Question about reverse tracing the Einstein field equations

[space-time]

From what I know, to get the reverse trace form of the Einstein field equations, you must multiply both sides by g^ab (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:

R_ab- \frac{1}{2}g_abR= kT_ab (where k= (8\piG)/c⁴)

into this:

R- \frac{1}{2}R = kT

which equals:

\frac{1}{2}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.

 

[Nugatory]

Your problem is that $g^{ab}g_{ab}=4$ when you do the summation.

 

[space-time]

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.

 

[pervect]

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).