Errata Sheet: Relativity Demystified (David McMahon, McGraw-Hill, 2006) Susan Larsen Friday, June 26, 2015

Transcript

1 In this fil you find all sorts of misprints. Thr ar quit many but it dos not man that th book is ntirly bad. On th contrary, it givs you th opportunity to carry out a lot of calculations, and it is a good starting point. If th misprint is markd by an astrix (*) it mans that you can find a mor thorough calculation in my fil lots of calculations on my hompag Th list is not complt, it covrs th chaptrs -3 and th final xam, but I will post mor on my hompag as I gt along. I hav mad this list ntirly on my own, and should you find that I hav mad somthing wrong or missd somthing, which I am bound to, plas do not hsitat to contact m: logik.susan@gmail.com Contnt Contnts... Chaptr... Chaptr... Chaptr 3... Chaptr 4... Chaptr Chaptr Chaptr Chaptr Chaptr Chaptr Chaptr... 7 Chaptr... 7 Chaptr Final Exam... 0 Quiz and Exam Solutions... Bibliografi... Contnts p.ix l.7 Eddington-Finklstin Coordinats Th Radial Null Godsic Chaptr p. l.0 oprator to th right-hand sid of (.), w obtain p.6 l.6-7 clock at tim t, and th rflctd bam arrivs at th location of clock at tim t₂.

2 p. l. ct = p.6 l.3 β (β₁ + β₂) 3 = ( + β₁β₂) l.5 β₁ + β₂ β 3 = + β₁β₂ Chaptr p.9 l.8 Vf = (V a a )f = V a a p.33 l.7 as shown in Fig. -5 p.34 l.4 ds = dr + r dθ + r sin θ dφ (.9) p.40 l.5 g yx x + g yy = 0 l.6 g yy = x g yx p.4 l. W x = x (0) + () = l. W y = (0) () = l.3 W a = (, ) p.4 l. S b = l.3 = g ac T b c = l.4 = g a R bcd p.44 l. r = x r x + y r y + z r z Chaptr 3 p.49 l.5 U. A point p that blongs to th manifold M (say in U ) p.55 l.3 Q b = S a b T b Chaptr 4 p.67 l. = ( Ab x a + Γ b ca l.5 b A a = Aa x p.7 l.9 Γ abc = ( g ab b +Γ bc A c ) b a A c (4.6) + g ac x a ) (4.5) l. Γ abc = ( g ab + g ac x a ) + C abc + C acb C bca p.73 l.7 Γ abc = ( g ab + g ac x a ) l.8 Γ bca = ( g bc x a + g ba g ca x b ) l.0 Γ abc + Γ bca = ( g ab + g ac x a ) + ( g bc x a + g ba g ca x b ) l. = ( g ac x b g ca x b + g ab + g ba g bc x a + g bc x a ) l. = ( g ab ) = g ab l.7 a Γ bc = gad ( g db + g dc x d ) (4.6)

