On Feb 20, 9:37 pm, Koobee Wublee wrote:
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

http://preposterousuniverse.com/grnotes/

Read chapter 2.

Wait....didn't I reference you these exact notes before?

http://groups.google.com/group/sci.p...f?dmode=source

Why, yes I did. I wonder it the arguments for why you won't click a
link and read what's contained will have changed.


(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

An excellent question...answered in any textbook on tensor analysis or
the introductory chapters of any relativity textbook. Apparently you
never looked at either.

On Feb 20, 5:55 pm, Edward Green wrote:
[...]


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)

That's what I found. I don't see any evidence that this is related to
Kerr in any way.


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).

What I mean by "Kerr is not a disk" is that the solution spans all 3
spatial dimensions and is not a metric describing a planar...anything.
Whereas the Neugebauer/Meiniel solution is, well, a disk of rotating
matter.

The only relation between the two seems to be that the rotating disk
is presumed to collapse into a rotating Kerr hole. Past that, I see no
relationship.


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?

Yes - because negative mass [which I have thought about but haven't
formed a solid opinion of] does not exist whereas negative energy
densities can and do exist in limited circumstances.

If we actually had some stuff with this intensive property, I'm not
sure we could tell the difference.

Depleted vacuum via Casimir effect.


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.

If you have a pure and very large spin source that is massless, maybe.


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 didn't think the wording mattered since I didn't think there'd be
confusion regarding quantum spin. I could have worded it better, I
suppose, but "extrinsic angular momentum" just seems awkward and
simply saying "angular momentum" doesn't seem quite good enough.


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!

The concept of a rotating object isn't classically dubious. The
concept of quantum spin, however, is. Keep in mind that this theory is
still classical.


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

There is a minor discussion of the Weyssenhoff [it has a "y", correct
spelling makes searching easier] in Hehl, et. al, Rev. Mod. Phys., Vol
48, No. 3, 1976 on page 408. If you care about the theory, you should
look it up. If you can't, I'll send you the dozen pages or so I took
photographs of [I lose paper so easily].

There are better discussions he

arXiv:gr-qc/0601089v2 - This is all about the Weyssenhoff fluid.
arXiv:gr-qc/9309027v1 - This is a survey of Einstein-Cartan theory.
Same basic content as Hehl, et. al, but slightly different in
presentation and certain subjective naming conventions for stuff.
Contains more stuff on the Weyssenhoff fluid.
arXiv:gr-qc/0606062 - Trautman's review of linearized Einstein-Cartan
theory
http://www.numdam.org/numdam-bin/fit...974__21_1_89_0 -
contains the derivation of the linearized Kerr metric using a
Weyseenhoff source.


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? :-)

No idea. OTOH, if Maclauren could get the entire Taylor series
consisting of expansions centered around zero just by setting the
expansion point to be zero, it doesn't strike me as /that/ odd.


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!

Oh. The blurb isn't followed up upon or explained so I'll ignore it as
something irrelevant from the author. I'll pay it more heed if I catch
it written in a paper book.


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.

Expunge all quantum theory from your mind. This is a classical theory,
so if I mention angular momentum I mean good ol' r x p.


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. ;-)

http://iopscience.iop.org/0264-9381/20/13/330/pdf
http://iopscience.iop.org/0264-9381/...ect=iopscience

The Einstein tensor for the most general axially symmetric metric
fills several Maple worksheet pages, is highly nonlinear, and is most
likely unsolvable analytically. So I'm going to be more restrictive
and examine an ellipsoidal symmetry instead, which is what I wanted
originally but didn't think I needed to be that restrictive.

Koobee Wublee wrote:
On Feb 20, 6:29 am, Tom Roberts wrote:
[to Koobee] You just do not understand tensors, and the distinction between
tensors and their components.

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.

OK, I'll try once more.

Let's drop down a level and discuss vectors, not rank-2 tensors, and use
3-space, not spacetime. This permits easy visualization of the issues.

Consider a vector of unit length pointing east.

