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Ricci Tensor of Diagonal Metric Abstract View metadata, citation and similar papers at core.ac.uk brought to you by CORE provided by CERN Document Server February 1996 Ricci Tensor of Diagonal Metric K. Z. Win Department of Physics and Astronomy University of Massachusetts Amherst, MA 01003 Abstract Ecient formulae of Ricci tensor for an arbitrary diagonal metric are presented. Intro duction Calulation of the Ricci tensor is often a cumb ersome task. In this note useful formulae of the Ricci tensor are presented in equations (1) and (2) for the case of the diagonal metric tensor. Application of the formulae in computing the Ricci tensor of n-sphere is also presented. Derivation The sign conventions and notation of Wald [1] will b e followed. Latin indices are part of abstract index notation and Greek indices denote basis comp onents. The Riemann d d curvature tensor R is de ned by r r ! r r ! = R ! . The Ricci tensor abc a b c b a c abc d + where the Christo el in a co ordinate basis is R = @ @ symb ols are = g ( @ g + @ g @ g ) =2. For the rest of this note we assume that g is diagonal unless otherwise indicated. Also to avoid having to write down \ 6= " rep eatedly we will always take 6= and no sum will b e assumed on rep eated and . We rst note that =0if 6=6=. It means that at least two indices of P n must b e the same for it to b e nonvanishing. Also ln jg j = ln jg j where g and n are =1 determinant of the metric and dimension of manifold, resp ectively. Then it is easy to show that 2 = @ ln jg j for all and that 2 = g @ g . (Again 6= and no sum on and .) These two equations give 4 = g (@ g ) g (@ g )=(@ ln jg j)(@ ln jg j)=4 Using ab ove facts, we get, with each sum having upp er limit n and lower limit 1, X R =@ + @ (1=2)@ @ ln jg j 6=; +(1=2) @ ln jg j +(1=2) @ ln jg j X g g j =(1=2)@ @ ln j g 6=; q q p p @ ln @ ln @ ln jg j + jg j + jg j + jg j @ ln + g g +(1=2) @ ln j j +(1=2) @ ln j j g g g g 1 X g g =(1=2)@ @ ln j j g 6=; + + + + g g j +(1=2) @ ln j j +(1=2) @ ln j g g g g X g g 1 1 g g =(1=2)@ @ ln j j + j j + @ ln j @ ln j g 2 g g 2 g g 6=; or (remember 6= ) X g 4R =(@ ln jg j@ )@ ln j j +( $ ) @ ln jg j@ ln jg j (1) g g 6=; where ( $ ) stands for preceding terms with and interchanged.* A numb er of useful facts follow from the ab ove formula. First of all R =0 if x is an ignorable coordina te b ecause @ is in each term of the formula. If there are m ignorable co ordinates then maximum numb er of nonzero o -diagonal comp onents of the Ricci tensor is (n m)(n m 1)=2. Thus a sucient condition for the Ricci tensor to b e diagonal is that the (diagonal) metric b e a function only of one co ordinate. The Rob ertson-Walker metric with at spatial 2 2 2 2 2 2 sections, ds = dt + a(t) (dx + dy + dz ), satis es this condition and its Ricci tensor is consequently diagonal. If the (diagonal) metric is a function only of two co ordinates, say x and x , then all o diagonal comp onents of the Ricci tensor except R are zero. 2 2 2 2 2 2 2 For example, for Schwartzschild metric ds = B (r )dt + A(r )dr + r d + sin d only R needs to b e explicitly calculated, all other b eing zero. It is easily checked using r equation (1) that R is also zero. Symmetry arguments can also b e giveny as to why the r Ricci tensor is diagonal in this case. Secondly second derivative terms are identical ly zero whenever each component of the metric is a product of functions of one single coordina te Q n l l i.e. if, for all , g = f (x ) where some of f may b e one. A nal trivial corollary l l=1 * Alternative forms of equation (1) are P 4R = [( @ ln jg j @ )@ ln jg j +( $ )@ ln jg j@ ln jg j] and 6=; h i 2 P g @ ln 2@ @ ln jg j +( $ ) 8R = 6=; jg j y Page 178 of Weinb erg [2]. 