Problemsets-General RelativitySolutionsbySergeiWinitzkiJanuary11,2007Contents1 Coordinatesand1-forms 21.1 Invertibletransformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Examplesofcoordinatetransformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.3 Basisintangent space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.4 Dierentialsoffunctionsas1-forms. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.5 Basisincotangent space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.6 Linearlyindependent1-forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.7 Transformationlawfor1-forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.8 Examplesoftransformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42 Tensors 72.1 Denitionoftensorproduct . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.3 Exampleoftensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.5 Contractionoftensorindices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.8 Examplesofspaceswithametric. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83 TheChristoelsymbol 93.1 Transformations 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93.2 Transformations 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103.3 Covariant derivatives. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103.4 TheLeibnitzrule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103.5 Locallyinertialreferenceframe . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 Geodesicsandcurvature 114.1 Geodesics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114.1.1 Firstderivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114.1.2 Secondderivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124.2 Commutatorofcovariant derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124.3 Paralleltransport. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134.4 Riemanntensor. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134.5 Lorentztransformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145 Gravitationtheoryapplied 145.1 Redshift . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145.2 Energy-momentum tensor1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155.3 Energy-momentum tensor2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155.4 Weakgravity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165.5 Equationsofmotionfromconservationlaw . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166 Thegravitational eld 176.1 Degreesoffreedom. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176.2 Sphericallysymmetricspacetime . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176.2.1 Straightforward solution. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176.2.2 Solutionusingconformaltransformation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206.3 Equationsofmotion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227 Weakgravitational elds 227.1 Gravitationalbendingoflight . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227.2 Einsteintensorforweakeld . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237.3 GravitationalperturbationsI . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247.4 GravitationalperturbationsII. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2418 Gravitational radiationI 268.1 Gaugeinvariant variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 268.2 Detectinggravitational waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 268.2.1 Usingdistancesbetweenparticles. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 268.2.2 Usinggeodesicdeviationequation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278.3 Poissonequation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278.4 Metricperturbations1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 288.5 Metricperturbations2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 289 Gravitational radiationII 299.1 Projectionofthemattertensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 299.2 Mattersources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 309.3 Energy-momentum tensorofgravitational waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 309.4 Powerofemittedradiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3210Derivation: gravitational wavesinatspacetime 33Agoodtextbookcorrespondingtothelevelofthiscourse: General Relativity: AnIntroductionforPhysicists byM.P.Hobson,G.P.Efstathiou,andA.N.Lasenby(CambridgeUniversityPress,2006).1 Coordinatesand1-forms1.1 InvertibletransformationsThe inverse function theorem guarantees that the equations = (x) are solvable near a point x0 if det_(x)/x_,=0atx0. Underthiscondition, thecoordinatetransformationisinvertibleatx0. Note: weareinvertingnotjustonefunction = (x),but we are determining x from a system of n equations, say (x) = C,where Care n given values.1.2 Examplesofcoordinatetransformations1. a) Sincex = u_1 +v2_+u3/3,itisclear thatxhasrange (, +) foranyxedvalueofvas uvariesintherange(, +). Similarly, yhastherange(, +). Toverifythatthecoordinatesystem(x, y)coverstheentireplane,itissucienttoshowthatxhasthefull rangeateveryxedvalueofy. Itissucienttoconsidery0> 0(elsechangev v). Atxedy=y0>0, wehavey0=v + vu2+ v3/3andthustheadmissiblevaluesofuarefrom to+,whiletheadmissiblevaluesofvarefrom0tov= vmaxsuchthaty0= vmax +13v3max. Thenu = y0v 13v3v(wehavey0v 13v3v 0 for0 < v < vmax),x = _1 +v2+y0v 13v33v_y0v 13v3v= _23+89v2+y03v_y0v 13v3v.Wehavenowexpressedxasafunctionof v, i.e. x=x(v). Whenvvariesfrom0tovmax, x(v)variesfrom to0.Sincex(v)isnonsingular forv > 0,itfollowsthatxhasthefullrange. Therefore, thecoordinates (x, y)cover theentiretwo-dimensional plane.b)Thecoordinatetransformation isnonsingular ifdet (x, y) (u, v) ,= 0.Compute:det_xuyuxvyv_ = det_1 +u2+v22uv2uv 1 +u2+v2_(1)= 1 + 2_u2+v2_+_u2v2_2> 0.Thereforetherearenosingularpoints.2. a) To determine the range, rst consider = 0. Then x = r sinh , y= 0, z= r cosh . It is clear that z2x2= r2.Sincer 0, thecoordinates(x, y, z)coveronlythedomain [z[> [x[. Witharbitrary, itisclearthatthecoordinates(x, y, z)coverthedomain [z[ >_x2+y2.2b)Computethedeterminant:det__sinh cos r cosh cos r sinh sin sinh sin r cosh sin r sinh cos cosh r sinh 0__ = r2sinh .Thecoordinates aresingularifr = 0or = 0.c)Thesingularityatr=0isduetothefactthattheset r = 0, , correspondstoasinglepointx=y=z=0.Thisissimilartothesingularityof thespherical coordinatesatr=0. Pointsalongthecone [z[ =_x2+y2arenotcoveredbecausetheycorrespondto , r 0. Thesingularityat=0, r ,=0isduetothefactthatthesetr, = 0, corresponds tothepointx = y= 0,z= ratxedr ,= 0. Thisissimilartothepolarcoordinate singularity.3. a) To determine the range, note that r sin 0 for the given range of and r. However, this is immaterial since thefactorscos andsin willmakex, ycoverthefullrange(, +). Thecoordinates(x, y, z)areaslightmodicationofthestandardsphericalcoordinates. Thesecoordinates coverthewholespace(x, y, z).b)Computethedeterminant:det__sin cos r cos cos r sin sin sin sin r cos sin r sin cos cos r sin 0__ = r2sin .Thisisnonzerounlessr = 0or = 0.c)Thesingularitiesarecompletelyanalogous tothoseinthesphericalcoordinates.1.3 BasisintangentspaceSuppose that the vectors e=xare linearly dependent, then there exist constants c, not all zero, such that the vectorceequalszero. Actwiththisvectoronthecoordinatefunctionx1:cex1= cxx1= c1.Byassumption,ce= 0,thereforec1= 0. Itfollowsthateverycequalszero,contradictingtheassumption.1.4 Dierentialsoffunctionsas1-formsd(x) = dx, d_x2_ = 2xdx,d (xy) = xdy +ydx, d (x +y) = dx +dy,d_4x2y +x3z_=_8xy + 3x2z_dx + 4x2dy +x3dz, d_3_x2+y2_ = 3xdx +ydy_x2+y2.Nowletuscomputedhbyrstndingd (arctan(x y)) =dx dy1 + (x y)2,d_arctan2xx2y21_ =11 +4x2(x2y21)2_2dxx2y21 4x(xdx ydy)(x2y21)2_=2_x2+y2+ 1_dx + 4xydy(x2y21)2+ 4x2.Addingthesetogetherandnotingthat_1 + (x +y)2__1 + (x y)2_ =_x2y21_2+ 4x2,wegetdh(x, y) = d_arctan(x + y) + arctan(x y) + arctan2xx2y21_ = 0.Thismeansthath(x, y)isaconstant. Byusingthetangentsumrule,wecaneasilyshowthath(x, y) = 0.1.5 BasisincotangentspaceNotethattherelation_dx,x_ = 3isthedenitionof howthe1-form dxacts onvectors /x. Now,it isclear that any 1-form isdecomposed as alinearcombinationof the1-formsdx1, ..., dxn. Itremainstoshowthatall theseformsarelinearlyindependent. If thiswerenotso,therewouldexistalinearcombinationcdx= 0suchthatnotallc= 0. Actwiththisonavector/x1andobtain0 =_0,x1_ =_cdx,x1_ = c1.Thereforec1= 0. Similarly,wendthateveryotherc= 0,whichcontradictstheassumption.1.6 Linearlyindependent1-forms1. Two1-formsd(excos y),d(exsiny)arelinearlyindependentforeveryx, ybecaused (excos y) = excos ydx exsinydy,d (exsin y) = exsinydx +excos ydy,andthefollowingdeterminantisalwaysnonzero,det_excos y exsinyexsiny excos y_ = e2x,= 0.2. Two1-forms(1 +y)dx 2xydy,8dxarelinearlyindependentifthefollowingdeterminantisnonzero,det_1 +y 2xy8 0_ = 16xy.Thishappensforxy ,= 0.3. Three1-formsdx +dy,dx +dz,dy +dzarealwayslinearlyindependent.4. Three1-formsdx dy,dy dz,dz dxarealwayslinearlydependent(theirsumiszero).1.7 Transformationlawfor1-formsThetransformation lawfor1-forms,d x= xxdx,underacoordinatetransformation x x= x(x),ismerelyadierentinterpretation ofthedenitionofthe1-formd x(seeProblem 1.4),where x(x)isconsideredascalarfunctioninthecoordinates_x_.1.8 Examplesoftransformationsa)Firstcomputedxanddy:dx =_1 +u2+v2_du + (2uv) dv,dy= (2uv) du +_1 +u2+v2_dv.Thenitiseasytocomputexdx +ydy,etc. Forinstance,d1x +y= dx +dy(x +y)2= (du +dv)_1 + (u +v)2_(u +v)2_1 +13 (u +v)2_2.b)ThecomponenttransformationmatrixisgiveninEq.(1).1ATangentplaneIf the tangent plane is at angle with thehorizontal, then theacceleration is g sin (from elementary mechanics). Since0 1. Nowweseethat(x, y)isobtainedfrom_2 sin u_1bymultiplyingwithcos vandsin v. Therefore,thefullsurfaceisarotationsurface,whereweneedtousethexcoordinateastheradius. Therefore,thegureinthe(x, z)planeneedstoberotated around thezaxis. Theresultingsurface isatorus. It maybedescribed bytheequation__x2+y22_2+z21 = 0.Notethat_x2+y2=12 sin u> 0.Also_x2+y22 =1 2_2 sinu_2 sin u=2 sin u 12 sin u.2. Since thesurface is now given by an equation of theform F(x, y, z) = 0,thenormal vector (upto a constant factorC)canbefoundas(nx, ny, nz) = C_Fx, Fy , Fz_ = C_2_x2+y22_x2+y2x, 2_x2+y22_x2+y2y, 2z_.MultiplyingbyC 12_x2+y2(thisfactorischosenforsimplicity),wehave(nx, ny, nz) =_x__x2+y22_, y__x2+y22_, z_x2+y2_.5Expressed throughthecoordinates (u, v),thisbecomesn= (nx, ny, nz) =_cos v2 sin u 1_2 sinu_2, sin v2 sinu 1_2 sin u_2,cos u_2 sin u_2_.Theequationofthetangentplaneatpointx0isn_xx(0)_ = 0,wherenmustbecomputedatx = x0.3. Wecomputexu=cos v cos u_2 sin u_2,xv= sin v2 sin u,yu=sin v cos u_2 sin u_2,yv=cos v2 sin u,zu=1 2 sin u_2 sin u_2,zv= 0.Nowwecanexpanddx =xudu +xvdv, etc.Thereforetheinducedmetricisg = dx2+dy2+dz2=___cos u_2 sinu_2_2+_1 2 sin u_2 sinu_2_2__du2+dv2_2 sin u_2=du2+dv2_2 sin u_2.ThevectorVa= (cos v, sin v)isnotaunitvectorbecauseg(V, V ) =cos2v + sin2v_2 sin u_2=1_2 sin u_2 ,= 1.TheCartesiancomponentsofthevectors/u,/varefoundfromu=x(u, v)ux+y(u, v)uy+z(u, v)uz,v=x(u, v)vx+y(u, v)vy+z(u, v)vz.Therefore,thevectorVahasthefollowingCartesiancomponents,V= cos v_sin v2 sin ux+cos v2 sin uy_+ sin v_cos v cos u_2 sin u_2x+sin v cos u_2 sin u_2y+1 2 sin u_2 sin u_2z_= sinv cos v_sin u + cos u 2_2 sinu_2_x+_sin2v cos u +2 cos2v sin u cos2v_2 sin u_2_y+ sin v1 2 sin u_2 sin u_2z.ThisvectoriswithinthetangentplanebecausenV= 0,nV= cos v2 sin u 1_2 sinu_2sin v cos v_sinu + cos u 2_2 sin u_2_+ sin v2 sinu 1_2 sin u_2_sin2v cos u +2 cos2v 2 sin u cos2v_2 sin u_2_+cos u_2 sin u_2sinv1 2 sin u_2 sin u_2(aftersimplication) = 0.62 Tensors2.1 Denitionoftensorproducta),b)Adirectcalculationusingthepropertydxi, /xk_ = ikgives:1, v1) =_dx + 2ydy, 3x_ = 3, etc.Theresults:T(v1, v1) = 0, T(v1, v2) = 6x.c) d) First, show that the sum of two linear functions is again a linear function: If A(x) and B(x) are linear functions,i.e.ifA(x +y) = A(x) +A(y)and likewise for B, then A+Bobviously has the same property. Now, since a tensor is dened as a multi-linear function,itisclearthattensorsformavectorspace.2.3 Exampleoftensora) An obvious example of such Tis the vector product, T(u, v) = uv, dened in three-dimensional space. To determinetherankofT,weneedtorepresentTasamultilinearnumber-valuedfunctionofsomenumberof1-formsandvectors,e.g. A(f1, ..., fr, v1, ..., vs). ItisclearthatT(v1, v2)itselfisnotsuchafunctionbecauseithasvector valuesinsteadofscalar(number)values. Soweneedtoadda1-form tothelistofarguments. WecandeneA(f1, v1, v2) = f1, T(v1, v2))andthenitsclearthatAismultilinear. ThereforeTisatensorofrank1+2.b)Thecalculationmaygoasfollows. WeneedtodeterminethecomponentsTsuchthat[T(u, v)]= Tuv.SowerewritethegivendenitionofT(u, v)intheindexnotation,e.g. likethis:[T(u, v)]= 2uvunv.Nowwewouldliketomoveuvoutof thebracketsandsodetermineT. However, theexpressionabovecontainsuvinsteadof uv. Therefore werenametheindextoandalsointroduceaKronecker symbol ,so astorewriteidenticallyunv= uvn.Therefore[T(u, v)]= 2uvuvn=_2 n_uv,T= 2 n.2.5 Contractionoftensorindicesa) Using the denition of a tensor as a multilinear function, it is easy to show that linear combinations of tensors are alsomultilinearfunctions. Tensorproductsandcontractionsarealsomultilinear. Theargumentsaremuchsimplerthantheproofoftensortransformation lawforcomponents.b) Contractingtwolower indices, e.g. T, gives components of aquantitywhichis not atensor becausethesecomponentsdonottransform correctlyunderchangesofbasis. IfTwereatensoritwouldtransform asT=x xT.However,thisdoesnotagreewiththecontractionofthetensorT,whichtransforms asT=x xx xx xT.Thecontraction over = yields

