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February 1996

Ricci of Diagonal Metric

K. Z. Win

Department of 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

. 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 and Greek indices denote 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 of , 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-. 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. Break up the sum into  =  1

2

and <1 and use g = g sin x . Then

 1;1 1



jg j

1;1

[ ] =4(n ) @ ln +2@ g @ g

1 1 1 

2

g



 

X

jg j



4@ ln j sin x j +2@ g @ g @ ln

 1    

2

g

1;1

<1

 

1;1

=4(n ) @ ln jg j2@ ln jg j +2@ g g sin 2x

1 1  1 1;1 1

 

X



jg j

2



+2@ g @ g sin x @ ln

  1;1 1 

2

g

1;1

<1

 

X



@ ln jg j 2  2 cot x +2@ =4(n ) sin 2x

1  1 1 1

>1

!

X

jg j

2



sin x @ ln +2@ g @ g

1    1;1

2

g

1;1

<1 5

=4(n ) [2 cot x (n  +1)4 cot x ] sin 2x 4 cos 2x

1 1 1 1

!

X

jg j

2



sin x @ ln g @ g +2@

1   1;1 

2

g

1;1

<1

h i

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 g @ g +2@

 1  1;1 

2

g

1;1

<1

h i

2

2 2

=4 n  cos x (n +1 ) + sin x + cos x

1 1 1

 

X

jg j

2 

@ ln sin x +2@ g @ g

 1   1;1

2

g

1;1

<1

 

X

jg j

2 2 

@ ln =4(n +1) sin x sin x +2@ g @ g

 1 1   1;1

2

g

1;1

<1

=4R



Next

h i  

P P

2

1

2

2

4R = 2@ ln jg j +(@ ln jg j) =4 cot x =4(n1)

2

11    1

1

>1 >1

sin x

1



All preceding calculations amount to proving that R =(n1)g . Let fv g b e the dual

 

a

basis of the tangent space where  lab els a particular basis vector. Then

   

R = R v v =(n1)g v v =(n1)g

ab   ab

a a

b b

This fact ab out n-sphere is also provable using the machinery of symmetric spaces. See

chapter 13 of [2].

Equations (1) and (2) were derived when attempt was made to establish any connec-

tion b etween diagonalityof R and that of g . Finally we remark that Dingle [3] has

 



listed values of T for the case of 4-dimensional diagonal metric.



Acknowledgements

Iwould like to thank Professor Jennie Traschen for much encouragement and discus-

sion and for useful comments on the manuscript. 6

References

1. Wald, R. M. 1984, (Chicago: University of Chicago Press).

2. Weinb erg, S. 1972, Gravitation and Cosmology: Principles and Applications of The

General Theory of Relativity (New York: Wiley).



3. Dingle, H. 1933, \Values of T and the Christo el Symb ols for a Line Elementof



Considerable Generality," Proc. Nat. Acad. Sci., 19, 559-563. 7