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?

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.

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!?

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.

On Feb 13, 11:49*pm, Edward Green wrote:
[...]

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


Consider the solid ball in R3 of radius ð (that is,
all points of R3 of distance ð or less from the origin).
Given the above, for every point in this ball there is
a rotation, with axis through the point and rotation
angle equal to the distance of the point from the origin.
The identity rotation corresponds to the point at the
center of the ball. Rotation through angles between
0 and -ð correspond to the point on the same axis
and distance from the origin but on the opposite
side of the origin. The one remaining issue is that
the two rotations through ð and through -ð are the
same. So we identify (or "glue together") antipodal
points on the surface of the ball. After this
identification, we arrive at a topological space
homeomorphic to the rotation group.
http://en.wikipedia.org/wiki/Rotation_group

consider a closed Universe, that is, we take omega
= 1+å and either identify the antipodes after a
parity reversal, or consider a tunneling over the
extent of the hypersphere at a given time. In
either case, it is the red-shifted radiative
image of the local matter that appears spread
here in the far infrared while the tunneling
is assumed to be instantaneous from here to
here, through a distance. In other words, the
energy borrowed from the vacuum by virtual
photons of energy ~ mc2 is entirely returned
by their cosmological redshift, the process being
coherent. The scalar Riemannian curvature R thus
appears in the one-loop diagrams of QFT while
the Newtonian force emerges only at the two-loop
level. Now the range of the near field of our
coupled oscillators reaches to the Universe, within
which coherent modes can arise in which the local
oscillators fully partake if they are assumed to
lie entirely within their coherence domain. The
equivalence principle also emerges naturally
http://arxiv.org/abs/physics/0107015

IOW... "over all space" is code-speak for
the notions for of that nut E. Mach.


Sue...






- Hide quoted text -

- Show quoted text -

On Feb 14, 1:38 am, Eric Gisse wrote, in part:

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.

Yes, to me too.

I had intended to ask whether "interior solution" means interior to
the event horizon or interior to a surface within which the vacuum is
replaced with a mass density. It turns out, on some investigation,
both!

The vacuum solution inside the event horizon is said to be "unstable",
whatever that means, or else "unphysical" (that part of a respectable
theory which in lesser theories might be described as "wrong"). I also
learned, as you say, that there is no known solution including mass
density which can be smoothly joined to the outer Kerr solution
(although some treatment of a spinning dust disk reduces to Kerr
solution in a limiting case).

The evidence is firmly ambiguous: if we could show no non-vacuum
extension of the solution were possible inward, then I think we could
safely say that the hope that Kerr describes the exterior field of a
massive spinning body is doomed; however, the language is "not known"
-- and there is the tantalizing datum of a single known solution is a
single special case.

I am tempted to claim that the exterior vacuum Kerr solution, while a
valid solution of the field equations, cannot represent the external
field of a spinning body. As I mentioned, a rotating massive object
seems to transmit no intelligence of its sense of rotation to the GR
source term. If we are in fact able to tell from purely gravitational
observation which way a spheroid is spinning, then there must be some
additional factor which breaks the symmetry.

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.

In the Wikipedia article, I see a parameter "J" in the metric? If
that's the angular momentum, it's built in at the outset.

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.

I've long been suspicious of "intrinsic angular momentum". I've
grappled with the idea several times, and I seem to recall concluding
it doesn't make sense. IIRC, my objection was that holding the volume
density of angular momentum constant and shrinking the elementary
spinning bits required the angular speed of the bits to blow up.
OTOH, I can see the idea might make sense for a system with
significant unpaired spins: macroscopically it might be simplest to
treat the spins as a smooth volume density of intrinsic angular
momentum.

But you may agree we ought to be able to model a spinning star or
planet without appealing to "intrinsic angular momentum": the body's
angular momentum is out there in the open.

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

Interesting guess: I take it Einstein Cartan theory is an extension to
plain Einstein theory. Your guess means that to make sense of an
extended source supporting a Kerr metric, we must look outside vanilla
GR. That's interesting, because the exterior Kerr solution is by all
accounts a perfectly physical looking GR vacuum solution. If the
Gisse hypothesis were true we could say that GR is incomplete, in the
sense that it allows self-consistent solutions which can only be fully
understood by admitting more terms into the theory.

