The Relativistic Motion of a Binary System

Transcription

1 The Relativistic Motion of a Binay System Stephan Dinkgeve sdinkgeve@gmail.com Maste Thesis Theoetical Physics Supevised by Jan Piete van de Schaa Univesity of Amstedam May 8, 01 Abstact The Post-Newtonian expansion (PN) is used to find the fist elativistic coections to the equations of motion of a binay system of point paticles. The expansion assumes velocities to be small (v << 1) and the field to be weak (Φ << 1), yet allowing fo tems of highe ode in Φ with evey subsequent tem in the expansion. We will deive the obital equations fo the fist coection (1PN) following Damou and Deuelle []. These esults wee used in the calculation of the change of the obital peiod of the Hulse-Taylo pulsa, the atio of which compaed to the obseved value is an astonishing (1) [15]. We also deive effective potentials fo 1PN and PN, which will give us a measue fo the accuacy of the expansion. We will then follow Maggioe [9] and Poison [11] to show that the equations of motion at.5pn contain a back-eaction tem, which is equivalent to the powe of the gavitational adiation emitted fom the system. Knowing the equations of motion and thei coesponding gavitational wavefoms to high ode will be of majo impotance fo the extaction of gavitational waves fom the noise in the upcoming gavitational wave detectos. 1

2 Contents 1 Intoduction 3 The Newtonian Two-Body Poblem 4.1 The Radial Equation The Effective Potential The Newtonian Obits The Newtonian Appoximation The Lineaized Einstein Equations The Schwazschild Solution Deiving the Schwazschild Metic The Schwazschild Obits The Pecession of Mecuy Hamonic Coodinates Gavitational Waves Gavitational Waves in Linea Theoy The Enegy of Gavitational Waves The Quadupole Moment The Ode of Magnitude of Gavitational Radiation Gavitational Waves fom a Binay in Cicula Obit The Inspialing of a Binay System Gavitational Waves fom a Binay in Elliptical Obit The Fist Ode Post-Newtonian Appoximation Deiving the Fist Post-Newtonian Metic The Fist Ode Schwazschild Radial Equation The Fist Ode Post-Newtonian Lagangian The Relative Lagangian The Post-Newtonian Obits The Pseudo-Effective Potentials Visualizing the Accuacy of the PN Appoximation Applying 1PN: The Hulse-Taylo Pulsa Radiation Reaction The Landau-Lifshitz Field Equations The 1PN Potentials The Suface Integal Method The Radiation Reaction Potentials The.5PN Equations of Motion Conclusion 67 9 Appendix Noethe s Theoem

3 1 Intoduction In 1915 Einstein deived a wondeful set of fomulas called the Einstein Field Equations. In these equations the Newtonian foce of gavity is eplaced by the cuvatue of space-time and elated to the enegy and momentum in the univese. Unfotunately, these equations ae vey difficult to solve, even fo simple enegy-momentum configuations and we shall have to esot to appoximations. In this aticle we will study the elativistic motion of a two-body system. To find the equations of motion we use the post-newtonian expansion. This expansion is based on the assumption that the velocities of the paticles ae low (v << 1 ) and that the fields ae weak (Φ << 1), yet allowing fo tems of highe ode in Φ with evey subsequent tem in the expansion. Since ou stating point will be low velocities and weak fields, ou leading tem in the equation of motion will be Newtonian and the othe tems will be elativistic coections. The dawback of this appoach is that ou equations will take a deceptively Newtonian fom, which wipes out the conceptual famewok of geneal elativity. Despite this dawback, the obtained esults ae in stong ageement with obsevation and constitute stong evidence fo geneal elativity. The Einstein Equations suggest the existence of gavitational waves, which extact enegy and angula momentum fom a binay system, causing a change in the obital peiod. This change matches the obseved obital peiod change of the Hulse-Taylo pulsa with geat pecision (the atio between the theoetical and the obseved value is an astonishing (1) [15]). We stat in the fist chapte by deiving the main esults of the Newtonian binay system. Both the esults and the method to obtain them will have close analogies in late chaptes. Next we deive the Newtonian equations of motion fom Einstein s theoy, assuming low velocities and weak fields (v << 1 and Φ << 1). The esult is called the zeoth ode tem of the Post-Newtonian expansion (0PN). In the following chapte we will deive the main esults fom the Schwazschild binay system. In this case we shall neithe assume low velocities no a weak field. Howeve, we will assume that one of the paticles has negligible mass compaed to the othe (m << m 1 ). In the case of weak field and low velocities this system will seve as a limiting case to the Post-Newtonian expansion. Following this we study the gavitational adiation that is being geneated by the binay system. The adiation causes a back-eaction on ou system and thus influences the equations of motion. In this chapte we will conclude that the adiation will only contibute at.5pn. In the last two chaptes we study the fist elativistic coections to the equations of motion, stating with 1PN. This esult was fist deived by Einstein, Infeld and Hoffmann in 1938 [5]. We will then continue to deive the obital equations, which wee deived by Damou and Deuelle in 1985 []. In the last chapte we shall calculate the equations of motions coesponding to.5pn, which descibes the lagest contibution to the back-eaction on the system caused by the emission of gavitational waves. On vaious occasions we shall deive effective potentials, which give a visual epesentation of the possible obits associated with diffeent enegies and angula momenta. The Newtonian and Schwazschild effective potentials ae discussed in most intoductoy textbooks on the subject and will be discussed in chapte two and fou. In this aticle howeve, we will extend thei use to both 1PN and PN (chapte 6) and to a system emitting gavitational adiation (chapte 5). We shall use the 1PN and PN effective potentials to get an idea of the accuacy of the Post-Newtonian appoximation. In case of gavitational adiation we use it to visualize the ciculaization of the obit due to the back eaction. Let s stat. We have a lot to talk about. 3

