Friday, August 21, 2020

A Closer Look At The Ricci Tensor

First some preliminaries:  The Riemann curvature tensor was defined in my blog post of Dec. 27, 2014 as:   R n b m n

This tensor , as noted in that post, is of key importance in general relativity.   The Ricci tensor of the first kind is defined simply as a contraction of the Riemann tensor of the 2nd kind, i.e. the above  curvature tensor.  Thus:

 b m =   n b m n

Raising an index yields  the Ricci tensor of the second kind, e.g.  R b m 

It is completely determined by knowing the quantity  R b  m   for all vectors V i  of unit length. The tensor is obtained by defining a Ricci tensor of the 2nd kind thus:

  R b m   =  g bn  R  nm

The number of independent components of this tensor  in a space of N- dimensions is:
  ½ N (N + 1 ).  Where the  bn  denote the  g- tensor components with indices raised.

Hence, there will be three components if N = 2,  six components if N = 3 and ten  components if N = 4.  In the latter we have the case for relativistic 4 – dimensional space-time.

We consider here the simplified case for N = 2 and let the metric of interest be *:

g 11    =   1,     g 22 =  1 , 

For this (N=2) case:

R =  11  ( R’ 11 ) +  g  22    ( R’ 21   )


Where:  11      =   1,     g  22 =  1/  1 , 

(Note:  Value of  22   is not the same as  g  22  !)

 R 11  =  g  22   ( R’ 21   ) =    (1/  1  )( -1/ 1 )

 R 22 = 11  ( R’ 12   )  =  (1)( -1/ 1 ) 

  
R (Ricci) =  11  ( R  11 )  + g  22    ( R 22   )

R (Ricci) =   11  (1/  1  )( -1/ 1 )  + g  22    ( - 1/ 1 )     

R (Ricci) =  

= (1) [- 1/  (1 ) 2 ]  +  (1/  1  ) [- 1/ 1  ]

  = - 1/  (1 ) 2     - 1/  (1 ) 2      =   - 2 /  (1 ) 2      

The reader should bear in mind this is the simplest form of the calculation and while in this instance the components are proportional to the components of the metric tensor, this is not true for spaces of higher dimension.  For example, if N= 3, one has six components and the final equation is written:

R (Ricci) =  11  ( R 11 )  +  g  22  ( R 22 ) +  g   33   ( R 33 )


Where:  R 11 =    g  22     R 2112


R 22 =   11   R 1221 +    g   33    R 3223

R 33 =    g  22     R 2332

----------------------------
*  Assume a metric given by the 'g' values:


g 11    =   1,     g 22 1 ,    g 33 2

Then the nonzero Christoffel symbols have values:

G 1 22    =  -1 ,    G 2 12    G 2 21    = 1/ 1


G 3 23    =  G 3 32     =   - 1/ 2

E. g.   Find:   g 11   G 2 21     


g 11   G 2 21      =  (1) (1/ 1) =  1/ 1

Find:     g 22 G 3 32     


g 22 G 3 32     = 1  (- 1/ 2 ) =  - 1  /2

Recall the  relations of  Riemann tensors to Christoffel values, e.g.

  R 1  212  =  - 1    -  G 1 22   G 1 11   -  G 2 21   G 1 22   = 0

Suggested Problem:

Obtain the Ricci tensor for the metric : 

g 11    =   1,     g 22 =  1 ,   g 33 =  2.

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