdry conditions.

Hyperbolicity of PDEsExample: scalar wave in 1d.

nxAx =

0 0 00 0 10 1 0

Eigenvalues v(0) = 0, v(1) = 1, v(−1) = −1.

Eigenvectors ~e 0 =

, ~e 1 =

, ~e−1 =

01−1

Char fields u0 = ψ, u±1 = Π± Φ.

Strongly hyperbolic.

B.C. on Π− ΦB.C. on Π + Φ

ADM evolution equations are only weakly hyperbolic.

Usually ill-posed.No indication of which variables require BCs.

Fortunately, there are a few alternatives to ADM that work.We will discuss two:

BSSNGeneralized harmonic

BSSN FormulationShibata/Nakamura 1995, Baumgarte/Shapiro 1999

Define new variables

γij ≡ e−4φγij

Aij ≡ e−4φKij −1

3γijK

Γi ≡ γjkΓijk

Demand det γij = 1,→ ln γ = 12φ

Demand Aij = 0

Then can write Rij = Rφij + Rij , where

Rφij = terms with 2nd derivs of φ

Rij =− 1

2γ`mγij,`m + γk(iΓ

k,j) + terms without derivs

Now evolve γij , Aij , φ, K, and Γi instead of γij , Kij .

BSSN FormulationShibata/Nakamura 1995, Baumgarte/Shapiro 1999

Define new variables

γij ≡ e−4φγij

Aij ≡ e−4φKij −1

3γijK

Γi ≡ γjkΓijk

Demand det γij = 1,→ ln γ = 12φ

Demand Aij = 0

Then can write Rij = Rφij + Rij , where

Rφij = terms with 2nd derivs of φ

Rij =− 1

2γ`mγij,`m + γk(iΓ

k,j) + terms without derivs

Now evolve γij , Aij , φ, K, and Γi instead of γij , Kij .

BSSN Hyperbolicity

BSSN is strongly hyperbolic for pre-chosen α, βi

(Sarbach+ 02, Nagy+04).BSSN with dynamical α, βi strongly hyperbolic if shift is not toolarge (too close to 1) (Gundlach+ 06).

Gauge (coordinate) conditions for ADM/BSSN

Einstein’s Eqs. do notdetermine lapse α or shift βi.Lapse and shift equivalent tocoordinate choice.

How to choose coordinates?Simplest choice: α = 1, βi = 0. Bad. Why?

In vacuum, ∂tK = AijAij + 13K

2, so K →∞ at late times.Since ∂t ln

√γ = −K, if K →∞, then γ → 0. Coordinate singularity.

Gauge conditions: Maximal slicing

Figure: C. Ott.

Maximal slicing: K = ∂tK = 0.Why is this a gauge choice?

∂tK − βiK,i =− γijα;ij

+ α[KijK

ij + 4π(ρ+ S)]

K = ∂tK = 0

=⇒ γijα;ij = α[KijK

ij + 4π(ρ+ S)]

Properties:Singularity avoiding“Collapse of the lapse”Grid stetching

Figure: C. Ott.

+ α[KijK

ij + 4π(ρ+ S)]

K = ∂tK = 0

ij + 4π(ρ+ S)]

Figure: C. Ott.

+ α[KijK

ij + 4π(ρ+ S)]

K = ∂tK = 0

ij + 4π(ρ+ S)]

ADM/BSSN Gauge conditions: More lapse choices

Harmonic slicing:

�t = ∇α∇αt = 0

Reduces to

∂tα− βkα,k = −α2K

Choose initial lapse; thisdetermines lapse at t > 0

Shift still unspecified.

In practice, put arbitrary function f(α) on RHS:

∂tα− βkα,k = −α2f(α)K

f(α) = 1 ⇒ harmonic slicingf(α) = 0 ⇒ geodesic slicingf(α) =∞ ⇒ maximal slicingf(α) = 2

α ⇒ “1+log” slicing

Called “1+log” because for βi = 0, ∂tα = −2αK. So α = 1 + ln γ.