But even if all this were true I'm not sure it gets Kerr off the hook
as being the vacuum metric of a spinning body: as appealing as
Weyssenhoff fluid is, it seems unlikely we need it to model spinning
balls of rock and gas. Maybe the Kerr solution describes the
gravitational field of a giant collapsed globule of electrons, whose
gravitation has overcome their electrostatic repulsion. :-) I guess
then we need the charged version, too.

I'm willing to bet money also (maybe $5), that at least one of the
following is true: (1) The Kerr metric is not the vacuum solution
associated with ordinary rotating matter, or (2) GR requires an
extension to handle ordinary rotating bodies, but not the one you
propose. A role will appear for the four momentum density -- not just
for the the stress tensor, whose terms are sensitive to orientation or
surfaces, but not to the choice of positive normals.

Since our bets are not opposite sides of the trade, I'm not sure whom
we're betting!

On Feb 14, 6:11 pm, Edward Green wrote:
On Feb 14, 1:38 am, wrote, in part:

There are NO interior solutions toKerrin 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.

Yes, to me too.

I had intended to ask whether "interior solution" means interior to
the event horizon or interior to a surface within which the vacuum is
replaced with a mass density. It turns out, on some investigation,
both!

I should have been more specific.

There exists a coordinate chart for Kerr that extends past the
horizon. There does not exist, to my knowledge, a solution to the
field equations with some non-vacuum source that has Kerr as the
exterior solution. That's what I meant by "interior".


The vacuum solution inside the event horizon is said to be "unstable",
whatever that means, or else "unphysical" (that part of a respectable
theory which in lesser theories might be described as "wrong"). I also
learned, as you say, that there is no known solution including mass
density which can be smoothly joined to the outerKerrsolution

A Kerr black hole is unstable in the sense that it isn't the "ground
state" of a black hole. The Penrose process for extracting energy is
one reason, and super-radiant instability [nee black hole bomb] is
another. A Kerr hole is very ... obvious in its' environment. It can
impart energy to stuff, and drags stuff along its' direction of
rotation. I personally believe Kerr holes are behind quasars and the
more intense class of gamma ray bursts.

http://arxiv.org/abs/hep-th/0404096

(although some treatment of a spinning dust disk reduces toKerr
solution in a limiting case).

It isn't a GR solution though - its a solution to linearized Einstein-
Cartan theory. Unless you found something specific to GR, which I'd
encourage you to share.

I think Einstein-Cartan theory is very cute in that it not only
sidesteps a lot of the singularity and existence theorems that depend
on the strong energy condition [T_uv U^u U^v 0 iirc] because spin
counteracts positive stress-energy. Plus it has the same overall
structure as classical GR since you can cast the field equations in a
form that has everything packed into an effective stress-energy
tensor.


The evidence is firmly ambiguous: if we could show no non-vacuum
extension of the solution were possible inward, then I think we could
safely say that the hope thatKerrdescribes the exterior field of a
massive spinning body is doomed; however, the language is "not known"
-- and there is the tantalizing datum of a single known solution is a
single special case.

I have no idea how this would be done or even if it could be done.

I think the idea of the Kerr solution [rotating black hole] is very
beautiful and important, it's just that its' derivation is ugly as
hell. All sorts of boundary conditions and assumptions about algebraic
structure. Its' a mess.


I am tempted to claim that the exterior vacuumKerrsolution, while a
valid solution of the field equations, cannot represent the external
field of a spinning body. As I mentioned, a rotating massive object
seems to transmit no intelligence of its sense of rotation to the GR
source term. If we are in fact able to tell from purely gravitational
observation which way a spheroid is spinning, then there must be some
additional factor which breaks the symmetry.

Thus, Einstein-Cartan theory.


TheKerrsolution 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 aKerrhole,
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.

In the Wikipedia article, I see a parameter "J" in the metric? If
that's the angular momentum, it's built in at the outset.

Where...?




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.

I've long been suspicious of "intrinsic angular momentum". I've
grappled with the idea several times, and I seem to recall concluding
it doesn't make sense. IIRC, my objection was that holding the volume
density of angular momentum constant and shrinking the elementary
spinning bits required the angular speed of the bits to blow up.

A fair objection, but one rooted in classical mechanics. You and I
know it isn't that simple.

Classical E&M contains a similar internal inconsistency regarding the
self-energy of an electron.

