On May 15, 7:04*pm, JanPB wrote:
On May 14, 10:08*pm, Koobee Wublee wrote:

On May 14, 7:10 am, Mike wrote:

On May 13, 5:46 pm, Koobee Wublee wrote:
The spacetime described by the Schwarzschild metric only applied to
the spherically symmetric polar coordinate. *You are welcome to
transform to another coordinate system with a different metric.

No, the Schwarzschild metric IS the metric. There is no such thing as
Schwarzschild metric in a coordinate system with a metric.

Mathematically, you are just wrong. *For example, describing flat
spacetime using the linear rectangular (Euclidean) and using the
spherically symmetric polar coordinate systems require you to supply
different metric for each choice of coordinate system. *shrug

No, it's the same metric in both cases, e.g. in the plane the
following are equal:

* * dx^2 + dy^2

and:

* * dr^2 + r^2 dtheta^2

...where x = r cos(theta) and y = r sin(theta) (polar coordinates).

The above are two different coordinate decompositions of the same
metric. To see that they are equal, just evaluate them on an arbitrary
vector and see you get the same value in both cases.

This is where KW gets the small chicken bone stuck in his throat,
completely unable to swallow the distinction between a vector or
tensor and its coordinate decompositions. To him, if you have a
different component representation, you're talking about a physically
different thing, and he pretends to sniff that it is "mathematically
so".


For example, consider vector v defined in Cartesian coordinates like
this:

* * v is situated at point (1,1),
* * v = (1,2)

Then, applying ds^2 = dx^2 + dy^2:

* * ds^2(v) = v.v = dx(v) * dx(v) + dy(v) * dy(v) = 1^2 + 2^2 = 5

Now calculate using the polar representation of ds^2:

* * v is situated at (sqrt(2), pi/4),
* * v = (3/sqrt(2), -1/2)

Then, applying ds^2 = dr^2 + r^2 dtheta^2:

* * ds^2(v) = v.v = dr(v) * dr(v) + (sqrt(2))^2 * dtheta(v) *
dtheta(v) =

* * * * * * = (3/sqrt(2))^2 + 2 * (-1/2)^2 =

* * * * * * = 9/2 + 2 * 1/4 =

* * * * * * = 9/2 + 1/2 = 10/2 = 5

Same result! Yout hink that's an accident? :-) You'd obtain the
corresponding same results with any vector. It must be so because both
expressions for ds^2 are equal.

Same thing with Schwarzschild, etc.

The term "metric" as used by everyone refers to that mapping that
takes tangent vectors (like v) to their squared lengths (like v.v).
That mapping is coordinate-INdependent. OTOH its _coordinate
representation_ of course depends on coordinates (by definition). But
dx^2+dy^2 and dr^2 + r^2 dtheta^2 are two different coordinate
representations of the same metric (the same mapping taking v to v.v,
or equivalently, a mapping taking vector pairs v,w to v.w - that's
what metric _is_).

--
Jan Bielawski

On May 15, 5:04 pm, JanPB wrote:
On May 14, 10:08 pm, Koobee Wublee wrote:

Mathematically, you are just wrong. For example, describing flat
spacetime using the linear rectangular (Euclidean) and using the
spherically symmetric polar coordinate systems require you to supply
different metric for each choice of coordinate system. shrug

No, it's the same metric in both cases, e.g. in the plane the
following are equal:

dx^2 + dy^2

and:

dr^2 + r^2 dtheta^2

...where x = r cos(theta) and y = r sin(theta) (polar coordinates).

The above are two different coordinate decompositions of the same
metric. To see that they are equal, just evaluate them on an arbitrary
vector and see you get the same value in both cases.

You are making the same mistake again. What you refer to is the
geometry itself not the metric. The equations above represent the
same geometry, yes. They are equivalent. However, the coordinates
are different, and the metrics are different. The metric cannot
adequately describe the geometry despite your voodoo conjectures of
dot products, and the coordinates itself cannot adequately describe
the geometry. It takes both well specified coordinate systems and the
metrics to fully describe the geometries. shrug

We cannot go on without you understand my point of view, and I have
understood yours and pointed the errors in your logic. If you are not
malicious as Eric Gisse is, you need to understand my point of view.
shrug

On May 15, 10:24*pm, Koobee Wublee wrote:
[snip]

How can you be so confident in your assertions that the math supports
you when nobody else agrees with you along with your inability to
actually calculate anything?

