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Solutions to exercises

Solutions to exercises

Exercise 1.1 A‘stationary’ particle in anylaboratory on theEarth is actually subject to gravitationalforcesdue to theEarth andthe Sun. Thesehelp to ensure that theparticle moveswith thelaboratory.Ifstepsweretaken to counterbalance theseforcessothatthe particle wasreally not subject to anynet force, then the rotation of theEarth andthe Earth’sorbital motionaround theSun would carry thelaboratory away from theparticle, causing theforce-free particle to followacurving path through thelaboratory.Thiswouldclearly show that the particle didnot have constantvelocity in the laboratory (i.e.constantspeed in a fixed direction) andhence that aframe fixed in the laboratory is not an inertial frame.More realistically,anexperimentperformed usingthe kind of long, freely suspendedpendulum known as a Foucaultpendulum couldreveal the fact that a frame fixed on theEarth is rotating andthereforecannot be an inertial . An even more practical demonstrationisprovidedbythe winds,which do not flowdirectly from areas of high pressure to areas of lowpressure because of theEarth’srotation. - Exercise 1.2 TheLorentzfactor is γ(V )=1/ 1−V2/c2. (a) If V =0.1c,then 1 γ = - =1.01 (to 3s.f.). 1 − (0.1c)2/c2

(b) If V =0.9c,then 1 γ = - =2.29 (to 3s.f.). 1 − (0.9c)2/c2 Notethatitisoften convenient to write speedsinterms of c instead of writingthe values in ms−1,because of thecancellation between factorsofc. ? @ AB Exercise 1.3 2 × 2 M = Theinverse of a matrix CDis ? @ 1 D −B M −1 = AD − BC −CA. Taking A = γ(V ), B = −γ(V )V/c, C = −γ(V)V/c and D = γ(V ),and noting that AD − BC =[γ(V)]2(1 − V 2/c2)=1,wehave ? @ γ(V )+γ(V)V/c [Λ]−1 = . +γ(V)V/cγ(V) This is thecorrect form of theinverse Lorentztransformation matrix. Exercise 1.4 First computethe Lorentzfactor: - γ(V )=1/ 1−V2/c2 - - =1/ 1−9/25 =1/ 16/25 =5/4. Thus themeasured lifetimeisΔT =5×2.2/4µs=2.8µs. Note that not all muons live for thesametime; rather,theyhavearangeoflifetimes.But alarge group of muons travelling with acommonspeed doeshaveawell-defined mean lifetime,and it is thedilationofthisquantity that is easily demonstrated experimentally. 279 Solutions to exercises

Exercise 1.5 Thealternative definitionoflengthcan’t be used in the rest frame of therod as the rod does not move in its ownrestframe.The proper lengthis therefore defined as beforeand related to the positions of thetwo events as observedinthe restframe.(This works, because 1 andevent 2 still occurat theend-pointsofthe rod andthe rod nevermovesinthe restframe S%.) As before,itishelpful to write down all the intervals that areknown in atable.

EventS(laboratory) S% (restframe) % % 2(t2,0) (t2,x2) % % 1(t1,0) (t1,x1) % % % % Intervals (t2 − t1, 0) (t2 − t1,x2 −x1) ≡(Δt, Δx) ≡ (Δt%, Δx%) Relation to intervals (L/V,0) (?,LP)

By examiningthe intervals,itcan be seen that Δx, Δt and Δx% areknown. From theintervaltransformation rules,onlyEquation1.33 relates thethree known intervals.Substitutingthe known intervals into that equationgives LP = γ(V )(0 − V (L/V )).Inthisway,lengthcontractionispredicted as before:

L = LP/γ(V ).

Exercise 1.6 Thereceivedwavelength is lessthanthe emitted wavelength. This meansthatthe jet is approaching. We can therefore useEquation1.42 providedthatwec-hange thesignofV.Combining it with theformula fλ = c showsthat λ% = λ (c − V )/(c + V ).Squaring bothsides andrearranginggives (λ%/λ)2 =(c−V)/(c+V). Fromthisitfollows that (λ%/λ)2(c + V )=(c−V), so V (1 +(λ%/λ)2)=c(1 − (λ%/λ)2), thus V = c(1 − (λ%/λ)2)/(1 +(λ%/λ)2). Substituting λ% =4483 × 10−10 mand λ =5850 × 10−10 m, thespeed is found to be v =0.26c (to 2s.f.). Exercise1.7 Letthe spacestation be theoriginofframe S, andthe nearer of the spacecraftthe originofframe S%,which therefore moveswith speed V = c/2 as measured in S. Letthese twoframesbeinstandard configuration. Thevelocity of thefurtherofthe twospacecraft, as observedinS,isthen v =(3c/4, 0, 0).It follows from thevelocity transformation that the velocity of thefurtherspacecraft % % as observedfrom thenearer will be v =(vx,0,0),where v − V 3c/4 − c/2 v% = x = =2c/5. x 2 2 1 − vxV/c 1 − (3c/4)(c/2)/c

Exercise 1.8 Δx =(5−7) m = −2 mand c Δt =(5−3) m =2m. Sincethe spacetimeseparation is (Δs)2 =(cΔt)2−(Δx)2 in this case, it follows that 280 Solutions to exercises

(Δs)2 =(2m)2−(2 m)2 =0.The value (Δs)2 =0is permitted;itdescribes situations in which thetwo events couldbelinkedbyalight signal. In fact, any such separationissaid to be light-like. Exercise 1.9 Startwith (Δs%)2 =(cΔt%)2−(Δx%)2.The aim is to show that (Δs%)2 =(Δs)2. Substitute Δx% = γ(Δx − V Δt) and c Δt% = γ(c Δt − V Δx/c) so that 1 ) (Δs%)2 = γ2 c2(Δt)2 − 2V ΔxΔt + V 2(Δx)2/c2 1 ) − γ2 (Δx)2 − 2V ΔxΔt + V 2(Δt)2 .

Crossterms involving Δx Δt cancel. Collectingcommonterms in c2(Δt)2 and (Δx)2 gives (Δs%)2 = γ2c2(Δt)2(1 − V 2/c2) − γ2(Δx)2(1 − V 2/c2).

Finally,notingthat γ2 =[1−V2/c2]−1,there is acancellation of terms, giving (Δs%)2 = c2(Δt)2 − (Δx)2 =(Δs)2, thus showingthat (Δs%)2 =(Δs)2. Exercise 1.10 Since (Δs)2 =(cΔt)2−(Δl)2,and (Δs)2 is invariant, it follows that all inertial observers will find (c Δt)2 =(Δs)2+(Δl)2,where (Δl)2 cannot be negative.Since (Δl)2 =0in theframe in which theproper timeis measured, it follows that no otherinertial observercan findasmaller valuefor the timebetween the events. Exercise 1.11 In Terra’s frame,Stella’sshiphas velocity (vx,vy,vz)=(−V,0,0).Itfollows from thevelocity transformation that % in Astra’s frame,the velocity of Stella’sshipwill be (vx, 0, 0),where % 2 vx =(vx−V)/(1 − vxV/c ).Taking vx = −V gives (−V − V ) −2V v% = = . x (1 − (−V )V/c2) 1+V2/c2 Taking themagnitude of this singlenon-zero velocity component givesthe speed of approach, 2V/(1 + V 2/c2),asrequired. Exercise 1.12 In Terra’s frame,the signals wouldhaveanemitted frequency fem =1Hz.InAstra’s frame,the Doppler effect tells us that thesignals wouldbe receivedwith adifferent frequency frec.Onthe outward legofthe journey, the signals wouldberedshifted andthe receivedfrequency wouldbe - frec = fem (c − V )/(c + V ). On thereturnleg of thejourney, thesignals wouldbeblueshiftedand thereceived frequency wouldbe - frec = fem (c + V )/(c − V ).

Exercise 2.1 TheLorentzfactor is - - - γ =1/ 1−v2/c2 =1/ 1−16c2/25c2 =1/ 9/25 =5/3.

281 Solutions to exercises

Theelectron hasmass m =9.11 × 10−31 kg. Thus themagnitude of theelectron’s momentum is p =5/3×4c/5×m =(5/3)×(4×3.00×108 ms−1/5)×9.11×10−31 kg =3.6×10−22 kg ms−1.