T=

x xx xx xT ,=x x T.c)CalculationgivesT= 2 n= 3nbecause= 0and= 3.72.8 Examplesofspaceswithametrica)Weperformthecalculationincomponents,g(u, v) = uv unvn.Wewould like to write g(u, v) = guv,where gare thecomponents of themetric tensor. Using the known identityforthe-symbol,= ,wendg(u, v) = uv() unvn= uvuvnn+unvn.Denoten2 nn g(n, n),andthenweneedtorelabelindicessuchthatvucanbemovedoutsidetheparentheses.Theresult isg(u, v) =_ n2 +nn_vu.Thereforeg= n2 +nn.Toanalyzetheconditionsunderwhichdet g ,=0, wecanchooseanorthonormalbasissuchthatnisparalleltotherstbasisvector. Thenthecomponentsofthevectorninthisbasisare([n[ , 0, 0)andthematrixghasthefollowingsimpleform:g=__1 0 00 1 n200 0 1 n2__.Thenitisclearthatdet g=_1 n2_2. Therefore,thematrixgisnondegenerate ifn2,= 1.b)Similarcalculationsgiveg=_3 +n200 3_.Thereforethedeterminantofgisalways nonzero.c)Considerationsareanalogous tob),exceptthatthesizeofthematrixgislarger.d)Themetricis(dx)2+ (dy)2+ (dz)2,andweneedtoexpressdx,dy,dzthroughduanddv. Acalculationgivesg = dx2+dy2+dz2= R2_sinh2u + cosh2u_du2+R2cosh2u dv2.2AVectorequationsa)TheequationcontainstwogivenvectorsAandB. ThesolutionxcanbefoundasalinearcombinationofA,B,andthecrossproductABwithunknowncoecients. Usingvectornotation,wehavex = A+B+ (AB) .Substitutingthisexpression intothegivenequation,kx +x A = B,andusingtheknownidentity(AB) C = B(A C) A(B C) , (2)wendA(k A B) +B(k + A A1) + (A B) (k ) = 0.Onpurpose, wewritethisequationintheformoflinearcombinationofthethreevectorsA, B, andA B. Sinceweareconsideringthegenericcase,thesethreevectorsareindependentandsoeachofthecoecientsabovemustbezero:k A B = 0, k +A A1 = 0, k = 0.Solvingthissystemofequations,wend(assumingk ,= 0inthegenericcase) =A Bk, =1 A Ak, =1 A Ak2.b)Wehaveinvectornotationx A = B, xC = k.Multiply C:(x A) C = BC.8Simplifyusingtheidentity(2),A(xC) x(A C) = kAx(A C) = BC.Thereforex =kABCA C.c), d)TheequationshavetheformxM=A, whereMisamatrixandAisaknownvector. Thesolutionisx = M1A,whereM1istheinversematrix(itexistsinthegenericcase).2BTensorequationsa)Sincethevectors AandBareabasisintwo-dimensional space(inthegenericcasetheyarelinearlyindependent),thenthesymmetrictensorXcanbewrittenasX= fAA+g_AB+AA_+hBB,wherethecoecientsf, g, hareunknown. Itremainstodeterminethesecoecients. UsingAB=0anddenotingAA [A[2,wegetthesystemofequationsXA= B (f+g) [A[2= 1,XB= 0 (g +h) [B[2= 0,X= 0 f [A[2+h[B[2= 0.Theresult isX=1[A[2[B[2[B[2[A[2AA+1[B[2[A[2_BBABBA.b)AnantisymmetrictensorXinthreedimensionscanbealways expressedasX= u,whereuisanunknownvector thatweneedtodetermine. Wecannowrewrite theconditionsonXinavector form,XA= B u A = BXB= 0 u B = 0It follows that u is parallel to B and then the condition uA = B leaves the only solution u = 0, and therefore X= 0istheonlyadmissiblesolution.2CDegeneracyofthemetrica)Themetriccanbewritteninthebasis dx, dyasthematrixg=_y21 +x21 +x20_.Thedeterminant ofthismatrixis _1 +x2_2whichisalwaysnonzero.b) In the basis where Ais parallel to the rst basis vector, the vector Ahas components (A, 0, 0, 0, ...) and thereforethemetricghastheform_____r20 0 ...0 1 00 0 1......_____.Themetricisdegenerateifr = 0(i.e.attheorigin).3 TheChristoel symbol3.1 Transformations1WeareconsideringaatspacewhereEuclideancoordinatesexist. Suppose xand xaretwocoordinatesystems,while isthestandardEuclideancoordinatesystem. TheChristoelsymbolsaredenedas=2xxx ,=2 x x x .9The relationship between and can be found as follows. Assuming that the functions x( x) and also x(x) are known,wemayexpressthepartialderivativeoperators usingthechainrule,x= xx x, x=x xx.Alsowecanexpress x= xxx .Thereforewecancalculate(whenisknown)asfollows,= xxx_x xx__x xx_= xxx_x xx x2xx+2x x xx_= xxx xx x +2x x x xxxx= xxx xx x +2x x x xx. (3)NotethattheEuclideancoordinate systemisnotneededtodeterminethetransformationof.3.2 Transformations2Consideravectoreldu. Assumethat uobeysthecorrect transformation lawforrank(1,1)tensors,u=_ u_ x x xx ,andsubstituteuxu+ u, u x u+ u.Wecannowexpressthrough. Notethat xxx x= becausethematricesare x/xandx/ xareinversetoeachother. Theresultis= xx xxx x+2x x x xx xx.3.3 CovariantderivativesTheruleisthateveryupperindexgetsa+andeverylowerindexgetsa . Eachtermwithreplacesoneindexintheoriginaltensorbyoneoftheindicesin. ThereforewecanwritetheanswerasT;= T + T + TTTT3.4 TheLeibnitzrulePerform anexplicitcalculation,A;B+AB;= B_AA_+A_B+ B_;_AB_;= _AB_AB+ AB.Thisprovestherequiredproperty.103.5 Locallyinertial referenceframeInthisproblem (unlikeproblem 3.1)thecoordinatesystem isnotaatEuclideancoordinatesystem,butitisjustacoordinatesystemwhichislikeEuclideanatonepoint. Nowwewanttousetheformula(3), whichwill enableustocomputetheChristoel symbolinthecoordinatesystem, giventheChristoel symbol intheoriginalcoordinatesystem x. Tousethatformula,weneedtocomputesomederivatives. Denoting x ,wend xx= +_xx(0)_(0)= +_xx(0)_(0).The inverse derivative, x/ x,can befound by inverting thismatrix; the result can befound simply by assuming thatx x= +_xx(0)_A+O_(x x(0))2_,whereAisanunknownmatrix. Souptoquadratictermswendx x= _xx(0)_(0) +O_(x x(0))2_.Therefore2x x x= (0)+O_x x(0)_.Finally,wend= xxx xx x +2x x x xx= (0)+O_x x(0)_.Atx = x0wehave= (0). ItfollowsthatthenewChristoelsymbolisequaltozeroatx = x0.Alternatively, onecanusethetransformationlawfortheChristoel symbol intheinversedirection, =... + ...,i.e. one denotes x , x x. This has the advantage that only derivatives 2/xx need to be computed,andnotthederivatives2x/. Sinceall derivativesonlyneedtobeevaluatedatx=x0, therst-orderderivativesx/atx=x0canbefoundastheinversematrixto/x=, i.e.x/=. Thisconsiderablysimpliesthecalculations.4 Geodesicsandcurvature4.1 Geodesics(a) Notethat d/dsis theordinary(not covariant)derivativeinthedirectionof u. Thegeodesicequationcanberewrittenforthe1-formuasuu;= 0 =dudsuu.Anexplicitformulaforyieldsuu= uu=12 (g, +g,g,) uu.Notethatg, g,isantisymmetricin( ). Thereforethesetermswill cancel afteracontractionwithuu.Theremainingtermyieldsuu=12g,uu.b)Wegivetwoderivations;therstoneisdirectandthesecondoneusestheproperty(a).4.1.1 FirstderivationNotethatdds_guu_ = u_guu_,wherewemust useanordinary derivative insteadof thecovariant derivative (according tothedenitionofd/ds). Sowendu_guu_,= g,uuu+ 2guuu,.Nowweneedtosimplifyanexpressioncontainingu,. Byassumption,thederivativeofthevectoreldusatisesddsu+ uu= uu, + uu= 0= uu, +12g(g,+g,g,) uu.11Thereforeuuu,= 12ug(g,+g,g,) uu= 12u(g,+g,g,) uu= 12g,uuu.Whatremainsisastraightforward computation:u_guu_,= g,uuu+ 2guuu,= g,uuug,uuu= 0.4.1.2 SecondderivationWewritedds_guu_ =dds_guu_ = ug,uu + 2gu,uu.Nowweneedtoexpressu,u du/ds. Todothat,weusethepropertyderivedin(a),duds=12g,uu,andnddds_guu_ =dgdsuu +gg,uuu=dgdsuu +dgdsuu.Itremainstoexpressthederivativeofgthroughthederivativeofg. Weusetheidentitydds_gg_ =dds__ = 0,thusdgdsg= dgdsgdgds= dgdsgg.Thereforedgdsuu= dgdsgguu= dgdsuu,andthusdds_guu_ = 0.4.2 CommutatorofcovariantderivativesFirstcomputeu;= u, + u,u;=_u, + u_,+ _u, + u__u, + u_= u, + ,u+ u,+ u, + uu;.Sincewewanttocomputethecommutatoru; u;, wecanomitthetermsthataresymmetricin( ). Thesetermsarethefollowing:u,, u, + u,, u;.Theremainingtermsareu;= ,u+ u+ (symmetricin ) ,whichyieldsu;u;= u_, , + _ = uR.124.3 Parallel transportTheparallel-transported vectorcanberepresentedbya1-form A(s)suchthatdAdsgAu= 0, udxds.However,theclosedcurveisassumedtocoveronlyaverysmall neighborhoodofonepointx0, sowecanapproximateAbyaconstant, A(x0),alongthecurve. ThereforeA=_dAdsds =_(x)A(x)dx A(x0)_(x)dx.Now,inalocallyinertialsystematx0wehave(x0) = 0. ThereforewecanTaylorexpand(x)nearx0as(x) =_xx0_, +O_(x x0)2_.ThereforeA A(x0)_,_xx0_dx A(x0),(x0)_xdx,wherewehaveagainapproximated ,(x)byitsvalueatx = x0,andalsousedtheidentity_dx= 0. Further,_d_xx_ = 0 =_xdx+_xdx.ThereforewemayrewriteA A,_xdx=12A_,,__xdx=12AR_xdx.4.4 Riemanntensora)Itismoreconvenient toconsiderthefullycovariant tensorR. Thistensorhasthefollowingsymmetries,R= R= R, (4)R +R +R= 0, (5)R= R. (6)However,itisknownthattheproperty(6)followsfrom(4)-(5),thereforeitissucienttoconsiderthesetwoproperties[notethat(5)doesnotfollowfrom(4), (6)]. Letusrstconsidertheproperty(4). Forxed, , wehavethatRisanantisymmetricn nmatrix(indices, ). Thismatrixhas12n(n 1)independentcomponents. Likewiseforxed, . Therefore,thenumberofindependentcomponentsofRisreducedtoN1=_12n(n 1)_2.Now we use the property (5). Let us see whether theproperty (5) is nontrivial at xed . If = ,then theproperty (5)becomesR +R +R= 0 (nosummation),whichisalreadyaconsequenceof(4). Likewisefor= orfor=. Therefore,theproperty(5)isanewconstraintonlyifallthreeindices, , aredierent(i.e. ,= , ,= , ,= ). Supposethat, , aredierent. ThereareN2=16n(n 1)(n 2)choicesof such, , . Therefore, foreach=1, ..., nweobtainN2additional constraints. Finally, letuscheckthateverysuchconstraint isnontrivialforevery(evenifisequaltooneof, , ). Suppose= ,then(5)becomesR +R +R= 0 (nosummation).Thisisanontrivial constraint(equivalenttoR=R). Therefore, thenumberofconstraintsisnN2, andthusthetotalnumberofindependentcomponentsofRisN= N1 nN2 =n2_n21_12.b)Weinberg,Chapter6, 813c)ThereisonlyoneindependentcomponentofRintwodimensions. Forinstance, wecanchooseR1212astheindependentparameter. ThenwecanexpresstheRiccitensorasR= gR.Calculatingcomponentbycomponent,wendR11= g22R1212, R12= g12R1212, R22= g11R1212,R = gR=_2g11g222g12g21_R1212= 2gR1212.Notethatthematrix_g22g12g12g11_isequaltotheinverse matrixtog(whichisg),multipliedbythedeterminant det g;sincedet g= 1/g,wehave_g22g12g12g11_ = gg.ThereforeG= R12gR = ggR121212g2gR1212= 0.4.5 LorentztransformationsALorentztransformationisrepresentedbyamatrixsuchthatg= g.ConsideraninnitesimalLorentztransformation,= +H.ThenumberofparametersinLorentztransformationsisthesameasthenumberofparametersinH. TheconditionforHis( +H)_ +H_g= g.Disregarding termsoforder2,wend0 = Hg +Hg= H +H.Therefore, Hisanantisymmetricnnmatrix, whichhas12n(n 1)independentcomponents. Forn=4weget6components. ThesecanbeinterpretedasthreespatialrotationsandthreeLorentzrotations(boosts).5 Gravitationtheoryapplied5.1 RedshiftIntheweakeldlimit,theNewtoniangravitational potentialnearamassMis =GMr,whilethecomponentg00ofthemetricisg00= 1 +2c2.(Wewritetheunitsexplicitly.) Thereforetheredshiftfactorz(r)atdistancerfromthecenteroftheEarthisz(r) =_1 +2GMc2r 1 +GMc2r.TocomparetheredshiftfactorsatthesurfaceoftheEarth, denotebyREtheradiusoftheEarth. Weknowthatthegravitational acceleration atthesurfaceisgE=GMR2E 9.81ms2.Therefore,itisconvenient toexpressGM= gER2E. ForaverticaldistanceLbetweensenderandreceiver, wendz(RE)z(RE+L)=1 +gEREc21 +gEREc2RERE+L 1 +gEREc2_1 RERE +L_ = 1 +gERELc2(RE +L).SinceinourproblemL RE,wemayapproximatez(RE)z(RE +L) 1 +gELc2 1 + 1.1L1016m.145.2 Energy-momentumtensor1Inthenonrelativisticlimit,wemaydisregardgravitation; g= . TheEMTofanidealuidisT= p+ (p +) uu,whereuisthe4-velocityvectoroftheuidmotion. Inthenonrelativisticlimit,u (1, v),where visthe3-vectorofvelocityand [v[ 1intheunitswherec = 1.Theconservationlawis0 = T,= p,+ (p +), uu+ (p +) uu, + (p +) u,u.Letussimplifythisexpressionbyintroducingthetimederivativealongtheuidow,ddt u.Thenwend0 = p,+u_ p + + (p +) u,_+ (p +) u. (7)Contractingwithuandusingu u= 0,wend + (p +) u,= 0. (8)Thisistherelativisticcontinuityequation. Usingthisequation,wendfromEq.(7)that0 = p,+u p + (p +) u. (9)Nowletusapplythenonrelativisticlimit, u(1, v), toEqs.(8)and(9). Inournotation, foranyquantityXwehaveX ddtX tX + (v) X.Thecontinuityequation(8)givesddt+ (p +) divv = 0.Thisistheordinary, nonrelativisticcontinuityequation.1Finally,Eq.(9)gives

