Solutions for Problem Set for Ch. 25 A-D

Transcription

1 Solutions fo Poblem Set fo Ch. 5 compiled by Nate Bode) May 0, 009 A-D 5.4 Behavio of h + and h unde otations and boosts [by Xinkai Wu 0] a) Quantities with a tilde denote those in the new basis, and those without tilde in the old basis. As suggested in equation 5.51 we pefom a change of basis: ẽ x + iẽ y = e x + ie y )e iψ. Then, ẽ x = e x cos ψ e y sin ψ, ẽ y = e y cos ψ + e x sin ψ. 1) Plugging the above tansfomation matix into equation 5.41 we find the components of Riemann in the new basis R x0 x0 = cos ψr x0x0 + sin ψr y0y0 cos ψ sin ψr x0y0 ) = cos ψ 1 ) sin ψ 1 ) 3) ḧ+ ḧ on the othe hand R x0 x0 = 1 h +, thus we get h + = cos ψ)h + sin ψ)h 4) Similaly, by looking at R x0ỹ0, we find h = cos ψ)h + sin ψ)h + 5) Tanslated into complex numbes, this is just h + + i h = h + + ih )e iψ 6) b) The desied boost is a boost along the z diection, which gives ẽ 0 = e 0 cosh + e z sinh, ẽ z = e 0 sinh + e z cosh 7) 1

2 with e x, e y unchanged. And the coesponding tansfomation fo the coodinates is t = t cosh z sinh, z = t sinh + z cosh 8) which gives t z = cosh + sinh )t z) 9) with x, y unchanged. Look at components of Riemann in the new basis using the above tansfomation matix, we find ) 1 R = cosh sinh x 0x 0 ) R x0x0 = cosh sinh ) ḧ + t z) 10) on the othe hand R = ) 1 x 0x 0 h+ t z) = 1/) h+ t z)/cosh +sinh ). Equating this with eq. 10) we find h + t z) = ḧ+t z) h + = h + 11) By looking at R x 0y 0 one can show the invaiance of h in a simila manne. 5.5 Enegy-momentum consevation in geometic optics limit [by Alexande Putilin 00] Stating fom equation 5.58: In the geometic optics limit: T GW α = 1 16π h +,αh +, + h,α h, 1) h + = Q +τ ; θ, φ), h = Q τ ; θ, φ) The wave vecto k τ is null, and we have k k = 0, k = 1 k). GW To show Tα = 0, it s sufficient to pove that h+ = h = 0. We ll follow the patten used in Sec 5.3.6, 13) h +, = h +,τ τ x + h +, = h +,τ k + h +, 14) whee pime denotes deivatives at fixed τ i.e. h +, h +, ) x Diffeentiating the second time we get θ φ = h +,θ + h x +,φ + x h +, = h+,τ k h +,τ k ) + h+ 15) = h +,ττ k k h +,τ k k)h +,τ h +,τ k + h+ 16) = k h +,ττ k h +,τ k)h +,τ + h+ 17)

3 k = 0 so the fist tem vanishes. Notice that in geometic optics limit h + is a fast vaying function of τ and slowly vaying function of θ, φ,. It means that deivatives of h + w..t. τ ae much lage than deivatives at fixed τ and we can neglect the last tem. h+ k h +,τ k)h +,τ 18) k h +,τ is the diectional deivative along k at fixed τ ). Since θ, φ ae constant along a ay, only 1/ factos can vay ) 1 k h +,τ = Q + k = Q + 1 ) τ, τ k = 1 k h +,τ 19) Finally The equality h using k = 1 k) 0) = 1 k)h +,τ 1) h+ = k)h +,τ k)h +,τ = 0 ) = 0 can be deived in exactly the same way. 5.6 Tansfomation to TT gauge [by Alexande Putilin 99 and Keith Matthews 05] a) Conside the gauge tansfomation geneated by ξ α : h α h α = h α ξ α, ξ,α, 3) o, Then h α h α = h α ξ α, ξ,α + η α ξ,µ µ. 4) h,, α = h α = h, α ξ α, ξ,α + ξ µ µ,α 5) ξα, 6) = ξα, 7) h, α whee to get the last expession we ve used the fact that = 0, since h α is in Loentz gauge. If we want h α to emain in Loentz gauge, we see that the geneatos ξ α should satisfy wave equation: ξα, = 0 The geneal solution of this equation can be witten as a sum of plane waves: d 3 k [ ξ α t, x) = π) 3 A α k)e ik x ωt) + B α k)e ik x+ωt)] 8) 3

