I see that the discussion of whether the metric qualifies as a tensor
or not has splintered into many different localized groups which are
very confusing. Thus, I would like to put everything into a nutshell
of the pros and cons. Although the tensors can cover more ranks, the
discussion should only be limited what GR covers. That is rank 2
tensors or the 4-by-4 matrices.

The definition of a tensor as agreed by almost everyone says that any
matrix invariant under a coordinate transformation is a tensor while
the elements to the matrix can change under the same transformation.

**** Pros' argument:

The geometry described is invariant under any transformation. Thus, it
can be represented by an operator operating on the same vector twice.

ds^2 = f(dq,dq)

Where

** ds^2 = the invariant geometry
** f() = the operator
** dq = the coordinate vector

Of course, we all know that

ds^2 = g_ij dq^i dq^j = g'_mn dq'^m dq'^n

Where

** dq' = Different coordinate system

The question is the matrices ([g] = [g']).

**** Cons's argument:

According to the pros, although the elements of the matrices (g_ij =/=
g'_ij), the martrix ([g] = [g']). In doing so, they can never
describe what constitutes how these 2 matrices are identical. They
avoid it as if a plague in fact. They can only hand-wave it by saying
over and over again that they are indeed identical.

**** What is at stake?

Does it matter if the metric is a tensor or not for the sake of the
mathematics involved? Yes, it does. The interpretation to the
infinite number of solutions to the field equations is at stake. The
existence of the black holes is at stake.

Apparently, the pros have never followed through the derivation of the
solutions to the field equations. Each solution in terms of g_ij is
only valid to the choice of coordinate system where each solution very
different from the others must describe a different geometry using the
same coordinate system. This means the field equations do have an
infinite number of solutions in which the Schwarzschild metric,
Schwarzschild's original metric, or any other is just as valid as any
other where each describes a different geometry. This would shatter
the general theory of relativity.

The field equations in free space are

R_ij(q^0, q^1, q^2, q^3) = 0

Where

** R_ij(q) = Ricci tensor as a function of q
** g1_ij(q) = Solution as function of q
** g2_ij(q) = Solution as function of q
** g3_ij(q) = Solution as function of q
** ...

And all these different geometries described by each solution with the
same coordinate system.

** ds1^2 = g1_ij dq^i dq^j
** ds2^2 = g2_ij dq^i dq^j
** ds3^2 = g3_ij dq^i dq^j
** ...

The silly claim by the pros is that ([g1] = [g2] = [g3] = ...). Thus,
(ds1^2 = ds2^2 = ds^3 = ...). They are wrong. The simple mathematics
shows so. They cannot prove why ([g1] = [g2] = [g3] = ...). The
Schwarzschild metric is not unique. The existence of black holes is
based on a non-unique solution to the field equations. 100 years of
physics is totally BS based on this wrong concept of linear algebra.

Koobee Wublee wrote:
I see that the discussion of whether the metric qualifies as a tensor
or not has splintered into many different localized groups which are
very confusing. Thus, I would like to put everything into a nutshell
of the pros and cons. Although the tensors can cover more ranks, the
discussion should only be limited what GR covers. That is rank 2
tensors or the 4-by-4 matrices.

The definition of a tensor as agreed by almost everyone says that any
matrix invariant under a coordinate transformation is a tensor while
the elements to the matrix can change under the same transformation.

That isn't what a tensor is. A tensor is _NOT_ a matrix that is
invariant under a coordinate transformation. Stop trying to shoehorn
what you learned in linear algebra into a different subject.

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

[...]

The silly claim by the pros is that ([g1] = [g2] = [g3] = ...). Thus,
(ds1^2 = ds2^2 = ds^3 = ...). They are wrong. The simple mathematics
shows so. They cannot prove why ([g1] = [g2] = [g3] = ...). The
Schwarzschild metric is not unique. The existence of black holes is
based on a non-unique solution to the field equations. 100 years of
physics is totally BS based on this wrong concept of linear algebra.

Jesus christ more of this tripe.

You continually make mistake after mistake because you have no training
in differential geometry. Your argument is DOA because differential
geometry does not obey the rules of linear algebra. All that crap you
learned in there is _NOT APPLICABLE_.

