On Feb 19, 9:15 pm, Koobee Wublee wrote:
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.

Always the "observed" qualifier, eh? The weasel words are one of the
many annoying aspects of you. When asked to calculate the surface area
of a sphere of constant radius at a specific slice of time, you weasel
out of the simple calculation by whining about "observed" radius.

Thus, the energy-momentum tensor
must be observer dependent as well.

False logic leading to a wrong conclusion from someone who doesn't
know any better. The stress-energy tensor for a lone particle isn't
made out of the particle's 3-momentum OR its' kinetic energy.

1) Its' a tensor, but we both know multilinear objects are beyond your
grasp but not your reach so understanding is not expected of you.

2) The STRESS-ENERGY tensor is not a function of observer dependent
quantities. Go look up a few.

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.

....and when asked "what is a tensor", will you finally provide an
answer that DOESN'T talk about matrices?

Please give me a date when you personally will be in peace with this
particular self-inconsistency.

Please give us a date when you will sit your ass down and start
learning the theories which you routinely criticize.


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.

Please stop talking about Lagrangians and actions until you actually
understand the concepts and mathematics relating to the concepts.


T_ij oc rho g_ij

You pulled this straight out of your ass. Plus it has the added
stupidity of not only being written wrong, but is also deliberately
non tensorial by the *******ized inclusion of the metric.


Where

** g_ij = Elements of the metric, the matrix

The metric is not a matrix you ignorant ****.

** rho = mass density

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

Pragmatic reason? The number of rank two tensors that you can build
out of a particle's four-velocity is rather limited. Kinetic energy is
not frame invariant and momentum is a function of its' four-velocity,
so you don't have any choices.

Technical reason?

The definition of the stress-energy tensor for a single particle is
T^uv = m U^u U^v. Test the definition for simple cases. It easily
generalizes to a density.

Try U^i = (c,0,0,0) - the particle is at rest. So the only nonzero
component of the tensor is T^00 = mc^2. Remember SR?


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

Why do you care? It isn't as if you can use this knowledge, or even
repeat what you are told without butchering it.


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.

YAWN. This is explained in every relativity textbook that has the
Lagrangian derivation of the field equations - there is a specific
component that is simply labeled the stress-energy tensor.


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

Covariant, dumb****. COVARIANT.