Final Exam. Bad Tensor Analysis Problem 1. a. A i + B j. b. A ij B j C ji. c. A ijk B k + C ijk B j. Answer 1a. Answer 1b.

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1 Finl Exm Plese ttempt ll of the following problems before the ue te. All problems count the sme even though some re more complex thn others. B Tensor Anlysis Problem Inex nottion, with the Einstein Summtion Convention is extremely exible n cn express ny kin of tensor opertion. However it cn be too exible becuse it cn express opertions tht o not exist. Here re few exmples of "B Tensor Anlysis." Explin wht is wrong with ech exmple. Do more thn just stte wht rule is violte n inicte why the opertion cnnot rise from vli geometricl opertions.. A i + B j Answer The violte rule here is tht you hve i erent free inexes in i erent terms. Tht correspons to ing i erent components of vectors together the x-component of one to the y-component of nother so it cnnot hppen. b. A ij B j C ji Answer b Here the inex j is repete more thn twice. Legitimte tensor opertions lwys involve just two inexes, one up n one own. The two types of inexes trnsform in opposite wys so the result in then invrint when the bsis vectors chnge. The expression given here woul epen on which bsis is use. c. A ijk B k + C ijk B j Answer c The i culty here is ctully just the sme s in prt. The free inexes on the rst term re i; j with k summe over while the free inexes on the secon term re i; k with j summe over. Tht correspons to ing i erent tensor components together n cnnot hppen.

2 . A bc K c = B bc J n Answer Here we hve i erent free inexes n lso hve i erent rnk tensors set equl to one nother. Tht cnnot hppen.

3 Metric Tensor Follies The following problems refer to the following sitution: Strt with the spcetime line element s = e (r) t + B (r) t ' + e r + R (r) + sin ' with coorintes x 0 = t; x = r; x = ; x 3 = ': Notice tht this metric is speci e by three functions ; b; R; of the single vrible, r. The worl line of prticulr observer is given by t = e r = r 0 = ' = p where is the curve prmeter n r 0 n p re constnts. Problem :. Use the line element to clculte how much proper time elpses for this observer for ech unit of coorinte time t. Answer The elpse proper time is given by the integrl Z t= = s t=0 v Z t= u = t e t t ' B t=0 e r R + sin!! ' Since r n re not chnging n = simpli es to v Z t= u = t e t t=0 ll long the worl-line, the expression t ' B! ' R 3

4 Evlute the erivtives from the equtions of the worl-line. t = e ; ' = p n = Z t= t=0 p Be p R p = e t: Nothing is chnging s function of t becuse the worl-line ws t constnt r, so the integrl is trivil. Z t= = e t p Be p R p t=0 = e p Be p R p b. Disply the metric tensor components g tht re inicte by this line element. Answer b The line element is given by the oubly summe expression: s = g x x Write the ouble sum out to see wht we hve here: s = g 00 t + g 0 tr + g 0 t + g 03 t' +g 0 rt + g r + g r + g 3 r' +g 0 t + g r + g + g 3 ' +g 30 't + g 3 'r + g 3 ' + g 33 ' Notice tht the cross terms ll pper twice, so we get s = g 00 t + g r + g + g 33 ' +g 0 tr + g 0 t + g 03 t' + ::: Compre this to the line element tht we hve: s = e t + B t ' + e r + R + R sin ' 4

5 n get the non-zero metric components: g 00 = e ; g = e ; g = R ; g 33 = R sin g 03 = B c. Fin the four-velocity vector of this observer. Answer c Metho : First n tngent vector to the worl line by just using the chin rule: = t t + r r + + ' ' From the equtions for the worl-line, t = e ; r = 0; = 0; ' = p so tht tngent vector is = e t + p ' The four-velocity will be in the irection of this tngent vector, but it must be normlize so tht u u = Tke so tht u = N u u = N e t + p N e ' t + p ' = N e t t + pe t ' + p ' ' But t t = g x 0 ; x 0 = g 00 5