I repeat because it bears repeating: Nobody says that the components of
the metric tensor are invariant under a coordinate transformation.
Nobody says that a particular representation of the metric tensor is
invariant under a coordinate transformation.

I am simply amazed that you even talk about the uniqueness of the
Schwarzschild solution when you admit not having even seen the proof of
Birkhoff's theorem.

If you want to continue 'debating', I would like to remind you that you
once asserted that you could introduce curvature into a manifold
through a coordinate transformation. You really do not know what you
are talking about and should best study a subject before you attempt to
**** all over it.

Koobee Wublee says...

The geometry described is invariant under any transformation. Thus, it
can be represented by an operator operating on the same vector twice.

ds^2 = f(dq,dq)

Where

** ds^2 = the invariant geometry
** f() = the operator
** dq = the coordinate vector

Of course, we all know that

ds^2 = g_ij dq^i dq^j = g'_mn dq'^m dq'^n

Where

** dq' = Different coordinate system

The question is the matrices ([g] = [g']).

According to the pros, although the elements of the matrices (g_ij =/=
g'_ij), the martrix ([g] = [g']).

Nobody has said that the *matrices* are the same. What they have
told you is that a tensor is *not* a matrix. A tensor is a bilinear
mapping on vectors and 1-forms.

In doing so, they can never
describe what constitutes how these 2 matrices are identical.

They're *not* identical. Nobody said they were. What they said
was that the *tensor* is not equal to the matrix. The tensor is
not changed by changing coordinate systems, although the corresponding
matrix is.

They avoid it as if a plague in fact. They can only hand-wave it
by saying over and over again that they are indeed identical.

Nobody has said it even once. What they have told you, over and
over again is that the metric is a tensor, and a tensor is independent
of coordinates, and a tensor is *not* a matrix, it is a bilinear
operator.

--
Daryl McCullough
Ithaca, NY

You would need 50 more IQ points to realize
that you're incompetent.

Koobee Wublee wrote:
I see that the discussion of whether the metric qualifies as a tensor
or not has splintered into many different localized groups which are
very confusing. Thus, I would like to put everything into a nutshell
of the pros and cons. Although the tensors can cover more ranks, the
discussion should only be limited what GR covers. That is rank 2
tensors or the 4-by-4 matrices.

The definition of a tensor as agreed by almost everyone says that any
matrix invariant under a coordinate transformation is a tensor while
the elements to the matrix can change under the same transformation.

**** Pros' argument:

The geometry described is invariant under any transformation. Thus, it
can be represented by an operator operating on the same vector twice.

ds^2 = f(dq,dq)

Where

** ds^2 = the invariant geometry
** f() = the operator
** dq = the coordinate vector

Of course, we all know that

ds^2 = g_ij dq^i dq^j = g'_mn dq'^m dq'^n

Where

** dq' = Different coordinate system

The question is the matrices ([g] = [g']).

**** Cons's argument:

According to the pros, although the elements of the matrices (g_ij =/=
g'_ij), the martrix ([g] = [g']). In doing so, they can never
describe what constitutes how these 2 matrices are identical. They
avoid it as if a plague in fact. They can only hand-wave it by saying
over and over again that they are indeed identical.

**** What is at stake?

Does it matter if the metric is a tensor or not for the sake of the
mathematics involved? Yes, it does. The interpretation to the
infinite number of solutions to the field equations is at stake. The
existence of the black holes is at stake.

Apparently, the pros have never followed through the derivation of the
solutions to the field equations. Each solution in terms of g_ij is
only valid to the choice of coordinate system where each solution very
different from the others must describe a different geometry using the
same coordinate system. This means the field equations do have an
infinite number of solutions in which the Schwarzschild metric,
Schwarzschild's original metric, or any other is just as valid as any
other where each describes a different geometry. This would shatter
the general theory of relativity.

The field equations in free space are

R_ij(q^0, q^1, q^2, q^3) = 0

Where

** R_ij(q) = Ricci tensor as a function of q
** g1_ij(q) = Solution as function of q
** g2_ij(q) = Solution as function of q
** g3_ij(q) = Solution as function of q
** ...