p +v p + (p +) v = 0.(Notethatp,j= jp.) ThisistheEulerequation,dvdt=1p +_

p vdpdt_.5.3 Energy-momentumtensor2Computethecovariant derivative,T;=_;;12g;;_;= ;;;+ ;;;g;;= ;;;.Hereweused;= ;whichfollowsfrom= (notethatisascalar;covariant derivativesdonotcommutewhenappliedtovectors!)andalsotheproperty;;X= ;X,whichisduetog;= 0. Therefore,weget;;;= 0;thisentails;;=0(since;=0isaweakerconditionthan;;=0, i.e. if ;=0thenalso;;=0, soitissucienttowritethelatter).1Notethat intheusual, nonrelativisticcontinuityequations as theyarewritteninmost books, thereis no p +- just . This is sobecauseinmostcasesthematterisnonrelativistic,sop andp + . Thisis,however,nottrueforrelativisticmatter,suchasphotons(electromagneticradiation)forwhichp =13.155.4 WeakgravityAveryshortsolutionistowriteR00directlythroughandnotethatonly00,comesin(if wedisregardtermsofsecondorder). Thencompute00explicitlythrough. (Assumethatg,0= 0.) Wemaydisregardtermsoforderbecauseisoforder,andalsowemayraiseandlowerindicesusinginsteadofg. (Thisissomewhatheuristic;seebelow.) Thecalculationgoeslikethis:R00= 00,0,0 +O( ),00=12(g0,0 +g0,0g00,) = ,,therefore(using,0= 0)R00= ,= = ,00 + ,11 + ,22 + ,33= .Hereisanother,somewhatmorecomprehensive solution. Intheweakeldlimit,wewriteg= +h.Then we only compute everything up to rst order in h. Therefore, we may raise and lower indices using the Minkowskimetricinsteadofg.Note: theNewtonian limitdoes not determinethecomponents of gexcept for g00= 1 +2. Theactualmetric gis equal to plus a small rst-order deviation, h,but this deviation cannot beexpressed just through the Newtonianpotential 12h00. Inprinciple,oneneedstosolvethefullEinsteinequationstondh;inotherwords,oneneedstodetermineother, post-Newtonianpotentials andnotjusttheNewtonianpotential . However, whenoneonlywantstocomputeeectsofgravitationonmotionofslowbodies, onlyg00isnecessary. SoitissucienttocomputejusttheNewtonian potential . But e.g. trajectories of light rays cannot be computed accurately in the Newtonian limit (becauselightdoesnotmoveslowly). Tocomputetrajectoriesoflightrays,oneneedsallcomponentsofh,notjusth00.Let us do the computation through hin a more general way. First we compute the Christoel symbol and the Riccitensor:=12(h, +h,h,) ; O();R= R= ,,+ ,,.(Wemaydisregard thetermssincetheyaresecondorderin.) Nowwecompute(againuptorstorderin),=12(h, +h,h,) =12h,,R=12(h, +h,h,h,) .LetusnowcomputejustthecomponentR00,recallingthathistime-independent(soh,0= 0)andh00= 2:R00=12(h0,0 +h0,0 h00,h,00) = 12h00,= ,= .5.5 EquationsofmotionfromconservationlawWewouldliketorewritethecovariant conservationlawT;= 0throughordinaryderivatives. Thegivenrelationsareuseful;letsderivethemrst.xg = 12g_xg_ = 12g_ggg,_=12ggg,;=12g(g,+g,g,) =12gg,=1gxg.Nowrewritethecovariant derivativeofTexplicitly:T;= T,+ T+ T=1g_gT_, + T.Applythistothegiven T:0 = T;=m0g__dsdxdsdxds(4)(xx(s))_,+m0g_dsdxdsdxds(x)(4)(xx(s)).16Sinceinthersttermthedependenceonxisonlythrough(4),wecanrewritedxdsx_(4)(xx(s))_ = dds_(4)(xx(s))_(thisiseasilyunderstoodifreadfromrighttoleft)andthenintegratebyparts,_dsdxdsdds(4)(xx(s)) = _dsd2xds2(4)(xx(s)).Finally,0 =gm0T;=_ds_d2xds2+ dxdsdxds_(4)(xx(s)).Thisisafunctionof xwhichshouldequal zero everywhere. Therefore, theintegrand shouldvanishfor every valueof s,d2xds2+ dxdsdxds= 0.Remark: ingeneral,equationsofmotiondonotfollowfromconservationlaw,buttheydofollowifthereisonlyoneeld. (e.g. one uid, or one scalar eld, or some number of point particles). The situation in ordinary mechanics is similar:e.g.theequationofmotionforaparticlefollowfromtheconservationofenergyonlyifthemotionisinonedimension:E=mv22+V (x) = const,0 =dEdt= (m v +V(x)) v m v = V(x).However, equationsof motiondonotfollowfromconservationof energyif thereismorethanonedegreeof freedom.Similarly, equationsofmotionforsaytwoscalarelds, donotfollowfromtheconservationoftheircombinedT.Theseeldshavetwodierentequationsofmotion,andonecannothopetoderivethemfromasingleconservationlaw.6 Thegravitational eld6.1 DegreesoffreedomTheelectromagneticeldisdescribedbya4-vectorpotential A(x). Thiswouldgive4degreesof freedom. However,thereisalsoagaugesymmetry,A A +,,where (x)isanarbitrary functionof spacetime. Usingthisgauge symmetry, wemaye.g. set thecomponent A0(x) = 0.Thenonlythreefunctionsofspacetime(A1, A2, A3)areleft. Hencetheelectromagneticeldhas3degreesoffreedom.Thereareadditionalgaugesymmetriesinvolvingfunctions(x)thatdonotdependontime. Sincethesefunctions(x)arefunctionsonlyofthreearguments,theydonotchangethenumberofdegreesoffreedom.6.2 Sphericallysymmetricspacetime6.2.1 StraightforwardsolutionA direct computation listing all thepossible Christoel symbols and components of theRicci tensor is certainly straight-forward butverylong. Hereisaway tocomputethecurvaturetensorwithoutwritingindividualcomponents. Sincethemetrichasadiagonalform,letusdenoteg= A, g= 1A(nosummation!), (10)whereA _eN, eL, r2, r2sin2_(11)is axed array of four functions. For this calculation, we do not use theEinstein summation convention any more; everysummationwill bewrittenexplicitly. However, wemakeheavyuseof thefactthat ,=0onlyif =, andthat=. AttheendofthecalculationoftheRicci tensorR, weshall substitutetheknownfunctionsAandusetheresultingsimplications.WebeginwiththecalculationoftheChristoelsymbols,=

121A[A,+A,A,]=12A,A+12A,A12A,A. (12)17Note that thesummation over results insetting = due to,and that we have relations suchas = and= ,whichholdwithoutsummation. Forconvenience,werewriteEq.(12)as=12_B,+B,AAB,_,wherewedenedtheauxiliary functionB ln A.Asacheck,wecomputethe trace oftheChristoelsymbolsandcomparewiththeknownformula,

=12_B, +

B,AAB,_ =12

B,=_lng_,.LetusalsodenoteforbrevityC lng =12

B;

= C,.WeproceedtothecomputationoftheRiccitensor. Weusetheformula(withLandau-Lifshitzsignconventions)R=

_,,_+

,_ _.Wenowcomputethenecessary terms:

,=12B, +12B,12

_AAB,_,,

,= C,,

,=

C,12_B, +12B,12AAB,_=12_C,B, +C,B,

C,AAB,_,

,=14

,_B, +B,AAB,_ _B, +B,AAB,_=14

B,_B, +B,AAB,_(hereset = )+14

B,_B,+B,AAB,_(hereset = )14

AAB,_B, +B,AAB,_(hereset = )=14B,B, +14B,B,14

AAB,B,+14

B,B, +14B,B,14B,B,14B,B,14

AAB,B, +14B,B,=12B,B,12

AAB,B, +14

B,B,.Finally,weputallthetermstogether:R=_12B +12B C_,12A

__B,A_,+C,B,A_+12C,B, +12C,B,12B,B,14

B,B,.18Wecansimplifythisexpressionbyconsideringseparatelydiagonal ando-diagonal components:R= (BC),12A