In coordinates with x headed east, y headed north, and z headed up, it
has components (1,0,0). Those three real numbers are NOT the vector, but
they are related to it, and with proper additional structure they can be
used to compute the vector. That additional structure is the BASIS
VECTORS of the coordinate system on which the vector was projected to
obtain those components. In this case the basis vector for x is a unit
vector pointed east, and similarly for y and z.

In coordinates with x' headed north, y' headed west, and z' headed up,
that SAME VECTOR has components (0,-1,0). Those three real numbers are
NOT the vector, but they are related to it, and with proper additional
structure can be used to generate the vector. That additional structure
is the BASIS VECTORS of the coordinate system on which the vector was
projected to obtain those components. In this case the basis vector for
y' is a unit vector headed west, and similarly for x' and z'.

It should be clear that by computing the vector sum of the components of
the vector times the corresponding basis vectors, one obtains the vector
itself:
v = v_i e^i (here the {v_i} are the components of vector v, and
the {e^i} are the corresponding basis vectors.
The {v_i} are each a real number; the {e^i} are
each a vector, and the sum over i is a vector sum.)
The VECTOR is independent of the coordinate system, but the COMPONENTS
are not.


Applied to a rank 2 tensor, the "matrix" you keep referring to is a
matrix of the tensor's COMPONENTS. To obtain the tensor itself, they
must be composed with the basis vectors of the coordinate system onto
which the tensor was projected to obtain the components:
T = T_ij e^i (X) e^j (here (X) indicates a tensor product.
The {T_ij} are a matrix of real numbers
corresponding to the COMPONENTS of the
tensor T; they are not T itself. The
sum over i and j is a rank-2 tensor sum.)
The result is the same: the TENSOR is independent of the coordinate
system, but the COMPONENTS (the "matrix" you mention) are not.


There is no "matheMagic" here, just very basic tensor math. Stuff you
repeatedly refuse to learn, and/or are congenitally unable to
understand. shrug


Tom Roberts

On Feb 22, 2:00*pm, Tom Roberts wrote:
Koobee Wublee wrote:
On Feb 20, 6:29 am, Tom Roberts wrote:
[to Koobee] You just do not understand tensors, and the distinction between
tensors and their components.

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. *

OK, I'll try once more.

Let's drop down a level and discuss vectors, not rank-2 tensors, and use
3-space, not spacetime. This permits easy visualization of the issues.

Consider a vector of unit length pointing east.

In coordinates with x headed east, y headed north, and z headed up, it
has components (1,0,0). Those three real numbers are NOT the vector, but
they are related to it, and with proper additional structure they can be
used to compute the vector. That additional structure is the BASIS
VECTORS of the coordinate system on which the vector was projected to
obtain those components. In this case the basis vector for x is a unit
vector pointed east, and similarly for y and z.

In coordinates with x' headed north, y' headed west, and z' headed up,
that SAME VECTOR has components (0,-1,0). Those three real numbers are
NOT the vector, but they are related to it, and with proper additional
structure can be used to generate the vector. That additional structure
is the BASIS VECTORS of the coordinate system on which the vector was
projected to obtain those components. In this case the basis vector for
y' is a unit vector headed west, and similarly for x' and z'.

It should be clear that by computing the vector sum of the components of
the vector times the corresponding basis vectors, one obtains the vector
itself:
* * v = v_i e^i * (here the {v_i} are the components of vector v, and
* * * * * * * * * *the {e^i} are the corresponding basis vectors.
* * * * * * * * * *The {v_i} are each a real number; the {e^i} are
* * * * * * * * * *each a vector, and the sum over i is a vector sum.)
The VECTOR is independent of the coordinate system, but the COMPONENTS
are not.

