THE ENERGY-MOMENTUM PSEUDOTENSOR T µν of matter satisfies the (covariant) divergenceless equation

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1 THE ENERGY-MOMENTUM PSEUDOTENSOR T µν of matter satisfies the (covariant) divergenceless equation T µν ;ν = 0 (3) We know it is not a conservation law, because it cannot be written as an ordinary divergence. In a locally inertial frame (LIF): eq. (3) becomes T µν = 0, (4) x ν this means that T µν can be written as: T µν = x αηµνα, (5) where η µνα is antisymmetric in ν and α; INDEED 2 η µνα x ν x = 0, α because the derivative operator is symmetric in ν and α We want to find the expression of η µνα : write Einstein eqs. G µν = 8πG c 4 T µν T µν = c4 8πG In a LIF R µν is: R µν = 1 2 gµα g νβ g γδ 2 g γβ x α x + 2 g αδ δ x γ x β R µν 1 2 gµν R. (6) 2 g γδ x α x 2 g αβ β x γ x δ By replacing in eq. (6), T µν becomes T µν = c 4 1 [ ( x α ( g) g µν g αβ g µα g νβ)] 16πG ( g) x (7) β The part within { } is antisymmetric in ν and α, symmetric in µ and ν, and it is the quantity η µνα we were looking for..

2 T µν = x αηµνα, η µνα = c 4 1 [ ( ( g) g µν g αβ g µα g νβ)] 16πG ( g) x β 1 Since in a LIF g µν,α = 0 we can extract ( g) write this equation as and where ζ µνα = ( g)η µνα = ( g)t µν = ζµνα x α, (8) c4 [ ( ( g) g µν g αβ g µα g νβ)]. (9) 16πG x β EQ. (8) has been derived in a locally inertial frame. In any other frame ( g)t µν will not equate ζµνα x α, therefore, in a generic frame ζ µνα x α ( g)tµν 0. We shall call this difference ( g)t µν, i.e. ( g)t µν = ζµνα x α ( g)tµν. The quantities t µν are symmetric, because T µν and ζ µνα x α are symmetric in µ and ν. It follows that ( g) (T µν + t µν ) = ζµνα x α, x ν [( g) (T µν + t µν )] = 0, this is THE CONSERVATION LAW OF THE TOTAL EN- ERGY AND MOMENTUM OF MATTER + GRAV- ITATIONAL FIELD VALID IN ANY REFERENCE FRAME.

3 ( g)t µν = ζµνα x α ( g)tµν. If we express T µν in terms of g µν and by using Einstein s eqs. G µν = 8πG c T µν T µν = c4 R µν 1 4 8πG 2 gµν R and eq. (9) it is possible to show that t µν can be written as follows. t µν = c4 {( 2Γ δ αβ Γ σ δσ Γ δ ασγ σ βδ Γ δ αδγ σ ( βσ) g µα g νβ g µν g αβ) 16πG + g µα g βδ (Γ ν ασγ σ βδ + Γ ν βδγ σ ασ Γ ν δσγ σ αβ Γ ν αβγ σ δσ) + g να g βδ (Γ µ ασγ σ βδ + Γ µ βδγ σ ασ Γ µ δσγ σ αβ Γ µ αβγ σ δσ) + g αβ g δσ (Γ µ αδγ ν βσ Γ µ αβγ ν δσ) } This is the stress-energy pesudotensor of the gravitazional field. t µν it is not a tensor because : 1) it is the ordinary derivative, (not the covariant one) of a tensor 2) it is a combination of the Γ s that are not tensors. However, as the Γ s, it behaves as a tensor under a linear coordinate transformation.

4 Let us consider an emitting source and the associated 3-dimensional coordinate frame O (x, y, z). Be an observer located at P = (x1, y1, z1) at a distance r = x1 2 + y1 2 + z1 2 from the origin. The observer wants to detect the wave coming along the direction identified by the versor n = r r. y y P x n z x z Consider a second frame O (x, y, z ), with origin coincident with O, and having the x -axis aligned with n. Assuming that the wave traveling along x direction is linearly polarized and has only one polarization, the corresponding metric tensor will be g µ ν = (t) (x) (y) (z) [1 + h TT + (t, x )] [1 h TT + (t, x )], The observer wants to measure the energy which flows per unit time across the unit surface orthogonal to x,

5 i.e. t 0x, therefore he needs to compute the Christoffel symbols i.e. the derivatives of h T µ T ν. The metric perturbation has the form h T T (t, x ) = const x f(t x c ), the only derivatives which matter are those with respect to time and x h TT t h TT ḣ TT = const x h TT = const x x 2 f, f + const x f 1 c const x f = 1 cḣt T, where we have retained only the dominant 1/x term. Thus, the non-vanishing Christoffel symbols are: Γ 0 y y = Γ0 z z = 1 2 Γ x y y = Γx z z = 1 2c ḣt + T Γ y 0y = Γz 0z = 1 2 ḣtt + ḣt T + Γ y y x = Γz z x = 1 2c ḣt T +. By substituting Christoffel s symbols in t µν, we find t 0x de GW dx 0 ds = c2 16πG dh TT (t, x ) 2. Thus de GW ds = c3 16πG dh TT (t, x ) 2.

