Revisiting The Ricci Tensor

 The Riemann curvature tensor was defined in an earlier blog post, e.g.

A Look At General Relativity and Tensors (Part 3)

 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:

R  _b m =   R 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 2^nd 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  g ^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.

Consider first the simplest case for N = 2 and let the metric of interest be *:

g ₁₁    =   1,     g ₂₂ =  x ¹ , 

For this (N=2) case:

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

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

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

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

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

  

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

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

R (Ricci) =  

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

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

More complex, if N= 3, one has six components and the final equation is written:

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

Where:  R ₁₁ =    g  ²²     R ₂₁₁₂

R ₂₂ =   g 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 =  x ¹ ,    g ₃₃ =  x ²

Then the nonzero Christoffel symbols have values:

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

G ³ ₂₃    =  G ³ ₃₂     =   - 1/ x ²

Example (1):   Find:   g ₁₁   G ² ₂₁     

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

Example (2):  Find:     g ₂₂ G ³ ₃₂     

g ₂₂ G ³ ₃₂     = x ¹  (- 1/ x ² ) =  - x ¹  / x 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 =  x ¹ ,   g ₃₃ =  x ².