Discussion:
The Absurd Claim of the Metric as a Tensor
Koobee Wublee
2007-01-15 07:36:01 UTC
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.
Eric Gisse
2007-01-15 13:27:06 UTC
Post by Koobee Wublee
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

[...]
Post by Koobee Wublee
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
shit all over it.
Koobee Wublee
2007-01-15 22:05:12 UTC
Post by Eric Gisse
Post by Koobee Wublee
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
[...]
Post by Koobee Wublee
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
shit all over it.
Koobee Wublee
2007-01-15 22:12:01 UTC
Post by Eric Gisse
Post by Koobee Wublee
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
Post by Eric Gisse
Post by Koobee Wublee
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.
Post by Eric Gisse
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.
Post by Eric Gisse
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.
Post by Eric Gisse
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.
Post by Eric Gisse
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.
Post by Eric Gisse
You really do not know what you
are talking about and should best study a subject before you attempt to
shit all over it.
Open your mouth wide. Here it comes. Does it still taste the curry
flavor?
Eric Gisse
2007-01-16 01:58:09 UTC
Post by Koobee Wublee
Post by Eric Gisse
Post by Koobee Wublee
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?
Post by Koobee Wublee
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?
Post by Koobee Wublee
Post by Eric Gisse
Post by Koobee Wublee
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.

Post by Koobee Wublee
Post by Eric Gisse
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.
Post by Eric Gisse
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?
Post by Koobee Wublee
The Birkhoff's theorem is so wrong.
You have never even seen the proof of Birkhoff's theorem.
Post by Koobee Wublee
Post by Eric Gisse
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 shit arguments like
"well there could be more than one solution to that ODE!".

"Again, foliation? There is not even such a word in the English

Post by Koobee Wublee
Post by Eric Gisse
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?

You never did compute the curvature scalar of that manifold to prove
that is curved. Or even the Riemann tensor.
Post by Koobee Wublee
Post by Eric Gisse
You really do not know what you
are talking about and should best study a subject before you attempt to
shit all over it.
Open your mouth wide. Here it comes. Does it still taste the curry
flavor?
Pmb
2007-01-16 04:04:13 UTC
Post by Eric Gisse
Post by Koobee Wublee
Post by Eric Gisse
Post by Koobee Wublee
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?
Post by Koobee Wublee
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?
Post by Koobee Wublee
Post by Eric Gisse
Post by Koobee Wublee
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.
Post by Eric Gisse
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!
Post by Eric Gisse
Post by Koobee Wublee
Post by Eric Gisse
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.
Post by Eric Gisse
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.
Post by Eric Gisse
Post by Koobee Wublee
The Birkhoff's theorem is so wrong.
You have never even seen the proof of Birkhoff's theorem.
Post by Koobee Wublee
Post by Eric Gisse
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
Koobee Wublee
2007-01-16 06:16:26 UTC
Post by Pmb
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.
How pathetic? You are just as low as the rest of cranks. You had me
fooled for a few days.
Post by Pmb
Post by Eric Gisse
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!
You should not have listened to Sal. Now, you are burying your head in
a hole on the ground. Your arse is also exposed. Didn't Confucius
say "only a fool can convince another fool to buy into the most
foolish thing"?
Post by Pmb
You still have a lot to learn especially hitting these middle-age
years.
Post by Pmb
Post by Eric Gisse
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.
You'll be deeply surprised.
Post by Pmb
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.
Have you asked yourself if Koobee Wublee really know what he is talking
Post by Pmb
Post by Eric Gisse
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
Please chime in. I will at least get a laugh out of it, too. It
breaks up the seriousness in the discussions of very serious stuff. I
know you are very capable of sinking down to their level.
Koobee Wublee
2007-01-16 06:07:48 UTC
Post by Eric Gisse
Post by Koobee Wublee
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.
I pick them because they are involved in GR.
Post by Eric Gisse
How about the Reimann tensor? Weyl tensor? Levi-Citva tensor? What is
the matrix representation of them?
Also 4-by-4 matrices.
Post by Eric Gisse
Post by Koobee Wublee
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?
Just trying to make a point.
Post by Eric Gisse
So you still think linear algebra == differential geometry?
I never did. Linear algebra was brought up by others, not I. Linear
algebra however is the basis of differential geometry though.
Post by Eric Gisse
Looks like sal still has you pegged. Four months later and you still
repeat the same mistakes.
No, Sal is burying his head in a hold with his arse exposed. I can
stick a dynamite in it.
Post by Eric Gisse
He was displaying his ignorance in which you also sing in the same
tune.
Post by Eric Gisse
Post by Koobee Wublee
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.
Infinite geometries are represented by infinite solutions where the
metric, the Ricci tensor, and the Einstein tensor are all valid to one
particular choice of coordinate system. This is what you have a lot of
trouble to understand.
Post by Eric Gisse
If I start off with the Minkowski metric and transform to cylindrical
coordinates, do I have a different manifold?
No, you don't.
Post by Eric Gisse
If I perform the inverse
transformation on that then transform to spherical coordinates, do I
have a different manifold?
No, you don't.
Post by Eric Gisse
If I transform to prolate spherical
coordinates, do I have a different manifold?
Don't understand this one.
Post by Eric Gisse
If I perform all those
transformations without inverting, then transform back, do I have a
different manifold?
If I understand what you are saying, it should be the same manifold.
Post by Eric Gisse
Post by Koobee Wublee
The Birkhoff's theorem is so wrong.
You have never even seen the proof of Birkhoff's theorem.
Yes, I have. Ask Reverend Bielawski the voodoo mathemagician. He
showed it to me once.
Post by Eric Gisse
Post by Koobee Wublee
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?
You said so yourself.
Post by Eric Gisse
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.
Ask you professor to do that. After all, you are paying him.
Post by Eric Gisse
I don't really know why you are so arrogant.
I am not. You are just ignorant.
Post by Eric Gisse
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 shit arguments like
"well there could be more than one solution to that ODE!".
You are confused. <shrug>
Post by Eric Gisse
"Again, foliation? There is not even such a word in the English
So! <shrug>
Post by Eric Gisse
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.
Post by Koobee Wublee
No, I never said that.
Never?
That is correct as what I wrote. If I have a linear metric such as the
Minkowski one and introduce a non-linear coordinate system, I get a
non-linear geometry. On the other hand, if I introduce a non-linear
metric such as the one for the spherically symmetric polar coordinate
system and a non-linear coordinate such as the spherically polar
coordinate system, I get a linear geometry. I meant what I said.
<shrug>
Post by Eric Gisse
You never did compute the curvature scalar of that manifold to prove
that is curved. Or even the Riemann tensor.
Do you want to lead the way for the Minkowski metric?
Post by Eric Gisse
Post by Koobee Wublee
Open your mouth wide. Here it comes. Does it still taste the curry
flavor?
Well, what flavor is it?
Eric Gisse
2007-01-16 07:18:40 UTC
Post by Koobee Wublee
Post by Eric Gisse
Post by Koobee Wublee
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.
I pick them because they are involved in GR.
...and because they fit nicely into what you think a tensor is, while
ignoring other examples that destroy your position.
Post by Koobee Wublee
Post by Eric Gisse
How about the Reimann tensor? Weyl tensor? Levi-Citva tensor? What is
the matrix representation of them?
Also 4-by-4 matrices.
The Riemann tensor has 4x4x4x4 = 256 components. A 4x4 matrix has 16
components.

A tensor is not a matrix. A tensor does not transform like a matrix.
Not all tensors can even be represented by a matrix.
Post by Koobee Wublee
Post by Eric Gisse
Post by Koobee Wublee
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?
Just trying to make a point.
Which point would that be?

That you do not read what you cite? That you like to cite web pages
that explicitly disagree with you?
Post by Koobee Wublee
Post by Eric Gisse
So you still think linear algebra == differential geometry?
I never did. Linear algebra was brought up by others, not I. Linear
algebra however is the basis of differential geometry though.
Well, it looks like you just did say that linear algebra is the basis
of differential geometry rather than something more accurate like
analysis on Riemannian manifolds.

But you want to assert that this is the case, find a resource that
agrees with you. You had to learn this stuff from somewhere so you
should be able to point me to the spot that says what you say.
Post by Koobee Wublee
Post by Eric Gisse
Looks like sal still has you pegged. Four months later and you still
repeat the same mistakes.
No, Sal is burying his head in a hold with his arse exposed. I can
stick a dynamite in it.
Post by Eric Gisse
He was displaying his ignorance in which you also sing in the same
tune.
The tune rings true.
Post by Koobee Wublee
Post by Eric Gisse
Post by Koobee Wublee
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.
Infinite geometries are represented by infinite solutions where the
metric, the Ricci tensor, and the Einstein tensor are all valid to one
particular choice of coordinate system. This is what you have a lot of
trouble to understand.
You continue to not understand.

Writing out the tensor in a coordinate system does _not_ mean it is
only expressible in that coordinate system.
Post by Koobee Wublee
Post by Eric Gisse
If I start off with the Minkowski metric and transform to cylindrical
coordinates, do I have a different manifold?
No, you don't.
Post by Eric Gisse
If I perform the inverse
transformation on that then transform to spherical coordinates, do I
have a different manifold?
No, you don't.
Post by Eric Gisse
If I transform to prolate spherical
coordinates, do I have a different manifold?
Don't understand this one.
http://mathworld.wolfram.com/ProlateSpheroidalCoordinates.html
Post by Koobee Wublee
Post by Eric Gisse
If I perform all those
transformations without inverting, then transform back, do I have a
different manifold?
If I understand what you are saying, it should be the same manifold.
Congratulations - you just feigned understanding of something we have
been trying to show you all along. It doesn't matter what coordinate
system you write the tensor in - the manifold, thus the geometry, does
not change just because you transform to a different coordinate system.
Post by Koobee Wublee
Post by Eric Gisse
Post by Koobee Wublee
The Birkhoff's theorem is so wrong.
You have never even seen the proof of Birkhoff's theorem.
Yes, I have. Ask Reverend Bielawski the voodoo mathemagician. He
showed it to me once.
Have you ever read it in a book and attempted to follow through?
Post by Koobee Wublee
Post by Eric Gisse
Post by Koobee Wublee
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?
You said so yourself.
Post by Eric Gisse
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.
Ask you professor to do that. After all, you are paying him.
Post by Eric Gisse
I don't really know why you are so arrogant.
I am not. You are just ignorant.
Post by Eric Gisse
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 shit arguments like
"well there could be more than one solution to that ODE!".
You are confused. <shrug>

You not knowing that ODE existence theorems even _existed_ before JanPB
pointed them out to you?
Post by Koobee Wublee
Post by Eric Gisse
"Again, foliation? There is not even such a word in the English
So! <shrug>
Post by Eric Gisse
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.
Post by Koobee Wublee
No, I never said that.
Never?
That is correct as what I wrote. If I have a linear metric such as the
Minkowski one and introduce a non-linear coordinate system, I get a
non-linear geometry. On the other hand, if I introduce a non-linear
metric such as the one for the spherically symmetric polar coordinate
system and a non-linear coordinate such as the spherically polar
coordinate system, I get a linear geometry. I meant what I said.
<shrug>
Post by Eric Gisse
You never did compute the curvature scalar of that manifold to prove
that is curved. Or even the Riemann tensor.
Do you want to lead the way for the Minkowski metric?
"Thus, the only logical conclusion is that the described spacetime must
be curved despite the metric being Minkowski."

The connection coefficients are all zero so Riemann is zero, Ricci is
zero, and the Ricci scalar is zero. No point in writing out a bunch of
zeros. Plus *you* are the one who made the assertion. Either support it
or retract it.
Post by Koobee Wublee
Post by Eric Gisse
Post by Koobee Wublee
Open your mouth wide. Here it comes. Does it still taste the curry
flavor?
Well, what flavor is it?
Koobee Wublee
2007-01-16 07:34:47 UTC
Post by Eric Gisse
You only pick those two tensors because they fit your predetermined
notion of what a tensor is.
Post by Koobee Wublee
I pick them because they are involved in GR.
...and because they fit nicely into what you think a tensor is, while
ignoring other examples that destroy your position.
The bottom line is that we are talking about GR. So, get used to it.
Post by Eric Gisse
The Riemann tensor has 4x4x4x4 = 256 components. A 4x4 matrix has 16
components.
Yes, you are correct. That is why Levi-Civita invented the Ricci
tensor to reduce it into a 4-by-4 matrix. Thanks for point that out.
Post by Eric Gisse
A tensor is not a matrix. A tensor does not transform like a matrix.
Not all tensors can even be represented by a matrix.
I am not contesting a tensor is not a matrix in general. The metric
tensor and the Ricci tensor, however, are matrices.
Post by Eric Gisse
Post by Koobee Wublee
Just trying to make a point.
Which point would that be?
Post by Eric Gisse
That you do not read what you cite? That you like to cite web pages
that explicitly disagree with you?
No, that you do not read what I wrote. <shrug>
Post by Eric Gisse
So you still think linear algebra == differential geometry?
Post by Koobee Wublee
I never did. Linear algebra was brought up by others, not I. Linear
algebra however is the basis of differential geometry though.
Well, it looks like you just did say that linear algebra is the basis
of differential geometry rather than something more accurate like
analysis on Riemannian manifolds.
<shrug>
Post by Eric Gisse
But you want to assert that this is the case, find a resource that
agrees with you. You had to learn this stuff from somewhere so you
should be able to point me to the spot that says what you say.
Can't you read what I wrote? I said I am pointing out an error made
Post by Eric Gisse
Post by Koobee Wublee
No, Sal is burying his head in a hold with his arse exposed. I can
stick a dynamite in it.
He was displaying his ignorance in which you also sing in the same
tune.
The tune rings true.
Yes, I agree. You are Sal's eyes and ears while he chose to bury his
Post by Eric Gisse
Post by Koobee Wublee
Infinite geometries are represented by infinite solutions where the
metric, the Ricci tensor, and the Einstein tensor are all valid to one
particular choice of coordinate system. This is what you have a lot of
trouble to understand.
You continue to not understand.
No, it is you who does not understand.
Post by Eric Gisse
Writing out the tensor in a coordinate system does _not_ mean it is
only expressible in that coordinate system.
Have you not been reading what I wrote? I said categorizing the metric
as a tensor is wrong. The metric is not a tensor because the metric is
only valid to a very specific coordinate system.
Post by Eric Gisse
Post by Koobee Wublee
Post by Eric Gisse
If I start off with the Minkowski metric and transform to cylindrical
coordinates, do I have a different manifold?
No, you don't.
Post by Eric Gisse
If I perform the inverse
transformation on that then transform to spherical coordinates, do I
have a different manifold?
No, you don't.
Post by Eric Gisse
If I transform to prolate spherical
coordinates, do I have a different manifold?
Don't understand this one.
http://mathworld.wolfram.com/ProlateSpheroidalCoordinates.html
<shrug>
Post by Eric Gisse
Congratulations - you just feigned understanding of something we have
been trying to show you all along. It doesn't matter what coordinate
system you write the tensor in - the manifold, thus the geometry, does
not change just because you transform to a different coordinate system.
That is fine as long as you don't categorize the metric as a tensor.
Post by Eric Gisse
You have never even seen the proof of Birkhoff's theorem.
Post by Koobee Wublee
Yes, I have. Ask Reverend Bielawski the voodoo mathemagician. He
showed it to me once.
Have you ever read it in a book and attempted to follow through?
Post by Eric Gisse
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 shit arguments like
"well there could be more than one solution to that ODE!".
Post by Koobee Wublee
You are confused. <shrug>
ODE's and the whatever thereom.
Post by Eric Gisse
You not knowing that ODE existence theorems even _existed_ before JanPB
pointed them out to you?
So what! The field equations don't even apply. <shrug>
Post by Eric Gisse
"Thus, the only logical conclusion is that the described spacetime must
be curved despite the metric being Minkowski."
That is correct if your choice of coordinate system is very non-linear.
Post by Eric Gisse
The connection coefficients are all zero so Riemann is zero, Ricci is
zero, and the Ricci scalar is zero. No point in writing out a bunch of
zeros. Plus *you* are the one who made the assertion. Either support it
or retract it.
Again, have you not reading what I wrote? I have said the metric
tensor, the Riemann tensor, and the Ricci tensor are all not qualified
to be tensors because they are only valid to only one and only
coordinate system of choice.
Post by Eric Gisse
Post by Koobee Wublee
Open your mouth wide. Here it comes. Does it still taste the curry
flavor?
Well, what flavor is it?
Eric Gisse
2007-01-16 08:58:51 UTC
Post by Koobee Wublee
Post by Eric Gisse
You only pick those two tensors because they fit your predetermined
notion of what a tensor is.
Post by Koobee Wublee
I pick them because they are involved in GR.
...and because they fit nicely into what you think a tensor is, while
ignoring other examples that destroy your position.
The bottom line is that we are talking about GR. So, get used to it.
Post by Eric Gisse
The Riemann tensor has 4x4x4x4 = 256 components. A 4x4 matrix has 16
components.
Yes, you are correct. That is why Levi-Civita invented the Ricci
tensor to reduce it into a 4-by-4 matrix. Thanks for point that out.
WOW. Amazing argument!

