On Feb 16, 12:45 pm, Tom Roberts wrote:
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?

The energy-momentum tensor automatically includes the motion of an
object. One must include both its kinetic energy and its momentum, in
addition to its mass.

Kinetic energy is observer dependent, and so is the momentum. This is
also true for the observed mass. Thus, the energy-momentum tensor
must be observer dependent as well. Since it is the energy-momentum
tensor that equates with the Einstein tensor, the field equations
allow an observer dependent quantity to shape the curvature of
spacetime that is supposed to be observer independent or invariant.
Please give me a date when you personally will be in peace with this
particular self-inconsistency.

In relativity, a pointlike mass m contributes to
the energy-momentum tensor:
T^ij(x) = m delta^4(x-X(t))) U^i U^j
Where U is the 4-momentum of the object and X(t) is its trajectory; x
represents the 4 coordinates on spacetime.

Properly identified according to Hilbert's Lagrangian that satisfies
as an density to the Einstein-Hilbert action, the energy-momentum
tensor is described as follows.

T_ij oc rho g_ij

Where

** g_ij = Elements of the metric, the matrix
** rho = mass density

So, where did you get that energy-momentum tensor?

When projected onto the locally inertial frame in which it is at rest,
only U^00 is nonzero, and corresponds to its mass density.

Follow on question: assuming the answer is that we may treat each
mass as contributing its rest mass alone, [...]

You cannot do so, except in its rest frame.

The verdict depends on how you pull out that energy-momentum tensor
from. shrug

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. [...]

Yes. A collection of non-interacting pointlike masses is called "dust",
and the energy-momentum tensor for dust with a mass density \rho is:
T^ij(x) = \rho(x) U^i U^j
Where U is the 4-momentum of the infinitesimal element at x.

This still does not make much sense. It still has nothing to do with
the field equations that Hilbert derived before 1916.

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?

Not if you do it correctly.

It is less contradictory if you allow an invariant quantity to decide
how spacetime should be curved. shrug