2 Exercise 2.2 Thekinetic energy is EK =(γ−1)mc .Taking the speed to be 9c/10,the Lorentzfactor is - - γ =1/ 1−v2/c2 =1/ 1−(9/10)2 =2.29.

Notingthat m =1.88 × 10−28 kg, thekinetic energy is −28 8 −1 2 −11 EK =(2.29 − 1) × 1.88 × 10 kg × (3.00 × 10 ms ) =2.2×10 J.

Exercise 2.3 v =3c/5 corresponds to aLorentzfactor - - γ(v)=1/ 1−v2/c2 =1/ 1−9/25 =5/4.

−27 Theproton hasmass mp =1.67 × 10 kg, therefore thetotal energy is E = γ(v)mc2 =(5/4)×1.67×10−27 kg×(3.00×108 ms−1)2 =1.88×10−10 J.

2 Exercise 2.4 Sincethe total energy is E = γmc ,itisclear that th-etotal energy is twice themassenergywhen γ =2.Thismeansthat 2=1/ 1−v2/c2. 2 2 2 2 Squaring andinverting boths√ides, 1/4=1−v /c ,sov /c =3/4.Taking the positive square root, v/c = 3/2. Exercise 2.5 (a) TheenergydifferenceisΔE=Δmc2,where Δm =3.08 × 10−28 kg. Thus ΔE =3.08 × 10−28 kg × (3.00 × 108 ms−1)2 =2.77 × 10−11 J. Convertingtoelectronvolts,thisis 2.77 × 10−11 J/1.60 × 10−19 JeV−1 =1.73 × 108 eV =173 MeV.

(b) From ΔE =Δmc2,the mass difference is Δm =ΔE/c2.Now, ΔE =13.6eV or,convertingtojoules, ΔE =13.6eV × 1.60 × 10−19 JeV−1 =2.18 × 10−18 J. Therefore Δm =2.18 × 10−18 J/(3.00 × 108 ms−1)2 =2.42 × 10−35 kg. Note that themassesofthe electron andproton are 9.11 × 10−31 kg and 1.67 × 10−27 kg, respectively,sothe mass difference from chemical binding is small enough to be negligible in most cases.However,mass–energy equivalence is not unique to nuclear reactions. % Exercise 2.6 Thetransformationsare E = γ(V )(E − Vpx√)and % 2 2 2 px = γ(V )(px − VE/c ).Inthiscase, E =3mec and px = 8mec .For relative-speed V =4c/5 between the twoframes, theLorentzfactor is γ =1/ 1−(4/5)2 =5/3.Substitutingthe values, √ % 2 2 E =5/3(3mec − 4c/5 × 8mec)=1.23mec

282 Solutions to exercises and √ % 2 2 p =5/3( 8mec − 4c/5 × 3mec /c )=0.714mec.

Exercise 2.7 (a) Foraphoton m =0,so 6.63 × 10−34 Js×5.00 × 1014 s−1 p=E/c = hf/c = =1.11×10−27 kg ms−1. 3.00 × 108 ms−1 (b) Usingthe Newtonian relationthatthe force is equaltothe rateofchange of momentum (weshall have more to sayabout this later), themagnitude of the forceonthe sail will be F = np,where n is therate at which photons are absorbedbythe sail (number of photons persecond). Thus n = F/p =10N/1.11 × 10−27 kg ms−1 =9.0×1027 s−1.

Exercise 2.8 To be avalid energy/momentum combination, the 2 2 2 2 4 energy–momentum relationmustbesatisfied, i.e. Ef − pf c = mf c .For the givenvalues of energy andmomentum, 2 2 2 2 4 2 4 2 4 2 4 Ef − pf c =9mfc −49mf c = −40mf c 4= mf c . So they arenot valid values. Exercise 2.9 It follows directly from thetransformation rules for thelastthree componentsofthe four-force F µ that 8 C % % 2 γ(v )fx = γ(V ) γ(v)fx − Vγ(v)f·v/c , % % γ(v )fy = γ(v)fy, % % γ(v )fz = γ(v)fz.

Note that thetransformation of fx involves boththe speed of theparticle v as measured in frame Sand thespeed V of frameS%as measured in frame S. Both γ(v) and γ(V ) appearinthe transformation. Exercise 2.10 Sincethe four-vector is contravariant, it transformsjustlikethe four-displacement. Thus % cρ = γ(V )(cρ − VJx/c), % Jx = γ(V )(Jx − V (cρ)/c), % Jy = Jy, % Jz = Jz, where V is the speed of frameS%as measured in frame S.

Thecovariant counterpart to (cρ,Jx,Jy,Jz)is (cρ, −Jx, −Jy, −Jz). Exercise 2.11 Thecomponentsofacontravariantfour-vector transform differently from thoseofacovariantfour-vector.The formertransform likethe µ componentsofadisplacement, according to thematrix [Λ ν] that implements the Lorentztransformation.The latter transform likederivatives, according to the −1 ν inverseofthe Lorentztransformation matrix, [(Λ )µ ].Since one matrix ‘undoes’the effect of theother in thesense th,at their product is theunitmatrix, it 3 µ is to be expected that combinations such as µ=0 JµJ will transform as 283 Solutions to exercises

, , 3 3 µ µ invariants,while othercombinations,suchas µ=0 JµJµ and µ=0 J J ,will not. Exercise 2.12 Theindices must balance. They do this in bothcases,but in the former casethe lowering of indices can be achievedbythe legitimate process of multiplyingbythe Minkowskimetric andsumming overacommonindex. In the latter caseanadditionalstep is required,the replacementofFµν by Fνµ.This wouldbeallowableif[Fνµ] wassymmetric —thatis, if Fµν = Fνµ for all values of µ and ν —but it is not.Making such an additionalchange will alter some of thesigns in an unacceptable way. Thegeneral lessonisclear:indices maybe raised andlowered in abalanced way, butthe order of indices is important and shouldbepreserved.Thisiswhy elements of themixed versionofthe field tensor µ ν µ maybewritten as F ν or Fµ butshouldnot be written as Fν . Exercise 2.13 Thefield component of interestisgiven by cF%10,soweneed to evaluate A %10 1 0 αβ F = Λ αΛ βF . α,β

1 0 Λ α is non-zero onlywhen α =0and α =1.Similarly, Λ β is non-zero only when β =0and β =1.Thismakes the summuchshorter, so it can be written out explicitly: %10 1 0 00 1 0 01 1 0 10 1 0 11 F =Λ 0Λ 0F +Λ 0Λ 1F +Λ 1Λ 0F +Λ 1Λ 1F . Since F00 =0and F11 =0,the sumreduces to %10 1 0 01 1 0 10 F =Λ 0Λ 1F +Λ 1Λ 0F . 10 01 It is nowamatter of substitutingknown values: F = −F = Ex/c, 0 1 0 1 Λ 0 =Λ 1=γ(V)and Λ 1 =Λ 0=−Vγ(V)/c,which leads to % 2 2 2 Ex/c = γ (1 − V /c )Ex/c. Since 1 − V 2/c2 = γ−2,wehave % Ex = Ex, as required. With patience, all theother field transformation rules can be determinedinthe same way. A3 % µ ν ρ η Exercise 2.14 Hαβγδ = Λα Λβ Λγ Λδ Hµνρη. µ,ν,ρ,η=0

Exercise 3.1 (a) Youcouldnotethat y/x =4/3for all values of u,and also u =0gives y = x =0,sothisisthe part of thestraight linewith positive u values andgradient 4/3 through theorigin. Or youcouldwork out x and y for a fewvalues of u,asshown in thetable below. u 01 23 x031227 y041636 284 Solutions to exercises

Either way, your sketch shouldlook likeFigure S3.1.

y 40 u=3

30

20

u=2

10

u =1

0 10 20 30 x

FigureS3.1 Sketch of theline x =3u2,y=4u2. (b) We have dx dy =6u and =8u, du du so * * 3 1 ) 3 8 C L = (6u)2 +(8u)2 1/2 du= 10u du = 5u2 3 =45. 0 0 0

Exercise 3.2 Since r = R and φ = u,wehave dr =0and dφ =du,so * 2π * 2π * 2π C= dl= (dr2 + r2 dφ2)1/2 = (02 + R2 du2)1/2 0 0 0 * 2π 2π = R du =[Ru]0 =2πR. 0