4 The Newtonian Two-Body Poblem We shall begin studying the two-body poblem in the case of Newtonian mechanics. This is necessay, because a lot of the techniques used in this chapte will have thei analogies in late chaptes. Also, the Newtonian theoy will seve as a limiting case of the geneal theoy of elativity. Moe explicitly, the Newtonian theoy should eappea when velocities ae negligible (v << 1) and the fields ae weak (Φ << 1). We ll stat by deiving the adial equation, which descibes the change of the distance between the two bodies. This expession can be witten in the fom ṙ = E + V, whee V is the effective potential. A plot of this potential gives a visual epesentation of the types of obits that ae possible in ou system (cicula, elliptical, paabolic and hypebolic obits). Finally we shall calculate the paths of the elliptical obits explicitly. In this chapte we ve often used [], [6] and [16]..1 The Radial Equation If we want to descibe the location of two paticles in thee-dimensional space we geneally need six quantities. We can fo instance pick the two thee-component position vectos 1 and. Altenatively, we can use the cente of mass vecto R 1 m 1+m (m m ) and a vecto eaching fom one paticle to the othe 1. This choice is depicted in the fist figue on the ight. Using Noethe s theoem (appendix) we can find conseved quantities coesponding to invaiance unde esp. spatial tanslations and Galileo tanslations: P = m 1 v 1 + m v K = m1 1 + m t P (1) The fist equation expesses consevation of the total momentum. Consequently, we can pick an inetial fame in which the total momentum is zeo. This also simplifies the second equation, which now implies the consevation of the cente-of-mass vecto R. We can theefoe pick the fame in which R = K = 0. We end up with: 0 = m 1 v 1 + m v 0 = m m () This simplification is depicted on the ight side of the pevious figue and is called the cente-ofmass fame. We ewite the last two equations to obtain: 1 = µ/m 1 = µ/m (3) v 1 = µ/m 1 v v = µ/m v (4) Hee we have use the educed mass (µ = m 1 m /(m 1 + m )) and the diffeence vecto v = v v 1. The Lagangian of ou two body system is: L = 1 m 1v m v + m 1m In tems of the cente-of-mass coodinates this can be ewitten to obtain the so-called elative Lagangian: (5) L = 1 µv + m 1m (6) 4

5 Obseve that the initial two-body lagangian now has the simplified shape of a one-body lagangian. The lagangian is invaiant unde time tanslation and unde otation. Again using Noethe s theoem, these invaiances can be show to coespond to the consevation of enegy and momentum (esp. E = v dl/d v L and J = dl/d v). We find: E = 1 µv m 1m J = µ v (7) It is now convenient to switch to pola coodinates. Because of the consevation of angula momentum, the motion of the binay system takes place in a plane. This means that the pola angle ϕ is constant. The velocity in pola coodinates theefoe becomes: v = ṙ + θ Also using v = θ, we find: J = µ θ J = θ (9) E = 1 µṙ + J µ m 1m Ẽ = 1 ṙ + J M Hee we ve used M = m 1 + m, Ẽ = E/µ and J = J/µ. Fom the last equation we find the adial equation of motion: ṙ M = (Ẽ + J ). The Effective Potential The thid tem in equation (11) is a centifugal tem. Howeve, since this tem is dependent on, we can petend it to be a pat of the potential enegy. We call this fictitious potential the effective potential V: ) ṙ = (Ẽ V V = M + J The effective potential is plotted in figue to the ight. Notice that the fist paticle is always positioned at the oigin of this gaph, since that is whee the vecto stats. The second paticle is at vaiable distance. Fo lage the gaph stats to look moe and moe like the gavitational potential, yet fo small the angula momentum tem stats to dominate, peventing the second paticle fom closing in too much. Since the left hand side of equation (1) is always zeo o positive, the enegy must always be highe than the effective potential. Imagine the second paticle coming in fom infinity towad the paticle at the oigin. Since the potential is zeo at infinity, this must mean Ẽ > 0. At some point the paticle eaches a point whee Ẽ = V. The adial equation tells us, that this is the point whee the velocity of the paticle is zeo. (8) (10) (11) (1) 5

6 But what about the acceleation at that point? To answe this we diffeentiate the adial equation with espect to time: dtṙ d = ṙ = dv d ṙ = dv d The expession dv/d will only be zeo at a minimum o maximum of the last gaph, so the acceleation of ou paticle with Ẽ > 0 will be non-zeo at Ẽ = V. The paticle thus closes in to the fist paticle, until its enegy is equal to the effective potential and then goes back to infinity. In the next paagaph we will show that these tajectoies coespond to the hypebolic and paabolic obits. If Ẽ < 0 the paticle is stuck between two tuning points and will thus move back and foth between these points. The paticle thus oscillates between two fixed adii. These tajectoies will tun out to be the elliptical obits. In the case whee the enegy is equal to the the minimum of the potential, the adius of the paticle stays constant. This coesponds to cicula motion..3 The Newtonian Obits In this paagaph we ll deive the exact shapes of the obits by discoveing how the angle θ and the adius ae elated. We use: dθ = dθ dt θ J d = d dθ = d ṙ ṙ d (13) ṙ = d J dθ (14) In the second step we have used (9). Inseting this into the adial equation (11) we obtain: θ() = ± J/ d (15) J (Ẽ + M/ ) The solution to this integal is: J = M(1 + e cos θ) e = 1 + Ẽ J M (16) In case of the cicula obits we have e = 0. Fo elliptical obits we have 0 < e < 1. Fo e = 1 and e > 1 we have the paabolic and hypebolic obits espectively. The solutions fo diffeent e ae gaphically pesented in the figue to the ight. In case of elliptical obits we can ewite the last equation in tems of the semi-majo axis a (the longest adius of the ellipse). This is the famous obital equation fo the ellipse: = a(1 e ) 1 + e cos θ J a = M(1 e ) (17) Witing the enegy and the angula momentum in tems of the semimajo axis and the eccenticity, we find: Ẽ = M a J = Ma(1 e ) (18) An ellipse is called closed when afte a otation of θ = θ 0 + π the paticle ends up at the same adius. If not, it is said to be open. Since cos(θ) = cos(θ+π) we must conclude fom equation (17) 6

7 that Newtonian ellipses ae closed. Newton aleady ealized this and noted that if the elliptical obits wee found to be open (fo which thee was no evidence at the time), it would equie a foce law diffeent fom his invese-squae law [16]. With equation (17) we can find the shape of the obits, but we do not yet know how the adius is dependent on time. To deive this elation we again stat with the adial equation (11): d dt = ± µ (E + m 1m Flipping the faction on the left and integating we find: µ d t = 0 µ ) J µ (19) (E + m1m ) l µ (0) In case of elliptical motion this is most conveniently integated though an auxiliay vaiable u(t), called the eccentic anomaly. u(t) is defined by the following elation: = a(1 e cos u) (1) Inseting this into the obital equation (17) we find: cos θ = cos u e 1 e cos u 1 + e cos θ = 1 e 1 e cos u We can ewite this, using tigonometic identities, as: θ = A e (u) [ 1 + e A e actan 1 e tan u ] () (3) Rewiting the integal in tems of u, a and e we find: µa t = 3 u (1 e cos u)du (4) m 1 m 0 Integating this fom 0 to π we find Keple s law: ω 0 = π T = m1 + m a 3 (5) Integating without fixing u in the integal we aive at the Keple equation: ω 0 t = u e sin u (6) Combining this equation with equation (1) we can elate to t. Similaly, combining the Keple equation with equation () gives us the elation between θ and t. The solution to the Keple equation can be expessed in tems of Bessel functions J n : u = ω 0 t + /nj n (ne) sin ω 0 t (7) n=1 7