If I contract the Ricci tensor with the metric, I get a real number.

Does that mean all tensors are scalars? Stop trying to shoehorn linear
algebra onto tensors. Tensors are _bilinear_ operators, matrices are
_linear_ operators.
Post by Koobee Wublee
Post by Eric Gisse
A tensor is not a matrix. A tensor does not transform like a matrix.
Not all tensors can even be represented by a matrix.
I am not contesting a tensor is not a matrix in general. The metric
tensor and the Ricci tensor, however, are matrices.
They can be represented as matrices in a particular coordinate system
but that does not mean linear algebra is in play.
Post by Koobee Wublee
Post by Eric Gisse
Post by Koobee Wublee
Just trying to make a point.
Which point would that be?
Post by Eric Gisse
That you do not read what you cite? That you like to cite web pages
that explicitly disagree with you?
No, that you do not read what I wrote. <shrug>
Post by Eric Gisse
So you still think linear algebra == differential geometry?
Post by Koobee Wublee
I never did. Linear algebra was brought up by others, not I. Linear
algebra however is the basis of differential geometry though.
Well, it looks like you just did say that linear algebra is the basis
of differential geometry rather than something more accurate like
analysis on Riemannian manifolds.
<shrug>
Post by Eric Gisse
But you want to assert that this is the case, find a resource that
agrees with you. You had to learn this stuff from somewhere so you
should be able to point me to the spot that says what you say.
Can't you read what I wrote? I said I am pointing out an error made
Except you are the *only one* who thinks it is an error.
Post by Koobee Wublee
Post by Eric Gisse
Post by Koobee Wublee
No, Sal is burying his head in a hold with his arse exposed. I can
stick a dynamite in it.
He was displaying his ignorance in which you also sing in the same
tune.
The tune rings true.
Yes, I agree. You are Sal's eyes and ears while he chose to bury his
Post by Eric Gisse
Post by Koobee Wublee
Infinite geometries are represented by infinite solutions where the
metric, the Ricci tensor, and the Einstein tensor are all valid to one
particular choice of coordinate system. This is what you have a lot of
trouble to understand.
You continue to not understand.
No, it is you who does not understand.
Post by Eric Gisse
Writing out the tensor in a coordinate system does _not_ mean it is
only expressible in that coordinate system.
Have you not been reading what I wrote? I said categorizing the metric
as a tensor is wrong. The metric is not a tensor because the metric is
only valid to a very specific coordinate system.
DO YOU EVEN KNOW WHAT A TENSOR IS?!
Post by Koobee Wublee
Post by Eric Gisse
Post by Koobee Wublee
Post by Eric Gisse
If I start off with the Minkowski metric and transform to cylindrical
coordinates, do I have a different manifold?
No, you don't.
Post by Eric Gisse
If I perform the inverse
transformation on that then transform to spherical coordinates, do I
have a different manifold?
No, you don't.
Post by Eric Gisse
If I transform to prolate spherical
coordinates, do I have a different manifold?
Don't understand this one.
http://mathworld.wolfram.com/ProlateSpheroidalCoordinates.html
<shrug>
Post by Eric Gisse
Congratulations - you just feigned understanding of something we have
been trying to show you all along. It doesn't matter what coordinate
system you write the tensor in - the manifold, thus the geometry, does
not change just because you transform to a different coordinate system.
That is fine as long as you don't categorize the metric as a tensor.
Why?

It acts upon vectors & covectors like a tensor. It acts like a bilinear
function. It is a tensor according to the definition of a tensor.
Post by Koobee Wublee
Post by Eric Gisse
You have never even seen the proof of Birkhoff's theorem.
Post by Koobee Wublee
Yes, I have. Ask Reverend Bielawski the voodoo mathemagician. He
showed it to me once.
Have you ever read it in a book and attempted to follow through?
Post by Eric Gisse
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 shit arguments like
"well there could be more than one solution to that ODE!".
Post by Koobee Wublee
You are confused. <shrug>
ODE's and the whatever thereom.
Post by Eric Gisse
You not knowing that ODE existence theorems even _existed_ before JanPB
pointed them out to you?
So what! The field equations don't even apply. <shrug>
As you have been told before, the field equations reduce from partial
differential to ordinary differential equations in this situation.
Post by Koobee Wublee
Post by Eric Gisse
"Thus, the only logical conclusion is that the described spacetime must
be curved despite the metric being Minkowski."
That is correct if your choice of coordinate system is very non-linear.
Who cares if it is nonlinear?
Post by Koobee Wublee
Post by Eric Gisse
The connection coefficients are all zero so Riemann is zero, Ricci is
zero, and the Ricci scalar is zero. No point in writing out a bunch of
zeros. Plus *you* are the one who made the assertion. Either support it
or retract it.
Again, have you not reading what I wrote? I have said the metric
tensor, the Riemann tensor, and the Ricci tensor are all not qualified
to be tensors because they are only valid to only one and only
coordinate system of choice.
Where were you taught this stuff? I gotta know because I'm tired of
seeing you repeat the same incorrect babble over and over again.
Post by Koobee Wublee
Post by Eric Gisse
Post by Koobee Wublee
Open your mouth wide. Here it comes. Does it still taste the curry
flavor?
Well, what flavor is it?
Koobee Wublee
2007-01-16 18:52:31 UTC
Post by Eric Gisse
Post by Koobee Wublee
Yes, you are correct. That is why Levi-Civita invented the Ricci
tensor to reduce it into a 4-by-4 matrix. Thanks for point that out.
WOW. Amazing argument!
Post by Eric Gisse
If I contract the Ricci tensor with the metric, I get a real number.
Ricci scalar. <shrug>
Post by Eric Gisse
Does that mean all tensors are scalars?
No. <shrug>
Post by Eric Gisse
Stop trying to shoehorn linear algebra onto tensors.
I don't. Mr. Bielawski, Mr. Brown, and others did.
Post by Eric Gisse
Tensors are _bilinear_ operators, matrices are
_linear_ operators.
OK. <shrug>
Post by Eric Gisse
Post by Koobee Wublee
I am not contesting a tensor is not a matrix in general. The metric
tensor and the Ricci tensor, however, are matrices.
They can be represented as matrices in a particular coordinate system
but that does not mean linear algebra is in play.
OK. <shrug>

In the meantime, the metric should not be a tensor. The metric is not
necessarily unitless.
Post by Eric Gisse
Post by Koobee Wublee
Can't you read what I wrote? I said I am pointing out an error made
Except you are the *only one* who thinks it is an error.
Thus, I am the only one who understands the curvature in space and in
spacetime since Riemann.
Post by Eric Gisse
DO YOU EVEN KNOW WHAT A TENSOR IS?!
YES!
Post by Eric Gisse
Post by Koobee Wublee
That is fine as long as you don't categorize the metric as a tensor.
Why?
Because the metric is not a tensor.
Post by Eric Gisse
It acts upon vectors & covectors like a tensor. It acts like a bilinear
function. It is a tensor according to the definition of a tensor.
No, the metric represents a set of correction factors to interpret an
invariant line segment using your choice of coordinate system.
Post by Eric Gisse
As you have been told before, the field equations reduce from partial
differential to ordinary differential equations in this situation.
I know that. I was the one who showed Mr. Bielawski that.
Post by Eric Gisse
Post by Koobee Wublee
That is correct if your choice of coordinate system is very non-linear.
Who cares if it is nonlinear?
I do. The Minkowski metric with a non-linear coordinate system means
the spacetime is non-linear or curved. You should too.
Post by Eric Gisse
Where were you taught this stuff? I gotta know because I'm tired of
seeing you repeat the same incorrect babble over and over again.
Why don't you guess? Hint: Not Alaska.
Eric Gisse
2007-01-17 01:04:33 UTC
Koobee Wublee wrote:

[...]

Since you are the only one who truly understands Riemanniean geometry,
I think it is time for you to write out the definition of a tensor.
Post by Koobee Wublee
Post by Eric Gisse
Post by Koobee Wublee
That is correct if your choice of coordinate system is very non-linear.
Who cares if it is nonlinear?
I do. The Minkowski metric with a non-linear coordinate system means
the spacetime is non-linear or curved. You should too.
Prove it.

You know, with math. Since you are the only one since Riemann to
understand Riemannian geometry, it should be easy for you to _prove_
what you are saying is true.
Post by Koobee Wublee
Post by Eric Gisse
Where were you taught this stuff? I gotta know because I'm tired of
seeing you repeat the same incorrect babble over and over again.
Why don't you guess? Hint: Not Alaska.
I repeat: Where were you taught this stuff? Where was the person who
truly understands Riemann taught his knowledge?
Pmb
2007-01-17 02:04:59 UTC
Post by Eric Gisse
[...]
Since you are the only one who truly understands Riemanniean geometry,
I think it is time for you to write out the definition of a tensor.
Post by Koobee Wublee
Post by Eric Gisse
Post by Koobee Wublee
That is correct if your choice of coordinate system is very non-linear.
Who cares if it is nonlinear?
I do. The Minkowski metric with a non-linear coordinate system means
the spacetime is non-linear or curved. You should too.
Prove it.
You know, with math. Since you are the only one since Riemann to
understand Riemannian geometry, it should be easy for you to _prove_
what you are saying is true.
Post by Koobee Wublee
Post by Eric Gisse
Where were you taught this stuff? I gotta know because I'm tired of
seeing you repeat the same incorrect babble over and over again.
Why don't you guess? Hint: Not Alaska.
I repeat: Where were you taught this stuff? Where was the person who
truly understands Riemann taught his knowledge?
Eric - You have to admit. He sure knows how to keep us asking. He has led us
to believe that we haven't asked the correct question to get the answer we
seek. The answer being "I'm looking at/I read textbook X". That's far too
easy. Yet he's thanked me for the work I put into my web pages for him so in
that sense he's polite. Then he starts insulting me. I've gotta stop getting
into discussions with people like KW. I plonked him but will respond in
special cases where I see fit to do so.

What is is about KW that keeps you going? With me its usually boredom. How

Best wishes Eric

Pete
Koobee Wublee
2007-01-17 05:13:57 UTC
Post by Eric Gisse
Post by Koobee Wublee
The Minkowski metric with a non-linear coordinate system means
the spacetime is non-linear or curved. You should too.
Prove it.
You know, with math. Since you are the only one since Riemann to
understand Riemannian geometry, it should be easy for you to _prove_
what you are saying is true.
Yes, it is exceedingly easy.

Say we have the Euclidean rectangular coordinate system (x, y, z) ---
the usual. With the Minkowski metric, we have the following flat
spacetime.

ds^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2

Then, we have another coordinate system (x', y', z') where

** t' = t
** x' = f(x)
** y' = y
** z' = z

Then, the Minkowski spacetime with this set of highly non-linear
coordinate system is

ds'^2 = c^2 dt'^2 - dx'^2 - dy'^2 - dz'^2

This spacetime translate into the usual rectangular coordinate is the
following.

ds'^2 = c^2 dt^2 - (df/dx)^2 dx^2 - dy^2 - dz^2

You can easily see that ds'^2 is distorted along the x axis.

QED.

This sh*t is so easy. I am very surprised that you have so much
trouble understanding this. <shrug>
Post by Eric Gisse
I repeat: Where were you taught this stuff? Where was the person who
truly understands Riemann taught his knowledge?
Not Mr. Brown for sure.
Eric Gisse
2007-01-17 05:45:09 UTC
Post by Koobee Wublee
Post by Eric Gisse
Post by Koobee Wublee
The Minkowski metric with a non-linear coordinate system means
the spacetime is non-linear or curved. You should too.
Prove it.
You know, with math. Since you are the only one since Riemann to
understand Riemannian geometry, it should be easy for you to _prove_
what you are saying is true.
Yes, it is exceedingly easy.
Say we have the Euclidean rectangular coordinate system (x, y, z) ---
the usual. With the Minkowski metric, we have the following flat
spacetime.
ds^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2
Then, we have another coordinate system (x', y', z') where
** t' = t
** x' = f(x)
** y' = y
** z' = z
Then, the Minkowski spacetime with this set of highly non-linear
coordinate system is
ds'^2 = c^2 dt'^2 - dx'^2 - dy'^2 - dz'^2
This spacetime translate into the usual rectangular coordinate is the
following.
ds'^2 = c^2 dt^2 - (df/dx)^2 dx^2 - dy^2 - dz^2
You can easily see that ds'^2 is distorted along the x axis.
QED.
This sh*t is so easy. I am very surprised that you have so much
trouble understanding this. <shrug>
Gee.

I asked earlier if a coordinate transformation from cartesian to
cylindrical coordinates changes the manifold. You said "no". Yet here
you are saying "yes" in a different example. The true disciple of
Riemann appears to have a difficult time remaining consistent.

Oh true disciple of Riemann, why do you struggle to answer a simple
question? I asked for you to _prove_ that this 'new manifold' is
curved, yet all you did was do a coordinate transformation and say
"SEE?! IT'S CURVED" without actually having computed anything relevant
to curvature.

