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Distance to Ringing BHs: New GW Astronomy and Gravity Test Kunihito Ioka (KEK) Hiroyuki Nakano (RIT) Hirotaka Takahashi (Nagaoka U. Tech.)
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Flanagan &Hughes(98) S/N~100 S/N~10 5 Black Hole Coalescence E GW /M~1% Pretorius(05) ~h GW @1Gpc
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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
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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
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BH Quasi-Normal Mode l=2 l=3 a=0 Kokkotas&Schmidt(99) purely outgoing Oscillation+ Exp-decay Re ,Im →M,J purely ingoing
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Amplitude is Finite Amplitude BH deforms appreciably ⇒ Higher-order QNM In the BH coalescence ※ c=1 h (2) ~0.1 h (1)
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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
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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
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Source term
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=(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
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KI & Nakano 07
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Numerically confirmed Initial Value Problem Follow the time evolution With 1 st order subtracted Okuzumi, KI & Sakagami 08
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Basic Idea Distance r But … not so easy
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From theory of 2 nd order QNM i=1, 2 次 (m=0,±4) Waveform
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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
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Observable Waveform Antenna pattern Sky positionPolarization angle 10 Unknown parameters Label for detectors Ready to be determined
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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 )
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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
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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
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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
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LISA は 実は3台 ⇒ QNM 1 st だけで, 方向が 分かる. (New?)
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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
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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
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Baker & Centrella 04
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Berti+07
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