1/30/17 Lecture 6 outline

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1 1/30/17 Lectue 6 outline Recall G µν R µν 1 2 Rg µν = 8πGT µν ; so R µν = 8πG(T µν 1 2 Tg µν). (1) g µν = η µν +h µν G (1) µν (h) = hµν ρ µ hνρ ρ ν hµρ 1 2 η µν ρ σ hρσ = 8πT µν. Choose an x µ x µ + ξ µ gauge such that ρ hρσ = 0, which eliminates all but the fist tem in G (1) µν. Last time: Poduction of gavitational waves: want to solve G (1) µν = 8πGT µν which in the µ hµν = 0 gauge choice becomes 2 hµν = 16πGT µν. We know how to solve this equation, using 2 (1/) = 4πδ 3 ( x), just like the Lienad-Wiechet potential in E&M : h µν (t, x) = 4G d 3 1 y x y T µν(t x y, y). Fa away fom the souce, do a multipole expansion. The leading tem is the quadupole tem: h ij 2G d 2 I ij dt 2 (t ), I ij (t) = d 3 yy i y j T 00 (t, y). Hee is how to show it: the y integal above is ove the past light cone of the space-time point x µ. Let s F.T. t ω: h µν (ω, x) = 4G d 3 1 y x y T µν(ω, y)e iω x y. The gauge condition gives iωh 0µ (ω, x) = i h iµ (ω, x), so we only need to solve fo the i,j 0 space components of the metic wave: µ i and ν j. In the fa zone, to leading ode, we eplace x y and can take e iω / out of the integal. Now use d 3 yt ij = ( k (T kj y i ) k T kj y i ) = iω T 0(j y i) = = 1 2 ( k (T 0k y i y j ) k T 0k y i y j ) = 1 2 ω2 T 00 y i y j d 3 y (using consevation of T µν in leading, appoximately flat-space fom, and dopping suface tems since we take the souce to be compact). So the leading tem is quadupole adiation, 1

2 unlike E&M whee the leading tem is dipole adiation. Thee is no dipole adiation hee because consevation of momentum implies that the dipole tem is a constant in time. E.g. two stas of mass M, sepaated by distance 2R in the weak field, non-elativistic limit have I ij MR 2 and d 2 I ij /dt 2 Ω 2 MR 2 with Ω = 2π/T = v/r = GM/4R 3. How much powe is caied away in the gavitational adiation of the quadupole? Recall fom last time we can wite G (1) µν = 8πG(T µν +t µν ), t µν 1 8πG (G µν G (1) µν) and we can sot-of intepet t µν as the enegy-momentum of the gavitational fields. To leading, quadatic ode in h: t µν 1 ( ) 1 8πG 2 h µνη ρλ R (1) ρλ η µνh ρλ R (1) ρλ +R(2) µν 1 2 η µνη ρλ R (2) ρλ. As we discussed last time, this is not gauge invaiant so not fully koshe. Nevetheless, it has some meit. If we integate it, the gauge vaiations dop out to leading ode in small h petubation theoy aound flat space, so this is a sensible way to compute the powe adiated. Since G (1) µν satisfies the leading ode Einstein s equations, we have t µν 1 8πG (R(2) µν 1 2 η µνη ρλ R (2) ρλ ). Upshot: the adiated powe is elated to the deivative of h ij squaed. As above, h ij is elated to a 2nd time deivative of the quadupole moment. So the powe is elated to the thid deivative of the quadupole squaed (the details and indices ae tedious to wok out): P = G 5c 9 d 3 I ij (t ) dt 3 d 3 I ij (t ) dt 3. (check units and put back in c s: [G] = M 1 (L/T) 3 T, [ d3 I dt 3 ] = M(L/T) 4 T 1, P = M(L/T) 2 T 1 ). See e.g. Weinbeg ch 10.5 fo details, e.g. the facto of 1/5 is elated to the integal of fou unit vectos ove all solid angle. Example (details in Weinbeg ch 10.5): Webe s attempt to diectly obseve gavitational adiation fom a sound vibation in an lage aluminum cylinde. The density in the cylinde is ρ = ρ 0 + ρ 1, whee ρ 1 is the vibating sound wave fluctuation: ρ 1 = ǫρ 0 sin(kz)cos(ωt). Thee is a quadupole tem I zz = A L 0 ρ 1z 2 dz. Find enegy loss: Γ gav = P E = 64GMv4 s 15L 2 c 5, 2