Also, why is it you wont write down your definition of a tensor, oh
true disciple of Riemann? You continually assert that the metric is not
a tensor, but you refuse to explain what you think a tensor is.
Post by Koobee Wublee
Post by Eric Gisse
I repeat: Where were you taught this stuff? Where was the person who
truly understands Riemann taught his knowledge?
Not Mr. Brown for sure.
Is the true disciple of Riemann [rightfully] afraid when he reveals the
source of his knowledge that he will be mercilessly mocked for being so
stupid?
Koobee Wublee
2007-01-17 06:13:32 UTC
Post by Eric Gisse
Post by Koobee Wublee
Say we have the Euclidean rectangular coordinate system (x, y, z) ---
the usual. With the Minkowski metric, we have the following flat
spacetime.
ds^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2
Then, we have another coordinate system (x', y', z') where
** t' = t
** x' = f(x)
** y' = y
** z' = z
Then, the Minkowski spacetime with this set of highly non-linear
coordinate system is
ds'^2 = c^2 dt'^2 - dx'^2 - dy'^2 - dz'^2
This spacetime translate into the usual rectangular coordinate is the
following.
ds'^2 = c^2 dt^2 - (df/dx)^2 dx^2 - dy^2 - dz^2
You can easily see that ds'^2 is distorted along the x axis.
QED.
This sh*t is so easy. I am very surprised that you have so much
trouble understanding this. <shrug>
Gee.
I asked earlier if a coordinate transformation from cartesian to
cylindrical coordinates changes the manifold. You said "no". Yet here
you are saying "yes" in a different example. The true disciple of
Riemann appears to have a difficult time remaining consistent.
They are two very different events you asked. <shrug>
Post by Eric Gisse
Oh true disciple of Riemann, why do you struggle to answer a simple
question? I asked for you to _prove_ that this 'new manifold' is
curved, yet all you did was do a coordinate transformation and say
"SEE?! IT'S CURVED" without actually having computed anything relevant
to curvature.
You do not understand the very simple math. It is not my fault.
<shrug>
Post by Eric Gisse
Also, why is it you wont write down your definition of a tensor, oh
true disciple of Riemann? You continually assert that the metric is not
a tensor, but you refuse to explain what you think a tensor is.
I did. You just did not listen. Again, it is not my fault. <shrug>
Post by Eric Gisse
Is the true disciple of Riemann [rightfully] afraid when he reveals the
source of his knowledge that he will be mercilessly mocked for being so
stupid?
This is so laughable. hanson is going to have a good time on this one.
Keep up the absurd clownish doo-dah. We need a laughter discussing
among these series subjects.
Eric Gisse
2007-01-17 08:09:55 UTC
Post by Koobee Wublee
Post by Eric Gisse
Post by Koobee Wublee
Say we have the Euclidean rectangular coordinate system (x, y, z) ---
the usual. With the Minkowski metric, we have the following flat
spacetime.
ds^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2
Then, we have another coordinate system (x', y', z') where
** t' = t
** x' = f(x)
** y' = y
** z' = z
Then, the Minkowski spacetime with this set of highly non-linear
coordinate system is
ds'^2 = c^2 dt'^2 - dx'^2 - dy'^2 - dz'^2
This spacetime translate into the usual rectangular coordinate is the
following.
ds'^2 = c^2 dt^2 - (df/dx)^2 dx^2 - dy^2 - dz^2
You can easily see that ds'^2 is distorted along the x axis.
QED.
This sh*t is so easy. I am very surprised that you have so much
trouble understanding this. <shrug>
Gee.
I asked earlier if a coordinate transformation from cartesian to
cylindrical coordinates changes the manifold. You said "no". Yet here
you are saying "yes" in a different example. The true disciple of
Riemann appears to have a difficult time remaining consistent.
They are two very different events you asked. <shrug>
Post by Eric Gisse
Oh true disciple of Riemann, why do you struggle to answer a simple
question? I asked for you to _prove_ that this 'new manifold' is
curved, yet all you did was do a coordinate transformation and say
"SEE?! IT'S CURVED" without actually having computed anything relevant
to curvature.
You do not understand the very simple math. It is not my fault.
<shrug>
Too bad the true disciple of Riemann finds it difficult to answer the
question that is asked rather than one that is not.

I asked for you to prove the manifold becomes curved upon a coordinate
transformation. Looks like you can't prove what you say.
Post by Koobee Wublee
Post by Eric Gisse
Also, why is it you wont write down your definition of a tensor, oh
true disciple of Riemann? You continually assert that the metric is not
a tensor, but you refuse to explain what you think a tensor is.
I did. You just did not listen. Again, it is not my fault. <shrug>
Post by Eric Gisse
Is the true disciple of Riemann [rightfully] afraid when he reveals the
source of his knowledge that he will be mercilessly mocked for being so
stupid?
This is so laughable. hanson is going to have a good time on this one.
Keep up the absurd clownish doo-dah. We need a laughter discussing
among these series subjects.
Pmb
2007-01-17 08:32:42 UTC
Post by Eric Gisse
Post by Koobee Wublee
Post by Eric Gisse
Post by Koobee Wublee
Say we have the Euclidean rectangular coordinate system (x, y, z) ---
the usual. With the Minkowski metric, we have the following flat
spacetime.
ds^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2
Then, we have another coordinate system (x', y', z') where
** t' = t
** x' = f(x)
** y' = y
** z' = z
Then, the Minkowski spacetime with this set of highly non-linear
coordinate system is
ds'^2 = c^2 dt'^2 - dx'^2 - dy'^2 - dz'^2
This spacetime translate into the usual rectangular coordinate is the
following.
ds'^2 = c^2 dt^2 - (df/dx)^2 dx^2 - dy^2 - dz^2
You can easily see that ds'^2 is distorted along the x axis.
QED.
This sh*t is so easy. I am very surprised that you have so much
trouble understanding this. <shrug>
Gee.
I asked earlier if a coordinate transformation from cartesian to
cylindrical coordinates changes the manifold. You said "no". Yet here
you are saying "yes" in a different example. The true disciple of
Riemann appears to have a difficult time remaining consistent.
They are two very different events you asked. <shrug>
Post by Eric Gisse
Oh true disciple of Riemann, why do you struggle to answer a simple
question? I asked for you to _prove_ that this 'new manifold' is
curved, yet all you did was do a coordinate transformation and say
"SEE?! IT'S CURVED" without actually having computed anything relevant
to curvature.
You do not understand the very simple math. It is not my fault.
<shrug>
Too bad the true disciple of Riemann finds it difficult to answer the
question that is asked rather than one that is not.
I asked for you to prove the manifold becomes curved upon a coordinate
transformation. Looks like you can't prove what you say.
Post by Koobee Wublee
Post by Eric Gisse
Also, why is it you wont write down your definition of a tensor, oh
true disciple of Riemann? You continually assert that the metric is not
a tensor, but you refuse to explain what you think a tensor is.
I did. You just did not listen. Again, it is not my fault. <shrug>
Post by Eric Gisse
Is the true disciple of Riemann [rightfully] afraid when he reveals the
source of his knowledge that he will be mercilessly mocked for being so
stupid?
This is so laughable. hanson is going to have a good time on this one.
Keep up the absurd clownish doo-dah. We need a laughter discussing
among these series subjects.
The problem with his defining tensor is that it is so vauge and flightly
that it is impossible to recognize it as a definition. I imagine that he
says that a tensor is a matrix whose components don't change under a
coordinate transformation. People who truly understand tensors pass by this
bit of nonsense and keep looking for a real definition. If I were you I'd
have him place it in an undeniable and highly visible form such as
=======================================
Definition

Tensor - A object whose components transfrom from one set of coordinates
according to a set of coordinate transformtions.

=======================================

But he'll never do that since it would pin him down to a definition and
that's when he'd get into trouble he'd be impossible to back away from.

Pete
Pmb
2007-01-17 06:16:04 UTC
Post by Eric Gisse
Post by Koobee Wublee
Post by Eric Gisse
Post by Koobee Wublee
The Minkowski metric with a non-linear coordinate system means
the spacetime is non-linear or curved. You should too.
Prove it.
You know, with math. Since you are the only one since Riemann to
understand Riemannian geometry, it should be easy for you to _prove_
what you are saying is true.
Yes, it is exceedingly easy.
Say we have the Euclidean rectangular coordinate system (x, y, z) ---
the usual. With the Minkowski metric, we have the following flat
spacetime.
ds^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2
Then, we have another coordinate system (x', y', z') where
** t' = t
** x' = f(x)
** y' = y
** z' = z
Then, the Minkowski spacetime with this set of highly non-linear
coordinate system is
ds'^2 = c^2 dt'^2 - dx'^2 - dy'^2 - dz'^2
This spacetime translate into the usual rectangular coordinate is the
following.
ds'^2 = c^2 dt^2 - (df/dx)^2 dx^2 - dy^2 - dz^2
You can easily see that ds'^2 is distorted along the x axis.
QED.
This sh*t is so easy. I am very surprised that you have so much
trouble understanding this. <shrug>
Gee.
I asked earlier if a coordinate transformation from cartesian to
cylindrical coordinates changes the manifold. You said "no". Yet here
you are saying "yes" in a different example. The true disciple of
Riemann appears to have a difficult time remaining consistent.
Oh true disciple of Riemann, why do you struggle to answer a simple
question? I asked for you to _prove_ that this 'new manifold' is
curved, yet all you did was do a coordinate transformation and say
"SEE?! IT'S CURVED" without actually having computed anything relevant
to curvature.
Also, why is it you wont write down your definition of a tensor, oh
true disciple of Riemann? You continually assert that the metric is not
a tensor, but you refuse to explain what you think a tensor is.
Post by Koobee Wublee
Post by Eric Gisse
I repeat: Where were you taught this stuff? Where was the person who
truly understands Riemann taught his knowledge?
Not Mr. Brown for sure.
Hmmm. I'm not even participating in this conversation and KW still manages
to get a little dig in. I find it abhorent to "toot my own horn" but you
really love to push people don't ya KW? It is overly to everyone here that I
know differential geometry and tensor analysis to a much greater quality
than you have demonstrated. If it needs to be said (and I hate saying this)
but I have a BA in both physics and math from Merrimack College in Andover
MA. I did some graduate work at Northeastern University in Boston and I
studied GR at MIT back in 1999 with Dr. Edmund Bertschinger. I was layed off
at the time and Alan Guth sugggested that I ask Ed if I could sit in on his
class. Ed said sure so I sat in on the course. After that I kept up self
study of this material and presently am taking a directed study with a group
at Merrimack College in Andover MA. The group consists of three people. The
math chair, a math prof and myself. We are using Schutz's text on the
subject and we are doing it at our own leasurely pace. We're about half way
through the text I'd say. But during the last 8 years I've had many mentors
on this subject matter and that includes Edwin F. Taylor, Hans C. Ohanian,
Robert W. Brehme and John Stachel.

But from what KW has displayed here its doubtful that he's ever read the
definition of tensors and the terminology. His lack of understanding of how
a geometric object can remain the same geometrical object even when the
components of the object change with the basis/coordinate system. This is
taught in Calc III in into to vector analysis yet KW seems completely
ignorant of these simple and basic facts, facts a sophmore math major would
understand.

Its okay to be ignorant KW. We are all ignorant in some areas of all aspects
of life and areas of study. Nobody knows everything about even a single
subject. So you don't have to pretend that you didn't make a mistake. We'll
understand. What is shameful is that you refuse to admit your mistake and
move on. And in case this is unclear - We are right and you are wrong -
Period! There is no room for misunderstanding here.
Post by Eric Gisse
Is the true disciple of Riemann [rightfully] afraid when he reveals the
source of his knowledge that he will be mercilessly mocked for being so
stupid?
I truly feel sorry for KW. On the one hand he seems like a nice person and
on the other hand he's an insulting and condescending person.

Then again its not my job in life to understand KW!

Best wishes KW - I really hope the best for you. I'll pray for you to let
God open your mind enough to see the elements of your error.

Pete
Pete
Koobee Wublee
2007-01-17 06:44:07 UTC
Post by Pmb
Hmmm. I'm not even participating in this conversation and KW still manages
to get a little dig in. I find it abhorent to "toot my own horn" but you
really love to push people don't ya KW? It is overly to everyone here that I
know differential geometry and tensor analysis to a much greater quality
than you have demonstrated. If it needs to be said (and I hate saying this)
but I have a BA in both physics and math from Merrimack College in Andover
MA. I did some graduate work at Northeastern University in Boston and I
studied GR at MIT back in 1999 with Dr. Edmund Bertschinger. I was layed off
at the time and Alan Guth sugggested that I ask Ed if I could sit in on his
class. Ed said sure so I sat in on the course. After that I kept up self
study of this material and presently am taking a directed study with a group
at Merrimack College in Andover MA. The group consists of three people. The
math chair, a math prof and myself. We are using Schutz's text on the
subject and we are doing it at our own leasurely pace. We're about half way
through the text I'd say. But during the last 8 years I've had many mentors
on this subject matter and that includes Edwin F. Taylor, Hans C. Ohanian,
Robert W. Brehme and John Stachel.
Oh, man. Your resume only fits in one short paragraph. Mine is as
long as a roll of toilet paper in which I am not very proud of.

Professor Taylor's derivation of photon bending abides to the
principle of minimum accumulated time while his derivation on
Mercury's orbital anomaly abides to the absurd assumption where the
geodesic paths follow the least amount of accumulated spacetime. His
derivation of GPS time difference is baloney.

Mr. Stachel worships Einstein. That is understandable since without
Einstein being elevated into godhood Mr. Stachel's job would be in
jeopardy.
Post by Pmb
But from what KW has displayed here its doubtful that he's ever read the
definition of tensors and the terminology. His lack of understanding of how
a geometric object can remain the same geometrical object even when the
components of the object change with the basis/coordinate system. This is
taught in Calc III in into to vector analysis yet KW seems completely
ignorant of these simple and basic facts, facts a sophmore math major would
understand.
Ho, ho, ho. Christmas is over. It is time to get back to reality.
Koobee Wublee's remarks cannot be taken lightly. This is reality.
Post by Pmb
Its okay to be ignorant KW. We are all ignorant in some areas of all aspects
of life and areas of study. Nobody knows everything about even a single
subject. So you don't have to pretend that you didn't make a mistake. We'll
understand. What is shameful is that you refuse to admit your mistake and
move on. And in case this is unclear - We are right and you are wrong -
Period! There is no room for misunderstanding here.
Who are you referring to? The punk who has been a multi-year junior at
the university of Alaska for a while?
Post by Pmb
I truly feel sorry for KW. On the one hand he seems like a nice person and
on the other hand he's an insulting and condescending person.
Have you prayed for me lately?
Post by Pmb
Then again its not my job in life to understand KW!
Thank god for that. Please don't ask me which god because I have no
clue.
Post by Pmb
Best wishes KW - I really hope the best for you. I'll pray for you to let
God open your mind enough to see the elements of your error.
Thanks. Best wishes to you too. In time, your god will reveal to you
that what I have talked about is correct. Hopefully, you will not
behave like an infant trapped in a man's body then.
JanPB
2007-01-17 08:21:09 UTC
Pmb wrote:
[...]
Post by Pmb
Its okay to be ignorant KW. We are all ignorant in some areas of all aspects
of life and areas of study.
Exactly. I know very little about Kant but I do not parade around
philosophy groups telling everyone there they are idiots who
misunderstood the last 100 years of some Kantian idea.

"pompous fools drive me up the wall. Ordinary folks are all right; you
can talk to them, and try to help them out. But pompous fools - guys
who are fools and are covering it all over and impressing people as to
how wonderful they are with all this hocus pocus - THAT, I CANNOT
STAND!"

Richard Feynman

--
Jan Bielawski
Koobee Wublee
2007-01-17 18:11:27 UTC
Post by JanPB
Exactly. I know very little about Kant but I do not parade around
philosophy groups telling everyone there they are idiots who
misunderstood the last 100 years of some Kantian idea.
"pompous fools drive me up the wall. Ordinary folks are all right; you
can talk to them, and try to help them out. But pompous fools - guys
who are fools and are covering it all over and impressing people as to
how wonderful they are with all this hocus pocus - THAT, I CANNOT
STAND!"
Richard Feynman
You are exactly what Feynman was referring to as a pompous fool.
<shrug>
Daryl McCullough
2007-01-17 19:37:07 UTC
Koobee Wublee says...
Post by Koobee Wublee
Post by JanPB
Exactly. I know very little about Kant but I do not parade around
philosophy groups telling everyone there they are idiots who
misunderstood the last 100 years of some Kantian idea.
"pompous fools drive me up the wall. Ordinary folks are all right; you
can talk to them, and try to help them out. But pompous fools - guys
who are fools and are covering it all over and impressing people as to
how wonderful they are with all this hocus pocus - THAT, I CANNOT
STAND!"
Richard Feynman
You are exactly what Feynman was referring to as a pompous fool.
No, he was talking about people like you.