And all these different geometries described by each solution with the
same coordinate system.

** ds1^2 = g1_ij dq^i dq^j
** ds2^2 = g2_ij dq^i dq^j
** ds3^2 = g3_ij dq^i dq^j
** ...

The silly claim by the pros is that ([g1] = [g2] = [g3] = ...). Thus,
(ds1^2 = ds2^2 = ds^3 = ...). They are wrong. The simple mathematics
shows so. They cannot prove why ([g1] = [g2] = [g3] = ...). The
Schwarzschild metric is not unique. The existence of black holes is
based on a non-unique solution to the field equations. 100 years of
physics is totally BS based on this wrong concept of linear algebra.

Koobee Wublee wrote:
The definition of a tensor as agreed by almost everyone says that any
matrix invariant under a coordinate transformation is a tensor

That is not the definition of tensor. Not even close. You need to LEARN
about the subject before attempting to write about it.


Tom Roberts

Koobee Wublee wrote:
I see that the discussion of whether the metric qualifies as a tensor
or not has splintered into many ....

Actually two...


....different localized groups which are
very confusing. Thus, I would like to put everything into a nutshell
of the pros and cons. Although the tensors can cover more ranks, the
discussion should only be limited what GR covers. That is rank 2
tensors or the 4-by-4 matrices.

Do ypou know what a rank is? Do you know the difference between the
type of a tensor and the rank of the tensor?

The definition of a tensor as agreed by almost everyone ....

This has never been demonstrated by you. In fact it would be impossible
for you to do because you don't know what a tensor is. This I have
gleened from the posts of yours that I've readon this newsgroup. That
you refuse to refer us to a math text which would back your assertion
speaks volumes to this end.

...says that any
matrix invariant under a coordinate transformation is a tensor while
the elements to the matrix can change under the same transformation.

And here is the problem. You constantly confuse matrices with tensors.
While a tensor can be *represented* by matrices in some cases (e.g. in
a given coordinate system) a tensor is not a matrix. Not all matrices
represent tensors either.


**** Pros' argument:

The geometry described is invariant under any transformation. Thus, it
can be represented by an operator operating on the same vector twice.

ds^2 = f(dq,dq)

Where

** ds^2 = the invariant geometry

ds^2 is the invariant *interval*.

** f() = the operator
** dq = the coordinate vector

There is zero need for dq to be a coordinate vector at all. It may only
need to be the geometric object that is the vector without any
coordinate representation.

Of course, we all know that

ds^2 = g_ij dq^i dq^j = g'_mn dq'^m dq'^n

That we agree on.

Where

** dq' = Different coordinate system

dq is the differential of a coordinate.

The question is the matrices ([g] = [g']).

That is a question brought to this newsgroup by you and for which you
invented your own definition of "tensor".


**** Cons's argument:

According to the pros, although the elements of the matrices (g_ij =/=
g'_ij), the martrix ([g] = [g']). In doing so, they can never
describe what constitutes how these 2 matrices are identical. They
avoid it as if a plague in fact. They can only hand-wave it by saying
over and over again that they are indeed identical.

If all tensors were matrices and nothing else then you might have
something there. But you're wrong. Tensors and matrices are totally
different things. If I give you a matrix then there it would be
impossible for you to tell me if it represented a tensor.

So I repeat my challenge to you - Post a reference to a math or GR
textbook which uses your definition of tensor.

The challenge is made.

Will the challenge be accepted?

Best wishes

Pete

"Koobee Wublee" wrote in message ps.com...
I see that the discussion of whether the metric qualifies as a tensor
or not has splintered into many different localized groups which are
very confusing. Thus, I would like to put everything into a nutshell
of the pros and cons. Although the tensors can cover more ranks, the
discussion should only be limited what GR covers. That is rank 2
tensors or the 4-by-4 matrices.

The definition of a tensor as agreed by almost everyone says that any
matrix invariant under a coordinate transformation is a tensor while
the elements to the matrix can change under the same transformation.

