Post-Newtonian limit of general relativity

Transcription

1 Post-Newtonian limit of general relativity g 00 = 1+ 2 c 2 U + 2 c ttx U 2 + O(c 6 ), g 0j = 4 c 3 U j + O(c 5 ), g jk = jk 1+ 2 c 2 U + O(c 4 ), 0 U(t, x) :=G x x 0 d3 x 0, (t, x) :=G v02 U p 0 / 0 x x 0 X(t, x) :=G 0 x x 0 d 3 x 0, U j 0 v 0j (t, x) :=G x x 0 d3 x 0 d 3 x 0,

2 Post-Newtonian Hydrodynamics From r T =0 Post-Newtonian equation of hydrodynamics dvj dt jp j U + 1 apple 1 c 2 2 v2 + U + + j p v t p + 1 c 2 h (v 2 4U)@ j U v j 3@ t U +4v k U i +4@ t U j +4v k U j U k j + O(c 4 ) = + 1 ttx

3 N-body equations of motion Main assumptions: Bodies small compared to typical separation (R << r) isolated -- no mass flow ignore contributions that scale as R n assume bodies are reflection symmetric mass : m A d 3 x A position : r A (t) 1 xd 3 x m A A velocity : v A (t) 1 m A acceleration : a A (t) 1 m A A A vd 3 x = dr A dt ad 3 x = dv A dt r A (t) x A B x x r A (t)+ x C

4 N-body equations of motion A v j Ud 3 x = va 2 +2v A v + v j U A d 3 x A + X va 2 +2v A v + v j U B d 3 x B A =2H jk A vk A + X X m A va@ 2 j U B +2T j U B B B

5 N-body equations of motion Dependence on internal structure? T A 1 v 2 d 3 x, P A pd 3 x, 2 A A A 1 2 G 0 x x 0 d3 x 0 d 3 x, EA int d 3 x A Use the virial theorem: 2T A + A +3P A =0 Then all structure integrals for body A disappear and those for the external bodies can be absorbed into a single total mass: A M B m B + 1 c 2 T B + B + E int B + O(c 4 ) This is a manifestation of the Strong Equivalence Principle, satisfied by GR, but not by most alternative theories. The motions of all bodies, including NS and BH, are independent of their internal structure in GR!

6 N-body equations of motion a A = X B6=A GM B r 2 AB + 1 c 2 8> : X B6=A n AB GM B r 2 AB apple v 2 A 4(v A v B )+2v 2 B 3 2 (n AB v B ) 2 + X B6=A + X B6=A 7 2 X GM B r 2 AB X C6=A,B X B6=A C6=A,B h 5GM A r AB i n AB (4v A 3v B ) G 2 M B M C r 2 AB G 2 M B M C r AB r 2 BC apple 4 r AC + 1 r BC 4GM B r AB (v A v B ) r AB 2r 2 BC n BC 9> ; + O(c 4 ). n AB (n AB n BC ) n AB

7 Post-Newtonian limit of general relativity g 00 = 1+ 2 c 2 U + 2 c 4 U 2 + O(c 6 ), g 0j = 4 c 3 U j + O(c 5 ), g jk = jk 1+ 2 c 2 U + O(c 4 ), = + 1 ttx 0 U(t, x) :=G x x 0 d3 x 0, (t, x) :=G v02 U p 0 / 0 x x 0 X(t, x) :=G 0 x x 0 d 3 x 0, U j 0 v 0j (t, x) :=G x x 0 d3 x 0 d 3 x 0,

8 Post-Newtonian geodesic equation d 2 r d 2 + dr d dr d =0 Use 0 component d 2 t/dλ 2 to change from λ to t dv dt = v c 0 v v λ = proper time (timelike) = affine parameter (null) v α = (c,v) Timelike test body v 2 ~ U dv j dt ju + 1 apple c 2 (v 2 4U)@ j U 4v k U +3@ t U v j 4v j U k U j +4@ t U j j + O(c 4 )

9 Lightlike test body v 2 ~ c 2 ds 2 =0= Post-Newtonian geodesic equation 1 v 2 = c c 2 U + O(c 4 ) c 2 dt c 2 U + O(c 3 ) 1+ 2 c 2 U + O(c 4 ) v 2 dt 2 8 c 2 U j v j dt 2 ) v = c 1 2 c 2 U n Geodesic equation with v ~ c dv j dt = 1+ v2 c j U 4 c 2 vj v k U + O(c 3 ) dn j dt = 2 c jk n j n k U + O(c 2 ) These will be used for the deflection of light, Shapiro time delay and gravitational lenses

10 N-body equations of motion a A = X B6=A GM B r 2 AB + 1 c 2 8> : X B6=A n AB GM B r 2 AB apple v 2 A 4(v A v B )+2v 2 B 3 2 (n AB v B ) 2 + X B6=A + X B6=A 7 2 X GM B r 2 AB X C6=A,B X B6=A C6=A,B h 5GM A r AB i n AB (4v A 3v B ) G 2 M B M C r 2 AB G 2 M B M C r AB r 2 BC apple 4 r AC + 1 r BC 4GM B r AB (v A v B ) r AB 2r 2 BC n BC 9> ; + O(c 4 ). n AB (n AB n BC ) n AB