Applied to a rank 2 tensor, the "matrix" you keep referring to is a
matrix of the tensor's COMPONENTS. To obtain the tensor itself, they
must be composed with the basis vectors of the coordinate system onto
which the tensor was projected to obtain the components:
* * *T = T_ij e^i (X) e^j * *(here (X) indicates a tensor product.
* * * * * * * * * * * * * * * The {T_ij} are a matrix of real numbers
* * * * * * * * * * * * * * * corresponding to the COMPONENTS of the
* * * * * * * * * * * * * * * tensor T; they are not T itself. The
* * * * * * * * * * * * * * * sum over i and j is a rank-2 tensor sum.)
The result is the same: the TENSOR is independent of the coordinate
system, but the COMPONENTS (the "matrix" you mention) are not.

There is no "matheMagic" here, just very basic tensor math. Stuff you
repeatedly refuse to learn, and/or are congenitally unable to
understand. shrug

Tom Roberts

Well constructed, very basic lesson. Nice work.
I have no doubt this will sail 4 inches above and half a foot to one
side of KW's head.

PD

On Feb 20, 10:37 pm, Koobee Wublee wrote:
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.

The matrix - yes, obviously. But not the tensor each matrix represents
in this case.

Again, it's like vectors: a vector is not an n-tuple of numbers. If I
say "let the vector v be given as (1,2)", I'm not saying anything -
it's just a pair of numbers. Without specifying a basis, (1,2) does
not specify a unique vector:

- is it 1 unit in the x-direction and 2 units in the y-direction?
- is it the opposite?
- is it 1 unit Northeast and 2 units Southeast?
- is it 1 radian clockwise from the direction West and 2 units counted
along that direction?
- etc. etc.?

OTOH if I say "let v be (1,2) in polar coordinatese" then I'm
specifying a vector. This vector - which is a 1 x 2 matrix - has
different matrix elements in the Cartesian frame:
(cos(2), sin(2)).

Two different matrices:

(1,2)

and:

(cos(2), sin(2))

yet one vector represented by both.

Same with tensors.

--
Jan Bielawski

On Feb 22, 12:50 pm, JanPB wrote:
On Feb 20, 10:37 pm, Koobee Wublee wrote:

Given a matrix with its elements being observer dependent, the matrix
must be observer dependent as well.

The matrix - yes, obviously. But not the tensor each matrix represents
in this case.

Thus, how do you turn a matrix into a tensor?

Again, it's like vectors: a vector is not an n-tuple of numbers. If I
say "let the vector v be given as (1,2)", I'm not saying anything -
it's just a pair of numbers. Without specifying a basis, (1,2) does
not specify a unique vector:

Yes, agreed. I would like to emphasize what you said below.

"Without specifying a basis, (1,2) does not specify a unique vector"

- is it 1 unit in the x-direction and 2 units in the y-direction?
- is it the opposite?
- is it 1 unit Northeast and 2 units Southeast?
- is it 1 radian clockwise from the direction West and 2 units counted
along that direction?
- etc. etc.?

OTOH if I say "let v be (1,2) in polar coordinatese" then I'm
specifying a vector. This vector - which is a 1 x 2 matrix - has
different matrix elements in the Cartesian frame:
(cos(2), sin(2)).

Yes, agreed. The basis vectors are very much an integral to the
coordinate system.

Two different matrices:

(1,2)

and:

(cos(2), sin(2))

yet one vector represented by both.

Yes, agreed.

Same with tensors.

Bullsh*t.

Given the vector 1, 2 with the Euclidean unit vectors x^ and y^ as
the basis vectors, another vector 4, 7 using the same basis vectors
is not the same vector as 1, 2. Do you not know that? At the
moment of presenting the field equations, the basis vectors are
already cast in concrete. Therefore, any solutions to the field
equations using the same basis vectors *do not* represent the same
metric. shrug

On Feb 22, 12:00 pm, Tom Roberts wrote:

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.

In coordinates with x headed east, y headed north, and z headed up, it
has components (1,0,0). Those three real numbers are NOT the vector, but
they are related to it, and with proper additional structure they can be
used to compute the vector. That additional structure is the BASIS
VECTORS of the coordinate system on which the vector was projected to
obtain those components. In this case the basis vector for x is a unit
vector pointed east, and similarly for y and z.

Yes, agreed.

