Gravitational waves generation

Transcription

1 Chapte 4 Gavitational waves geneation In this hapte we will intodue the quadupole fomalism whih allows to estimate the gavitational enegy and the wavefoms emitted by an evolving physial system desibed by the stess-enegy tenso T µ. We shall solve eq. (3.7) unde the following assumption: we shall assume that the egion whee the soue is onfined, namely is muh smalle than the wavelenght of the emitted adiation,! x i <, T µ 6=0, (4.)!!! v typial, GW =!.Thisimpliesthat i.e. the veloities typial of the physial poesses we ae onsideing ae muh smalle than the speed of light; fo this eason this is alled the slow-motion appoximation. Let us onside the fist equation in (3.7) whee F hµ = 6 G 4 T µ, (4.) " h µ = h µ µ h and F = # t +. By Fouie-expanding both h µ and T µ T µ (t, x i ) = h µ (t, x i ) = + + T µ (!, x i )e i!t d!, (4.3) h µ (!, x i )e i!t d!, i =, 3 eq. (4.) beomes " whee +! # h µ (!, x i )= KT µ (!, x i ) (4.4) K = 6 G 4. (4.5)

2 We shall solve eq. (4.4) outside and inside the soue, mathing the two solutions aoss the soue bounday. The exteio solution Outside the soue T µ =0andeq. (4.4)beomes " +! In pola oodinates, the Laplaian opeato is = " # + " sin # + sin # h µ (!, x i )=0. (4.6) sin. We shall onside the simplest solution of this equation, i.e. one whih does not depend on and : h µ (!, ) = A µ (!) e i! + µ (!) e i!. This solution epesents a spheial wave, with an ingoing pat ( e i! ), and an outgoing ( e i! )pat;indeed,substitutingintheseondeq.(4.3) h µ (!, x i )by e ±i! the esult of the integation ove! gives a funtion of (t )espetively. Sine we ae inteested only in the wave emitted fom the soue, we shall set µ =0, and onside the solution h µ (!, ) = A µ (!) e i!. (4.7) This is the solution outside the soue and on its bounday, whee T µ vanishes as well. A µ is the wave amplitude to be found by solving the equations inside the soue. The wave equation " +! The inteio solution # h µ (!, x i )= KT µ (!, x i ) (4.8) an be solved fo eah assigned value of the indies µ,. Tosolveeq.(4.8)letusintegate ove the soue volume " # +! h µ (!, x i )d 3 x = K T µ (!, x i )d 3 x. The fist tem an be expanded as follows hµ (!, x i ) d 3 x = div[ hµ ] d 3 x = S hµ k dsk (4.9) whee hµ is the gadient of hµ, S is the sufae suounding the soue volume, and we have applied Gauss theoem to hµ. Using eq. (4.7) the sufae integal an be appoximated as follows

3 S =4 hµ k dsk ' 4 d d apple A µ e i! + A µ i! A µ e i!! e i! if we keep the leading tem and disad tems of ode, we find and eq. (4.8) beomes The seond tem 4 A µ + satisfies the following inequality hµ (!, x i ) d 3 x ' 4 A µ (!),! h µ (!, x i ) d 3 x = K! h µ (!, x i ) d 3 x = ; = T µ (!, x i ) d 3 x. (4.0)! h µ (!, x i ) d 3 x < h µ max! 4 3 3, (4.) whee h µ max is the maximum eahed by h µ in the volume,andsinetheight-hand side of eq. (4.) is of ode 3 it an be negleted. Consequently eq. (4.0) beomes 4 A µ (!) = K T µ (!, x i ) d 3 x (4.) i.e. A µ (!) = 4G T 4 µ (!, x i ) d 3 x. Thus, the solution of the wave equation inside the soue gives the wave amplitude A µ (!) as an integal of the stess-enegy tenso of the soue ove the soue volume. Knowing A µ (!) we finally find h µ (!, ) = 4G ei! 4 o, by the invese Fouie tansfom T µ (!, x i ) d 3 x, (4.3) h µ (t, ) = 4G T 4 µ (t,xi ) d 3 x. (4.4) This is the gavitational signal emitted by the soue. The integal in (4.4) an be futhe simplified, but in the meantime note that: It should be noted that e i! sine we have assumed that GW >>.

