"Pmb" wrote in message ...

"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.

You can get KW to see his errors alright. He couldn't care less.
He is just like Androcles, although slightly more intelligent.
He doesn't care that he is (shown) wrong - he is anonymous, so
he can just pick another name again and start all over.


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. :-)

You keep saying this, and I do recall that we had some flame
exchanges, but I don't recall having a fumble of yours on my site.
Perhaps I had one for a while, but I don't have any now afaics.
Do you have a pointer to a post I made?


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.)

No problem.
I'm not religious but I am very tolerant to religious people - as
long as they don't try to violently impose their views upon others :-)

Dirk Vdm

Dirk Van de moortel wrote:
aka. Sperm Lover

(1) he *does* understand that the link contradicts him,

The sperm lover does not understand that link.

(2) he is not willing or able to find anything that agrees with him,

The sperm lover started learning about SR 10 years ago. With one of
the most prolific posters, the sperm lover is still clueless about GR.
The sperm lover is still stuck in SR. In fact, the sperm lover has not
fully understood SR. shrug

(3) the only thing he is after, is to keep you busy fighting him.

The sperm lover loves to post nonsense and garbage.

He is a troll :-)

The sperm lover is a troll. That is why I ignore the sperm lover most
of the time. shrug

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

"Pmb" wrote in message
...

"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.

You can get KW to see his errors alright. He couldn't care less.
He is just like Androcles, although slightly more intelligent.
He doesn't care that he is (shown) wrong - he is anonymous, so
he can just pick another name again and start all over.

Yipes! Another Andocles? Oy vey! But it seems to be true.


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. :-)

You keep saying this, and I do recall that we had some flame
exchanges, but I don't recall having a fumble of yours on my site.

Could be my imagination. I think it was in the days when mike varney was
around and he was hitting below the belt, but with my memory I'm sure to be
wrong.

No problem.
I'm not religious but I am very tolerant to religious people - as
long as they don't try to violently impose their views upon others :-)

I couldn't concieve of forcing my religion on others. As Jesus once said, if
you walk into a town and that town will not hear you then go to the edge of
the town and kick the dirt of that town off your feet. He never said to blow
it up with a scud missle. :-)

Best regards

Pete

Pmb wrote:

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.

Here we go again about that asinine challenge of yours.

We still have no idea if
(1) he understands that the link he just gave contradicts him

Yes, that is correct.

and (2) he's willing and able to find anything that agrees with him.

No, I cannot because I am pointing out a misconception existed for 100
years regarding this subject in linear algebra.

Or at least a very
very good reason why the rest of the world should all change their
definitions to meet his needs.

I am not asking the world to change the definition.

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.

The definition of tensor is not the argument. It is if the metric is a
tensor or not. I have already explained why the metric is not a
tensor.

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?

So, it is OK for you to insult others but not the other way around. Is
that a fine trait of a Christian?

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.

I did not find it, and I think you are just blowing smoke.

And you have yet to tell us what field it is
you want derived.

Reread what I wrote then.

Care to give us the Lagrangian to start with?

That would be a start? How do you derive that Lagrangian?

bergeron wrote:

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?

If you don't understand the subject matter and the stakes at hand,
why do pretend you know something by making these unworthy of comments?

Pmb wrote:

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

So, it is OK for you to insult me by challenging me with an asinine
research project, and you get bent out of shape when I challenge you
with the derivation of the field equations and the Christoffel symbols
of the second kind. I have treated you with respect by giving you
rightful challenges which is unlike your insult on me. What type of
Christians are you anyway? Everyone else's god is false while yours
is the true god?

That is fine with me if you cannot take this challenge. If you don't
know, just say so. Don't call this an insult and use it as an excuse
out of this one.

Koobee Wublee wrote:
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?

You only pick those two tensors because they fit your predetermined
notion of what a tensor is.

How about the Reimann tensor? Weyl tensor? Levi-Citva tensor? What is
the matrix representation of them?


What I say about what a tensor is remains correct.

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

....what is up with people continually quoting links that explicitly
disagree with them?


