Analytical and numerical treatment of black holesin horizon-penetrating coordinates

Maitraya K. Bhattacharyya1,2

withDr. David Hilditch3,4, Prof. Rajesh K. Nayak1,2

Dr. Hannes Ruter5,6 and Prof. Bernd Brugmann6

1Indian Institute of Science Education and Research Kolkata2Center of Excellence in Space Sciences India

3CENTRA, Instituto Superior Technico, Portugal4 Inter-University Centre for Astronomy and Astrophysics, Pune

5Max Planck Institute for Gravitational Physics, Germany6Theoretical Physics Institute, University of Jena, Germany

Southampton

Motivation

I Bridge gap between NR and perturbation theory →quantifying deviations important for waveform modeling

I QNM calculations are in Schwarzschild/Regge-Wheelercoordinates (Leaver, Andersson, Berti+) hence incompatiblewith horizon penetrating coordinates.

I Computation of the excitation amplitudes and tails from anyinitial configuration of the scalar field for all observers.

I Overtone excitation and detection for different types of initialdata.

I Compare with numerical results using bamps (Brugmann,Hilditch+).

QNMs and the confluent Heun equation

I Solve �Φ = 0 on a Schwarzschild background in Kerr-SchildI Φ(t, r , θ, φ) =

∑l ,m Kl ,m(t, r)Yl ,m(θ, φ).

I QNM boundary conditions (ω = ωM):

Kl ,m ∼ 1

re−iω(t+r), r → 2M, Kl ,m ∼ 1

re−iω(t−r)

( r

2M

)4iω, r → ∞.

I Frequency domain Green’s function (for single l):

Gl ,m(ω, r , r′) =

p(ω, r ′)

A(ω)

{f−(ω, r)f+(ω, r

′), r ≤ r ′,

f−(ω, r′)f+(ω, r), r ′ ≤ r ,

where p = r2(r − 2M)−4iωM and A = w(f−f

′+ − f ′−f+

).

I f± = e−iωrH(r/2M) with H being two linearly independent solutionsof the confluent Heun equation (Heun, Ronveaux, Fiziev), r = 2Mx :

d2H(x)

dx2+

(α+

β + 1

x+γ + 1

x − 1

)dH(x)

dx+

(µ

x+

ν

x − 1

)H(x) = 0.

I Using analytic continuation (Slavyanov+, Philipp+) and U-series solutions(Leaver):

f− =1

2Me4iωe iωr

( r

2M

)−1+4iω∞∑n=0

an

(r − 2M

r

)n

,

f+ =1

2M(−4i ω)−1+4iωΓ(1− 4i ω)e iωr

∞∑n=0

anΓ(1 + n − 4i ω)U(1 + n − 4i ω, 1,−2iωr).

I Both {an} → same three term recurrence → QNMs continuous fractionequation → same result as Leaver!

I Initial data evolves according to:

Kl ,m(t, r) =

∫Gl ,m(t, r , r

′)∂tKl ,m(0, r′)dr ′ +

∫∂tGl ,m(t, r , r

′)Kl ,m(0, r′)dr ′.

I Calculate separately individual contribution of poles and branch cut:

Gl ,m = GQl ,m + GB

l ,m + . . .

I Respect causality! Integration limits by solving:

r ′ + 4M log(r ′ − 2M) = r + 4M log(r − 2M)− t,

and r ′ = r + t for the upper limit.

Exact solution tests

0 5 10 15 20 25 30

x

−0.004

−0.002

0.000

0.002

0.004

ΦTime: 0.0 M

0 5 10 15 20 25

x

−0.004

−0.002

0.000

0.002

0.004

Φ

Time: 7.5 M

0 5 10 15 20

x

−0.004

−0.002

0.000

0.002

0.004

Φ

Time: 15.0 M

bamps data exact data

0 2 4 6 8 10 12time

1012

1010

108

106

104

102

erro

r

9 11 13 15 17 19

0 2 4 6 8 10 12time

1013

1011

109

107

105

103

101

erro

r

I Very accurate QNM extraction: 0.11043074− 0.10485913i (< 0.01%error) for n = 0 and 0.0857− 0.3472i (< 0.1% error) for n = 1.

Branch cut contributionI Branch cut contribution:

GB(t, r , r ′) =1

2π

∫ −i∞

0p(ω, r ′)f−(ω, r

′)

[f+(ωe

2πi , r)

A(ωe2πi )− f+(ω, r)

A(ω)

]e−iωtdω.

