Eric Gisse wrote:
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.

Daryl McCullough wrote:
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.

Apparently, you are trying to find a way out of your past mistake. See
how a metric tensor is defined. It is a 4-by-4 matrix.

http://mathworld.wolfram.com/MetricTensor.html

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.

A particular metric is only valid to a particular choice of coordinate
system. Thus, the metric is only a matrix. It is not a tensor.

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.

That is because what I said is correct. You et al cannot produce
anything to prove the metric is invariant. The metric directly lead to
the solutions to the field equations.

Tom Roberts wrote:
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.

You are the one who needs to learn what a tensor is. Here is the
definition of a metric.

http://mathworld.wolfram.com/MetricTensor.html

pmb wrote:
Koobee Wublee wrote:

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

Yes. Do you know how to derive the field equations? Do you know how
to derive the Christoffel symbols of the second kind? Show me.

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.

You are totally confused. You need to read my posts more carefully.

...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.

I maintain that the metric cannot be a tensor.

**** 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*.

Yes, I agree.

** 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.

Without dq, you can never describe ds^2. You are confused.

Of course, we all know that

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

That we agree on.

Good.

Where

** dq' = Different coordinate system

dq is the differential of a coordinate.

I meant dq' is of a different coordinate system from dq.

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".

Oh, no. You agreed on the following, remember?

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

So, what is your problem?

**** 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.

I am not saying all matrices are tensors. I maintain the metric matrix
is not a tensor.

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

Fine, if it would make an insecure middle aged man feel secure.

http://mathworld.wolfram.com/MetricTensor.html

The challenge is made.

Consider you silly challenge replied.

Will the challenge be accepted?

I don't even call your childish demand a challenge. It was so
infantile that I had to ignore such silly demand. So, next time if you
really want to challenge me with something, please keep it
intellectually stimulating and don't use your level as the standard.

Eric Gisse wrote:
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 metric and Ricci tensors are merely 4-by-4 matrices. Do you know
what matrices are?

What I say about what a tensor is remains correct.

http://mathworld.wolfram.com/MetricTensor.html

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.

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_.

There is no mistake. You still have no argument and ranting like a
buffoon.

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

I repeat. You changed your mind when I pointed out the obvious.

Nobody says that a particular representation of the metric tensor is
invariant under a coordinate transformation.

Then, what is your problem? You should then agree with me that each
solution to the field equations must describe a different geometry.
The Birkhoff's theorem is so wrong.

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.

I am amazed that you have so much to talk about without understanding
the calculus of variations, differential geometry, field equations,
solutions to the field equations, and the Birkhoff's theorem itself.

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.

No, I never said that.

You really do not know what you
are talking about and should best study a subject before you attempt to
**** all over it.

Open your mouth wide. Here it comes. Does it still taste the curry
flavor?

Koobee Wublee wrote:
Tom Roberts wrote:
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.

You are the one who needs to learn what a tensor is. Here is the
definition of a metric.

http://mathworld.wolfram.com/MetricTensor.html

If you agree with your reference, why did you
post the non-sense in your orginal article and
then argue with everyone who called it non-sense?
If you don't agree with your reference why did you
bother posting the link?

"Koobee Wublee" wrote in message
oups.com...
pmb wrote:
Koobee Wublee wrote:

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


[snipped]

When you can respond without being insulting and condescending then I will
resume this discussion.

Regards

Pete

"bergeron" wrote in message
oups.com...

Koobee Wublee wrote:
Tom Roberts wrote:
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.

You are the one who needs to learn what a tensor is. Here is the
definition of a metric.

http://mathworld.wolfram.com/MetricTensor.html

If you agree with your reference, why did you
post the non-sense in your orginal article and
then argue with everyone who called it non-sense?
If you don't agree with your reference why did you
bother posting the link?

Actually he failed here in this challenge. First of all I asked him to
produce a *source* of the *definition* he adheres to. I didn't ask him to
give me a reference to an *example* of a tensor. We still have no idea if
(1) he understands that the link he just gave contradicts him and (2) he's
willing and able to find anything that agrees with him. Or at least a very
very good reason why the rest of the world should all change their
definitions to meet his needs.

KW - We have not yet argued a point of logic since we have not yet agreed on
a definition of tensor, one that you failed to provide and one which I wrote
an entire web page to explain to you. Since you insult those who you're
disccussing physics with then can you think of one such person who would
want to discuss physics with you? Oh yes. By the way. The derivations (i.e.
the work you wanted me to do) has been on my web site for years. If you look
for it you will easily find it. And you have yet to tell us what field it is
you want derived. Care to give us the Lagrangian to start with?

Pete

"Pmb" wrote in message . ..

"bergeron" wrote in message
oups.com...

Koobee Wublee wrote:
Tom Roberts wrote:
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.

You are the one who needs to learn what a tensor is. Here is the
definition of a metric.

http://mathworld.wolfram.com/MetricTensor.html

If you agree with your reference, why did you
post the non-sense in your orginal article and
then argue with everyone who called it non-sense?
If you don't agree with your reference why did you
bother posting the link?

Actually he failed here in this challenge. First of all I asked him to
produce a *source* of the *definition* he adheres to. I didn't ask him to
give me a reference to an *example* of a tensor. We still have no idea if
(1) he understands that the link he just gave contradicts him and (2) he's
willing and able to find anything that agrees with him. Or at least a very
very good reason why the rest of the world should all change their
definitions to meet his needs.

Pete,
(1) he *does* understand that the link contradicts him,
(2) he is not willing or able to find anything that agrees with him,
(3) the only thing he is after, is to keep you busy fighting him.
He is a troll :-)

Dirk Vdm

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

"Pmb" wrote in message
. ..

"bergeron" wrote in message
oups.com...

Koobee Wublee wrote:
Tom Roberts wrote:
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.

You are the one who needs to learn what a tensor is. Here is the
definition of a metric.

http://mathworld.wolfram.com/MetricTensor.html

If you agree with your reference, why did you
post the non-sense in your orginal article and
then argue with everyone who called it non-sense?
If you don't agree with your reference why did you
bother posting the link?

Actually he failed here in this challenge. First of all I asked him to
produce a *source* of the *definition* he adheres to. I didn't ask him to
give me a reference to an *example* of a tensor. We still have no idea if
(1) he understands that the link he just gave contradicts him and (2)
he's willing and able to find anything that agrees with him. Or at least
a very very good reason why the rest of the world should all change their
definitions to meet his needs.

Pete,
(1) he *does* understand that the link contradicts him,
(2) he is not willing or able to find anything that agrees with him,
(3) the only thing he is after, is to keep you busy fighting him.
He is a troll :-)

Dirk Vdm

Thanks Dirk

I guess that's one problem being a Christian. When you give someone every
benefit of the doubt they'll still let ya down. I guess that's why I used to
be combative. I thought I could actually get people to see their errors. Now
as I grow as a Christian I see that its also important to know when to quit.

By the way, I'm still hoping you recall a fumble that you have on your web
site that I was in. I'm feeling a bit sentimental for the old flame-war
days. :-)

KW - Consider yourself Plonked

Per KWs challenge to me - it will be fulfilled at anyone elses request. Just
to be sure that someone really gives a hoot.

Best Regards

Pete

ps - Dirk - Sorry for all the religious talk. I was merely letting people
get an idea of how I've changed from past flame-war days. It took a deep
look inside plus a lot of drugs. (anti-depressants, anti-anxiety etc.)