Exercise 3.3 (a) Like thecylinder, thecone can be formedbyrollingupa region of theplane. Once again this won’tchange thegeometry; thecircles and triangles will have thesameproperties as theyhaveonthe plane. So thecone has flatgeometry. (b) In this case, distances for thebugs areshortertowards theedge of thedisc, so theshortest distance from P to Q,asmeasured by thebugs,will appear to us to curveoutwards. Theangles of thetriangle PQR adduptomore than 180◦,as showninFigure 3.12, so for this inversehotplate the results arequalitatively similar to thegeometryofthe sphere, andthe hotplate again hasintrinsically curved geometrydespite thelack of anyextrinsic curvature. Exercise 3.4 FromEquation3.10,wehave dl2 = R2 dθ2 + R2 sin2 θ dφ2. 285 Solutions to exercises

Again thereare only squared coordinate differentials,sogij =0for i =4 j.We 2 2 2 1 can alsosee that g11 = R and g22 = R sin x ,so ? @ R2 0 [g ]= . ij 0 R2sin2 x1

Exercise 3.5 In this caseweonlyhavesquared coordinate differentials,so 1 2 1 2 2 2 gij =0for i =4 j.Also, g11 =1,g22 =(x ) ,g33 =(x ) sin x ,and therefore   10 0 12 [gij]=0(x) 0  . 00(x1)2sin2 x2 Note that thefinalentry involves thecoordinate x2, not x squared. Exercise 3.6 Defining x1 = r and x2 = φ,wehave ? @ 10 [g ]= . ij 0(x1)2

Exercise 3.7 (a) Sincethe lineelementisdl2=(dx1)2+(dx2)2,wehave ? @ 10 [g ]= . ij 01 FromEquation3.23,the connection coefficients aredefinedby ? @ 1 A ∂g ∂g ∂g Γi = gil lk + jl − jk , jk 2 ∂xj ∂xk ∂xl l k i andsince ∂gij/∂x =0for all values of i, j, k,itfollows that Γ jk =0for all i, j, k. Comment:Thisargumentgeneralizes to any n-dimensionalEuclidean space; consequently,whenCartesian coordinates areused, such spaces have vanishing connection coefficients. (b) From Exercise3.4, themetric is ? @ R2 0 [g ]= , ij 0 R2sin2 x1 andthe dua?lmetric is the inversem@ atrix 1/R2 0 [gij]= . 01/R2 sin2 x1 But in this case R =1,so ? @ 10 [gij]= . 01/sin2 x1

Since ? @ 1 A ∂g ∂g ∂g Γi = gil lk + jl − jk , jk 2 ∂xj ∂xk ∂xl l thereare sixindependent connection coefficients: 1 1 1 1 Γ 11, Γ 12(= Γ 21), Γ 22, 2 2 2 2 Γ 11, Γ 12(= Γ 21), Γ 22.

286 Solutions to exercises

However, ∂g ∂g 22 =2sin x1 cos x1, ij =0 ∂x1 while ∂xk for all othervalues of i, j, k.Also, gil =0for i =4 l,from which we can seethat ? @ 1 ∂g ∂g ∂g Γ1 = g11 11 + 11 − 11 =0, 11 2 ∂x1 ∂x1 ∂x1 ? @ 1 ∂g ∂g ∂g Γ1 = g11 12 + 11 − 12 =0, 12 2 ∂x1 ∂x2 ∂x1 ? @ 1 ∂g ∂g ∂g 1 ∂g Γ1 = g11 12 + 21 − 22 = − g11 22 , 22 2 ∂x2 ∂x2 ∂x1 2 ∂x1 ? @ 1 ∂g ∂g ∂g Γ2 = g22 21 + 12 − 11 =0, 11 2 ∂x1 ∂x1 ∂x2 ? @ 1 ∂g ∂g ∂g 1 ∂g Γ2 = g22 22 + 12 − 12 = g22 22 , 12 2 ∂x1 ∂x2 ∂x2 2 ∂x1 ? @ 1 ∂g ∂g ∂g Γ2 = g22 22 + 22 − 22 =0. 22 2 ∂x2 ∂x2 ∂x2 Consequently,the onlynon-zero values of thesix independent connection coefficients listed above are 1 ∂g 1 ∂g cos x1 Γ1 = − g11 22 = −sin x1 cos x1 Γ2 = g22 22 = =cot x1. 22 2 ∂x1 and 12 2 ∂x1 sin x1 2 2 (The onlyother non-zero connection coefficientisΓ 21 =Γ 12.) i Exercise3.8 From Exercise3.7(a), Γ jk =0for all i, j, k in this metric, so Equation3.27 reduces to d2xi =0, dλ2 i giving thesolutions x = aiλ + bi for constants ai,bi.Writingthisas x(λ)=aλ + b and y(λ)=cλ + d,wesee that theseequations parameterize the straight linethrough (b, d) with gradient c/a. Exercise 3.9 Using our usualcoordinates for thesurface of asphere, x1 = θ, x2 = φ,and theresults of Exercise3.7(b)for theconnection coefficients, Equation3.27 becomes ? @ d2θ dθ 2 − sin θ cos θ =0 dλ2 dλ (3.69) and d2φ cos θ dθ dφ +2 =0. dλ2 sin θ dλ dλ (3.70) (a)The portion of ameridian A can be parameterized by π θ(λ)=λ, 0 ≤ λ ≤ 2 , φ(λ)=0,

287 Solutions to exercises so we have dθ d2θ d2φ dφ =1, = = =0, dλ dλ2 dλ2 dλ sin θ =sin(λ), cos θ =cos(λ). Equation3.69 becomes 0 − sin(λ)cos(λ) × 0=0, andEquation3.70 becomes 0+2cot(λ) × 1 × 0=0. So A satisfiesthe geodesicequations andisageodesic. Comment:Thisiswhatwewouldexpect, because A is part of agreat circle. (b) B can be parameterized by π θ(λ)= 2, φ(λ)=λ, 0 ≤ λ<2π. So we have dφ d2φ d2θ dθ =1, = = =0, dλ dλ2 dλ2 dλ sin θ =1, cos θ =0. Equation3.69 becomes 0 − 1 × 0 × 1=0,and Equation3.70 becomes 0+2×0×1×0=0.SoBsatisfiesthe geodesicequations andisageodesic. (c) C can be parameterized by π θ(λ)= 4, φ(λ)=λ, 0 ≤ λ<2π. So we have dφ d2φ d2θ dθ =1, = = =0, 2 2 dλ dλ √ dλ dλ sin θ =cos θ = 2. √ √ Equation3.69 becomes 0 − 2 × 2 × 1=−24=0,and Equation3.70 becomes 0+2×1×0×1=0.SoCis not ageodesicbecause it doesn’t satisfy both geodesicequations. Exercise 3.10 (a) Since k is constantatevery pointonthe curveand k =1/R,wehave 1 1 R = = =5cm. k 0.2 cm−1 So thebestapproximating circle at everypoint on thecurveisacircle of radius 5 cm,and thecurveitself is acircle of radius 5 cm. (b) Here again k will be constant, as thestraight linehas constant‘curvature’. Howeverbig we drawthe circle, alargercircle will approximate the straight linebetter,sothe curvatureofastraight linemustbesmaller than 1/R for all possible R.Hence k must be zero. In otherwords, 1 k =lim =0. R→∞ R 288 Solutions to exercises