On May 16, 1:24*am, Koobee Wublee wrote:
On May 15, 5:04 pm, JanPB wrote:



On May 14, 10:08 pm, Koobee Wublee wrote:
Mathematically, you are just wrong. *For example, describing flat
spacetime using the linear rectangular (Euclidean) and using the
spherically symmetric polar coordinate systems require you to supply
different metric for each choice of coordinate system. *shrug

No, it's the same metric in both cases, e.g. in the plane the
following are equal:

* * dx^2 + dy^2

and:

* * dr^2 + r^2 dtheta^2

...where x = r cos(theta) and y = r sin(theta) (polar coordinates).

The above are two different coordinate decompositions of the same
metric. To see that they are equal, just evaluate them on an arbitrary
vector and see you get the same value in both cases.

You are making the same mistake again. *What you refer to is the
geometry itself not the metric.

Then you are suffering from a basic misalignment on terminology. The
metric *is* the geometry. That's the point.

*The equations above represent the
same geometry, yes. *They are equivalent. *However, the coordinates
are different, and the metrics are different. *The metric cannot
adequately describe the geometry despite your voodoo conjectures of
dot products, and the coordinates itself cannot adequately describe
the geometry. *It takes both well specified coordinate systems and the
metrics to fully describe the geometries. *shrug

We cannot go on without you understand my point of view, and I have
understood yours and pointed the errors in your logic. *If you are not
malicious as Eric Gisse is, you need to understand my point of view.
shrug

On May 16, 8:45*am, PD wrote:
On May 16, 1:24*am, Koobee Wublee wrote:





On May 15, 5:04 pm, JanPB wrote:

On May 14, 10:08 pm, Koobee Wublee wrote:
Mathematically, you are just wrong. *For example, describing flat
spacetime using the linear rectangular (Euclidean) and using the
spherically symmetric polar coordinate systems require you to supply
different metric for each choice of coordinate system. *shrug

No, it's the same metric in both cases, e.g. in the plane the
following are equal:

* * dx^2 + dy^2

and:

* * dr^2 + r^2 dtheta^2

...where x = r cos(theta) and y = r sin(theta) (polar coordinates)..

The above are two different coordinate decompositions of the same
metric. To see that they are equal, just evaluate them on an arbitrary
vector and see you get the same value in both cases.

You are making the same mistake again. *What you refer to is the
geometry itself not the metric.

Then you are suffering from a basic misalignment on terminology. The
metric *is* the geometry. That's the point.


You are wrong once more along with your wannabees friends.

You confuse metrics with metric spaces. A metric is just a function
that defines distance between elements of a set. A metric space is a
set WITH a metric defined.

A metric induces a topology on a set but it may not be sufficient to
generate a topology.

Listen you morons: KW is right, you have no freaken idea of what you
are talking about. My suspicion that you are just cranks has been
confirmed many times.

Hey stupidos, I can have a Manhattan metric in Euclidean space. But
given just the Manhattan metric, how would anyone know the geometry is
Euclidean?

Hey stupid, I tell you my metric is abs(x1-x2)+abs(y1-y2)

what tha fok do you know from that? Pther than how I measure
distance.

The metric, especially a strange one like the ones imagined in GR,
does not tell you much about the topology. OFTEN THE UNDERLINE
TOPOLOGY IS ASSUMED, AND THIS AMOUNTS TO SCIENTIFIC CON.