__B,A_,+C,B,A_+_C,12B,_B,14

B,B,;R=12_B +B

B_,+12 (C,B, +C,B,B,B,)14

B,B,. (onlyfor ,= )Nowweneedtosimplifythisexpressionfurtherbyusingthespecicformofthemetric(10)-(11). WehaveAt= eN, Ar= eL, A= r2, A= r2sin2;Bt= N, Br= L, B= 2 lnr, B= 2 lnr + 2 ln sin ;C= lng =N+L2+ 2 lnr + ln sin ;C,t=N+L2, C,r=N+L2+2r, C,= cot , C,= 0.Notethattheterm12_B +B

B_,(13)alwaysvanisheswhen ,= . Wend(aftersomeomittedalgebra):R=12r2sin2

____ln_r2sin2_,A_,+C,_ln_r2sin2_,A__= r2sin2__NL2r+1r2_eL1r2_;R= 1 r2eL_NL2r+1r2_;Rt= Rr= R= Rt= 0;Rr=12C,B,r14B,rB,=12 cot 2r 142r2 cot = 0;Rtr=12_N+L2N+_N+L2+2r_L NL_14NN14LL=Lr ;Rrr= N2+2r2+NLN4+12eLN_L +L N2L_+1rL;Rtt= L2+LN L4+12eNL_N+_NL2+2r_N_.Finally,wecomputetheRicciscalar,R =

1AR= eN_L2+LN L4_+12eL_N+_NL2+2r_N_eL_N2+2r2+NLN4+1rL_12eN_L +L N2L_1r2+eL_NL2r+1r2_+_NL2r+1r2_eL1r2= eN_L +LN L2_eL_2NLrN+NLN2_2r2.19Hence,thenonzerocomponentsoftheEinsteintensorareGtt= eNLLr+eNr2;Gtr=Lr ;Grr=2r2+NreLr2 ;G= r2eL_NL2r+1r2+N2+NLN4_+r2eN2_L +LN L2_;G= 12r2sin2_eN_L +LN L2_eL_NLr+2r2 N+NLN2__.6.2.2 Solutionusingconformal transformationAmorecleverwaytoreducetheamountofcomputationistonoticethatthemetricgissimpliedafteraconformaltransformation,g= r2h= r2____r2eL0r2eN10 sin2____.Themetrichseparatesintother tcomponentsandthe components. WeshallrstcomputetheRiccitensorforthemetrichandthendeterminehowRchangesunderaconformaltransformation. ThecalculationofRforthemetrichismuchsimplerbecausehisadirect(block) sumoftwometricsdenedon2-dimensionalspaces. ItisclearthatRwillalsobeadirectsumofthecorresponding two-dimensional Riccitensors.Let us rst compute the Ricci tensor for a diagonal metric ab= diag_eA, eB_ in two dimensions; set A A1, B A2,andindicesa, b, c,... rangefrom1to2. Foratwo-dimensional metricab,weknowthattheRiccitensorisproportionaltoab,namely(seeProblem 4.4c)Rab=12abR, R abRab= 2 (det ab) R1212.SoitissucienttocomputesayR11,R11= a11,aa1a,1 + abab11b1aa1b,andafterwardswewillhaveRab= abR1111; R = abRab= 2R1111.Thenecessary Christoelsymbolsarefoundas(noimplicitsummationfromnowon!)a1b=

c12ac(1c,b + bc,11b,c) =12_A1,ba1+abAb,11beA1AaA1,a_;112=12A,2; 212=12B,1;a11= A1,1a1 12eA1AaA1,a; 111=12A,1; 211= 12eABA,2.

