Gravitation: Perturbation Theory & Gravitational Radiation

Transcription

1 Gravitation: Perturbation Theory & Gravitational Radiation An Introduction to General Relativity Center for Relativistic Astrophysics School of Physics Georgia Institute of Technology Notes based on textbook: Spacetime and Geometry by S.M. Carroll Spring 2013

2 Linearized Gravity & Gauge Transformations Recall g µν = η µν + h µν, h µν << 1 and obtain g µν = η µν h µν where h µν = η µρ η νσ h ρσ. That is, indices are raised and lowered using η µν and η µν. Linearized general relativity can be viewed as the theory of a symmetric tensor field h µν propagating on a flat background spacetime. Linearized general relativity is Lorentz invariant in the sense of special relativity; that is, under a Lorentz transformation x µ = Λ µ µx µ, the perturbation transforms as h µ ν = Λ µ µ Λ ν ν h µν Notice: one can also consider small perturbations about an arbitrary background spacetime g µν = g (0) µν + h µν, and arrive to a theory of a symmetric tensor propagating on the curved space with metric g (0) µν.

3 We derive now the equations of motion for h µν. Start with the Christoffel symbols Γ ρ µν = = 1 2 gρλ ( µg νλ + ν g λµ λ g µν ) 1 2 ηρλ ( µh νλ + ν h λµ λ h µν ) The only contribution to the Riemann tensor will come from the derivatives of the Γ and not the products; thus, R µνρσ = η µλ ργ λ νσ η µλ σγ λ νρ 1 = ( ρ ν hµσ + σ µhνρ σ ν hµρ ρ µhνσ). 2 The Ricci tensor reads R µν = 1 2 ( σ ν hσ µ + σ µh σ ν µ ν h h µν ), where h = η µν h µν = h µ µ, and the D Alembertian is simply the one from flat space, = 2 t + 2 x + 2 y + 2 z. The Ricci scalar reads R = µ ν h µν h.

4 The Einstein tensor reads G µν = R µν 1 2 ηµν R = 1 2 ( σ ν hσ µ + σ µh σ ν µ ν h h µν η µν µ ν h µν + η µν h). The linearized field equation are G µν = 8πT µν, with G µν and T µν the energy-momentum tensor, calculated to zeroth order in h µν. Notice that the conservation law to lowest order is simply µt µν = 0.

5 Gauge Conditions Demanding g µν = η µν + h µν does not completely specify the coordinate system on spacetime Thus, the decomposition of the metric into a flat background plus a perturbation is not unique. Consider linearized theory as a background spacetime M b, a physical spacetime M p, and a diffeomorphism φ : M b M p. Manifolds M b and M p are the same (since they are diffeomorphic), but we imagine that they possess some different tensor fields On M b there is the flat Minkowski metric η µν equipped with coordinates x µ On M p one has the metric g αβ which obeys the full Einstein s equations and is equipped with coordinates y α The diffeomorphism φ allows us to move tensors back and forth between M b and M p. The perturbation h µν is then defined as h µν = (φ g) µν η µν. h µν at this point are not necessarily small; if the gravitational fields on M p are weak, then for some diffeomorphisms φ we will have h µν << 1. Then since g αβ obeys Einstein s equations on the physical spacetime means that h µν will obey the linearized equations on the background spacetime.

6 Mb M p Mb ( ) M p µ ( * g) µ g µ * ( ) * Gauge invariance means that there are a large number of diffeomorphisms between M b and M p Consider a vector field ξ µ (x) on the background spacetime generating a one-parameter family of diffeomorphisms ψ ɛ : M b M b. For ɛ sufficiently small, (φ ψ ɛ) will also be a diffeomorphism such that: h (ɛ) µν = [(φ ψ ɛ) g] µν η µν = [ψ ɛ (φ g)] µν η µν = ψ ɛ (h + η) µν η µν = ψ ɛ (h µν ) + ψ ɛ (η µν ) η µν [ ] ψɛ (η µν ) η µν = ψ ɛ (h µν ) + ɛ ɛ = h µν + ɛl ξ η µν = h µν + 2ɛ (µ ξ ν). where we have used that ψ ɛ (h µν ) = h µν since ɛ is small and also that the Lie derivative of the metric along a vector field is L ξ g µν = 2 (µ ξ ν)

