On Feb 13, 7:49 pm, Edward Green wrote:
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

Not quite.

The scalar product g_uv u^u u^v, for a time-like path in a region in
locally inertial coordinates, is always -c^2 [or -1 if you like c =
1]. Keep in mind that this is a covariant expression so it will be
true regardless of your coordinate system.

Covariance is important, yo. Keep in mind something that would be true
for 3-velocity [can always find a rest frame] isn't true for 4-
velocity.


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?

Overcounting with the knowledge you are overcounting will result in
overcounting. No surprise for all involved.

Take a stress energy tensor in a Cartesian coordinate system. Then in
that case, T_ij can be interpreted as the pressure on the i'th surface
in the j'th direction. That's how I remember it, at any rate. When one
of the i,j are time, then you have energy density instead of pressure.

Whether or not the stress-energy tensor has a trace, there are still
physical effects associable with that specific tensor component - it
just isn't covariant. OTOH, I cannot think of a direct physical
consequence of trace of a stress tensor other than how [as I remember]
it enters into the covariant equations of motion that are an analogue
to the classical Eulerian hydrodynamic equations. Take that last part
with a grain of salt - I'm generalizing from my brief recalling of a
perfect fluid and how it behaves, and I'm not getting up to dig up
MTW.

For fun, imagine a traceless tensor that has only off-diagonal
components. I see no physical reason why you can't have one - it'd
just be weird. I am thinking of an analogy between situations in the
Earth's upper atmosphere from magnetic effects which have an effective
pressure on a charged particle that is one value along a field line
and an entirely different value orthogonal to the field line.


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.


This knowledge is independent of Kerr's derivation.

http://xxx.lanl.gov/abs/0706.1109

For fun, explain how to write the contribution angular momentum gives
a stress-energy tensor while retaining the requirement that the tensor
remain antisymmetric.


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).

I tend to think of pressure as a statistical quantity though I have no
trouble reversing the generalization to a singular particle.


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.

Therein lies the crux. I wrote the "for fun.." part before seeing this
paragraph.

I don't actually know how to answer this. My understanding is that GR
cannot easily use source terms that incorporate angular momentum.
Uncle Al would have more profound words to say on this exact subject
since he's the one who introduced me to Einstein-Cartan theory which /
does/ handle angular momentum naturally.


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

A bull**** evasive answer would be "Well the Kerr solution is a vacuum
solution so there /is/ no source term."

There are two issues that immediately come to mind.

There are NO interior solutions to Kerr in general relativity to my
knowledge. That raises a red flag - I don't believe it is proven they
don't exist, but it's suggestive to me.

The Kerr solution still retains a very definable sense of rotation
through its' behavior at infinity [ala Schwarzschild & the Komar
integral definitions of mass] and its' behavior in the strong field as
its' effect on photons and massive particles.

Except I don't know how to write the angular momentum of a Kerr hole,
and I don't think you can - at all. The four dimensional & /covariant/
way to write angular momentum of a particle is the generalization of r
x p : L^uv = x^u /\ p^v - p^u /\ x^v. Except both x^u and p^v are both
singular as hell at the point [ok, annulus] and everything is vacuum
except at the singularity.

Since you might appreciate this, I'll tell you a little more that I
have found in my researching of Einstein-Cartan theory.

Basically Einstein-Cartan theory is what happens when you allow a
direct coupling with angular momentum that happens by allowing
torsion. I'm butchering this slightly, but in effect you have the same
field equations as general relativity except there is a source term
that couples to spin/angular momentum.

What's really cool is this:

Look at the dust solution for GR. Let it be spherically symmetric n'
****, and the exterior solution is Schwarzschild.

Try the same thing but different in Einstein-Cartan theory. Use
something called the Wessenhoff fluid as a source - it is a fluid with
intrinsic angular momentum. The /approximate/ exterior solution to the
Weyssenhoff fluid is...get this...the linearized Kerr solution.

I'm willing to bet money that the exact exterior solution for a
Weyssenhoff source will either be the full Kerr solution or will have
the full Kerr solution as a subset.