1+log slicing & variations used in most BSSN codes.

ADM/BSSN Gauge conditions: More lapse choices

Harmonic slicing:

�t = ∇α∇αt = 0

Reduces to

∂tα− βkα,k = −α2K

Choose initial lapse; thisdetermines lapse at t > 0

Shift still unspecified.

In practice, put arbitrary function f(α) on RHS:

∂tα− βkα,k = −α2f(α)K

f(α) = 1 ⇒ harmonic slicingf(α) = 0 ⇒ geodesic slicingf(α) =∞ ⇒ maximal slicingf(α) = 2

α ⇒ “1+log” slicing

Called “1+log” because for βi = 0, ∂tα = −2αK. So α = 1 + ln γ.

1+log slicing & variations used in most BSSN codes.

BSSN Gauge conditions: shift choices

Demand ∂tΓi = 0:

0 = ∂tΓi = +

3γ`iβk,kl + γkjβi,kj + terms without 2nd derivs of shift

Elliptic eq. for shift, similar to “minimal distortion”.

Easier to use “Gamma driver” to drive Γi to zero:

∂tβi =

∂tBi = ∂tΓ

i − 1

Gamma driver & variations used in most BSSN codes.

BSSN Gauge conditions: shift choices

Demand ∂tΓi = 0:

0 = ∂tΓi = +

3γ`iβk,kl + γkjβi,kj + terms without 2nd derivs of shift

Elliptic eq. for shift, similar to “minimal distortion”.Easier to use “Gamma driver” to drive Γi to zero:

∂tβi =

∂tBi = ∂tΓ

i − 1

Gamma driver & variations used in most BSSN codes.

Generalized Harmonic FormulationDifferent approach than ADM/BSSN

Basic idea: Choose coordinates xµ satisfying

�xµ = 0 “harmonic coordinates”

You are free to choose Hµ

Then:1 Einstein equations become wave equations for gµν :

gλρ∂λ∂ρgµν + 2∇(µHν) + (ΓΓ terms) = (matter terms)

2 Constraints Cµ ≡ Hµ −�xµ involve only 1st derivatives of gµν .Can add constraint damping (Gundlach+ 2005, Pretorius 2005).

gλρ∂λ∂ρgµν = . . .

⇓�Cµ = . . .

3 Strongly hyperbolic with char speeds on light cone, indep. of Hµ.

�xµ = Hµ(xµ, gµν) “generalized harmonic coordinates”You are free to choose Hµ

⇓�Cµ = . . .

2 Constraints Cµ ≡ Hµ −�xµ involve only 1st derivatives of gµν .

Can add constraint damping (Gundlach+ 2005, Pretorius 2005).

⇓�Cµ = . . .

gλρ∂λ∂ρgµν = . . . + γ0QλµνCλ

⇓�Cµ = . . . + ????

⇓�Cµ = . . . + γ0∂tCµ

Generalized Harmonic: specification of coordinates

Different way of choosing coordinates than ADM/BSSN:

ADM/BSSN: Choose α, βi GH: Choose Hα

Relation between two methods:

∂tα− βkα,k = −α(Ht − βkHk + αK)

∂tβi − βkβi,k = αγij

[α(Hj + γk`Γjk`)− α,j

Choice of Hα determines evolution of lapse and shift.Expect any ADM/BSSN gauge approachable by choice of Hα.

Generalized Harmonic: How to choose H?

Damped harmonic gauge works well for BHs:Spatial coords xi obey damped wave eqn.γ1/2/α driven towards unity.

To accomplish this,

Hµ = η0 ln

(γ1/2

)nµ − µ0

αγµkβ

η0 and µ0 are constants.

For even more damping,

choose µ0 = η0 =[ln γ1/2

Generalized Harmonic: How to choose H?

Damped harmonic gauge works well for BHs:Spatial coords xi obey damped wave eqn.γ1/2/α driven towards unity.