I Simplify expression for asymptotic observers → use Whittakersolutions.

I Late times → low frequency expansion of Whittaker functions →BesselJ!

I Approximate Green’s function for asymptotic observers:

GB(t, r , r ′) ≈ C1r l r ′l+2

η(t, r , r ′)2l+3FΛ

({l};− 2r

η(t, r , r ′),− 2r ′

η(t, r , r ′)

), η = t − f1(r , r

′).

I Alternative expression:

GB(t, r , r ′) ≈∞∑

m=0

C2r′l+2r l+2mΓ(2l + 2m + 3) 2F1

({l ,m}; r ′2

r2

)m!Γ

(l +m + 3

2

)(t − f2(r , r ′))2l+2m+3

I Obtain familiar power law at late times:

GB ∼ 1

t2l+3.

Tail tests

600 650 700 750 800 850 900 950 1000time

10-3

10-2

10-1

|Φ|

1 2 3 4 5 15

1000 1500 2000 2500 3000 3500time

10-12

10-11

10-10

10-9

10-8

10-7

10-6

10-5

10-4

10-3

|Φ|

0 1 2 3 4

500 1500 2500 350010-6

10-5

10-4

10-3

10-2

10-1

101 102 103

time

10-9

10-8

10-7

10-6

10-5

10-4

10-3

10-2

10-1

100

erro

r

A B C D E F G

500 1000 1500 2000 2500 3000 3500time

2

3

4

5

6

7

8

9

10

11

λ

H D E F G

A B C D E F G

10-4

10-3

QNM excitation factors

I QNM contribution to Green’s function:

GQ(t, r , r ′) =∞∑l=0

∞∑n=0

Bl ,ne−iωl,nt

{p(ω, r ′)f−(ωl ,n, r)f+(ωl ,n, r

′), r ≤ r ′,

p(ω, r ′)f−(ωl ,n, r′)f+(ωl ,n, r), r ′ ≤ r .

I Can now calculate excitation amplitudes for all observers.I Assume near the poles:

A(ωl ,n) ≈ (ω − ωl ,n)A′(ωl ,n).

I Some QNEfs:

l n A′(ωl ,n) Bl ,n

0 0 1.32962 + 3.01240i 0.55566 + 0.24526i1 4.37158 + 0.92283i 0.09244 + 0.43798i

Results

40 60 80 100 120 140time

10-5

10-4

10-3

10-2

10-1

100

101

102

103

104

|Φ|

data n = 0 n = 1 n = 2 tail

70 80 90 100 110 120 130 140time

0

10

20

30

40

Φ

data overtones overtones + tail

70 80 90 100 110 120 130 140time

10-2

10-1

100

101

erro

r

I Overtones become more important at earlier times → n = 1 just anorder of magnitude less than n = 0 near start.

I Summing overtones not sufficiently good at intermediate and lateringing.

I Late time tail result works remarkably well for earlier times → errorbecomes one order less at intermediate and late time ringing.

I Need contribution from high frequency arc to explain the beginning→ unfortunately not so easy!

Results

I l = 0 ringdown short in general → branch cut contribution dominatesafter a few cycles

I n = 0 dominates for all ID → can recover ωQNM within ∼ 1% error.I Longer ringdown with sine-Gaussian ID → more reliable ωQNM

extracted (∼ 0.03%)I n = 1 for sine-Gaussian ID → larger errors (∼ 10%).

Ongoing workI Solve the conformal transverse traceless form of constraints for Ψ and

V i .I Hyperbolic relaxation (Ruter+): ∂2t ψ + ∂tψ = ∆ψI Conformal quantities → Schwarzschild quantities in Kerr-SchildI Evolve using generalized harmonic coordinates (Lindblom+)I Event horizon locator: Integrate ‘outgoing’ null geodesic backwards

(Bohn+)

1.8 2.0 2.2 2.4areal radius (M)

0

10

20

30

40

50

60

70

80

time

(M)

Variation of the horizon areal radius with time

1.8 2.0 2.2 2.4x (M)

0

10

20

30

40

50

60

70

80

time

(M)

Variation of horizon position with time

with scalarfield Awith scalarfield Bwithout scalarfield

0 50 100 150time (M)

107

106

105

104

103

102

101

||

Scalarfield A response at horizon

I Excision fails for ‘large’ perturbations → fix!

Thanks!