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Question about reverse tracing the Einstein field equations

  1. Aug 18, 2014 #1
    From what I know, to get the reverse trace form of the Einstein field equations, you must multiply both sides by gab (I didn't have a lot of time to make this thread so I did not spend time finding the Greek letters in the latex).

    This turns:

    Rab- [itex]\frac{1}{2}[/itex]gabR= kTab (where k= (8[itex]\pi[/itex]G)/c4)

    into this:

    R- [itex]\frac{1}{2}[/itex]R = kT

    which equals:

    [itex]\frac{1}{2}[/itex]R = kT

    which yields

    R= 2kT= LT (I set L= 2k).

    However, many sources say that the reverse trace is:

    R= -LT

    Why is it negative? Is it based on sign signature?

    Note that I general, I use (- + + +) signature.
     
  2. jcsd
  3. Aug 18, 2014 #2

    Nugatory

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    Staff: Mentor

    Your problem is that ##g^{ab}g_{ab}=4## when you do the summation.
     
    Last edited: Aug 18, 2014
  4. Aug 18, 2014 #3
    So then this yields:

    R-2R= kT

    which equals

    -R = kT

    which yields:

    R = -kT

    I presume that this is the correct process. Is this what you are getting at? If so then I get it now. Thank you.
     
  5. Aug 18, 2014 #4

    pervect

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    Staff Emeritus
    Science Advisor

    It looks right. I think Baez has some lecture notes that do the derivation online that go through this, if you haven't seen them already (I suspect it could be your source).
     
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