7 Then h (ɛ) µν = hµν + 2ɛ (µ ξ ν) represents the change of the metric perturbation under an infinitesimal diffeomorphism along a vector ɛξ µ also known as a Gauge Transformation. It is very important to keep in mind that ψ ɛ provides a different representation of the same physical situation such that h (ɛ) µν << 1 Therefore, h (ɛ) µν = h µν + 2ɛ (µ ξ ν) tells us what kind of metric perturbations denote physically equivalent spacetimes those related to each other by 2ɛ (µ ξ ν), for some vector ξ µ. This invariance is analogous to the gauge invariance of electromagnetism A µ A µ + µλ It is not difficult to show that h (ɛ) µν = h µν + 2ɛ (µ ξ ν) changes the linearized Riemann tensor by 1 δr µνρσ = ( ρ ν µξσ + ρ ν σξµ + σ µ ν ξρ + σ µ ρξν 2 σ ν µξ ρ σ ν ρξ µ ρ µ ν ξ σ ρ µ σξ ν ) = 0

8 Gauge invariance can also be understood in terms of infinitesimal coordinate transformations. That is, ψ ɛ as the generator of coordinates transformations from x µ to x µ ɛξ µ. We need to address now the issue of gauge fixing. For instance, in harmonic coordinates we require x µ = 0, which is equivalent to g µν Γ ρ µν = 0 In the weak field limit, the harmonic gauge becomes this becomes 1 2 ηµν η λρ ( µh νλ + ν h λµ λ h µν ) = 0, or µh µ λ 1 2 λh = 0. Harmonic gauge is also known as Lorentz, Einstein Hilbert, de Donder or Fock gauge. In this harmonic gauge, the linearized Einstein equations G µν = 8πGT µν simplify to h µν 1 ηµν h = 16πGTµν, 2 while the vacuum equations R µν = 0 take on the elegant form h µν = 0, which is simply the conventional relativistic wave equation. Together, µh µ λ 1 2 λh = 0 and h µν = 0 determine the evolution of h µν in vacuum in the harmonic gauge.

9 It is often convenient to define the trace-reversed perturbation h µν by h µν = h µν 1 2 ηµν h. In terms of h µν the harmonic gauge condition becomes µ hµ λ = 0. The full field equations are and in vacuum h µν = 16πGT µν, h µν = 0.

10 Degrees of Freedom Let s write h µν as h 00 = 2 Φ h 0i = w i h ij = 2 s ij 2 Ψ η ij such that Ψ = 1 6 ηij h ij 1 s ij = (h ij 13 ) 2 ηkl h kl η ij Notice that η ij = δ ij Φ and Ψ are scalar perturbations, w i are vector perturbations, and h ij tensor perturbations. The metric is then given by ds 2 = (1 + 2 Φ) dt w i dx i ] dt + [(1 2 Ψ)η ij + 2 s ij dx i dx j

11 The Christoffel symbols read Γ 0 00 = 0 Φ Γ i 00 = i Φ + 0 w i Γ 0 j0 = j Φ Γ i j0 = [j w i] h ij Γ 0 jk = (j w k) h jk Γ i jk = (j h k)i 1 2 i h jk Consider the motion of a test particle p µ = m dxµ dτ = ( p 0, p i ) = = dx µ dλ (E, E v i ) where λ = τ/m, E = dt/dλ and v i = dx i /dt

12 The geodesic equation reads dp µ dλ + Γµ ρσ pρ p σ = 0 dp µ dt = Γ µ p ρ p σ ρσ E The time component yields de dt [ = E 0 Φ + 2 v k k Φ ( (j w k) 12 ) ] 0h jk v j v k The spatial component yields dp i dt = E [ i Φ + 0 w i + 2 v j ( ) ( [i w j] + 0 h ij + (j h k)i 1 ) ] 2 i h jk v j v k Let s introduce the following definitions G i i Φ 0 w i gravito-electric vector H i ( w ) i = ɛ ijk j w k gravito-magneticvector

13 Thus dp i dt [ = E G i ( ( + v H) i 2 v j 0 h ij (j h k)i 1 ) ] 2 i h jk v j v k ( The term G i + v H) i describe the analogue of the Lorentz force in electromagnatism. The other terms are the response to tensor perturbations.