4 ) The solution (4.4) fo h µ automatially satisfies the seond eq. (3.7), i.e. the hamoni gauge ondition x h µ µ =0. To pove this, we fist notie that the solution (4.4) is equivalent to the expession (3.9) h µ (t, x) = 4G 4 T µ (t x-x 0, x 0 ) x-x 0 d 3 x 0 ; (4.5) indeed, sine then By defining the following funtion g (~x x 0 <, and, (4.6) x ' x-x. (4.7) ~x 0 ) 4G 5 x-x 0 " t 0 t x-x 0!#, (4.8) whee ~x =(t, x) and~x 0 =(t 0, x 0 ), eq. (4.5) an be witten as a fou-dimensional integal as follows h µ (~x )= T µ (~x 0 ) g (~x ~x 0 ) d 4 x 0, (4.9) whee I, andi is the time inteval to be taken suh that g(~x ~x 0 )vanishesatthe extema of I; thishappensifi is so lage that, fo all x 0,theexpessiont x-x 0 is inside I; indeed,fomthedefinition(4.8)g is di eent fom zeo only fo t 0 = t x-x 0. Sine g is a funtion of the di eene (~x ~x 0 ), then Consequently, x µ [g (~x ~x 0 )] = x h µ (~x )= T µ (~x 0 ) µ x g (~x ~x 0 ) d 4 x 0 = µ The last tem an be integated by pats and gives T µ (~x 0 ) x g (~x ~x 0 ) d 4 x 0 = µ 0 x µ 0 [g (~x ~x 0 )]. (4.0) T µ (~x 0 ) x g (~x ~x 0 ) d 4 x 0. (4.) µ 0 d 4 x 0 x [T µ (~x 0 )g (~x ~x 0 )] " µ 0 # d 4 x 0 x T µ (~x 0 )g (~x ~x 0 ) µ 0 d 4 x 0 =0. The fist integal vanishes sine T µ =0ontheboundayof and g =0onthebounday of I, theseondbeausethestess-enegytensosatisfiestheonsevationlawt µ, =0. Consequently x µ h µ (~x )=0.

5 Q.E.D. ) In ode to extat the physial omponents of the wave we still have to pojet h µ on the TT-gauge. 3) Eq. (4.4) has been deived on two vey stong assumptions: weak field (g µ = µ +h µ ) and slow motion (v typial << ). Fo this eason that expession has to be onsideed as an estimate of the emitted adiation by the system, unless the two onditions ae eally satisfied. 4. The Tenso iial Theoem In ode to simplify the integal in eq. (4.4) we shall use the onsevation law that T µ satisfies (see hapte 7) T µ x = 0,! T µ0 t = T µk, µ =0..3, k =..3. (4.) xk Let us integate this equation ove the soue volume, assuming the index µ is fixed t T µ0 d 3 x = T µk x k d3 x. By Gauss theoem, the integal ove the volume is equal to the flux of T µk aoss the sufae S enlosing that volume, thus the ight-hand-side beomes T µk x k d3 x = T µk ds k. S By definition, on S T µ =0andonsequentlythesufaeintegalvanishes;thus t Fom eq. (4.4) it follows that T µ0 d 3 x =0,! h µ0 = onst, µ =0..3, T µ0 d 3 x = onst. (4.3) and sine we ae inteested in the time-dependent pat of the field we shall put h µ0 =0, µ =0..3. (4.4) (Indeed, in the TT-gauge h µ0 =0.) WeshallnowpovetheTenso-iial Theoem whih establishes that t T 00 x k x n d 3 x = Let us onside the spae-omponents of the onsevation law (4.) T n0 = Tni, i, n =..3; x 0 xi T kn d 3 x, k, n =..3. (4.5)

6 multiply both membes by x k and integate ove the soue volume (emembe that xk x i = = t T n0 x k d 3 x = T ni x k 4 d 3 x x i T ni x k ds i + S = i k ). As befoe t S T ni x i x k d 3 x ni xk T x i T nk d 3 x, 3 d 3 x5 T ni x k ds i =0, theefoe T n0 x k d 3 x = T nk d 3 x. Sine T nk is symmeti we an ewite this equation in the following fom t T n0 x k + T k0 x n d 3 x = Let us now onside the 0 omponent of the onsevation law multiply by x k x n and integate ove = = t T 00 x k x n d 3 x = T 0i x k x n 4 d 3 x x i T 0i x k x n ds i + S T 00 + T0i =0, i =..3 t x i T 0i x i the fist integal vanishes and this equation beomes t T 00 x k x n d 3 x = If we now di eentiate with espet to x 0 we find t T 00 x k x n d 3 x = and using eq. (4.6) we finally find t T 00 x k x n d 3 x = x k x n d 3 x 0i xk T x i T nk d 3 x. (4.6)! x n + T 0i x k xn x i T 0k x n + T 0n x k d 3 x t T 0k x n + T 0n x k d 3 x. T 0k x n + T 0n x k d 3 x, 3 d 3 x5 T kn d 3 x, k, n =, 3. (4.7)

