B Appendix B Useful results B.1 Linear algebra The inverse of a 2 × 2 matrix can be found from (B.1). Note the final factor in parentheses is the determinant. a b d −b 1 0 = (ad − bc). (B.1) c d −c a 0 1 B.2 Series approximations The follow series are obtained via a Taylor series expansion about x = 0. The first one appears very often so we christen it “binomial series”: 1 1 (1 + x)a = 1 + ax + a(a − 1)x 2 + a(a − 1)(a − 2)x 3 · · · 2 6 1 = 1 − x + x2 − x3 + · · · 1+x binomial series (B.2) (B.3) Letting x → −x in (B.3) immediately gives 1 = 1 + x + x2 + x3 + · · · 1−x √ 1 1 1 1 + x = 1 + x − x2 + x3 + · · · 2 8 16 (B.4) (B.5) Letting x → −x in (B.5) immediately gives √ 1 1 1 1 − x = 1 − x − x2 − x3 + · · · 2 8 16 5 35 4 63 5 3 1 1 x − x ··· = 1 − x + x2 − x3 + √ 2 8 16 128 256 1+x (B.6) (B.7) Letting x → −x in (B.7) immediately gives the series we will use repeatedly to approximate the Lorentz factor; just let (x = v 2 ): √ 295 1 5 35 4 63 5 3 1 x + x ··· = 1 + x + x2 + x3 + 2 8 16 128 256 1−x (B.8) 296 Useful results x2 x3 x4 + − + ··· 2 3 4 (B.9) a2 2 a3 3 a4 4 x + x + x + ··· 2 6 24 (B.10) ln(1 + x) = x − eax = 1 + ax + B.3 Transformations between spherical polar and Cartesian coordinates B.3.1 Upper indices: Cartesian to polar (lower indices polar to Cartesian) r= x 2 + y 2 + z2 θ = arccos(z/r) φ = arctan(y/x) (B.11) ∂θ 1 = cos θ cos φ ∂x r ∂θ 1 = cos θ sin φ ∂y r ∂θ 1 = − sin θ ∂z r ∂r = sin θ cos φ ∂x ∂r = sin θ sin φ ∂y ∂r = cos θ ∂z ∂φ 1 sin φ =− ∂x r sin θ ∂φ 1 cos φ = ∂y r sin θ ∂φ =0 ∂z (B.12) B.3.2 Upper indices: polar to Cartesian (lower indices Cartesian to polar) x = r sin θ cos φ y = r sin θ sin φ z = r cos θ ∂x = sin θ cos φ ∂r ∂y = sin θ sin φ ∂r ∂z = cos θ ∂r ∂x = r cos θ cos φ ∂θ ∂y = r cos θ sin φ ∂θ ∂z = −r sin θ ∂θ (B.13) ∂x = −r sin θ sin φ ∂φ ∂y = r sin θ cos φ ∂φ ∂z =0 ∂φ (B.14) B.4 Selection of spacetimes Below we summarize the important spacetimes studied herein by listing their line element, Christoffel symbols, and important tensors. Components that are not listed and are not related by symmetry to one of those listed are zero. 297 Selection of spacetimes B.4.1 Rindler spacetime The line element is ds 2 = −a 2 dλ2 + da 2 . (B.15) This metric applies in flat (Minkowski) spacetime and was derived in Exercise 5.21 for a set of uniformly accelerating observers. Because it’s a flat spacetime it follows that the Riemann tensor vanishes, and thus the Ricci tensor and scalar and Einstein tensor all vanish. Christoffel symbols λ λa = 1 a a λλ = a. (B.16) B.4.2 Static spherically symmetric spacetimes See Exercise 6.35. The line element is ds 2 = −e2(r) dt 2 + e2Λ(r) dr 2 + r 2 dθ 2 + r 2 sin2 θ dφ 2 . (B.17) Christoffel symbols = e−2Λ e2 r φφ = −e−2Λ r sin2 θ θ φ rφ = tr = θθ = −e−2Λ r r φφ = − sin(θ ) cos(θ ) t r θ r tt 1 r rr rθ φ θφ =Λ 1 = r cos(θ ) = , sin(θ ) (B.18) where ≡ d/dr and Λ ≡ dΛdr. Ricci tensor and Ricci scalar See SP6.8: 2 Rtt = −e(2φ−2Λ) Λ − 2 − − r −2Λ [1 − r(Λ − )] Rθθ = − −1 + e 2Λ Rrr = − −Λ + 2 + − r 2 −2Λ Rφφ = − sin θ e [1 + r( − Λ )] − 1 (B.19) and 2( − Λ ) 1 − e2Λ + . R = −2e−2Λ −Λ + 2 + + r r2 (B.20) 298 Useful results Einstein tensor 1 2φ d −2Λ e ) , r(1 − e r2 dr Λ 2 −2Λ 2 =r e −Λ − , + ( ) + r r 1 2 (1 − e2Λ ) + , 2 r r Gtt = Grr = Gθθ Gφφ = sin2 θ Gθθ . (B.21) as in Schutz Eq. (10.14)–(10.17). B.4.3 Schwarzschild spacetime The line element is 2M 2M −1 2 dr + r 2 d2 , dt 2 + 1 − ds 2 = − 1 − r r Schutz Eq. (10.36) (B.22) with d2 = dθ 2 + sin2 θ dφ 2 (same as metric (ii) of Exercise 7.7). This metric applies in the vacuum around a static spherically symmetric source. Because it’s a vacuum spacetime it follows from the Einstein equations (see SP9.2) that Rαβ = Gαβ = 0 and R = 0. Christoffel symbols M r2 2M r −1 2M r tr θθ = −r + 2M r φφ = (−r + 2M) sin2 θ φφ = − sin(θ ) cos(θ ) φ rφ = r θ 1− r tt = M r2 = t 1− rr =− rθ = r θ 1 r φ θφ M r2 1− 2M r 1 r cos(θ ) = . sin(θ ) −1 (B.23) B.4.4 Weak gravitational field See Exercise 7.2. The line element is: ds 2 = −(1 + 2φ)dt 2 + (1 − 2φ) dx 2 + dy 2 + dz2 . Schutz Eq. (7.8) (B.24) Christoffel symbols i tt i jk = φ,i + O(φ 2 ), = δj k δ il φ,l − δji φ,k − δki φ,j + O(φ 2 ), where i, j , k ∈ {x, y, z}. i tj = −φ,t δ i j + O(φ 2 ), (B.25) 299 Selection of spacetimes Ricci tensor and Ricci scalar Rtt = 3φ,tt + φ,xx + φ,yy + φ,zz + O(φ 2 ) Rti = 2φ,ti + O(φ 2 ) Rii = −φ,tt + φ,xx + φ,yy + φ,zz + O(φ 2 ) Rij = 0 + O(φ 2 ) when i j . (B.26) R = −6φ,tt + 2(φ,xx + φ,yy + φ,zz ) + O(φ 2 ). (B.27) Einstein tensor Gtt = 2(φ,xx + φ,yy + φ,zz ) + O(φ 2 ) Gti = 2φ,it + O(φ 2 ) Gii = 2φ,tt + O(φ 2 ) Gij = 0 + O(φ 2 ) when i j . (B.28) B.4.5 Post-Newtonian spherical rotating star The line element was derived in Exercise 8.19, see eqn. (8.59): 2M sin2 θ 2M 2 2 dt − 4J dt dφ + 1 + (dr 2 + r 2 dθ 2 + r 2 sin2 θ dφ 2 ). ds = − 1 − r r r (B.29) This metric applies in the vacuum around a spherical source that rotates. Because it’s a vacuum spacetime it follows from the Einstein equations (see SP9.2) that Rαβ = Gαβ = 0 and R = 0. Christoffel symbols t tr t rφ r r r tt rr φφ θ rθ φ tr φ rφ 2J 2 cos2 θ + 2M 2 r 2 + r 3 M − 2J 2 r(4J 2 cos2 θ + 4M 2 r 2 − r 4 − 4J 2 ) rJ sin2 θ (3r + 4M) = 4J 2 cos2 θ + 4M 2 r 2 − r 4 − 4J 2 M = r(r + 2M) M =− r(r + 2M) r(M + r) sin2 θ =− r + 2M M +r = r(r + 2M) J = 2 2 4(J sin θ − M 2 r 2 ) + r 4 2J 2 sin2 θ + 2M 2 r 2 + r 3 M − r 4 =− r[4(J 2 sin2 θ − M 2 r 2 ) + r 4 ] =− tθ =− tφ =− t r r θθ θ tφ 4J 2 sin θ cos θ 4J 2 cos2 θ + 4M 2 r 2 − r 4 − 4J 2 J sin2 θ r(r + 2M) r(M + r) =− r + 2M 2J sin θ cos θ = 2 r (r + 2M) φφ = − sin θ cos θ φ tθ = φ θφ = cot θ θ 