6 In general, if both polarization are present g µ ν = (t) (x) (y) (z) [1 + h TT + (t, x )] h TT (t, x ) 0 0 h TT (t, x ) [1 h TT + (t, x )], t 0x = c2 16πG dh TT dh TT 2 = c2 32πG dh TT 2. This is the energy per unit time which flows across a unit surface orthogonal to the direction x. However, the direction x is arbitrary; if the observer il located in a different position and computes the energy flux he receives, he will find formally the same but with h T T referreed to the TT-gauge associated with the new direction. Therefore, if we consider a generic direction r = r n t 0r = c2 32πG dh TT (t, r) 2. (10) Since in GR the energy of the gravitational field cannot be defined locally, to find the GW-flux we need to average over several wavelenghts, i.e. de GW ds = ct 0r c 3 = 32πG dh TT 2. We shall now express the energy flux directly in terms of the quadrupole moment.

7 Since h TTµ0 = 0, µ = 0, 3 h TT 2G ik (t, r) = c 4 r d 2 2 QTT ik (t r c ) by direct substitution we find de GW ds = c 3 32πG G = 8πc 5 r 2 G = 8πc 5 r 2 dh TT 2 [ Q TT ( t r )] 2 c [P mn Q mn ( t r c)] 2. From this formula we can compute the gravitational luminosity L GW = de GW L GW = de GW ds ds = = G 2c 5 1 4π dω de GW ds r2 dω ( (P c)) mn Q mn t r 2. Let us compute this integral. By using the properties of P mn we find P mn Q P mnrs Q rs = P rs Q Q rs = ( δ jr δ ks n j n r δ ks n k n s δ jr n ) jn k n r n s Q Q rs = Q Q 2n k Q kr Q rs n s n jn k n r n s Q Q rs. The integrals of the n s over the solid angle are: 1 4π 1 dωni n j = 1 4π 3 δ ij dωni n j n r n s = 1 15 (δ ijδ rs + δ ir δ js + δ is δ jr )

8 so that 1 4π dω ( Q Q 2n k Q kr Q rs n s n jn k n r n s Q Q rs ) = 2 Q 5 Q and, finally, the power emitted in gravitational waves by an evolving source is L GW = de GW = G 5c 5 ( Q t r ) ( Q c t r ) c. where Q ij = q ij 1 3 δ ij q k k

9 We shall now compute the GW-luminosity of a binary system We shall use the formula: L GW = G 5c 5 Q Q. where Q ij = q ij 1 2 δ ijδ kl q kl = µ 2 l2 0 cos 2ωω K t sin 2ωω K t 0 sin 2ωω K t cos 2ωω K t and Q ij = µ 2 l2 0 8 ω K ω K 3 sin 2ωω K t cos 2ωω K t 0 cos 2ωω K t sin 2ωω K t ω K = GM l 3 0 Q Q = 32 µ 2 l 4 0 ω 6 K = 32 µ 2 G 3 M3 l 5 0 By direct substitution we find L GW de GW = 32 µ 2 M 3 5 c 5 l 5 0 and if m 1 = m 2 = m and consequently µ = m 2 and M = 2m L GW = 64G4 m 5 5c 5 l 5 0 G 4 For the binary pulsar PSR L GW = erg/s

10 Since we know how much energy is radiated in GW, we can compute the consequent variation of the orbital period The total energy of the system is E orb = 1 2 µω ω 2 K l 2 0 U, where U = Gm 1m 2 therefore and Since de orb E orb = 1 GµM 2 l 0 = 1 GµM 1 dl 0 2 l 0 l 0 l 0 = E orb 1 dl 0 l 0 = GµM l 0 ω 2 K = GMl 3 0 2lnωω K = lngm 3lnl 0 1 dωω K ω K = 3 2 and since 1 dωω K ω K and de orb = 1 dt T = 2 E orb dt 3 T ω K 1 dl 0 l 0 = T dt = 3 2 T E orb dt ω K de orb The energy lost in GWs must be compensated by a variation of orbital energy (adiabatic approximation 1 dl 0 l 0 de orb + L GW = 0 de orb = L GW we find dt = 3 2 T E orb L GW

11 dt = 3 2 T E orb L GW For PSR T = s, with these data we find E orb ergs, L GW erg/s dt By refining the calculations, using the equations of motion appropriate for an eccentric orbit with ɛ one finds dt = The observed value is dt = (± ) FIRST INDIRECT EVIDENCE OF THE EXISTENCE OF GRAVITATIONAL WAVES For PSRJ T = 8640 s, E orb ergs, L GW erg/s dt

12 ORBITAL EVOLUTION moreover, dt = 3 2 T 1 dωω K L GW E orb ω K ω K ω K = 1 dt T L GW de GW = 32 5 G 4 c 5 µ 2 M 3 l 5 0 E orb = 1 GµM 2 l 0 therefore ω K 1 dωω K ω K = 96 5 When integrated it gives G 3 µm 2. c 5 l 4 0 ω K (t) = ω KinK t 3/8 coal [t coal t] 3/8 ω in K = ω K (t = 0) (11) ( l in 0 ) 4 where t coal = 5 c 5 and l in 256 G 3 µm 2 0 = l 0 (t = 0) (12) The orbital frequency, and consequently the frequency of the emitted wave ν GW = ω K /π, change accordingly: the frequency increases with time l 3 0 ν GW (t) = νin GW t 3/8 coal [t coal t] 3/8 (13) Since ω K = GM, using eq. (11) the orbital separation is l 0 (t) = lin 0 t 1/4 coal [t coal t] 1/4 (14) the orbital distance decreases with time.