4 The fist tem descibes the wave popagating in the k diection and second one in k diection. In ou cases we need only the fist tem since we conside a gavitational wave popagating in some paticula diection). So ξ α t, x) = d 3 k π) 3 A αk)e ik x ωt) 9) At time t = 0: ξ α 0, x) = d 3 k π) A 3 α k)e ik x, o A α k) = d 3 xξ α 0, x)e ik x. We see that ξ α x) ae completely detemined by fou functions of thee spatial coodinates: ξ α 0, x). These functions give initial conditions fo the wave equation at t = 0. b) Conside a plane gavitational wave popagating in the z-diection. h α is in Loentz gauge, i.e. h α = h α t z) = h α τ), τ t z 30) h,,t α = h αt +,z h αz = h αt,t + h αz,z = h αt h αz 31) = 0 3) Whee pime denotes deivatives with espect to τ as defined in the pevious poblem. Integating: h αz = h αt + const. The constant is ielevant and we can set it to zeo, thus h αz = h αt. These fou gauge conditions educe the numbe of independent components of h α fom 10 to 6: h tt, h tx, h ty, h xx, h xy, h yy. Now make additional gauge tansfomation with ξ α = ξ α τ) = ξ α t z), ξα, = 0 33) α h α ξ α, ξ,α + η α ξ µ,µ 34) We note that ξ,µ µ = ξ t,t + ξ z,z = ξ t ξ z. new We want to choose ξ α so that h α satisfy additional constaints: hnew new new new tx = h ty = 0, h xx + h yy = 0. tt = tt = h tt ξ t,t + ξ t + ξ z) = h tt + ξ z ξ t 35) tx = h tx ξ t,x ξ x,t = h tx ξ x 36) ty = h ty ξ t,y ξ y,t = h ty ξ y 37) xx = h xx ξ x,x ξ t ξ z = h xx ξ t ξ z 38) yy = h yy ξ t ξ z 39) xy = h xy 40) 4

5 This gives the system of equations: ξ x = h tx 41) ξ y = h ty 4) ξ t = h tt + 1 h xx + h yy ) ξ z = h tt + 1 h xx + h yy ) 43) 44) These equations have unique solutions up to an additive constant) given by simple integations. c) We apply eqns 5.94 and 5.95 whee we use z k = e k ẑ fo nk. Then P l j P m k h lm = h jk z j h kz z k h jz + z j z k h zz and P jk P lm h lm = δ jk z j z k )h xx + h yy ). Hee ae the esults: xx = h xx h xx + h yy ) = 1 h xx h yy ) yy = h yy h xx + h yy ) = 1 h xx h yy ) zz = h zz h zz h zz + h zz 0 = 0 xy = h xy 1 0) = h xy xz = h xz 0 h xz ) = 0 yz = h yz h yz = 0 We still have h T µν, T ν = h T yz T = 0. ty = 0 so tt = tz = zz = 0, xt h T jk T = 0 1 h xx h yy ) h xy 0 0 h xy 1 h xx h yy ) = xz = 0 and So we find that h T xx T should. = yy = h +, xy = yx = h and t ) = 0 as they 5.9 Enegy emoved by gavitational adiation eaction [by Keith Matthews 05] dm = 1 5 ρ dxj 5 ) I lm t 5 xl x m d 3 x =, j 5 I5 jl ρ dxj xl) d 3 x 45) 5