3 p.75 l. th last trm φ l.3 Γ bc φ l.5 Γ φθ = gφφ ( g φb = gφφ ( g φφ θ + g φc x b ) + g φθ φ ) = p.76 l.3 Γ abc = ( cg ab + b g ac a g bc ) l.6-7 Looking at Γ abc = ( cg ab + b g ac a g bc ) p.77 (*) l. Γ xxu = ( ug xx + x g xu x g xu ) = ug xx = [( u)] = u l.4 Γ uxx = ( xg ux + x g ux u g xx ) = ug xx = [( u)] = + u l.8 Γ uyy = + u l.9 Γ yyu = u l. thir last two l.3 Γ abc = Γ acb p.80 l.6 (aα + bβ) γ = aα γ + bβ γ p.8 l.7 L V W = V b b W W b b V or altrnativly (L V W) a = V b b W a W b b V a (4.7) l.9 L V T ab = V c c T ab + T cb a V c + T ac b V c (4.8) p.8 l.0- p.83 (*) l.8 This should b omittd: W can writ (4.33) in th mor convnint form: d x a ds + Γ a bc d φ ds + r τ p.85 (*) l.3 Γ ξτ dr ds dx b dx c ds ds = 0 dφ ds = 0 = ξ = Γ τ τξ p.86 l.6 R abcd = g a R bcd (*) l.8 R abcd = c Γ abd d Γ abc Γ ac Γ bd + Γ ad Γ bc (4.4) l. Γ ac Γ bd + Γ ad Γ bc (4.43) p.87 l.3 All togthr, in n dimnsions, thr ar n (n )/ indpndnt nonzro componnts l.0 R ; R 33 ; R 33 ; R 3 ; R 3 = R 3 ; R 33 = R 33 p.88 l. of th Rimann tnsor in two dimnsions ar p.9 (*) Q Γ θφφ = ( θg φφ ) = ( θ(r sin θ)) = r cos θ sin θ p.9 (*) Q7 v Γ xx = + u Chaptr 5 p.05 l.7 α β = ( ) pq β α (5.3) p.0 (*) l.4 R (*) Q3 r Γ φφ = a + k a = (Λ ) r Γ (*) Q4 u v Γ v v = Γ u v = + u p. Q5 Whr ar th hats? φλ φ Λ = ( r ) (r sin θ) = r sin θ 3

4 (*) Q7 G = R [ RΨ (R Ψ + R + R R ) R Ψ R + + R ] Chaptr 6 p.30 l.5 Cass and 3 in our p.33 l.5 L u η a = u b b η a η b b u a = 0 l.9 c a S b = a S c c b + Γ ads d d c b Γ bas d l.4 b a V c = b ( a V c + p.34 l.4 a ( b V c c Γ bv ) b ( a V c + (6.) p.35 l. c = R dabv d (6.3) p.36 l. Howvr, sinc u a is th tangnt p.37 l. a Γ bc = gad ( g db + g dc x d ) (4.6) p.39 l. ds = dt + b(t,r) dr + R (t, r)dφ p.4 (*) l.8 = R ω ω R R t R r b(t,r) ω ω p.43 l.3 Γ = l.7 = R a b c d ω c ω d p.44 l.0 Ω = p.46 l.4 w writ out Ω = l.6 Ω = l. R = l. All togthr, th indpndnt nonzro componnts p.47 l. Using th rsults of Exampl 6- (*) l.5 R = R c l.8 R = R c = R R + R + R b (6.) (6.9) p.49 (*) l.5 G = b(t,r) R t r + b(t,r) R R t r l.8 ds = dt + b(t,r) dr + R (t, r)dφ, this is asy nough: 0 0 a (*) l.9 Λ b = ( 0 b(t,r) 0 ) (6.33) 0 0 R(t, r) (*) l.5 G tt = Λ tλ tg = ()()G (6.35) (*) l.8 ()( b(t,r) )( b(t,r) R + b(t,r) b R ) (6.36) R t r R t r p.53 l.0 Γ l.5 Γ Chaptr 7 p.55 l.9 tnsor and nrgy-momntum tnsor intrchangably. p.59 l.4 γ = ( v c ) 4

