On Feb 20, 3:31*pm, "Ken S. Tucker" wrote:
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?


A ~Machian~ backgroud does not have a direction
however a pair of molecules interacting as
as induction partners has a direction wrt
each other.

For pairs of hydrogen:

+- -+ Repelling, unstable

-+ -+ Attracting, stable

+- +- Attracting, stable

I suppose you could say the directions are
inward and outward.

... unlike the case of Newtonian mechanics, we have to
be careful about the presuppositions of such a
coordinate system, since in general relativity various
reference frames are possible and allowed. How should
we choose axes of a coordinate system, and how should
we obtain a value on such an axis?
http://www.bun.kyoto-u.ac.jp/~suchii/schwarzschild.html

Sue...




Regards
Ken S. Tucker

Hi Sue,
This is a good thread and I wish
to stay on topic.
so let me put you on hold pending
the replies of others, with all due
respect.
Ken S. Tucker


On Feb 20, 1:07 pm, "Sue..." wrote:
On Feb 20, 3:31 pm, "Ken S. Tucker" wrote:



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?

A ~Machian~ backgroud does not have a direction
however a pair of molecules interacting as
as induction partners has a direction wrt
each other.

For pairs of hydrogen:

+- -+ Repelling, unstable

-+ -+ Attracting, stable

+- +- Attracting, stable

I suppose you could say the directions are
inward and outward.

... unlike the case of Newtonian mechanics, we have to
be careful about the presuppositions of such a
coordinate system, since in general relativity various
reference frames are possible and allowed. How should
we choose axes of a coordinate system, and how should
we obtain a value on such an axis? http://www.bun.kyoto-u.ac.jp/~suchii/schwarzschild.html

Sue...

Regards
Ken S. Tucker

On Feb 19, 10:04 pm, Koobee Wublee wrote:
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?

It's so obvious that it doesn't need to be written. You append
"observed" whether it should be there or not.


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'?

It allows you to deliberately not discuss more subtle issues by
shoveling everything into an "observed" qualifier.


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

The answer is worthless whether or not it is correct because you can't
actually /calculate/ the answer. You repeatedly fail to do the simple
calculation.


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 isn't a function of kinetic energy.


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

Since both he and I gave you two stress-energy tensors - neither of
which are functions of kinetic energy, I fail to see what your whining
about.


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

Royal we - tensors aren't matrices.


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

Nope.


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.

Nope - one occasion, and I had it prefaced with a qualifier. OTOH I
fail to see why I should have to rationalize myself to you.


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.

Yet you fail to understand the basic concepts related to the method.
You are encouraged to show your calculation that proves the Einstein-
Hilbert action doesn't give the field equations. Then again, I have
asked this before and you ignored the request.


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

The Lagrangian was specifically chosen such that varying with respect
to it gives the field equations. OTOH your definition is utterly
meaningless.


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.

Prove it is a tensor, then. Since you claim to know all about the
subject, it should be a simple exercise for you to explicitly write
the expression for transforming your candidate stress-energy tensor to
a different coordinate system and seeing if it satisfies the
definition of a tensor.

Put up or shut up.


Where

** g_ij = Elements of the metric, the matrix

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

The metric is merely matrix. shrug

Nope.


** 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

Just because you don't like the reason doesn't make it foaming or an
"excuse".


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?

Do you do the same whine when presented with F = ma? You can derive F
= ma from the Lagrangian method that has an arbitrary part that's
called "force", and quantity called force is rather arbitrary.

Or do you not see why I make the argument?


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

Not my fault you can't follow something incredibly simple or point out
where the supposed error is.


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

Yet you have spent several years whining about it. Find retirement
lonely?


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

So where did you learn GR?


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.

In a way, it does. The Einstein-Hilbert action is int R*sqrt(-|g|)
d^4x - the entire quantity is invariant and is built out of invariant
quantities.

On Feb 19, 10:06 pm, Eric Gisse wrote:
On Feb 19, 8:53 am, Edward Green wrote:

...