13 WAVEFORM: AMPLITUDE AND PHASE We have seen that if a binary system moves on a circular orbit and we look, for example, in the direction orthogonal to the orbital plane, the wave we would detect is h TT ij = 4µMG2 rl 0 c 4 A TT kl A TT ij = cos 2ωω K t sin 2ωω K t 0 sin 2ωω K t cos 2ωω K t and ω K = GM l 3 0 = π ν GW. We shall model the waveform emitted in the inspiralling as follows: 1) an instantaneous amplitude h 0 (t) = 4µMG2 rl 0 (t)c 4 = 4µMG2 rc 4 ω 2/3 K (t) G 1/3 M 1/3 where = 4π2/3 G 5/3 M 5/3 c 4 r ν 2/3 GW(t) M 5/3 = µ M 2/3 M = µ 3/5 M 2/5 = chirp mass Since the frequency increases in time the amplitude increases too. 2) Since the wave frequency is changing in time, the phase appearing in A TT ij, i.e. 2ωω K t, will be replaced by Φ(t) = t 2πνGW (t) + Φ in, where Φ in = Φ(t = 0)

14 The phase Φ(t) = t 2πνGW (t) + Φ in, where Φ in = Φ(t = 0) Since and ν GW (t) = νin GW t 3/8 coal [t coal t] 3/8 ν in t 3/8 coal = ( 5 3/8) 1 8π c 3 GM 5/8 then ν GW (t) = 1 8π c 3 GM 5/8 5 t coal t 3/8 and the integrated phase will be c 3 5/8 Φ(t) = 2 + Φ in 5GM (t coal t) If we know the phase we can measure the chirp mass Thus, the signal emitted during the inspiralling will be h TT ij = 4π2/3 G 5/3 M 5/3 c 4 r νgw(t)a 2/3 TT ij where A TT ij = cos Φ(t) sin Φ(t) 0 sin Φ(t) cos Φ(t)

15 LIGO[40 Hz 1 2 khz] VIRGO[10 Hz 1 2 khz] LISA[ ] Hz 1e-20 S h 1/2 [Hz -1/2 ] 1e-21 1e-22 GEO VIRGO LIGO 1e log ν GW (Hz) Let us consider 3 binary system a) m 1 = m 2 = 1.4 M b) m 1 = m 2 = 10 M c) m 1 = m 2 = 10 6 M Let us first calculate what is the orbital distance between the two bodies on the innermost stable circular orbit (ISCO) and the corresponding emission frequency l I 0SCO 6GM, ω c 2 K = GM l 3 0 = π ν GW ν ISCO GW = 1 π GM (l ISCO 0 ) 3 a) l 0 ISCO = 24, 8 km ν GW = Hz b) l 0 ISCO = 177, 2 km ν GW = Hz c) l 0 ISCO = , 3 km ν GW = Hz a) and b) are interesting for LIGO and VIRGO, c) will be detected by LISA

16 Let us consider LIGO and VIRGO, and let us compute the time a given signal spends in the detector bandwih before coalescence. From we get ν GW (t) = νin GW t3/8 coal [t coal t] 3/8 t = t coal 1 ν in GW ν GW (t) 8/3. Putting : ν in GW = lowest frequency detectable by the antenna, and ν max GW = ν ISCO GW we find (LIGO) (VIRGO) a) [ Hz] [ khz] t = s t = 16.7 m b) [ Hz] [ khz] t = 0.93 s t = s VIRGO catches the signal for a longer time!

17 WHAT ABOUT LISA? LISA [ ] Hz 1e-20 LISA 1-yr observation Detection threshold 1e-21 1e-22 1e-23 1e ν [ Hz ] Let us consider 2 BH-BH binary systems a) m 1 = m 2 = 10 2 M b) m 1 = m 2 = 10 6 M Orbital distance between the two bodies on the innermost stable circular orbit (ISCO) and the corresponding emission frequency l I 6GM 0SCO, ω c 2 K = GM l 3 0 = π ν GW ν ISCO GW = 1 π GM (l ISCO 0 ) 3 a) l ISCO 0 = 1772 km ν GW = Hz b) l ISCO 0 = , 3 km ν GW = Hz

18 Time a given signal spends in the detector bandwih before coalescence. a) m 1 = m 2 = 10 2 M b) m 1 = m 2 = 10 6 M t = t coal 1 ν in ν GW (t) 8/3 LISA a) [ Hz] t = years b) [ Hz] t = 0, 12 years = 43 d 18 h 43 m 24 s