7 The left-hand-side of this equation is the seond time deivative of the quadupole moment tenso of the system q kn (t) = whih is a funtion of time only. Thus, in onlusion T 00 (t, x i ) x k x n d 3 x, k, n =, 3, (4.8) T kn (t, x i ) d 3 x = By using eqs. (4.4) and (4.4) we finally find 8 h µ0 =0, µ =0..3 >< >: h ik (t, ) = G " d 4 dt qik (t d dt qkn (t). #. (4.9) ) This is the gavitational wave emitted by a gavitating system evolving in time. It an be omposed of masses o of any fom of enegy, beause mass and enegy ae both soues of the gavitational field. NOTE THAT. G s /g m : this is the eason why gavitational waves ae extemely weak!!. In ode to make the physial degees of feedom expliitely manifest we still have to tansfom to the TT-gauge. 3. These equations have been deived on vey stong assumptions: one is that T µ, =0, i.e. that the motion of the bodies is dominated by non-gavitational foes. Howeve, and emakably, the esult (4.9) depends only on the soues motion and not on the foes ating on them. 4. A system of aeleated haged patiles has a time-vaying dipole moment ~d EM = X i q i ~ i and it will emit dipole adiation, the flux of whih depends on the seond time deivative of ~ d EM. Fo an isolated system of masses we an define a gavitational dipole moment ~d G = X i m i ~ i, whih satisfies the onsevation law of the total momentum of an isolated system d dt ~ d G = ~0.

8 Fo this eason, gavitational waves do not have a dipole ontibution. It should be stessed that fo a spheial o axisymmeti, stationay distibution of matte (o enegy) the quadupole moment is a onstant, even if the body is otating. Thus, aspheialoaxisymmetistadoesnotemitgavitationalwaves;similalyasta whih ollapses in a pefetly spheially symmeti way has a vanishing d q ik and does dt not emit gavitational waves. To podue these waves we need a etain degee of asymmety, as it ous fo instane in the non-adial pulsations of stas, in a non spheial gavitational ollapse, in the oalesene of massive bodies et. 4. How to tansfom to the TT-gauge The solution (4.9) desibes a spheial wave fa fom the emitting soue. Loally, it looks like a plane wave popagating along the dietion of the unit veto othogonal to the wavefont n =(0,n i ), i =.., 3 (4.30) whee n i = xi. (4.3) In ode to expess this wavefom in the TT-gauge we shall make an infinitesimal oodinate tansfomation x µ 0 = x µ + µ and hoose the veto µ whih satisfies the wave equation F µ =0, so that the hamoni gauge ondition is peseved, as explained in hapte 4. The onditions to impose on the petubed meti ae h 0 n =0, tasvese wave ondition h 0 =0, vanishing tae. It should be mentioned that the tansvese-wave ondition implies that h µ0 =0, µ =0, 3 as equied in eq. (4.4). Indeed, given the wave-veto k µ =(k 0,k 0 n i )weknowbyeq. (3.4) that k µ h0 µ =0, i.e. k 0 h0 0 + k 0 n i h0 i =0. The seond tem vanishes beause of the tasvese wave ondition, theefoe h 0 0 =0. We emind hee that, as shown in eq. (3.6), in the TT-gauge h µ and h µ oinide. To heeafte, we shall wok in the 3-dimensional eulidean spae with meti ij. Consequently, thee will be no di eene between ovaiant and ontavaiant indies. We shall now desibe a poedue to pojet the wave in the TT-gauge, whih is equivalent to pefom the oodinate tansfomation mentioned above. As a fist step, we define the opeato whih pojets a veto onto the plane othogonal to the dietion of n P jk jk n j n k. (4.3)