Exercise 3.11 Theparabolacan be parameterized by x(λ)=λand y(λ)=λ2. Consequently, x˙ =1, x¨=0, y˙=2λ, y¨ =2, andfor λ =0we have x˙ =1, x¨=0, y˙=0, y¨=2. So thecurvatureatλ=0is |x˙y¨ − y˙x¨| |1×2−0×0| k = = =2, (˙x2+˙y2)3/2 (12 +02)3/2 andthe approximating circle hasthe radius 1 1 R = = . k 2 Thecentreofthe circle is at x =0,y=0.5. Exercise 3.12 Thederivativesofxand y aregiven by x˙ = −a sin λ, x¨ = −a cos λ, y˙ = b cos λ, y¨ = −b sin λ, so thecurvatureisgiven by |x˙y¨ − y˙x¨| ab sin2 λ + ab cos2 λ ab k = = = . (˙x2+˙y2)3/2 (a2 sin2 λ + b2 cos2 λ)3/2 (a2 sin2 λ + b2 cos2 λ)3/2 Forthe circle of radius R we have a = R and b = R,so ab R2 1 k= = = , (a2 sin2 λ + b2 cos2 λ)3/2 (R2 sin2 λ + R2 cos2 λ)3/2 R which is as expected. Exercise 3.13 Interchanging the j, k indices in Equation3.35,weget ∂Γl ∂Γl A A Rl = ij − ik + Γm Γl − Γm Γl . ikj ∂xk ∂xj ij mk ik mj m m Swapping thefirstand second terms, andthe thirdand fourth terms, leadsto ∂Γl ∂Γl A A Rl = − ik + ij − Γm Γl + Γm Γl . ikj ∂xj ∂xk ik mj ij mk m m Comparison with Equation3.35 showsthatthe expression on theright-hand side l l l of this equationis−Rijk,hence proving that R ijk = −R ikj. Exercise 3.14 FromExercise3.7(a), all connection coefficients for this space arezero, andhence from Equation3.35,wehave l R ijk =0. Sincethe connection coefficients alsovanishfor an n-dimensionalEuclidean space, it follows that theRiemanntensoriszerofor such spaces. Exercise 3.15 FromEquation3.35 andExercise3.7(b), we have ∂Γ1 ∂Γ1 A A R1 = 22 − 21 + Γλ Γ1 − Γλ Γ1 212 ∂x1 ∂x2 22 λ1 21 λ2 λ λ ∂Γ1 ∂Γ1 = 22 − 21 +Γ1 Γ1 +Γ2 Γ1 − Γ1 Γ1 − Γ2 Γ1 . ∂x1 ∂x2 22 11 22 21 21 12 21 22 289 Solutions to exercises

But from Exercise3.7(b), 1 1 1 2 2 Γ 11 =Γ 12 =Γ 21 =Γ 11 =Γ 22 =0, so ∂Γ1 R1 = 22 − Γ2 Γ1 212 ∂x1 21 22 ∂ cos x1 = (−sin x1 cos x1) − (− sin x1 cos x1) ∂x1 sin x1 = − cos2(x1)+sin2(x1)+cos2(x1) =sin2 x1.

Exercise 3.16 Fromthe earlier in-text question, we knowthat K = a−2,and from Exercise3.15, 1 2 1 R 212 =sin x . However, from Exercise3.7(b), ? @ a2 0 [g ]= , ij 0 a2sin2 x1 so 4 2 1 g =det[gij]=a sin x . 1 Also,from Chapter2we knowthatlowering the first indexonR 212 gives A2 i 1 2 R1212 = g1iR 212 = g11R 212 + g12R 212. i=1 However, g12 =0,hence R a2 × sin2 x1 1 1212 = = , g a4 sin2 x1 a2 which is thesameasK. Exercise 3.17 (a) Just as in Exercise3.7(a), theconnection coefficients are zerosince themetric is constant. (b) Sincethe connection coefficients for aMinkowskispacetimeare zero, as showninpart(a),and each terminthe RiemanntensordefinedbyEquation3.35 involves at leastone connection coefficient, it follows that all componentsofthe Riemanntensorare zero. Exercise 3.18 (a) Themetric is ? @ c2 0 [g ]= ij 0 −f2(t) andthe dua?lmetric is @ 1/c2 0 [gij]= . 0 −1/f 2(t) As in Exercise3.7(b), thereare only sixindependent connection coefficients: 0 0 0 0 Γ 00, Γ 01(= Γ 10), Γ 11, 1 1 1 1 Γ 00, Γ 01(= Γ 10), Γ 11.

290 Solutions to exercises

Moreover, ∂g df(t) 11 = −2ff,˙ f˙ ≡ , ∂x0 where dt and ∂g ij =0 ∂xk for all othervalues of i, j, k.Also, gil =0for i =4 l,from which we can seethat ? @ 1 ∂g ∂g ∂g Γ0 = g00 00 + 00 − 00 =0, 00 2 ∂x0 ∂x0 ∂x0 ? @ 1 ∂g ∂g ∂g Γ0 = g00 01 + 00 − 01 =0, 01 2 ∂x0 ∂x1 ∂x0 ? @ 1 ∂g ∂g ∂g 1 ∂g Γ0 = g00 01 + 10 − 11 = − g00 11 , 11 2 ∂x1 ∂x1 ∂x0 2 ∂x0 ? @ 1 ∂g ∂g ∂g Γ1 = g11 10 + 01 − 00 =0, 00 2 ∂x0 ∂x0 ∂x1 ? @ 1 ∂g ∂g ∂g 1 ∂g Γ1 = g11 11 + 01 − 01 = g11 11 , 01 2 ∂x0 ∂x1 ∂x1 2 ∂x0 ? @ 1 ∂g ∂g ∂g Γ1 = g11 11 + 11 − 11 =0. 11 2 ∂x1 ∂x1 ∂x1 Consequently,the onlynon-zero values of thesix independent connection coefficients listed above are 1 ∂g 1 1 ff˙ Γ0 = − g00 11 = − × × (−2ff˙)= 11 2 ∂x0 2 c2 c2 and 1 ∂g 1 −1 f˙ Γ1 = g11 11 = × × (−2ff˙)= . 01 2 ∂x0 2 f2 f 1 1 Theonlyother non-zero connection coefficientisΓ 10 =Γ 01. (b) As in Exercise3.15, ∂Γ0 ∂Γ0 A A R0 = 11 − 10 + Γλ Γ0 − Γλ Γ0 101 ∂x0 ∂x1 11 λ0 10 λ1 λ λ ∂Γ0 ∂Γ0 = 11 − 10 +Γ0 Γ0 +Γ1 Γ0 − Γ0 Γ0 − Γ1 Γ0 . ∂x0 ∂x1 11 00 11 10 10 01 10 11 0 0 0 1 1 Since Γ 00 =Γ 01 =Γ 10 =Γ 00 =Γ 11 =0,wehave ' < ∂Γ0 ∂ ff˙ f˙ ff˙ R0 = 11 − Γ1 Γ0 = − × 101 ∂x0 10 11 ∂x0 c2 f c2 1 ∂ % : f˙2 1 % : f˙2 = ff˙ − = f˙f˙ + ff¨ − c2 ∂t c2 c2 c2 ff¨ = . c2

291 Solutions to exercises

Exercise 4.1 (a) Supposethatthe separationislandthe distance from the centreofthe EarthisR,asshown in Figure S4.1. Then the magnitude of thehorizontal acceleration of each object is g sin θ ≈ gθ, so thetotal (relative)acceleration is g2θ.However, 2θ = l/R,sothe magnitude of thetotal acceleration, a,isgiven by gl 9.81 × 2.00 a = = −2 =3.08 × 10−6 −2. R 6.38 × 106 ms ms

(b) Suppose that one object is adistance l vertically above theother object. Since Newtonian is an inversesquare law, themagnitudesofacceleration at R and R + l arerelated by ? @ g(R) (R + l)2 l 2 2l = = 1+ ≈1+ . g(R + l) R2 R R Hence Δg,the difference between themagnitudesofacceleration at R and R + l, is givenby 2gl 2 × 9.81 × 2.00 Δg= = −2 =6.15 × 10−6 −2. R 6.38 × 106 ms ms

Exercise 4.2 (a) As indicated by Figure S4.2,the coordinates arerelated by x = r cos θ, y = r sin θ. Setting (x%1,x%2)=(x, y) and (x1,x2)=(r, θ),wehave ∂x%1 ∂x ∂x%1 ∂x = =cos θ, = = −r sin θ ∂x1 ∂r ∂x2 ∂θ and ∂x%2 ∂y ∂x%2 ∂y = =sin θ, = = r cos θ. ∂x1 ∂r ∂x2 ∂θ In this case, thegeneral tensor transformation lawreduces to A ∂x%1 A ∂x%2 A%1 = Aν, A%2 = Aν. ∂xν and ∂xν ν ν This meansthat A%µ and Aµ must be related by A%1 =cos θA1−rsin θA2, and A%2 =sin θA1+rcos θA2.

