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2 Newtonian (Newton ) Post-Newtonian Post-Newton () Newtonian Post-Newtonian NewtonianPost-Newtonian

3 7 37 3

4 1 1.1 (cosmic censorship hypothesis) Post-Newton

5 1: 1 2:

6 2 [1, 2] m I d 2 r dt 2 = f (2.1) (2.1) m I (2.1) f = Gm Gm G r 2 r r (2.2) (2.2) m G (2.1) m G m I r g(r) r = r 1 2 g(r)t2 (2.3) m I d 2 r dt 2 = m d 2 r I dt m Ig(r) = (m 2 G m I )g(r) = 0 (2.4) r ( ) 2.2 6

7 2 x µ x µ + dx µ ds ds 2 = 3 µ=0 ν=0 3 g µν (x)dx µ dx ν (2.5) x i (i = 1, 2, 3) x 0 t x µ (µ = 0, 1, 2, 3) ds 2 4 (2.5) g µν (x) (2.5) ds 2 = g µν (x)dx µ dx ν (2.6) 1 x g µν (2.6) x g µν(x ) ds 2 = g µν(x )dx µ dx ν (2.7) 2 g µν (x)dx µ dx ν = g µν(x )dx µ dx ν (2.8) g αβ(x ) = g µν (x) xµ x α x ν x β (2.9) 2.3 : x S(x) S (x ) S(x) = S (x ) (2.10) : (2.11) A µ (x ) = xµ x α Aα (x) (2.11) 7

8 : (2.12) B µ (x ) = xα x µ B α (x) (2.12) : n m ( ) ( ) ( T µ 1 µ m x µ 1 x µ m x β 1 ) ( x β n ) ν 1 ν n (x ) = T α 1 α m x α β1 β 1 x αm x ν n (x) (2.13) n x ν φ(x) φ(x) x µ V µ ν V µ = V µ x ν Γα µνv α (2.14) ν V µ = V µ x ν + Γµ ανv α (2.15) Γ α µν Γ α µν = 1 ( gµβ 2 gαβ x + g νβ ν x g ) µν (2.16) µ x β β T µ ν = T µ ν x β + Γµ αβt α ν Γ α νβt µ α (2.17) 2.5 α β V µ α β V µ = α ( β V µ ) + Γ µ λα( β V λ ) Γ ν βα( ν V µ ) ( α β β α )V µ (2.18) = α β V µ + α Γ µ λβv λ + Γ µ λβ α V λ + Γ µ λα( β V λ ) Γ ν βα( ν V µ ) = α β V µ + ( α Γ µ λβ + Γ µ ραγ ρ λβ)v λ + (Γ µ λβ α V λ + Γ µ λα β V λ ) Γ ν βα( ν V µ ) 8 (2.19)

9 α β 2 ( α β β α )V µ = ( α Γ µ λβ β Γ µ λα + Γ µ ραγ ρ λβ Γ µ ρβγ ρ λα)v λ (2.20) (2.20) R µ λαβ α Γ µ λβ β Γ µ λα + Γ µ ραγ ρ λβ Γ µ ρβγ ρ λα (2.21) R αβ R µ αµβ (2.22) R R µ µ = g αβ R µ αµβ (2.23) 2.6 g µν G µν (g αβ ) = κt µν (2.24) κ T µν (2.24) ν T µν = 0 ν G µν = 0 G µν γ R µ ναβ + α R µ νβγ + β R µ νγα = 0 (2.25) (2.25) g γν ν γ β µ γ R µγ αµ + α R µγ µγ + µ R µγ γα = 0 (2.26) 2 α R R αβµν = R βαµν (2.27) R αβµν = R αβνµ (2.28) 3 1 β (R αβ 1 ) 2 gαβ R = 0 (2.29) G µν = R µν 1 2 gµν R (2.30) 9

