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Einstein Tensor

  1. Dec 12, 2008 #1
    G_ab = R_ab - 1/2 * R * g_ab

    and

    R = g^(ab) * R_ab

    then

    G_ab = R_ab - 1/2 * R_ab * (g^(ab) * g_ab)
    =1/2 * R_ab

    for this abomination someone please correct me.
     
  2. jcsd
  3. Dec 12, 2008 #2

    George Jones

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    No. a and b are free indices on the right, so a and b have to be free indices on right, i.e., they cannot be repeated and summed over.

    Use [itex]R = g^{cd} R_{cd}[/itex] instead of [itex]R = g^{ab} R_{ab}[/itex].

    Also, what does [itex]g^{ab} g_{ab}[/itex] equal?
     
  4. Dec 12, 2008 #3
    g^(ab) * g_ab = I, the Identity Matrix only if a square matrix and when dotted with R_ab gives you back R_ab.
     
  5. Dec 12, 2008 #4

    George Jones

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    No.

    It is true that

    [tex]g^{ab} g_{bc} = \delta^a_c.[/tex]

    Think about it some more.
     
  6. Dec 12, 2008 #5
    That term doesn't appear in the equation. What you have is:

    [tex]g^{cd} g_{ab}[/tex]

    Where the left metric tensor is contracted with the Ricci tensor. Remember not to repeat indices more than twice. It messes up the summation convention. Use different letters like George Jones said.
     
    Last edited: Dec 12, 2008
  7. Dec 17, 2008 #6
    If you use the equation:

    G_ab = R_ab - 1/2 * R _cd

    You will get all the elements of the Ricci Tensor in each cell. Is really what the equation is?
     
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