--
Daryl McCullough
Ithaca, NY
JanPB
2007-01-17 08:04:05 UTC
Post by Koobee Wublee
Post by Eric Gisse
Post by Koobee Wublee
The Minkowski metric with a non-linear coordinate system means
the spacetime is non-linear or curved. You should too.
Prove it.
You know, with math. Since you are the only one since Riemann to
understand Riemannian geometry, it should be easy for you to _prove_
what you are saying is true.
Yes, it is exceedingly easy.
Say we have the Euclidean rectangular coordinate system (x, y, z) ---
the usual. With the Minkowski metric, we have the following flat
spacetime.
ds^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2
Then, we have another coordinate system (x', y', z') where
** t' = t
** x' = f(x)
** y' = y
** z' = z
Then, the Minkowski spacetime with this set of highly non-linear
coordinate system is
ds'^2 = c^2 dt'^2 - dx'^2 - dy'^2 - dz'^2
This is not the Minkowski metric on the (t,x,y,z)-manifold. You've
changed coordinates but _kept the matrix unchanged_ - this CHANGES THE
METRIC. For the Nth time: in order NOT to change the metric while
switching coordinates, you have to CHANGE the matrix of metric
coefficients g_ij according to the usual formula (X-transpose)...etc.

If you change the coordinates but not the metric coefficients g_ij OR
vice-versa, you CHANGE THE METRIC.

The new metric you wrote above is flat and hence isometric to Minkowski
but it's not _the_ Minkowski metric on (t,x,y,z). The original
Minkowski metric written in the (t',x',y',z') is:

ds^2 = c^2 dt'^2 - (dh/dx')^2 dx'^2 - dy'^2 - dz'^2 (*)

...where by h(x') I denoted the inverse of f(x).
Post by Koobee Wublee
This spacetime translate into the usual rectangular coordinate is the
following.
ds'^2 = c^2 dt^2 - (df/dx)^2 dx^2 - dy^2 - dz^2
You can easily see that ds'^2 is distorted along the x axis.
QED.
You've changed the metric, so naturally having performed a (correct!)
coordinate transformation from x' to x you ended up with the same
(i.e., changed!) metric. Had you started from the correct formula (*)
you would've obtained while transforming from x' to x:

ds^2 = c^2 dt^2 - (dh/dx')^2 (df/dx)^2 dx^2 - dy^2 - dz^2 =

= c^2 dt^2 - dx^2 - dy^2 - dz^2

...since (dh/dx')^2 (df/dx)^2 = 1 (f and h are mutual inverses). Same,
original, Minkowski metric.
Post by Koobee Wublee
This sh*t is so easy. I am very surprised that you have so much
trouble understanding this. <shrug>
It has lots of subtleties though. You are much too careless.

[...]

--
Jan Bielawski
Koobee Wublee
2007-01-17 18:10:44 UTC
Post by JanPB
Post by Koobee Wublee
Say we have the Euclidean rectangular coordinate system (x, y, z) ---
the usual. With the Minkowski metric, we have the following flat
spacetime.
ds^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2
Then, we have another coordinate system (x', y', z') where
** t' = t
** x' = f(x)
** y' = y
** z' = z
Then, the Minkowski spacetime with this set of highly non-linear
coordinate system is
ds'^2 = c^2 dt'^2 - dx'^2 - dy'^2 - dz'^2
This is not the Minkowski metric on the (t,x,y,z)-manifold.
That is correct. But ds'^2 still enjoys a Miknkowski metric on
(t', x', y', z').
Post by JanPB
You've
changed coordinates but _kept the matrix unchanged_ - this CHANGES THE
METRIC.
No, the metric remains the same. The Minkowski meric is now describing
a different spacetime.
Post by JanPB
For the Nth time: in order NOT to change the metric while
switching coordinates, you have to CHANGE the matrix of metric
coefficients g_ij according to the usual formula (X-transpose)...etc.
For the n'th time, [g] is observer dependent that is why you have so
much trouble with.
Post by JanPB
If you change the coordinates but not the metric coefficients g_ij OR
vice-versa, you CHANGE THE METRIC.
The new metric you wrote above is flat
Minkowski metric is flat, yes. <shrug>
Post by JanPB
and hence isometric to Minkowski
but it's not _the_ Minkowski metric on (t,x,y,z). The original
ds^2 = c^2 dt'^2 - (dh/dx')^2 dx'^2 - dy'^2 - dz'^2 (*)
...where by h(x') I denoted the inverse of f(x).
That is correct, yes. However, ds'^2 is not the same as ds^2.
Post by JanPB
Post by Koobee Wublee
This spacetime translate into the usual rectangular coordinate is the
following.
ds'^2 = c^2 dt^2 - (df/dx)^2 dx^2 - dy^2 - dz^2
You can easily see that ds'^2 is distorted along the x axis.
QED.
You've changed the metric, so naturally having performed a (correct!)
coordinate transformation from x' to x you ended up with the same
(i.e., changed!) metric. Had you started from the correct formula (*)
No, I have not changed the metric. The metric is still Minkowski. I
have demonstrated the geometry depends on the choice of coordinate
system if the metric is kept the same.
Post by JanPB
ds^2 = c^2 dt^2 - (dh/dx')^2 (df/dx)^2 dx^2 - dy^2 - dz^2 =
= c^2 dt^2 - dx^2 - dy^2 - dz^2
...since (dh/dx')^2 (df/dx)^2 = 1 (f and h are mutual inverses). Same,
original, Minkowski metric.
Yes, but

ds'^2 = c^2 dt'^2 - dx'^2 - dy'^2 - dz'^2
= c^2 dt^2 - (df/dx)^2 dx^2 - dy^2 - dz^2

Where

** ds'^2 =/= ds^2
Post by JanPB
Post by Koobee Wublee
This sh*t is so easy. I am very surprised that you have so much
trouble understanding this. <shrug>
It has lots of subtleties though. You are much too careless.
It is just elementary school mathematics. You have so much trouble
understanding basic mathematics. <shrug>
JanPB
2007-01-17 18:37:38 UTC
Post by Koobee Wublee
Post by JanPB
Post by Koobee Wublee
Say we have the Euclidean rectangular coordinate system (x, y, z) ---
the usual. With the Minkowski metric, we have the following flat
spacetime.
ds^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2
Then, we have another coordinate system (x', y', z') where
** t' = t
** x' = f(x)
** y' = y
** z' = z
Then, the Minkowski spacetime with this set of highly non-linear
coordinate system is
ds'^2 = c^2 dt'^2 - dx'^2 - dy'^2 - dz'^2
This is not the Minkowski metric on the (t,x,y,z)-manifold.
That is correct. But ds'^2 still enjoys a Miknkowski metric on
(t', x', y', z').
Post by JanPB
You've
changed coordinates but _kept the matrix unchanged_ - this CHANGES THE
METRIC.
No, the metric remains the same.
Not on the (t,x,y,z)-space.
Post by Koobee Wublee
The Minkowski meric is now describing
a different spacetime.
You just defined a mapping and then demanded that it be an isometry.
Post by Koobee Wublee
Post by JanPB
For the Nth time: in order NOT to change the metric while
switching coordinates, you have to CHANGE the matrix of metric
coefficients g_ij according to the usual formula (X-transpose)...etc.
For the n'th time, [g] is observer dependent that is why you have so
much trouble with.
This is irrelevant to my remark above.
Post by Koobee Wublee
Post by JanPB
If you change the coordinates but not the metric coefficients g_ij OR
vice-versa, you CHANGE THE METRIC.
The new metric you wrote above is flat
Minkowski metric is flat, yes. <shrug>
Post by JanPB
and hence isometric to Minkowski
but it's not _the_ Minkowski metric on (t,x,y,z). The original
ds^2 = c^2 dt'^2 - (dh/dx')^2 dx'^2 - dy'^2 - dz'^2 (*)
...where by h(x') I denoted the inverse of f(x).
That is correct, yes. However, ds'^2 is not the same as ds^2.
No - as I said it's either a different metric on (t,x,y,z) or the
Minkowski metric on a different manifold altogether (the
(t',x',y',z')-space). Either way ds'^2 is different.
Post by Koobee Wublee
Post by JanPB
Post by Koobee Wublee
This spacetime translate into the usual rectangular coordinate is the
following.
ds'^2 = c^2 dt^2 - (df/dx)^2 dx^2 - dy^2 - dz^2
You can easily see that ds'^2 is distorted along the x axis.
QED.
You've changed the metric, so naturally having performed a (correct!)
coordinate transformation from x' to x you ended up with the same
(i.e., changed!) metric. Had you started from the correct formula (*)
No, I have not changed the metric. The metric is still Minkowski.
On a different manifold.
Post by Koobee Wublee
I have demonstrated the geometry depends on the choice of coordinate
system if the metric is kept the same.
You have demonstrated that if you change coordinates while keeping
g_ij's fixed, you change ds^2.
Post by Koobee Wublee
Post by JanPB
ds^2 = c^2 dt^2 - (dh/dx')^2 (df/dx)^2 dx^2 - dy^2 - dz^2 =
= c^2 dt^2 - dx^2 - dy^2 - dz^2
...since (dh/dx')^2 (df/dx)^2 = 1 (f and h are mutual inverses). Same,
original, Minkowski metric.
Yes, but
ds'^2 = c^2 dt'^2 - dx'^2 - dy'^2 - dz'^2
= c^2 dt^2 - (df/dx)^2 dx^2 - dy^2 - dz^2
Where
** ds'^2 =/= ds^2
Again, you change the metric, you change ds^2. The metric *is* ds^2 in
general. That's how the object called "the metric" or "the metric
tensor" is defined.
Post by Koobee Wublee
Post by JanPB
Post by Koobee Wublee
This sh*t is so easy. I am very surprised that you have so much
trouble understanding this. <shrug>
It has lots of subtleties though. You are much too careless.
It is just elementary school mathematics. You have so much trouble
understanding basic mathematics. <shrug>
Differential geometry is hardly "elementary school mathematics". It was
enough to confuse the best minds not too long ago. Now it's very well
understood.

--
Jan Bielawski
Koobee Wublee
2007-01-17 19:15:53 UTC
Post by JanPB
Again, you change the metric, you change ds^2.
That is correct. This is exactly what I am trying to show.
Post by JanPB
The metric *is* ds^2 in general.
And this is wrong. This is exactly what I am trying to show as well.
The metric is not ds^2.
Post by JanPB
That's how the object called "the metric" or "the metric
tensor" is defined.
And this is exactly what I am trying to tell you. 100 years of
nonsense of referring to the metric as ds^2 is wrong.
Post by JanPB
Differential geometry is hardly "elementary school mathematics". It was
enough to confuse the best minds not too long ago. Now it's very well
understood.
No, this stuff is very trivial. It was made it this way in order for
the ignorant public to buy into GR.

The simplest relevant mathematical form is as follows.

Given,

A = B C

B is not necessarily equivalent to A.

In the mathematical model of spacetime,

ds^2 = g_ij dq^i dq^j

[g], the metric or a matrix, is not ds^2 period.

Saying the spacetime is flat given the Minkowski metric is wrong. You
need also to find out what the coordinate system is. The metric along
can never tell you about the geometry it is describing. You also need
the choice of coordinate system to complete the observation.

I don't think you are capable of understanding this very trivial
concept. <shrug> Thus, what Feynman was saying about the pompous
fools that he could not stand include you. It is enough to beat this
simple mathematical concept into pulp.

"Pompous fools drive me up the wall. Ordinary folks are all right; you
can talk to them, and try to help them out. But pompous fools - guys
who are fools and are covering it all over and impressing people as to
how wonderful they are with all this hocus pocus - that, I cannot
stand!"
--- Richard Feynman
JanPB
2007-01-17 21:25:34 UTC
Post by Koobee Wublee
Post by JanPB
Again, you change the metric, you change ds^2.
That is correct. This is exactly what I am trying to show.
Post by JanPB
The metric *is* ds^2 in general.
And this is wrong. This is exactly what I am trying to show as well.
The metric is not ds^2.
It's a definition. The metric is defined as a certain tensor denoted by
"ds^2".
Post by Koobee Wublee
Post by JanPB
That's how the object called "the metric" or "the metric
tensor" is defined.
And this is exactly what I am trying to tell you. 100 years of
nonsense of referring to the metric as ds^2 is wrong.
It's a definition. Your definition doesn't model real world correctly
because you think of "metric" as a matrix which means that - for
example - in the 2D plane the following two "metrics" are different:

[1 0]
[0 1] in rectangular coordinates

and

[1 0 ]
[0 r^2] in polar coordinates.

Since these two are the same 2D plane, one cannot sensibly define
"metric" as a "matrix".
Post by Koobee Wublee
Post by JanPB
Differential geometry is hardly "elementary school mathematics". It was
enough to confuse the best minds not too long ago. Now it's very well
understood.
No, this stuff is very trivial. It was made it this way in order for
the ignorant public to buy into GR.
It's not very trivial at all - you are confused by the very basics of
it. And I know of better way to con the public than talk about general
relativity (of all things). Ludicrous.
Post by Koobee Wublee
The simplest relevant mathematical form is as follows.
Given,
A = B C
B is not necessarily equivalent to A.
In the mathematical model of spacetime,
ds^2 = g_ij dq^i dq^j
[g], the metric or a matrix, is not ds^2 period.
Sure, a tensor is not a matrix.
Post by Koobee Wublee
Saying the spacetime is flat given the Minkowski metric is wrong. You
need also to find out what the coordinate system is.
No. Flatness is an abstraction of a property that exists before
introduction of any man-made labels. A given spacetime just *is*
certain way whether it's measured or not. Metric tensor is a
mathematical model representing this.

Coordinate systems OTOH are a means to turn certain pre-existing
physical manifestations into numbers or some other quantification
means.
Post by Koobee Wublee
The metric along
can never tell you about the geometry it is describing. You also need
the choice of coordinate system to complete the observation.
Yes but that's different than saying that metric is
coordinate-dependent. What's coordinate-dependent is the details of
relevant quantification procedures. Metric is an abstraction which is
fixed because it represents a fixed physical structure.

[...]

--
Jan Bielawski
Koobee Wublee
2007-01-18 05:15:06 UTC
Post by JanPB
The metric *is* ds^2 in general.
Post by Koobee Wublee
And this is wrong. This is exactly what I am trying to show as well.
The metric is not ds^2.
It's a definition. The metric is defined as a certain tensor denoted by
"ds^2".
The definition is absurd. ds^2 is an invariant scalar. The metric is
a matrix.
Post by JanPB
That's how the object called "the metric" or "the metric
tensor" is defined.
Post by Koobee Wublee
And this is exactly what I am trying to tell you. 100 years of
nonsense of referring to the metric as ds^2 is wrong.
It's a definition. Your definition doesn't model real world correctly
because you think of "metric" as a matrix which means that - for
[1 0]
[0 1] in rectangular coordinates
and
[1 0 ]
[0 r^2] in polar coordinates.
Since these two are the same 2D plane, one cannot sensibly define
"metric" as a "matrix".
These are matrices, are they not? They are two very different
matrices. Each one does not describe a geometry until you also define
what coordinate system you have.
Post by JanPB
Post by Koobee Wublee
The simplest relevant mathematical form is as follows.
Given,
A = B C
B is not necessarily equivalent to A.
In the mathematical model of spacetime,
ds^2 = g_ij dq^i dq^j
[g], the metric or a matrix, is not ds^2 period.
Sure, a tensor is not a matrix.
This matrix [g] is observer dependent. It would lead to an infinite
number of solutions to the Einstein field equations where each solution
describes a unique geometry. Are you now in agreement?
Post by JanPB
Post by Koobee Wublee
Saying the spacetime is flat given the Minkowski metric is wrong. You
need also to find out what the coordinate system is.
No. Flatness is an abstraction of a property that exists before
introduction of any man-made labels. A given spacetime just *is*
certain way whether it's measured or not. Metric tensor is a
mathematical model representing this.
Let me rephrase what I said. The Minkowski metric does not necessarily
imply spacetime is flat.
Post by JanPB
Coordinate systems OTOH are a means to turn certain pre-existing
physical manifestations into numbers or some other quantification
means.
The matrix [g] is useless without knowing what coordinate system one
chooses.
Post by JanPB
Post by Koobee Wublee
The metric along
can never tell you about the geometry it is describing. You also need
the choice of coordinate system to complete the observation.
Yes but that's different than saying that metric is
coordinate-dependent.
Why not? The metric is [g] which is a matrix.
Post by JanPB
What's coordinate-dependent is the details of
relevant quantification procedures.
Metric is an abstraction which is
fixed because it represents a fixed physical structure.
But, the metric is not an abstract mathematical quantity. It is the
matrix [g].
Daryl McCullough
2007-01-18 12:19:14 UTC
Koobee Wublee says...
Post by Koobee Wublee
The definition is absurd. ds^2 is an invariant scalar. The metric is
a matrix.
Look at the equation for ds^2 in, say, good-old 2D Euclidean space
in Cartesian coordinates:

ds^2 = dx^2 + dy^2

ds^2 depends on dx and dy. If you make dx twice as large, and
you make dy twice as large, then ds^2 will be four times as
large.