9 Indeed, it is easy to veify that fo any veto j, P jk k is othogonal to n j, i.e. (P jk k )n j = 0, and that P j kp k l l = P j l l. (4.33) Note that P jk = P kj, i.e. P jk is symmeti. Thepojetoistansvese, i.e. n j P jk =0. (4.34) Then, we define the tansvese taeless pojeto: P jkmn P jm P kn P jkp mn. (4.35)! 0 whih extats the tansvese-taeless pat of a tenso. In fat, using the definition (4.35), it is easy to see that it satisfies the following popeties P jklm = P lmjk P jklm = P kjml and it is tansvese: it is taeless: P jkmn P mns = P jks ; (4.36) n j P jkmn = n k P jkmn = n m P jkmn = n n P jkmn =0; (4.37) jk P jkmn = mn P jkmn =0. (4.38) Sine h jk and h jk di e only by the tae, and sine the pojeto P jklm extats the taeless pat of a tenso (eq. 4.38), the omponents of the petubed meti tenso in the TT-gauge an be obtained by applying the pojeto P jkmn eithe to h jk o to h jk By applying P on h jk defined in eq. (4.9) we get 8 >< h TT µ0 =0, µ =0, 3 >: h TT jk (t, ) = G " d 4 dt QTT jk (t h TT jk = P jkmn h mn = P jkmn hmn. (4.39) # ) (4.40) whee Q TT jk P jkmn q mn (4.4) is the tansvese taeless pat of the quadupole moment. Sometimesitisusefulto define the edued quadupole moment Q jk Q jk q jk 3 jk qm m (4.4) whose tae is zeo by definition, i.e. and fom eq. (4.38), it follows that jk Q jk =0, (4.43) Q TT jk = P jkmn q mn = P jkmn Q mn. (4.44)

10 4.3 Gavitational wave emitted by a hamoni osillato Let us onside a hamoni osillato omposed of two equal masses m osillating at a fequeny =! with amplitude A. Bel 0 the pope length of the sting when the system is at est. Assuming that the osillato moves on the x-axis, the position of the two masses will be ( x = l 0 A os!t x =+ l. 0 + A os!t The 00-omponent of the stess-enegy tenso of the system is y z x T 00 = X n= p 0 (x x n ) (y) (z); and sine v<<,!! p 0 = m, it edues to T 00 = m X n= (x x n ) (y) (z); the xx-omponent of the quadupole moment q ik (t) = R T 00 (t, x) x i x k dx 3 is apple q xx = q xx = m + (x x ) x dx (y) dy (z) dz (4.45) (x x ) x dx (y) dy (z) dz = m h x + x i = m apple l 0 +A os!t +Al 0 os!t = m h ost + A os!t +Al 0 os!t i,

11 whee we have used the tigonometi expession os =os. The zz-omponent of the quadupole moment is apple q zz = = m (x x ) dx (y) dy (z) z dz + (x x ) dx (y) dy (z) z dz =0 beause z (z) dz = 0. Sine the motion is onfined to the x-axis, all emaining omponents of q ij vanish. We shall ompute, as an example, the wave emeging in the z-dietion; in this ase n = x! (0, 0, ) and B C P jk = jk n j n k = 0 0A By applying to Q ij the tansvese-taeless pojeto P jkmn onstuted fom P jk,wefind Q TT xx = P xm P xn P xxp mn q mn = P xx P xx P xx q xx = q xx, (4.46) Q TT yy = P ym P yn P yyp mn q mn = P yyp xx q xx = q xx, Q TT xy = P xm P yn P xyp mn q mn =0, Q TT zz = P zm P zn P zzp mn q mn =0. In addition Q TT zx = Q TT zy = 0. Using these expessions eqs. (4.40) beome 8 h TT µ0 =0 >< h TT zi =0, h TT xy =0 >: h TT xx(t, z) = h TT yy(t, z) = G d (4.47) 4 z dt q z xx(t ), and using eq. (4.45) h TT xx = h TT yy = G 4 z " d dt q xx(t apple A os!(t # z ), (4.48) z )+Al z 0 os!(t ). Gm = 4 z! Thus, adiation emitted by the hamoni osillato along the z-axis is linealy polaized. If, fo instane, we onside two masses m =0 3 kg, with l 0 =m,a =0 4 m, and! =0 4 ad/s, the tem [A os!t] is negligible, and the dominant tem is at the same fequeny of the osillations: h TT xx Gm 35 z z! Al 0 os!(t ), z whih is, as expeted, vey vey small. It should be notied that due to the symmety of the system, the wave emitted along y will be the same. To find the wave emitted along x, we hoose n =(, 0, 0) and use the same poedue: no adiation will be found.