(b) In thecaseofthe infinitesimal displacement, this general transformation rule implies that dx =cos θ dr − r sin θ dθ, and dy =sin θ dr + r cos θ dθ. But this is exactly the relationshipbetween thesedifferent sets of coordinates givenbythe chain rule, so theinfinitesimal displacementdoestransform as a contravariantrank 1 tensor.

292 Solutions to exercises

l g sin θ g sin θ

g cos θ g cos θ y

r R R

θ θ θ x

FigureS4.2 Polar coordinates. FigureS4.1 Accelerationsofhorizontally separated masses in afreely fallinglift.

Exercise 4.3 We knowthat A3 α Aµ = gµα A . α=0 Multiplyingbygνµ andsumming over µ,wehave A3 A3 A3 νµ νµ α g Aµ = g gµα A . µ=0 µ=0 α=0 Reversingthe order in which we do thesummationonthe right-hand side of this equationenables us to write it as A3 A3 A3 νµ α νµ g Aµ = A g gµα. µ=0 α=0 µ=0 However, A3 νµ ν g gµα = δ α. µ=0 ν ν Since δ α =1when ν = α and δ α =0when ν 4= α,wehave A3 νµ ν g Aµ = A . µ=0

Exercise 4.4 (a) Thereare tworeasons.The µ indexisuponAµbutdown on Bµ.The K termhas no µ index. (b) The ν indexcannot be up on both Y µν and Zν ;itmustbeuponone termand down on theother. (c)There cannot be threeinstances of the ν indexonthe right-hand side of this equation.

293 Solutions to exercises

Exercise 4.5 Beingascalar,thisquantity hasnocontravariantorcovariant indices.Sointhisparticular case, covariantdifferentiationsimplygives ∂S ∇ S = . λ ∂xλ Exercise 4.6 Weknowthat  1000 0−100 [η ]=[ηµν ]=  µν 00−10 000−1 and ? @ dx1 dx2 dx3 [U µ]=γ(v)(c, v)=γ(v) c, , , . dt dt dt Since U 0 = c in theinstantaneous restframe,wehave T 00 = ρc2.Also, T 0i =0 since η0i =0and U i =0in this frame.Likewise, ! p B T ii = ρ + U iU i + p = p. c2 Finally,for i 4= j, ! p B T ij = ρ + U iU j − pηij =0 c2 since ηij =0for i =4 j and U i =0in theinstantaneous restframe. Exercise 4.7 MultiplyingEquation4.34 by gµν andsumming overboth indices,weobtain A A A µν 1 µν µν g Rµν − 2 Rgµν g = −κg Tµν. µ,ν µ,ν µ,ν Nowusing thefact that A A µν ν g gµν = δ ν =4, µ,ν ν this becomes R − 2R = −κT. Hence R = κT ,which we can substitute in Equation4.34 to obtain Equation4.35: 1 Rµν − 2 κT gµν = −κTµν , so 1 ) 1 Rµν = −κ Tµν − 2 gµν T .

Exercise 5.1 Fromthe definitionofthe , 1 G00 = R00 − 2 g00R andwehave ? @ 2A% R = −e2(A−B) A%% +(A%)2−A%B%+ , 00 r 2A g00 =e

294 Solutions to exercises and ? @ 2 1 2 R = −2e−2B A%% +(A%)2−A%B%+ (A%−B%)+ + . r r2 r2 So G = R − 1 g R 00 00 2 00? @ 2A% = −e2(A−B) A%% +(A%)2−A%B%+ r ? @ 2 1 e2A +e2(A−B) A%% +(A%)2−A%B%+ (A%−B%)+ − r r2 r2 ? @ 2B% 1 e2A =−e2(A−B) − − , r r2 r2 as required. Exercise 5.2 (a) Theonlyplace wherethe coordinate φ appearsinthe 2 2 2 % Schwarzschild line elementisinthe term r sin θ (dφ) .But since φ = φ + φ0, thedifferenceinthe φ-coordinates of anytwo events will be equaltothe difference in the φ%-coordinates of thoseevents, andinthe limit, for infinitesimally % separated events, dφ =d(φ+φ0)=dφ.Sothe Schwarzschild line elementis unaffected by thechange of coordinates apartfrom thereplacementofφby φ%. This establishesthe form-invariance of themetric underthe change of coordinates. (b) In asystem of spherical coordinates,agivenvalue of thecoordinate φ corresponds to ameridian of thekindshown in Figure S5.1.

θ r

φ

FigureS5.1 Radial coordinates with a(meridian)lineofconstant φ. Thereplacementofφby φ% effectively shifts everysuchmeridian by thesame angle φ0.Since thebody that determinesthe is spherically symmetric, the displacementofthe meridianswill have no physical significance. Moreover, sinceeach meridian is replaced by another, all that really happens in this caseisthateach meridian is relabelled,and this will not even change theform of themetric. Exercise 5.3 We require dτ ≤ 1 − 10−8. dt 295 Solutions to exercises

With dr =dθ=dφ=0themetric reduces to ? @ dτ 2GM 1/2 GM GM = 1 − ≈ 1 − , ≤ 10−8. dt c2r c2r so c2r Rearranging gives GM r ≥ =1.5×1011 . c2 × 10−8 metres We have not yetfound therelationship between theSchwarzschild coordinate r andphysical (proper) distance —thatisthe subject of thenextsection. Nonethelessitisinteresting to notethataproper distance of 1.5 × 1011 metres is about thedistance from theEarth to the Sun. Exercise 5.4 Theproper distance dσ between twoneighbouring events that happenatthe same time(dt=0)isgiven by themetric viathe relationship (ds)2 = −(dσ)2.Thus (dr)2 (dσ)2 = + r2(dθ)2 + r2 sin2 θ (dφ)2. 1 − 2GM c2r Forthe circumference at agiven r-coordinate in the θ = π/2 plane, dr =dθ=0, hence (dσ)2 = r2(dφ)2.

So * 2π dσ = r dφ andtherefore C = r dφ =2πr, 0 as required. Exercise 5.5 It follows from thegeneral equationfor an affinely parameterized geodesicthat d2x0 A dxν dxρ + Γ0 =0. dλ2 νρ dλ dλ ν,ρ Sincethe onlynon-zero connection coefficients with araised index 0 are 0 0 Γ 01 =Γ 10,the summay be expandedtogive d2x0 dx0 dx1 +2Γ0 =0. dλ2 01 dλ dλ 0 1 0 GM Identifying x = ct, x = r and Γ 01 = ! B ,wesee that r2c2 1− 2GM c2r d2t 2GM dr dt + 1 ) =0, dλ2 c2r2 1 − 2GM dλ dλ c2r as required. Exercise 5.6 Forcircular motionatagiven r-coordinate in theequatorial plane, u is constant, so du d2u dr = =0 =0. dφ dφ2 andalso dτ 296 Solutions to exercises

(a) It followsfrom theorbital shapeequation(Equation5.36)thatfor acircular orbit with J 2/m2 =12G2M2/c2, ? @ 3GMu2 12G2M 2 −1 − u + GM =0, c2 c2 that is 3GMu2 c2 − u + =0. c2 12GM Solvingthisquadratic equationinugives u = c2/6GM,sor=6GM/c2 is the minimumradius of astable circular orbit. (b) Thecorresponding valueofEmaybedeterminedfrom theradial motion equation(Equation5.32), remembering that dr/dτ =0: ? @ ? @ '? @ < dr 2 J2 2GM 2GM E 2 + 1− − = c2 − 1 . dτ m2r2 c2r r mc2

So ? @ ? @ 12G2M 2 c2 2 2GM c2 c2 0+ 1 − − 2GM c2 6GM c2 6GM 6GM '? @ < E 2 = c2 − 1 . mc2

Simplifying this,wehave ? @ '? @ < c2 2 c2 E 2 1 − − = c2 − 1 3 6 3 mc2 that is '? @ < c2 E 2 − = c2 − 1 , 9 mc2 √ which can be rearrangedtogive E = 8mc2/3.

Exercise 6.1 (a) Forthe Sun, RS =3km.Sofor ablack holewith threetimes theSun’smass, theSchwarzschild radius is 9 km.Substitutingthisvalue into Equation6.10,wefind that theproper timerequired for thefall is just 3 8 −5 τfall =6×10 /(3 × 10 ) s =2×10 s.

