On Feb 13, 5:50 pm, Eric Gisse wrote:
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.

I looked at that and thought "but for one mass, can't we chose a
coordinate system where the mass is at rest, and therefore, can't we
zero this expression out?"

But of course, even in that coordinate system, there is a c in the
four velocity, so we have...

mc^2

Cool. Shows how much I know.

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.- Hide quoted text -

That's interesting. First, it seems there is energy, and there is
pressure, and if both occur in an expression, we should be able to
blindly plug both in without reasoning whether one accounts for the
other. Second, I wonder if this amounts to a redundancy in the
theory: do we _ever_ see physical effects from the distribution of
terms along the diagonal, or is the trace all that matters?

One possibility: even focusing on the diagonal terms, pressure
(stress) can be in principle be anisotropic -- a swarm of particles
may be traveling parallel to the x-axis in either direction, with zero
yz velocity components. We can impute a "p_x" to this, while p_y,p_z
= 0. This describes a mass density moving (in a suitable coordinate
system) in a locally uniform direction -- such as part of a rotating
body. This "of course" (affecting knowledge) leads to the Kerr
metric.

Before you object that translating mass does not have "pressure", I
note that most fundamentally, pressure or stress is momentum flux, and
mass moving past a plane in a fixed coordinate system implies momentum
flux. The zinger, which you may not have thought about, is that the
term has the same value whether the mass is coming or going! (Positive
x momentum moving across a yz plane in the x+ direction is equivalent
to negative x momentum flowing in the x- direction).

Yet the Kerr solution supposedly entrains mass to move in the same
sense of rotation, which it could not do if the GR field only saw
"momentum flux", and didn't see the direction of motion. A sphere with
a marked pole rotating in a right hand sense around an axis passing
through that pole has exactly the same momentum flux tensor (stress
tensor) as a similar sphere rotating in a left hand sense.

How is information of the sense of rotation communicated to the field,
when apparently nothing in the source term admits it!?