Question about reverse tracing the Einstein field equations

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  • #1
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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.
 

Answers and Replies

  • #2
Nugatory
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Your problem is that ##g^{ab}g_{ab}=4## when you do the summation.
 
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  • #3
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Your problem is that ##g^{ab}g_{ab}=4## when you do the summation.
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.
 
  • #4
pervect
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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.
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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