Distance to Ringing BHs: New GW Astronomy and Gravity Test Kunihito Ioka (KEK) Hiroyuki Nakano (RIT) Hirotaka Takahashi (Nagaoka U. Tech.)

Flanagan &Hughes(98) S/N~100 S/N~10 5 Black Hole Coalescence E GW /M~1% Pretorius(05) ~h

今回のような連星を サーチできる範囲 連星の活動期間 連星 NS を 検出できる距離

Distance Measurements Distance is important: ・ Andromeda galaxy ・ Gamma-Ray Burst ・ Dark Energy … Inspiral Ringdown Distance Energy E QNM and Distance r are degenerated ⇒ No r? However, with 2nd order QNM ⇒ We can measure r

BH Quasi-Normal Mode l=2 l=3 a=0 Kokkotas&Schmidt(99) purely outgoing Oscillation+ Exp-decay Re ,Im  →M,J purely ingoing

Amplitude is Finite Amplitude BH deforms appreciably ⇒ Higher-order QNM In the BH coalescence ※ c=1 h (2) ~0.1 h (1)

Anharmonic oscillation Let us go back to “Mechanics” by Landau & Lifshitz 2 nd order ∝a2∝a2 (  )(  ) The 2 nd order oscillation always exists and is completely determined by the 1 st order

Second-order QNM ~0.1 Prediction ~e -i  t ・  2nd =2 x  1st ・ 2 nd QMN always exists ・ (2 nd ) is determined by (1 st )  2nd =2 x  1st

Source term

  =(Special solution)+(Homogeneous solution) ・ The equation is the same as the 1 st one  2nd =  1st ・ Pure (2 nd ) separates from (1 st )x(1 st ) in the frequency domain ・ (2 nd ) is independent of the initial condition  (l=2,m=2) x  (l=2,m=2)  (l=2,m=2) x  (l=2,m=-2)  (l=2,m=-2) x  (l=2,m=2)  (l=2,m=-2) x  (l=2,m=-2) m=4,-4,0 for 2 nd QNM m=0 mode is “no oscillation” ∵ Re(  l,m +  l,-m )=0 ⇒ A Test of the General Relativity ※ Only polar (even) mode for simplicity The most dominant (1 st ) terms are

KI & Nakano 07

Numerically confirmed Initial Value Problem Follow the time evolution With 1 st order subtracted Okuzumi, KI & Sakagami 08

Basic Idea Distance r But … not so easy

From theory of 2 nd order QNM i=1, 2 次 (m=0,±4) Waveform

1 st : even: (l=2,m=±2) 2 nd : even: (l=4,m=±4), (l=4 or 2,m=0) どちらも sin 2  倍

Observable Waveform Antenna pattern Sky positionPolarization angle 10 Unknown parameters Label for detectors Ready to be determined

答えは Ａ．２台の検出器 m=±4, m=0 の両方の２次を 使えば解ける  r/r~1/SNR min (2 nd ) Ｂ．３台の検出器 m=±4 の２次を使えば解ける  r/r~1/SNR max (2 nd )+1/SNR min (1 st )

２次の m=±4 モードだけを使うと， C の数は [(1),(2)] ⊗ [+,×] ⊗ [  =1,2]=8 未知数は 8 しかし，縮退が起こって解けない。 ⇒ ２次の m=0 or ３台目

1. t 0 : arbitrary (  (2) =2  (1) 、 A (2) ∝ [A (1) ] 2 、  (2) =2  (1) +…) 2. C 1+ (1) /C 1+ (2) =C 2+ (1) /C 2+ (2) ⇒   （ + ⇔ × でもＯＫ = Gravity test ） 3. C 1+ (1) / C 1+ (2) （ = C 1× (1) / C 1× (2) ） C 1+ (1) / C 1+ (2;m=0) ⇒ A (1),  4. C 1+ (1) / C 1× (1), C 2+ (1) / C 2× (1), C 1+ (1) / C 2+ (1) ⇒  S,  S,  S 5. ⇒ r 6. + Inspiral = Gravity test

3 台あると， 1. t 0 : arbitrary 2. C 1+ (1) / C 2+ (1), C 2+ (1) / C 3+ (1), C 1× (1) / C 2× (1), C 2× (1) / C 3× (1) ⇒ ,  S,  S,  S 3. C 1+ (1) / C 1× (1) ⇒  4. C 1+ (1) / C 1+ (2) ⇒ A (1) 5. ⇒ r 6. + Inspiral = Gravity test ここまで １次のみ

LISA は 実は３台 ⇒ QNM 1 st だけで， 方向が 分かる． (New?)

まとめ ２次 QNM LISA や DECIGO では必ず必要 LIGO や LCGT でも必要になる ? 数値的にも確認 距離指標，重力テスト 偏光，アンテナパターンを考慮 2 台でもいけるが 3 台以上が良い 課題 : Kerr, m=0, odd

23 １次と２次で full non-linear の結果をほぼ再現 前半２次ＱＮＭが現れ、後半は２次 tail が卓越

Baker & Centrella 04

Berti+07