elfmotat

Each object measures their own clock as running normally and the others' as running slow.

I think everyone would benefit from reading this: http://www.colorado.edu/physics/phys1110/phys1110_fa10/Feynman_energy.pdf .

Yes, this is true with appropriately chosen units. First off all, the (inverse) metric is not simply diag(1,-1,-1,-1) in general. The metric is related to local matter/energy distribution by the field equations. The Minkowski metric, ημν, which represents flat spacetime is given by diag(1,-1,-1,-1). The Minkowski metric is a special case of the general metric in a vacuum (i.e. Tμν=0). Now, on to a bigger point: the signs actually depend on your choice of metric signature. You can use the signature (+,-,-,-) as the above author did, or you can use the signature (-,+,+,+). If you use (-,+,+,+), the Minkowski metric becomes: ημν=diag(-1,1,1,1). The choice is completely arbitrary. As long as you pick one and stick with it, all of the consequences work out the same. Most people working with General Relativity tend to use the (-,+,+,+) signature simply because it's easier to deal with one minus than than three. People doing more advanced work, such as work involving spinors, tend to use (+,-,-,-) because it's easier to work with in those fields. T isn't diag(Tμμ), T=Tμμ. The "diag()" would imply that the trace of the SET is still a tensor, which it is not. T is a scalar. The sign of T and R depend on your metric signature. Using (+,-,-,-), T>0 and R<0. Using (-,+,+,+), T<0 and R>0. There is no "correct" choice. I'm not sure what your question is here.