14 Let s consider now the field equations for the h ij tensor perturbations. Riemann tensor R 0j0l = j l Φ 0 (j w l) h jl R 0jkl = j [k w l] 0 [k h l]j R ijkl = j [k h l]i i [k h l]j Ricci tensor R 00 = 2 Φ + 0 k w k Ψ R 0j = w j j k wk j Ψ + 0 k S j k R ij = i j (Φ Ψ) 0 (i w j) + Ψη ij s ij + 2 k (i s j) k Einstein tensor G 00 = 2 2 Ψ + k l s kl G 0j = w j j k wk k j Ψ + 0 k S j G ij = (η ij 2 ) i j (Φ Ψ) + η ij 0 k w k 0 (i w j) +2 η ij 2 0 Ψ s k ij + 2 k (i s j) ηij k l s kl

15 Let s investigate now how from G µν = 8 π T µν which are the true degrees of freedom Consider first the (0, 0) component of the Einstein equations 2 Ψ = 4 π T k l skl Ψ is determined from T 00 and s ij. Consider first the (0, j) component of the Einstein equations (η jk 2 + j k )w k = 16 π T 0j j Ψ k s j k w j is determined from T 0j, s ij and Ψ. Consider first the (i, j) component of the Einstein equations (η ij 2 ) ( i j Φ = 8 π T ij + η ij 2 i j 2 η ij 2 ) 0 Ψ η ij 0 k w k k + 0 (i w j) + s ij 2 k (i s j) + ηij k l s kl Φ is determined from T ij, Ψ, w j and s ij. Notice that all the equations for Ψ, w i and Φ are non-propagating equations.

16 Notice: The only propagating degrees of freedom are those in s ij or equivalently in h ij Recall: gauge transformation h µν h µν + µξ ν + ν ξ µ. In term of the metric perturbation, it reads Φ Φ + 0 ξ 0 w i w i + 0 ξ i i ξ 0 Ψ Ψ 1 3 i ξi s ij s ij + (i ξ j) 1 3 k ξk η ij Transverse gauge (four conditions) i w i = 0 i w i + 0 i ξ i 2 ξ 0 = 0 i s ij = 0 i s ij + i (i ξ j) 1 3 j i ξ i = 0 Thus 2 ξ 0 = i w i + 0 i ξ i 2 ξ j j i ξi = 2 i s ij yield solutions for ξ 0 and ξ i

17 In the transverse gauge, Einstein equations become 2 Ψ = 4 π T 00 2 w j = 16 π T 0j j Ψ (η ij 2 ) i j (Φ Ψ) 0 (i w j) + 2 η ij 2 0 Ψ s ij = 8 π T ij Synchronous gauge Φ = 0 Φ + 0 ξ 0 = 0 w i = 0 w i + 0 ξ i i ξ 0 = 0 thus 0 ξ 0 = Φ 0 ξ i = w i + i ξ 0 = 0 this yields a metric ds 2 = dt 2 + (η ij + h ij )dx i dx j

18 Gravitational Wave Solutions We will consider the case of freely-propagating degrees of freedom of the gravitational field in the transverse gauge without local sources (i.e. T µν = 0). Thus, from 2 Ψ = 4 π T 00 2 w j = 16 π T 0j j Ψ (η ij 2 ) i j (Φ Ψ) 0 (i w j) + 2 η ij 2 0 Ψ s ij = 8 π T ij we get, with the appropriate boundary conditions that 2 Ψ = 0 Ψ = 0 2 w j = 0 w j = 0 2 Φ = 0 Φ = 0 s ij = 0 Therefore, the metric perturbation h µν can be written in this transverse traceless gauge as h TT µν = s ij 0 and satisfies the following equation h TT µν = 0

19 From now on we concentrate on hµν TT µ keeping in mind that htt 0µ = 0, htt µ = 0, µh µν TT = 0 We consider plane-wave solutions h TT xσ µν = Cµν eikσ such that C µν and k µ are the constant amplitude tensor and wave vector, respectively. Thus, 0 = h TT µν = η ρσ ρ σh TT µν = η ρσ ρ(ik σh TT µν ) = η ρσ k ρk σh TT µν = k σk σ h TT µν which implies that k σk σ = 0; that is, the wave vector k µ is null. ( Also, k µ = ω, ) k with ω the frequency of the wave. From k σk σ = 0, we have that ω 2 = η ij k i k j An observer with 4-velocity U µ will measure a frequency ω = k µu µ