3 whee v s is the sound velocity. Plug in Webe s values: L = 1.53m, v s = m/s, M = kg, get Γ gav = s 1. It poved to be too small fo him to eliminate all systematics and accuately and pecisely measue. Instead, tied to use the bas as detectos of gavity waves poduced in space. Found some possible effects, but could not eliminate the systematics to show they wee actually gavity waves; so did not succeed. Anothe example (again, Weinbeg ch 10.5). Gavitational adiation of a lage, otating body. If otation fequency is Ω, the adiation has fequency peaked at 2Ω (since we squae h). The esult is P(2Ω) = 32GΩ5 I 2 e 2 5c 5, whee I = I 11 +I 22 and e = (I 11 I 22 )/I, whee the otation is aound the 3 axis. Only adiates if not axially symmetic aound axis of otation. Apply e.g. to the otation of Jupite aound the sun, take e = 1, I = m 2, Ω = s 1, m = kg, = m, get P 5.3kW, tiny. Conside two stas, each of mass M, sepaated by distance 2R, otating with fequency Ω. Evaluate I xx = 2MR 2 cos 2 (Ωt), I xy = MR 2 sin(2ωt), I yy = 2MR 2 sin 2 (Ωt). Let Ω = 2π/T with T the peiod and use Keple s law V 2 /R = GM/(2R) 2 to get R = (GMT 2 /16π 2 ) 1/3. Aveage ove a peiod. Get P = 128 5c 5GM2 R 4 Ω 6 = 128 ( ) 10/3 πgm 5 41/3c5 G c 3. T ( ) 10/3 M = h W. M sun T By compaison, the sun adiates electomagnetic adiation W and a lage galaxy about W and a bight gamma-ay bust about W. The LIGO event of Sept 14, 2015 was estimated to have a peak gavitational powe adiation of W, which fo that bief peiod was geate than all light adiated by all stas in the visible univese. The appoximations that led to the above fomula ae of couse invalid fo getting the powe adiated by black-hole meges. The event was intepeted as an inspial of a 35 sola mass black hole and a 30 sola mass black hole, esulting in a 62 sola mass black hole. The mass and otational enegy diffeence was adiated pimaily in gavity waves. The GR analysis could not be done analytically; it was done numeically on computes. 3

4 Conside isotopic and homogeneous space times: looks the same in all diections and the same unde tanslations. A maximally symmetic space-time of dimension n has R ρσµν = R n(n 1) (g ρµg σν g ρν g σµ ), with R the Ricci scala that is constant ove the space-time. The Weyl tenso fo these spaces is C ρσµν = 0. Thee ae thee possibilities: R = 0: flat; R > 0: de Sitte; R < 0: anti-de Sitte. These ae solutions of Einstein s equations fo T µν g µν, with zeo, positive, and negative CC espectively. Fo n = 4 we have R µν = 3κg µν, whee R = 12κ and Einstein s equations ae satisfied if ρ = p = 3κ/8πG. One waytogetdesittespaceistostatin5d, withds 2 5 = du2 +dx 2 +dy 2 +dz 2 +dw 2 and estict to a hypeboloid u 2 + x 2 + y 2 + z 2 + w 2 = C 2, whee C is the de Sitte adius. By taking u = Csinh(t/C), and w,z,y,z Ccosh(t/C) times S 3 coodinates, the metic is ds 2 = dt 2 +C 2 cosh 2 (t/c)dω 2 3, whee dω 2 3 is the solid angle on an S 3. These ae geodesically complete coodinates, so the topology is R S 3. Likewise anti-de Sitte stats with ds 2 5 = du2 dv 2 +dx 2 +dy 2 +dz 2 and esticts to a hypeboloid u 2 +v 2 x 2 y 2 z 2 = C 2. Wite u = Csin(t )coshρ, v = Ccos(t )coshρ, and x,y,z Csinhρ times S 2 coodinates, gives ds 2 = C 2 ( cosh 2 ρdt 2 +dρ 2 +sinh 2 ρdω 2 2). The t coodinate is a closed time-like cuve, which is bad, so conside the coveing space whee t is not identified with t +2π. Recall Fiedmann Robetson Walke: ds 2 = dt 2 +a 2 (t)dσ 2, and conside case whee the 3d space dσ 2 is maximally symmetic. Again, thee possibilities: the 3d space can have k = R 3d /6 negative (open), positive (flat), o positive (closed). By a choice of coodinates, dσ 2 = d2 1 k 2 +2 dω 2 with k = 0,1, 1. These spaces solve Einstein s equations fo a fluid T µν = (p+ρ)u µ U ν + pg µν. Consevation of enegy equies ρ/ρ = 3(1+w)ȧ/a, whee w p/ρ. Fo constant 4

