Curved spacetime tells matter how to move

Transcription

1 Curved spacetime tells matter how to move Continuous matter, stress energy tensor Perfect fluid: T 1st law of Thermodynamics Relativistic Euler equation Compare with Newton =( c p)u u /c 2 + pg j = u r T =0, r j =0 u r T =0= d" d (µ + p) Du d dv dt ru = ρ = rest mass density ε = energy density p = pressure u α = four velocity +(" + p)r ~u d("v)+pdv =0 = c 2 g + u u /c 2 r p rp

2 Matter tells spacetime how to curve Riemann tensor + µ µ µ Ricci tensor R = R µ µ Ricci scalar R = g R Einstein tensor G = R 1 2 g R Bianchi identities Action S = Einstein s equations: r G =0 c3 16 G G p = 8 G c 4 grd 4 x + S matter T

3 Landau-Lifshitz Formulation of GR Post-Newtonian and post-minkowskian theory start with the Landau-Lifshitz formulation Define the gothic metric density g p gg Then Einstein s equations can be written in the µ H µ = 16 G c 4 ( g) T + t LL H µ g g µ g g Antisymmetry of H αµβν implies the conservation equation h ( g) T + t LL i =0 () r T =0

4 Landau-Lifshitz Formulation of GR ( g)t LL The Landau-Lifshitz pseudotensor := c4 16 G µ g µ g µ g g g µ g g g µ g g g µ + g µ g µ g g µ g g µ 2g g g µ g )

5 Landau-Lifshitz Formulation of GR Conservation equation allows the formulation of global conservation laws: E de dt = I ( g) T 00 + t 00 LL ( g)t 0j LL d2 S j d 3 x Similar conservation laws for linear momentum, angular momentum, and motion of a center of mass, with P j 1 c J j 1 c jkl X j 1 E ( g) T j0 + t j0 LL d3 x ( g)x k T l0 + t l0 LL ( g)x j T 00 + t 00 LL d 3 x d 3 x

6 The relaxed Einstein equations Define potentials h g Impose a coordinate condition (gauge): Harmonic or dedonder gauge Matter tells spacetime how to curve Spacetime tells matter how to h =0 g x ( ) =0 h = 16 G c 4 2 c 2 ( g) T [m,g]+t LL [h]+t H [h] ( g)t c4 H µ h µ h µ h 16 =0 Still equivalent to the exact Einstein equations

7 The relaxed Einstein equations h = 16 G c =0 Solve for h as a functional of matter variables Solve for evolution of matter variables to give h(t,x)

8 Iterating the Relaxed Einstein Equations Assume that h αβ is small, and iterate the relaxed equation: h N+1 = 16 G c 4 (h N ) h N+1 = 4G (h N )(t x x 0 /c, x 0 ) c 4 x x 0 d 3 x 0 Start with h 0 = 0 and truncate at a desired N Yields an expansion in powers of G, called a post-minkowskian expansion Find the motion of matter (h N )=0

9 Solving the Relaxed Einstein Equations = 4 µ =) = C µ(t x x 0 /c, x 0 ) x x 0 d 3 x 0 N : r 0 < R, W : r 0 > R R wavelength s/v = N + W

10 Solving the Relaxed Einstein Equations: Far zone Near zone integral: µ(t x x 0 /c, y) x x 0 N (t, x) = 1X `=0 N For x >> x, Taylor expand x-x = 1X `=0 apple ( 1)` L `! r ( 1)` `! M x 0L µ(t r/c, L. r µ(,x 0 )x 0L d 3 x 0 A multipole expansion = t R/c Integrals depend on R

11 Solving the Relaxed Einstein Equations: Far zone Far zone integral: W Since contributions to µ in the far zone come from retarded fields, they have the generic form µ f( 0, 0, 0 )/r 0n Change variables from (r, θ, φ ) to (u, θ, φ ), where u = cτ = ct -r u 0 + r 0 = ct x x 0

12 Solving the Relaxed Einstein Equations: Far zone Far zone integral: W W = 1 4 u 1 du 0 I S(u 0 ) f(u 0 /c, 0, 0 ) r 0 (u 0, 0, 0 ) n 2 d 0 ct u 0 n 0 x Integral also depends on R But = N + W is independent of R

13 Gravity as a source of gravity and gravitational tails

14 Solving the Relaxed Einstein Equations: Near zone Near zone integral: For x ~ x, Taylor expand about t 1X µ(t x x 0 ( /c) = N (t, x) = 1X `=0 N ( 1)` ` M ` µ(t, x 0 ) x x 0 ` µ(t, x 0 ) x x 0 ` 1 d 3 x 0 A post-newtonian expansion in powers of 1/c Instantaneous potentials Must also calculate the far-zone integral W

15 Post-Newtonian approximation: Near zone Newtonian Illustration: plus expand h 00 in the near zone: corrections up to 2.5 PN order within τ 00 No 0.5 PN term: conservation of M 1 PN correction d 2 X/dt 2 h 00 N = 4G 00 c 4 M x x 0 d3 x c 2 00 x x 0 d 3 x 0 M 3 6c 3 00 x x 0 2 d 3 x c 4 00 x x 0 3 d 3 x c 5 M M 00 x x 0 4 d 3 x 0i + O(c 6 ) M 2 PN term Pure function of time a coordinate effect 2.5 PN term Gm rc 2 v2 c 2

16 Near zone physics; Motion of extended fluid bodies Matter variables: rescaled mass density : p proper pressure : p internal energy per unit mass : four g(u 0 /c) velocity of fluid element :u = u 0 (1, v/c) r ( + r( v)=0 Slow-motion assumption v/c << 1: T 0j /T 00 v/c, T jk /T 00 (v/c) 2 h 0j /h 00 v/c, h jk /h 00 (v/c) 2

17 Post-Newtonian approximation: Near zone Recall the action for a geodesic 2 S = mc 2 d 1 r = mc = mc We need to calculate 1 g dr dt dr dt dt Gm rc Uc 2 g 00 2 vj c g v 2 v i v j 0j c 2 c 2 ε ε 2 ε 2 ε ε 2 v2 c 2 g ij 1/2 dt g 00 to O( 2 ) g 0j to O( 3/2 ) g ij to O( ) Two iterations of the relaxed equations required

18 Post-Newtonian approximation: Near zone Conversion between h and g ( g) =1 h h2 2 hµ h µ + O(G 3 ), 1 g = + h 2 h + h µ h µ 1 + To 1PN order: 1 8 h2 1 4 hµ h µ 2 hh + O(G 3 ), g 00 = h hkk 3 8 h O(c 6 ), g 0j = h 0j + O(c 5 ), g jk = jk apple h00 + O(c 4 ),

19 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,

20 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 )

21 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

22 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 A Then all structure integrals can be absorbed into a single total mass: M A m A + 1 c 2 T A + A + E int A + 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!

23 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