5 l.6 = γ ρ t + (ρux ) γ + γ (ρuy ) + γ (ρuz ) x y z l.7 = γ ρ t + γ (ρu ) p.60 l.8 T ab = Au a u b + Bg ab (7.8) l. T ij = δ ij P (7.9) p.6 l.3 T ij = Bη ij l.6 = Au 0 u 0 + Bη 00 = l.9 = (ρ + P)u a u b Pη ab (7.0) l. = (ρ + P)u a u b Pg ab (7.) l.4 = (ρ + P)u a u b + Pη ab l.5 = (ρ + P)u a u b + Pg ab (7.) p.6 l.5 G a b Λη a b = 8πT a b (7.4) l.3 Putting this togthr with (7.4) and (7.3), w hav p.65 l.0 worldlinss Chaptr 8 p.68, xampl 8-: Altrnativ solution (*): Calculat th Li-drivativ of g ab (4.8) and using (4.8) you gt: L X g ab = X c c g ab + g cb a X c + g ac b X c = a (g cb X c ) + b (g ac X c ) = a X b + b X a If L X g ab = 0 this implis that a X b + b X a = 0, which is th Killing quation. p.7 l.6 c θ X θ = θ X θ Γ θθx c = θ X θ Γ θθx θ Γ θθ p.7 l.7 θ X φ = θ [ sin θ cos θ f(φ )dφ + g(θ)] p.74 l.8 P(t) = p(t )dt l.0 y(t) = P(t) P(t ) r(t )dt + C P(t) l.4 P(θ) = l.6 P(θ) = l.7 From this w dduc that P(θ) = p.76 l.0 X θ = g θθ X θ = a X θ l. X φ = g φφ X φ = a sin θ X φ p.77 l.6 = a (A cos φ + B sin φ) θ + a [C cot θ (A sin φ B cos φ)] φ l.7 = A cos φ θ A cot θ sin φ φ + B sin φ θ + B cot θ cos φ φ + C φ l.8 = +A L x + B L y + C L z l.0 L x = + cos φ cot θ sin φ θ φ p.78 l.5 b T ab = 0 φ X φ Chaptr 9 p.88 l. = r sin θ 5

6 p.90 l. m m = m a m a = l.3 δ = m a a p.98 (*) l. v Γ uu = gvd ( u g du + u g ud d g uu ) = gvu ( u g uu + u g uu u g uu ) = gvu ( u g uu ) = () ( H u ) = + H u p.99 l.8 Now b l a = and so w can immdiatly l. Having a look at th Christoffl symbols (9.9), w s (9.9) Chaptr 0 p.04 l.0 This tlls us that th mtric lin lmnt will not p.05 l. = Cr dθ + Cr sin θ dφ n l.6 = C ( r dc r + ) dr c dr p.07 (*) l.9 Γ = Γ θ = Γ = 0 (0.6) θ p.08 (*) l.5 Γ = Γ θ θ = Γ θ = 0 (0.7) ϕ (*) l.9 Γ ϕ = Γ = 0 (0.8) p.0 l. = R l.7 All togthr, th indpndnt nonzro componnts of th l.9 = R p. l.4 c R a b = R a c b p.6 l.3 largrangian p.7 (*) l.8 = 4m m r + (r m) (r ) r p.8 (*) l. = m r (t ) + ( m r ) ( m r ) (r ) r(θ ) r sin θ (φ ) p.0 l.9 = r sin θ dφ dφ = r dτ dτ p.6 l. rspct to φ r a scond tim l.9 and using yr 0 = r sin φ l.0 = r sin φ = yr A 0 l. th constant /A rprsnts th p.7 l.5 u p = D sin φ + E cos φ l.7 u p = l.8 u p = l.0 This lin should b omittd: u p + u p = D sin φ + 4D sin φ = 3D sin φ l. p u + p u = l.4 p u + p u = A sin φ l.7 p u + p u = A sin φ p.8 l.7 and stting this u qual to zro p.30 l. so w put ct in plac of t and G m c r r (0.48) 6

