On Feb 13, 5:10 am, Edward Green wrote:
Suppose we had two masses in relative motion, so that at least one
mass must be in motion in any coordinate system (at least in
approximately locally Lorentz coordinates which encompass both
masses).

At least one of the masses therefore has kinetic energy. Must we
include this kinetic energy in the stress energy tensor, or can we
simply use the rest mass of each as the energy, so long as we treat
the masses discretely?

Follow on question: assuming the answer is that we may treat each
mass as contributing its rest mass alone, provided we track the masses
independently in the solution, we now begin to multiply the number of
independent bodies moving in some volume of (approximately flat)
space, while we hold total mass fixed.

Eventually our independent masses will appear to be a swarm of dust
particles, and we will elect to treat then on average, rather than
discretely. At this point, presumably, we must include their kinetic
energy as a component of their energy in the stress energy tensor. We
also must presumably include their pressure -- as a pressure is
imputed to a swarm of particles sufficiently small so that we do not
resolve them.

Since both pressure and kinetic energy stem from the same root: that
the particles are in relative motion, are we over-counting if we
include both?

Yea.

The stress energy tensor for one mass is T_uv = m u_u u_v wnere u is
the [contravariant] four velocity.

To get the dust stress energy tensor, you sum over all the masses and
then approximate by using the bulk behavior of the dust. If you
include the pressure - a statistical quantity that is a function of
kinetic energy, you are already counting the kinetic contribution.