In coordinates with x' headed north, y' headed west, and z' headed up,
that SAME VECTOR has components (0,-1,0). Those three real numbers are
NOT the vector, but they are related to it, and with proper additional
structure can be used to generate the vector. That additional structure
is the BASIS VECTORS of the coordinate system on which the vector was
projected to obtain those components. In this case the basis vector for
y' is a unit vector headed west, and similarly for x' and z'.

Yes, agreed. Notice that the basis vectors are very much part of the
coordinate system.

It should be clear that by computing the vector sum of the components of
the vector times the corresponding basis vectors, one obtains the vector
itself:
v = v_i e^i (here the {v_i} are the components of vector v, and
the {e^i} are the corresponding basis vectors.
The {v_i} are each a real number; the {e^i} are
each a vector, and the sum over i is a vector sum.)
The VECTOR is independent of the coordinate system, but the COMPONENTS
are not.

Yes, agreed. What you are describing, v, is the invariant geometry.
It should be observer independent. On the other hand, I can write the
same geometry into many different coordinate systems.

v = v_i e^I = v'_i e'^I = ...

Applied to a rank 2 tensor, the "matrix" you keep referring to is a
matrix of the tensor's COMPONENTS. To obtain the tensor itself, they
must be composed with the basis vectors of the coordinate system onto
which the tensor was projected to obtain the components:
T = T_ij e^i (X) e^j (here (X) indicates a tensor product.
The {T_ij} are a matrix of real numbers
corresponding to the COMPONENTS of the
tensor T; they are not T itself. The
sum over i and j is a rank-2 tensor sum.)
The result is the same: the TENSOR is independent of the coordinate
system, but the COMPONENTS (the "matrix" you mention) are not.

Yes, agreed. Your T can also be written into as followed.

T = T_ij e^i (x) e^j = T'_ij e'^i (x) e'^j = ...

There is no "matheMagic" here, just very basic tensor math.

Yes, you are still playing with matheMagic. What you refer to as the
tensor is T which is a scalar. Are the two following scalars the
same?

** T_ij e^i (x) e^j
** T'_ij e^i (x) e^j

The basis vectors are the same. They can only be the same if (T_ij =
T'_ij). Come on. This mathematics is very basic. So, when you
finally reach the field equations, what basis vectors are you working
with?

Stuff you
repeatedly refuse to learn, and/or are congenitally unable to
understand. shrug

It looks like you are still trapped in your matheMagic realm. Of
course, 'trapped' means your own indulgence in that make-believe world
of matheMagic realm. shrug

On Feb 23, 12:36 am, Koobee Wublee wrote:
On Feb 22, 12:00 pm, Tom Roberts wrote:

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.
In coordinates with x headed east, y headed north, and z headed up, it
has components (1,0,0). Those three real numbers are NOT the vector, but
they are related to it, and with proper additional structure they can be
used to compute the vector. That additional structure is the BASIS
VECTORS of the coordinate system on which the vector was projected to
obtain those components. In this case the basis vector for x is a unit
vector pointed east, and similarly for y and z.

Yes, agreed.

In coordinates with x' headed north, y' headed west, and z' headed up,
that SAME VECTOR has components (0,-1,0). Those three real numbers are
NOT the vector, but they are related to it, and with proper additional
structure can be used to generate the vector. That additional structure
is the BASIS VECTORS of the coordinate system on which the vector was
projected to obtain those components. In this case the basis vector for
y' is a unit vector headed west, and similarly for x' and z'.

Yes, agreed. Notice that the basis vectors are very much part of the
coordinate system.

It should be clear that by computing the vector sum of the components of
the vector times the corresponding basis vectors, one obtains the vector
itself:
v = v_i e^i (here the {v_i} are the components of vector v, and
the {e^i} are the corresponding basis vectors.
The {v_i} are each a real number; the {e^i} are
each a vector, and the sum over i is a vector sum.)
The VECTOR is independent of the coordinate system, but the COMPONENTS
are not.