9 (b) Fora10 M) galactic-centreblack hole, the Schwarzschild radius andthe in-fall timeare both greater by afactor of 109/3.Acalculationsimilar to that in part (a) therefore givesafreefall timeof6700 s, or about 112 minutes.(Note that theseresults applytoabody that starts its fall from faraway, not from the horizon.) Exercise 6.2 According to Equation6.12,for events on theworld-lineofan outward radially travelling photon, dr = c(1 − R /r). dt S 297 Solutions to exercises

Forastationarylocal observer, i.e. an observeratrestatr,wesaw in Chapter 5 that intervals of proper timeare related to intervals of coordinate timeby 1/2 dτ=dt(1 − RS/r) ,while intervals of proper distance arerelated to intervals −1/2 of coordinate distance by dσ =dr(1 − RS/r) .Itfollows that thespeed of light as measured by alocal observer, irrespective of their location, will always be dσ dr 1 = . dτ dt 1 − RS/r So,inthe casethatthe intervals beingreferredtoare thosebetween events on the world-lineofaradially travelling photon, we seethatthe locally observedspeed of thephoton is dσ 1 = c(1 − RS/r) = c. dτ 1 − RS/r

Exercise 6.3 According to thereciprocal of Equation6.17,for events on the world-lineofafreely fallingbody, ? @ dr 1 − R /r r − r 1/2 = −cR1/2 S 0 . S 1/2 dt (1 − RS/r0) rr0 We alreadyknowfrom thepreviousexercisethatfor astationary local observer, dσ dr 1 = . dτ dt 1 − RS/r So,inthe caseofafreely fallingbody,the measured inward radial velocity will be ? @ ? @ dσ 1 − R /r r − r 1/2 1 1 r − r 1/2 = −cR1/2 S 0 = −cR1/2 0 S 1/2 S 1/2 dτ (1 − RS/r0) rr0 1 − RS/r (1 − RS/r0) rr0 ? @ R r − r 1/2 = −c S × 0 . (r0 − RS) r

In thelimit as r → RS,the locally observedspeed is givenby|dσ/dτ|→c. Exercise 6.4 Initially,the fall wouldlook fairly normal with theastronaut apparently gettingsmaller andpickingupspeed as the distance from theobserver increased.Atfirstthe frequencyofthe astronaut’s waveswouldalsolook normal, though detailed measurements wouldreveal asmall decreasedue to theDoppler effect. As thedistance increased,the astronaut’s speed of fall wouldcontinue to increaseand thefrequency of waving woulddecrease. This wouldbe accompanied by asimilar change in thefrequency of light receivedfrom the fallingastronaut, so theastronautwouldappear to become redder as well as more distant. Thereddening wouldbeincreaseddue to gravitationalredshift, though theastronaut’smotionwouldcontinue to contribute. As theastronautapproached theevent horizon, theeffect of spacetimedistortionwouldbecome dominant. The astronaut’s rate of fall wouldbeseen to decrease, butthe imagewouldbecome very redand wouldrapidly dim, causing thedeparting astronaut to fade away. Though somethingalong theselines is theexpected answer, thereisanother effect to takeintoaccount,thatdepends on themassofthe . This is a consequenceoftidal forces andwill be discussedinthe next section. 298 Solutions to exercises

Exercise 6.5 Theincreasing narrowness andgradual tipping of thelightcones as theyapproachthe eventhorizon indicates the difficulty of outward escapefor photons and, by implication, for anyparticles that travel slower than light. This effect reaches acritical stageatthe eventhorizon, wherethe outgoing edge of the lightcone becomesvertical, indicating that even photons emitted in the outward direction areunabletomakeprogress in that direction. Adiagrammatic study of lightconesalone is unabletoprove theimpossibility of escapefrom within the eventhorizon, butthe progressive narrowingand tipping of lightconesinthat region is at leastsuggestiveofthe impossibility of escape, anditisindeed a fact that all affinely parameterized geodesics that enter the eventhorizon of a non-rotating black holereach the central singularity at some finite valueofthe affineparameter. Exercise 6.6 Thetime-likegeodesicfor theSchwarzschild casehas already been giveninFigure 6.11. Thenatureofthe lightconesisalsorepresented in that figure, so theexpected answerisshown in Figure S6.1a. In thecaseof Eddington–Finkelstein coordinates,Figure 6.13 playsasimilar role, suggesting (rather than showing) theform of thetime-likegeodesicand indicating the form of thelightcones. Theexpected answerisshown in Figure S6.1b.

ct ct" orizon orizon ty ty th th en en ev ev singulari singulari

0 RS r 0 RS r (a) (b)

FigureS6.1 Lightconesalong atime-likegeodesicin(a) Schwarzschild and (b) advanced Eddington–Finkelstein coordinates.

2 2 Exercise 6.7 (a) When J = GM /c,wehave a = J/Mc = GM/c = RS/2. InsertingthisintoEquations 6.32 and6.33,the second termvanishesand we find 299 Solutions to exercises

r± = RS/2.

(b) When J =0,wehave a =0andweobtain r+ = RS, r− =0. In both cases (a)and (b),there is onlyone eventhorizon as the inner horizon vanishes. Exercise 6.8 (a) Thepath indicated by thedashedlineinFigure 6.20 shows no change in angleasitapproachesthe static limit. Space outside thestatic limit is alsodragged around, even though rotationisnolongercompulsory.However,a particle in freefall must be affected by this dragging, andsoaparticle in freefall couldnot fall in on thedashedline. Thepath of free fall wouldhavetocurvein thedirectionofrotation of theblack hole. (b) It is possibletofollowthe dashed path,but thespacecraftwouldhavetoexert thrust to counteract the effects of thespacetimecurvatureofthe rotatingblack holethatmakethe pathsoffreefall have adecreasing angular coordinate. (c)The dotted path representsanimpossibletripfor thespacecraft. Inside the ergosphere, no amount of thrust in theanticlockwise directioncan make the spacecraftmaintain aconstantangular coordinate while decreasing the radial coordinate. Exercise 6.9 Thediscovery of aminiblack holewouldimply (contrarytomost expectations) that conditions during theBig Bang were such as to lead to the productionofminiblack holes.Thiswouldbeanimportant developmentfor . Such adiscovery wouldalsoopenupthe possibility of confirmingthe existence of Hawkingradiation,thus giving some experimental support to attemptstoweld together quantum theory andgeneral relativity,suchasstringtheory. Exercise 7.1 We first need to decide howmanydaysmakeupacentury. This is not entirely straightforward because leap yearsdon’tsimplyoccurevery 4 years in the Gregoriancalendar.However,itisthe Julian year that is used in astronomy andthisisdefinedsothatone year is precisely 365.25 days.Consequently we have 36 525 days percentury, which we denotebyd.Ifweuse T to denotethe period of theorbitin(Julian)days, then thenumberoforbits percenturyis d/T .Equation7.1 givesthe angleinradians,but it is more usualtoexpressthe observations in seconds of arcsoweneed to usethe fact that π radiansequals 180 × 3600 seconds of arc. Puttingall this together,wefind that thegeneral relativistic contributiontothe mean rate of precession of theperihelioninseconds of arcper centuryisgiven by dφ d 6πGM 648 000 dGM = × ) × = ) × 3888 000 dt T a(1 − e2)c2 π seconds of arc Ta(1 − e2)c2 seconds of arc 36 525 × 6.673 × 10−11 × 1.989 × 1030 × 3888 000 = 87.969 × 5.791 × 1010 × (1 − (0.2067)2) × (2.998 × 108)2 seconds of arcper century =42%%.99 percentury.

Exercise 7.2 Forraysjustgrazing the Sun, b is theradius of theSun, which is 8 30 R) =6.96 × 10 m, and M is M) =1.989 × 10 kg. Hencethe deflection in

300 Solutions to exercises seconds of arcisgiven by 4GM 648 000 6.674 × 10−11 × 1.989 × 1030 2592 000 Δθ = ) × = × c2b π seconds of arc (2.998 × 108)2 × 6.96 × 108 π seconds of arc =1%%.75.