aaba=12b ln (det cd) =12 (A1,b +A2,b) =12 (A +B),b.ThenthecomponentR11oftheRiccitensorisR11= a11,aa1a,1 + abab11b1aa1b= 111,1 + 211,212 (A+B),11 +12 (A +B),1 111 +12 (A +B),2 2111111112112211212212=12A,1112_eABA,2_,212 (A+B),11 +12 (A+ B),112A,112 (A +B),212eABA,214A,1A,1 + A,212eABA,214B,1B,1= 12B,1112eABA,2214A,2 (A B),2eAB+14B,1(AB),1.Wealsond(notethesymmetryapparent inthisformula;thisshowsthatatleastthisisnotobviously wrong)R = 2R1111= eAB,11eBA,2212A,2 (A B),2eB12eAB,1 (B A),1.20Well,wearenotgoingtonishthiscalculationhere, anyway. Butthisisroughlyhowitgoes. Letusatleastderivetheusefulformulabelow.Therelationship betweenthecurvature tensors underaconformal transformation isfoundasfollows. First wedenetheconformallytransformedmetricforconvenienceasfollows, g= e2g.ThentheChristoelsymbolsreceiveacorrection whichisatensorB,= +B; B , +,g,.TheRiemannandtheRiccitensorsaredened(inLandau-Lifshitzsignconvention) byR= ,,+ ,R= R= ,,+ .The same relation holds forRand Rthrough (note that these relations do not involve the metric gexplicitly).ThereforeRR= B,B, +B + B +BBBBBB;R R= B,B, +B + B +BBBBBB.WeshallonlycomputetheexpressionfortheRiccitensorR. Asapreparation, wecomputeB= , N,,where N= = ggis the number of spacetime dimensions. We shall always raise and lower indices using the originalmetricg. Sowecomputetermbyterm,RR=_, +,g,_,N, +N,+ _, +,g,_+N,_, +,g,__, +, g,_ (, +,g,) _,+,g,_(, +,g,)= 2,g,,g,,N, +N, + , + ,g,+ 2N,,Ng,,,, +g, ,, +g,(2 +N) ,, + 2g,,= g,,g, Ng,,+ 2g,,= (N 2)_,,+ (N 2) ,,g_(N 2) ,,+ ,, + ,_+_g,+g,g,,.Nowwenotethatsomeof thetermscanbeabsorbedintocovariantderivatives, andalsothatthetermsinthelastbracket cancel,_g,+g,g,, = 0,sotheresultingformulacanbewrittenmoreconciselyasRR= (N 2) [,,;] g_(N 2) ,,+ ;;.ThemodiedRicciscalar isR = g R= e2gR +e2g_(N 2) [,,;] g_(N 2) ,,+ ;;_= e2R +e2_(N 2) [,,;;] N_(N 2) ,,+ ;;_= e2R (N 2) (N 1) ,,2 (N 1) ;; .TheEinsteintensorismodiedasfollows,G=R 12 g R = R + (N 2) [,,;] g_(N 2) ,,+ ;;12g [R (N 2) (N 1) ,,2 (N 1) ;;]= G+ (N 2) [,,;] +g_(N 2) (N 3)2,,+ (N 2) ;;_.NotethatthereisnochangeinGintwodimensions(sincetheEinsteintensorisalwaysequaltozero).216.3 EquationsofmotionTheequationforthecovariant componentu1(s)isdu1ds12uur(g) = 0.Using the metric g= diag_f, 1/f, r2, r2sin2_,where f 1 rg/r,and u=_t, r,,_,where d/d,we nddd_f1 r_12_dfdrt2ddr_1f_ r22r22r 2sin2_ = 0. (14)Toderivethisequationfromotherequationsgiveninthelecture,wetransform inacleverwaytheexpression0 =dd/ =dd_f t2f1 r2r2 2r2 2sin2_.Namely,wetrytoseparatetermsoftheformdd (u)outofthetermsoftheformdd (uu)inthefollowingway,dd_u1u1_ =dd_g11u1u1_ = 2u1dd_g11u1_u1u1ddg11(nosummation!).Forexample,dd_f t2_ = 2tdd_f t_t2ddf, etc.Wend0 =dd/ =dd_f t2f1 r2r2 2r2sin2 2_= 2tdd_f t_t2 dfd 2 rdd_f1 r_+ r2ddf12dd_r2 _+2dd_r2_2 dd_r2sin2 _+2dd_r2sin2_.Nowwesubstitutethegivenequations(2)-(4),andalsoevaluatederivativesofthemetric,e.g.df/d = f r,so0 = t2f r 2 rdd_f1 r_ r2 ff2 r 2r2 2sin cos +22r r +22r sin2 r + 2 2r2 sin cos = r_dd_2f1 r_ft2ff2 r2+ 2r2+ 2r 2sin2_.Thisisobviouslyequivalent toEq.(14).Note: the reason one of the equations follows from other equations is that the equation uu= const is a consequenceofthefourgeodesicequations,uu;=0, andthefactthatg;=0. Therefore,whenweconsiderthefourgeodesicequationsandtheequationuu= const,anyoneoftheseveequationsisaconsequenceoffourothers.7 Weakgravitational elds7.1 Gravitational bendingoflightInthelectureitwasshownthatthetrajectory ofalightrayinpolarcoordinates satisestheequationd2d2_1r_+1r=32rgr2, rg 2GMc2 3km,whereMisthemassoftheSun. Introduceanauxiliaryvariable v() r1andsolvetheequationv+v =32rgv2perturbatively,assumingthatvissmall,v() = v0() +v1() +...Theunperturbedsolutionisv0() =1R0cos ,whereR0isthedistanceofclosestapproach totheSun. Thenv1+v1=32rgR20cos2 =34rgR20(1 + cos 2) .22Thesolutionisfoundwithundeterminedcoecients,v1() = A+Bcos 2, A =34rgR20, B= 14rgR20.The total deection angle is found as = 12, where 1,2are xed by the condition v() = 0. We nd a quadraticequationcos2 2R0rgcos 2 = 0, cos =R0rgR20r2g+ 2.Only the solution withthe minus sign is meaningful (cos < 1). Since rg R0,we may expand thisin Taylor series andndcos rgR0+O(r3g/R30).Therefore,theangleisverycloseto/2,1,2= _2+_, rgR0, = 2 =2rgR0.Thisformulacanberewrittenas=2rg/RR0/R=_2rgR_ RR0.FortheSunwehaveR= 6, 96 105kmandrg= 2, 954 km,therefore2rg/R= 8, 489 106=__8, 489 106360/23600[arcseconds]= 1, 751(seeR.Olo GeometriederRaumzeit, 2ndGermanedition,page151).7.2 EinsteintensorforweakeldFor this problemChapter 4fromthebookNorbert StraumannGeneral RelativityandRelativisticAstrophysics isuseful. Wehaveg= +handR= ,,,where(...),denotesaderivative(...). Hereonecanaskthestudentsaboutthesymmetryofthistensor.Furthermore=12[h, +h,h,] =12_h,+h,h,, (15)where as usual weuse theconvention that indicesare raised or lowered with ;thus e.g.h h. Using Eq. (15)wehaveR=12_h, +h,h h,,where = andh = h= h. AndfortheRicciscalarweobtainR = R= h,h.ThusinthelinearapproximationwehaveG= R 12R =12_h, +h,hh, h, +h_.Letusintroduceanewvariable h 12h. Thetracesoftwotensorshandarerelatedby= h,thush 12. InsertingthelastexpressionforhinG,wehaveG=12_, +, ,_ ==12_,,+,, ,_,ornallyG=12_,,+,, ,,_.237.3 Gravitational perturbationsIThemetriciswrittenasg= +g,i.e.g00= 1 + 2, g0i= B,i +Si, gij= ij + 2ij + 2E,ij +Fi,j+Fj,i +hij, (16)where S,ii= F,ii= h,iij= hijij= 0, hij= hji. We shall use the formula for Gderived in Problem 7.2. All 3-dimensionalindicesareraisedandloweredusingij,sowecanwritetheseindicesinanyposition,asconvenient:g0j= g0j= gj0, gji= gij.AlsonotethatforanyquantityXwehaveX,j=_X X,j_.Weneedtowritethecomponentsof= g 12h,h g,usingthe3+1decomposition:h = g= g= 2 ( 3E) ,00= + 3 + E, 0j= B,j +Sj= j0= 0j,ij= ( E) ij2E,ij Fi,j Fj,ihij= ji.Nowwecompute0,= 00 0j,j= + 3 + _E B_;j,= j,= 0jji,i= B,jSj(( E) ij 2E,ij Fi,j Fj,ihij),i= B,jSj + ( + E),j+ Fj;,,=_0,_,0+_j,_,j= + 3 + _E B_+_B,jSj+ ( + E),j + Fj_,j= + 3 + E 2B + ( + E) .ThenwecomputeeachcomponentofGseparately:2G00= 20,,0 00 00,,= 2_ + 3 + _E B__00 ( + 3 + E) + ( + 3 + E)_ + 3 + E_+ 2B ( + E)= 4;2G0j= 0,,j+,0j, 0j 0j,,=_ + 3 + _E B__,j_B,jSj +_ + E_,j+ Fj__B,j+Sj_+ (B,j +Sj) = 4 ,j + Sj Fj,2Gij= i,,j+j,,i ijij,,=_B,jSj + ( + E),j + Fj_,i+_B,iSi + ( + E),i + Fi_,j+(( E) ij + 2E,ij +Fi,j+Fj,i +hij) ij_ + 3 + E 2B + ( + E)_= 2_ B +E_,ij+Fi,j+Fj,iSi,jSj,i +hij 2ij_2 + _ B +E__.*-theoriginoftheminussignhereis,0j,= j,,0.7.4 Gravitational perturbationsIIUnderaninnitesimaltransformation xx+,themetricchangesasg gg,g,= g,,. (17)(This canbe easilyfoundfromthe standardformulafor the change of coordinages, involving x/x.) Nowletus writeEq. (17) infull, usingtheperturbationvariables(16), thecovariantcomponents, andthedecomposition=_0, i +,i_. We can write the transformation of gcomponent by component using the 3+1 decomposition, andweusethefactthatthebackgroundmetricisdiagonal,g00 g0020,0; g0i g0i0,ii,0; gij gij i,j j,i.24Tosimplifycalculations, weadopt theconventionof raisingandloweringthespatial indices i, j, ... bytheEuclideanspatial metric ijrather thanby ij. This willget rid of some minus signs. Wealso denote 0 tby theoverdot. Thuswehaveg00 g0020; g0i g0i0,ii; gij gij i,jj,i.SubstitutingtheperturbationvariablesfromEq.(16),weget 0, (18)B,i +Si B,i +Si0,ii,i, (19)2ij + 2E,ij+Fi,j+Fj,i +hij 2ij + 2E,ij+Fi,j+Fj,i +hij i,j j,i2,ij. (20)Nowweneedtoseparatetheseequationsandderivethetransformationlawsfortheindividual perturbationvariables.This is easy to do if we perform a Fourier transform of Eqs. (18)-(20) and pass to theFourier space (where every variableisafunctionofa3-vector k). AvectorViisdecomposedintoscalarandvectorcomponentsasfollows,Vj= ikjV(S)+V(V )j; V(S)Vlklk2, V(V )j Vj ikjVlklk2= VjikjV(S). (21)Theideaisrst, toprojectthegivenvectorVi(k)ontothedirectionofki, andsecond, tosubtracttheprojectionfromViandtoobtainthecomponentofViwhichistransversaltoki. Theimaginaryunitfactorsareaddedascoecientsatkjforconvenience: withthesefactors,thedecomposition(21)translatestorealspaceasVj= jV(S)+V(V )j.The same procedureappliedtoasymmetric tensor Tijleads toadecompositionintoscalar, vector, andtensorcomponents. Letusgothroughthisprocedureinmoredetail. First,wesubtractthetraceandobtainthetracelesspartT(1)ofthetensorT,T(1)ij Tij13ijTll; T(1)ii= 0.Notethecoecient13thatdependsonthenumberof spatial dimensions(three). NowweprojectT(1)ijontokikjandobtainthescalarcomponentT(S)proportional tokikjandthetensorT(2)ijorthogonal tokikj:T(1)ij_kikj +13ijk2_T(S)+T(2)ij; T(S) 32kikjT(1)ijk4; T(2)ijkikj= 0.Note that T(2)ijis again a trace-free tensor, T(2)ii= 0, due to the subtraction of13k2ijin the rst term. Finally, we projectT(2)ijonto kiand kjseparately, to obtain a vector part T(V )jsuch that T(V )jkj= 0, and a completely traceless (tensor)partT(T)ijsuchthatT(T)ijkj= 0andT(T)ii= 0:T(2)ij= ikiT(V )j+ ikjT(V )i+T(T)ij; T(V )j iklk2T(2)jl, T(T)ij T(2)iji_kiT(V )j+kjT(V )i_.Inrealspace,thefulldecompositionisTij=13Tllij+_ij13ij_T(S)+iT(V )j+jT(V )i+T(T)ij.T(S)3212ij_Tij13Tllij_; T(2)ij_Tij 13Tllij__ij 13ij_T(S);T(V )j=1iT(2)ij, T(T)ij= T(2)ijiT(V )jjT(V )i.Itmaybeconvenienttogatherthe trace terms(thetermscontainingij)asoneterm,Tij= T(tr)ij+ijT(S)+iT(V )j+jT(V )i+T(T)ij, T(tr)13Tll13T(S).Note that the perturbationvariables , E, Fi, hijare obtainedbythis decompositionmethod, startingfromthesymmetric perturbation tensor gij, with slight modications: there are some cosmetic factors of 2 and some minus signs.ApplyingthedecompositionmethodtoEqs.(18)-(20),wegetthefollowingtransformation lawsfortheperturbationvariables, 0, B B 0, Si Sii,E E , , Fi Fii, hij hij.25Remarks:1. ItisclearthatonecansetFi=0, B=E=0withacoordinatetransformation. Othercomponentswillthenshowwhetherthegeometryisreallyperturbedoritsjustacoordinatetransformationofaatspace. Ingeneral, therewillremain6independentcomponentsofperturbations(, , Si, hij).2. Theseconsiderationsdependrathercrucially onthesilentlymadeassumptionthatall themetricperturbationsvanish, g 0, at spatial innity. These boundary conditions are implicitly used when dening the Fourier transformsnecessary for the tensor/vector/scalar decompositions (a Fourier transform is undened without this boundary condition).Alternatively,onemaydowithoutFouriertransformsbutthenonestillneedsboundaryconditionstosolvetherelevantPoisson equationsforcomponents. Withoutboundaryconditions,thereisnouniquedecompositionoftheformXi= A,i +Bi, A =1Xi,i,becausethefunctionAis deneduptosolutions of A=0. Sothetensor/vector/scalardecompositionis actuallyundened withoutaxedassumptionabouttheboundaryconditions. Theboundaryconditionsg 0atspatialinnityisanatural,physically motivatedset ofboundary conditions. Anexplicitcounterexample wheretheseboundaryconditions arenot satised: g=diag (A, B, B, B), where A ,=1, B,=1areconstants. This metric is atbutonecannotseethisbyusingtheperturbationformalism! (Thecomponent ,=0cannotberemovedbyagaugetransformation.)Thereason isthatthisgisa perturbation of at metricwithandthat donot decay tozero atspatialinnity. Soacoordinatetransformation withdecayingtozerocannotbringthismetricto.8 Gravitational radiationI8.1 GaugeinvariantvariablesUsingtheequationsderivedinProblem7.4,itisveryeasytoverifythatD= B +EandSiFiareinvariantunderinnitesimalchangesofcoordinates(i.e.invariantunderinnitesimalgaugetransformations).8.2 Detectinggravitational waves8.2.1 Usingdistancesbetweenparticles(ThissolutionfollowsHobson-Efstathiou-Lasenby [2006], 18.4.)Consideraplanewavemovinginthezdirection,(allothercomponentsofharezero)hxx= hyy= A+ei(tz), hxy= hyx= Aei(tz). (22)To detect the presence of this gravitational wave, let us imagine a cloud of particles initially at rest at dierent positions.The 4-vectors describing the particles are u= (1, 0, 0, 0), so one can easily see that these particles move along geodesics:uu;= uu,+ uu= 00,=12(h, +h, h,) ,00=12(h0,0 +h0,0h00,) = 0.Thereforethecoordinatesxof eachparticleremainconstantwithtime. However, thedistancebetweeneachpairofparticlesisdeterminedthroughthespacelikevectorx x(1)x(2)asL2 (+h) xxandwill changewithtimebecauseof thedependenceonh. Sincetheonlynonzerocomponentsof harethex, ycomponents,itisclear thatonly changing lengthsarebetweenparticles thathave someseparation inthex, ydirections.Thereforeitissucienttoconsideraringof particlessituatedinthex yplane. Thephysicallymeasureddistancesbetweentheparticlesintheringwillchangewithtime,i.e.theringwillexperienceadeformation.Tovisualizethedeformation, itisconvenienttomakealocal coordinatetransformation(local intheneighborhoodofthering)suchthatthemetricbecomesat,gxx= x x(uptosecond-order terms). Thetrickthatperformsthistransformation isthefollowing, x= x+12hx x+12hx.Itiseasytocheckthatgxx (+h) xx= x x+O(h2).Therefore, xcanbeunderstoodas the(approximate)Cartesiancoordinateswherethelengthis givenbytheusualPythagoreanformula. Nowif wecomputetheshapeof theringinthesecoordinates, itwill beeasytointerpretthisshapeinastraightforward way.26Consideraparticlewithconstant3-coordinates (x, y, z). Afterthecoordinatetransformation, wehave x = x +12 (A+x +Ay) ei(tz), y= y +12 (Ax A+y) ei(tz), z= z.Tovisualizethedeformation, itisconvenienttoconsiderrstthecaseA+ ,=0, A=0andthentheoppositecase.Thedeformationoftheringissqueezinginonedirectionandexpansionintheorthogonaldirection. ItfollowsthatA+describesadeformationinthetwoverticaldirections,whileAdescribesadeformationinthedirectionsat45.Notethatthedeformationschangetheshapeoftheringinthesameway, exceptfortherotatedorientation. Thiscanbeveriedbyperformingarotationby4,_ x y_12_1 11 1__ x y_,andthenitisstraightforward toseethatthiswillexchange A+andA.8.2.2 UsinggeodesicdeviationequationPLEASENOTE: Thecommonlyfoundargumentsthatusethegeodesicdeviationequationaresuspectbecausethegeodesicdeviationequationusescoordinates ratherthangauge-invariantquantities. Acloudof particlesatrestinthegravitational eldhdescribedbyEq. (22) will stayindenitelyat rest inthecoordinatesystem(=const)eventhoughthedistancesbetweenparticleswill changewithtime. Seearxiv:gr-qc/0605033 forniceexplanations. Thesolutiongivenaboveissimpleandstraightforward. Theargumentusingthegeodesicdeviation(seeCarroll,Chapter6,p. 152-154) goeslikethis:ThegeodesicdeviationequationcanbesimpliedforadeviationvectorScorrespondingtononrelativistic(almoststationary) particlesmovingwith4-velocityapproximately equalto(1, 0, 0, 0),d2Sdt2= R00S.TheRiemanntensortorstorderinhcanbeexpressedasR00= h(notethath0= 0). Therefore,thegeodesicdeviationequationbecomesS= hS,Sx= 2(hxxSx+hxySy) = 2(A+Sx+ASy) ei(tz),Sy= 2(hyxSx+hyySy) = 2(ASxA+Sy) ei(tz),andthereisnochangeinthezdirection.8.3 PoissonequationThegeneral solutionofthePoissonequation, = 4,withboundaryconditions 0atinnity,iseasytondusingtheFouriertransform:k2(k) = 4(k),(x) = _d3k(2)3eikx4(k)k2= _d3k(2)3eikx4k2_d3yeiky(y) =_d3y (y)G(x y),whereG(x)istheGreensfunction,G(x) = _d3k(2)3eikx4k2= 1_0dk_0d sin eik|x| cos = 2 [x[_0dkksinkx = 1[x[.Hereweusedtheknownintegral_0sinzzdz=12_+sin zzdz=12.Therefore(x) = _d3y[x y[(y). (23)27Onecandenotethisintegralmoreconcisely, = 41,wheretheoperator1isjustashorthandnotationfortheintegralinEq.(23).Notethatthefunctionmustfallosucientlyrapidlyas [x[ orelsetheintegral(23)willnotconverge. Itissucientthat [(x)[ [x[2atlarge [x[(where > 0).8.4 Metricperturbations1Anarbitrary 3-vector Xi(suchasT0i)isdecomposedintoscalar andvectorpartsasfollows,Xi= a,i +bi, bi,i= 0.Todetermineanexplicitexpressionfora,letuscomputethedivergence ofXi,Xi,i= a,ii= a.Thereforea(x) = 14_d3y[x y[Xi,i(y).Onecanwritemoreconciselya =1Xi,i.8.5 Metricperturbations2Theenergy-momentum tensorTisdecomposedasT0i= ,i +i, 1T0k,k, i T0i _ 1T0k,k_,i;Tik= ik +,ik +i,k +k,i +T(T)ik,i,i= i,i= 0, T(T)ii= 0, T(T)ik,i= 0.Weneedtoverifythattheequation116G( SiFi) = i, (24)which represents the vector part of the spatial Einstein equation (here iis the vector part of the spatial Tij), also followsfromtheconservationofTandfromtheotherEinsteinequations.Tocalculatethecomponents, oftheEMT,wecomputeTii= + 3,Tik,i= ,k +,k + k,Tik,ik= + .Nowwesolvethissystemofequationsandnd =12_Tii 1Tik,ik_, =321Tik,ik12Tii,j=1Tij,i1_ 1Tik,ik_,j,(T)Tik= Tikik,iki,kk,i.Notethattheoperator12appliedtoafunctionf(x)isdenedonlyif thefunctionf hasasucientlyfastdecayat[x[ . Itissucientthat [f(x)[ [x[3with>0atlarge [x[. Thisisafasterdecaythanthatrequiredbytheoperator1.TheEinsteinequationsare2 = 8GT00,2 ,i +12Si= 8GT0i= 8G(,i +i) ,D,ijij_D + 2_12_ Si,j+Sj,i_+12hij= 8GTij= 8G_ik +,ik +i,k +k,i +T(T)ik_,wherewehavedenotedforbrevitySi SiFi, D +B E,28whicharegauge-invariant variables. Inthe3+1decomposition,theEinsteinequationsbecome = 4GT00, (25) = 4G, (26)Si= 16Gi, (27)D = 8G, (28)D + 2 = 8G, (29)Si= 16Gi, (30)hij= 16GT(T)ij. (31)Theconservation lawoftheEMTin3+1decompositionlookslikethis,T00,0 +Tj0,j= 0, T0i,0 +Tji,j= 0. (32)ThisgivesT00= , ,i +i + ,i +,i + i= 0,thereforeT00= , + + = 0,i + i= 0. (33)Thenit iseasytoseethatEqs. (26), (29), and(30)areconsequencesof Eqs. (25), (27), (28), andtheconservationlaws(33). Inparticular,Si= t116Gi= 116Gi= 16Gi.9 Gravitational radiationII9.1 Projectionofthemattertensora)FirstnotethatPabisaprojector,PabPbc= Pac,anditsimagehasdimension2,thatis,thetraceofPabis2,Pii= 3 nini= 2.Therefore,foranyXabwehave(T)Xii= PiaXabPbi12PiiPabXab= PabXab122PabXab= 0.b)Wecompute(T)Xik,i=_PiaXabPbk12PikPabXab_,i=_PiaPbk12PikPab_,iXab +_PiaPbk12PikPab_Xab,i. (34)NotethattheprojectionkillsanycomponentproportionaltoRibecausePiaRi= 0. Atthesametime,Xab,iispropor-tionaltoRibecauseXab,i=_X(t [