10 G µν (2.24) κ (2.24) (2.31) R µν 1 2 g µνr = 8πT µν (2.31) 10

11 3 [1, 2, 3] 3.1 x 0 = t, x 1 = r, x 2 = θ, x 3 = φ (3.1) ds 2 = f 1 (r)dt 2 + f 2 (r)dr 2 + f 3 (r)(dθ 2 + sin 2 θdφ 2 ) (3.2) dtdθdtdφdrdθdrdφ dtdrdtdθdtdφ r 2 = f 3 (r) r ds 2 = h 1 (r )dt 2 + h 2 (r )dr 2 + r 2 (dθ 2 + sin 2 θdφ 2 ) (3.3) h 1 (r )h 2 (r ) r 2 = f 3 (r) r r = g(r ) f 1 (r)f 2 (r) f 1 (g(r ))f 2 (g(r )) h 1 (r ) e ν(r ) h 2 (r ) e λ(r ) ν(r )λ(r ) ds 2 = e ν(r) dt 2 + e λ(r) dr 2 + r 2 (dθ 2 + sin 2 θdφ 2 ) (3.4) r r (3.4) 0 Γ 0 01 = Γ 0 10 = 1 2 ν, Γ 1 00 = 1 2 eν λ ν Γ 1 11 = 1 2 λ, Γ 1 22 = re λ, Γ 1 33 = e λ r sin 2 θ Γ 2 12 = Γ 2 21 = 1 r, Γ 3 13 = Γ 3 31 = 1 r, Γ2 33 = sin θ cos θ Γ3 23 = Γ 3 32 = cot θ r ( ) ν R 00 = e ν λ 2 + (ν ) 2 ν λ ν (3.6) r R 11 = ν 2 (ν ) 2 + ν λ λ (3.7) ( r ) λ R 22 = 1 e λ + re λ 2 ν (3.8) 2 R 33 = R 22 sin 2 θ (3.9) (3.5) 11

12 (2.31) g µν R νµ 1 2 gµν g νµ R = 8πg µν T νµ (3.10) R = 8πT (3.11) (3.11) (2.31) R µν = 8π ( T µν 1 ) 2 T g µν (3.12) R µν = 0 (3.6)(3.7) ν + λ = 0 ν + λ = 0 (3.13) (3.4) r 0 (3.13) (3.8) ν 1 + d dr (re λ ) = 0 e λ = 1 C r (3.14) ds 2 = ( 1 C ) ( dt C ) 1 dr 2 + r 2 (dθ 2 + sin 2 θdφ 2 ) (3.15) r r (3.15) C m M g 00 (3.17) (3.16) φ(r) = M r (3.16) g 00 = 1 2φ (3.17) g 00 = 1 + 2M r (3.18) (3.15) g 00 = (1 C/r) r (3.18) C = 2M ( ds 2 = 1 2M ) ( dt M ) 1 dr 2 + r 2 (dθ 2 + sin 2 θdφ 2 ) (3.19) r r (3.19) r 2M r = 0 r r = 2M r r < 2M g tt > 0g rr < 0 r = 2M ( r s ) 12

13 3.2 r s λ dθ dλ = dφ dλ = 0 (3.20) (3.19) (6) dt ( dr = ± 1 r ) 1 s (r > r s ) (3.21) r t = r + r s ln r r s + constant (3.22) t = (r + r s ln r r s + constant) (3.23) (6) r = r 2 (> r s ) r = r 1 (< r 2 ) r1 ( t = 1 r ) 1 r2 ( s dr = 1 + r ) s/r dr r 2 r r 1 1 r s /r = r 2 r 1 + r s ln r 2 r s r 1 r s (3.24) (3.24) r 2 r s r s r 2 r = r s 3.3 (event horizon) 3.3 (3.19) r = r s (trθφ) t t t = t + r s ln(r r s ) (3.25) ( trθφ) (3.25) (3.23) d t dr t = r + constant (3.26) = 1 (3.27) 45 (3.25) d t = dt + 13 r s r r s dr (3.28)

14 (3.19) ( ds 2 = 1 2M ) d t 2 + 4M ( r r d tdr M ) dr 2 + r 2 (dθ 2 + sin 2 θdφ 2 ) (3.29) r (3.29) r = r s 3.4 r = r s r = 0 R µνλσ R µνλσ = 48M 2 r 6 (3.30) r = ( [4]) 14

15 4 ( c G 1 ) Einstein Newton Newtonian Newton ( NewtonianPost-Newtonian ) 4.1 Newtonian (Newton ) F Mm r F = G Mm r 2 r r (4.1) r d 2 x i dt 2 = N j i,1 j N m j (x j x i ) (x j x i ) 3 (4.2) N x j m j (4.2) (4.2) dx i dt dv i dt = v i (4.3) = N j i,1 j N m j (x j x i ) (x j x i ) 3 (4.4) G = c = 1L = [m] = 1AU M = L c2 G [kg] (4.5) T = L c [s] (4.6) 15