Exercise 7.3 (a)Let R⊕ =6371.0 km be themean radius of theEarth, 24 M⊕ =5.9736 × 10 kg be themassofthe Earth, and h =20200 km be the height of thesatellite above theEarth. FromEquation5.14,the intervalatR⊕andthe coordinate timeintervalatR⊕+harerelated by  −1/2 2M⊕G Δt 1 − 2 R⊕+h =  c (R⊕+h)  . Δt 2M⊕G R⊕ 1 − 2 c R⊕ Sincethe timedilationissmall, we can usethe first fewterms of a 2 Taylor expansiontoevaluate this.Putting 2M⊕G/c (R⊕ + h)=xand 2 −1/2 1/2 2M⊕G/c R⊕ = y,the right-hand side above becomes (1 − x) × (1 − y) . x y x y By aTaylor expansion, this is approximately (1 + 2 )(1 − 2 ) ≈ 1+ 2 − 2.Sowe have ? @ M G M G M Gh Δt ≈ 1+ ⊕ − ⊕ Δt =Δt − ⊕ Δt . R⊕+h 2 2 R⊕ R⊕ 2 R⊕ c (R⊕+h) c R⊕ c R⊕(R⊕ + h) Thediscrepancy over 24 hours is givenby 5.9736 × 1024 × 6.673 × 10−11 × 2.02 × 107 Δt −Δt =− × 24 × 3600 R⊕+h R⊕ (2.998 × 108)2 × 6.371 × 106 × (6.371 +20.2) × 106 s = −45.7 µs. Thenegative sign indicates that the effect of is that thesatellite clock runsmore rapidlythanaground-basedone. (b) relates atimeinterval Δt for aclock moving at speed v with thetimeinterval Δt0 for oneatrestby ? @ v2 −1/2 Δt= 1− Δt . c2 0 Forasatellite of mass m orbitingthe Earthatadistance h from theEarth’s surface, its speed v is givenby GM m mv2 GM ⊕ = v2 = ⊕ 2 therefore (R⊕ + h) R⊕ + h R⊕ + h andhence ? @ ? @ GM −1/2 GM Δt = 1 − ⊕ Δt ≈ 1+ ⊕ Δt . 2 0 2 0 c (R⊕ + h) 2c (R⊕ + h) Hencethe discrepancyover 24 hours between satellite- andground-basedclocks is GM 6.673 × 10−11 × 5.9736 × 1024 Δt − Δt ≈ ⊕ Δt = × 24 × 3600 0 2 0 8 2 6 s 2c (R⊕ + h) 2 × (2.998 × 10 ) × (6.371 +20.2) × 10 =7.2µs. 301 Solutions to exercises

Thepositive result indicates that the effect of special relativity is that thesatellite clock runsslowerthanaground-basedone. (c) Thetotal effect of theresults obtained in parts (a) and(b) is adiscrepancy between ground-basedand satellite-basedclocksof(−45.7+7.2) = −38.5 µs perday.Since thebasis of theGPS is theaccurate timingofradio pulses, over 24 hours this couldlead to an error in distance of up to 8 −6 c(Δt − Δt0)=2.998 × 10 × 38.5 × 10 m =11.5km.

Exercise 7.4 We can approximate the radius of thesatellite’sorbitbythe Earth’sradius.Hence the period of theorbit, T ,isgiven by = R3 T =2π ⊕ . GM⊕ Since GM ⊕ ≈ 10−9 ( 1, 2 c R⊕ Equation7.13 can be approximated by 7 ? @F 3GM GM α ≈ 2π 1 − 1 − ⊕ ≈ 3π ⊕ . 2 2 2c R⊕ c R⊕ After atime Y ,the numberoforbits is Y/T andthe total precession is givenby ? @ = Y GM Y GM 1/2 GM 3Y G3M 3 α = × 3π ⊕ = ⊕ × 3π ⊕ = ⊕ . total 2 3 2 2 5 T c R⊕ 2π R⊕ c R⊕ 2c R⊕ Convertingfrom radianstoseconds of arc, we findthatthe total precession angle for one year is = 3 × 365.25 × 24 × 3600 (6.673 × 10−11)3 × (5.974 × 1024)3 180 × 3600 α = × × =8%%.44. total 2 × (2.998 × 108)2 (6.371 × 106)5 π

Exercise 7.5 We have previously carried out asimilar calculationfor low Earthorbit, theonlydifferenceherebeing that the radius of theorbitisnow R =(6.371 × 106 m)+(642 × 103 m) instead of 6.371 × 106 m. Consequently, theexpected precession is ? @ 6.371 5/2 8%%.44 × =6%%.64. 7.013

Exercise 7.6 When consideringlight rays travelling from adistantsource to a detector, it is not just one ray that travels from thesource to thedetector,but a cone of rays.Gravitationallensing effectively increases thesize of thecone of rays that reach the detector. Thelight is not concentrated in the same wayasin Figure 7.15, butitisconcentrated. Exercise 8.1 (i)Onsize scales significantly greater than 100 Mly,the large-scale structure of voidsand superclusters(i.e. clustersofclustersof galaxies)doesindeed appear to be homogeneous andisotropic. 302 Solutions to exercises

(ii)After removing distortions due to local motions,the mean intensity of the cosmicmicrowavebackground radiation differsbylessthanone part in ten thousandindifferent directions. This tooisevidenceofisotropyand homogeneity. (iii) Theuniformity of themotionofgalaxies on large scales,known as the Hubbleflow,isathirdpiece of evidence in favour of ahomogeneous andisotropic . Exercise 8.2 Geodesics arefound usingthe geodesicequation. Thefirststep is to identifythe covariantmetric coefficients of therelevant space-likehypersurface (only g11, g22 and g33 will be non-zero).The contravariantform of themetric coefficients will followimmediately from therequirementthat [gij] is thematrix inverseof[gij].The covariantand contravariantcomponentscan then be used to i determine theconnection coefficients Γ jk.Oncethe connection coefficients for thehypersurface have been determined, thespatial geodesics maybefound by solvingthe geodesicequationfor thehypersurface. At that stageitwould be sufficient to demonstrate that aparameterized path of theform r = r(λ), θ = constant, φ = constantdoesindeed satisfy thegeodesicequationfor the hypersurface. Exercise 8.3 TheMinkowskimetric differsinthatitdoesnot feature thescale factor R(t).Itistrue that k =0for both cases,and this meansthatspace is flat. Butthe presence of thescale factor in the Robertson–Walker metric allows spacetimetobenon-flat. Exercise 8.4 We startwith theenergyequation ? @ 1 dR 2 8πG kc2 = ρ − , R2 dt 3 R2 (Eqn 8.27) anddifferentiate it with respect to time t.Weuse theproductruleonthe left-hand side andobtain ? @ ? @ ? @ ? @ ? @ dR 2 d 1 1 d dR 2 8πG dρ d 1 + = − kc2 . dt dt R2 R2 dt dt 3 dt dt R2 1 ) d dR d We then usethe chain ruletoreplace dt with dt dR ,which gives ? @ ? @ ? @ ? @ ? @ ? @ ? @ ? @ dR 2 dR d 1 2 dR d dR 8πG dρ dR d 1 + = −kc2 . dt dt dR R2 R2 dt dt dt 3 dt dt dR R2 Then carrying outthe variousdifferentiations with respect to R and t,weget ? @ ? @ ? @? @ ? @ ? @ 2 dR 2 dR 2 dR d2R 8πG dρ 2kc2 dR − + = + . R3 dt dt R2 dt dt2 3 dt R3 dt 1 ) 1 dR 2 We then substitute back in for R2 dt in thefirsttermonthe left-hand side, usingthe energy equationagain,toget ? @? @ ? @? @ ? @ ? @ 2 dR 8πGρ kc2 2 dR d2R 8πG dρ 2kc2 dR − − + = + . R dt 3 R2 R2 dt dt2 3 dt R3 dt ! B 1 d2R We nowsubstitute for R dt2 in thesecond termonthe left-hand side,using the accelerationequation(Equation8.28), to get ? @? @ ? @7 ? @F ? @ ? @ 2 dR 8πGρ kc2 2 dR 4πG 3p 8πG dρ 2kc2 dR − − + − ρ + = + . R dt 3 R2 R dt 3 c2 3 dt R3 dt 303 Solutions to exercises

1 ) 1 dR Nowwecollect all terms with R dt as acommonfactor,toget ? @ ? @7 F 8πG dρ 1 dR 2kc2 16πGρ 2kc2 8πGρ 8πGp + + − + + =0. 3 dt R dt R2 3 R2 3 c2 2 2 8πG Theterms in 2kc /R cancel out,and dividing through by 3 gives ? @ ? @7 F dρ 1 dR 3p + 2ρ+ρ+ =0, dt R dt c2 which clearly yields the fluid equationasrequired: dρ ! p B 3 dR + ρ + =0. dt c2 R dt (Eqn 8.31)