Yes, agreed. What you are describing, v, is the invariant geometry.
It should be observer independent. On the other hand, I can write the
same geometry into many different coordinate systems.

v = v_i e^I = v'_i e'^I = ...

Applied to a rank 2 tensor, the "matrix" you keep referring to is a
matrix of the tensor's COMPONENTS. To obtain the tensor itself, they
must be composed with the basis vectors of the coordinate system onto
which the tensor was projected to obtain the components:
T = T_ij e^i (X) e^j (here (X) indicates a tensor product.
The {T_ij} are a matrix of real numbers
corresponding to the COMPONENTS of the
tensor T; they are not T itself. The
sum over i and j is a rank-2 tensor sum.)
The result is the same: the TENSOR is independent of the coordinate
system, but the COMPONENTS (the "matrix" you mention) are not.

Yes, agreed. Your T can also be written into as followed.

T = T_ij e^i (x) e^j = T'_ij e'^i (x) e'^j = ...

There is no "matheMagic" here, just very basic tensor math.

Yes, you are still playing with matheMagic. What you refer to as the
tensor is T which is a scalar.

Geez, what a goofball. T = T = T_ij e^i (X) e^j is a scalar. Geez,
what a goofball.

Are the two following scalars the
same?

** T_ij e^i (x) e^j
** T'_ij e^i (x) e^j

The basis vectors are the same. They can only be the same if (T_ij =
T'_ij). Come on. This mathematics is very basic. So, when you
finally reach the field equations, what basis vectors are you working
with?

Stuff you
repeatedly refuse to learn, and/or are congenitally unable to
understand. shrug

It looks like you are still trapped in your matheMagic realm. Of
course, 'trapped' means your own indulgence in that make-believe world
of matheMagic realm. shrug

On Feb 21, 5:59 am, Eric Gisse wrote:

On Feb 20, 5:55 pm, Edward Green wrote:

...

We seem to have developed a few miscommunications. You are certain
that I am incorrigibly assuming some elements of quantum mechanics in
a classical discussion, which I claim is not the case, and secondly, I
was using "intrinsic angular momentum" in a possibly non-standard
sense, which may need to be fixed. I'm beginning to think you
understood the intended sense of "intrinsic angular momentum", but
think that there is no point in considering such a thing unless we
admit quantum mechanical spin to our world.

Regarding the latter problem, I had forgotten that the angular
momentum of any finite object -- a globule of fluid, a rigid body, a
swarm of particles, etc. -- can be decomposed into "intrinsic" and
"extrinsic" angular momentum relative to any origin. I hadn't
intended to invoke this decomposition, but in hindsight I was not that
far off, because what I _was_ trying to invoke would form a
_component_ of "intrinsic angular momentum". We merely need one more
adjective: _specific_ intrinsic angular momentum, as in "specific
heat" (a term per unit mass or volume). In other words, analyzing the
angular momentum of a fluid globule, say, we have J_tot = J_ext +
J_int where J_ext = R x M V (motion of total mass taken at center of
mass), and J_int = Int[r x rho v]dr + Int[j(r)]dr, which is the
integral of the (intrinsic) angular momentum attributed to the
velocity and mass of the fluid, plus an integral of a residual term,
which is not attributable to velocity and mass of the fluid.

We are allowed to add such a term to classical continuum mechanics
without any physical motivation at all, just to see where it leads
us. But as motivation I suggested a few model systems, one of which
(unfortunately) included qm spin in an otherwise classical model
Another model system I suggested included spinning pebbles (definitely
non-qm!), on a scale where we did not resolve the pebbles, but
nonetheless had to include the contribution of their (non-qm!) spin in
J_tot.

[While is was undoubtedly clever for the Fathers to label
whatever_it_is_that_an_electron_has as "spin", it becomes very
annoying if the consequence of this labeling deny us any other use of
the ordinary English word in other physical contexts. "Spin" and
"rotate" seem to divide up the territory according to size and angular
velocity -- merry-go-rounds rotate while tops spin: merry-go-rounds
only spin prior to Playground Horror!, when, following a failure of
the steam kalliope regulator valve, they eject children like water
from a salad spinner, break free of their moorings, and kill and
injure scores in a wild careen before smashing into the Boat House.]