20 From the transverse condition 0 = µh µν TT = i C µν ikσ xσ k µ e which implies that k µ C µν = 0 Assume the wave traveling along the x 3 -direction. Thus, k µ = (ω, 0, 0, k 3 ) = (ω, 0, 0, ω), which implies that C 0ν = 0 and C 3ν = 0 Therefore the only non-vanishing components are C 11, C 12, C 21, C 22. Together with the conditions that C µν is traceless and symmetric, one gets C µν = 0 C 11 C C 12 C The quantities C 11 and C 12 encapsulate the true degrees of freedom

21 Let s consider the effect that the gravitational wave has on test particles. To measure the effects on a coordinate-independent fashion, we need to focus on relative effects on nearby particles. Thus, we consider the geodesic deviation equation D 2 dτ 2 Sµ = R µ νρσ Uν U ρ S σ where U µ is the 4-velocity and S µ the separation vector. Recall that the Riemann tensor is already O(h); thus, U ν = (1, 0, 0, 0) and only R µ00σ is needed. That is R µ00σ = 1 2 ( 0 0 h TT µσ + σ µhtt 00 σ 0h TT µ0 µ 0h TT σ0 ). But h TT µ0 = 0, so Rµ00σ = h TT µσ. Also, for slowly-moving particles τ = x 0 = t, so the geodesic deviation equation becomes 2 t 2 Sµ = Sσ t 2 htt µ σ. Since the wave is traveling in the x 3 -direction, only S 1 and S 2 are affected; that is, only the directions perpendicular to the wave propagation are disturbed.

22 Let s define h + = C 11 and h = C 12. Consider first h + = C 11 2 t 2 S1 = S1 t 2 (h+eikσ x σ ) 2 t 2 S2 = S2 t 2 (h+eikσ x σ ) To lowest order, ( S 1 = ) 2 h+eikσ xσ S 1 (0) ( S 2 = 1 1 ) 2 h+eikσ xσ S 2 (0). Thus, particles initially separated in the x 1 direction will oscillate back and forth in the x 1 direction, and likewise for x 2 separation. That is, consider a ring of stationary particles in the x-y plane, as the wave passes they will bounce back and forth in the shape of a +: y x

23 y x An equivalent analysis for h 0 yields S 1 = S 1 (0) h e ikσ xσ S 2 (0) S 2 = S 2 (0) h e ikσ xσ S 1 (0). In this case the circle of particles would bounce back and forth in the shape of a : h + and h measure the two independent modes of linear polarization of the gravitational wave. One could consider right- and left-handed circularly polarized modes by defining h R = h L = 1 2 (h + + ih ), 1 2 (h + ih ). y x

24 Polarization states and quantum particles In quantum mechanics, the spin S of a massless particle is related to the angle θ under which the polarization modes are invariant by S = 360 /θ E&M: two polarizations invariant under 360 rotations; thus, the photon has spin S = 1. Neutrinos: described by a field invariant under 720 ; thus, S = 1/2. Gravitational Field: polarization modes invariant under 180 rotations; thus, gravitons have S = 2.

25 Production of Gravitational Waves We are now going to consider the sources of gravitational radiation. That is, T µν 0 and we can not assume that h µν takes the transverse-traceless form as before with Ψ = Φ = w i = 0 Define trace-reversed perturbation h µν = h µν 1 2 h ηµν. Notice that h = h TT Far from the sources, h µν = htt µν Also, under a gauge transformation h µν h µν + 2 (µ ξ ν) λ ξ λ η µν One can then choose ξ µ to satisfy µ hµν + 2 µ (µ ξ ν) ν µξ µ = 0 µ µ hµν + ξ = 0 and therefore set µ hµν = 0; a.k.a. Lorentz gauge. Notice that under this gauge h µν is not transverse since µh µν = 1 2 ν h

26 Substitution of h µν = h µν 1 h ηµν into 2 G µν = 1 2 ( σ ν hσ µ + σ µh σ ν µ ν h h µν η µν µ ν h µν + η µν h) yields G µν = 1 2 h µν. Thus, together with G µν = 8 π T µν we get h µν = 16 π T µν The Green s function G(x σ y σ ) for the D Alembertian operator is x G(x σ y σ ) = δ (4) (x σ y σ ) where x denotes the D Alembertian with respect to the coordinates x σ The general solution to h µν = 16 π T µν is then h µν (x σ ) = 16π G(x σ y σ )T µν (y σ ) d 4 y, Solving x G(x σ y σ ) = δ (4) (x σ y σ ) yields retarded or advanced solutions. We are interested in the retarded solution G(x σ y σ 1 ) = 4π x y δ[ x y (x0 y 0 )] θ(x 0 y 0 ).

