Physics 411 Lectue 34 Souced Radiation Lectue 34 Physics 411 Classical Mechanics II Novembe 21st, 2007 We ae eady to move on to the souce side of lineaized waves. The point of this whole section has been to show the similaity between gavitational waves and the familia E&M waves, just an exta index, but it does lead to some inteesting physics. We have seen that if you want to detect gavitational waves, you need to measue a sepaation, and indeed, this is fundamentally how cuent detectos wok thee ae two test masses (mios) sepaated by a few kilometes, in the shape of an L, and you have effectively a giant intefeomete, lases ae shot at the mios, you cancel them pefectly at the output and wait fo a lack of cancellation, indicating that the mios have moved. Sounds easy enough. The poblem, as we shall see, is twofold: 1. Gavitational waves ae incedibly weak, both because of the coupling, and the quadupole natue of the adiation, and 2. The fequency ange of inteest is on the ode of one hetz tucks, fo example, ae an issue. 34.1 Souces We begin by etuning to the wave equation, this time leaving the stesstenso intact on the ight-hand side. Let s e-state the poblem hee to set the notation: ρ ρ h µν = 16 π T µν ρ (34.1) h νρ = 0. Stat with the top equation, this is a job fo the Geen s function. gauge condition amounts to stess-tenso consevation. The 1 of 12
34.1. SOURCES Lectue 34 34.1.1 Aside: Geen s Functions Hee we will develop the Geen s function fo the D Alembetian, the solution to ρ ρ φ = δ. Fom this, we can build up solutions to the wave equation that ae diven by souces, clealy the next step given ou lineaized left-hand side. To a cetain extent, classical physics, aside fom intepetation, is the study of ODEs and PDEs. Most physical laws ae expessed as deivative opeatos on some object of inteest tajectoy coodinates, fields, potentials, etc. We have seen this in ou discussion of geneal elativity ove and ove Einstein s equation is an example, the lineaized Einstein equation anothe, the equations of motion themselves ae an example. Paticulaly fo fields, though, the main tool is the equation of motion deived fom an action these will be, unless thee ae simplifying assumptions, patial deivative opeatos in all fou dimensions. With each PDE opeato (and bounday conditions) comes an associated Geen s function the question is, how do we geneically solve a PDE like: α α φ = ρ (34.2) if someone hands us a souce function ρ? The somewhat counteintuitive answe is: Solve the poblem α α G = δ(x µ x µ ), (34.3) and then the solution to (34.2) φ(x µ ) = dτ ρ(x µ ) G(x µ, x µ ). (34.4) That s a bunch of symbols, let me motivate with the thee-dimensional Laplacian. We have 2 φ = ρ, (34.5) which means, accoding to ou pesciption, we ae inteested in 2 G = δ(x x ), (34.6) 2 of 12
34.1. SOURCES Lectue 34 and you should think of G as a eal potential fo a point souce located at x as measued at x (that s what the souce ρ = δ(x x ) tells us). In foming the integal φ = dτ ρ G, we ae taking the eal distibution of chage (say) and viewing it as a set of points, each point contibutes G to the potential and so the total potential is just the sum (integal) of all the points with the appopiate potential a statement about supeposition. To pove that this φ actually solves Poisson s equation, we use Geen s theoem (φ 2 G G 2 φ ) dτ = (φ G G φ) da (34.7) whee the pimes tell us the integation is ove x. The integation occus ove whateve space we like, although it should include the x of the distibution itself if we want to get an inteesting answe and unde usual assumptions about the fields (that they fall off faste than a sphee), we take the suface at infinity, this just kills the ight-hand side of the above, and then we can evaluate the left because it aleady involves the Laplacian: dτ ( φ δ(x x ) + G ρ ) = 0 φ(x) + dτ G ρ = 0 (34.8) and we e done. Thee ae a couple of impotant facts about Geen s functions that one should keep in mind: 1. They depend on a given diffeential opeato and bounday conditions the whole of the specified poblem (we ignoed that above), 2. They depend on the dimension of space (o spacetime), 3. They ely on supeposition that is, they only apply to linea (o lineaized) PDEs. A moe useful (and elatively geneic) popety of Geen s functions is thei symmety, G(x, x ) = G(x, x) and indeed, fo ou poblems, which have been specified only with the implicit bounday condition at x (G vanishes thee), one can go even futhe: G(x, x ) = G( x x ). The Geen s function appoach, while it looks like a cueall, equies us to compute the appopiate G fo ou poblem. Because of the δ-function on the ight-hand side, this is not tivial, and the calculation of these functions is sometimes ticky. Fo example, one way to appoach the Laplacian solution is to specialize to spheical coodinates, and set the 3 of 12