If I refer to a "spinning pebble" or even, incautiously, to the "spin
of a pebble", the ordinary meaning should be clear in context.

Now, supposing we are on the same page about
whatever_we_should_call_it, the side issue became, does a We(y)sshoff
fluid have "specific intrinsic angular momentum"? You say no, and I
say it looks that way, but that of course the references will govern.

What I mean by "Kerr is not a disk" is that the solution spans all 3
spatial dimensions and is not a metric describing a planar...anything.
Whereas the Neugebauer/Meiniel solution is, well, a disk of rotating
matter.

I take it, if the Neubegauer/Meiniel solution is a solution in GR --
as our reference

http://xrl.us/bgic4

seems to imply, that the solution also spans all 3 spatial dimensions.
The _source_ may be a two-dimensional disk, but the field is three
dimensional: like the electric field of a charged disk. The claim by
the Wikipediast was that some limiting form of the Neubegauer/Meiniel
solution converged to some form or part of the Kerr solution. We
don't have to take his word for it, but this assertion is prima facie
plausible: it may reflect a case where the radius of the disk shrinks
to zero, holding its angular momentum constant, or else in the far
field, where the disk looks like a point source, etc. -- similar to
the far field of a charged disk.

The only relation between the two seems to be that the rotating disk
is presumed to collapse into a rotating Kerr hole. Past that, I see no
relationship.

You've asserted that you believe the exterior part of the Kerr
solution describes the gravitational field of rotating massive bodies,
and a rotating disk of dust would seem to qualify? The only
distinction is the geometry, if you are thinking of rotating spherical
bodies, but under some limiting conditions the cases should coincide
-- certainly they seem related.

...

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 didn't think the wording mattered since I didn't think there'd be
confusion regarding quantum spin.

Well there wasn't, on my part: I merely cited a fluid with a
distribution of unpaired spins as a possible _example_ of a fluid
which might be treated as posessing (what I was calling) intrinsic
angular momentum. I went on to give other examples.

I could have worded it better, I
suppose, but "extrinsic angular momentum" just seems awkward and
simply saying "angular momentum" doesn't seem quite good enough.

Ok... I think we may be clear on our claims he you are saying that
a Weyssenhoff is simply a rotating globule of fluid, having intrinsic
angular momentum only in the sense that any other rotating body has
intrinsic angular momentum, and I _thought_ you were saying (without
any necessary reference to qm!) that it has _specific_ intrinsic
angular momentum.

If I had to bet a dollar, I'd still put it on "specific intrinsic
angular momentum"; it's my dollar. :-) Now that you've corrected the
spelling, it indeed becomes easier to search...

http://arxiv.org/abs/0706.2367

" The Weyssenhoff fluid is a perfect fluid with spin where the spin of
the matter fields is the source of torsion in an Einstein-Cartan
framework."


http://arxiv.org/abs/gr-qc/0601089 (one of your references)

See especially section 3, involving the "intrinsic angular momentum
tensor", S_ij, and the "spin tensor", S^ij


http://wwwitp.physik.tu-berlin.de/~h...on/node38.html

"There are two spin fluids which are typically different from each
other: The Weyssenhoff fluid which is a classical one, and the Dirac-
electron fluid which represents the classical description of a quantum-
fluid."


I'm not going to pretend I can fully understand all this, but it
certainly sounds like the Weyssenhoff fluid has "specific intrinsic
angular momentum", which may even be called "spin", although
differentiated from "quantum spin". [Those Skilled in the Art seem to
be perfectly comfortable with using "spin" in a classical context.]

Lots else to reply to, but that's enough for one post.

On Feb 21, 12:34 am, Koobee Wublee wrote:
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

Sure it can. You've ONCE AGAIN confused the metric with the
*components* of the metric. You have considered the matrix of
components, but seem to have lost track of the basis vectors.


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