The whole point of this equation is to describe a function;
a quadratic function from vectors to reals. ds^2 is not a
fixed number, it is a function of the displacement vector.

When people talk about the metric, they mean the mapping
from vectors to reals, and ds^2 specifies that mapping.
Post by Koobee Wublee
Post by JanPB
It's a definition. Your definition doesn't model real world correctly
because you think of "metric" as a matrix which means that - for
[1 0]
[0 1] in rectangular coordinates
and
[1 0 ]
[0 r^2] in polar coordinates.
Since these two are the same 2D plane, one cannot sensibly define
"metric" as a "matrix".
These are matrices, are they not?
Yes, but they are *not* metrics. That's the point---a metric
is not a matrix.
Post by Koobee Wublee
Post by JanPB
Post by Koobee Wublee
[g], the metric or a matrix, is not ds^2 period.
Sure, a tensor is not a matrix.
This matrix [g] is observer dependent.
Yes, and [g] is not the metric.
Post by Koobee Wublee
Let me rephrase what I said. The Minkowski metric does not necessarily
imply spacetime is flat.
Yes, it certainly does.
Post by Koobee Wublee
Post by JanPB
Coordinate systems OTOH are a means to turn certain pre-existing
physical manifestations into numbers or some other quantification
means.
The matrix [g] is useless without knowing what coordinate system one
chooses.
That's why [g] is not the metric. The metric does not depend
on coordinate systems.
Post by Koobee Wublee
Post by JanPB
Post by Koobee Wublee
The metric along
can never tell you about the geometry it is describing. You also need
the choice of coordinate system to complete the observation.
Yes but that's different than saying that metric is
coordinate-dependent.
Why not? The metric is [g] which is a matrix.
The metric is *not* [g]! Why do you keep saying that?
Post by Koobee Wublee
But, the metric is not an abstract mathematical quantity. It is the
matrix [g].
No, it's not. The metric is the mapping from pairs of vectors to reals.
[g] is the representation of the metric in a particular coordinate
system.

--
Daryl McCullough
Ithaca, NY
Mike
2007-01-18 14:40:36 UTC
Post by Daryl McCullough
Koobee Wublee says...
Post by Koobee Wublee
The definition is absurd. ds^2 is an invariant scalar. The metric is
a matrix.
Look at the equation for ds^2 in, say, good-old 2D Euclidean space
ds^2 = dx^2 + dy^2
ds^2 depends on dx and dy. If you make dx twice as large, and
you make dy twice as large, then ds^2 will be four times as
large.
The whole point of this equation is to describe a function;
a quadratic function from vectors to reals. ds^2 is not a
fixed number, it is a function of the displacement vector.
You should learn that "scalar" does not mean a "fixed number".

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

An inner product is a vector multiplication that results in a scalar.
ds^2 is a function, you are correct, which computes distance between
two points in space (time). The result is a scalar.
Post by Daryl McCullough
When people talk about the metric, they mean the mapping
from vectors to reals, and ds^2 specifies that mapping.
I believe KW is not a mathematician and often assumes people understand
what he means. However, experience shows that in these groups people
will not deal with the essence of things but try to grap an irrelevant
issue, such as strict mathematical notation or terminology, to attack
another poster.

Fundamentally, all the points KW makes are correct even though he often
misses strict terminology.

Try to concentrate on his points. Men do that.

Mike
Post by Daryl McCullough
Post by Koobee Wublee
Post by JanPB
It's a definition. Your definition doesn't model real world correctly
because you think of "metric" as a matrix which means that - for
[1 0]
[0 1] in rectangular coordinates
and
[1 0 ]
[0 r^2] in polar coordinates.
Since these two are the same 2D plane, one cannot sensibly define
"metric" as a "matrix".
These are matrices, are they not?
Yes, but they are *not* metrics. That's the point---a metric
is not a matrix.
Post by Koobee Wublee
Post by JanPB
Post by Koobee Wublee
[g], the metric or a matrix, is not ds^2 period.
Sure, a tensor is not a matrix.
This matrix [g] is observer dependent.
Yes, and [g] is not the metric.
Post by Koobee Wublee
Let me rephrase what I said. The Minkowski metric does not necessarily
imply spacetime is flat.
Yes, it certainly does.
Post by Koobee Wublee
Post by JanPB
Coordinate systems OTOH are a means to turn certain pre-existing
physical manifestations into numbers or some other quantification
means.
The matrix [g] is useless without knowing what coordinate system one
chooses.
That's why [g] is not the metric. The metric does not depend
on coordinate systems.
Post by Koobee Wublee
Post by JanPB
Post by Koobee Wublee
The metric along
can never tell you about the geometry it is describing. You also need
the choice of coordinate system to complete the observation.
Yes but that's different than saying that metric is
coordinate-dependent.
Why not? The metric is [g] which is a matrix.
The metric is *not* [g]! Why do you keep saying that?
Post by Koobee Wublee
But, the metric is not an abstract mathematical quantity. It is the
matrix [g].
No, it's not. The metric is the mapping from pairs of vectors to reals.
[g] is the representation of the metric in a particular coordinate
system.
--
Daryl McCullough
Ithaca, NY
Eric Gisse
2007-01-18 14:47:20 UTC
Post by Mike
Post by Daryl McCullough
Koobee Wublee says...
Post by Koobee Wublee
The definition is absurd. ds^2 is an invariant scalar. The metric is
a matrix.
Look at the equation for ds^2 in, say, good-old 2D Euclidean space
ds^2 = dx^2 + dy^2
ds^2 depends on dx and dy. If you make dx twice as large, and
you make dy twice as large, then ds^2 will be four times as
large.
The whole point of this equation is to describe a function;
a quadratic function from vectors to reals. ds^2 is not a
fixed number, it is a function of the displacement vector.
You should learn that "scalar" does not mean a "fixed number".
http://mathworld.wolfram.com/Scalar.html
An inner product is a vector multiplication that results in a scalar.
ds^2 is a function, you are correct, which computes distance between
two points in space (time). The result is a scalar.
Post by Daryl McCullough
When people talk about the metric, they mean the mapping
from vectors to reals, and ds^2 specifies that mapping.
I believe KW is not a mathematician and often assumes people understand
what he means. However, experience shows that in these groups people
will not deal with the essence of things but try to grap an irrelevant
issue, such as strict mathematical notation or terminology, to attack
another poster.
Fundamentally, all the points KW makes are correct even though he often
misses strict terminology.
None of the 'points' KW makes are correct. He makes the same
fundamental mistakes over and over even after they are painstakingly
corrected by others.
Post by Mike
Try to concentrate on his points. Men do that.
Mike
Post by Daryl McCullough
Post by Koobee Wublee
Post by JanPB
It's a definition. Your definition doesn't model real world correctly
because you think of "metric" as a matrix which means that - for
[1 0]
[0 1] in rectangular coordinates
and
[1 0 ]
[0 r^2] in polar coordinates.
Since these two are the same 2D plane, one cannot sensibly define
"metric" as a "matrix".
These are matrices, are they not?
Yes, but they are *not* metrics. That's the point---a metric
is not a matrix.
Post by Koobee Wublee
Post by JanPB
Post by Koobee Wublee
[g], the metric or a matrix, is not ds^2 period.
Sure, a tensor is not a matrix.
This matrix [g] is observer dependent.
Yes, and [g] is not the metric.
Post by Koobee Wublee
Let me rephrase what I said. The Minkowski metric does not necessarily
imply spacetime is flat.
Yes, it certainly does.
Post by Koobee Wublee
Post by JanPB
Coordinate systems OTOH are a means to turn certain pre-existing
physical manifestations into numbers or some other quantification
means.
The matrix [g] is useless without knowing what coordinate system one
chooses.
That's why [g] is not the metric. The metric does not depend
on coordinate systems.
Post by Koobee Wublee
Post by JanPB
Post by Koobee Wublee
The metric along
can never tell you about the geometry it is describing. You also need
the choice of coordinate system to complete the observation.
Yes but that's different than saying that metric is
coordinate-dependent.
Why not? The metric is [g] which is a matrix.
The metric is *not* [g]! Why do you keep saying that?
Post by Koobee Wublee
But, the metric is not an abstract mathematical quantity. It is the
matrix [g].
No, it's not. The metric is the mapping from pairs of vectors to reals.
[g] is the representation of the metric in a particular coordinate
system.
--
Daryl McCullough
Ithaca, NY
Pmb
2007-01-18 17:45:34 UTC
Post by Mike
Post by Daryl McCullough
Koobee Wublee says...
Post by Koobee Wublee
The definition is absurd. ds^2 is an invariant scalar. The metric is
a matrix.
It is unclear if you understand that "function" and "operator" often mean
the same thing!
Post by Mike
Post by Daryl McCullough
Look at the equation for ds^2 in, say, good-old 2D Euclidean space
That statement lacks precision. A space is neither Euclidean nor MMinkowski
etc. until a metric is defined on that space.
Post by Mike
Post by Daryl McCullough
ds^2 = dx^2 + dy^2
ds^2 depends on dx and dy. If you make dx twice as large, and
you make dy twice as large, then ds^2 will be four times as
large.
The whole point of this equation is to describe a function;
a quadratic function from vectors to reals. ds^2 is not a
fixed number, it is a function of the displacement vector.
You should learn that "scalar" does not mean a "fixed number".
http://mathworld.wolfram.com/Scalar.html
Actually that definition is not quite right either. What it refers to are
Cartesian scalars, aka affine scalars. What you're speaking about is a
general scalar which is a quantity *(a number) which remains unchanged by a
change in the coordinate system
Post by Mike
I believe KW is not a mathematician and often assumes people understand
what he means. However, experience shows that in these groups people
will not deal with the essence of things but try to grap an irrelevant
issue, such as strict mathematical notation or terminology, to attack
another poster.
That is strange behaviour in that I personally would never attack a person
in a field of expertise if I disagree with them. I may have strong feelings
but I try to keep them my own, and that is a battle in and of itself!
Post by Mike
Fundamentally, all the points KW makes are correct even though he often
misses strict terminology.
That I haven't seen yet Mike no matter how hard I've tried!

Best wishes

Pete
Daryl McCullough
2007-01-18 19:33:50 UTC
Pmb says...
Post by Pmb
Post by Daryl McCullough
Look at the equation for ds^2 in, say, good-old 2D Euclidean space
That statement lacks precision. A space is neither Euclidean nor MMinkowski
etc. until a metric is defined on that space.
Calling it "Euclidean" or Minkowskian implies that it has a particular
metric, as I understand it.

--
Daryl McCullough
Ithaca, NY
Pmb
2007-01-18 19:50:58 UTC
Post by Daryl McCullough
Pmb says...
Post by Pmb
Post by Daryl McCullough
Look at the equation for ds^2 in, say, good-old 2D Euclidean space
That statement lacks precision. A space is neither Euclidean nor MMinkowski
etc. until a metric is defined on that space.
Calling it "Euclidean" or Minkowskian implies that it has a particular
metric, as I understand it.
Yes. That is correct.

Pete
Mike
2007-01-18 14:49:46 UTC
Post by Daryl McCullough
Koobee Wublee says...
Post by Koobee Wublee
The definition is absurd. ds^2 is an invariant scalar. The metric is
a matrix.
Look at the equation for ds^2 in, say, good-old 2D Euclidean space
ds^2 = dx^2 + dy^2
ds^2 depends on dx and dy. If you make dx twice as large, and
you make dy twice as large, then ds^2 will be four times as
large.
The whole point of this equation is to describe a function;
a quadratic function from vectors to reals. ds^2 is not a
fixed number, it is a function of the displacement vector.
When people talk about the metric, they mean the mapping
from vectors to reals, and ds^2 specifies that mapping.
Post by Koobee Wublee
Post by JanPB
It's a definition. Your definition doesn't model real world correctly
because you think of "metric" as a matrix which means that - for
[1 0]
[0 1] in rectangular coordinates
and
[1 0 ]
[0 r^2] in polar coordinates.
Since these two are the same 2D plane, one cannot sensibly define
"metric" as a "matrix".
These are matrices, are they not?
Yes, but they are *not* metrics. That's the point---a metric
is not a matrix.
Post by Koobee Wublee
Post by JanPB
Post by Koobee Wublee
[g], the metric or a matrix, is not ds^2 period.
Sure, a tensor is not a matrix.
This matrix [g] is observer dependent.
Yes, and [g] is not the metric.
Post by Koobee Wublee
Let me rephrase what I said. The Minkowski metric does not necessarily
imply spacetime is flat.
Yes, it certainly does.
Post by Koobee Wublee
Post by JanPB
Coordinate systems OTOH are a means to turn certain pre-existing
physical manifestations into numbers or some other quantification
means.
The matrix [g] is useless without knowing what coordinate system one
chooses.
That's why [g] is not the metric. The metric does not depend
on coordinate systems.
Post by Koobee Wublee
Post by JanPB
Post by Koobee Wublee
The metric along
can never tell you about the geometry it is describing. You also need
the choice of coordinate system to complete the observation.
Yes but that's different than saying that metric is
coordinate-dependent.
Why not? The metric is [g] which is a matrix.
The metric is *not* [g]! Why do you keep saying that?
Post by Koobee Wublee
But, the metric is not an abstract mathematical quantity. It is the
matrix [g].
No, it's not. The metric is the mapping from pairs of vectors to reals.
[g] is the representation of the metric in a particular coordinate
system.
[g] is also called the metric.

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

The POINT is that GRists have developed their own terminology which
within the context of what they mean is correct but it is confusing to
innocent people passing by. They also use the bastardized terminology
to attack people and divert attention from the REAL ISSSUES.

and you do a good job Daryl.

Ask anyone, the metric is [g]. ds^2 is the line element.

Mike
Post by Daryl McCullough
--
Daryl McCullough
Ithaca, NY
Daryl McCullough
2007-01-18 14:04:07 UTC
JanPB says...
Post by JanPB
It's a definition. The metric is defined as a certain tensor denoted by
"ds^2".
Actually, it's a little confusing because ds^2 is a *quadratic*
function of one vector, while the metric is a *bilinear* function
of two vectors. Of course, they are computable from one another:

ds^2 = g(dx,dx)

this gives ds^2 as a function of an infinitesimal displacement
vector dx. The other way around, if U and V are displacement
vectors, then we can compute g(U,V) as follows:

g(U,V) = (ds^2(U+V) - ds^2(U) - ds^2(V))/2

--
Daryl McCullough
Ithaca, NY
JanPB
2007-01-18 20:18:12 UTC
Post by Daryl McCullough
JanPB says...
Post by JanPB
It's a definition. The metric is defined as a certain tensor denoted by
"ds^2".
Actually, it's a little confusing because ds^2 is a *quadratic*
function of one vector, while the metric is a *bilinear* function
of two vectors.
It can be treated as either, as you point out. As a quadratic form,
terms like "dx dy" are interpreted as multiplication of functions:

(dx dy)(U) =(by definition)= dx(U) * dy(U).