34.1. SOURCES Lectue 34 souce coodinates x at the oigin then the Geen s function depends only on, and away fom the oigin, we have: G + 2 G = 0 G() = α + β. (34.9) The ODE at the oigin has an infinite ight-hand side to get id of this, we can integate aound an infinitesimal ball enclosing = 0 δ(x) 2 sin θ dθ dφ d = 1, (34.10) a fundamental popety of the δ function fo the left-hand side, we have: dτ 2 G = dτ ( G) = G ˆn da = 4 π α = 1 (34.11) so we set the nomalization to α = 1 4π the above says nothing about β, but in E&M at least, we know that β just sets the zeo of the potential at some point. Ou final solution is: G() = 1 4 π (34.12) and if we had the souce position at x athe than the oigin, 2 G = δ(x x ) G( x x ) = 1 4 π (34.13) with R x x. The Geen s function appoach is useful, because in the lineaized Einstein equations with souce, we have: α α h µν = 16 π T µν. (34.14) So we ask the point-souce question, what is the solution to α α Q µν = δ(x µ x µ ) (34.15) (I m using Q µν as the Geen s function athe than G µν ). Notice that, when we expand out the flat space D Alembetian, we get 2 t + 2, so this is 4 of 12
34.1. SOURCES Lectue 34 almost a Laplacian, and we can make it even moe like a Laplacian by a tempoal Fouie tansfom multiply by e i ω t and integate ove t: dt ( t 2 Q µν (t, x, x )) e i ω t + 2 dt Q µν (t, x, x ) e i ω t = δ(x x ) e i ω t dt ( ω 2 ) Q µν (t, x, x ) e i ω t + 2 dt Q µν (t, x, x ) e i ω t = δ(x x ) e i ω t (34.16) whee we have integated-by-pats twice to get the second line. Now to set the Fouie tansfom nomalization, define Q µν (ω, x, x ) = 1 2 π Q µν (t, x, x ) = 1 2 π and ou D Alembetian becomes dt Q µν (t, x, x ) e i ω t dω Q µν (ω, x, x ) e i ω t (34.17) (ω 2 + 2 ) Q µν (ω, x, x ) = 1 2 π δ(x x ) e i ω t. (34.18) The diffeential opeato on the left-hand side of this is called the Helmholtz opeato, and its Geens function is known, but let s continue in ou deivation. We know, because this PDE comes to us with the implicit condition that h µν 0 at infinity, that the Geen s function depends only on the magnitude of the diffeence: x x between the (flat-space) obsevation point x and a souce point x. Then the spheical Laplacian may be used as usual fo 2 Qµν (ω, x x ), (ω 2 + d2 d 2 ) [ e i ω t Q µν(ω, )] = 2 π δ(r ). (34.19) This equation is only valid away fom = 0, since we ve multiplied though by to get it in this fom but then we can solve the homogenous pat immediately, it s just a hamonic oscillato e i ω t Q µν (ω, ) = α e i ω +β e i ω Q µν (ω, ) = ei ω t ( α e i ω + β e i ω ). (34.20) We need to check the spatial pat does this educe to a δ function, and what nomalization should we use? The quickest way to do this is to look at the limit of Q µν (ω, ) as 0 lim Q µν (ω, ) = α 0 + α (i ω) + β β (i ω) + O( ) (34.21) 5 of 12
34.2. SOURCE APPROXIMATIONS AND MANIPULATION Lectue 34 so that the behavio of this function nea the oigin is identical to the theedimensional Geen s function fo the Laplacian (34.13), modulo constants. 1 Then we should set, α + β = 2 π 4 π, and now we can wite the final fom fo the tansfomed Geen s function: Q µν (ω, ) = 1 4 π 2 π ei ω (t ± ), (34.22) whee I ve witten this as two sepaate solutions athe than a linea combination. Now we Fouie tansfom back to t Q µν (t, ) = 1 ( ) dω e i ω t 1 2 π 4 π 2 π ei ω (t ± ) and note that so = 1 4 π (2 π) 1 2 π dω e i ω (t t± ), (34.23) dω e i ω (t p) = δ(t p) (34.24) Q µν (t, ) = 1 4 π δ( t t ± ) = 1 4 π δ( t t ) (34.25) is the appopiate Geen s function in tems of time. 34.2 Souce Appoximations and Manipulation We have not made any assumptions about the souces themselves yet and because of the spatial dependence on time (though t = t ), we need to be caeful to sepaate out all the spatial dependence befoe making fa away types of appoximation. The easiest way to disentangle is to Fouie tansfom both sides of 2 Tµν (t, x ) h µν = 16 π T µν h µν = 4 δ(t t ) dτ dt (34.26) whee dτ = dx dy dz (just the spatial potion of the integation). Let s focus on the ight-hand side fist multiply by e i ω t and integate ove t, using the δ to pefom the dt integal, we have: 4 dτ i ω (t ) ei ω dt e T µν (t, x ), (34.27) 6 of 12
