Koobee Wublee says...

Daryl McCullough wrote:
GR does not deal with arrays.

Let me rephrase. GR deals with *matrices*.

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

Koobee Wublee says...

The definition is absurd. ds^2 is an invariant scalar. The metric is
a matrix.

You are extremely confused about this subject. ds^2 is a *function*.
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.

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".

These are matrices, are they not?

Yes, but they are *not* metrics. That's the point---a metric
is not a matrix.

[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.

Let me rephrase what I said. The Minkowski metric does not necessarily
imply spacetime is flat.

Yes, it certainly does.

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.

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?

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

JanPB says...

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

Daryl McCullough wrote:
Koobee Wublee says...

The definition is absurd. ds^2 is an invariant scalar. The metric is
a matrix.

You are extremely confused about this subject. ds^2 is a *function*.
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.


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.


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





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".

These are matrices, are they not?

Yes, but they are *not* metrics. That's the point---a metric
is not a matrix.

[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.

Let me rephrase what I said. The Minkowski metric does not necessarily
imply spacetime is flat.

Yes, it certainly does.

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.

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?

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 wrote:
Daryl McCullough wrote:
Koobee Wublee says...

The definition is absurd. ds^2 is an invariant scalar. The metric is
a matrix.

You are extremely confused about this subject. ds^2 is a *function*.
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.


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.


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.


Try to concentrate on his points. Men do that.

Mike





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".

These are matrices, are they not?

Yes, but they are *not* metrics. That's the point---a metric
is not a matrix.

[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.

Let me rephrase what I said. The Minkowski metric does not necessarily
imply spacetime is flat.

Yes, it certainly does.

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.

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?

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

Daryl McCullough wrote:
Koobee Wublee says...

The definition is absurd. ds^2 is an invariant scalar. The metric is
a matrix.

You are extremely confused about this subject. ds^2 is a *function*.
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.

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".

These are matrices, are they not?

Yes, but they are *not* metrics. That's the point---a metric
is not a matrix.

[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.

Let me rephrase what I said. The Minkowski metric does not necessarily
imply spacetime is flat.

Yes, it certainly does.

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.

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?

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 *******ized 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









--
Daryl McCullough
Ithaca, NY

"Mike" wrote in message
oups.com...

Daryl McCullough wrote:
Koobee Wublee says...

The definition is absurd. ds^2 is an invariant scalar. The metric is
a matrix.

You are extremely confused about this subject. ds^2 is a *function*.

It is unclear if you understand that "function" and "operator" often mean
the same thing!

Look at the equation for ds^2 in, say, good-old 2D Euclidean space
in Cartesian coordinates:

That statement lacks precision. A space is neither Euclidean nor MMinkowski
etc. until a metric is defined on that 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

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

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!

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

Pmb says...

Look at the equation for ds^2 in, say, good-old 2D Euclidean space
in Cartesian coordinates:

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

"Daryl McCullough" wrote in message
...
Pmb says...

Look at the equation for ds^2 in, say, good-old 2D Euclidean space
in Cartesian coordinates:

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

Daryl McCullough wrote:
JanPB says...

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