On Feb 14, 10:39*pm, Eric Gisse wrote:

On Feb 14, 6:11 pm, Edward Green wrote:

...

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" .... or else "unphysical" ...

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

I wouldn't say a flywheel is unstable, though it may run down, but I
would if the flywheel were liable to explode!

OTOH, the author of the wikipedia article goes beyond "unstable" to
"unphysical" ... that sounds like the case where one solution decays
at infinity, and the other blows up -- something that is physically
unreasonable. Of course I don't know what he had in mind.

(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.

My sole reference (Wikipedia) mentions the the "Neugebauer/Meinel
disk".

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.

You mean that intrinsic angular momentum shows up as "negative mass"?
Does this mean if we had a mass with sufficient intrinsic angular
momentum at the Earth's surface it would float away like a balloon?

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.

With the result surmised above, that it's equivalent to allowing
"negative energy"? How curious that would be.

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.

I'm still not sure I see the relevance of a theory which takes "a
Wessenhoff fluid" as its source (unless I misunderstood your
remarks). Ordinary rotating massive objects do not resemble
Wessenhoff fluids -- they have angular momentum solely as a result of
the global configuration of many bits of perfectly ordinary matter
moving in a particular configuration.

...

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...?

First sentence and third equation of section "Mathematical form".


...

... 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.

Well, perhaps you are saying we have "spin" in the quantum mechanical
sense. Even then we really don't have "intrinsic angular momentum" in
the continuum sense, but more like a distribution of rapidly rotating
particles below our level of resolution: whether we have tiny rapidly
spinning (but still classical) dust particles, or quantum particles
possessing angular momenta through quantum spin, doesn't really matter
-- as I've said (too many times by now), if we have a distribution of
such sources, I can see the idea might apply.

We might have a role for an intrinsic angular momentum term in a
modified GR, I just don't see how it can have a big role in the
graviational field of objects which are not terribly exotic, like
planets and stars. Maybe neutron stars?

OTOH, I can see the idea might make sense for a system with
significant unpaired spins ...

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.

I think I read you that way, but we are tripped up by language, since,
when I actually want to talk about qm spin, I have to use the word.
Perhaps we should agree to use "rotating" when we do not specifically
want to invoke the qm concept.

...

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.

So perhaps Kerr stumbled on a physically significant solution of the
GR field equations which cannot be easily justified within the frame
of GR: so the theory is incomplete. But I'm repeating what we've
suggested several time. You suggest the completion is Einstein Cartan
theory; since my understanding of both theories is even more
superficial than yours, but I have my doubts, because of the
appearence of "intrinsic angular momentum". However, maybe the
astronomical objects which seem to be manifesting the external Kerr
solution are really so exotic that this is the right contruction. But
that still leaves ordinary rotating massive bodies out -- unless we
look at the source from so far away that it appears to be a point
source of angular momentum. Maybe that's the trope.

On Feb 19, 8:53 am, Edward Green wrote:
On Feb 14, 10:39 pm, Eric Gisse wrote:

On Feb 14, 6:11 pm, Edward Green wrote:

...



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" ... or else "unphysical" ...
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

I wouldn't say a flywheel is unstable, though it may run down, but I
would if the flywheel were liable to explode!

OTOH, the author of the wikipedia article goes beyond "unstable" to
"unphysical" ... that sounds like the case where one solution decays
at infinity, and the other blows up -- something that is physically
unreasonable. Of course I don't know what he had in mind.

....now the question is: Does anyone really care what Wikipedia says? I
tend not to listen to Wikipedia anymore since the explanations in
Wikipedia tend to be not as good as they could be.

I have my meaning for unstable, MTW has another, etc.


(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.

My sole reference (Wikipedia) mentions the the "Neugebauer/Meinel
disk".

The Kerr solution isn't a disk though. From my searching, it seems
that the two are unrelated. Can you find a non-Wiki reference?


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.

You mean that intrinsic angular momentum shows up as "negative mass"?
Does this mean if we had a mass with sufficient intrinsic angular
momentum at the Earth's surface it would float away like a balloon?

No. I specifically do not mean negative mass. The best phrasing I can
think of at the moment is negative energy density.


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.

With the result surmised above, that it's equivalent to allowing
"negative energy"? How curious that would be.

Yup. That's another reason why I think EC-GR is cute - it's the second
way I know of, other than vacuum energy which is a quantum field
theory thing, to circumvent positive energy conditions.


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.

I'm still not sure I see the relevance of a theory which takes "a
Wessenhoff fluid" as its source (unless I misunderstood your
remarks). Ordinary rotating massive objects do not resemble
Wessenhoff fluids -- they have angular momentum solely as a result of
the global configuration of many bits of perfectly ordinary matter
moving in a particular configuration.

A Weyssenhoff fluid is a perfect fluid that is _rotating_. It is
simply one source among many that can be put into the field equations,
it's just the most relevant one that I know about. The reason I care
is that when the Weyseenhoff fluid is a source in the linearized
equations, the metric you get is the weak field Kerr solution. To me
that is tantalizing.

Once again - there is no claim about intrinsic angular momentum being
made here. These are bulk properties.