6 Whee I 5 indicates 5 I and it comes out of the integal because it is not a t 5 function of x. To evaluate the integal on the ight we conside Ijk 1 = t I jk = d I jk = ρ dxj xl) d 3 x 1 3 δjl ρ d ) d 3 x 46) Because δ ij x i x j = and δ ij δ ij = 3, as applied below, we find that the second integal above doesn t contibute. I ij I ij = ρ x)ρ x )x i x j 1 3 δij )x i x j 1 3 δij )d 3 x d 3 x 47) = ρ x)ρ x )x i x j x i x j δij δ ij )d 3 x d 3 x 48) = ρ x)x i x j 1 ) ) 3 δij )d 3 x ρ x )x i x j d 3 x 49) So dm = d M = 1 I 5 5 jl Ijl 1 which we integate by pats twice to give 50) which is the desied esult. dm = d M = 1 5 I 3 jl I 3 jl 51) 5.10 Popagation of waves though an expanding univese [by Alexande Putilin 00] ds = b [ dη + dχ + χ dθ + sin θdφ )], whee b = b 0 η. 5) a) We can pove that cuves of constant θ, φ, η χ satisfy geodesic equation by explicit calculation of connection coefficients. But the easie way is to use symmety. Spheical symmety implies that a adial cuve η = ηζ), χ = χζ), θ, φ = const must be a geodesic fo some paamete ζ. Since a geodesic is null we have dη dζ ) + dχ dζ ) = 0, dη dζ = dχ dζ, η χ = const along a geodesic. b) Symmety also helps hee. Spheical symmety guaantees that k eˆθ cannot point in χ o φ diection. So k eˆθ = a eˆθ + b k. k = k η eˆη + k χ e ˆχ = k η eˆη + e ˆχ ), since k = 0 implies k η = k χ. But eˆθ k eˆθ = a = 1 k eˆθ eˆθ) = 1 k 1) = 0 gives a = 0. and k eˆθ = b k = bk η eˆη + e ˆχ ). Take a dot poduct of this eqn with e ˆχ : k ˆα eˆθ;ˆα = k ˆα Γˆµ ˆθ ˆα eˆµ = bk ˆη eˆη + e ˆχ ) 53) 6

7 bk ˆη = k ˆα Γˆµ ˆθ ˆα η ˆχˆµ = k ˆα Γ ˆχˆθ ˆα = k ˆη Γ ˆχˆθˆη +Γ ˆχˆθ ˆχ ), so b = Γ ˆχˆθˆη +Γ ˆχˆθ ˆχ. Now we need only to calculate two connection coefficients to veify that Γ ˆχˆθˆη = Γ ˆχˆθ ˆχ = 0, so that b = 0 k eˆθ = 0. The poof that k e ˆφ = 0 is vey simila. c) The geneal solutions ae, in the geometic optics limit: h + = Q +τ, θ, φ), h = Q τ, θ, φ) 54) whee k = τ and k = 1 k). To fix τ = τ η χ) notice that at χ = 0 o coespondingly = 0) τ = t = t, whee t is a pope time at χ = 0. = ds = b dη = b 0η 4 dη = b 0 η dη t = 1 3 b 0η 3 so τ η) = t = 1 3 b 0η 3 τ η χ) = 1 3 b 0η χ) 3 k = τ k η = k χ = η χ) b 0 η 4 55) k) = 1 g g k α ),α = 1 g [ g k η ),η + g k χ ),χ ] 56) = η χ) η + χ) b 0 η 5 χ afte some calculations) 57) Then k = 1 k) = k η,η +,χ ) 58) educes to η + ) 1 = χ χ + ) 59) η changing vaiables: a = η χ, b = η + χ, we get: 1 b = b a + ) 60) b + a a, b) = Ca)e R db 1 b a + b+a) = Ca)b a)b + a) 61) χ, η) = Cη χ)χη 6) 7