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.

So you still think linear algebra == differential geometry?

Looks like sal still has you pegged. Four months later and you still
repeat the same mistakes.

http://groups.google.com/group/sci.p...0?dmode=source



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.

One solution, infinitely many representations. That seems to be the
thing you have the most trouble with.

If I start off with the Minkowski metric and transform to cylindrical
coordinates, do I have a different manifold? If I perform the inverse
transformation on that then transform to spherical coordinates, do I
have a different manifold? If I transform to prolate spherical
coordinates, do I have a different manifold? If I perform all those
transformations without inverting, then transform back, do I have a
different manifold?


The Birkhoff's theorem is so wrong.

You have never even seen the proof of Birkhoff's theorem.


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.

Since you are completely unwilling to test me, how do you know I don't
understand the calculus of variations? As I told you before, submit a
sample problem you want me to solve. All I ask is that you be able to
solve it yourself.

I don't really know why you are so arrogant.

You are unfamiliar with even ODE existence theorems. I don't know them
by name but I know they exist so I don't make dumb **** arguments like
"well there could be more than one solution to that ODE!".

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

"Again, foliation? There is not even such a word in the English
language. Please stop talking in riddles to justify your points."

http://groups.google.com/group/sci.p...6?dmode=source



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.

Never?

http://groups.google.com/group/sci.p...4?dmode=source

You never did compute the curvature scalar of that manifold to prove
that is curved. Or even the Riemann tensor.


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:
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.

Write out the matrix representation of the Riemann curvature tensor
since you are so sure that tensors are matricies.

Koobee Wublee says...

Daryl McCullough wrote:

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.

Well, you were not correct when you said

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

Nobody ever said that the matrices were equal. So either you
are completely misunderstanding what is being said to you, or
you are lying about it.

--
Daryl McCullough
Ithaca, NY

"Eric Gisse" wrote in message
oups.com...

Koobee Wublee wrote:
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?

You only pick those two tensors because they fit your predetermined
notion of what a tensor is.

How about the Reimann tensor? Weyl tensor? Levi-Citva tensor? What is
the matrix representation of them?


What I say about what a tensor is remains correct.

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

...what is up with people continually quoting links that explicitly
disagree with them?


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.

Don't worry Eric. We know how silly this comment to you looks. He seems to
do that when he is stuck in a corner. Poor guy, doesn't know when to quit.

So you still think linear algebra == differential geometry?

Looks like sal still has you pegged. Four months later and you still
repeat the same mistakes.

I should have listened to sal too. He's always right on these things. I
guess that's because sal is a pretty darn sharp man himself. Ooooo. If only
I had his mind for math!



http://groups.google.com/group/sci.p...0?dmode=source



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.

One solution, infinitely many representations. That seems to be the
thing you have the most trouble with.

If I start off with the Minkowski metric and transform to cylindrical
coordinates, do I have a different manifold? If I perform the inverse
transformation on that then transform to spherical coordinates, do I
have a different manifold? If I transform to prolate spherical
coordinates, do I have a different manifold? If I perform all those
transformations without inverting, then transform back, do I have a
different manifold?

I don't think KW understands that the metric defines the geometry of a
manifold. At least he doesn't display that knowledge.

What I can't grasp is why everyone in the world who knows math knows that
something as simple as a vector from intro calc has a geometric meaning
independant of coordinates and yet when we use them we almost always use
them in component form in which the components themselves differ from
coordinate system to coordinate system. Now what's so hard to understand
about that KW???? You claim to know math well that's about as simple as math
can get and a vector is a tensor of rank 1.


The Birkhoff's theorem is so wrong.

You have never even seen the proof of Birkhoff's theorem.


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.

Since you are completely unwilling to test me, how do you know I don't
understand the calculus of variations? As I told you before, submit a
sample problem you want me to solve. All I ask is that you be able to
solve it yourself.

I don't really know why you are so arrogant.

I'd chime in on that and add to it but the Christian side of me is holding
me back. Its a struggle. LOL

Best wishes to all

Pete