Re: tensor "reverse trace" operation?

In article <1170177482.331004.253440@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>, Mike says...

The wikipedia article on the Einstein field equations makes the
statement that
R_uv = (1/2)*R*g_uv
reduces to
R_uv = 0
by performing the "reverse trace". Can someone explain this to me?

Hmm. I'm not sure what is meant by "reverse trace", but I do
understand the conclusion:

Start with
R_uv = 1/2 R g_uv

multiplying both sides by g^uv and summing gives

sum over u,v of R_uv g^uv = sum over u,v 1/2 R g_uv g^uv

The left side gives, by definition, R.
The right side gives 2R (because sum over u,v g_uv g^uv = 4).

so we have

R = 2R

which implies R=0.

--
Daryl McCullough
Ithaca, NY

.

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