8 whee Cη χ) is an abitay function. Conside the egion η = η 0, χ << η 0. In this egion we should have: χ, η) = 63) whee d = ds = b dχ = b 0η0dx 4, = b 0 η0χ, Cη 0 )χη0 Cη 0 ) = b 0. So finally we get = b 0 η 0χ, = b 0 η χ 64) To detemine Q +, Q, compae them to the solution of gavitational wave eqn. in the nea zone: η η 0, χ << η 0 τ = t ) h + = ] T T [Ïˆθ ˆθ t ) = h = ] T T [Ïˆθ ˆφ t ) = ] T T [Ïˆθ ˆθ τ ) 65) ] T T [Ïˆθ ˆφ τ ) 66) Q + τ, θ, φ) = [Ïˆθ ˆθ τ )] T T 67) Q τ, θ, φ) = [Ïˆθ ˆφ τ )] T T 68) 5.11 Gavitational waves emitted by a linea oscillato [by Alexande Putilin 00] Since the mass is moving along the z diection the second moment of mass distibution has only a zz-component. o and we have which gives I zz t) = mz t) = ma cos Ωt 69) It) = ma cos Ωt e z e z 70) jk = ] T T [Ïjk t ) 71) = ma 4Ω cos Ωt ) [ e z e z ] T T 7) = 4mΩ a cos Ωt )) [ e z e z ] T T 73) 8

9 To pefom TT-pojection notice that e z = cos θ eˆ sin θ eˆθ, and thus e z e z = cos θ eˆ eˆ cos θ sin θ eˆ eˆθ + eˆθ eˆ ) + sin θ eˆθ eˆθ 74) TT-pojection on eˆθ, e ˆφ) plane gives: [ e z e z ] T T = 1 ) sin θ eˆθ eˆθ e ˆφ e ˆφ 75) so = mω a ) sin θ cosωt )) eˆθ eˆθ e ˆφ e ˆφ 76) = mω a sin θ cosωt )) e + 77) It follows immediately fom the esult above that: In conventional units h + t,, θ, φ) = mω a sin θ cosωt )) 78) h t,, θ, φ) = 0 79) h + t,, θ, φ) = GmΩ a c 4 sin θ cosωt )) 80) 5.1 Gavitational waves fom waving ams [by Xinkai Wu 0] a) Take the fequency to be f Hz, then the wavelength is λ c/f m. The majo contibution to the gavitational wave comes fom the mass quadupole moment and is given by equation 5.11: h + h G Mv c 4, and we take M 10kg, v Lf 1m Hz m/swhee L is the length of the am), and λ 10 8 m, this gives h + h b) The total powe de de G M v 6 c 5 L 1 de πf is given by equation Restoing G, c, we get J/s, and the numbe of gavitons emitted pe second is Hz, which means the gavitational waves emitted ae so weak that they can t eally be teated as classical waves Light in an intefeometic gavitational wave detecto in TT gauge [by Xinkai Wu 0] a) This expession fo φ gives φ t = ω 0 [1 + 1 ht x) 1 ] ht) φ x = ω 0 [ 1 1 ] ht x) 81) 8) 9

10 Ignoing tems quadatic and highe) in h, we find ) ) φ φ + [1 ht)] = ω t x 0[1 + ht x) ht)] +[1 ht)]ω 0[1 + ht x)] 83) = 0 84) b) Setting x = 0, we get φ t = ω 0. c) k k)ν = k µ µ k ν 85) = k µ µ ν φ 86) = k µ ν µ φ 87) = k µ ν k µ 88) = 1 νk µ k µ ) 89) = 0 90) d) The null geodesic of the photon is given by 0 = ds = + [1 + ht)]dx, which gives dx = 1 1 ht). Now p x = φ/ x = ω 0 [1 + 1 ht x)], and along the null geodesic we have dp x = ω 1 0 x) 1 dx ) ) 1 1 = ω 0 x) ḣt ḣt ht) = 0 91) up to linea ode in h. Thus we see p x is indeed conseved along the geodesic. e) The obseve at est has 4-velocity u = 1, 0, 0, 0), thus the photon s enegy measued by him is ω = k u = k α u α = k t = φ/ t = ω 0 [1 + 1 ht x) 1 ht)]. dω = ω t + ω dx ω x. Since x is aleady of ode h, we can appoximate dx dω as unity, and thus getting t + x )ω. And this is dω t + ) ω 9) x [ 1 = ω 0 ḣt x) 1 ḣt) 1 ] ḣt x) 93) as desied. = 1 ω0ḣt) 94) 10

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