...

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...?

First sentence and third equation of section "Mathematical form".

What article?


...

... 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.

Well, perhaps you are saying we have "spin" in the quantum mechanical
sense. Even then we really don't have "intrinsic angular momentum" in
the continuum sense, but more like a distribution of rapidly rotating
particles below our level of resolution: whether we have tiny rapidly
spinning (but still classical) dust particles, or quantum particles
possessing angular momenta through quantum spin, doesn't really matter
-- as I've said (too many times by now), if we have a distribution of
such sources, I can see the idea might apply.

There is no intrinsic angular momentum here!

Think bulk properties! Bulk properties!


We might have a role for an intrinsic angular momentum term in a
modified GR, I just don't see how it can have a big role in the
graviational field of objects which are not terribly exotic, like
planets and stars. Maybe neutron stars?

Doubt it. Two binary neutron star systems perform exactly to spec.
Corrections from spin are going to be small.


OTOH, I can see the idea might make sense for a system with
significant unpaired spins ...
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.

I think I read you that way, but we are tripped up by language, since,
when I actually want to talk about qm spin, I have to use the word.
Perhaps we should agree to use "rotating" when we do not specifically
want to invoke the qm concept.

k


...

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.

So perhaps Kerr stumbled on a physically significant solution of the
GR field equations which cannot be easily justified within the frame
of GR: so the theory is incomplete.

Could be. I'm only thinking that because of the convoluted means Kerr
used to reach the solution and because Einstein-Cartan theory and GR
have the same effective field equations.

But I'm repeating what we've
suggested several time. You suggest the completion is Einstein Cartan
theory; since my understanding of both theories is even more
superficial than yours, but I have my doubts, because of the
appearence of "intrinsic angular momentum". However, maybe the
astronomical objects which seem to be manifesting the external Kerr
solution are really so exotic that this is the right contruction. But
that still leaves ordinary rotating massive bodies out -- unless we
look at the source from so far away that it appears to be a point
source of angular momentum. Maybe that's the trope.

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

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.

On Feb 19, 10:47 pm, Eric Gisse wrote:
On Feb 19, 9:15 pm, Koobee Wublee wrote:

Kinetic energy is observer dependent, and so is the momentum. This is
also true for the observed mass.

Always the "observed" qualifier, eh?

Do you not agree that the kinetic energy is observer dependent?

The weasel words are one of the
many annoying aspects of you.

What weasel words? Do you not understand the meaning of the word
'observed'?

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.

I did give you a straight answer. You were the one who chose to
weasel your way out ignoring my answer. shrug

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.

How can it be false logic if the kinetic energy is observer dependent?

The stress-energy tensor for a lone particle isn't
made out of the particle's 3-momentum OR its' kinetic energy.

Then, you disagree with Professor Roberts. You two need to make up.
shrug

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.

We know the tensor is merely a matrix. shrug

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

According to Professor Roberts, it consists of the kinetic energy and
the momentum. shrug

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.

I have already done so. When are you getting your BS degree? You
have missed your promise date in several occasions so far, Mr. super-
super-super-super senior.

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.

Well, I have understood about Lagrangians and the calculus of
variations. This means I have the right to use the Euler-Lagrange
equation in a timely situation where the Lagrangian actually satisfies
the mathematical constraints according to the Lagrangian method.

T_ij oc rho g_ij

You pulled this straight out of your ass.

Well, since Hilbert did pull out that Lagrangian out of his ass, and
the above is merely a derivation from it, so technically you are
correct, but so. shrug

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.

The plot gets more intrigued. I thought you believe in the divine
tensority of the metric.

Where

** g_ij = Elements of the metric, the matrix

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

The metric is merely matrix. shrug

** 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.

Foaming in the mouth without any answer but excuses, I see. shrug

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.

Hey, why stuff the field equations with definitions after Hilbert
derived them?

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?

And, the joke continues. shrug

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.

You are correct. I do not care about bullsh*t. shrug

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

YAWN.

Yes, it is getting late.

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.

It sounds like you need to fabricate toilet papers with all these
books that are so wrong. shrug

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

Covariant, dumb****. COVARIANT.

There is no more excuse. It is less contradictory if you allow an
invariant quantity to decide how spacetime should be curved.

Koobee Wublee wrote:
On Feb 16, 12:45 pm, Tom Roberts wrote:
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.

(sigh) You just do not understand tensors, and the distinction between
tensors and their components. Kinetic energy, momentum, etc. are related
to COMPONENTS of the energy-momentum tensor, and THEY are observer
(coordinate) dependent. The tensor itself is not coordinate dependent --
that's why we use tensors. shrug


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.

Perhaps if you had READ WHAT I WROTE you would not be so confused. I
said nothing at all about the field equation, I was talking about the
energy-momentum tensor. shrug


[I only occasionally respond to Koobee, because it is useless to do so.
I do it only to prevent others from being fooled by his idiocies, and
thinking that no response implies he is correct. He isn't.]


Tom Roberts

Hi Edward and all.

On Feb 14, 7:11 pm, Edward Green wrote:
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.