To accomplish this,

Hµ = η0 ln

(γ1/2

)nµ − µ0

αγµkβ

η0 and µ0 are constants.

For even more damping,

choose µ0 = η0 =[ln γ1/2

Handling singularities inside black holesMethod 1: Excision

Problem: black holes contain physical singularities.

Here beDragons

Horizon

Excision Boundary

Horizon Excision: Solve equations only in the exterior region.

Causality: interior does not affect exterior.If all characteristic speeds subluminal (like in generalizedharmonic), no boundary condition needed.

Here beDragons

Horizon

Excision Boundary

Horizon Excision: Solve equations only in the exterior region.

Causality: interior does not affect exterior.If all characteristic speeds subluminal (like in generalizedharmonic), no boundary condition needed.

Here beDragons

Horizon

Excision Boundary

Horizon Excision: Solve equations only in the exterior region.Causality: interior does not affect exterior.If all characteristic speeds subluminal (like in generalizedharmonic), no boundary condition needed.

Apparent horizons

Event horizon (EH)Boundary of region where photons can escape to infinity.Nonlocal in time.

Apparent horizon (AH)Smooth closed surface of zero null expansion. ∇µkµ = 0Local in time.

grey=EH; red,green=AH

Theorem: if an AH exists,it cannot be outside an EH.

Handling singularities inside black holesMethod 2: Moving punctures

What are ’punctures’?

Singularity

Puncture Initial Data

Excision Initial Data"other universe"

our universe

Numerical

Embedding

Diagram

Singularity moves through the grid during evolution!Key to success is carefully-chosen coordinate conditions.

Singularity

our universe

Numerical

Embedding

Diagram

Singularity

our universe

Numerical

Embedding

Diagram

Current status of numerical relativity

About a dozen working codes in existence

BSSN Generalized Harmonic

Finite Differencing

Moving Punctures

Form of Equations:

Singularity Treatment:

Numerical Methods:

Spectral

Excision

Most codes BAM

Princeton, AEI Harmonic Code SXS Collaboration (SpEC)

Comparing different codes improves confidence in results.

Finite Differencing

Moving Punctures

Form of Equations:

Numerical Methods:

Spectral

Excision

Most codes

Finite Differencing

Moving Punctures

Form of Equations:

Numerical Methods:

Spectral

Excision

Most codes BAM

Finite Differencing

Moving Punctures

Form of Equations:

Numerical Methods:

Spectral

Excision

Most codes BAM

Princeton, AEI Harmonic Code

SXS Collaboration (SpEC)

Finite Differencing

Moving Punctures

Form of Equations:

Numerical Methods:

Spectral

Excision

Most codes BAM

Finite Differencing

Moving Punctures

Form of Equations:

Numerical Methods:

Spectral

Excision

Most codes BAM

BSSN and GH codes can run 3D GR simulations forBinary black holesBinary neutron starsBH/NS binariesSupernovae

. . . and compute waveforms.

Challenges:Simulate more orbits for binaries.Extreme parameters (e.g. large spins) still difficult.Better accuracy/resolution.Interpretation of results (coordinates are just coordinates!)Include more physics in NSNS/BHNS/CCSNe simulations.

NeutrinosMagnetic fieldsNuclear reaction rates

BBH example Color = Vorticity(a measure of spin)

Mass ratio 6Large hole spin ∼ 0.91Small hole spin ∼ 0.3

Event Horizon: Andy Bohn Movie: Curran MuhlbergerMark A. Scheel Numerical Relativity July 7, 2015 32 / 34

Simulation catalogs

Goal: cover 7D parameter space of BBHs.

(179 simulations,SXS collaboration)

Movie:Haroon Khan

Several groups have catalogs.

We have not covered these important topics:

Initial data (solving constraints)Boundary conditionsFinding (apparent and event) horizonsExtracting gravitational waves from simulationsMeasuring spins, masses, energy, angular momentumHydrodynamicsNumerical methods (see hands-on exercises)

Lecture on Numerical Relativity

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