34.2. SOURCE APPROXIMATIONS AND MANIPULATION Lectue 34 whee we multiply and divide by the spatial tem so that we can define the etaded time t t. A change of vaiables gives: = 4 dτ e i ω dt e i ω t T µν (t, x ) = 4 2 π dτ ei ω T µν (ω, x ). (34.28) We make the usual appoximation, that = x x is dominated by the obsevation distance x, i.e. we ae fa away and the oigin is centeed inside the mass distibution. Then appoximately, we have: 4 2 π ei ω dτ T µν (ω, x ) 4 2 π ei ω x x dτ Tµν (ω, x ). (34.29) Now we can attack the Fouie tansfom of the souce tem. Witing the full (lineaized) Einstein equation fo distant souces, h µν (ω, x) 4 ei ω x x dτ T µν (ω, x ), (34.30) the fist thing to bea in mind is that we need only deal with the spatial components on eithe side this comes fom the gauge condition µ h µν = 0, and the equivalent statement fo the stess tenso (lineaized consevation): µ h µν = 0 ( i ω) h 0ν + j hjν = 0 µ T µν = 0 ( i ω) T 0ν + j T jν = 0. (34.31) The condition allows us to solve fo the space-space components in tems of the othes. So efeing only to (i, j) spatial indices, we can ewite the integand on the ight as a total divegence and a emaining piece: dτ T ( jk = dτ s ( T sk x j ) x j s T sk), (34.32) whee the total divegence is tuned into a suface integal, and as long as the distibution is not infinite, can be set to zeo (extend the volume of integation to infinity, T jk will be zeo outside of some finite egion, then the souce at infinity is zeo). Replacing the spatial deivatives with the momenta (using consevation fo Fouie modes, as shown above) and 7 of 12
34.2. SOURCE APPROXIMATIONS AND MANIPULATION Lectue 34 noting that h jk is symmetic in (j, k), we can ewite and epeat the pocess: dτ T jk = i ω ( dτ x k T 0j + x j T ) 0k 2 = i ω ( dτ s (x k x j T ) 0s x k x j s T 0s) (34.33) 2 (i ω)2 = dτ x k x j T 00. 2 The above is just what we would call the (Fouie tansfom of) an integal elated to the quadupole moment of the souce in E&M. Again, we see the impotance of a second index in E&M, all of this machiney is applied to the wave equation fo the potential α α A µ = j µ. The move fom two spatial indices to one via a total deivative and the consevation law fo T µν is the mathematical statement of chage consevation (enegy density consevation hee). That is identical to E&M, and thee, foces us to look at dipole adiation as the fist adiating tem. With the move fom one spatial index to none, as above, we ae effectively using momentum consevation in GR, so thee is no dipole adiation, in addition to no monopole, and we ae left with the quadupole as the fist contibuting tem. We have, on the Fouie side, the elation: h jk (ω, x) = 4 ei ω x x (i ω) 2 2 dτ x j x k T 00 (ω, x ). (34.34) Multiplying by e i ω t and integating, we can etun to time, h jk (t, x) = 2 d x dt 2 dω e i ω (t x ) dτ x j x k T 00 (ω, x ) = 2 d 2 ( x dt 2 dτ x j x k T 00 (t, x )). t=t t x (34.35) Defining the enegy density quadupole to be thee times the quantity in paenthesis (a nomalization), h jk (t, x) = 2 3 x qjk t=t q jk 3 dτ T 00 (t, x ) x j x k. (34.36) 8 of 12
34.3. EXAMPLE CIRCULAR ORBITS Lectue 34 34.3 Example Cicula Obits Let s take a simple example to get a geneal pictue of the connection between metic petubations and souces. Ou model system is a small body in a cicula obit about a moe massive one, as shown in Figue 34.1. M m R Figue 34.1: A cicula obit fo a massive body M and a smalle body m obseved on-axis fom a distance R away. Remembe the Hamiltonian fom of Newtonian obits, ṙ 2 = 2 H 2 ( 1) ( 1 ) 1 = 1 2 H ( M + 2 H J 2 z + M 2 ) 2 = 1 2 H ( M 2 H J 2 z + M 2 ) (34.37) whee H is the Hamiltonian (the enegy in this context), J z is the angula momentum of the obiting body and 1, 2 define the tuning points of the motion. Fo a cicula obit, what we mean is that 1 = 2, i.e. the point of closest appoach is identical to the point of futhest appoach. This gives us a elation fo the obital enegy H: H = M 2. In addition, we can 2 Jz 2 set the (t) coodinate to have the value 1 = 2 = J z 2 M fo all time that s consistent. Then the ight-hand side of the adial equation above is zeo, as is the left. Now we can ask, what is the peiod of the motion? The answe is povided by φ(t), we integate ove one full peiod, φ = J z M 2π T M 2 = 3 dφ = dt 0 0 3 T = 2 π 3 M, (34.38) 9 of 12