8 3 The Newtonian Appoximation In this chapte we will simplify the Einstein equations by assuming the metic to be nealy minkowskian (g µν = η µν + h µν with h µν 1). The esulting equation is called the lineaized Einstein equations, since we keep only the tems linea in h µν. The lineaized Einstein equations have the fom of a wave function, which suggests the existence of gavitational waves. If we also assume low velocities (v << 1) we obtain the Newtonian equations of motion. In this chapte we have often used [9] and [13]. 3.1 The Lineaized Einstein Equations When the metic is nealy flat (nealy Minkowskian), we ae able to pick a coodinate system in which the metic can be witten in this fom: g µν = η µν + h µν h µν 1 (8) Since h µν is aleady vey small, quadatic tems in h µν can be neglected. We find the invese of the metic using g µν g να = δ α µ. When guessing: g µν = η να h να (9) we indeed find to fist ode: g µν g να = η µν η να + η µν h να h µν η να = δ α µ (30) In its full fom the Einstein equations ae invaiant unde a huge symmety goup, namely the goup of all possible coodinate tansfomation x µ x µ. Choosing a nealy flat metic means we ve boken this symmety. We can howeve still pefom the gauge tansfomations x µ x µ = x µ + ξ µ (x) which keeps the weak-field metic unchanged. To find how h µν tansfoms unde this gauge tansfomation, we use the geneal metic tansfomation ule, g µν (x ) = xρ x µ x σ x ν g ρσ(x) (31) The tansfomation matices become: x µ x ρ = [xµ + ξ µ ] x ρ = δ µ ρ + µ ξ µ x µ x ρ = δµ ρ µ ξ µ (3) In the last step we ve used that the tansfomation metic times its invese gives the identity matix. Filling this into the metic tansfomation ule we find: h µν (x) h µν (x ) = h µν (x) µ ξ ν ν ξ µ (33) When µ ξ ν is of same ode o smalle then h µν, the condition h µν 1 is indeed peseved. In case of the nealy flat metic (8) the Riemann tenso educes to: R µνρσ = 1 ( ν ρ h µσ + µ σ h νρ µ ρ h νσ ν σ h µρ ) (34) Inseting equation (33) in the last equation we find that the lineaized Riemann tenso stays invaiant unde gauge tansfomations. We now define: h µν = h µν 1 η µνh h = η µν h µν (35) 8

9 Obseve that h η µν h µν = h h = h, having used η µν η µν = 4. We can use this to invet the last equation: h µν = h µν 1 η µν h (36) In this notation the Einstein equations become: h µν + η µν ρ σ hρσ ρ ν hµρ ρ µ hνρ = 16πT µν (37) This equation would dastically simplify if we could set: ν hµν = 0 (38) This is called the hamonic gauge condition. Luckily, it is always possible to pefom a gauge tansfomation to a system whee this condition is satisfied. In tems of h µν the gauge tansfomation (33) becomes: Theefoe: h µν (x) h µν (x ) = h µν (x) µ ξ ν ν ξ µ + η µν ρ ξ ρ (39) ν hµν (x) ν hµν (x ) = ν hµν (x) ξ µ (40) So if the initial nea-flat field is such that ν h µν (x) = f µ (x), to obtain the hamonic gauge condition ν h µν (x ) = 0 in the new fame, we much choose ξ µ, such that ξ µ = f µ (x). This equation always has a solution: ξ µ (x) = d 4 xg(x y)f µ (y) x G(x y) = δ 4 (x y) (41) With the hamonic gauge condition, the esulting lineaized Einstein equations take the fom of a wave equation: h µν = 16πT µν (4) Diffeentiation both sides and using the hamonic gauge condition, we find an expession fo the consevation of enegy-momentum µ T µν = 0. The physical solutions to lineaized Einstein equations ae: h µν = 16π d 3 x G(x x )T µν (x ) G(x x 1 ) = 4π x x δ(t et t ) (43) Hee t et = t x x. This etaded time eflects the fact that gavitational waves fom souce point x need a cetain time to aive at the obseve at position x. Fom the expession of the etaded time we see the velocity of the gavitational waves is equal to the speed of light. The othe possible solution to the lineaized Einstein equations is witten in tems of the advanced Geen s function, but these give unphysical solutions, since this would mean that the obseve might notice the effect of gavitational waves befoe they ae emitted. We wite the solution as: h µν (t, x) = 4 d 3 x 1 x x T µν(t x x, x ) (44) In contast Newton s theoy is govened by Poisson s equation, which has the solution: ρ Φ = x x d3 x (45) Notice thee is no time delay in the Newtonian case, which means the potential acts instantly. Einstein showed that in case of gavity thee was no mysteious action at a distance. 9

10 When x = is much geate than the souce d, we can expand x x as: x x = x n +... (46) So fa away fom the souce we aive at: h µν (t, x) = 4 d 3 x T µν (t + x n, x ) (47) In the Newtonian limit we will also assume the velocities to be small (v << 1). In this limit the enegy-momentum tenso will be dominated by T 00 = ρ, since the othe components ae dependent on the velocity. Looking at the lineaized Einstein equations (4), we see that the dominance of T 00 immediately tanslates to the dominance of h 00. Since the velocities ae small, the metic only changes slowly, so we can also neglect the time deivative of the metic. Theefoe we ae only left with: h 00 = 16πρ (48) Compaing this to Poisson s equation of Newtonian mechanics we find h 00 = 4Φ. When h 00 is dominant we have h = h 00 = 4Φ and fom equation (35) we find h 00 = Φ. Using equation (35) fo µ = ν = i we find h ii = Φ. The line element is theefoe given by: ds = (1 + Φ)dt + (1 Φ)(dx + dy + dz ) (49) 10