OTOH, I can see the idea might make sense for a system with
significant unpaired spins: macroscopically it might be simplest to
treat the spins as a smooth volume density of intrinsic angular
momentum.

But you may agree we ought to be able to model a spinning star or
planet without appealing to "intrinsic angular momentum": the body's
angular momentum is out there in the open.

When I say spin, I do not specifically refer to the quantized angular
momentum of a particle in quantum theory. I treat, in this context,
spin and angular momentum to be the exact same thing. They are as such
in quantum mechanics, but "spin" has a specific connotation that I
don't mean to imply.


The /approximate/ exterior solution to the
Weyssenhoff fluid is...get this...the linearizedKerrsolution.

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

Interesting guess: I take it Einstein Cartan theory is an extension to
plain Einstein theory. Your guess means that to make sense of an
extended source supporting aKerrmetric, we must look outside vanilla
GR. That's interesting, because the exteriorKerrsolution is by all
accounts a perfectly physical looking GR vacuum solution. If theGissehypothesis were true we could say that GR is incomplete, in the
sense that it allows self-consistent solutions which can only be fully
understood by admitting more terms into the theory.

I do believe GR is incomplete in the sense that the theory handles
angular momentum poorly. I am not saying I believe it to be wrong [I
have no evidence] or that it can't, it's just that /I don't see how/,
and I know enough to be able to differentiate between the two. To me,
Einstein-Cartan theory handles it /better/.

Look very carefully at what Roy Kerr had to do to obtain the Kerr
solution. He had to go digging very deep into the algebraic structure
of general relativity to find something that could be interpreted as a
spinning black hole.

In my superficial understanding of both theories, the structures you
can have are /exactly the same/. The only significant difference
between the two is that in GR you make the explicit choice to forbid
torsion and in Einstein-Cartan [EC-GR?] theory, torsion is allowed.

At the moment it is my personal belief that Kerr engineered a solution
that would be a natural solution in EC-GR by digging through the
overall structure that is common to both theories.


But even if all this were true I'm not sure it getsKerroff the hook
as being the vacuum metric of a spinning body: as appealing as
Weyssenhoff fluid is, it seems unlikely we need it to model spinning
balls of rock and gas. Maybe theKerrsolution describes the
gravitational field of a giant collapsed globule of electrons, whose
gravitation has overcome their electrostatic repulsion. :-) I guess
then we need the charged version, too.

Ooowww. Charged.


I'm willing to bet money also (maybe $5), that at least one of the
following is true: (1) TheKerrmetric is not the vacuum solution
associated with ordinary rotating matter, or (2) GR requires an
extension to handle ordinary rotating bodies, but not the one you
propose. A role will appear for the four momentum density -- not just
for the the stress tensor, whose terms are sensitive to orientation or
surfaces, but not to the choice of positive normals.

I think you'll lose #1, and that you stand a good chance of making
money with the first part of #2.

At the moment I am satisfied that the Kerr metric is an adequate
description of the exterior metric of a rotating body, Lense-
Thirring effect, geodetic precession, and the observation of near-
extremal Kerr black holes.


Since our bets are not opposite sides of the trade, I'm not sure whom
we're betting!

We are betting on angular momentum.

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. 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.

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.


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.


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.


Tom Roberts

On Feb 13, 11:49*pm, Edward Green wrote:



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!?- Hide quoted text -

- Show quoted text -

xxein: This is a longer post than even I expected it to be. Be
patient.

Time for a new theory, huh? The fact is that present theories are
primarily for the purpose of utilization and manipulation. They were
offspring of coarser attempts and did not alter their dna to an
understanding of the physic, itself.

As you noticed, our theories simply do not address enough. We will
probably never have a TOE as a physical theory, but more of an
"understanding theory" that is not in a different class than theories
we presently accept today. The former is rather impossible and the
latter will certainly have variations and be necessarily less
definitive.

So what do we do to help get a better understanding? Well, first of
all we must be willing to shed some past, strongly held beliefs on the
subject. I have a personal account of how I unwittingly did this -
giving me first-hand knowledge of how it might be done by others. No
guarantees though. In it's own way, it's tougher than understanding
the complete Einstein theory with all the i's dotted and t's crossed.
But in the end (after a successful divorce from Einstein), we can find
that it is a simpler way to understanding the physic and that both can
co-exist (as Newton-Einstein do).