Exercise 8.5 Thedensity andpressure terminthe originalversion of the second of theFriedmann equations (Equation8.28)may be written as 3p 3 ρ + = ρ + ρ + ρ + (p + p + p ) . c2 m r Λ c2 m r Λ Thedarkenergydensity termisconstant(ρΛ), andthe otherdensity terms maybe written as 7 F 7 F R 3 R 4 ρ = ρ 0 ,ρ=ρ 0 . m m,0 R(t) r r,0R(t) Thepressure duetomatter is assumedtobezero(i.e. dust),the pressure dueto 2 2 radiation is pr = ρr c /3,and thepressure duetodarkenergyispΛ=−ρΛ/c . Puttingall this together,wehave 7 F 7 F ? @ 3p R 3 R 4 3 ρc2 ρ ρ + = ρ 0 + ρ 0 + ρ + 0+ r − Λ c2 m,0 R(t) r,0 R(t) Λ c2 3 c2 7 F 7 F 6 7 F 9 R 3 R 4 3 ρ c2 R 4 ρ =ρ 0 +ρ 0 +ρ + r,0 0 − Λ m,0 R(t) r,0 R(t) Λ c2 3 R(t) c2 7 F 7 F R 3 R 4 =ρ 0 +2ρ 0 −2ρ , m,0 R(t) r,0 R(t) Λ as required.

Exercise 8.6 (a) Substitutingthe proposed solutionintothe differential equation, we have # d ! B 8πG R2 R (2H t)1/2 = ρ 0 . 0 0 r,0 1/2 dt 3 R0(2H0t) Evaluatingthe derivative,weget # 1 8πG R R (2H )1/2 = ρ 0 . 0 0 1/2 r,0 1/2 1/2 2t 3 (2H0) t 1/2 Cancelling the factor R0/t on bothsides andcollectingterms in H0,thisyields # 8πG H = ρ , 0 3 r,0 as required.

(b) Usingthe definitionofthe Hubbleparameter, 1 dR H(t)= , R dt 304 Solutions to exercises we substitute in for R(t) from theproposedsolutiontoget ? @ ? @ 1 d ! B 1 R (2H )1/2 1 H(t)= R (2H t)1/2 = 0 0 = , 1/2 0 0 1/2 1/2 R0(2H0t) dt R0(2H0t) 2t 2t as required.Hence H0 =1/2t0,and substitutingthisintothe proposed solution gives ? @1/2 1/2 2t0 R(t0)=R0(2H0t0) = R0 = R0, 2t0 again as required.

Exercise 8.7 Setting dR/dt =0and ρm,0 =0in thefirstFriedmann equation implies that ' 7 F < 8πG R 3 kc2 0= ρ 0 + ρ − . 3 m,0 R(t) Λ R2

ρ ρ But we alreadyknowfrom Equation8.50 that Λ and m,0 must have thesame sign in this case. Consequently, k must be positive andhence equalto+1.Using Equation8.50,and thefirstFriedmann equationatt=t0,wecan therefore write 7 F 2 8πG 3ρm,0 c = 2 , 3 2 R0 leadingimmediately to the required result ? @ c2 1/2 R0 = . 4πGρm,0 Insertingvalues for G and c,along with thequoted approximate valuefor the 26 current mean cosmicdensity of matter,gives R0 =1.8×10 m. Since 15 1 ly =9.46 × 10 m, it follows that, in round figures, R0 =20000 Mly in this static model. Recallingthataparsec is 3.26 light-years,wecan alsosay,roughly speaking,thatinthe Einstein model, for thegiven matter density, R0 is about 6000 Mpc. Exercise 8.8 Theconditionfor an expanding FRWmodeltobeaccelerating at t 1 d2R time 0 is that R dt2 shouldbepositive at that time. We alreadyknowfrom Equation8.50 that theconditionfor it to vanish is that Ω Ω = m,0 . Λ,0 2 Examiningthe equation, it is clear that theconditionthatwenow seek is Ω Ω ≥ m,0 . Λ,0 2

Exercise 8.9 In the ΩΛ,0–Ωm,0 plane, the dividing linebetween the k =+1 and k = −1 models corresponds to theconditionfor k =0.Thisisthe condition 2 that thedensity shouldhavethe critical value ρc(t)=3H (t)/8πG,and maybe expressedinterms of ΩΛ,0 and Ωm,0 as

Ωm,0 +ΩΛ,0 =1.

305 Solutions to exercises

(i) ThedeSitter modelisatthe point Ωm,0 =0,ΩΛ,0=1.

(ii) TheEinstein–de Sitter modelisatthe point Ωm,0 =1,ΩΛ,0=0.

(iii) TheEinstein modelhas alocation that depends on thevalue of Ωm,0,sointhe ΩΛ,0–Ωm,0 planeitisrepresented by theline ΩΛ,0 =Ωm,0/2,which coincides with thedividinglinebetween accelerating anddecelerating models.

Exercise 8.10 Thescale change R(tob)/R(tem) showsupinextragalactic measurements because thelight hasbeen ‘intransit’ for along timeas space hasexpanded. To measure changesinR(t)locally requires our measuring equipmenttobeinfreefall, farfrom anynon-gravitationalforcesthatwouldmask the effects of general relativity.However,the large aggregates of matter within our galaxydistort spacetimelocally andcreate agravitationalredshiftthatwould almostcertainly mask theeffects of cosmicexpansiononthe wavelengthoflight. Nearby starssimplywill not participate in the cosmicexpansiondue to theselocal effects.Thus alocal measurementwouldnot be expected to reveal the changing —any more than asurvey of theirregularities on your kitchen floor wouldreveal the curvatureofthe Earth. Exercise 8.11 Thefigure of 5 billionlight-years relates to the proper distances of sourcesatthe timeofemission. Forsourcesatredshifts of 2 or 3,asinthe case of Figure 8.2, thecurrentproper distances of thesourcesare between about 16 and 25 billionlight-years.The distances quoted in Figure 8.2 indicate that, in a field such as relativistic cosmology wherethere aremanydifferent kindsof distance, thereisaproblemofconvertingmeasured quantities such as into ‘deduced’quantities such as distances.Whensuchdeduced quantities are used,itisalwaysnecessary to provide clear information about their precise meaning if theyare to be properlyinterpreted. Exercise 8.12 Historically,the discovery of theFriedmann–Robertson–Walker models wasarather tortuous process. Einstein initiated relativistic cosmology with his1917 proposal of astatic cosmological model. Einstein’s modelfeatured apositively curved space (k =+1)and used the repulsiveeffect of apositive Λ to balance thegravitationaleffect of ahomogeneous distributionofmatter of density ρm.Later in the same year, introduced the first modelofanexpanding Universe,effectively introducingthe scale factor R(t),though he didnot presenthis modelinthatway.DeSitter’s modelincludedflat space (k =0), andacosmological constantbut no matter,so therewas nothing to opposeacontinuously acceleratingexpansionofspace. In 1922,AlexanderFriedmann, amathematician from St Petersburg, publisheda general analysis of cosmological models with k =+1and k =0,showing that the models of Einstein anddeSitter were special cases of abroadfamily of models. He publishedasimilar analysisofk=−1models in 1924. Together,these two publications introduced all the basicfeatures of theRobertson–Walker buttheywerebased on some specificassumptions that detracted from their appeal. In 1927 Lemaˆıtreintroducedamodelthatwas supportedbyEddington, in which expansioncouldstartfrom apre-existingEinstein model. Lemaˆıtrelater (1933)proposedamodelthatwouldbecategorized nowadays as avariant of theory andhebecameinterested in models that started from R =0. By 1936 Robertson and Walker hadcompleted their essentially mathematical investigations of homogeneous relativistic ,givingFriedmann’sideas a

306 Solutions to exercises more rigorousbasis andassociating their nameswith themetric. This setthe scenefor thenaming of theFriedmann–Robertson–Walker models.(Sometimes they arereferredtoasLemaˆıtre–Friedmann–Robertson–Walker models.)

307