6 n so forth, so n thus (with = =) = N e g 00 + pe g 03 + p g 33 = N e e + pe B + p R sin = N + pe B + p R sin N = The four velocity vector is then Use n u = p pe B p R e p pe B p R t + p p pe B p R ' Metho : Just plug stright into the e nition u = = = e t + p ' = s = e t + B t ' + e r + R + sin ' = e t B t ' e r R + sin ' = e t B t ' From the equtions of motion for the prticle, so n t = e ; r e R ' = p; r = = 0 = e e Be p R sin p = Be p R sin p = p Be p R sin p which gives the nl result: u = p e Be p R sin p t + p ' + sin!! ' 6

7 Problem 3. Wht motions leve this metric unchnge? (Note: There is nothing to clculte here. Just look t the metric.) Answer 3 Since nothing epens on t, trnsltions in the t coorinte leve the metric unchnge. The sme thing works for the ' coorinte. With the t' term present, prticulr xis for polr coorintes is picke out, so we o not hve sphericl symmetry. b. Note the vector els whose integrl curves correspon to the motions escribe in prt. (Agin, nothing to clculte. Just write own the nswer.) Answer 3b The metric is invrint uner trnsltions long the integrl curves of the vector els t n ' c. Is this spcetime sttic or sttionry? Answer 3c It is sttionry becuse it is invrint uner time trnsltion. However the t' term prevents it from being invrint uner time reversl, so it is not sttic. 7

8 Problem 4. The coorinte (or holonomic) bsis vectors t ; r ; ; ' re not orthonorml. Fin set of bsis vectors e tht re orthonorml Answer 4 First notice tht the only pir of vectors tht fil to be orthogonl to ech other re t n ' so strt by ning liner combintions of those two vectors tht re orthogonl to ech other. Try v = t + n ' n require it to be orthogonl to '. This is n rbitrry choice, by the wy. We coul just s esily n combintion orthogonl to t. Tht woul give i erent, but eqully vli orthonorml frme. Tht gives the result or t + n ' v ' = 0 ' = 0 g 03 + ng 33 = 0 n = g 03 =g 33 = B= R sin The vectors v; r ; ; ' to n their norms to get them normlize. B v v = t R sin ' v = B t R sin ' re ll orthogonl to ech other now n we just nee t B R sin ' = t B t R sin t ' + B R 4 sin 4 ' ' = g 00 B R sin g B 03 + R 4 sin 4 g 33 = e B R sin + B R 4 sin 4 R sin = e B R sin + B R sin = e B R sin 8

9 r r = g = e = g = R ' ' = g 33 = R sin so our orthonorml frme vectors re e 0 = q e + B R sin t B R sin ' e = e r e = R e 3 = R sin ' b. The coorinte (or holonomic) bsis forms t; r; ; ' re not orthonorml. Fin set of bsis forms! tht re orthonorml. Answer 4b We hve trnformtion from the holonomic bsis to the orthonorml bsis tht hs the mtrix form [e] = M x so the ul bsis will be given by the inverse mtrix [!] = [x] M In etil, we hve 0 0 e 0 B e C e A = B e 3 q e + B R sin 0 0 B R sin 0 e R R sin so tht the row of orthonorml bsis forms is given by q e + B R sin 0 C B A t r ' C A 9

10 ! 0!!! 3 = 0 = t r ' B q e + B R sin 0 0 B R sin 0 e R R sin q e + B R sin C A 0 = t r ' B q B R sin + e R sin B 0 0 R sin 0 e R R sin C A = q t B R sin + e R sin ; re B ; R; R' sin + t R sin or! 0 = t r! = e r e + B R sin! = R! 3 = B t + R sin ' R sin 0

11 The Conforml Trick The spcetime line element s = t + 4 x + y + z where is function of the spce coorintes x; y; z is not solution of Einstein s equtions, but it oes incorporte vluble trick the conforml fctor. Why the fourth power? For one thing it vois ipping either the signture of the metric or the orienttion of n orthonorml bsis. The following problems explore this metric. Problem 5 For this spcetime,. Wht woul be the e ective Newtonin grvittionl potentil governing the motion of slow-moving objects? Answer 5 From the equivlence principl lone, the motion of slow objects is governe entirely by the metric component g 00. Here tht component is the sme s for Minkowski spcctime, so there is no grvittionl potentil t ll n no Newtonin ccelertion of slow moving objects. b. Wht spcetime motions woul leve this spcetime metric unchnge? Is this spcetime sttic or sttionry? Answer 5b Since the function is n rbitrry function of the spce coorintes, there re not sptil motions tht leve the metric unchnge. However nothing epens on t so trnsltion n in t will leve the metric unchnge. Tht invrince mens the spcetime is t lest sttionry. There re no cross-terms in the metric to spoil time reversl, to the metric is lso sttic.