7 l.3 ct = dr r 0 r ( + m r G c mr 0 G r 3 c ) l.7 ct = r p r 0 r r 0 + m G c ln (r p + r p r 0 )(r + r r 0 ) r 0 l.0 m G c [ r p r 0 r p r r 0 ] r Using th variational mthod dscribd in Exampl 4-0, th nonzro Christoffl symbols for th gnral Schwarzschild mtric ar p.3 (*) Q (b) R rt = r (dλ) dt (*) Q3 (b) Γ 3m Λr = 3 r 3/ 9r 8m 3Λr 3 p.36 l.7 Eddington-Finklstin Coordinats Th Radial Null Godsic p.38 (*) l.0 t t 0 = 3 m (r3/ r 3/ 0 + 6m r 6m r 0 ) + m p.39 (*) l. r m = (r 0 m) (t t 0)/m Chaptr p.45 l.6 ds = ( m Σ ) dt + 4amr sin θ dtdθ (.9) Σ p.46 (*) l. g φφ = (Δ a sin θ) (.3) ΣΔ sin θ p.47 l.4 In this cas, θ = π,which l.5 mans that sin θ = and p.5 l.3 R = r + a + m3 r Chaptr p.58 l. Suppos that S rprsnts th spaclik p.6 l.4 f = (.9) C kr (.4) l.5 To find th constant C = C, w can us th l.3 = C + Kr Kr = C + Kr l.5 C + Kr = Kr or C = 0 C = p.68 l.3 Th dtail wr workd out in Exampl 7-3 P.73 l.0 ds = dt Λ 3 t [dr + r dθ r sin θ dφ ] p.74 Fig. -5 Th vrtical coordinat is a not R l. Th d Sittr solution rprsnts a univrs without mattr and without radiation. p.75 Fig. -6 Th vrtical coordinat is a not R da dτ l.0 = C sin τ cos τ dt dt p.76 Fig. -7 Th vrtical coordinat is a not R l.3 Intgrating th lft sid bcoms p.77 (*) Q r ν(r) = A + Br + 3 k Λr 3 7

8 (*) Q3 B = k (*) Q4 dl dr = m r + + r dθ + r sin θ dφ 3 Λr Chaptr 3 p.80 l.5 a Γ bc p.8 l.6 a Γ bc = gad ( g bd = gad ( g bd l.7 = ε gad ( h bd + g cd x d ) + g cd x d ) + h cd x b h bc x d ) l.8 = ε (ηad εh ad ) ( h bd l.9 = (εηad ε h ad ) ( h bd l. = εηad ( h bd + h cd x b h bc x d ) + h cd x b h bc x d ) p.83 l.3 = c [ εηa ( h b x d + h d x b h bd )] x l.4 d [ εηaf ( h bf l.5 = ε (ηa h b x + h cd x b h bc x d ) (3.3) + h cf x b h bc x f )] + h d ηa h bd ηa d b x l.6 ε h bf (ηaf x d + ηaf xc x d x l.7 Lt s rlabl it as f l. Th first trm will cancl h cf ηaf b x ) h bc x d x f) l.3 = εηa ( h bd x f + h df x b + h bc x d x f h cf x d xb) (3.4) c l.5 c R ab = c Γ ab b Γ ac p.87 l. ψ a a b,a = ψ b,a φ b b l.4 Wφ a = ψ a,b p.88 l.3 ψ a b,a l.6 εwψ ab l.8 Wψ ab = 0 = l.4 Wψ ab = W (h ab η abh ) = Wh ab Wη abh = 0 l.6 0 = Wψ ab = η ab Wψ ab l.9 Wh ab = 0 = W(η ab ψ ab ) = W(ψ b b ) = Wψ = Wh tnsor can b shown to b a function of h xx, h xy, h yx and h yy alon, in th Einstin gaug. p.89 l. drivativs with rspct to y and zx vanish p.93 (*) l. ds = d t d x d y d z + εh xy dxdy (3.9) p.94 (*) l. ds = dt ( εh xy )dx ( + εh xy )dy dz (3.0) p.99 l.5 (s Quiz) 8