R[)_,i= RiR X.ThereforethesecondterminEq.(34)vanishes:_PiaPbk12PikPab_Ri= 0.Soonlythersttermremains,(T)Xik,i=_PiaPbk12PikPab_,iXab.29However,thistermcontainsderivativesofPab,whicharealsosometimesproportional toRi. WecomputePik,a= ni,anknink,a; ni,a=_RiR_,a=Ri,aRRiR2 [R[,a=iaRRiRaR3=1RPia,Pik,a= 1R (Paink +Pakni) , Pik,i= 2Rnk, (notethatPaknk= 0)_PiaPbk12PikPab_,i= Pia,iPbk12Pik,iPab +PiaPbk,i12PikPab,i= 2RnaPbk +1RnkPab1RPia (Pbink +Piknb) +12RPik (Painb +Pbina)=1R_2naPbk +nkPabPabnkPaknb +12Paknb +12Pbkna_= 1R_32Pbkna +12Paknb_.Thisishigher-order in1/[

R[thanPab,asrequired.9.2 MattersourcesThequestionistoverifythefollowingproperty,(T)Xik=(T)Qik,whereQik= Xik_13ik r2T00d3r.ItiseasytoseethatXikdiersfromQikonlybyatermoftheformA(t, R)ij. Thetransverse-traceless partofijiszero,_PiaPbk12PikPab_ab= 0.Thereforethetransverse-traceless partsofXikandQikarethesame.9.3 Energy-momentumtensorofgravitational wavesSeeHobson-Efstathiou-Lasenby [2006], 17.11.Weneedtocomputethesecond-ordertermsintheEinsteintensor. Theideaistoseparatethesecond-ordertermsalreadyintheRiccitensor. Wewillalsotrytosimplifythingsbyusingthefactthathispurelytransverse-traceless;h0= 0,hii= 0,hik,i= 0. Itfollowsthath= 0, h,= 0.Also,itisgiventhattheEMTofmattervanishes,T= 0,whichwewillusebelow.Firstwedecomposethemetric,g= +h, g= h;notethatnowindicesarealways raisedandloweredusing. WeneedtocomputetheRicci tensortosecondorder.TheChristoelsymboluptosecondorderis=12_h_(h, +h, h,) = (1)+ (2),(1)=12(h, +h,h,) ,(2)= 12h(h, +h,h,) .TheRiccitensorisR= ,,+ = R(1)+R(2),R(1)= (1),(1),,R(2)= (2),(2),+ (1)(1)(1)(1).Letusnowevaluatetheseexpressionsandsimplifyasmuchaspossible,asearlyaspossible:(1)=12h,=12_h_,= 0,R(1)= (1),(1),=12(h, +h,h,) = 12h,30becauseofthetransverse tracelesspropertyofh. Now,sinceR(1)isfoundfromtherst-order EinsteinequationR(1) 12R(1)= 8GT,anditisgiventhatT= 0. Hence,wehaveh= 0.Letusnowevaluatederivativesofthesecond-order termsintheChristoelsymbols:(2)= 12h(h, +h,h,) = 12hh,,(2),=12hh,+12h, h,,(2),= 12h(h, +h,h,)(inthelastlineweusedh,= 0). Finally,wetackletheterm(1)(1). Inthisterm,ithelpstowrite(1)=12_h, +h,h,_,where again the indices are raised via since we only need this term to rst order. Then we can simplify this expressionbygroupingtogethertermswhere, appearinsimilarpositions:4(1)(1)=_h, +h, h,__h, +h,h,_(expandbrackets) = h,h, +h,h,h,h, +h,h, +h,h,h,h,h,h,h,h, +h,h,(move,rename, ) = h,h, +h,h, h,h, +h, h, +h, h,h, h, h,h,h,h,+h,h,(gatherterms) = 2h,h,+ (h, h,) h, 2h,h, +h,(h,h,) +h, h,(symmetryofh) = h,h,+ 2h,h, 2h,h,.Finally,weputtogethertheexpression forR(2):R(2)=12h(h,h, +h, +h,) +12h, h,14_h,h,+ 2h,h,2h,h,_=12h(h,h, +h, +h,) +14h, h, +12_h,h,h,h,_. (35)TheRicciscalarisR(2)= R(2)=12hh +14h, h, +12_h,h,h,h,_ =34h,h,12h,h,, (36)whereweagainusedthetransversetracelesspropertyofhandalsoh=0. Notethattherst-orderRicci scalaris zero, R(1)=0, sinceT=0. For this reasonwe mayuseinEq. (36), otherwisewe wouldhavetowrite( +h)_R(1)+R(2)_andpickupasecond-order termhR(1).Again,sinceR(1)= 0,wemayuseratherthangtocomputetheEinsteintensor:G(2)= R(2) 12R(2).WedonotwritetheanswerexplicitlysinceitisacombinationofEqs.(35)and(36).Nowletusperformanaveragingof thequantityG(2)overbothspaceandtime. Inotherwords, weintegrateG(2)overa4-dimensionalregionsuchthath= 0andh,= 0ontheboundaryofthatregion. Then (...)) = 0andsowemayintegratebyparts,forexampleAB,) = A,B) ,aslongasABcontainsrstpowersofhorh,,sothatboundarytermsvanish. Then,forexample,hh,_ = h, h,_, (37)hh,_ = _h, h,_ = 0,h,h,) = hh) = 0,byh=0andbythetransversetracelesspropertyofh. Manytermscancelinthisway;forinstance,R(2)_=0.Finally,weget_G(2)_ =_12h(h,h, +h, +h,) +14h, h, +12_h,h,h,h,__=12hh,_+14h, h,_ = 14h, h,_usingEq.(37). Finally,weobtaintherequiredequation,(GW)T= 18G_G(2)_ =132Gh, h,_ =132Ghij,hij,_.319.4 PowerofemittedradiationToderivetherelations_nlnmd4=13lm,_nlnmnknrd4=115(lmkr+lkmr+lkmr),letusconsiderthegeneratingfunctiong(ql) _d4exp_inlql,whichisafunctionof avector argument ql. Aftercomputingg(ql)itwillbeeasy toobtainintegrals suchastheabove:_nlnmd4=iqliqmg(qj)qj=0, etc.The computation is easy if we introduce spherical coordinates with the z axis parallel to the vector ql, then nlql= [q[ cos ,where [q[ qlql,andthenwehaveg(ql) =14_20d_0d sin exp [i [q[ cos ] =1422i sin [q[i [q[=sin [q[[q[= 1 13!qlql +15! (qlql)217! (qlql)3+...WehaveusedtheTaylorexpansionforconvenienceofevaluatingderivativesat [q[= 0. Thesederivativescanbefoundasfollows,gql= 23!ql +45!ql[q[2... =_13+130[q[2_ql,2gqkql=_13+130[q[2_kl +115qlqk,3gqjqkql=115 (qjkl +qkjl +qljk) ,4gqjqkqlqm=115 (jmkl +kmjl +lmjk) .Nowwecomputetheintensityofradiation. Theuxofradiationinthedirectionnkis(GW)T0knk, andweneedtointegratethisuxthroughasphereofradiusR:dEdt= R2_d2(GW)T0knk=R232G_d2_hij,0hij,k_nk.The perturbation hijis found from theEinstein equation. It was derived in thelecture that,in theleading order in1/R,wehavehij= 2G(TT) Qij(t [