As a symmetric bilinear form "dx dy" is the symmetric product:

dx dy =(by definition)= 1/2 * (dx X dy + dy X dx)

...where X is the tensor product:

(dx X dy)(U,V) = dx(U) * dy(V).

I prefer "direct access" to the dot product (independent of the
polarisation identity) which the second interpretation provides. It
also goes along with the anti-symmetric version (the wedge product) and
the Clifford version (not that they are relevant to this discussion
:-))

--
Jan Bielawski

bergeron
2007-01-17 09:16:59 UTC
Post by Koobee Wublee
Say we have the Euclidean rectangular coordinate system (x, y, z) ---
the usual. With the Minkowski metric, we have the following flat
spacetime.
ds^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2
Then, we have another coordinate system (x', y', z') where
** t' = t
** x' = f(x)
** y' = y
** z' = z
Then, the Minkowski spacetime with this set of highly non-linear
coordinate system is
You have to be one of the stupidest fucks
ever to post on this news group.

Pay attention, dip shit. ds^2 is an invariant,
so ds^2 = ds'^2.
Post by Koobee Wublee
ds'^2 = c^2 dt'^2 - dx'^2 - dy'^2 - dz'^2
This spacetime translate into the usual rectangular coordinate is the
following.
ds'^2 = c^2 dt^2 - (df/dx)^2 dx^2 - dy^2 - dz^2
You can easily see that ds'^2 is distorted along the x axis.
Congratulations, Mr. Self-Proclaimed Disciple of Riemann.
You have just made the remarkable discovery that stretching
something changes its length. WOW! I'm certain that
as his Prophet and Representitive on Earth.

Pretty soon, you might be able answer the more general question
of whether or not a blob of Play-doh will be deformed by
squishing it.
Post by Koobee Wublee
QED.
This sh*t is so easy. I am very surprised that you have so much
trouble understanding this. <shrug>
Well, it is easy to understand for most of us,
but you have a long way to go before you're only
the oven and spare those around you the burden of
caring for a oversized broccoli.
Koobee Wublee
2007-01-17 18:10:05 UTC
Post by bergeron
Post by Koobee Wublee
Say we have the Euclidean rectangular coordinate system (x, y, z) ---
the usual. With the Minkowski metric, we have the following flat
spacetime.
ds^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2
Then, we have another coordinate system (x', y', z') where
** t' = t
** x' = f(x)
** y' = y
** z' = z
Then, the Minkowski spacetime with this set of highly non-linear
coordinate system is
You have to be one of the stupidest fucks
ever to post on this news group.
You are talking about yourself, dude.
Post by bergeron
Pay attention, dip shit. ds^2 is an invariant,
so ds^2 = ds'^2.
What I have shown, ds'^2 =/= ds^2. You have trouble with elementary
school mathematics. You need to go back to your nearest elementary
<shrug>
Post by bergeron
Post by Koobee Wublee
ds'^2 = c^2 dt'^2 - dx'^2 - dy'^2 - dz'^2
This spacetime translate into the usual rectangular coordinate is the
following.
ds'^2 = c^2 dt^2 - (df/dx)^2 dx^2 - dy^2 - dz^2
You can easily see that ds'^2 is distorted along the x axis.
Congratulations, Mr. Self-Proclaimed Disciple of Riemann.
It is the only one who has understood the curvature in space or in
spacetime since Riemann.
Post by bergeron
You have just made the remarkable discovery that stretching
something changes its length. WOW!
Yes, wow! I am with you. I have just made the remarkable discovery.
<shrug>
Post by bergeron
I'm certain that
as his Prophet and Representitive on Earth.
Riemann would be proud of me, yes. Since Riemann was no god, he did
not need any prophet or representative on earth. You need too take
Post by bergeron
Pretty soon, you might be able answer the more general question
of whether or not a blob of Play-doh will be deformed by
squishing it.
Post by Koobee Wublee
QED.
This sh*t is so easy. I am very surprised that you have so much
trouble understanding this. <shrug>
Well, it is easy to understand for most of us,
but you have a long way to go before you're only
confused.
Didn't Confucius say "The confused is constantly accusing the wise
as down in his confused level"? Confucius must be talking about you
Post by bergeron
the oven and spare those around you the burden of
caring for a oversized broccoli.
No, thanks. Look at what that had done to yourself. You should not
have tried that. <shrug>
bergeron
2007-01-18 01:38:10 UTC
Post by Koobee Wublee
Post by bergeron
Post by Koobee Wublee
Then, the Minkowski spacetime with this set of highly non-linear
coordinate system is
You have to be one of the stupidest fucks
ever to post on this news group.
You are talking about yourself, dude.
Post by bergeron
Pay attention, dip shit. ds^2 is an invariant,
so ds^2 = ds'^2.
What I have shown,
What you have shown is that you don't know the difference
between an affine space and an affine space equipped with a
metric. That is your deficiency and has nothing to do with
general relativity, which is a metric theory.
Post by Koobee Wublee
ds'^2 =/= ds^2. You have trouble with elementary
school mathematics.
Any time you want to compare your mathematical
skills to mine, come up with a means of settling
the issue such that the outcome is indisputable and
everything gets posted to this newsgroup so that
I and everyone else can refer to it as proof that
you are an incompetent poseur for as long as this
newsgroup exists.

*snip*
Post by Koobee Wublee
Post by bergeron
You have just made the remarkable discovery that stretching
something changes its length. WOW!
Yes, wow! I am with you. I have just made the remarkable discovery.
I hate to break the news to you, but the
entire point of a metric space is that it has
a metric. Are you really as fucking stupid as

*snip*
Post by Koobee Wublee
Post by bergeron
Well, it is easy to understand for most of us,
but you have a long way to go before you're only
confused.
Didn't Confucius say "The confused is constantly accusing the wise
as down in his confused level"?
Koobee Wublee
2007-01-18 05:15:59 UTC
Post by bergeron
What you have shown is that you don't know the difference
between an affine space and an affine space equipped with a
metric. That is your deficiency and has nothing to do with
general relativity, which is a metric theory.
Unbelievable. You can come up with a theory on a theme based on the
following absurd mathematical claim.

If A = B C, then A = B.
Post by bergeron
Post by Koobee Wublee
ds'^2 =/= ds^2. You have trouble with elementary
school mathematics.
Any time you want to compare your mathematical
skills to mine, come up with a means of settling
the issue such that the outcome is indisputable and
everything gets posted to this newsgroup so that
I and everyone else can refer to it as proof that
you are an incompetent poseur for as long as this
newsgroup exists.
We have already done so. I am not impressed with your crap below.

If A = B C, then A = B.
Post by bergeron
I hate to break the news to you, but the
entire point of a metric space is that it has
a metric.
Post by bergeron
Are you really as fucking stupid as
Post by bergeron
Post by Koobee Wublee
Didn't Confucius say "The confused is constantly accusing the wise
as down in his confused level"?
Perhaps. Now, how do we find out?
bergeron
2007-01-18 05:51:19 UTC
Post by Koobee Wublee
Post by bergeron
What you have shown is that you don't know the difference
between an affine space and an affine space equipped with a
metric. That is your deficiency and has nothing to do with
general relativity, which is a metric theory.
Unbelievable. You can come up with a theory on a theme based on the
following absurd mathematical claim.
If A = B C, then A = B.
Post the reference to an article I posted
which contains that statement.
Post by Koobee Wublee
Post by bergeron
Post by Koobee Wublee
ds'^2 =/= ds^2. You have trouble with elementary
school mathematics.
Any time you want to compare your mathematical
skills to mine, come up with a means of settling
the issue such that the outcome is indisputable and
everything gets posted to this newsgroup so that
I and everyone else can refer to it as proof that
you are an incompetent poseur for as long as this
newsgroup exists.
We have already done so. I am not impressed with your crap below.
Excuse me, but unless you can find the article in which
you claim I posted that crap, it's _your_ crap. As the self-
proclaimed Sole Individual to Understand Riemann's work since
Riemann, don't you think it would be easier to prove it than
Post by Koobee Wublee
If A = B C, then A = B.
What makes you say that?
Post by Koobee Wublee
Post by bergeron
I hate to break the news to you, but the
entire point of a metric space is that it has
a metric.
If you knew that, then why did your post
something to the contrary? Perhaps you are
just modest and enjoy being a bufoon.
Post by Koobee Wublee
Post by bergeron
Are you really as fucking stupid as
Is that a yes?
Koobee Wublee
2007-01-18 06:00:23 UTC
Post by bergeron
Post by Koobee Wublee
Unbelievable. You can come up with a theory on a theme based on the
following absurd mathematical claim.
If A = B C, then A = B.
Post the reference to an article I posted
which contains that statement.
I don't need to. When you say the metric is invariant, you do.
<shrug>
Post by bergeron
Excuse me, but unless you can find the article in which
you claim I posted that crap, it's _your_ crap. As the self-
proclaimed Sole Individual to Understand Riemann's work since
Riemann, don't you think it would be easier to prove it than
Post by Koobee Wublee
If A = B C, then A = B.
What makes you say that?
This is basic theme of the ones claiming the metric as invariant.
<shrug>
Post by bergeron
Post by Koobee Wublee
Post by bergeron
I hate to break the news to you, but the
entire point of a metric space is that it has
a metric.
If you knew that, then why did your post
something to the contrary?
I did not. <shrug>
Post by bergeron
Perhaps you are
just modest and enjoy being a bufoon.
On the contrary, I am not. <shrug>
Post by bergeron
Are you really as fucking stupid as
Post by Koobee Wublee
Is that a yes?
No. <shrug>
bergeron
2007-01-18 09:11:41 UTC
Post by Koobee Wublee
Post by bergeron
Post by Koobee Wublee
Unbelievable. You can come up with a theory on a theme based on the
following absurd mathematical claim.
If A = B C, then A = B.
Post the reference to an article I posted
which contains that statement.
I don't need to. When you say the metric is invariant, you do.
<shrug>
In other words, you made it up.
Post by Koobee Wublee
Post by bergeron
Excuse me, but unless you can find the article in which
you claim I posted that crap, it's _your_ crap. As the self-
proclaimed Sole Individual to Understand Riemann's work since
Riemann, don't you think it would be easier to prove it than
Post by Koobee Wublee
If A = B C, then A = B.
What makes you say that?
This is basic theme of the ones claiming the metric as invariant.
<shrug>
Wow Mr. Disciple of Riemann, that is some awsome
mathematical talent you have there.
I'll bet you are really perplexed by
your inability to be taken seriously
by anyone with a positive IQ.
Pmb
2007-01-16 12:43:32 UTC
The connection coefficients are all zero so Riemann is zero, ..
This isn't true. The connection coefficients could all be zero and yet
still have a non-vanishing Riemann tensor. E.g. in a locally inertial frame
in a curved spacetime all the connection coefficients vanish at the origin,
however the Riemann tensor doesn't vanish because the derivatives of the
connection coefficients don't vanish.

Best Regards

Pete
Eric Gisse
2007-01-16 14:52:41 UTC
Post by Pmb
The connection coefficients are all zero so Riemann is zero, ..
This isn't true. The connection coefficients could all be zero and yet
still have a non-vanishing Riemann tensor. E.g. in a locally inertial frame
in a curved spacetime all the connection coefficients vanish at the origin,
however the Riemann tensor doesn't vanish because the derivatives of the
connection coefficients don't vanish.
Did I say "locally inertial" That is trivally true, so I did not
mention it.

I am talking about in general: If the coefficients are zero then the
manifold is not curved because the Riemann tensor is defined in terms
of the coefficients and their derivatives.
Post by Pmb
Best Regards
Pete
Pmb
2007-01-16 15:10:52 UTC
Post by Eric Gisse
Post by Pmb
The connection coefficients are all zero so Riemann is zero, ..
This isn't true. The connection coefficients could all be zero and yet
still have a non-vanishing Riemann tensor. E.g. in a locally inertial frame
in a curved spacetime all the connection coefficients vanish at the origin,
however the Riemann tensor doesn't vanish because the derivatives of the
connection coefficients don't vanish.
Did I say "locally inertial" That is trivally true, so I did not
mention it.
The point was that the vanishing of the Christoffel symbols is no gaurentee
that the Reimann tensor will vanish I merely used a locally inertial frame
**as an example** to illustrate what I meant. I appologize if I was unclear.
Post by Eric Gisse
I am talking about in general: If the coefficients are zero then the
manifold is not curved because the Riemann tensor is defined in terms
of the coefficients and their derivatives.
That is not true. The Riemann tensor is defined in terms of the connection
coefficients **and** their derivatives. Please see the defining equation
for the Riemann tensor in Eq. (21) at

http://www.geocities.com/physics_world/gr/geodesic_deviation.htm

Best Regards

Pete
Eric Gisse
2007-01-16 15:25:41 UTC
Post by Pmb
Post by Eric Gisse
Post by Pmb
The connection coefficients are all zero so Riemann is zero, ..
This isn't true. The connection coefficients could all be zero and yet
still have a non-vanishing Riemann tensor. E.g. in a locally inertial frame
in a curved spacetime all the connection coefficients vanish at the origin,
however the Riemann tensor doesn't vanish because the derivatives of the
connection coefficients don't vanish.
Did I say "locally inertial" That is trivally true, so I did not
mention it.
The point was that the vanishing of the Christoffel symbols is no gaurentee
that the Reimann tensor will vanish I merely used a locally inertial frame
**as an example** to illustrate what I meant. I appologize if I was unclear.
Post by Eric Gisse
I am talking about in general: If the coefficients are zero then the
manifold is not curved because the Riemann tensor is defined in terms
of the coefficients and their derivatives.
That is not true. The Riemann tensor is defined in terms of the connection
coefficients **and** their derivatives. Please see the defining equation
for the Riemann tensor in Eq. (21) at
zuh?

That is _exactly_ what I said.
Post by Pmb
http://www.geocities.com/physics_world/gr/geodesic_deviation.htm
Best Regards
Pete
Pmb
2007-01-16 15:40:40 UTC
Post by Eric Gisse
Post by Pmb
Post by Eric Gisse
Post by Pmb
The connection coefficients are all zero so Riemann is zero, ..
This isn't true. The connection coefficients could all be zero and yet
still have a non-vanishing Riemann tensor. E.g. in a locally inertial frame
in a curved spacetime all the connection coefficients vanish at the origin,
however the Riemann tensor doesn't vanish because the derivatives of the
connection coefficients don't vanish.
Did I say "locally inertial" That is trivally true, so I did not
mention it.
The point was that the vanishing of the Christoffel symbols is no gaurentee
that the Reimann tensor will vanish I merely used a locally inertial frame
**as an example** to illustrate what I meant. I appologize if I was unclear.
Post by Eric Gisse
I am talking about in general: If the coefficients are zero then the
manifold is not curved because the Riemann tensor is defined in terms
of the coefficients and their derivatives.
That is not true. The Riemann tensor is defined in terms of the connection
coefficients **and** their derivatives. Please see the defining equation
for the Riemann tensor in Eq. (21) at
zuh?
That is _exactly_ what I said.
What you *exactly* said was this - ..... uh oh! You're correct Eric. I
appologize for that mistake.

