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Reverse-tracing Einstein's field equation

  1. Feb 16, 2008 #1
    How do I rewrite Einsteins famous field equation

    [tex]R_{ab} - \frac{1}{2} g_{ab} R = \frac{8 \pi G}{c^4} T_{ab}[/tex]

    into:

    [tex]R_{ab} = \frac{8 \pi G}{c^4} (T_{ab} - \frac{1}{2} g_{ab} T)[/tex]

    I've tried experimenting with reverse-tracing the Ricci-scalar, but I just don't get the right equation. The trace from these equations would yield [tex]R=-T[/tex]. Is this really correct?
     
  2. jcsd
  3. Feb 17, 2008 #2

    Mentz114

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    Multiply both sides by [tex]g^{ab}[/tex] to get

    [tex]\frac{1}{2}R = \kappa T[/tex]

    then substitute back.
     
  4. Feb 17, 2008 #3
    Ok, I insert the expression and get:

    [tex]R_{ab} - \frac{8 \pi G}{c^4} g_{ab} T = \frac{8 \pi G}{c^4} T_{ab}[/tex]

    [tex]R_{ab} = \frac{8 \pi G}{c^4} \left(T_{ab} + g_{ab} T \right)[/tex]

    Which as you can see is not entirely right. Have I missed something very obvious?
     
  5. Feb 17, 2008 #4
    Take trace of both sides of Einstein eq. Trace of the metric is 4. Trace of Ricci tensor is R. You get

    R = - k T

    then substitute that back for R and take the metric term on the other side.
     
  6. Feb 17, 2008 #5

    Mentz114

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    I made a mistake, [tex]g^{ab}g_{ab} = -1 [/tex].
     
  7. Feb 17, 2008 #6
    Check for an errant sign in this equation.
     
  8. Feb 17, 2008 #7
    Okay, that fixes it.
     
  9. Feb 17, 2008 #8
    [tex]g^{ab}g_{ab} = \delta^a_a = 4[/tex]
     
  10. Feb 17, 2008 #9

    Mentz114

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    Yes, you're correct. It's |g| = -1.
     
    Last edited: Feb 17, 2008
  11. Feb 17, 2008 #10
    How can |g| = abs(g) = -1 ? An absolute value can't be negative. For example a radius cannot be negative.
     
  12. Feb 17, 2008 #11
    |g| = 1 only in Special Relativity. In GR, the value of |g| changes with the coordinate system used but it's signature (the signs of the eigenvalues when you diagonalize it) doesn't change, depending on the sign convention it's either (-, +, +, +) or (+, -, -, -).
     
  13. Feb 17, 2008 #12
    Got it right now, thanks everybody.
     
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