2 of equation (1) is that the Ricci tensor is diagonal in 2-dimensions. This fact also follows trivially from the fact that in 2-dimensions, the Ricci tensor is the metric tensor (not necessarily diagonal) up to a factor of a scalar function. When n = 3, it is easy to nd an example of a diagonal metric which results in a non-diagonal Ricci tensor. For example, 2 2 2 2 ds = dx + xdy + ydz gives 4R =1=xy . xy Wenow calculate the diagonal comp onents of Ricci tensor. Wehave p p 2 jg j + @ ln jg j R =@ @ ln q X p 2 2 =@ ln jg j + @ @ ln jg j 6= X X p p + @ ln jg j + @ ln jg j 6= 6= X X X g 2 2 =(1=2)@ ln j j + @ g 6= 6= 6= r q X X p g jg j + @ ln j j + jg j + @ ln @ ln g 6= 6= h X g 2 =(1=4) ( @ ln jg j2@ )@ ln j j + @ 2 g 6= i p + @ ln jg j X g =(1=4) ( @ ln jg j2@ )@ ln j j + (1=2)@ (g @ g ) g 6= q p 2 p jg j @ ln jg j (g =2)(@ g )@ ln jg j + g (@ g ) @ ln X 1 g 1 2 = (@ ln jg j2@ )@ ln j j 2@ (g @ g )+(@ ln jg j) 4 g 4 6= jg j + g (@ g ) @ ln 2 g or h i X jg j g 2 j (@ ln jg j) + @ ln +2@ g @ g 4R =(@ ln jg j2@ )@ ln j 2 g g 6= (2) where 2@ acts everything on its right. The eciency of this formula is obvious for Schwartzschild metric given earlier. To compute R we note that all terms with @ are zero. tt t 3 P jg j Thus 4R = +2@ (g @ g ). Because g = B (r ) only the = r @ ln tt tt tt 2 6=t g tt 0 0 0 0 00 B B A B B term survives and we get, after very little computation, R = + + tt Ar 4A A B 2A where prime indicates di erentiation with resp ect to r .z Other diagonal comp onents are just as easy to compute. Application to n-sphere To illustrate the use of ab ove formulae we compute Ricci tensor of n-sphere which 2 in spherical co ordinates has diagonal metric g = 1 and g = g sin x for 1 11 +1;+1 P 1 n 1. First note that ln jg j = 2lnjsin x j and thus @ g =0 ifn =1 1. These facts will b e used rep eatedly in what follows. We rst compute the o diagonal comp onents of Ricci tensor. According to the general argument given earlier second derivative terms will b e zero. Let >. Then equation (1) b ecomes X [(@ ln jg j) @ ln jg j(@ ln jg j) @ ln jg j] 4R = > X =4 ( cot x cot x cot x cot x )=0 > We next compute the diagonal comp onents. Wehave, from equation (2), h i X X jg j 2 2 4R = 2@ ln jg j +(@ ln jg j) @ ln +2@ g @ g 2 g > < X jg j 1;1 2 +2@ g @ g = 2 2@ cot x 4 cot x @ ln 1 1 1 2 g > X jg j 2 @ ln 4@ ln j sin x j +2@ g @ g sin x 1 1;1 1 2 g 1;1 <1 2 X cos x 4 4 = 2 2 sin x sin x > X 1;1 @ ln jg j 2@ ln jg j +2@ g @ g 1 1 1 1 >1 X jg j 2 @ ln sin x +2@ g @ g 1 1;1 2 g 1;1 <1 d z Cf. page 178 of Weinb erg and note that his R is minus of Wald's. Also see abc problem 6.2 of [1]. 4 i h X 1;1 4 cot x +2@ g g sin 2x =4(n ) 2 cot x 1 1 1;1 1 1 >1 X jg j 2 @ ln sin x +2@ g @ g 1 1;1 2 g 1;1 <1 =4(n ) [2 cot x (n +1)4 cot x ] sin 2x 2 2 cos 2x 1 1 1 1 X jg j 2 sin x @ ln +2@ g @ g 1 1;1 2 g 1;1 <1 2 2 2 2 =4(n ) 4 cos x (n +1) 2 cos x + cos x sin x 1 1 1 1 X jg j 2 @ ln sin x +2@ g @ g 1 1;1 2 g 1;1 <1 2 =4 n cos x (n +1)+1 1 X jg j 2 @ ln sin x +2@ g @ g 1 1;1 2 g 1;1 <1 h i X jg j 2 = sin x 4(n +1) @ ln +2@ g @ g 1 1;1 2 g 1;1 <1 Thus X jg j 2 4R = sin x 4(n ) @ ln +2@ g @ g +1;+1 2 g < Wenow rewrite the expression inside the square brackets.
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