27 The retarded solutions yields after integrating over y 0 h µν (t, x) = 4 1 x y Tµν (tr, y) d3 y, where t r = t x y. is the retarded time. h µν (t, x) give as above represents the disturbance in the gravitational field at (t, x) by summing the influences from the energy and momentum sources at the point (t r, x y) on the past light cone. t x i y i (t, y i r )

28 Consider the case where the gravitational radiation is emitted by an isolated, far away, nonrelativistic matter source Our Fourier transform convention, φ(ω, x) = φ(t, x) = 1 dt e iωt φ(t, x), 2π 1 dω e iωt φ(ω, x). 2π Then hµν (ω, x) = = = 1 dt e iωt hµν (t, x) 2π 4 dt d 3 y e iωt T µν (t x y, y) 2π x y 4 dt r d 3 y e iωtr iω x y Tµν (tr, y) e 2π = 4 d 3 y e iω x y T µν (ω, y). x y x y

29 observer R source R Since the source is isolated and far away, we can set the source to be centered at a (spatial) distance r, with the different parts of the source at distances r + δr such that δr << r. Since the source is slowly moving, most of the radiation emitted will be at frequencies ω sufficiently low that δr << ω 1. Thus, e iω x y x y eiωr r and hµν (ω, x) = 4 eiωr r d 3 y T µν (ω, y). Since the harmonic gauge condition µ hµν (t, x) = 0 in Fourier space implies h0ν = need to concern ourselves with hij (ω, x). i ω i hiν, we only

30 d 3 y T ij (ω, y) = k (y i kj 3 T ) d y y i kj 3 ( k T ) d y The first term in the r.h.s. vanishes because the source is isolated. The second can be rewritten in terms of T 0j with the Fourier-space version of µt µν = 0: k T kµ = iω T 0µ. thus d 3 y T ij (ω, y) = iω y i 0j T d 3 y iω = (y i 0j T + y j 0i 3 T ) d y 2 iω [ = l (y i y j ] 0l i j 0l T ) y y ( l T ) 2 = ω2 y i y j 00 3 T d y. 2 d 3 y Define the quadrupole moment tensor I ij (t) = y i y j T 00 (t, y) d 3 y Thus hij (ω, x) = 2ω 2 eiωr Ĩ ij (ω), r

31 Transforming back to t, h ij (t, x) = 1 2 dω e iω(t r) ω 2 Ĩ ij (ω) 2π r 1 2 d 2 = 2π r dt 2 dω e iωtr Ĩ ij (ω) and one arrives to the Quadrupole Formula h ij (t, x) = 2 r d 2 I ij dt 2 (tr ) That is, the gravitational wave produced by an isolated nonrelativistic object is proportional to the second time derivative of the quadrupole moment of the energy density at the point where the past light cone of the observer intersects the source. In contrast, the leading contribution to electromagnetic radiation comes from the changing dipole moment of the charge density.

32 Gravitational Waves from a Binary System x3 x2 M v r r M v x1 Consider two stars of mass M in a circular orbit in the x 1 -x 2 plane, at distance R from their common center of mass. Treat the motion of the stars in the Newtonian approximation, that is (gravitational force = centrifugal force), M 2 (2R) 2 = M Ω2 R so the angular frequency of the orbit is ( ) M 1/2 Ω = 4R 3

33 In terms of Ω, the path of star a is x 1 a = R cos Ωt, x2 a = R sin Ωt, x3 a = 0 and star b, x 1 b = R cos Ωt, x2 b = R sin Ωt, x3 b = 0 The corresponding energy density is T 00 (t, x) = Mδ(x 3 [ ) δ(x 1 R cos Ωt)δ(x 2 R sin Ωt) + δ(x 1 + R cos Ωt)δ(x 2 ] + R sin Ωt) From I ij (t) = y i y j T 00 (t, y) d 3 y The quadrupole moment is then: I 11 = 2MR 2 cos 2 Ωt = MR 2 (1 + cos 2Ωt) I 22 = 2MR 2 sin 2 Ωt = MR 2 (1 cos 2Ωt) I 12 = I 21 = 2MR 2 (cos Ωt)(sin Ωt) = MR 2 sin 2Ωt I i3 = 0