Best wishes

Pete

ps - I must need another percocet! LOL!
Daryl McCullough
2007-01-15 14:32:48 UTC
Koobee Wublee says...
Post by Koobee Wublee
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.
Post by Koobee Wublee
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.
Post by Koobee Wublee
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
Koobee Wublee
2007-01-15 22:07:27 UTC
Post by Daryl McCullough
Koobee Wublee says...
Post by Koobee Wublee
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
Post by Daryl McCullough
Post by Koobee Wublee
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.
Post by Daryl McCullough
Post by Koobee Wublee
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.
Eric Gisse
2007-01-16 02:00:24 UTC
Post by Koobee Wublee
Post by Daryl McCullough
Koobee Wublee says...
Post by Koobee Wublee
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
Post by Daryl McCullough
Post by Koobee Wublee
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.
Post by Daryl McCullough
Post by Koobee Wublee
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.
Daryl McCullough
2007-01-16 02:25:31 UTC
Koobee Wublee says...
Post by Koobee Wublee
Post by Daryl McCullough
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
Post by Koobee Wublee
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

--
Daryl McCullough
Ithaca, NY
Koobee Wublee
2007-01-16 07:19:21 UTC
Post by Daryl McCullough
Koobee Wublee says...
Well, you were not correct when you said
Post by Koobee Wublee
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
Well, your definition of the metric seems not to be [g]. Then, you are
way out in the left field. Thus, you have no right to contest my claim
that each solution to the field equations in a particular coordinate
system is unique and represents a very different geometry than the
others.
Daryl McCullough
2007-01-17 16:50:38 UTC
Koobee Wublee says...
Post by Koobee Wublee
Post by Daryl McCullough
Koobee Wublee says...
Well, you were not correct when you said
Post by Koobee Wublee
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
Well, your definition of the metric seems not to be [g].
So, are you agreeing that you are wrong to say "According
to the pros, although the elements of the matrices
(g_ij =/= g'_ij), the matrix ([g] = [g'])"? *Nobody* says
that.

--
Daryl McCullough
Ithaca, NY
Pmb
2007-01-17 17:47:49 UTC
Post by Daryl McCullough
Koobee Wublee says...
Post by Koobee Wublee
Post by Daryl McCullough
Koobee Wublee says...
Well, you were not correct when you said
Post by Koobee Wublee
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
Well, your definition of the metric seems not to be [g].
So, are you agreeing that you are wrong to say "According
to the pros, although the elements of the matrices
(g_ij =/= g'_ij), the matrix ([g] = [g'])"? *Nobody* says
that.
--
Daryl McCullough
Ithaca, NY
And yet another day with KW refusing to provide a textual source for his
definition. To be honest he sounds like a sophmore math student he knows
just enough math to hurt himself. That can be gleened by his attempts at
testing me. Only someone who is very insecure would do that. Don't worry
KW - I still think that there is hope for you since you have demonstrated
that its possible for you to be grateful for the work done for you by
others. I commend you on that.

Pete
Koobee Wublee
2007-01-17 18:09:14 UTC
Post by Daryl McCullough
Koobee Wublee says...
Post by Koobee Wublee
Well, your definition of the metric seems not to be [g].
So, are you agreeing that you are wrong to say "According
to the pros, although the elements of the matrices
(g_ij =/= g'_ij), the matrix ([g] = [g'])"?
No, I am not wrong by saying that. I was just saying about you. I
still don't know what you agenda is after botching the surface of a
sphere as it appears in curved space.
Post by Daryl McCullough
*Nobody* says that.
Did you talk to Mr. Bielawski, Mr. Brown, Dr. Roberts, Professor
Carlip? Have you not been reading what they wrote? <shrug>
Daryl McCullough
2007-01-17 19:31:04 UTC
Koobee Wublee says...
Post by Koobee Wublee
Post by Daryl McCullough
So, are you agreeing that you are wrong to say "According
to the pros, although the elements of the matrices
(g_ij =/= g'_ij), the matrix ([g] = [g'])"?
No, I am not wrong by saying that.
Then provide a single quote from a "pro" who says that.
Post by Koobee Wublee
Did you talk to Mr. Bielawski, Mr. Brown, Dr. Roberts, Professor
Carlip? Have you not been reading what they wrote?
Yes, but unlike you, I actually understand what they are talking
about. None of those people have *ever* said that the matrix
[g] = [g'].

Jan Bielawski wrote:

[g] is not "the metric". [g] is the "matrix of components
of the metric g wrt coordinate system (q)".

Peter Brown says:
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.

Tom Roberts says:

This is the essence of your confusion. A _tensor_ is _NOT_ a matrix or
determinant or collection of components; its _projection_ onto a given
coordinate system is.

Daryl McCullough says:
A metric tensor is *not* a matrix, it is a bilinear
function of two vectors.

All these people are saying the same thing, that the metric is
not a matrix, it is a tensor.

--
Daryl McCullough
Ithaca, NY
bergeron
2007-01-15 14:56:11 UTC
You would need 50 more IQ points to realize
that you're incompetent.
Post by Koobee Wublee
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 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']). 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.
Tom Roberts
2007-01-15 16:45:32 UTC
Post by Koobee Wublee
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

Tom Roberts
Koobee Wublee
2007-01-15 22:08:10 UTC
Post by Tom Roberts
Post by Koobee Wublee
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
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
bergeron
2007-01-15 22:27:46 UTC
Post by Koobee Wublee
Post by Tom Roberts
Post by Koobee Wublee
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
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
Pmb
2007-01-15 23:02:58 UTC
Post by bergeron
Post by Koobee Wublee
Post by Tom Roberts
Post by Koobee Wublee
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
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
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
Dirk Van de moortel
2007-01-15 23:06:44 UTC
Post by Pmb
Post by bergeron
Post by Koobee Wublee
Post by Tom Roberts
Post by Koobee Wublee
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
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
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,
(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
Pmb
2007-01-15 23:21:58 UTC
Post by Dirk Van de moortel
Post by Pmb
Post by bergeron
Post by Koobee Wublee
Post by Tom Roberts
Post by Koobee Wublee
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
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
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,
(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.)
Dirk Van de moortel
2007-01-15 23:34:56 UTC
Post by Pmb
Post by Dirk Van de moortel
Post by Pmb
Post by bergeron
Post by Koobee Wublee
Post by Tom Roberts
Post by Koobee Wublee
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
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
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,
(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.
Post by Pmb
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?
Post by Pmb
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
Pmb
2007-01-15 23:57:42 UTC
Post by Dirk Van de moortel
Post by Pmb
Post by Dirk Van de moortel
Post by Pmb
Post by bergeron
Post by Koobee Wublee
Post by Tom Roberts
Post by Koobee Wublee
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
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
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,
(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.
Post by Dirk Van de moortel
Post by Pmb
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.
Post by Dirk Van de moortel
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
Koobee Wublee
2007-01-15 23:56:42 UTC
Dirk Van de moortel wrote:
aka. Sperm Lover
Post by Dirk Van de moortel
The sperm lover does not understand that link.
Post by Dirk Van de moortel
(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>
Post by Dirk Van de moortel
(3) the only thing he is after, is to keep you busy fighting him.
The sperm lover loves to post nonsense and garbage.
Post by Dirk Van de moortel
He is a troll :-)
The sperm lover is a troll. That is why I ignore the sperm lover most
of the time. <shrug>
Koobee Wublee
2007-01-15 23:58:21 UTC
Post by Pmb
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.
Post by Pmb
We still have no idea if
Yes, that is correct.
Post by Pmb
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.
Post by Pmb
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.
Post by Pmb
KW - We have not yet argued a point of logic since we have not yet agreed on
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.
Post by Pmb
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?
Post by Pmb
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.
Post by Pmb
And you have yet to tell us what field it is
you want derived.
Post by Pmb
That would be a start? How do you derive that Lagrangian?
Pmb
2007-01-16 12:57:32 UTC
Post by Pmb
Post by bergeron
Post by Koobee Wublee
Post by Tom Roberts
Post by Koobee Wublee
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
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
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
Since I'm in such a good mood this morning I'm going to post those web pages
where I did all that work which now seems to fit KWs unreasonable challenge.
The challenge is unreasonable due to the amount of work that is required to
do this whereas my challenge required basically no work at all. Buy since

KW wanted me to derive the Christoffel symbols of the second kind. To derive
them means to derive the equation in which they appear and henceforth seemed
to require a seperate symbold and name. Here are several ways to get the
Christoffel symbols
http://www.geocities.com/physics_world/ma/chris_sym.htm
http://www.geocities.com/physics_world/gr/geodesic_equation.htm

Then KW wanted me to show him I could derived field equaitions. Of course he
didn't specify what field so since I've already derived Maxwell's equations
from the Lagrangian density here it is
http://www.geocities.com/physics_world/em/lagrangian_density.htm

(or course I already gave that to him and he ignored it but se la vie!)

Best Regards

Pete
Koobee Wublee
2007-01-16 19:54:42 UTC
Post by JanPB
And I say that you are an ignoramus.
KW wanted me to derive the Christoffel symbols of the second kind. To derive
them means to derive the equation in which they appear and henceforth seemed
to require a seperate symbold and name. Here are several ways to get the
Christoffel symbols
http://www.geocities.com/physics_world/ma/chris_sym.htm
OK. Now, you have got your hands dirty by working in the trench. Now,
I go back to the geodesic equations.

dq^n/ds^2 = Gamma^n_ij dq^i/ds dq^j/ds

Gamma^n_ij = g^nk (@g_ik/@q^j + @g_jk/@q^i - @g_ij/@q^k) / 2

I can group these connection coefficients into the following.

Eta^n_ij = g^nk (@g_ik/@q^j - @g_ij/@q^k / 2)

The same geodesic equations can also be expressed as follows.

dq^n/ds^2 = Eta^n_ij dq^i/ds dq^j/ds

Therefore, the covariant derivative below is just as valid as the one
you had to derive the Riemann curvature tensor.

Df^n/Ds = df^n/ds + Eta^n_ij df^i/ds f^j

Using this covariant derivative, you would get a very different Riemann
curvature tensor. On top of that, there are also several ways to
arrange the connection coefficients just like the Christoffel symbols
of the second.

I have not seen anything else on what you have at your website. So
far, I have pointed out these three mathemagic tricks which have been
around for 100 years so far.

** Metric [g] is not a tensor.

** Christoffel symbols do not represent the only unique way of
grouping these coefficients.

** The covariant derivative is just as valid using other connection
coefficients.

** There are also ways to arrange the coefficients from the geodesic
deviation equations to get different Riemann curvature tensors. This
means the Riemann curvature tensor is not unique.

The whole crap about GR is a total nonsense, and we have not addressed
the field equations yet. That is another absurdity in the making.
pmb
2007-01-17 00:42:10 UTC
Post by Koobee Wublee
Post by JanPB
And I say that you are an ignoramus.
KW wanted me to derive the Christoffel symbols of the second kind. To derive
them means to derive the equation in which they appear and henceforth seemed
to require a seperate symbold and name. Here are several ways to get the
Christoffel symbols
http://www.geocities.com/physics_world/ma/chris_sym.htm
OK. Now, you have got your hands dirty by working in the trench. Now,
I go back to the geodesic equations.
dq^n/ds^2 = Gamma^n_ij dq^i/ds dq^j/ds
I can group these connection coefficients into the following.
The same geodesic equations can also be expressed as follows.
dq^n/ds^2 = Eta^n_ij dq^i/ds dq^j/ds
Therefore, the covariant derivative below is just as valid as the one
you had to derive the Riemann curvature tensor.
Df^n/Ds = df^n/ds + Eta^n_ij df^i/ds f^j
Using this covariant derivative, you would get a very different Riemann
curvature tensor. On top of that, there are also several ways to
arrange the connection coefficients just like the Christoffel symbols
of the second.
I have not seen anything else on what you have at your website. So
far, I have pointed out these three mathemagic tricks which have been
around for 100 years so far.
** Metric [g] is not a tensor.
** Christoffel symbols do not represent the only unique way of
grouping these coefficients.
** The covariant derivative is just as valid using other connection
coefficients.
** There are also ways to arrange the coefficients from the geodesic
deviation equations to get different Riemann curvature tensors. This
means the Riemann curvature tensor is not unique.
The whole crap about GR is a total nonsense, and we have not addressed
the field equations yet. That is another absurdity in the making.
As I said, I'd be most glad to continue this conversation and follow
your mathematical arguements if you were willing to stop the insult
game. I am unable to comprehend why you think insulting me will help

Best wishes

Pete

ps - I got my "hands dirty" years ago. Let's see, I started college in
the early 80's. Had to take and extra two years to catch up on the
basics of math and physics, then 4 more years as a physics/math
undergrad and then two years graduate school part time. Then there was
my work on my personal projects (work was computational/experimental
physics which was boring) which drove me deeply into relativity back in
1991. From 1991 to 1998 I worked on solving what turned out to be a
trivial problem. But it was trivial in the amount of work, not in the
concepts. Even the EM geniuses at MIT couldn't solve the problem.
Eventually I came across a text which gave me the math I needed and I
figured out the actual physics. Then at that time some crackpot who
called himself "naysayer" claimed that I was the only person in the
entire earth who still used the concept of relativistic mass (which was
nonsense since I choose not to make a choice - but to decide on the
basis of what makes better physics. I could still care less what people
call what thogh). However to fully understand mass I had to start
studying general relativity. Lucky for me, at that time I was working
at MIT and I got to know Alan Guth. A real nice guy. To me he was just
another professor there. I didn't know he was famous. Then I got to
know more relativistists. But to communicate with them in e-mail proved
easier if I made a web site to work out the math. That's all there was
too it. Then May 2000 came along and my entire world fell apart. But
that's a personal story for another day.
Koobee Wublee
2007-01-15 23:59:11 UTC
Post by bergeron
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
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?
bergeron
2007-01-16 07:06:44 UTC
Post by Koobee Wublee
Post by bergeron
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
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?
be because I am part of the relativity conspiracy devoted to
preventing you, the One and Only, True Self-Proclaimed Disciple
of Riemann from establishing your Holy Reign of Enlightenment
as the Pope of All science on Earth. Or perhaps it's just to
keep you busy while we raffle off your wife for a six-pack of
malt liquor.
Koobee Wublee
2007-01-16 07:47:25 UTC
Post by bergeron
be because I am part of the relativity conspiracy devoted to
preventing you, the One and Only, True Self-Proclaimed Disciple
of Riemann from establishing your Holy Reign of Enlightenment
as the Pope of All science on Earth.
Wait a minute. I never said I am a disciple of Riemann. Although
Riemann's work remains a quantum leap from Gauss's proposal of
possible curvature in space, it is still very primitive. Despite being
primitive, it is not wrong. I said I am the only one since Riemann (or
maybe Christoffel) who has understood the curvature in space or in
spacetime thoroughly and rightfully. This is no understatement.
Post by bergeron
Or perhaps it's just to
keep you busy while we raffle off your wife for a six-pack of
malt liquor.
Oh, should I be offended since you are attacking my family members?

Although she does not understand anything about the plagiarist Einstein
and the crap that is supposed to be credited to him, she is smart
enough to out-raffle you, and if it is a six-pack of good malt beer,
through wits I will take it under anyone's nose including hers. She
thinks she can limit my intake of beer to one a day.
JanPB
2007-01-16 09:59:27 UTC
Post by Koobee Wublee
Post by bergeron
be because I am part of the relativity conspiracy devoted to
preventing you, the One and Only, True Self-Proclaimed Disciple
of Riemann from establishing your Holy Reign of Enlightenment
as the Pope of All science on Earth.
Wait a minute. I never said I am a disciple of Riemann. Although
Riemann's work remains a quantum leap from Gauss's proposal of
possible curvature in space, it is still very primitive. Despite being
primitive, it is not wrong. I said I am the only one since Riemann (or
maybe Christoffel) who has understood the curvature in space or in
spacetime thoroughly and rightfully. This is no understatement.
And I say that you are an ignoramus.

--
Jan Bielawski
Koobee Wublee
2007-01-16 19:02:06 UTC
Post by JanPB
And I say that you are an ignoramus.
And more.

ds^2 = g_ij dq^i dq^j = f(dq, dq)

Clearly, all the literatures in GR, g_ij are the elements to the metric
[g] which is a 4-by-4 matrix. The voodoo mathemagic comes from the
switch-and-bait tactics. You show ds^2 as invariant through f() and
weakly refer to that as the metric. Then, in all mathematical
calculations, you switch out f() and use [g] as if an invariant
quantity. Your mathemagic trick is exposed.