12 Problem 6 Clculte the connection coe cients, the Riemnn tensor components, the Ricci tensor components, n the Einstein tensor components for this spcetime. (Hint: You cn sve lot of e ort by expressing things in terms of Kronecker elts like ij inste of writing out components.) Wht hppens when you try to write Einstein s equtions with perfect ui stress-energy tensor? Answer 6 The strightforwr proceure is to just plug into the formuls. Strt by giving the non-zero metric tensor components g 00 = ; g ij = 4 ij g 00 = ; g ij = 4 ij It is useful to let Ltin inexes rnge from to 3 n use the Kronecker elt for the t spce metric components. Next, give the erivtives g ;. All of the ones with ny of the inexes equl to zero vnish, leving just g mn; = 4 3 ; mn Now work out the connection coe cients from the formul = g ( g + g g ) All of the terms with ny zero inexes vnish, leving just the sptil connection coe cients b = gr ( g rb + b g r r g b ) = 4 r 4 3 ; rb ;b r 4 3 ;r b = r ; rb + ;b r ;r b = ; b + ;b ; b Tht wsn t so hr. Now work out the Riemnn tensor components from the stnr formul: R = + Components with zero inexes ny where will vnish, so we just hve the spce components R bmn = m bn n bm + rm r bn rn r bm

13 Work on the erivtive terms rst. m bn = ;n b + ;b n ; bn ;m = ;m ;n b + ;b n ; bn + ;nm b + ;bm n ;m bn n bm = ;m b + ;b m ; bm ;n = ;n ;m b + ;b m ; bm + ;nm b + ;bn m ;n bm n the i erence: m bn n bm = ;m ;n b + ;b n ; bn + ;n ;m b + ;b m ; bm + ;nm b + ;bm n ;m bn ;nm b + ;bn m ;n bm = ;m ;n b + ;m ;b n ;m ; bn + ;n ;m b + ;n ;b m ;n ; bm + ;nm b + ;bm n ;m bn ;nm b ;bn m + ;n bm = ;n ;m b + ;n ;b m ;n ; bm ;m ;n b ;m ;b n + ;m ; bn + ;nm b + ;bm n ;m bn ;nm b ;bn m + ;n bm = ;n ;b m ;n ; bm ;m ;b n + ;m ; bn + ;bm n ;m bn ;bn m + ;n bm Now pull out fctor so tht we cn compre n cncel terms more esily. ( m bn n bm ) = ;n ;b m ;n ; bm ;m ;b n + ;m ; bn + ;bm n ;m bn ;bn m + ;n bm Next, strt on the prouct terms in the curvture. r rm bn = ;m r + ;r m ; rm ;n rb + ;b rn ;r bn = 4 ;m r + ;r m ; rm ;n rb + ;b rn ;r bn Pull out fctor so we cn see terms to cncel more esily. 3