34 Given I 11 = MR 2 (1 + cos 2Ωt) I 22 = MR 2 (1 cos 2Ωt) I 12 = MR 2 sin 2Ωt one gets Ï 11 = 4MR 2 Ω 2 cos 2Ωt Ï 22 = 4MR 2 Ω 2 cos 2Ωt Ï 12 = 4MR 2 Ω 2 sin 2Ωt Finally, from h ij (t, x) = 2 Ï ij r one gets h ij (t, x) = M2 cos 2Ωtr sin 2Ωtr 0 sin 2Ωt r cos 2Ωt r 0 r R where we used that Ω 2 = M/(4R 3 )

35 Gravitational Radiation Energy Loss We have seen that gravitational waves can put energy into things that they pass through. What is the energy emitted via gravitational radiation? There is however no true local measure of the energy in the gravitational field. In the weak field limit, where gravitation is viewed as a symmetric tensor propagating on a fixed background metric, it is possible to derive an energy-momentum tensor for the fluctuations h µν, although the definition is not unique. However, the different proposals in the literature for the energy-momentum tensor for gravitation in the weak field limit give the same answers for physically well-posed questions. Recall T µν O[( φ) 2 ] T µν O[(F µν ) 2 ] but in the weak field limit T µν O(h µν ) Hence, in order to keep track of the energy carried by the gravitational waves, we will have to extend our calculations to at least second order in h µν.

36 That is, g µν = η µν + h (1) µν + h(2) µν R µν = R (0) µν + R(1) µν + R(2) µν To zero-order R (0) µν [η µν ] = 0 because of the flat background spacetime condition. To first-order R (1) µν [h (1) µν ] = 0, which we have already seen it determines h (1) µν up to gauge transformations. To second-order, R (1) µν [h(2) ] + R (2) µν [h(1) ] = 0. where above R (1) µν = 1 2 ( σ ν hσ µ + σ µh σ ν µ ν h h µν ), with h µν = h (2) µν and R (2) µν = 1 2 hρσ µ ν h ρσ h ρσ ρ (µ h ν)σ ( µhρσ) ν hρσ + ( σ h ρ ν ) [σ h ρ]µ σ(hρσ ρh µν ) 1 4 ( ρhµν ) ρ h ( σh ρσ 1 2 ρ h) (µ h ν)ρ. with h µν = h (1) µν

37 Rewrite R (1) µν [h (2) ] + R (2) µν [h (1) ] = 0 as G (1) µν [h(2) ] = R (1) µν [h(2) ] 1 2 ηµν R(1)α α[h (2) ] = 8πt µν where t µν = 1 ( R (2) µν 8π [h(1) ] 1 ) 2 R(2)α α[h (1) ] One can then view t µν (Symmetric tensor, quadratic in h µν ) as an energy-momentum tensor representing how the perturbations affect the spacetime. Note that the Bianchi identity for G (1) µν [h (2) ] implies that t µν is conserved in the flat-space sense; that is, µt µν = 0. Note also that t µν is not a tensor at all in the full theory is not gauge invariant

38 Calculating tµν Assume the transverse-traceless gauge µ h TT µν = 0, htt = 0 Substitution of R (2)TT µν = 1 2 hρσ µ ν h ρσ ( µhρσ) ν hρσ + ( σ h ρ ν ) [σ h ρ]µ h ρσ ρ (µ h ν)σ hρσ σ ρh µν (with h µν understood to be h (1)TT µν ) into t µν = 1 ( R (2) µν 8π 1 ) 2 ηµν R(2)α α yields R (2)TT µν = 1 µh ρσ ν h ρσ + 2h ρ µ 4 hρν where denotes average over wavelengths to cure the problem of lack of gauge invariance. Recall: h ρν = 0 are the equations of motion for the perturbation; thus, R (2)TT µν = 1 4 µh TT ρσ ν hρσ TT

39 The equations of motion h ρν = 0 also imply (after integration by parts) that η µν R (2)TT µν = 0 Finally t = 1 µν 32 π µh TT ρσ ν hρσ TT