12 4.4 Gavitational wave emitted by a binay system in iula obit We shall now estimate the gavitational signal emitted by a binay system omposed of two stas moving on a iula obit aound thei ommon ente of mass. Fo simpliity we shall assume that the two stas of mass m and m ae point masses. Be l 0 the obital sepaation, M the total mass M m + m, (4.49) and µ the edued mass µ m m M. (4.50) Let us onside a oodinate fame with oigin oinident with the ente of mass of the system as indiated in figue (4.) and be l 0 = +, = m l 0 M, = m l 0 M. (4.5) The obital fequeny an be found fom Keple s law fom whih we find G m m l 0 = m! K m l 0 M, Gm m l0! K = s GM l 3 0 = m! K m l 0 M, (4.5) is the Kepleian fequeny. Be (x,x )and(y,y )theoodinatesofthemassesm and m y m x m Figue 4.: Two point masses in iula obit aound the ommon ente of mass on the obital plane x = m M 0 os! K t x = m M l 0 os! K t y = m M 0 sin! K t y = m M l 0 sin! K t. (4.53)

13 The 00-omponent of the stess-enegy tenso of the system is X T 00 = m n (x x n ) (y y n ) (z), n= and the non vanishing omponents of the quadupole moment ae q xx = m x (x x ) dx (y y ) dy (z) dz + m x (x x ) dx (y y ) dy (z) dz = m x + m x = µl 0 os! K t = µ l 0 os! K t + ost, q yy = m + m (x x ) dx y (y y ) dy (z) dz (x x ) dx y (y y ) dy (z) dz = m y + m y = µl 0 sin! K t = µ l 0 os! K t + ost, and q xy = m + m x (x x ) dx y (y y ) dy (z) dz x (x x ) dx y (y y ) dy (z) dz = m x y + m x y = µl 0 os!t sin! K t = µ l 0 sin! K t. (we have used os =os, sin = In summay os and m m = µm). and q xx = µ l 0 os! K t + ost µ q yy = l 0 os! K t + ost q xy = µ l 0 sin! K t, q k k = kl q kl = q xx + q yy =ostant. Theefoe, the time-vaying pat of q ij and of Q ij = q ij 3 ij q k k ae equal: and defining a matix A ij A ij (t) = q xx = q yy = µ l 0 os! K t (4.54) q xy = µ l 0 sin! K t, 0 B os! K t sin! K t 0 sin! K t os! K t C A (4.55)

14 we an wite q ij = µ l 0 A ij + onst. (4.56) Sine the wave emitted along a genei dietion n in the TT-gauge is h TT ij (t, ) = G d 4 dt apple Q TT ij (t using eq. (4.5) we find ) whee QTT ij (t )=P ijklq kl (t )=P ijklq kl (t ) h TT ij = G 4 µ l 0 (! K ) [P ijkl A kl ]= 4 µmg l 0 4 [P ijkl A kl ]. (4.57) By defining a wave amplitude h 0 = 4 µmg l 0 4 (4.58) we an finally wite the emitted wave as h TT ij (t, ) = h 0 ATT ij (t ), (4.59) whee apple A TT ij (t )= P ijkl A kl (t ) (4.60) depends on the oientation of the line of sight with espet to the obital plane. Fom these equations we see that the adiation is emitted at twie the obital fequeny. Fo example, if n = z, P ij = diag(,, 0) and A TT ij (t) = 0 B os! K t sin! K t 0 sin! K t os! K t h TT xx = h TT yy = h 0 z h TT h 0 xy = z sin! K(t os! K(t z ). C A (4.6) z ) (4.6) In this ase the wave has both polaizations, and sine h TT xx = h 0 /z < n e i!(t x )o and h TT xy = h 0 /z = n e i!(t x )o,thewaveisiulalypolaized. If n = x, P ij = diag(0,, ) and A TT ij = 0 B os! Kt os! Kt h TT yy = h TT zz =+ h 0 x os! K (t C A (4.63) x ), (4.64)