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 merely state my source, for what it's worth. Of course it needs
corroboration.

(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?

http://xrl.us/bgic4 (Link to books.google.com)

if you trust this link shortening service, or else google "neugebauer
meinel disk":

There were lots of references, this was the second.

(The web makes any dweeb look learned).

I'm not sure what you mean by "the solution is not a disk". Of course
the solution isn't a disk -- but it is axial, and has some
characteristic surfaces which resemble oblate spheroids. I would
expect solutions related to a disk to have such surfaces, and the
author did say this was a limiting case (of the Neugebaurer/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?

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

I did go on to think "negative energy" was probably better ... but
why? Do you want to emphasize this is not the rest mass of anything?
If we actually had some stuff with this intensive property, I'm not
sure we could tell the difference.

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.

Well, "mass" or "energy" aside, this naively suggests anti-gravity.

I'm still not sure I see the relevance of a theory which takes "a
Wessenhoff fluid" as itssource(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 onesourceamong 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 asourcein the linearized
equations, the metric you get is the weak field Kerr solution. To me
that is tantalizing.

Indeed. All the most interesting areas seem to have clues that are
tantalizing.

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

Hmm... on Feb. 13 you wrote:

"Wessenhoff fluid ... is a fluid with intrinsic angular momentum"

Perhaps you didn't mean to put it that way?

I take "intrinsic angular momentum" to mean an intensive quantity
which is integrated over a volume to find a total angular momentum --
similar to a mass density, or a momentum density. We mentioned that
the concept is classically dubious, but perhaps can be rescued by some
vague non-classical effects, in whom all things are possible :-), or
more concretely by a material including distributed microscopic
centers of angular momentum on an unresolved scale - distributed
pointlike sources. I mentioned as a possible example a material with
significant unpaired (qm) spins -- which occasioned a side discussion
how sometimes, when we say "spin" or "spining" we don't mean to invoke
qm; which wasn't essential here anyway, as an example of "distributed
pointlike sources of angular momentum" could be molecules in some
excited state, or even a collection of (classically) spinning pebbles!

Provided we don't resolve the pebbles.

The few web references I could find do not really clarify whether
Wessenhoff fluid has "intrinsic angular momentum" in this sense.

http://www.iop.org/EJ/abstract/0264-9381/11/9/017

mentions "self-consistent spinning fluid", which sounds like it may
have nothing to do with intrinsic angular momentum, but also cites
"the ad hoc Wessenhoff spin fluid"! A "spin fluid" sounds much more
exotic than a "spinning fluid", and leaves us in doubt.

http://arxiv.org/ftp/hep-th/papers/0309/0309108.pdf

contains a reference to:

Wessenhoff, J., Raabe, A., "Relativistic Dynamics of Spin-Fluids and
Spin-Particles," Acta Pol., 9
(1947), 8-53

which means someone must visit a library. But I think the title
suggests something more along the lines I had in mind, and that the
author of the later abstract may have been misleading to describe this
as a "spinning fluid". And besides, why would Wessenhoff get to
attach his name to the stuff, if it were just an ordinary kind of
fluid, gaining angular momentum solely by relative motion of its bulk
parts? :-)

I mention that AFAIK "bulk property", since you mention it, means
precisely something that is (ideally) distributed in this intensive
way -- in the smallest subdivided bit (as dielectric constant, mass
density, and so forth). You may have intended something else, but for
me the evidence points to a Wessenhoff fluid having "bulk angular
momentum" in just this sense, which is the same thing as "intrinsic
angular momentum" as I understand the term.


...

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?

Sorry... the one on the Kerr Metric!

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!

You had better clarify what you mean by "intrinsic angular momentum"
and "bulk properties", since it is evidently not what I mean.

I have some parting Big Picture thoughts on how I think this all may
hold together and pan out, or else hold out and pan together, but I'll
save them for now. ;-)

On Feb 20, 1:15 am, 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. 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.