But you want some answers right now, right? If I gave them to you,
you would think they were wacky. They are not sponsored by any
present theory. On the other hand, chances are 1:trillion that you
might discover them yourself. Our whole human population might have
to wait 50 to 100 yrs. before someone successfully puts this out. So
I should try, right?

It is assumed (in a relativity theory) that a clock falling directly
toward a gravity object will slow down. But there is no direct
evidence for that. We have only given a state being for Schrodinger's
cat. In this case, we simply say that we know a higher clock goes
faster than a lower clock. But we don't know the transition. We
simply don't know what it does "while" it is falling. We only have
what we think of as stationary measurements.

It wouldn't be that difficult to measure a clock's rate during its
'fall', but our measurement would be corrupted by the assumptions that
dictate the method used. A 'catch-22' of sorts. But drawing your
attention to that is only the minor purpose of this post. (we could
simply mount detectors along a vertical pole that were synchronised
(rate) in the same way we do GPS - and not like a Pound-Rebka
measurement)

If a clock is falling, there are limits. No, not what you are
thinking. The clock can fall from an infinitely far away position
with no initial velocity or it can fall from anywhere with an initial
velocity (even c if so imagined). This may sound inconsequential in
the Einstein world, but it has consequences beyond.

If a clock dropped from infinity did 'not' change in its timerate
during its fall, Einstein would have to go back to processing
patents. So here's the deal. I understand completely what Einstein
was thinking and doing (not the intricate math, but the idea behind
the formulation for it). I understand the need for the Kerr
solution. I understand the need for the E-C solution. But they are
not true solutions. They are band-aids.

So I said an 'if' above. But that 'if' cuts completely through our
theories and their band-aids. The reason I bring up the falling clock
and its timerate is because I already know the implications. I
already know that it would not affect our daily life and technology
except in small incremental terms and measures (just as with Newton to
Einstein). I also know that a paradigm shift is slow. I also know
that it is possible to be wrong and still command the scientific
community like Einstein did. But he wasn't exactly wrong - just
limited. And I'll give TR his "limited domain of applicability".

So? Now what? Do we still want to hide in the "catch-22" that binds
our physics into a circular argument or do we (rhet) wreck our long
fought careers and come to admit that the whole thing of physics was
just a 'man behind the curtain' scenario?

The clock was just an example. It wasn't the spark (although I hope
it will be a spark to you or anybody else with a brain that reads
this). It was part of the conclusion. I came to this conclusion by
examining exactly what you are talking about here (but I did mine in
about 1989-90).

I'll have to say this eventually, so I'll say it now. It's like a
"falling space theory". It is and isn't. I haven't met a falling
space theory that worked. None could capture the whole of its
conjectures or conclusions. Mine is of a different breed - born out
of the unadulterated logic that seems to be evaded through our belief-
centered thinking. I could tell I was doing that unadulterated logic
from the outset. (another story, another time)

I want to give you a hammer and anvil that you can crack your nuts
on. But first I have to explain why a relativity theory works. In
its simplest essence, it is what we measure. That means how things
affect us. I'm sorry but I can't call that physics. It is simply a
relative measure. I also hear that it is described that way by its
honest proponents. It took Einstein many years to incorporate gravity
into his SRT and make GRT. I had a similar but quicker experience (no
thanks to Einstein).

You don't really want to think that a relative measure can dictate an
objective sense of a physics, do you? Otoh, you want a GRT, E-C, Kerr
solution to be united in some way, right? And let's include Q
theories also, ok? For that matter, let's include chaos, Mach, Bohr
and the whole panalopy. Somehow they must make a sense, right?
Otherwise, they are all 'crackpots'. And who is calling them
'crackpots'? Why not?

The point is (in my specialty, gravity) that there are many maths that
can be applied to describe a situation. They all might give correct
results to the measurer. But each has the conotation through a
different notion of physics. I could give you my version of the
timerate of GPS clocks that gives exactly the same result derived by
Einstein. Who is right in this different method of achieving the same
result? Wouldn't it be of interest to science to find out why they
achieve the same resultant by using different notions of how things
work?

I'll continue with a post to myself.

"xxein" wrote in message
...
On Feb 13, 11:49 pm, Edward Green wrote:



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!?- Hide quoted text -

- Show quoted text -

xxein: This is a longer post than even I expected it to be.

Geez you stupid ****.