5 w this gives ρ a 3(1+w). Recall e.g. that the null dominant enegy condition conjectue is w 1. Einstein s equations lead to the Fiedmann equations: (ȧ a ) 2 = 8πG 3 ρ k a 2, ä a = 4πG 3 (ρ+3p). Get ρ a n with equation of state w = 1 n 1. Matte has n = 3 (so w = 0), adiation 3 has n = 4 (so w = 1/3), cuvatue has n = 2 (so w = 1/3), and vacuum has n = 0, so w = 1. Fo example, the Einstein static univese had ρ Λ = 1 2 ρ M; it is topologically R S 3. Recall Schwazschild solution, e.g. fo a T µν a spheically symmetic, static, delta function at the oigin. Away fom the oigin, T µν = 0, so we solve Einstein s equations in vacuum, R µν = 0. Thee is Bikhoff s theoem, that thee is a unique vacuum solution with spheical symmety, and it tuns out to be static. If we take 1 ds 2 = e 2α() dt 2 +e 2β() d dω 2, (2) can compute R µν and see that it vanishes only if α = β and (e 2α ) = 1, which gives the Schwazschild solution, e 2α = 1 R s /. Recall thatweknowfom thenewtonianlimitthat h 00 = 2Φ, so R s = 2GM. The Ricci tenso vanishes fo Schwazschild, but the Riemann tenso does not. Wite out some example components, e.g. Rφφ = e 2β sin 2 θ β, R t θtθ = GM/, etc. The non-zeo Riemann tenso will give e.g. the coect focusing of neaby geodesics, D 2 = R µ dx ν dx ρ dλ 2δxµ νρσ dλ dλ δxσ. Using the full metic, can exploe this beyond the lineaized limit discussed above. Can show R µνρσ R µνρσ = 48G2 M 2 6. We see that = 0 is eally a singulaity wheeas = R s is not a eal singulaity. Since ds 2 = (1 2GM )dt 2 +(1 2GM ) 1 d dω 2 1 a e 2γ() facto in font of the 2 dω 2 tem could be eliminated by a edefinition of e γ 5

6 is static and otationally invaiant, thee ae fou Killing vectos coesponding to H and L. Recall that Killing vectos K satisfy L K g µν = 0. (Recall that the Lie deivative of a function is L V f = V µ µ f, while that of tensos have additional tems, simila to the connection tems of covaiant deivatives but instead involving V µ, with same plus o minus signs depending on whethe the indices ae uppe o lowe, e.g. fo a vecto L V U µ = [V,U] µ V ν ν U µ U ν ν V µ and the Lie deivative of the metic along V is L V g µν = V σ σ g µν +( µ V λ )g λν +( ν V λ )g µλ = µ V ν + µ V ν ). If K µ is a Killing vecto, dx then K µ dxµ µ dλ is conseved if dλ solves the geodesic equation. If pµ is the 4-momentum of a test mass on a geodesic, (µ K ν) implies consevation of K ν p ν : the geodesic equation gives p λ λ p µ = 0 so p µ µ (K p) = p µ p ν (µ K ν) = 0. If e.g. the metic is independent of t then the Killing vecto is K µ = δ µ t = (1,0,0,0). So fo Schwazschild we have H t K µ = ( (1 2GM ),0,0,0). L z φ L µ = (0,0,0, 2 sin 2 θ). Taking θ = π/2, the conseved quantities ae E = (1 2GM ) dt dλ, L = 2dφ dλ. (Fo massive paticles, it is actually L z /m.) The obits can be witten as 1 2 (d dλ )2 +V eff () = 1 2 E2, dx with ǫ g µ µν dλ paticle. dx ν dλ V eff () = 1 2 ǫ ǫgm + L2 2 2 GML2 3, so ǫ = 1 fo a massive paticle with λ = τ and ǫ = 0 fo a massless 6