16 [kg] [s] 1 (4.5)(4.6) ( [kg])/( [kg]) = (4.7) ( [s])/( [s]) = (4.8) x = 1.0y = 0z = 0 t = ( 1) 3: (Newtonian) 1 1 p = N m i v i (4.9) i=1 N i=1 (%) = m iv i N i=1 m 100 (4.10) i v i 4.2 Post-Newtonian Post-Newton Post-Newton v c ε (v/c) g 00 = 1 + 2U (4.11) g ij = δ ij (i, j = 1, 2, 3) (4.12) 16

17 2 g 00 = 1 + 2U 2U 2 (4.13) g ij = (1 + 2U)δ ij (4.14) U = φ φ (3.16) (t, x, y, z) x y z x min x max φ(x, y, z) = N i=1 M i (x xi ) 2 + (y y i ) 2 + (z z i ) 2 (4.15) x i y i z i M i N (4.13)(4.14) g µν txyz xyz x f(x, y, z) = f(x + x, y, z) f(x x, y, z) 2 x (4.16) x f(x min, y, z) = 2 x f(x min + x, y, z) x f(x min + 2 x, y, z) (4.17) x f(x max, y, z) = 2 x f(x max x, y, z) x f(x max 2 x, y, z) (4.18) 17

18 t = 0 0 t = t f(t) f(t t) t f(t) = (4.19) t t = 2 t 2 t f(t) = 3f(t) 4f(t t) + f(t 2 t) 2 t (4.20) () d 2 x α dτ + dx µ dx ν 2 Γα µν dτ dτ = 0 (4.21) 8 (x p, y p, z p ) (x p, y p, z p ) 8 8 (x n, y n, z n )(x n + x, y n + y, z n + z) z 4 4 abcd a = x p x n (4.22) b = x a (4.23) c = y p y n (4.24) d = y c (4.25) Γ(x p, y p, z n ) ( Γ(x p, y p, z n ) = Γ(x, y, z n ) b x + Γ(x n + x, y n, z n ) a x + ) d y ( Γ(x n, y n + y, z n ) b x + Γ(x n + x, y n + y, z n ) a x ) c y (4.26) Γ(x p, y p, z n + z) ( Γ(x p, y p, z n + z) = Γ(x n, y n, z n + z) b x + Γ(x n + x, y n, z n + z) a ) d x y ( + Γ(x n, y n + y, z n + z) b x + Γ(x n + x, y n + y, z n + z) a ) c x y (4.27) 18

19 (4.26) (4.27) Γ(x p, y p, z p ) Γ(x p, y p, z p ) = Γ(x p, y p, z n ) z n + z z p z + Γ(x p, y p, z n + z) z p z n z (4.28) Γ(xn, y n+ y, zn) Γ(xn+ a, y n+ y, zn) Γ(xn+ x, y n+ y, zn) d (xp, y p, zn) c a b Γ(xn, y n, zn) Γ(xn+ a, y n, zn) Γ(xn+ x, y n, zn) 4: (z ) (4.21) 4 (4.21) dx α dt du α dt = u α (4.29) = Γ α µνu µ u ν (4.30) xyz u 1 u 2 u 3 xyz u 0 (4.31) u µ u µ = 1 (4.31) g 00 u 0 u 0 + g 11 u 1 u 1 + g 22 u 2 u 2 + g 33 u 3 u 3 = 1 (4.32) u 0 u 0 = 1 g 11 u 1 u 1 g 22 u 2 u 2 g 33 u 3 u 3 g 00 (4.33) Newtonian 5 ( 1) 19

20 5: (Post-Newtonian)

21 5 5.1 (event horizon) 3 (apparent horizon) M v = 2M r (5.1) r (5.1) v v c 5.3 ds 2 = 0 τ λ d 2 x α dλ + dx µ dx ν 2 Γα µν dλ dλ = 0 (5.2) 21

22 (3.29) dθ = dφ = 0 (3.29) ( 1 2m r ) d t 2 + 4m ( r d tdr m r ) dr 2 = 0 (5.3) (5.3) dλ 2 ( 1 2m r ) d t 2 dλ 2 + 4m r d t dr dλ dλ + ( 1 + 2m r ) dr 2 dλ 2 = 0 (5.4) d t dλ = u0 = 1 (5.4) dr dλ = u1 2 u 1 m = 1 6 6: 6 r = 2 r = 2 backward photon method( [5]) backward photon method backward photon method d t dλ = u0 =