11 4 The Schwazschild Solution In this chapte we will study the full geneal elativistic solution of a binay system whee one of the paticles has negligible mass compaed to the othe (m 1 >> m ). The metic, called the Schwazschild metic, is theefoe only detemined by the lage mass. We stat by finding the metic fo this binay system. Using this we deive the adial equation and find the effective potential, which we will compae to the Newtonian vesions fom chapte. We will then study the shape of the elliptical obits, which will not tun out to be closed like in the Newtonian case. Finally we will deive the Schwazschild metic fo weak fields in hamonic coodinates, which we will need in chapte 6. In this chapte we have often used [13] and [18]. 4.1 Deiving the Schwazschild Metic Since the motion of the massive paticle isn t influenced by the second paticle, we can feely pick a fame in which the massive paticle is at est. Since the paticle will not move, the metic will be independent of time. Fo a point paticle the metic can also be no diectional dependence, so it will be spheically symmetic. Theefoe, the metic has this geneal fom: ds = A()dt + B()d + C() dω dω dθ + sin θ dϕ (50) We can simplify this equation with a change in coodinates. We set C() ( ). To change fom d to d we wite: d d = d(c() ) d = dc() ( C() ) + C() d = d d + C() d (51) We can absob the exta tem that appeas befoe d in the function B(). The metic becomes: ds = A( )dt + B( )d + dω (5) Since T µν does not contibute except at the massive paticle the Einstein equations simplify to R µν = 0. Witing out these equations in tems of the metic we end up with only thee independent equations (the last two equations only diffe by a facto sin θ): 4ḂA ÄAB + Ḃ AA + A B = 0 (53) ḂA + B A AB ȦB = 0 (54) ÄAB + Ḃ AA + Ȧ B 4AAB = 0 (55) sin θ[ ÄAB + ḂȦA + Ȧ B 4 AAB] = 0 (56) Subtacting equation (53) fom equation (55) we find: ȦB + AḂ = d dt (AB) = 0 AB = C 1 (57) Hee C 1 is a constant. Inseting this in equation (54) we end up with: Ȧ = A(1 A) (58) The solution to this equation is: A = ( ) 1 C And the metic thus becomes: ( ds = C ) ( dt ) 1d + dω (60) C C 11 (59)

12 To find the constants we compae g tt with the Newtonian metic, which fo a point-mass is equal to: ( g tt = 1 m ) 1 (61) We find C 1 = 1 and C = 1/(m 1 ). Notice that tuns out to be equal to. So finally the Schwazschild metic becomes: ( ds = 1 m ) ( 1 dt + 1 m ) 1 1d + dω (6) 4. The Schwazschild Obits The second paticle with mass m << m 1 moves on a geodesic in the metic of paticle m 1. The geodesic equation is: du ρ dτ + Γρ µνu µ u ν Multiplying with g ρν we find: (63) du ν dτ = Γρ µνu µ u ρ = Γ ρ µνu µ ρ = 1 ( µg σν + ν g µσ σ g µν )u µ u σ (64) If you switch ν and σ in the fist and thid tem in the paentheses you can see this pai is antisymmetic. u σ u µ on the othe hand is symmetic. A symmetic times an antisymmetic tenso gives zeo, so we ae only left with: du ν dτ = 1 νg µσ u µ u σ (65) This equation tells us that if all components of g µσ ae independent of some x ν with a paticula ν, we have ν g µσ = 0. Fom equation (65) we see this implies that u ν is a constant along the entie geodesic. Since the Schwazschild metic is time-independent, we have a constant of motion u 0. Since u 0 = p 0 /m, this constant is equal to the enegy pe unit mass. The metic is also independent of the angle ϕ, giving us consevation of angula momentum. This implies the motion is confined to a plane, just as in the Newtonian case. Fo convenience we choose θ = π/. We find: p 0 = E p ϕ = J (66) Using p µ p ν = m we get: g µν p µ p ν = g 00 p 0 + g p + g ϕϕ p ϕ = m (67) Filling in the components we aive at: ( 1 m 1 ) 1E + ( 1 m 1 ) 1m ( d ) J + dτ = m (68) This equation can be ewitten to find the adial equation and the associated effective potential: ( d ) ( = Ẽ dτ ( V1 = 1 m 1 1 m 1 )(1 + J ) Hee we have used Ẽ = E/m. )(1 + J ) = Ẽ V 1 (69) (70) 1

13 To show that in the limit of weak fields and low velocities we find back the Newtonian adial equation (11), we use: E m m = p The adial equation becomes: ( d dτ ) ( p = + m 1 J + m J 1 ( p ) 3 ) = V V = m 1 + J m J 1 3 (73) In the weak field limit, the m 1 / 3 tem can be neglected. In the low velocity limit the tem p / becomes the kinetic enegy pe unit mass. Rewiting the adial equation we get: p = 1 ( d ) m 1 dτ + J m J 1 3 (74) The ight hand side of this equation educes to the total Newtonian enegy (10). The cuious fact that the kinetic enegy seems to incopoate the Newtonian potential eflects the fact that the paticle on a geodesic expeiences no potential enegy (yet it follows a cuved path as though it does). So taking p / as the Newtonian equivalent of the total enegy, we find back the Newtonian adial equation (11). The effective potential V is plotted in the next diagam. (71) (7) Instead of the effective potential going all the way up fo small, as in the Newtonian case, it goes all the way down. Notice that the left side of the adial equation is always positive o zeo, theefoe p / will always be highe than the V. The velocity of the paticle is zeo when p/ = V. Diffeentiating the adial equation with espect to τ we find: d d dτ dτ = dv d d dτ d dτ = dv d (75) Fom this we find that only at the minima o maxima both the velocity and the acceleation ae zeo, descibing cicula obits. At a maximum the cicula obit is unstable and theefoe of no physical significance. Fo cicula obits the last equation becomes: 0 = d d ( m 1 + J m J 1 ) 3 = J ( ) 1 ± 1 1m 1 m 1 J Fom this equation we see thee ae two cicula obits fo J > 1m 1. At J = 1m 1 thee is only one obit left and fo J < 1m 1 we have no cicula obits at all. If a paticle is stuck in the potential well ( p/ < 0), we have the Einsteinian equivalent of the Newtonian elliptical obits. If a paticle with 0 < p / < V max comes in fom a distance it will at some point be equal to the potential and tun aound. These tajectoies ae the Einsteinian equivalents to the Newtonian hypebolic obits. If the enegy is lage than the maximum of the potential ( p/ > V max ), the paticle dops ight towads = 0 neve to etun. (76) 13

14 4.3 The Pecession of Mecuy To find the obital equations fo the quasi-elliptical obits we stat with: dϕ dτ u ϕ = g ϕϕ p ϕ = J m (77) Dividing the squae of the adial equation (6) by the squae of the last equation we find: ( d dτ ) / ( dϕ dτ ) = ( d dϕ ) = Ẽ (1 m1 J )(1 + ) (78) J / 4 Solving this equation, while neglecting tems of ode 1/ 3 we find back the Newtonian ellipse (16): 1 = m 1 J + Ẽ + m 1 / J 1 cos(ϕ) (79) J To see how this equation is equal to the Newtonian equation we again have to equate p / with the Newtonian Ẽ. As noted befoe, this fomula descibed a closed ellipse. Fo a cicula obit we have 1/ = m 1 / J (see equation (16)), so we can define y = 1/ m 1 / J to be the deviation fom ciculaity. In tems of y the solution becomes: Ẽ y = + m 1 / J 1 cos(ϕ) (80) J In the elativistic case we will not neglect the 1/ 3 tems. We will assume the obit to be nealy cicula, which makes y small enough to be able to neglect tems of ode y 3. Fo nealy cicula Einsteinian obits we find this solution: y = 3m3 1 k J + 1 Ẽ + m 1 / J 1 ( + m4 1 3m 3 ) k J J 1 cos(kϕ) (81) 6 k J k = 1 6m 1 J Since k isn t equal to 1 the obit doesn t etun to the same point afte a otation of ϕ = π. This means the ellipse pecesses. Fom one peiaston to the next need a otation of ϕ = π/k instead. The angle fom one peiaston to the next theefoe given by: δϕ = π k = π ( 1 6m 1 J ) 1/ Fo nealy cicula obits we have 1/ m 1 / J. Fo weak fields we have m 1 / 1. Combining these appoximations we get m 1/ J 1. Taylo expanding this equation in this appoximation we find: ( ) δϕ π 1 + 3m 1 (84) J The deviation fom π is called the peiaston shift ϕ: (8) (83) ϕ 6πm 1/ J 6πm 1 a(1 e ) (85) In the last equation we used the Newtonian equation (18). The numeical value of this quantity fo Mecuy is 43 pe centuy, which is in accod with obsevation. 14