34.3. EXAMPLE CIRCULAR ORBITS Lectue 34 fom which we can calculate the angula fequency ω = 2 π T in Catesian coodinates, we can wite the tajectoy as x = cos(ω t) y = sin(ω t) z = 0. = M 3. Then (34.39) The enegy density is just that of a point mass m evaluated along its tajectoy: T 00 = m δ 3 ( ) = m δ(z ) δ(x cos(ω t)) δ(y sin(ω t)), (34.40) and we can calculate the quadupole diectly q jk = 3 dτ T 00 (x ) x j x k =3 m = 3 m dτ δ(z ) δ(x cos(ω t)) δ(y sin(ω t)) 2 cos 2 (ω t) 2 sin(ω t) cos(ω t) 0 2 sin(ω t) cos(ω t) 2 sin 2 (ω t) 0 0 0 0 x 2 x y x z x y y 2 y z x z y z z 2. (34.41) Using (34.36), the spatial potion of the metic petubation is h jk (t, x) = 4 2 m ω 2 cos(2 ω t ) sin(2 ω t ) 0 sin(2 ω t ) cos(2 ω t ) 0. (34.42) R 0 0 0 Fo cicula obits, then, we can eplace ω in the magnitude potion, the facto in font of the matix is 4 m M R. We have constucted the wave to be in the z-diection, and the fom of the spatial potion tells us that the petubation is tansvese (aleady) and taceless. One of the hallmaks of gavitational waves, owing to the two time-deivatives of the quadupole, is that oscillations occu with twice the fequency of the souce (fo cicula obits). Fom ou discussion of geodesic deviation of test masses, we know that the sepaation as a function of time will go oughly like h ij itself times the initial sepaation (s(t) βh(t)). Given the above fom fo h ij deived fom 10 of 12
34.3. EXAMPLE CIRCULAR ORBITS Lectue 34 a cicula obit, we can get an idea of the sensitivity equiements fo a gavitational wave detecto. Suppose we ae viewing a sola-like obit, with a massive body (the sun) and a much smalle body (the eath) in an appoximately cicula obit. If we view fom a platfom located, say R = 10 (ten times the distance to the sun), then we need to measue a sepaation given by s(t) β h (34.43) whee β is the initial sepaation of the test masses, and h is h 4 m M R = 2 m M 5 2. (34.44) If we put in M 1.5 km, and the eath has m 4.4 10 6 km, with distance between the two = 1 AU 1.5 10 8 km. Then h 2 (1.5 km) ( 4.4 10 6 km ) 5 (1.5 10 8 km) 2 1.2 10 22. (34.45) We take a easonable initial sepaation fo the test masses, say β 1 km, then ou instument needs to be sensitive to s hβ ( 1.2 10 22) (1 km) 1.2 10 22 km 1.2 10 9 Å. (34.46) And one can compae this with the adius of the Hydogen atom, appoximately.5 Å. Thee ae bette souces, of couse, the eath-sun system is a vey weak adiato, but while thee ae moe massive, faste systems aound, they also tend to be futhe away. The upshot is, we ae petty much stuck measuing (vey) small sepaations. Amazingly, the gound-based detectos (LIGO, fo example) ae getting close to the sensitivity limit in some fequency anges. One last point about gavitational waves. It is not possible to define local enegy in the gavitational field this is just the usual statement about equivalence, how do you sepaate the backgound space-time fom the potion that would cay enegy? To put it anothe way: in geneal, we know that space-time is locally flat, that s the stating point of all of this. So in the local fame, whee the metic and connection vanish in a small neighbohood, we would say thee is no enegy associated with the metic. One can define total enegy at cetain points (like infinity) but not local enegy density. 11 of 12
34.3. EXAMPLE CIRCULAR ORBITS Lectue 34 That s supising when we think of lineaized gavitational adiation, which has a flat backgound and waves popagating on it just like E&M why can t we mimic the definition of field enegy thee? We can, but this is not a self-consistent pocedue in GR, the lineaization effectively beaks down, so it s not coect. Howeve, if one insists, and poceeds caefully, it is possible to talk about the enegy adiated in gavitational waves fom a souce like the cicula one we have been consideing. Enegy loss fo a Newtonian obit implies a peiod shift, and in 1993, Hulse and Taylo won the Nobel pize fo obseving the shift associated with gavitational wave enegy loss (obseved ove 20 yeas) in a binay system. This is the fist indiect evidence fo gavitational adiation. Cuent expeiments like LIGO and the space-based LISA ae attempting a diect obsevation though a sepaation distance measuement. 12 of 12