Special Relativity and Riemannian Geometry. Department of Mathematical Sciences

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1 Tutoral Letter 06//018 Specal Relatvty and Reannan Geoetry APM3713 Seester Departent of Matheatcal Scences IMPORTANT INFORMATION: Ths tutoral letter contans the solutons to Assgnent 06. BAR CODE Learn wthout lts. unversty of south afrca

2 Meo for Assgnent 6 S 018 Equvalence and Tensor Algebra (Â 4) Queston 1: Tensor transforatons The transforaton equatons for transforng a contravarant tensor of rank one fro polar to Cartesan coordnates are A 01 = A 1 cos A r sn A 0 = A 1 sn + A r cos where x =(r, ) (derved n Exercse 4. n the textbook). Use these to transfor the tensor descrbed by [A ]=(1/ cos, r) to Cartesan coordnates x 0 =(x, y). Whatsthevalueof A 0? cos (1 + r ) tan + r cos * 1 r sn cos r cos r sn r tan The transforaton equatons for transforng a contravarant tensor of rank one fro polar to Cartesan coordnates are A 01 = A 1 cos A r sn A 0 = A 1 sn + A r cos The coponents of the tensor we want to transfor s gven as A 1 =1/ cos and A = r. Substtutng ths nto the above equatons gve A 01 = cos cos r sn 1

3 = 1 r sn A 0 = sn cos + r cos = tan + r cos Queston : Tensor expressons Whch of the followng tensor expressons s ncorrect? A = P j g j A j = P j, k g j g jk A k Ā kl = P p, r, x p x k x l A p rs = 1 P 3 =1 =3 jk = j k * + Opton (1): wrte Correct. The rules of rasng and lowerng an ndex s followed. We can also k A k j, k g j g jk A k = k = 1A 1 + A A +... na n In the last step the su has been expanded and sus k fro 1 to n. Reeberthat defned so that t s equal to 1 f = k, andzerof 6= k. So all the ters wll be equal to zero, except the ter where = k, andnthatcasewehave =1so that we can wrte g j g jk A k = A j, k k s Opton (): Ths s the correct transforaton for a tensor of ths for. Here bars were used n stead of pres to ndcate the other coordnate frae. The textbook uses pres, but ths can soetes becoe unclear, especally when wrtng by hand. Usng ether s fne, as long as you are consstent.

4 Opton (3): Ths s correct. Lowerng the frst ndex of the Chrstoffel coeffcent gves = g = 1 g = 1 1 g In the last step we used g g = The only non-vanshng ter wll be the one for whch = so that = 1 + Opton (4): Ths s correct. Expandng the su gves 3 = =1 = = 3 Opton (5): Ths s ncorrect. The Chrstoffel coeffcents are syetrc n ther lower ndces, but not n the upper and lower ndex. The syetry n ther lower ndces s a consequence of the syetry of the etrc tensor. We can wrte Snce g jk = 1 g l l lk + jl j k jk l = g, we can nterchange the ndces of all the etrcs wthn the su. jk = 1 g l l kl + lj j k kj l The frst two ters can also be swtched to gve jk = 1 g l l lj + kl k j 3 kj l

5 whch s exactly the defnton for kj. So therefore jk = kj, buttdoesnotholdthat jk = j k. Queston 3: Contracton Whch of the followng expressons are correct? l g k =3g k l g k = g kl l g k = g k l g k = g kl * l g k = g k In the su l g k all the ters n the su where 6= l, wll be zero (snce l =0f 6= l). The ter where = l wll be equal to g kl,.e. l g k = 0 lg k0 + 1 lg k l lg kl N l g kn = (0)g k0 +(0)g k (1)g kl +...+(0)g kn = g kl So the Kronecker delta can effectvely be used to replace one ndex wth another. Queston 4: Ensten feld equatons How any equatons does the followng expresson represent? R µ 1 Rg µ = applet µ 4

6 * The expresson represents 16 dfferent equatons. One for each possble cobnaton of µ and, where both ndces can have values fro 0 to 3, snce ths equaton s wrtten for four densonal spacete. The 16 possble cobnatons are # µ # µ # µ # µ