15 4.4 Hamonic Coodinates In chapte 6 we will study the fist elativistic effect of a binay system whee both masses influence the metic. To deive these effects it is vey convenient to use hamonic coodinates. Since the Schwazschild solution is a limiting case of this system, we shall need the Schwazschild solution in hamonic coodinates as well. In this paagaph we will show how the hamonic coodinates follow fom the hamonic coodinate conditions, which we intoduced in the pevious chapte. We will stat with a genealization of the hamonic coodinate conditions, which will tun out to be equivalent to the expession fom the last chapte in case of weak fields. The genealization is given by: g µν Γ λ µν = 0 (86) To show that it is always possible to pick a coodinate system in which this is tue, we stat with the tansfomation of the Chistoffel symbol: Γ λ µν = x λ x ρ x τ x σ x µ x ν Γρ τσ xρ x ν Multiplying both sides with g µν we find: g µν Γ λ µν = x λ x λ x σ x µ x λ x ρ x σ (87) x ρ gτσ Γ ρ τσ g ρσ x ρ x σ (88) So if the left hand side of the pevious equation doesn t aleady vanish, we can define new coodinates in which it does by solving the next equation fo x : g ρσ x λ λ x ρ x σ = x x ρ gτσ Γ ρ τσ (89) The contacted connection can also be witten as: g ρσ Γ ν ρσ = 1 ( gg µν ) (90) g x ν Theefoe the hamonic coodinate conditions ae often witten in this fom as well: x ν ( gg µν ) = 0 (91) Fo weak fields equation (89) becomes: (η ρσ + h ρσ ) x λ x ρ x σ = 1 x λ x ρ ηµν η ρα ( ν h αµ + µ h αν α h µν ) (9) Setting x j = x j + ξ j, the gauge tansfomation, this takes the fom: (η ρσ + h ρσ ) (x λ + ξ λ ) x ρ x σ = 1 (x λ + ξ λ ) x ρ η µν η ρα ( ν h αµ + µ h αν α h µν ) (93) ξ λ = 1 ηµν η λα ( ν h αµ + µ h αν α h µν ) = µ h λ µ 1 λ h (94) Multiplying both sides with η µλ we find: ξ µ = λ h λ µ 1 η µλ λ h = λ h λµ 1 η µλ λ h = λ hµλ (95) We saw this equation befoe in the pevious chapte (see the discussion afte equation (40)). This was exactly the condition that needed to be satisfied in ode fo the hamonic gauge condition to be satisfied. We also saw that this equation always has a solution. 15

16 To satisfy the hamonic coodinate conditions fo the Schwazschild metic we will need these coodinates: X 1 = R() sin θ cos ϕ X = R() sin θ sin ϕ X 3 = R() cos θ (96) Fo simplicity we will wite Schwazschild metic as ds = B()dt + A()d + dω. Denoting diffeentiation with espect to with a pime, we find fom equation (89): g µν X i x µ x µ X i x λ gµν Γ λ µν = X (( i B AR B + A A Fom this we find that the coodinates ae hamonic when R() satisfies: d ( B d A ) R + R A R ) = 0 (97) dr ) ABR = 0 (98) d Filling in A and B this equation becomes: d ( d ( 1 m 1 The solution to the equation is: ) dr ) R = 0 (99) d R = M (100) In tems of R the Schwazschild line element takes this fom: ds = 1 m 1/R 1 + m 1 /R dt 1 + m 1/R 1 m 1 /R dr (1 + m 1 /R) R dω (101) Using the following elations: We find: X X = R d X = dr + R dω RdR = X d X (10) ds = 1 m 1/R 1 + m 1 /R dt (1 + m 1 /R) dx 1 + m 1/R m 1 1 m 1 /R R 4 ( X dx) (103) The metic in hamonic coodinates thus becomes: g 00 = 1 m 1/R 1 + m 1 /R ( g ij = 1 + m 1 R g i0 = 0 ) δij + ( m1 R ) 1 + m1 /R X i X j 1 m 1 /R R Keeping only tems up to the second ode in Φ we find: (104) g 00 = 1 + m 1 g ij = δ ij + δ ij m 1 g 0i = 0 ( m1 ) (105) 16

17 5 Gavitational Waves Unlike the Newtonian equations of motion, the solution to the lineaized Einstein equation suggests the existence of gavitational waves (as we saw in chapte 3). Since these waves take away enegy and angula momentum fom the system, they influence the two-body motion and ae thus of inteest to us. In this chapte we will fist deive the metic of the gavitational waves in lineaized theoy. To do this it is most convenient to use the tansvese-taceless (TT) gauge, which we will intoduce. Using the metic to find the influence of the waves on test-paticles, we find that the waves do wok of these paticles and thus have enegy themselves. If the waves, descibed by some h µν, have enegy, this enegy will in tun influences the metic. Theefoe we can t hold on to the lineaized appoximation g µν (x) = η µν +h µν (x). Instead we have to use g µν (x) = ḡ µν (x)+h µν (x), whee ḡ µν can be diffeent fom the minkowski metic. Finally we will obtain an expession fo the enegy, which allows us to find the metic of the gavitational waves fo a given enegy-momentum configuation. This esult is called the quadupole fomula, which was fist deived by Einstein. In the last thee paagaphs we apply this fomula to the Newtonian binay system in the case of cicula and elliptical obits. Extacting the gavitational enegy fom these obits we find that the paticles stat to spial inwad. We also show, using effective potentials, that the inspialing ciculaizes the obits. In this chapte we have often used [9] and [13]. 5.1 Gavitational Waves in Linea Theoy The lineaized Einstein equations outside the enegy-momentum souce ae: h 00 = 0 (106) This is a wave-equation, the solution of which is: h µν = A µν e ik σx σ (107) Putting this solution in the lineaized Einstein equations we get: h 00 = α α hµν = k α k α hµν = 0 (108) This implies k α k α = 0, which means that the wave vecto is a null vecto (the waves tavel with the speed of light as mentioned in chapte 3). With the solution (107) the hamonic gauge condition becomes: µ hµν = ik µ A µν e ik σx σ = 0 (109) We thus find k µ A µν = 0, which means that the wave vecto is pependicula to A µν. The hamonic gauge condition doesn t fix the gauge completely. We have still some esidual gauge feedom left, since we haven t completely specified ξ µ yet. Setting ξ µ in equation (40) to zeo, we find the solution ξ α = B α e ik µx µ. Inseting this solution togethe with equation (107) into equation (40), we find: A µν = A µν ib µ k ν ib ν k µ + iη µν B ρ k ρ (110) This equation can be used to fully specify ξ µ. Multiplying with η µν we obtain: A µµ = A µ µ + ib µ k µ (111) We can take A µµ = 0 if we demand: i Aµ µ = B µ k µ (11) 17