R[)[

R[,(TT)Qij _PaiPbj 12PabPij_Qab.TheprojectiontensorPijisdenedinProblem 9.1. ThetensorQijisdenedbyQij(t) _d3x_xixj 13[x[2ij_T00(x, t)andisbydenitiontrace-free, Qii= 0. Thuswehavehij,k=16GR(TT) Qij(t [R[)RkR,dEdt=G8_d2_(TT)...Qij(TT)...Qij_.Itremainstocomputetheaverage overthesphereof(TT)...Qij(TT)...Qij.32Consideranysymmetric,trace-freetensorAijinsteadof...Qij;thetransverse-traceless partofAijisdenedby(TT)Aij _PaiPbj 12PabPij_Aab.SinceAii= 0,wehaveAabPab= Aabnanbandso(TT)Aij(TT)Aij=_PaiPbj 12PabPij__PciPdj12PcdPij_AabAcd=_PacPbd12PabPcd_AabAcd=_PacPbd12nanbncnd_AabAcd= AabAab2AacAbcnanb +12AabAcdnanbncnd.Afterintegrationoverthesphere,accordingtoformulasderivedabove,wehave(againnotethatabAab= 0andAab=Aba)14_d2AacAbcnanb=13AacAbcab=13AabAab,14_d2AabAcdnanbncnd=115AabAcd(abcd +acbd +adbc) =215AabAab,andthus14_d2(TT)Aij(TT)Aij= AabAab_1 23+12215_ =25AabAab.Finally,substituting...QijinsteadofAij,wenddEdt=G214_d2_(TT)...Qij(TT)...Qij_ =G5...Qij...Qij_. (38)The angular brackets ) indicate that we must perform an averaging over spacetime domains. This means, for us, that weneedtoaverage overtime(sinceQijisafunctiononlyoftime). Averagingisperformedovertimescaleslargerthanthetypical timescale of change in the source. For instance, if the source is a rotating body, then averaging must be performedoverseveral periodsofrotation.10 Derivation: gravitational wavesinatspacetimeThemetricisassumedtobeoftheformg= +h,where= diag(1, 1, 1, 1) istheMinkowski metricforat space and hisasmallperturbation whichisassumedtofall otozeroquicklyatinnity. Westartwitha3+1decompositionof themetricperturbationhandcomputetheEinsteintensor(seeProblems7.2,7.3,7.4)intermsoftheperturbationvariables,,etc. Wealsodecomposethematterenergy-momentum tensorTandobtaintheEinsteinequations separately foreachcomponent (8.5). Theresultisthat(a)thevariables E,B,Ficanbesettozerobychoosingacoordinatesystem;(b)ifthereisnomatter(vacuum)thescalar andvector components of themetricperturbation are equaltozero; (c)thetensor component hijsatises thewaveequation(31).Solutionsofthewave equationinfourdimensionswithretarded boundaryconditioncanbewrittenusingtheknownGreensfunction. Forinstance,iff(t, r) = A(t, r) f(t, R) = 14_d3rA(t [r R[ , r)[r R[.Wewillusethisformulaforf (T)hijandA 16G(T)Tij. Now,weareinterestedindescribingtheradiationsentfarawaybyamatterdistribution,sowetakethelimit [R[ [r[,andthenwecanapproximately set(T)hij 4G[R[_d3r(T)Tij(t [r R[ , r).Nowweuseatrick(SeeHobson-Efstathiou-Lasenby, 17.9)toexpressthecomponentsTjithroughT00;itismucheasiertocompute withT00becausethis is just the energydensityof matter. Consider rst the tensor Tijrather thanitstransverse-traceless part(T)Tij. Thetrickistowritetheintegral(outofsheerluck)_d3r ab_rirj_Tab= 2_d3r Tij.33Thenweintegratebypartsandusetheconservationlaws(32),Tij,i= T0j,0, Tij,ij= T0j,0j= T00,00 T00; Tij,ij= T00.2_d3r Tij=_d3r Tab,abrirj= _d3r rirj T00.Now, weneedtoobtainthetransverse-tracelesspart of thetensor. Inprinciple, wehavetheformulas for this (seeProblem8.5). But theyareverycomplicated. AshortcutistonoticethattheprojectionoperatorPabdoesthejob(Problems9.1and9.2),atleastintheleadingorderin1/[R[. (Weareonlyinterestedineverythingtoleadingorderin1/ [R[sinceallsmallertermswillnotgiveanyuxofradiatedenergy.)Theresultis(T)hik(R, t) =2G[R[(TT) Qik(t [R[),Qik(t) _d3r T00(r, t)_rirk13r2ik_. (39)ThetensorQikisthequadrupolemomentofenergydistribution;itisatracelessandsymmetrictensor. Inprinciple,wecouldjustusetheintegral_d3r T00(r, t)rirk, (40)becausethetransverse-traceless partsof(40)andofQikarethesame,butitismoreconvenient touseQik.Since we foundthe tensor perturbation(T)hij,now we would like to compute theenergy radiated inthegravitationalwaves. Forthisweneedtheenergy-momentumtensorof gravitational waves. Thisisarathernontrivial object, sinceingeneral thegravitational elddoes not haveanyenergy-momentumtensor. Inthecaseof gravitational waves inat backgroundspacetime, onecandene some quantity(GW)Twhichlooks like the energy-momentumtensor ofgravitationalwaves(butactuallyisnotevenagenerallycovarianttensor). Wewill computethisquantitybelow. Thisquantityisuseful becauseitgivesthecorrectvalueoftheenergyafteroneintegratesoveralargeregionofspacetime.Thereal justicationfor usingthis procedureis complicatedandis beyondthescopeof this introductorycourseofGeneral Relativity. Wewill onlyshowaheuristicjustication, whichisthefollowing. Gravitationissensitivetoeverykindof energy, becausetheenergy-momentumtensoractsasa source forgravity(itisontheright-handsideof theEinsteinequation). Sogravitationshouldbealsosensitivetotheenergyingravitationalwaves. Oneexpectsthattheenergy-momentumtensorforgravitational waves,(GW)T(if weknowhowtocomputeit), will actasanadditionalsourceforgravity, likeeveryotherenergy-momentumtensorforotherkindsof matter. Wewill guesstheformulafor(GW)Tasfollows. WecanwritetheEinsteinequationandexpanditinpowersoftheperturbationh:G [+h] = G(1)[h] +G(2)[h] +... = 8GT . (41)HereG(1)istherst-orderEinsteintensor, G(2)isthesecond-orderetc. Firstwesolveonlytorst-orderinh(thisiswhatwehavebeendoingsofar)andthenwewillgetanapproximatesolutionh(1):G(1)[h(1)] = 8GT . (42)Thissolutiondisregardstheeectof gravitationalwavesandonlytakesintoaccounttheeectofmatterT . Wecantrytogetamoreprecisesolutionbyusingthesecond-order termsinEq.(41). Thenwewillgetacorrectionh(2)tothesolution;thesolution g= +h(1)+h(2)willbemoreprecise. FromEq.(41)wendG(1)[h(1)+h(2)] +G(2)[h(1)] = 8GT .NowthisissimilartoEq.(42),butitlooksasifthereisanadditionaltermintheenergy-momentumtensor,whichwemayrewriteasG(1)[h(1)+h(2)] = 8G_T+(GW)T_,(GW)T 18GG(2)[h(1)].ThismotivatesustosaythattheEMTforgravitationalwavesisgivenbythisformula. Butofcoursethisisnotarealderivationbecausethisdoesnotshowwhythequantity(GW)Thasanythingtodowiththeenergycarriedbywaves.Thesecond-ordertermsG(2)arecomputedinProblem9.3. Theresultisusedtocomputethepowerradiatedingravitational waves(Problem 9.4). NotethatthecalculationofG(2)usesaveraging overspacetimeinanessentialway.Thus, theresultisanaveragedpowerradiatedduringalongtimemuchlongerthanthetypical timescaleof changeinthesourcesandaveraged overlargedistances,muchlarger thanthetypicallengthscaleofthesources. Thiskindofaveraging isassumedinEq.(38). Itremainsunclearexactlyhowoneperformsaveraging overspaceandtime.Onecanusetheformula (38) tocomputethegravitational radiation emittedbynonrelativistic matterfaraway fromthoseplaceswherethematteriscontained. Thedistributionoftheenergydensity,T00(r, t), shouldbegiven. ThenonecomputesthetensorQijaccordingtoEq. (39), byintegratingoverspacewherethematteriscontained. Finally, onecomputesthethirdderivative...Qij,thetrace,andaveragesoverlongtime,asindicatedinEq.(38). Ifwewanttoinsertfactorsofc,wereplaceGbyGc9.34WiSe06 T VI General Relativity Problem Sheet 1http://www.theorie.physik.uni-muenchen.de/~serge/WS06-T6/1 Coordinates and 1-forms1.1 Invertible transformationsUnder what conditions is a coordinate transformation = (x) invertible in a neighborhood of some pointx?1.2 Examples of coordinate transformationsThefollowingcoordinatetransformationsaregiven, mappingthestandardEuclideancoordinates(x, y)or(x, y, z) into new coordinates.1. In a two-dimensional plane, (x, y) (u, v), where < u, v< +:x = u + uv2+13u3,y= v + vu2+13v3.2. In a three-dimensional space, (x, y, z) (r, , ), where < r< +, 0 < +, 0 < 2:x = r sinh cos ,y = r sinh sin,z= r cosh.3. In a three-dimensional space, (x, y, z) (r, , ), where 0 r< +, 0 , 0 < 2:x = r sin cos ,y= r sin sin ,z= r cos .The following questions must be answered in all three cases:(a)Findthesubdomaincoveredbythenewcoordinates. Hint: Considere.g.therangeofxatconstantvalue of y.(b) Find the points where the new coordinates do not specify a one-to-one invertible transformation (sin-gular points).(c) If singular points exist, give a geometric interpretation.1.3 Basis in tangent spaceProve that the vectorse=xare linearly independent.1.4 Differentials of functions as 1-formsIff(x) is a function of coordinates x, then one denes the 1-form df(called the differential of the functionf) asdf

fxdx. (1)Computed(x),d(x2),d(xy),d(x + y). Compute the 1-forms df,dg, dh, where the functions f, g, h are denedas follows,f(x, y, z) =4x2y + x3z,g(x, y) =3

x2+ y2,h(x, y) =arctan(x + y) + arctan(x y)+ arctan2xx2y21.1.5 Basis in cotangent spaceShow that the 1-forms dx1, ..., dxncomprise a basis in the space of 1-forms at any point M. Show that< dx,x>= .1.6 Linearly independent 1-formsCheck whether the following sets of 1-forms are linearly independent at each point of the 2-dimensional or the3-dimensional space respectively. If not, determine the points where these sets are linearly dependent.1. Two 1-forms d(excos y), d(exsin y).2. Two 1-forms (1 + y)dx 2xydy, 8dx.3. Three 1-forms dx + dy, dx + dz, dy + dz.4. Three 1-forms dx dy, dy dz, dz dx.1.7 Transformation law for 1-formsDerive the transformation law for 1-forms,d x= xxdx, (2)under a coordinate transformation x x= x(x).1.8 Examples of transformationsConsiderthecoordinatetransformation(x, y) (u, v)denedinProblem1.2(1). Transformthefollowing1-form,= d1x + y,into the coordinates (u, v) in two ways:(a) By a direct substitution of the new coordinates.(b) By using the transformation law (2).2WiSe06TVIGeneralRelativity SupplementalProblemSheet1http://www.theorie.physik.uni-muenchen.de/~serge/WS06-T6/2Dsurfacesembeddedin3DEuclideanspace1ATangentplaneConsiderthesurfacegivenbyz= hexp

122(x2+ y2)

.Ifgravityactsinthenegativez-direction,atwhatpointswillaballrollingalongthissurfaceexperiencethegreatest acceleration?Findthetangentplaneatoneofthesepoints.1BInducedmetricFindthemetricforthesurfacegivenparametrically byx = a sin2 cos ,y = a sin2 sin,z = a cos sin,where,asusual, [0, )and [0, 2).Isthemetricwelldenedat= 0?Doyouthinkthesurfaceiswelldenedthere?1CEmbeddingwaves1. Sketchthesurfacegivenbyx =cos v2 sin u,y =sin v2 sin u,z =cos u2 sin u,whereu, v [0, 2).(Hint: Considertheintersectionofthesurfacewiththeplaney= 0.Whathappensforgeneralv?)2. Findthenormalvectorandthetangentplanetothissurfaceatpoint(u, v).3. Determinetheinducedmetriconthesurface. Thenconsiderthe2DvectorVa= (cos v, sin v),i.e.V= cos vv+ sinvu,denedwithinthesurface. IsVaaunitvector?Whatarethe3DEuclideancomponentsofthevectorVinthe3Dspace? Showthatthe3DcomponentsofthevectorV everywherelieinthetangentplanetothesurface.WiSe06TVIGeneralRelativity ProblemSheet2http://www.theorie.physik.uni-muenchen.de/~serge/WS06-T6/2 Tensors2.1 DenitionoftensorproductIf1and2are1-forms,theirtensorproduct12isdenedasafunctiononpairsofvectors:(12)(v1, v2) = 1, v1 2, v2 . (1)Let1= dx + 2ydy,2= 2dybe1-forms ona2-dimensionalspaceandv1= 3/x,v2= x(/x + /y)bevectorelds(alsodenedinthis2-dimensionalspace).Justforthisproblem,letusdenoteT 12.(a)ComputeT(v1, v1).(b)ComputeT(v1, v2).(c)ShowthatT(a + b, u) = T(a, u) + T(b, u), (2)wherea, b, uarevectorsandisanumber). Thesamepropertyholdsforthesecondargumentof T. SuchfunctionsTarecalledbilinear.(d) Showthat all bilinear functions of pairs of 2-dimensional vectors belongtoavector space of suchfunctions.Showthatthetensorproductsdx dx,dx dy,dy dx,dy dyformabasisinthatspace.(Thatspaceiscalledthespaceoftensorsofrank0 + 2.)2.2 General tensors(a)Ageneraltensorofrankr + sisdenedasamultilinearfunctiononsetsofr1-formsfjandsvectorsvj(multilinearmeanslinearineveryargument). Anexampleofatensorofrankr + sisatensorproductofrvectorse1, ..., er, ands1-forms1, ..., s, denotedbye1 ... er 1 ... s. Thistensorisafunctionthatactsonasetofr1-formsfjandsvectorsvjviatheformulae1... er1... s(f1, ..., fr, v1, ..., vs)= f1, e1 ... fr, er 1, v1 ... s, vs .(This is a generalization of Eq. (1) to tensors of rank r +s.) Show that this function is linear in every argument.Suchfunctionsarecalledr + s-linearfunctions.Showthatallr + s-linearfunctionsformavectorspace.Thisvectorspaceiscalledthespaceoftensorsofrankr + s.(b)Letej, j=1, ..., N, andj,j=1, ..., Narebasesinthespaceofvectorsandinthespaceof1-formsrespectively(bothspaceshavedimensionN).Showthatthesetoftensorse1... er1... s(3)formabasisinthespaceofr + s-tensors(wherejandjexhaustallpossiblecombinationsofindices).Notethatthissetcontainsnr+sbasistensors.2.3 Example(a) Let T be a bilinear function of two vectors with vector values, i.e. T(v1, v2) is a vector if v1, v2are vectors.GiveasimpleexampleofsuchTasatensoranddetermineitsrank.(b)AparticularexampleofsuchatensorTin3-dimensionalEuclideanspaceisthefollowingfunction,T(v1, v2) = 2v1v2v1(n v2), (4)where nis axed vector. Show that thefunction T isbilinear in v1, v2.Determinethe components Tof thetensorTinanorthogonal basiswherethevectornhasthecomponentsn (n1, n2, n3).12.4 TransformationlawDerivethetransformation lawforthecomponentsT1...r1...sofatensorofrankr + s.2.5 Contractionsoftensorindices(a)Showthattheresultsofaddition,multiplicationbyscalar,tensormultiplication,andindexcontractionoftensorsareagaintensors.UsethedenitionoftensorfromProblem2.2.(b) Show that a contraction of indices in the same position (e.g. lower indices with lower indices, T) doesnotgenerallyyieldatensor.(c)ConsiderthetensorTdenedinProblem2.3(b)andcomputethecontractionT. Istheresultatensor?Ifso,determineitsrank.2.6 InvarianceoftheintervalShowthatthespacetimeintervalds2gdxdxisinvariantundercoordinatetransformationsx xifgarecomponentsofatensortransformingaccording tothetensortransformation lawg g=x xx x g. (5)2.7 Correspondencebetweenvectorsand1-formsForagivenmetricg, eachvectorvhasacorresponding1-formwhichweshall denotev. This1-formisdenedbyitsactiononanarbitrary vectorxasfollows,v, x = v x, (6)wherethescalarproductv xisdenedthroughthemetricg.Showthatthecomponentsofthe1-formvinthebasisdxarerelatedtothecomponentsofthevectorvinthebasis/xby= gv. (7)2.8 Examplesofspaceswithametric(a)Considertheusual,Euclidean3-dimensionalspacewiththemetricg(v1, v2) = v1 v2(n v1) (n v2), (8)where v1 v2is the usual scalar product, ab is the cross product, and n is a xed vector with components n.Compute the components of the tensor g. For which vectors n is the metric g nondegenerate (i.e. det g= 0)?(b)Answerthesamequestionsforthe2-dimensionalEuclideanspacewiththemetricg(v1, v2) = 3v1 v2 + (n v1)(n v2). (9)Notethatthecrossproductisundenedinthe2-dimensional space.(c)*Answerthesamequestionsforthemetric(9)nowdenedinanr-dimensionalEuclideanspace, r 3.(d)*Considera2-dimensionalsurfaceembeddedinthe3-dimensionalEuclideanspace,x = Rcoshu cos v, (10)y= Rcoshu sin v, (11)z= Rsinh u. (12)Determinethe2-dimensional metricginthebasisdu, dv.2WiSe06TVIGeneralRelativity SupplementalProblemSheet2http://www.theorie.physik.uni-muenchen.de/~serge/WS06-T6/Calculationswithtensorindices1AVectorequationsInthefollowingequations, thevectorxisunknownandallotherquantitiesareknown. Thesymbol denotesthecompletelyantisymmetrictensor. Determinetheunknownvectorxfrom the given data. In everycase,assume the generic choiceof data. Thismeans that everygivenscalar,vectorandtensorisnonzero(k, A, B,...),therearenoaccidentalcancellationsorlineardependencebetweengivenvectors,matricesarenondegenerate,etc.(a)kx+ xA=B(3-dimensionalvectors). Theassumptionofthe generic caseisk= 0andAandBlinearlyindependent.(b)xA= B,xC= k(3-dimensionalvectors).(c)xA= k,xB= l(2-dimensionalvectors).(d)xA= B(3-dimensionalvectorsandagiventensorA).1BTensorequationsIn the following equations, the tensorXis unknown and all other quantities are known. Thedimensionalityof the(Euclidean) spaceisindicated. DetermineXunder theassumptionthatallgivenquantitiesaregeneric.(a)X= X,XA= B,X= 0,XB= 0,whereAB= 0(2-dimensional).(b)X= X,XA= B,XB= 0,whereAB= 0(3-dimensionalvectors).1CDegeneracyofthemetric(a)Atwo-dimensionalspacewithcoordinates(x, y)hasthemetricgivenasabilinearformg= y2dx dx + (x2+ 1)(dx dy + dy dx). (1)Isthemetricnondegenerateatallpoints(x, y)?(b)Thesamequestionforthen-dimensionalmetricoftheformg=