9 p.300 l. a a XdudX l. = b ( b b Ydu + b dy) = b b YdudY p.303 l.4 Th curvatur tnsor Wyl scalar p.304 l. ds = δ(u)(x Y )du + dudr dx dy (3.4) l.3 (s Problm 7) p.305 l.5 (3.45) p.306 l.3 = νg xx l.4 = ν ( [ νθ(ν)] ) l.5 = ( νθ(ν)) ν ( νθ(ν)) l.6 = ( νθ(ν)) ( Θ(ν) ν dθ dν ) l.7 = ( νθ(ν))(θ(ν) νδ(ν)) l.9 f(ν)δ(ν)dν = f(0) p.307 l.7 sinc n u is null l.8 n a = (,0,0,0) p.308 l.3 m a = ( 0, 0, ( νθ(ν)), i ( + νθ(ν)) ) l.4 m a = ( 0, 0, ( νθ(ν)), i ( + νθ(ν)) ) p.309 l. m xm x = ( ( νθ(ν)) ) ( ( νθ(ν)) ) p.3 fig 3- u + v = p.35 (*) l.6 m a = (0, 0, cos aν, +i cosh aν) (3.56) p.37 l. σ = a (tan aν + tanh aν) l.6 Ψ 0 = +Dσ σρ (3.57) l. = ν [a (tan aν + tanh aν)] (*) l. = a ( + tan aν + tanh aν) l.4 Ψ 0 = +Dσ σρ l.5- (*) 8 = a ( + tan aν + tanh aν). = a p.38 l.7 Nariai l.8 R ab = Λg ab (*) l. if R ab = g ab Λ Ω should rightfully b Ω = + Λ 4 (x + y ) (*) l.8 l.6 g νν = +Λν ν Γ uν u = Γ uu ν = Λν; Γ uu y = Γ yy = Λy Ω = Λ ν 3 x ; Γ xx y = Γ yx x = Γ yy = Λx Ω ; Γ x y xy = Γ xx 9

10 l.3 m a = ( 0, 0, Ω, i Ω ) ; m a = ( 0, 0, Ω, i Ω ) l.4- Ths lins should b omittd 5 p.39 l.3 n a = (, + Λν, 0, 0) l.4 m a = (0, 0, Ω, iω); m a = (0, 0, Ω, iω) l. Λν l. l a;b n a n b = l u;u n u n u = Λν l.9 m x = i Ω l. γ = l u;un u n u = Λν An xrcis shows that all rmaining nonzro spin cofficints vanish ar α = p.30 (*) l. Λ (x iy); β = Λ (x + iy) l.5 = ρμ σλ + αα + ββ αβ + γ(ρ ρ ) + ε(μ μ ) Ψ + Λ NP + Φ (3.59) l.8 = ν ( Λν) = Λ l.0 = Λ (3.6) l.8 Λ = l.9 = Λ (3.64) p.3 l. Ψ = 6 Λ and Φ = 4 Λ (3.65) p.3 Q l. usd in Exampl 3- x Q l.3 = μ ( x + y + i y x + y ) Q3 l.5 usd in Exampl 3- l.8 Φ = δν Δμ μ λλ μ(γ + γ ) + ν π ν(τ 3β α ) (9.4) Final Exam p.33 E (*) l. Δs. E = (,3,,4) (*) l.3 E = (4,0,,) p.34 E6 (*) l.5 Γ aba = Γ aab = g aa x b E7 (*) u u ν Th nonzro Christoffl symbols ar: Γ uu = Γ νν = Γ uν = u ; Γ ν (u +ν ) νν = ν u Γ uu = Γ νu = ν ; Γ u (u +ν ) θθ = uν ; Γ ν (u +ν ) θθ = u ν ; Γ θ (u +ν ) uθ = ; Γ θ u νθ = ν ψ p.35 E9 (*) l. Γ θθ ψ = cosh ψ sinh ψ, Γ φφ = cosh ψ sinh ψ sin θ E0 (*) ψ Th non-zro Ricci rotation cofficints ar: Γ θ θ p.36 E (*) coth ψ Γ ψ =, Γ θ sin θ = Γ θ l. Ψ = r 4rm + 3 r 4 = cot θ sinh ψ θ = Γ ψ θ ψ = coth ψ, Γ = 0

11 E4 (*) (*) l.6 ds = ω ((dt + x dz) dx dy x dz ) l.8 T ab = ρ 0 0 x ω { } x 0 0 x Quiz and Exam Solutions p.330 l.3 (*) 7.b p.33 l.7 (*) 5.c l.9 (*) 3.a Bibliografi McMahon, D. (006). Rlativity Dmystifid. McGraw-Hill Companis, Inc.