18 This implies that h µ µ = h = 0, which in tun gives us h µν = h µν. We secondly impose A 0i = 0. Unde these two estiction the hamonic coodinate condition fo µ = 0 simplifies to: 0 h00 = 0 (113) Hee we see that h 00 is independent of time. Theefoe h 00 is the static pat of the metic, the Newtonian potential of the souce. The gavitational wave is obviously the time-dependent pat, so we can set this tem to zeo. A 0i = 0 theefoe implies A 0ν = 0. Fist we investigate the case ν = 0, which will give us the equation fo B 0 with which we satisfy A 00 = 0: A 00 ik 0 B 0 ik µ B µ = 0 (114) A 00 ik 0 B 0 1 Aµ µ = 0 B 0 = i ( A ) k 0 Aµ µ Now fo ν = i, finding an expession fo B j to satisfy A 0i = 0, we obtain: A 0i ik 0 B j ik j B 0 = 0 (115) A 0i ik 0 B j ik j [ i (A /A µ k µ)] = 0 0 B j = i [ ( k 0 A 0i + k j A )] Aµ µ k 0 If we plug these choices fo B µ in equation (11), using k α k α = 0 and k µ A µν = 0, we find it immediately satisfied. Note that we oiginally had 10 independent components in the metic. The hamonic gauge fixed 4 degees of feedom, leaving only six. Choosing B µ as we did above fixes all components of the distance vecto ξ µ, leaving only two degees of feedom. Note that equation (11) does not equie exta degees of feedom, since it is satisfied by ou choice of B µ without exta modifications. So we have now used up all ou gauge feedom. Having set all the gauge dependent components to zeo, all that is left physical components (as opposed to coodinate dependent contibutions). The hamonic gauge with these gauge estictions now simplifies to j h ij = 0. Summaizing this, we have: h 0µ = 0 h i i = 0 j h ij = 0 (116) Fom these equations we see that the tempoal pat of the metic is zeo, leaving us with only the spatial pat h ij. The metic is now witten in tansvese taceless (TT) fom. It is called taceless since h i i = 0 and tansvese since k µa µν = 0). If we fo instance pick a momentum vecto pointing in the z-diection, the eal of h T ij T becomes: h T T ij = h + h 0 h h cos[ω(t z)] (117) Notice that we ve demanded that A ij is taceless and, as usual, symmetic. We ve also used k = ω. It is useful to invent an opeation Lambda that can change a metic just satisfying the hamonic gauge conditions, into a metic in the tansvese-taceless fom. h T T ij = Λ ij,kl h kl (118) We do this with the following opeato: Λ ij,kl = P ik P jl 1 P ijp kl P ij = δ ij n i n j (119) 18

19 So what fame did we actually pick when we chose the TT gauge? To undestand this we have to look at the motion of test paticles. We fist have to take the diffeence between the geodesic equation fo a paticle on the geodesic x µ and one on the neaby geodesic x µ + ξ µ. The latte geodesic equation takes the fom: d (x µ + ξ µ ) dτ + Γ µ νρ(x + ξ) d(xν + ξ ν ) dτ d(x ρ + xi ρ ) dτ = 0 (10) The diffeence between these two equations gives us the geodesic deviation equation: d xi µ dτ + Γ µ νρ(x) xν ξ ρ dτ dτ + ξσ σ Γ µ νρ(x) xν x ρ dτ dτ = 0 (11) Fo a test-paticle tempoaily at est at τ = 0 the geodesic equation becomes: d x i ( dx 0 ) dτ = Γi 00 (at τ = 0) (1) dτ Expanding the Chistoffel symbol we find: Γ i 00 = 1 ( 0h oi i h 00 ) (13) In the TT gauge this immediately vanishes, since h 00 and h 0i ae both zeo in this fame. So if at moment τ = 0 the velocity is zeo (by choice), the last equation tells us that the acceleation must also be zeo. This means that if the velocity is zeo at time τ = 0 it will be zeo always. The coodinates of the TT fame thus seem to stetch themselves in such a way that the position of fee masses at est do not change as the wave passes by. This is of couse not the fame used by expeimentalists. A simple laboatoy would by one which is feely falling. Fo a sufficiently small laboatoy we have a flat metic, even in the pesence of gavitational waves. This way we can set Γ λ µν to zeo. Assuming the detecto moves non-elativistically, the spatial component of the geodesic deviation equation (11) becomes: d ξ i dτ + ξi i Γ i 00 = 0 (14) Since R i 0j0 = jγ i 00 in this appoximation, we have: ξ i = R i 0j0ξ j (15) We wote this equation in this fom, because the Riemann tenso is invaiant in lineaized theoy. This means we can compute it in any fame we want to with the same esult. Picking the TT gauge fo convenience we find: R i 0j0 = R i0j0 = 1 ḧt T ij (16) So the geodesic deviation equation in this fame becomes: ξ i = 1 ḧt ij T ξ j (17) This equation states that the effect of gavitational waves can be descibed in tems of a Newtonian foce: F i = m ḧt ij T ξ j (18) Let s conside a ing of test paticles in the (x,y) plane centeed aound the oigin and have the gavitational wave come in fom the z-diection. Fo this wave the TT metic components with 19