14 4 rm r bn = = ;m r + ;r m ; rm ;n rb + ;b rn ;r bn = ;m r ;n rb + ;m r ;b rn ;m r ;r bn + ;r m ;n rb + ;r m ;b rn ;r m ;r bn ; rm ;n rb ; rm ;b rn + ; rm ;r bn = ;m ;n b + ;m ;b n ;m ; bn + ;b ;n m + ;n ;b m ;r ;r m bn ; ;n mb ; ;b mn + ; ;m bn Now put in the other prouct term n see wht cncels. r rn bm ) = 4 ( rm r bn = ;m ;n b + ;m ;b n ;m ; bn + ;b ;n m + ;n ;b m ;r ;r m bn ; ;n mb ; ;b mn + ; ;m bn ;n ;m b ;n ;b m + ;n ; bm ;b ;m n ;m ;b n + ;r ;r n bm + ; ;m nb + ; ;b mn ; ;n bm = ;n ;b m ;r ;r m bn + ; ;m bn ;m ;b n + ;r ;r n bm ; ;n bm = ;n ;b m ;m ;b n + ; ;m bn ; ;n bm + ;r ;r ( n bm m bn ) Multiply this result by so we get the sme fctor outsie s for the erivtive terms. ( rm r r bn rn bm ) = ;n ;b m ;m ;b n + ; ;m bn ; ;n bm + ;r ;r ( n bm m bn ) n put the erivtive n prouct terms together to get R bmn = ;n ;b m ;n ; bm ;m ;b n + ;m ; bn + ;bm n ;m bn ;bn m + ;n bm + ;n ;b m ;m ;b n + ; ;m bn ; ;n bm + ;r ;r ( n bm m bn ) R bmn = 3 ;n ;b m 3 ;n ; bm 3 ;m ;b n + 3 ;m ; bn + ;bm n ;m bn ;bn m + ;n bm + ;r ;r ( n bm m bn ) 4

15 so tht our result for the non-zero Riemnn tensor components is: R bmn = 3 ;n ;b m ;n ; bm ;m ;b n + ;m ; bn + ;bm n ;m bn ;bn m + ;n bm + ;r ;r ( n bm m bn ) Now go for the Ricci tensor. Contrct on ; m R bn = R bn = 3 ;n ;b ;n ; b ; ;b n + ; ; bn + ;b n ; bn ;bn + ;n b + ;r ;r ( n b bn ) R bn = 3 3 ;n ;b ;n ;b ;n ;b + ;r ;r bn + ;bn ;rr bn 3 ;bn + ;bn + ;r ;r ( nb 3 bn ) R bn = 3 3 ;n ;b ;n ;b ;n ;b + ;r ;r bn + ;bn ;rr bn 3 ;bn + ;bn + ;r ;r ( nb 3 bn ) = 3 ;n ;b + ;r ;r bn + ;rr bn ;bn 4 ;r ;r nb = 3 ;n ;b + 3 ;r ;r bn ;rr bn ;bn 4 ;r ;r nb = 3 ;n ;b ;r ;r bn ;rr bn ;bn Behol the Ricci Tensor components: R bn = Contrct gin on n; b. R bn = 3 ;n ;b ;r ;r bn ;rr bn ;bn Remember tht we hve to use the ctul inverse metric tensor g bn = 4 bn to rise n inex to o this contrction. R = R bng bn = R = 4 4 ;rr R bn 4 bn R = 8 5 ;rr = 8 r 5 5

16 Now go for the Einstein Tensor G = R g R The timelike components of the tensor re then G 0 = g 0R or G 00 = R = 4r 5 G 0i = 0 The spcelike components re G bn = = = R bn g bnr R bn 4 R bn + 4 g bn R 4 bn 8 r 5 = R bn + r bn or G bn = 3 ;n ;b ;r ;r bn ;rr bn ;bn + rr bn = 3 ;n ;b ;bn + rr bn ;r ;r bn G bn = 3 ;n ;b ;bn + r r r bn To nish the Einstein equtions, note tht the contrvrint stress-energy tensor components for n isotropic ui re The covrint components re then or T 00 = ; T 0n = 0; T bn = p bn T = g g T T 00 = g 00 g 00 T 00 = T 00 = T 0n = 0 T bn = g br g ns T rs = 8 br ns rs = 8 p bn T bn = 0 p bn 6

17 The Einstein equtions re then G 00 = 4 r 5 = 8 or G bn = 3 ;n ;b ;bn r r r bn = r = 5 0 p bn 3 ;n ;b ;bn r r r bn = 0 p bn It is pretty cler tht there re no nontrivil solutions of the secon set of (six inepenent) equtions, but tht is no surprise since our originl guess for the metric tensor correspone to hving no Newtonin grvittionl el t ll. 7