You have been practicing voodoo mathemagic for 100 years. In doing so,
you build up that nonsense call the general theory of relativity. It
is time to tell the truth.
JanPB
2007-01-16 20:07:54 UTC
Post by Koobee Wublee
Post by JanPB
And I say that you are an ignoramus.
And more.
ds^2 = g_ij dq^i dq^j = f(dq, dq)
Clearly, all the literatures in GR, g_ij are the elements to the metric
[g] which is a 4-by-4 matrix.
No matter how many times you repeat it, it won't make it true. Metric
is not a matrix. It's a tensor.
Post by Koobee Wublee
The voodoo mathemagic comes from the
switch-and-bait tactics. You show ds^2 as invariant through f() and
weakly refer to that as the metric. Then, in all mathematical
calculations, you switch out f() and use [g] as if an invariant
quantity. Your mathemagic trick is exposed.
When a coordinate system is fixed, a specific matrix [g] of g_ij
describes the tensor fully, yes. Just like a vector in - let's say - 2D
plane is described fully by a pair of real numbers once a coordinate
system is fixed.
Post by Koobee Wublee
You have been practicing voodoo mathemagic for 100 years. In doing so,
you build up that nonsense call the general theory of relativity. It
is time to tell the truth.
But you are not telling the truth.

--
Jan Bielawski
Koobee Wublee
2007-01-16 22:08:40 UTC
Post by JanPB
Post by Koobee Wublee
ds^2 = g_ij dq^i dq^j = f(dq, dq)
Clearly, all the literatures in GR, g_ij are the elements to the metric
[g] which is a 4-by-4 matrix.
No matter how many times you repeat it, it won't make it true. Metric
is not a matrix. It's a tensor.
No matter how many times you played this mathemagic trick, you still
get caught. The metric is only valid to the choice of coordinate
system.
Post by JanPB
Post by Koobee Wublee
The voodoo mathemagic comes from the
switch-and-bait tactics. You show ds^2 as invariant through f() and
weakly refer to that as the metric. Then, in all mathematical
calculations, you switch out f() and use [g] as if an invariant
quantity. Your mathemagic trick is exposed.
When a coordinate system is fixed, a specific matrix [g] of g_ij
describes the tensor fully, yes. Just like a vector in - let's say - 2D
plane is described fully by a pair of real numbers once a coordinate
system is fixed.
In the following post, you did agree that the specified field equations
are only valid to the same choice of coordinate system. Thus, any
solution found must be only valid to this choice of coordinate system
and thus represent a different geometry each time. With the
Schwarzschild not unique, the field equations are totally useless and
nonsense.

Post by JanPB
Post by Koobee Wublee
You have been practicing voodoo mathemagic for 100 years. In doing so,
you build up that nonsense call the general theory of relativity. It
is time to tell the truth.
But you are not telling the truth.
I am not doing the telling. I am exposing your lies for almost 100
years.
pmb
2007-01-17 00:46:16 UTC
Post by JanPB
Post by Koobee Wublee
Post by JanPB
And I say that you are an ignoramus.
And more.
ds^2 = g_ij dq^i dq^j = f(dq, dq)
Clearly, all the literatures in GR, g_ij are the elements to the metric
[g] which is a 4-by-4 matrix.
No matter how many times you repeat it, it won't make it true. Metric
is not a matrix. It's a tensor.
Post by Koobee Wublee
The voodoo mathemagic comes from the
switch-and-bait tactics. You show ds^2 as invariant through f() and
weakly refer to that as the metric. Then, in all mathematical
calculations, you switch out f() and use [g] as if an invariant
quantity. Your mathemagic trick is exposed.
When a coordinate system is fixed, a specific matrix [g] of g_ij
describes the tensor fully, yes. Just like a vector in - let's say - 2D
plane is described fully by a pair of real numbers once a coordinate
system is fixed.
Post by Koobee Wublee
You have been practicing voodoo mathemagic for 100 years. In doing so,
you build up that nonsense call the general theory of relativity. It
is time to tell the truth.
But you are not telling the truth.
--
Jan Bielawski
I really don't think that you'll get anywhere like this. I think the
best way to explain it to him, if he's willing to listen, is to use a
vector as an example. A vector is a geometric object just as a tensor
is, in fact a vector is a tensor of rank 1. The only time you see
numbers like A = <1,5,3> is when the vector is projected onto a set of
basis vectors. The numbers are then known as the components of the
vector. We know the array of numbers is a matrix yet we also understand
that the matrix is not the geometric object that we're modeling, i.e.
the vector = directed line segment.

I explained this already. I believe he ignored it though.

Best wishes

Pete
pmb
2007-01-15 16:46:19 UTC
Post by Koobee Wublee
I see that the discussion of whether the metric qualifies as a tensor
or not has splintered into many ....
Actually two...
Post by Koobee Wublee
....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?
Post by Koobee Wublee
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.
Post by Koobee Wublee
...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.
Post by Koobee Wublee
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*.
Post by Koobee Wublee
** 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.
Post by Koobee Wublee
Of course, we all know that
ds^2 = g_ij dq^i dq^j = g'_mn dq'^m dq'^n
That we agree on.
Post by Koobee Wublee
Where
** dq' = Different coordinate system
dq is the differential of a coordinate.
Post by Koobee Wublee
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".
Post by Koobee Wublee
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.

Will the challenge be accepted?

Best wishes

Pete
Koobee Wublee
2007-01-15 22:08:55 UTC
Post by pmb
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.
Post by pmb
Post by Koobee Wublee
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.
Post by pmb
Post by Koobee Wublee
...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.
Post by pmb
Post by Koobee Wublee
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.
Post by pmb
Post by Koobee Wublee
** 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.
Post by pmb
Post by Koobee Wublee
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.
Post by pmb
Post by Koobee Wublee
Where
** dq' = Different coordinate system
dq is the differential of a coordinate.
I meant dq' is of a different coordinate system from dq.
Post by pmb
Post by Koobee Wublee
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

Post by pmb
Post by Koobee Wublee
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.
Post by pmb
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
Post by pmb
Consider you silly challenge replied.
Post by pmb
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.
Pmb
2007-01-15 22:34:57 UTC
Post by pmb
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
Koobee Wublee
2007-01-16 00:00:05 UTC
Post by Pmb
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.
Mike
2007-01-17 16:47:11 UTC
Post by pmb
Post by Koobee Wublee
I see that the discussion of whether the metric qualifies as a tensor
or not has splintered into many ....
Actually two...
Post by Koobee Wublee
....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?
Post by Koobee Wublee
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.
Post by Koobee Wublee
...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.
Post by Koobee Wublee
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*.
Post by Koobee Wublee
** 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.
Post by Koobee Wublee
Of course, we all know that
ds^2 = g_ij dq^i dq^j = g'_mn dq'^m dq'^n
That we agree on.
Post by Koobee Wublee
Where
** dq' = Different coordinate system
dq is the differential of a coordinate.
Post by Koobee Wublee
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".
Post by Koobee Wublee
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.
Correct, but if I give you a second rank tesnsor, then it is a matrix.
I do not hink that KW said all matrices are tensors. I think he said
some tensors are matrices which is correct.
Post by pmb
So I repeat my challenge to you - Post a reference to a math or GR
textbook which uses your definition of tensor.
"Objects that transform like zeroth-rank tensors are called scalars,
those that transform like first-rank tensors are called vectors, and
those that transform like second-rank tensors are called matrices. "

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

KW is somewhat right because the type of tesors he is talking about are
glorified matrices.

Mike
Post by pmb
Will the challenge be accepted?
Best wishes
Pete
Eric Gisse
2007-01-17 17:08:33 UTC
Post by Mike
Post by pmb
Post by Koobee Wublee
I see that the discussion of whether the metric qualifies as a tensor
or not has splintered into many ....
Actually two...
Post by Koobee Wublee
....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?
Post by Koobee Wublee
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.
Post by Koobee Wublee
...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.
Post by Koobee Wublee
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*.
Post by Koobee Wublee
** 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.
Post by Koobee Wublee
Of course, we all know that
ds^2 = g_ij dq^i dq^j = g'_mn dq'^m dq'^n
That we agree on.
Post by Koobee Wublee
Where
** dq' = Different coordinate system
dq is the differential of a coordinate.
Post by Koobee Wublee
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".
Post by Koobee Wublee
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.
Correct, but if I give you a second rank tesnsor, then it is a matrix.
I do not hink that KW said all matrices are tensors. I think he said
some tensors are matrices which is correct.
Post by pmb
So I repeat my challenge to you - Post a reference to a math or GR
textbook which uses your definition of tensor.
"Objects that transform like zeroth-rank tensors are called scalars,
those that transform like first-rank tensors are called vectors, and
those that transform like second-rank tensors are called matrices. "
http://mathworld.wolfram.com/Tensor.html
KW is somewhat right because the type of tesors he is talking about are
glorified matrices.
KW is incredibly wrong. Second rank tensors can be represented as
matrices but that does _not_ mean all the lessons from linear algebra
apply.
Post by Mike
Mike
Post by pmb
Will the challenge be accepted?
Best wishes
Pete
Pmb
2007-01-17 17:44:46 UTC
Post by Mike
Post by pmb
Post by Koobee Wublee
I see that the discussion of whether the metric qualifies as a tensor
or not has splintered into many ....
Actually two...
Post by Koobee Wublee
....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?
Post by Koobee Wublee
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.
Post by Koobee Wublee
...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.
Post by Koobee Wublee
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*.
Post by Koobee Wublee
** 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.
Post by Koobee Wublee
Of course, we all know that
ds^2 = g_ij dq^i dq^j = g'_mn dq'^m dq'^n
That we agree on.
Post by Koobee Wublee
Where
** dq' = Different coordinate system
dq is the differential of a coordinate.
Post by Koobee Wublee
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".
Post by Koobee Wublee
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.
Correct, but if I give you a second rank tesnsor, then it is a matrix.
I do not hink that KW said all matrices are tensors. I think he said
some tensors are matrices which is correct.
It can be represented by a matrix. It should never be said that it is a
matrix because one would expect the matrix and the tensor to be identical in
all concerns which they are not.
Post by Mike
Post by pmb
So I repeat my challenge to you - Post a reference to a math or GR
textbook which uses your definition of tensor.
"Objects that transform like zeroth-rank tensors are called scalars,
those that transform like first-rank tensors are called vectors, and
those that transform like second-rank tensors are called matrices. "
http://mathworld.wolfram.com/Tensor.html
First off that is not a math or GR textbook. Its a webpage which is not
perfect in its definitions. Its very good as a resource, don't get me wrong.
But the author slips up some times and this is one of those times.

That web page is very wrong. The author screwed up big time. If you notice
the list of references he has, i.e. the textbooks at the bottom, then you'll
see some very well known books on the subject and none that I recognize
gives that as a definition of "tensor". If you point that out to Eric
Wienstein he may fix it. He's good like that.

Pete
Koobee Wublee
2007-01-17 18:30:58 UTC
Post by Pmb
Post by Mike
Correct, but if I give you a second rank tesnsor, then it is a matrix.
I do not hink that KW said all matrices are tensors. I think he said
some tensors are matrices which is correct.
It can be represented by a matrix. It should never be said that it is a
matrix because one would expect the matrix and the tensor to be identical in
all concerns which they are not.
Mike is right. All the arrays GR deal with can be done with matrices.
This includes the Riemann curvature tensor after allowing me to pioneer
into the mathematical domain of a rank-4 matrix as a 4-by-4-by-4-by-4
matrix.

You can come out now and stop hiding a kindergarten kid under someone
else's skirt. After all, you are a middle aged adult now. So,
Daryl McCullough
2007-01-17 19:35:57 UTC
Koobee Wublee says...
Post by Koobee Wublee
Mike is right. All the arrays GR deal with can be done with matrices.
GR does not deal with arrays. It deals with *tensors*. Do you know what
a tensor is? Do you know why the metric tensor is a tensor, but the
Christoffel symbols are not?

--
Daryl McCullough
Ithaca, NY
Koobee Wublee
2007-01-18 05:54:19 UTC
Post by Daryl McCullough
Koobee Wublee says...
Post by Koobee Wublee
Mike is right. All the arrays GR deal with can be done with matrices.
GR does not deal with arrays.
Let me rephrase. GR deals with *matrices*.
Post by Daryl McCullough
It deals with *tensors*.
None of the mathematics proves so.
Post by Daryl McCullough
Do you know what a tensor is?
Yes, GR deals with matrices. <Shrug>
Post by Daryl McCullough
Do you know why the metric tensor is a tensor, but the
Christoffel symbols are not?
I really don't care because GR deals with matrices. However, if you
think you know, show me mathematically. Don't make any claims.
<shrug>
Daryl McCullough
2007-01-18 11:57:11 UTC
Koobee Wublee says...
Post by Koobee Wublee
Post by Daryl McCullough
GR does not deal with arrays.
Let me rephrase. GR deals with *matrices*.
Post by Daryl McCullough
It deals with *tensors*.
None of the mathematics proves so.
The point is that you have no understanding of the
mathematics of General Relativity, and at least
part of your confusion is based on not understanding
tensors and the distinction between tensors and matrices.
This distinction is *crucial* to understanding General
Relativity.

--
Daryl McCullough
Ithaca, NY
Dirk Van de moortel
2007-01-15 17:42:51 UTC
Post by Koobee Wublee
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/dirk/Physics/Fumbles/AbsurdClaim.html

Dirk Vdm
bergeron
2007-01-15 18:06:56 UTC
Post by Dirk Van de moortel
Post by Koobee Wublee
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.
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/AbsurdClaim.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.
Dirk Van de moortel
2007-01-15 18:16:19 UTC
Post by bergeron
Post by Dirk Van de moortel
Post by Koobee Wublee
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.
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/AbsurdClaim.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.

Dirk Vdm
Pmb
2007-01-15 18:47:33 UTC
Post by Dirk Van de moortel
Post by bergeron
Post by Dirk Van de moortel
Post by Koobee Wublee
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.
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/AbsurdClaim.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.
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
Koobee Wublee
2007-01-16 06:27:25 UTC
Post by Pmb
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.
There is another possibility. How about "the concept that you are
quoting from a book has been wrong for a long time, and here is why it
is wrong"?
Post by Pmb
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.
If I want text, I can look it up. Thank you very much. I don't need
a pastor to point out the exact passage.
Post by Pmb
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.
You are wrong on this one and many others.
Post by Pmb
Hey Dirk - I thought I was in your list of Fumbles?? Did I loose my proud
position of recognition there? LOL!
Keep on sucking. Maybe the sperm lover will remove you from his
asinine fumble list. <shrug>
Russell
2007-01-16 15:10:20 UTC
Post by Pmb
Post by Dirk Van de moortel
Post by bergeron
Post by Dirk Van de moortel
Post by Koobee Wublee
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.
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/AbsurdClaim.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.
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.
Hey, I think you left out a comma before "Dirk". In this case
it's rather crucial to your meaning, as (in the form you posted)
it looks like the subject of your paragraph, and the antecedent
of the word "his", is Dirk. But I'm pretty sure you mean Koobee.

[snip rest]
Pmb
2007-01-16 15:39:01 UTC
Post by Russell
Post by Pmb
Post by Dirk Van de moortel
Post by bergeron
Post by Dirk Van de moortel
Post by Koobee Wublee
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.
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/AbsurdClaim.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.
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.
Hey, I think you left out a comma before "Dirk". In this case
it's rather crucial to your meaning, as (in the form you posted)
it looks like the subject of your paragraph, and the antecedent
of the word "his", is Dirk. But I'm pretty sure you mean Koobee.
Oops! Yes. Thank you for pointing that out. Its a mistake I make too often.
Much appreciated.

Best wishes

Pete