20 eithe i = 3 o j = 3 ae zeo. Theefoe we see fom the geodesic deviation equation (17) that a paticle that is initially in the z = 0 plane will stay thee. Since h T ij T sin ωt we can pick t = 0 at a moment when h T ij T is zeo. If at fist we only look at the plus-polaization h +, we have: h T ij T = h + sin ωt (19) Let s conside a paticle at the small distance fom the oigin ξ a = (x 0 + δx(t), y 0 + δy(t)), whee x 0 and y 0 is the initial position of the paticle and δx(t) and δy(t) epesent the displacement due to the gavitational wave. In that case the geodesic deviation equation (17) becomes: δẍ = h + (x 0 + δx)ω sin ωt δÿ = h + (y 0 + δy)ω sin ωt (130) Since δx x 0 we can neglect it on the ight side of both equations. Integating we find: δx = h + x 0 sin ωt δy = h + y 0 sin ωt (131) Similaly fo h we get: δx = h y 0 sin ωt δy = h x 0 sin ωt (13) The defomation of the ing of test-paticles is depicted on the ight fo diffeent moments in time. Fom the shape it is clea whee the teminology plus and coss polaization came fom. 5. The Enegy of Gavitational Waves We ve seen that the motion of test paticles by a gavitational wave can be descibed with a Newtonian foce. If a gavitational wave momentaily passes two test paticle connected by a sping, the sping will stat to oscillate. Fiction will cause this motion to subside afte a while, ceating heat. This pocess suggests that gavitational waves can do wok and theefoe they must have enegy. The enegy-momentum tenso of fo instance electomagnetism is quadatic in the elevant fields (which ae descibed by F µν ). In ou weak field limit we only kept linea tems, so we might need to take the expansion a step futhe to find an expession fo this gavitational enegy. The fist ode vacuum Einstein equations can be used to find h µν : R (1) µν = 0 (133) Hee R µν (n) contain of tems fom the Ricci tenso of n-th ode in h. Although h µν is a solution of the fist ode, it is geneally not a solution of the full Einstein equations and not even the second ode, so R µν () 0. The lineaized expansion is theefoe not sufficient fo highe odes. To be able to set the Einstein equations of second ode to zeo again, we need an exta petubation tem, R µν of ode h : R µν + R (1) µν + R () µν = 0 (134) To solve this poblem we have to incopoate the effect of gavitational wave on the backgound space-time. Accoding to geneal elativity enegy influences the metic. So if gavitational waves 0

21 have enegy, they must cuve space-time as well. Up until now ou backgound space-time was assumed flat (o Minkowskian). To have a fixed backgound excludes the possibility that the gavitational waves can affect it. So instead we have to take: g µν = ḡ µν (x) + h µν (135) Fa away fom any souces ḡ µν (x) tuns into the minkowskian metic. If gavitational waves ae also absent we also have h µν = 0. The question becomes how to distinguish between the pats that belong to ḡ µν (x) and the pats that belong to h µν (x). If we knew the wavefom aleady, this would have been an easy task. Howeve, fo ealistic astophysical souces these have not been found. Thee ae howeve some conditions in which the distinction is possible. A natual way to do this is by assuming that the typical scale of vaiation L B of ḡ µν is much lage than the educed wavelength λ of the gavitational waves: λ L B (136) Altenatively, the backgound can have a maximum fequency that is much smalle than the typical fequency ange of the waves. In this case the backgound is slowly vaying, while the waves ae high-fequency petubations: f B,max f GW (137) The fist tem of equation (134) is only dependent on ḡ µν. It theefoe contains only low fequency modes. R µν (1) on the othe hand only contains high fequency modes, since it is dependent on tems linea in h µν. R µν () can depend on both high and low fequency modes. Conside fo example a high wave-vecto k 1 combined with anothe high wave-vecto k k 1. Recalling that h µν e ikx the quadatic tems might be in the low fequency ange. The Einstein equations can theefoe be witten as: ( R µν = [R µν () ] low + 8π T µν 1 ) low g µνt (138) ( R µν (1) = [R µν () ] high + 8π T µν 1 ) high g µνt (139) Low and high denote espectively low and high fequencies o high o low wavelengths. We now intoduce a length l such that λ l L. Aveaging ove spatial volumes l would leave L unaffected, since they ae basically constant ove the volume. Howeve, the quick petubations λ ae aveaged to zeo. Altenatively, if we distinguish by fequency, we use a tempoal aveage ove seveal peiods of the gavitational wave. This means that we can wite the low mode Einstein Equations as: R µν = R () µν + 8π T µν 1 g µνt (140) We also wite T µν = T µν. Since T µν coesponds to the low fequency pat, it is a smooth vesion of T µν. Since in typical situations macoscopic matte distibutions ae aleady quite smooth we have T µν T µν. At lage distances fom the souce covaiant deivatives tun into nomal deivatives. R () µν at lage distances at lage distances becomes: R () µν = 1 [ 1 µh αβ ν h αβ + h αβ ν µ h αβ h αβ ν β h αµ h αβ µ β h αν + (141) h αβ α β h µν + β h α ν β h αµ β h α ν α h βµ β h αβ ν h αµ + β h αβ α h µν β h αβ µ h αν 1 α h α h µν + 1 α h ν h αµ + 1 α h µ h αν ] 1

22 Picking the TT gauge, which includes the hamonic gauge condition µ h µν = 0, we can dastically simplify the aveage of R µν (). To do this we need integation by pats. Note that geneally integation by pats of t is possible only if we have pefomed an integal ove time as well. Similaly, i equies a spatial integal. Since h µν is a function of the combination t z (see equation (117)), the spatial integal is equal to minus the tempoal integal, so integation by pats can be safely applied. Because we ae aveaging ove all the diections at each event, the aveage of any deivative tem vanishes, µ X = 0, we theefoe have A( µ B) = ( µ A)B. We immediately see that all tems of R µν () disappea, accept fo the fist two which ae elated by integation by pats. We find: R () µν = 1 4 µh αβ n uh αβ (14) Without a matte contibution the low-mode Einstein equations become: R µν = 1 4 µh αβ n uh αβ (143) This expesses the fact that the deivatives of the petubation affects the cuvatue of the backgound metic ḡ µν. We also define: t µν = 1 8π R() µν 1 ḡµνr () (144) The tace of t µν is: t = g µν t µν = 1 8π R() (145) Hee we used that g µν R () µν = ḡ µν R () µν = ḡ µν R () µν since ḡ µν by definition is a puely low fequency quantity. Inseting the tace into the equation fo t µν we find: R () µν = 8π(t µν 1 ḡµνt) (146) The low fequency Einstein Equations become: R µν = 8π(t µν 1 ḡµνt) + 8π( T µν 1 ḡµν T ) (147) O equivalently: R µν 1 ḡµν R = 8π(T µν + t µν ) (148) This looks like the egula Einstein equations, but now we have an exta tem t µν which is not dependent on matte, but only on the gavitational field and is quadatic in h µν. Using the Bianchi identity we have µ ( R µν 1 ḡµν R) = 0. This implies: µ ( T µν + t µν ) = 0 (149) The enegy of the matte and the gavitational waves ae togethe conseved. This eflects the fact that enegy can be exchanged between the two. With T µν = 0 the last equation simplifies to: µ t µν = 0 (150) The aveage in the definition of t µν is necessay, since without it the expession would not be gauge invaiant (and theefoe contain some fame dependent tems). Because of the ability to