40 Example: Plane-wave Recall that h TT µν = Cµν sin (k λx λ ) Then t µν = 1 32 π kµkν CρσCρσ cos 2 (k λ x λ ) Since cos 2 (k λ x λ 1 ) = 2 k λ = ( ω, 0, 0, ω) C ρσc ρσ = 2 (h h2 ) we get t µν = π f 2 (h h2 ) Typical sources have 10 4 Hz f 10 4 Hz and amplitudes h thus t 0z 10 4 ( f Hz ) 2 (h h 2 ) (10 21 ) 2 erg cm 2 s large when compared to E&M flux from supernova 10 9 erg/cm 2 /s

41 Back to the Quadrupole Formula Total energy on a surface Σ of constant time, E = t 00 d 3 x Σ Total energy radiated through to infinity, E = P dt where P is the power radiated P = t tµ n µ r 2 dω S where the integral is taken over a timelike surface S made of a spacelike two-sphere at infinity and some interval in time, and n µ = (0, 1, 0, 0) is a unit spacelike vector normal to S. Introduce the spatial projection operator P ij = η ij n i n j For an arbitrary, symmetric spatial tensor X ij one has that X TT ij = (P ki P lj 12 ) P ij Pkl X kl

42 From the quadrupole formula we have that h TT ij TT = h ij = 2 r d 2 I TT ij (t r) dt2 Introduce the reduced quadrupole moment J ij = I ij 1 3 η ij Ik k which for a Newtonian potential Φ = M r 1 r 3 D i xi 3 2 r 5 J ij xi x j +... Then h TT ij = 2 r d 2 J TT ij dt 2 (t r)

43 Their derivatives read t h TT ij = 2 r h TT ij = 2 r r 2 r d 3 Jij TT dt 3 d 3 Jij TT dt d J TT ij r 2 dt 2 d 3 J TT ij dt 3 Therefore yields t = 1 tr 32 π t h TT ρσ t tr = 1 d3 Jij TT 8 π r 2 dt 3 r hρσ TT d3 J ij TT dt 3 It is straightforward to show that J TT ij J ij TT = J ij Jij j ik 2J i J nk n j Jij J kl n i n j n k n l

44 Thus the power reads P = 1 8 π S d 3 J ij dt 3 3 d J ij dt 3 2 d 3 j 3 Ji d J ik dt 3 dt 3 n k n j + 1 d 3 J ij 3 d J kl 2 dt 3 dt 3 n i n j n k n l dω The quadrupole terms are independent of the angular coordinates; thus, dω = 4 π n i n j dω = n i n j n k n l dω = 4π 3 η ij 4π 15 (η ij η kl + η ik η jl + η il η jk ) Then P = 1 5 d 3 Jij dt 3 3 d J ij dt 3

45 For binary systems, J ij = M R2 3 (1 + 3 cos 2Ωt) 3 sin 2Ωt 0 3 sin 2Ωt (1 3 cos 2Ωt) therefore d 3 J sin 2Ωt cos 2Ωt 0 ij dt 3 = 8MR 2 Ω 3 cos 2Ωt sin 2Ωt The power radiated by the binary is thus P = 27 5 M2 R 4 Ω 6 or, using Ω = (M/4R 3 ) 1/2, P = 2 M 5 5 r 5 For binary systems with arbitrary masses m 1 and m 2 P = 32 5 G 4 2 (m 1 m 2 ) (m1 + m 2 ) c 5 r 5

46 Generation of Gravitational Waves In E&M the electric-dipole radiation dominates. Its luminosity is given by L em = 2 3 d 2 with d = q r The analog dipole radiation in GR is L gr = 2 3 d 2 with d = m r but d = ṗ = 0 because of conservation of momentum. Therefore there can be no mass dipole radiation in gravitational physics. Next in E&M is magnetic-dipole moment, with in GR has the following analogue: µ = r (m v) = J, namely the angular momentum. Thus, from conservation of angular momentum, we have that no radiation is possible. The first contribution comes from quadrupole radiation. L gr = Q ij... Q ij

47 Power of Gravitational Waves Consider a system of mass M, size R, moving in time-scales T :... (moving mass) (size of object)2 Q ij (moving time-scale) 3 T 3 = M (R/T ) 2 (non-axisymmetric kinetic energy) T T = M R 2 also thus and the GW luminosity is then ( ) R T mean velocity R R 3 1/2 (M/R) 1/2 = M... Q ij ( ) M 5/2 R ( ) M 5 L gw R For a binary system, M = m 1 + m 2 and separation R = a, the non-axisymmetric kinetic energy is given by its potential energy m 1 m 2 /a, and from Kepler s law a 3 /T 2 = M, so L gw µ2 M 3 a 5 with µ = m 1 m 2 /M