2J cos θ (2M − r) sin θ[4(J 2 sin2 θ − M 2 r 2 ) + r 4 ] (B.30) 300 Useful results B.4.6 Kerr spacetime The line element is: − a 2 sin2 θ 2 2Mr sin2 θ dt − 2a dt dφ ρ2 ρ2 (r 2 + a 2 )2 − a 2 sin2 θ ρ2 2 2 2 + sin θ dφ + dr + ρ 2 dθ 2 , ρ2 ds 2 = − (B.31) where M and a are constants and ≡ r 2 − 2Mr + a 2 , ρ 2 ≡ r 2 + a 2 cos2 θ. (This is metric (iii) of Exercise 7.7). This metric applies in the vacuum around a source that rotates. Because it’s a vacuum spacetime it follows from the Einstein equations (see SP9.2) that Rαβ = Gαβ = 0 and R = 0. Christoffel Symbols t tr t tθ t rφ t θφ r r tφ r r r r tt rr rθ θθ φφ θ tt θ tφ M(r 2 − a 2 cos2 θ )(r 2 + a 2 ) ρ4 2 2a sin θ cos θ Mr =− ρ4 2 aM sin θ [a 4 cos2 θ − a 2 r 2 (1 + cos2 θ ) − 3r 4 ] = ρ4 2 sin3 θ cos θ Ma 3 r = ρ4 M(r 2 − a 2 cos2 θ ) = ρ6 = = −a sin2 θ r tt M(a 2 cos2 θ − r 2 ) + a 2 r sin2 θ ρ2 a 2 sin θ cos θ =− ρ2 r =− 2 ρ sin2 θ [rρ 4 + M(a 2 cos2 θ − r 2 )a 2 sin2 θ] =− ρ6 2Ma 2 r cos θ sin θ =− ρ6 2 2 a +r θ =− tt a = −pb2 qa 2 sin(2θ ) = a sin2 θ = (pb2 + 2qr) qa 3 sin2 θ sin(2θ ) =− p =− ap sin2 θ = r M −r = + 2 a sin(2θ ) =− 2 r =− sin2 θ =− (r + pa 2 sin2 θ ) qa 2 sin(2θ ) = 2 qab2 sin(2θ ) =− 2 = 301 Selection of spacetimes θ rr θ rθ θ θθ θ φφ a 2 sin θ cos θ ρ2 r = 2 ρ a 2 sin(2θ ) 2 r = = = r = = rθ sin θ cos θ 4 2 =− [ρ (r + a 2 ) ρ4 + 2Mra 2 sin2 θ (2ρ 2 + a 2 sin2 θ )] aM(r 2 − a 2 cos2 θ ) ρ4 φ tr = φ tθ =− φ rφ = rρ 4 − 2Mr 2 ρ 2 + a 2 M sin2 θ (a 2 cos2 θ − r 2 ) ρ4 φ θφ = cot θ 4 (ρ + 2Mra 2 sin2 θ ) ρ4 r sin(2θ ) 2 =− b − 2a 2 sin2 θ q 2 a 2 sin2 θ × 2+ =− 2aMr cot θ ρ4 rθ = ap 2qa cot θ sin2 θ r (1 + 2q) + a 2 p cot θ = (1+2q)(b2 −2qa 2 sin2 θ ) 2qa 2 b2 sin2 θ − = (B.32) The second equality for each αμν in eqn.(B.32) uses the notation of Frolov and Novikov (1998, Appendix D), with = ρ 2 , q = −Mr/, p = M(a 2 cos2 θ −r 2 )/ 2 , b2 = r 2 +a 2 when the total charge Q = 0. B.4.7 Robertson–Walker spacetime The line element is: 1 ds = −dt + R (t) dr 2 + r 2 (dθ 2 + sin2 θ dφ 2 ) 1 − kr 2 2 2 2 Schutz Eq. (12.13) (B.33) with three possible values of k: k=1 “closed” or “spherical” universe k=0 spatially “flat” universe k = −1 “open” or “hyperbolic” universe (This is metric (iv) of Exercise 7.7.) 302 Useful results Christoffel symbols t rr r r tr φφ θ tθ φ tφ R Ṙ 1 − kr 2 Ṙ = R = −r(1 − kr 2 ) sin2 θ = t θθ rr = rθ = r Ṙ R Ṙ = R = = R Ṙr 2 θ φ rφ t φφ kr 1 − kr 2 θθ = −r(1 − kr 2 ) φφ = − sin θ cos θ r 1 r 1 = r = R Ṙr 2 sin2 θ θ φ θφ = cot θ (B.34) Ricci tensor and Ricci scalar Ricci tensor: 3R̈ R 2 = r R R̈ + 2Ṙ 2 + 2k R R̈ + 2Ṙ 2 + 2k 1 − kr 2 Rtt = − Rrr = Rθθ Rφφ = Rθθ sin2 θ. Ricci scalar: R αα Einstein tensor: R̈ k Ṙ 2 =6 + 2+ 2 R R R . Ṙ 2 k Gtt = 3 + 2 R2 R 2 Gθθ = −r 2R R̈ + Ṙ 2 + k (B.35) Grr = (B.36) −2R R̈ − Ṙ 2 − k 1 − kr 2 Gφφ = Gθθ sin2 θ. (B.37)