15 i.e. the wave is a linealy polaized wave. If n = y, P ij = diag(, 0, ) and 0 A TT ij = B and again the wave is linealy polaized os! Kt os! Kt h TT xx = h TT zz = h 0 y os! K(t C A (4.65) y ). (4.66) Eqs. (4.58) an be used to estimate the amplitude of the gavitational signal emitted by the binay system PSR 93+6 disoveed in 975, (R.A. Hulse and J.H. Taylo, Disovey Of A Pulsa In A Binay System, Astophys. J. 95, L5, 975) whih onsists of two neuton stas obiting at a vey shot distane fom eah othe. The data we know fom obsevations ae: y l 0 x Figue 4.: Two equal point masses in iula obit m m.4m, l 0 =0.9 0 m (4.67) T =7h 45m 7s, K =! K Hz whee T is the obital peiod. Note that the two stas have nealy equal masses: they ae ompaable to that of the Sun, and thei obital sepaation is about twie the adius of the Sun! The obit is eenti with eentiity ' 0.67, howeve we shall assume it is iula and apply eqs. (4.58). Fo this system the emission fequeny is GW = K Hz, (4.68) theefoe the wavelenght of the emitted adiation is GW = 0 4 m i.e. GW >> l 0. (4.69) GW

16 Thus, the slow-motion appoximation, on whih the quadupole fomalism is based, is etainly satisfied in this ase even though the two neuton stas ae obiting at suh small distane fom eah othe. The distane of the system fom Eath is =5kp, and sine p = m,! =.5 0 m. The wave amplitude is h 0 = 4 µmg l Anewbinaypulsahasmoeeentlybeendisoveed(M. Bugayet al., An ineased estimate of the mege ate of double neuton stas fom obsevations of a highly elativisti system Natue 46, 53,003)whih has an even shote obital peiod and it is lose than PSR 93+6: it is the double pulsa PSR J , whose obital paametes ae In this ase the obit is nealy iula, m =.337M, m.50m (4.70) T =.4h, e =0.08 =500p l 0.R. µ = m m m + m =0.646M! h 0 = 4µMG l , and waves ae emitted at the fequeny GW = K = Hz. In this setion we have onsideed only iula obits; the alulations an be genealized to the ase of eenti o open obits by eplaing the equation of motion of the two masses (4.53) by those appopiate to the hosen obit. By this poedue it is possible to show that when the obits ae ellipses, gavitational waves ae emitted at fequenies multiple of the obital fequeny K, and that the numbe of equally spaed spetal lines ineases with the eentiity. 4.5 Enegy aied by a gavitational wave In ode to evaluate how muh enegy is adiated in gavitational waves by an evolving system, we need to define a tenso that popely desibes the enegy ontent of the gavitational field. Ou e ot will not be ompletely suessful, sine we will be able to define a quantity whih behaves like a tenso only unde linea oodinate tansfomations. Howeve, this pseudo-tenso will be useful fo the pupose we have in mind.

17 4.5. The stess-enegy pseudotenso of the gavitational field In Chapte 7 we have shown that the stess-enegy tenso of matte satisfies a divegeneless equation T µ ; =0. (4.7) If we hoose a loally inetial fame (LIF), the ovaiant deivative edues to the odinay deivative and eq. (4.7) beomes T µ x =0. (4.7) We shall now ty to find a quantity, µ, suh that T µ = x µ ; (4.73) In this way, if we impose that µ is antisymmeti in the indies and, the onsevation law (4.7) will automatially be satisfied. The poblem now is: an we find the expliit expession of µ? Fom Einstein s equations we know that T µ = 4 R µ 8 G gµ R ; (4.74) sine we ae in a loally inetial fame, the Riemann tenso, whose genei expession is R = " g x x + g g # g (4.75) x x x x x x +g, edues to the tem in squae bakets sine all Rii tenso beomes s vanish; it follows that in this fame the R µ = g µ g R = g µ g g R (4.76) = gµ g g g x x + g g! g. x x x x x x By using this equation, afte some umbesome alulations eq. (4.74) beomes T µ = ( 4 h ( g) g µ g g µ g i). (4.77) x 6 G ( g) x The tem in paentheses is antisymmeti in the indies and and it is the quantity we wee looking fo: µ = 4 6 G ( g) x If we now intodue the quantity h ( g) g µ g g µ g i. (4.78) µ =( g) µ = 4 h ( g) g µ g g µ g i, (4.79) 6 G x