1 + r2

AA, (2)whereAisagivenvectorandr2 xxisthesquaredEuclideandistance.WiSe06TVIGeneralRelativity ProblemSheet3http://www.theorie.physik.uni-muenchen.de/~serge/WS06-T6/3 TheChristoel symbol 3.1 Transformations1InatspacewithstandardEuclideancoordinatesandarbitrarycoordinatesx=x(), theChristoelsymbolcanbefoundas=2xxx .Derivethetransformation lawforbetweenarbitrary coordinatesystemsxand x:= xxx xx x+2x x x xx. (1)3.2 Transformations2Show that the Christoel symbol must transform according to Eq. (1) not only in at space but also in arbitraryspace. Hint: considerthecovariant derivativeofavectoreld,A;=AxA,anddemandthatthecomponentsA;transform asatensor.3.3 CovariantderivativesDerivetheexplicitformofthecovariant derivativeT;foratensoreldT.3.4 TheLeibnitzruleProvetheLeibnitzruleinthefollowingspeciccase,

AB

;= A;B+AB;.3.5 Locallyinertial referenceframeSupposethattheChristoel symbolatapointx(0)insomecoordinatesystemxhasthevalue(0)andissymmetric,(0)= (0).Thenalocallyinertialsystematpointx0canbeconstructedbydeningthenewcoordinates(x) = xx(0) +12

xx(0)

xx(0)

(0).Thepointx0inthenewcoordinatesistheorigin=0. ProveexplicitlythattheChristoel symbol, whentransformedtothenewcoordinates,isequaltozeroatthepoint= 0.WiSe06TVIGeneralRelativity ProblemSheet4http://www.theorie.physik.uni-muenchen.de/~serge/WS06-T6/4 Geodesicsandcurvature4.1 Geodesics(a)Showthatthegeodesicequationcanbewritteninthefollowingform,duds12gxuu= 0. (1)(b)Showthatguuisconstantalongageodesic.4.2 CommutatorofcovariantderivativesShowthatu;;u;;= Ru, (2)wheretheRiemanntensorisdenedbyR=xx+ . (3)4.3 Parallel transportConsider a vectorAparallel-transported along a small closed curvex(s). Show that the changeinAaftertheparalleltransport canbeapproximatelyexpressedasA

(x)Adx12RA

xdx, (4)whereitisassumedthattheareawithintheclosedcurveisverysmall.Hint: Usealocallyinertialcoordinatesystemwhere= 0atonepoint. Also,showthat

xdx=

xdx. (5)4.4 Riemanntensor(a) Using the symmetry properties of the Riemann tensor R, compute the number of indepen-dentcomponentsofRinann-dimensionalspace(n 2).(b)ProvetheBianchiidentity: R; + R; + R;= 0.(c) Compute the Einstein tensor Gin an arbitrary two-dimensional space. Hint: First determinetheindependentcomponentsofR.4.5 LorentztransformationsDeterminethenumberofindependentparametersinLorentztransformations x=x, givenby matrices , and interpret these parameters. Hint: It is easier to consider innitesimal Lorentztransformations= + H,where 1andso2canbedisregarded.2WiSe06TVIGeneralRelativity ProblemSheet5http://www.theorie.physik.uni-muenchen.de/~serge/WS06-T6/5 Gravitationtheoryapplied5.1 RedshiftCalculatethegravitational redshiftatthesurfaceoftheEarthfortheverticaldistanceof1mbetweenthesenderandthereceiver. Samequestionfor1km.5.2 Energy-momentumtensor1Rewrite the conservation law T;= 0 explicitly in the nonrelativistic limit for an ideal uid, and show that theseequationscoincidewiththecontinuityequationandtheEulerequation.5.3 Energy-momentumtensor2TheEMTforamasslessscalar eldisT= ;;12;; .Show(usingtheconservationlaw)thattheequationofmotiontheeldis;;= 0.5.4 WeakgravityShowthatinthelimitof weakstaticgravitational eld(g00=1 + 2(x, y, z), andgisindependentof t)thefollowingrelationholds,R00 + O(2) ,whereistheordinary Laplaceoperator, xx + yy + zz.5.5 EquationsofmotionfromconservationlawTheEMTforapointparticleofmassm0movingalongaworldlinex(s)canbeexpressedasT=1gm0

dsdxdsdxds(4)(xx(s)) .ShowthattheconservationlawT;= 0impliesthegeodesicequationforx(s).Hint: Firstderivetherelations=1gx(g) ,T;=1gx (gT) + T.WiSe06TVIGeneralRelativity ProblemSheet6http://www.theorie.physik.uni-muenchen.de/~serge/WS06-T6/6 Thegravitational eld6.1 DegreesoffreedomUsingthescheme developedinthelecture, computethenumber of degrees of freedomintheelectromagneticeld,takingintoaccountthepresenceofchargesandcurrents.6.2 SphericallysymmetricspacetimeCompute the Ricci tensor Rand the curvature scalar R for a spherically symmetric gravitationaleld. Assumethatthemetrichastheformg=e(t,r)0 0 00 e(t,r)0 00 0 r200 0 0 r2sin2.WritethecorrespondingEinsteinequationsinvacuum(T= 0).6.3 MotioninSchwarzschildspacetimeDerive the equation for the covariant component u1of the 4-velocity of a particle in Schwarzschildspacetime (u1() f1(r) r, f(r) = 1rg/r). Verify that this equation follows from Eqs. (1)(4)giveninthelecture:f t2f1 r2r2 2r2sin2 2= K (= uu), (1)dd

f t

= 0, (u0) (2)dd

r2

+ r2sin cos 2= 0, (u2) (3)dd

r2sin2

= 0, (u3) (4)wheretheoverdot()denotesd/d andthesphericalcoordinatesare{x0, x1, x2, x3} {t, r, , }.WiSe06TVIGeneralRelativity ProblemSheet7http://www.theorie.physik.uni-muenchen.de/~serge/WS06-T6/7 Weakgravitational elds7.1 Gravitational bendingoflightVerifythatthegravitational bendingoflightpassingneartheSunis= 1.75RRwhereRisthe distanceat whichthe light passesfrom the centeroftheSun andRis theradiusoftheSun.7.2 EinsteintensorforweakeldDerivethefollowingexpressionfortheEinsteintensorduetoaweakgravitationaleld,G=12

h,,h,, + h,, + h,,

+ O(h2), (1)whereh= h12 h.7.3 Gravitational perturbationsIDerivetheexpressions(showninthelecture)fortheEinsteintensorGintermsofthescalar,vector,andtensorperturbationsofthegravitational eld. ThebackgroundistheatMinkowskispacetime,(0)g= ,andthemetricisg00= 1 + 2, g0i= B,i + Si, gij= ij + 2ij + 2E,ij + Fi,j + Fj,i + hij. (2)7.4 Gravitational perturbationsIIDerive the transformation laws for the scalar, vector, and tensor perturbations of the gravitationaleld,underaninnitesimalchangeofthecoordinates, x= x+ (x). (3)Note: Itisconvenienttodecomposeas= (0, i + ,i),where0andarescalarfunctionsandi,i = 0.WiSe06TVIGeneralRelativity ProblemSheet8http://www.theorie.physik.uni-muenchen.de/~serge/WS06T6/8 Gravitational radiationI8.1 GaugeinvariantvariablesVerify that the following combinations of metric perturbations, D = B+Eand SiFi, are gauge-invariant.8.2 Detectinggravitational wavesLight noninteracting particles are situated in thexyplane in free space. A plane gravitational wave propagatinginthezdirectionpasses through thering. Themetric isof theformg= +h,wherehncontains onlythepuretensorcomponent,h=0 0 0 00 A+A00 AA+00 0 0 0exp [i (t z)] . (1)Describethedeformationoftheshapeoftheringduetothegravitationalwave. ConsidercasesA+=0, A=0andA+= 0, A= 0.8.3 PoissonequationDerivethesolutionofthefollowingdierentialequation,(x) = 4(x), (2)withboundaryconditions 0at|x| .8.4 Metricperturbations1DetermineanexplicitexpressionforthroughT0i,i,whereirepresents thescalarpartofT0i.8.5 Metricperturbations2Verifythattheequation12( SiFi) = 8Gi(3)which follows from vector part of thespatial Einstein equation, also follows from other components of the Einsteinequationandfromtheconservation law(asderivedinthelecture).WiSe06TVIGeneralRelativity ProblemSheet9http://www.theorie.physik.uni-muenchen.de/~serge/WS06-T6/9 Gravitational radiationII9.1 ProjectionofthemattertensorTheprojectionoperatorPijisdenedbyPij= ijninj, nini= 1, niRiR . (1)Showthattheprojectedtensor(T)Xik(t, |

R|)denedby(T)Xik= PiaXabPbk12PikPabXab, Xik(t, R)

d3r rirkT00(t |R| , r), (2)hasthefollowingproperties,a)(T)Xii= 0; (3)b)(T)Xik,i= O(X |R|1), (4)thatis,(T)Xikistransverse-traceless uptotermsoforder|R|1.9.2 MattersourcesVerifythat(T)Xik=(T)Qik,where(T)Xik(t |

R|)istheprojectedtensordenedinProblem 9.1and(T)Qik= PiaQabPbk12PikPabQab, (5)Qik

(rirk13ik r2) T00d3r. (6)9.3 Energy-momentumtensorofgravitational wavesComputethesecond-order termsG(2),i.e.terms quadratic inh,of theEinstein tensorGfor small perturba-tionsinat space,g= +h,whereonlythetransverse andtraceless part(T)hikisnonzero. Verifythattheenergy-momentum tensorofgravitational wavesinvacuum(T= 0formatter)is(GW)T 18G

G(2)

=132G

(T)hik,(T)hik,

. (7)9.4 PowerofemittedradiationShowthattherateofenergyloss(energylostperunittime)isdEdt= G8

d2(T)...Qik(T)...Qik= G5...Qik...Qik. (8)Here the integration goes over all directions niin 2-sphere. In the calculation, derive and use the following relations,

nlnmd24=13lm, (9)

nlnmnknrd24=115(lmkr+lkmr+lkmr). (10)