23 choose local minkowskian coodinates in any space-time, it is impossible to define a local measue of gavitational enegy-momentum, we could have expected that some aveaging pocedue was necessay. Aveaging ove seveal wavelengths o peiods we can pick up the physical cuvatue to obtain a gauge invaiant measue. Filling in equation (144) and using R () = 0, we get: t µν = 1 3π µh αβ ν h αβ (151) In paticula the enegy-density becomes: t 00 = 1 3π ḣt ij T ḣ T ij T = 1 16π ḣ + + ḣ (15) The gavitational enegy inside a volume is: E V = d 3 xt 00 = d 3 x 1 3π ḣt ij T ḣ T ij T (153) V V Fa fom mateial souces we can use 0 t 00 = i t 0i to find an expession fo the enegy flux: Ė V = d 3 x 0 t 00 = d 3 x i t i0 = dan i t 0i (154) V V S In the last step we ve used the Gauss s theoem. If S is a spheical suface, we have da = dω and n = ˆ: Ė V = dωt 0 (155) S As we saw at the end of chapte 3, at lage distances we have: h T T ij = 1 f ij(t ) (156) Fom this we find: h T T ij = 1 f ij + 1 f ij = 1 f ij 1 0f ij 0 h T T ij = 0 h T T ij (157) So at lage distances we have t 0 = t 00. We find: Ė V = dω h T ij T h T ij T 3π (158) The same equation fo t µν can be obtained fom Noethe s theoem. We shall apply Noethe s theoem in exactly the way we would electomagnetic theoy. In that case we usually use: L EM = 1 4 F µνf µν (159) We find: T µν EM = L EM ( µ A ρ ) ν A ρ + η µν L EM (160) T µν EM = F µρ F ν ρ 1 4 η µνf + ρ (F µν A ν ) (161) 3

24 The familia expession doesn t contain the last tem, so we would like to get id of it. If we take an aveage ove a volume lage enough that all bounday tems vanish, this tem indeed disappeas. So we need the aveage to get the coect answe: T µν EM = F µρ F ν ρ 1 4 η µνf (16) In case of electomagnetism this final aveaged expession also happens to be locally tue, but this fact cannot be fom Noethe s theoem and equies expeimental veification. Similaly, staight fom the full Hilbet-Einstein action to second ode we find: L E = 1 64π [ µh αβ µ h αβ µ h µ h + µ h µν ν h µ h µν ρ h ρ ν] (163) t µν = L ( µ h αβ ) ν h αβ η µν L (164) We obtain: t µν = 1 3π µ h αβ ν h αβ (165) Although the electomagnetic expession is also tue locally, we have seen that fo gavitational waves this is the best we can do. 5.3 The Quadupole Moment We have seen that the solution to the lineaized Einstein equations fa away fom the matte souces is: h µν (t, x) = 4 d 3 x T µν (t + x n, x ) (166) Using h T T ij = Λ ij,kl h kl = Λ ij,kl hkl (since Λ ii,kl = Λ ij,kk = 0), we can wite this in the TT gauge: d 3 x T kl (t + x n, x ) (167) h T ij T (t, x) = 4 Λ ij,kl Notice that we changed T µν to T kl. This is possible since the T 00 and T 0k ae elated to each othe by 0 T 00 = i T i0 and 0 T 0k = i T ik. Applying a Fouie tansfomation we find: T kl (t + x n, x d 4 k ) = (π) T 4 kl (ω, k)e iω(t + x n)+i k x (168) The fequency of the motion inside the souce of size d is ω s v/d. As we shall see late in this chapte, the fequency of the souce is of the same ode as the fequency of the gavitational adiation, we thus have ω GW v/d. Fo non-elativistic souces this becomes ω s d 1. The integal in the last equation is esticted to the small matte egion x d. Theefoe we have: ω x n ω s d 1 (169) To see this it is useful to look at the figue at the end of chapte 3. With this we can simplify the exponent in the last integal: e iω(t + x n) = e iω(t )[ 1 iωx i n i + 1 ] ( iω) x i x j n i n j +... (170) Putting this back in the last integal and invesing the Fouie tansfomation we get: h T ij T (t, x) = 4 Λ ij,kl d 3 x (T kl (t, x ) + x i n i 0 T kl + 1 x i x j n i n j 0T kl+... ) (171) 4

25 We now define: S ij = d 3 xt ij S ij,k = We find: d 3 xt ij x k S ij,kl = d 3 xt ij x k x l (17) h T µν T (t, x) = 1 [ 4Λ ij,kl S kl + n m Ṡ kl,m + 1 ] n mn Skl,mp p +... et (173) Whee the subscipt et means that the expession is evaluated at the etaded time t. The fomula becomes moe physically meaningful if we intoduce: I = d 3 xt 00 I i = d 3 xt 00 x i I ij = d 3 xt 00 x i x j (174) P i = d 3 xt 0i P i,j = d 3 xt 0i x j P i,jk = d 3 xt 0i x j x k (175) Imagine a box with a volume V lage than the souce. In that case the enegy momentum tenso at the bounday is zeo. Using 0 T 00 = i T 0i we find: I = d 3 x 0 T 00 = d 3 x i T 0i = ds i T 0i = 0 (176) V V S This implies the consevation of enegy T 00. At fist sight this might seem supising, since the system loses enegy by gavitational adiation. The consevation does hold howeve in the lineaized appoximation, whee the effect of adiation can still be neglected on the equations of motion. A simila calculation fo I i gives: I i = d 3 xx i 0 T 00 = d 3 xx i i T 0j = d 3 x( j x i )T 0j = d 3 xδjt i 0j = P i (177) V Continuing we find: I ij = P i,j + P j,i P i = 0 V P i,j = S ij V Iijk = P i,jk + P j,ki + P k,ij (178) Notice that P i implies convesation of momentum. Finally we deive: V P i,jk = S ij,k + S ik,j (179) S ij = 1 Ïij (180)... I ijk = (Ṡij,k + Ṡik,j + Ṡjk,i ) (181) Ṡ ij,k = 1... I ijk ( P i,jk + P j,ik P k,ij ) (18) In this notation the leading tem in the metic becomes: h T T µν = 1 Λ ij,klïkl (t ) (183) Notice that the leading tem is aleady a quadupole tem. The monopole tem Ï and the dipole tem Ïi = P i ae zeo because of the consevation of mass and momentum. When n = z we have: P ij = δ ij n i n j = (184) 5