18 sine we ae in a loally inetial fame x =0, theefoe we an wite eq. (4.77) as ( g) µ x =( g)t µ. (4.80) This equation has been deived in a LIF, whee all fist deivatives of the meti tenso vanish, but in any othe fame this will not be tue and the di eene µ ( g)t µ will x not be zeo, but a quantity whih we shall all ( g)t µ i.e. ( g)t µ = µ x ( g)t µ. (4.8) t µ is symmeti beause both T µ and µ x ae symmeti in µ and. The expliit expession of t µ an be found by substituting in eq. (4.8) the definition of µ given in eq. (4.79), and T µ omputed in tems of the Rii tenso fom eq. (4.74) in an abitay fame (i.e. stating fom the full expession of the Riemann tenso given in eq. 4.75): afte some aeful manipulation of the equations it is possible to show that 4 n t µ = 6 G + g µ g ( + + g g ( µ + µ + g g ( µ µ µ ) o g µ g g µ g ) µ ) This is the stess-enegy pseudotenso of the gavitational field we wee looking fo. Indeed we an ewite eq. (4.8), valid in any efeene fame, in the following fom and sine µ is antisymmeti in µ and ( g)(t µ + T µ )= µ x, (4.8) x µ µ x =0, and onsequently x [( µ g)(tµ + T µ )] = 0. (4.83) This equation expesses a onsevation law, beause, as explained in hapte 7, it has the fom of a vanishing odinay divegene of the quantity [( g)(t µ + T µ )]. Sine t µ when added to the stess-enegy tenso of matte (o fields) satisfies a onsevation law, and sine it vanishes only in a loally inetial fame whee gavity is suppessed, we intepet t µ as the entity that ontains the infomation on the enegy and momentum aied by the gavitational field. Thus eq. (4.83) expesses the onsevation law of the total enegy and momentum. Unfotunately, t µ is not a tenso; indeeditisaombinationofthe sthat ae not tensos. Howeve, as the s, it behaves as a tenso unde linea oodinate tansfomations.

19 4.5. The enegy flux aied by a gavitational wave Let us onside an emitting soue and the assoiated 3-dimensional oodinate fame O (x, p y, z). Be an obseve loated at P = (x,y,z) as shown in figue 4.3. Be = x + y + z its distane fom the oigin. The obseve wants to detet the wave oming along the dietion identified by the veso n =. As a pedagogial tool, let us onside a seond fame O (x 0,y 0,z 0 ), with oigin oinident with O, and having the x 0 -axis aligned with n. Assuming that the wave taveling along x 0 dietion is linealy polaized and has only one polaization, the oesponding meti tenso will be 0 g µ 0 0 = B (t) (x 0 ) (y 0 ) (z 0 ) [+h TT + (t, x 0 )] [ h TT + (t, x 0 )] y The obseve wants to measue the enegy whih flows pe unit time aoss the unit suy P x. C A n z x z Figue 4.3: A binay system lies in the z-x plane. An obseve loated at P wants to detet the enegy flux of gavitational waves emitted by the system. fae othogonal to x 0,i.e. t 0x0,theefoeheneedstoomputetheChisto elsymbolsi.e. the deivatives of h TT µ 0 0. Aoding to eq. (4.40) the meti petubation has the fom h TT (t, x 0 )= onst x f(t 0 ), and sine the only deivatives whih matte ae those with x 0 espet to time and x 0 h TT t ḣtt = onst x 0 h TT h TT 0 = x 0 onst x 0 f, f + onst f 0 onst x 0 x 0 f = ḣtt,

20 whee we have etained only the dominant /x 0 tem. Thus, the non-vanishing Chisto el symbols ae: 0 y 0 y 0 = 0 z 0 z 0 = ḣtt + y 0 0y 0 = z 0 0z 0 = ḣtt + (4.84) x 0 y 0 y 0 = x 0 z 0 z 0 = ḣtt + y 0 y 0 x 0 = z 0 z 0 x 0 = ḣtt +. By substituting the Chisto el symbols in t µ we find t 0x0 = de GW dtds = 3 6 G 4 dhtt (t, x 0 ) dt! 3 5. If both polaizations ae pesent 0 g µ 0 0 = B (t) (x 0 ) (y 0 ) (z 0 ) [+h TT + (t, x 0 )] h TT 0 0 h TT (t, x 0 ) [ h TT (t, x 0 ) + (t, x 0 )], C A and t 0x0 = de GW dtds = = 3 3 G 4 X jk 3 6 G dh TT 4 dhtt jk (t, x 0 ) dt + (t, x 0! ) + dhtt (t, x 0! 3 ) 5 (4.85) dt dt! 3 5. This is the enegy pe unit time whih flows aoss a unit sufae othogonal to the dietion x 0. Howeve, the dietion x 0 is abitay; if the obseve is loated in a di eent position and omputes the enegy flux he eeives, he will find fomally the same eq. (4.85) but with h TT jk efeed to the TT-gauge assoiated with the new dietion. Theefoe, if we onside ageneidietion = n t 0 = 3 G 4 X jk dh TT jk (t, ) dt! 3 5. (4.86) In Geneal Relativity the enegy of the gavitational field annot be defined loally, theefoe to find the GW-flux we need to aveage ove seveal wavelenghts, i.e. de GW dtds = D t 0E = 3 * X 3 G jk dh TT jk (t, ) dt! +.

