Re: tensor "reverse trace" operation?



On Jan 30, 12:18 pm, "Mike" <michael.h.william...@xxxxxxxxx> wrote:
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?

-Mike

No performing involved. The trace of the Einstein tensor is the same
as the trace of the Ricci tensor, except with the opposite sign.


.

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