I agree, because the Guv=Tuv is symmetrical,
furthermore I think AE made a mistake in Ch2
of this ref,
http://www.alberteinstein.info/galle...lish_pp146-200...

AE has S1 a as sphere and S2 as an ellipsoid,
then he relies on Mach's *principle* to
justify the differing shapes, (something
about rotation) but it's unnecessary in GR.
The internal "stress-energy" tensor shapes
differences work quite well, for example see,
http://en.wikipedia.org/wiki/Piezoelectricity

If S1 is composed of a Piezoelectrical
crystal, with no interior voltage diffs,
but S2 has interior voltage diffs to
force the sphere into a ellpsoidal shape,
as in the wiki ref, where the North and
South Pole are pushed together, the
reciprocal effect is to apply a voltage
and vary the geometry, as a smoke detector
squeaker does.

As Edward points out,(and I agree),the
fact of an ellpsoidal shape does NOT
provide any direction of rotation into
the g-field, and there is no need to
presume rotation is the cause of the
ellipsoid.
Regards
Ken S. Tucker
[snip]

On Feb 19, 10:15 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. Thus, the energy-momentum tensor
must be observer dependent as well.

No - the correct statement would be "thus, the energy-momentum tensor
_components_ must be observer dependent as well".
(The distinction is analogous to the difference between "vector" and
"vector components".)

--
Jan Bielawski

On Feb 20, 2:03*pm, "Ken S. Tucker" wrote:
Hi Edward and all.

On Feb 14, 7:11 pm, Edward Green wrote:





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.

I agree, because the Guv=Tuv is symmetrical,
furthermore I think AE made a mistake in Ch2
of this ref,

http://www.alberteinstein.info/galle..._pp146-200.pdf


AE has S1 a as sphere and S2 as an ellipsoid,
then he relies on Mach's *principle* to
justify the differing shapes, (something
about rotation) but it's unnecessary in GR.

A page number would be helpful.

Hmmm...

These molecules are ~ellipsoidal~

http://www.chem.purdue.edu/gchelp/liquids/inddip.html

but a cloud of zillions of them acting as an inertial
background (Machian) would exert even force and
so a test mass would be spherical.

The internal "stress-energy" tensor shapes
differences work quite well, for example see,

http://en.wikipedia.org/wiki/Piezoelectricity


If S1 is composed of a Piezoelectrical
crystal, with no interior voltage diffs,
but S2 has interior voltage diffs to
force the sphere into a ellpsoidal shape,
as in the wiki ref, where the North and
South Pole are pushed together, the
reciprocal effect is to apply a voltage
and vary the geometry, as a smoke detector
squeaker does.

As Edward points out,(and I agree),the
fact of an ellpsoidal shape does NOT
provide any direction of rotation into
the g-field, and there is no need to
presume rotation is the cause of the
ellipsoid.

That may be the case for you piezoelectric analog
but but the mechanism Koruopolis describes

--Koruopolis
http://arxiv.org/abs/physics/0107015

DOES have a directed orientation along the lines of force.

Sue...


Regards
Ken S. Tucker
[snip]- Hide quoted text -

- Show quoted text -

On Feb 20, 12:10 pm, "Sue..." wrote:
On Feb 20, 2:03 pm, "Ken S. Tucker" wrote:



Hi Edward and all.

On Feb 14, 7:11 pm, Edward Green wrote:

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.

I agree, because the Guv=Tuv is symmetrical,
furthermore I think AE made a mistake in Ch2
of this ref,

http://www.alberteinstein.info/galle...lish_pp146-200...



AE has S1 a as sphere and S2 as an ellipsoid,
then he relies on Mach's *principle* to
justify the differing shapes, (something
about rotation) but it's unnecessary in GR.

A page number would be helpful.

Hmmm...

These molecules are ~ellipsoidal~

http://www.chem.purdue.edu/gchelp/liquids/inddip.html

but a cloud of zillions of them acting as an inertial
background (Machian) would exert even force and
so a test mass would be spherical.

The internal "stress-energy" tensor shapes
differences work quite well, for example see,

http://en.wikipedia.org/wiki/Piezoelectricity





If S1 is composed of a Piezoelectrical
crystal, with no interior voltage diffs,
but S2 has interior voltage diffs to
force the sphere into a ellpsoidal shape,
as in the wiki ref, where the North and
South Pole are pushed together, the
reciprocal effect is to apply a voltage
and vary the geometry, as a smoke detector
squeaker does.

As Edward points out,(and I agree),the
fact of an ellpsoidal shape does NOT
provide any direction of rotation into
the g-field, and there is no need to
presume rotation is the cause of the
ellipsoid.

That may be the case for you piezoelectric analog
but but the mechanism Koruopolis describes

--Koruopolishttp://arxiv.org/abs/physics/0107015

DOES have a directed orientation along the lines of force.

Sue...

Regards
Ken S. Tucker
[snip]- Hide quoted text -

- Show quoted text -

Our Mr. Green is surveying an object
and finds it is an ellipsoid.
Now how can Mr. Green find the direction
of rotation of said ellipsoid?
Regards
Ken S. Tucker