21 Sine 8 >< h TT µ0 =0, µ =0, 3 >: h TT ik (t, ) = G " d 4 by diet substitution we find de GW dtds = G 8 5 * X jk dt QTT ik Q TT jk t t # +. (4.87) As explained in setion 4.39, Q TT jk P jkmn q mn is the quadupole tenso pojeted onto the TT-gauge; moeove, we intodued the edued quadupole moment Q jk q jk 3 jk qm m (4.88) whih is taeless by definition, and onsequently Q TT jk = P jkmn q mn = P jkmn Q mn. (4.89) In ode to obtain the gavitational luminosity of a soue L GW = de GW, i.e. the gavitational enegy emitted by the soue pe unit time, it is moe onvenient to use the edued dt quadupole moment, theefoe we shall wite Eq. (4.87) in tems of Q jk,i.e. de GW dtds = G 8 5 The gavitational luminosity theefoe is L GW = = G 5 4 * X jk P jkmn Q mn t +. (4.90) degw dtds ds = degw dtds d (4.9) d * X jk P jkmn Q mn t +, whee d =(d os )d is the solid angle element. This integal an be omputed by using the popeties of P jkmn : X jk Pjkmn Q mn = X 3 jk P jkmn Q mn P jks Q s = (4.9) = 4 X P mnjk P jks 5 Q mn Q s = P mns Q mn Q s jk apple = ( m n m n )( ns n n n s ) ( mn n m n n )( s n n s ) Q mn Q s. If we expand this expession, and emembe that

22 Q mn = s Q s =0 beause the tae of Q ij vanishes by definition, and mn n m n ns Q mn Q s = n n n s m beause Q ij is symmeti, Q mn Q s at the end we find X Pjkmn Q mn = Q n Q n n m Q ms Q s n + n mn n n n s Q mn Q s. (4.93) jk By substituting this expession in eq. (4.9) we find apple Q n Q n L GW = G 5 4 d Q ms Q s n m n d + Q mn Q s n m n n n n s d. (4.94) Thus, the integals to be pefomed ove the solid angle ae: n m n d, and n m n n n n s d. (4.95) 4 4 Let us ompute the fist. In pola oodinates, the veso n an be witten as n i =(sin os, sin sin, os ). (4.96) Thus, fo paity easons d n m n =0 whenm 6=. (4.97) 4 Futhemoe, sine thee is no pefeed dietion in the integation (isotopy), it must be d n = d n = d n 3! d n m n = onst m. (4.98) 4 Fo instane, d (n 3 ) = 4 4 d os d os = d 4 0 d os os = 3, (4.99) and onsequently d n m n = 4 3 m. (4.00) The seond integal in (4.95) an be omputed in a simila way and gives d n m n n n n s = 4 5 ( mn s + m ns + ms n ). (4.0) By substituting Eqs. (4.00) and (4.0) in Eq. (4.94), we find L GW = G 5 = G 5 = G 5 apple Q n apple Q n Q n Q n 3 Q n 3 Q n apple Q ms Q s m + 30 Q s Q s = G 5 Q mn Q s ( mn s + m ns + ms n ) Q mn mn Q s s + Q n Q n + Q sn Q ns Q n Q n 5 = G 5 5 Q n Q n,

23 whee we have used the popety Q mn mn = Q s s = 0 due to the fat that the edued quadupole tenso is taeless. Finally, the gavitational wave luminosity is L GW = G * 3X Q 5 5 kn t k,n= Q kn t +. (4.0) This fomula was deived by A. Einstein in the pape Übe Gavitationswellen published in 98. The oiginal atile an be found on the website