11 PN 2-body equations of motion Define: r r 1 r 2 m M 1 + M 2 a = v v 1 v 2 a a 1 a 2 Gm r 2 n Gm apple c 2 r 2 (1 + 3 )v ṙ2 2(2 + ) Gm r 2(2 )ṙv + O(c 4 ), n r/r M 1 M 2 (M 1 + M 2 ) 2 ṙ dr/dt = n v n q Geodesic equation: η = 0, more general potentials U, ψ, X q PN 2-body equation: general masses, ignore moments

12 PN 2-body equations of motion: conserved quantities Conserved energy: dot v into the equation of motion 1 2 d dt v2 = = d dt Gm r 2 ṙ + Gm c 2 r 2 ṙ Gm d r dt apple (3 5 )v ṙ2 + 2(2 + ) Gm r Gm c 2 r apple(3 4 )v Gm 2 r ṙ2 (Gm) 2 r 3 ṙ = 1 2 d Gm 2 dt r Gm r 2 ṙv2 = d apple Gm Gm dt r r Gm r 2 ṙ3 = 1 apple d Gm 3 Gm 3 dt r r v 2 2v 2 ṙ 2

13 PN 2-body equations of motion: conserved quantities Conserved energy: dot v into the equation of motion 1 d 2 dt v2 = Gm r 2 ṙ + Gm apple c 2 r 2 ṙ (3 5 )v ṙ2 + 2(2 + ) Gm r = d Gm d Gm apple(3 dt r dt c 2 4 )v 2 9 Gm 1 r 2 r ṙ2 " := 1 2 v2 Gm r 3 2c 2 (1 3 ) d dt 2 1 Gm 2 v2 r c 2 8 (1 3 )v4 + Gm apple (3 + )v 2 + ṙ 2 + Gm 2r r + O(c 4 ) E SR = 1 m 1 c m 2 c 2

14 PN 2-body equations of motion: conserved quantities Conserved angular momentum: cross x with the equation of motion d (r v) =(4 dt h = Gm 2 ) c 2 ṙ(r v) r2 = d apple (4 2 ) Gm dt c 2 (r v) r apple d 1 3 dt c 2 1 Gm 2 v2 c 2 r 1+ 1 c 2 apple 1 2 (1 3 )v2 +(3+ ) Gm r (r v) (r v)+o(c 4 ) L SR = 1 m 1 (x 1 v 1 )+ 2 m 2 (x 2 v 2 )

15 a = Gm r 2 n Gm c 2 r 2 PN circular orbits 2(2 )ṙv + O(c 4 ), apple (1 + 3 )v ṙ2 2(2 + ) Gm r n dn/dt = d /dt = n v =ṙn + r a =( r r 2 )n +(2ṙ + r ) ṙ =0=) ( r =0 = = const 2 = Gm apple r 3 1 (3 ) Gm c 2 r + O(c 4 ) apple " = Gm 2r h = p Gmr 1 (7 )Gm 4 c 2 r + O(c 4 ) 1 apple 1+2 Gm c 2 r + O(c 4 ) Remark: test body in Schwarzschild coords (η = 0, r= r S - Gm/c 2 ) 2 = Gm r 3 S

16 N-body equations of motion: Worked example: 2 bodies and the perihelion shift Components of the disturbing force R = Gm apple c 2 r 2 (1 + 3 )v (8 )ṙ2 + 2(2 + ) Gm, r S = Gm h c 2 r 2 2(2 )ṙ(r ) i, W = /c for Mercury Integrate the Lagrange planetary equations: e = a =0 = =0! = 6 G(M 1 + M 2 ) a(1 e 2 )c O /yr for PSR

17 Mercury s Perihelion: Trouble to Triumph 1687 Newtonian triumph 1859 Leverrier s conundrum 1900 A turn-of-the century crisis 575 per century Planet Advance Venus Earth 90.0 Mars 2.5 Jupiter Saturn 7.3 Total Discrepancy 42.9 Modern measured value ± General relativity prediction 42.98

18 v = c 1 2 c 2 U n Deflection of light dn j = 2 dt c jk n j n k U U = G 0 r(t) x 0 d3 x 0 d j eroth order trajectory: r(t) =r e + ck(t t e )+O(c 2 ) To PN order, let n = k + + O(c 4 ) dt = 2G c = 2G c = 2G c 2 jk k j k k 0 b s 3 d3 x 0 0 b b 2 d dt 0 r k e + ck k (t t e ) x 0 r e + ck(t t e ) x 0 3 d3 x 0 s k d 3 x 0 s s e = r e x 0 s = r(t) x 0 b = s e k(s e k) k r e b r(t) Φ(t) α (t) = 2G c 2 (t, x 0 ) b b 2 s k s s e k s e d 3 x 0

19 For a point mass at the origin Deflection of light (t) = 2GM c 2 apple b r(t) k r(t) apple cos b 2 4GM c 2 b ˆb (t)+1 2 r e k r e For Φ 0 = 4GM c 2 b M/M = b/r k r e b r(t) Φ(t) α

20 Deflection of light: The 1919 Eclipse A. S. Eddington Sobral site Photo from Principe

21 Deflection of light: Results (1+γ)/ Optical Radio DEFLECTION OF LIGHT VLBI 2X (1+γ)/ Hipparcos Galactic Lensing 1.05 SHAPIRO TIME DELAY PSR Voyager Viking Cassini (1X10-5 ) YEAR OF EXPERIMENT 2010