Even I can answer that like an establishment pro!

The _components_ of the tensor may be observer dependent, but the
abstract tensor itself is independent of any coordinate system.

That's just how tensors are. Are you saying the whole machinery was
put together wrong?

...

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

I'm still unsatisfied with this one. It doesn't seem to me to be
"contradictory" as much as "insuifficiently specified".

A relativistic ideal gas has, even in its rest frame, a pressure and a
an energy density, the latter including an increment to the rest
energy. Both of these terms happen to involve the motion of the
molecules, but they are observationally distinct, and don't require us
to know anything about any damn molecules. In treating such a gas in
field equations which include energy and pressure, if we can't simply
plug both the observable values in, but invoke extra accounting
principles -- that's cheating.

Either we can blindly plug in both terms, in or there is something
imperfect about the formulation.

On Feb 20, 7:17 pm, Edward Green wrote:
On Feb 20, 1:15 am, Koobee Wublee wrote:

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.

Even I can answer that like an establishment pro!

The _components_ of the tensor may be observer dependent, but the
abstract tensor itself is independent of any coordinate system.

Can you prove that mathematically? Proof by faith does not count.

That's just how tensors are.

Well, in that case, the metric cannot be a tensor. shrug

Are you saying the whole machinery was
put together wrong?

Yes.

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

I'm still unsatisfied with this one. It doesn't seem to me to be
"contradictory" as much as "insuifficiently specified".

A relativistic ideal gas has, even in its rest frame, a pressure and a
an energy density, the latter including an increment to the rest
energy. Both of these terms happen to involve the motion of the
molecules, but they are observationally distinct, and don't require us
to know anything about any damn molecules. In treating such a gas in
field equations which include energy and pressure, if we can't simply
plug both the observable values in, but invoke extra accounting
principles -- that's cheating.

You need to think about this scenario more carefully. The pressure as
you describe still is relative. shrug.

Either we can blindly plug in both terms, in or there is something
imperfect about the formulation.- Hide quoted text -

Just a misunderstanding on your part. shrug

On Feb 20, 11:50 am, JanPB wrote:
On Feb 19, 10:15 pm, Koobee Wublee wrote:

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".

Given a matrix with its elements being observer dependent, the matrix
must be observer dependent as well. I challenge you to prove your
point mathematically. shrug

(The distinction is analogous to the difference between "vector" and
"vector components".)

If the vector components are observer dependent, the vector itself
must be observer dependent. Again, I challenge you to prove
otherwise. shrug

On Feb 20, 6:29 am, Tom Roberts wrote:
Koobee Wublee wrote:

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

sigh You are still practicing matheMagic. After creating a matrix
where all of its elements are observer dependent, the matrix cannot
become a tensor just by waving your matheMagic wand over it a few
times. shrug

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

So, what is the significance of your energy-momentum tensor if you do
not allow it to be related to the Einstein tensor?

[I only occasionally respond to Koobee, because it is useless to do so.

Correction, it is embarrassing to do so for you since you have no
challenges to my postings. shrug

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.]

You do it just to keep you satisfied with the religion of SR and GR.
It is faith that you responded. shrug

On Feb 20, 5:01 pm, Eric Gisse wrote:
On Feb 19, 10:04 pm, Koobee Wublee wrote:

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

It's so obvious that it doesn't need to be written. You append
"observed" whether it should be there or not.

It is so obvious that you do not know what you are talking about. So,
the rest of crap is snipped unread.

On Feb 20, 10:48 pm, Koobee Wublee wrote:
On Feb 20, 5:01 pm, Eric Gisse wrote:

On Feb 19, 10:04 pm, Koobee Wublee wrote:
Do you not agree that the kinetic energy is observer dependent?

It's so obvious that it doesn't need to be written. You append
"observed" whether it should be there or not.

It is so obvious that you do not know what you are talking about. So,
the rest of crap is snipped unread.

How would you survey an ellipsoid?
Ken