23 7: backward photon method 7 forward time 6 backward photon method 23

24 t = 0 1 r = M = r = 2M r = free-fall time free-fall time 0 t ff t ff = 3π 32Gρ (6.1) t ff Newtonian ( ρ 0 t = 0 ) 8: ( r = 0.06) 24

25 図 9: ニュートンの運動方程式を用いて時間発展したときの粒子のスナップショット (粒子分布を xy 座標に射影したもの) 25

26 10 t = 1.05t ff r = 0.2 r = 0.2 t ff 10: t = 1.05t ff 11 11: Newtonian 26

27 12 t = 1.05t ff r > : t = 1.05t ff ( r = 0.06) 27

28 6.1.2 Post-Newtonian (x n, 0, 0) g xx 13 t = 1.05t ff t = 1.29t ff t = 1.29t ff 13: g xx t = 1.29t ff r = 0.3 ( 14) 14: t = 1.29t ff 28

29 図 15: 測地線方程式を用いて時間発展したときの粒子のスナップショット (粒子分布を xy 座標に射影したもの) 29

30 r = 0.3 t = 1.29t ff 16 backward photon method x = 0 backward photon method x = 1 x 16: 16 r = 0.3 backward photon method Post-Newton 17 t = 1.29t ff r > t ff 30% Post-Newton Post-Newton 17: t = 1.29t ff ( r = 0.06) 18: 1 30

31 6.1.3 NewtonianPost-Newtonian NewtonianPost-Newtonian 20 r = 0.3 Newtonian Post-Newtonian Post-Newtonian r = Newtonian r = 0.3 Post-Newtonian r = 0.1 Newtonian 14 Post-Newtonian u 0 u 0 u 0 dt dτ dt u0 dτ 19: r : r

32 Post-Newtonian Post-Newton x = 0.06 x = v x v y v z ( (4.10)) 21: 21 free-fall time free-fall time Post-Newton Post-Newton 32

33 6.2 モデル 2 ドーナツ型分布モデル 2 では図 2 のように粒子 2500 個をドーナツ型に分布させたものを時間発展させた現在 5 次元時空でのブラックリング解の存在が知られており将来的にこのブラックリングの性質を研究するためにモデル 2 ではドーナツ型に分布した粒子を考えたこのシミュレーションは測地線方程式を用いたプログラムで時間発展を行いモデル 1 と同様にブラックホール形成の判定を行う粒子の質量はモデル 1 と同じ M = とし粒子の初速度は 0 としたまず始めに総質量をドーナツ型の体積で割った密度を式 (6.1) に代入しその free-fall time を tf f 0 とする結果として図 22 の粒子のスナップショットと図 23 の座標 (xn, 0, 0) での計量 gxx を示す図 22: ドーナツ型分布を時間発展したときの粒子のスナップショット (線状に崩壊するまで) 33

34 23: g xx () 22 t ff0 23 g xx 1 free-fall time t ff1 2 free-fall time t ff = t ff0 + t ff (x n, 0, 0) g xx 24: g xx () 34

35 図 25: ドーナツ型分布を時間発展したときの粒子のスナップショット (一点に崩壊するまで) 図 25 のスナップショットと図 24 の計量 gxx のグラフからもわかるように時刻 tf f 付近で原点に粒子が集中していることがわかるまた図 26 は時刻 t = 1.06tf f での脱出速度を示しているが r = 0.3 付近で脱出速度の値が大きくなっている尚モデル 2 の時間発展で脱出速度によるブラックホール形成の判定を行ったが脱出速度が光速を超えることはなかった図 26: t = 1.06tf f での脱出速度 35

36 27: 28: 27 2 t ff 30% 1 28 t ff t ff 0 t ff Post-Newton 36

37 7 2 Post-Newton 1 Post-Newton Post-Newton Post-Newton Post-Newton Post-Newton [1] (1996) [2] (2005) [3] Ray d InvernoIntroducing Einstein s RelativityClarendon Press(1992) [4] S. L. Shapiro and S. A. Teukolsky, Phys. Rev. Lett. 66, 994(1991) [5] Peter Anninos et al, Phys. Rev. Lett. 74, 630(1995) 37

gr09.dvi

gr09.dvi .1, θ, ϕ d = A, t dt + B, t dtd + C, t d + D, t dθ +in θdϕ.1.1 t { = f1,t t = f,t { D, t = B, t =.1. t A, tdt e φ,t dt, C, td e λ,t d.1.3,t, t d = e φ,t dt + e λ,t d + dθ +in θdϕ.1.4 { = f1,t t = f,t {