"The Absurd Claim of the Imbecile about the Metric as a Tensor":
http://users.telenet.be/vdmoortel/di...surdClaim.html

Dirk Vdm

Dirk Van de moortel wrote:
"Koobee Wublee" wrote in message ps.com...
I see that the discussion of whether the metric qualifies as a tensor
or not has splintered into many different localized groups which are
very confusing. Thus, I would like to put everything into a nutshell
of the pros and cons. Although the tensors can cover more ranks, the
discussion should only be limited what GR covers. That is rank 2
tensors or the 4-by-4 matrices.

The definition of a tensor as agreed by almost everyone says that any
matrix invariant under a coordinate transformation is a tensor while
the elements to the matrix can change under the same transformation.

"The Absurd Claim of the Imbecile about the Metric as a Tensor":
http://users.telenet.be/vdmoortel/di...surdClaim.html

You ought to consider having a separate page
for classics like this one. I think that putting
a gem like this amongst fumbles due to the mere
willful stupidity of the standard kook doesn't
do justice to the self-delusion required to write
something so stupid while simultaneously believing
he is the one and only disciple of Riemann.
This one is just too funny.

"bergeron" wrote in message ups.com...

Dirk Van de moortel wrote:
"Koobee Wublee" wrote in message ps.com...
I see that the discussion of whether the metric qualifies as a tensor
or not has splintered into many different localized groups which are
very confusing. Thus, I would like to put everything into a nutshell
of the pros and cons. Although the tensors can cover more ranks, the
discussion should only be limited what GR covers. That is rank 2
tensors or the 4-by-4 matrices.

The definition of a tensor as agreed by almost everyone says that any
matrix invariant under a coordinate transformation is a tensor while
the elements to the matrix can change under the same transformation.

"The Absurd Claim of the Imbecile about the Metric as a Tensor":
http://users.telenet.be/vdmoortel/di...surdClaim.html

You ought to consider having a separate page
for classics like this one. I think that putting
a gem like this amongst fumbles due to the mere
willful stupidity of the standard kook doesn't
do justice to the self-delusion required to write
something so stupid while simultaneously believing
he is the one and only disciple of Riemann.
This one is just too funny.

Paradoxically painfully funny :-|

Dirk Vdm

"Dirk Van de moortel" wrote
in message ...

"bergeron" wrote in message
ups.com...

Dirk Van de moortel wrote:
"Koobee Wublee" wrote in message
ps.com...
I see that the discussion of whether the metric qualifies as a tensor
or not has splintered into many different localized groups which are
very confusing. Thus, I would like to put everything into a nutshell
of the pros and cons. Although the tensors can cover more ranks, the
discussion should only be limited what GR covers. That is rank 2
tensors or the 4-by-4 matrices.

The definition of a tensor as agreed by almost everyone says that any
matrix invariant under a coordinate transformation is a tensor while
the elements to the matrix can change under the same transformation.

"The Absurd Claim of the Imbecile about the Metric as a Tensor":

http://users.telenet.be/vdmoortel/di...surdClaim.html

You ought to consider having a separate page
for classics like this one. I think that putting
a gem like this amongst fumbles due to the mere
willful stupidity of the standard kook doesn't
do justice to the self-delusion required to write
something so stupid while simultaneously believing
he is the one and only disciple of Riemann.
This one is just too funny.

Paradoxically painfully funny :-|

Dirk Vdm

Wanna know what really irks me about threads like this? Its those people who
say things like
"You don't understand the book you're quoting" or that "You quote books
because you don't know what's in them" etc.

When in fact, and Koobee is a good example, is that I almost only refer to
texts when someone is claiming that something is defined other than the
claim it is, as in this case.

But watch Dirk. I placed a challege to post a textual source in which the
authors definition is identical to his. Think he'll answer the call? I say
noway Jose. He's commited himself far too greatly than to be able to admit
his mistake now. On the other hand I try to always admit my mistakes, even
when others have forgotten them. In fact I start threads which seem to come
out of nowhere or to rehash a dead subject when actually it was to correct
an error I made a while before that.

Hey Dirk - I thought I was in your list of Fumbles?? Did I loose my proud
position of recognition there? LOL!

Pete