48 The Hulse-Taylor Binary Pulsar Gravitational waves have been already detected indirectly. In 1974 Hulse and Taylor discover a binary system PSR in which one of the objects was a pulsar, providing a very accurate clock to monitor the decay of the orbit due to the emission of gravitational radiation. P = 8 5 M2 (2 R) 4 Ω 6 4 (MΩ) 10/3 E = M Ω 2 R 2 M2 2 R = M 2 4 R 0.4M5/3 Ω 2/3 1 de E dt dt dt = = 2 1 dω = 2 1 dt 3 Ω dt 3 T dt 3 T P 2 E 15 T M (M Ω)8/ s yr 1 taking into account eccentricity (e = 0.617) dt dt s yr 1

49 Astrophysical Sources of Gravitational Radiation BINARY SYSTEMS Collisions of supermassive black holes, BH-BH, BH-NS and NS-NS binaries. The lifetime of an equal-mass binary is τ gw = E ( ) 5/3 ( ) P = M 5/3 f 8/3 M f 8/3 = 2.43 s M 100 Hz The coalescence of a binary systems consist of three stages: Inspiral, Merger, Ringdown

50 Astrophysical Sources of Gravitational Radiation SPINNING NEUTRON STARS Neutron stars (NS) spin at frequencies above 20 Hz It the NS is not axi-symmetric about the rotation axis, it will emit GWs. E.g. NS with mountains or pimples Another sources of asymmetry are imperfections in the crust of a NS. Gravitational-wave driven instabilities (r-modes) can also produce radiation

51 Astrophysical Sources of Gravitational Radiation GRAVITATIONAL COLLAPSE A supernova marks the end of a star when it exhausts its supply from nuclear reactions The star collapses inward some of the energy into an explosion If the collapse is not axi-symmetric, gravitational waves are produced The details of the gravitational radiation produced are not well understood

52 Astrophysical Sources of Gravitational Radiation GRAVITATIONAL WAVES FROM THE BIG BANG The very early universe was a source of a random sea of gravitational radiation The radiation originated from various events too numerous to count. The waves arrive to us superimposed, as a background of radiation. It is estimated that the energy density from this radiation is around of the total energy density.

53 Detection of Gravitational Waves The direct detection of gravitational waves is one of the highest priorities in gravitational physics and astrophysics Gravitational wave observations will open a new window to the Universe The main difference between the sources of electromagnetic and gravitation radiation is that gravitational waves are produced by bulk motion of large masses and electromagnetic waves by incoherent excitations of individual particles Consider a binary system with mass components M separated a distance R The relevant scale is the Schwarzschild radius R S = 2 M and the distance r from us to the binary. Recall that h = 8M r Ω 2 R 2 cos ωt r = 2 M2 r R cos ωtr where Ω = (M/4R 3 ) 1/2 is the orbital frequency and ω = 2 Ω the frequency of the GWs. Thus h R2 S r R and f = ω 2 π R S R 3

54 Consider the following case ( ) M R S = 2 M 10 6 cm 10 M ( ) M R = 20 M 10 7 cm 10 M r 100 Mpc cm The GW source has characteristic amplitude and frequency h and f = 10 2 Hz Detecting GW amplitudes at the level of h is challenging but feasible. Interferometry is one of the most promising techniques that have been developed.

55 Recall: the effect of a GW is to perturb the relative positions of freely-falling masses. That is, S i S i (0) h S i (0) If two test masses are separated by a distance L, the passage of a GW will change the separation by δl such that h δl/l If h and the test masses are separated by L few km, then ( ) ( ) δl h L cm km The challenge is then to detect changes of separation much smaller than the typical size of an atom (i.e. Borhr radius a cm) or of a nucleus (i.e. Fermi 1 fm = cm)

56 Gravitational Wave Interferometers Consider an interferometer with a laser with characteristic wavelength λ 10 4 cm with mirrors separated a distance L The mirrors near the beamsplitter are partially-reflective, so the light travels approximately 100 times the length of the arm. The beams are then combined to produce, in the absence of GWs, a destructive interference pattern at the photodetector. The effect of a gravitational wave is to perturb orthogonal lengths in opposite senses, thus leading to a phase shift that disturbs the destructive interference.