Mike






*The equations above represent the
same geometry, yes. *They are equivalent. *However, the coordinates
are different, and the metrics are different. *The metric cannot
adequately describe the geometry despite your voodoo conjectures of
dot products, and the coordinates itself cannot adequately describe
the geometry. *It takes both well specified coordinate systems and the
metrics to fully describe the geometries. *shrug

We cannot go on without you understand my point of view, and I have
understood yours and pointed the errors in your logic. *If you are not
malicious as Eric Gisse is, you need to understand my point of view.
shrug- Hide quoted text -

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- Show quoted text -

On May 16, 6:10*am, Mike wrote:

[snip]

Wow, the first person who actually agrees with ol' kooby is an idiot.
Didn't see that one coming.

On May 16, 10:40*am, Eric Gisse wrote:
On May 16, 6:10*am, Mike wrote:

[snip]

Wow, the first person who actually agrees with ol' kooby is an idiot.
Didn't see that one coming.

YOu stupid dropout. What do you know about metric spaces when you
think that a = GM.r^2 is an Euler ODE?

http://groups.google.gr/group/sci.ph...b?dmode=source

How do you have the nerve to discuss such issues when you javen't gone
past high school algebra?

NOw, send email to your friend DIrt to come and throw his Dirt. Cry
stupid to Dirt, cry.

"Dirt, Dirt please wake up, please, wake up, come and post, Mike is
again showing that ODE thing... Please Dirt, when I come to Belgium I
buy you a six pack... please Dirt. You can do to me anything you
want...."

Moron....

Mike

On May 16, 9:10*am, Mike wrote:
On May 16, 8:45*am, PD wrote:



On May 16, 1:24*am, Koobee Wublee wrote:

On May 15, 5:04 pm, JanPB wrote:

On May 14, 10:08 pm, Koobee Wublee wrote:
Mathematically, you are just wrong. *For example, describing flat
spacetime using the linear rectangular (Euclidean) and using the
spherically symmetric polar coordinate systems require you to supply
different metric for each choice of coordinate system. *shrug

No, it's the same metric in both cases, e.g. in the plane the
following are equal:

* * dx^2 + dy^2

and:

* * dr^2 + r^2 dtheta^2

...where x = r cos(theta) and y = r sin(theta) (polar coordinates).

The above are two different coordinate decompositions of the same
metric. To see that they are equal, just evaluate them on an arbitrary
vector and see you get the same value in both cases.

You are making the same mistake again. *What you refer to is the
geometry itself not the metric.

Then you are suffering from a basic misalignment on terminology. The
metric *is* the geometry. That's the point.

You are wrong once more along with your wannabees friends.

You confuse metrics with metric spaces. A metric is just a function
that defines distance between elements of a set. A metric space is a
set WITH a metric defined.

A metric induces a topology on a set but it may not be sufficient to
generate a topology.

But the topology doesn't tell you (much of) anything about the *shape*
of the space as is relevant for physics, where the relevant feeds from
shape pertain to *interval* and, for a particular observer, perhaps
space and time coordinates, and these have everything to do with the
metric. Moreover, the notion of a geodesic, which is a path of
extremal measure between two points in the manifold, has everything to
do with the metric. Likewise, acceleration due to non-gravitational
forces is defined in terms of the metric.


Listen you morons: KW is right, you have no freaken idea of what you
are talking about. My suspicion that you are just cranks has been
confirmed many times.

Hey stupidos, I can have a Manhattan metric in Euclidean space. But
given just the Manhattan metric, how would anyone know the geometry is
Euclidean?

Because it is a constant metric with a zero curvature. That's what
flat means.


Hey stupid, I tell you my metric is abs(x1-x2)+abs(y1-y2)

what tha fok do you know from that? Pther than how I measure
distance.

The metric, especially a strange one like the ones imagined in GR,
does not tell you much about the topology. OFTEN THE UNDERLINE
TOPOLOGY IS ASSUMED, AND THIS AMOUNTS TO SCIENTIFIC CON.

Mike



*The equations above represent the
same geometry, yes. *They are equivalent. *However, the coordinates
are different, and the metrics are different. *The metric cannot
adequately describe the geometry despite your voodoo conjectures of
dot products, and the coordinates itself cannot adequately describe
the geometry. *It takes both well specified coordinate systems and the
metrics to fully describe the geometries. *shrug

We cannot go on without you understand my point of view, and I have
understood yours and pointed the errors in your logic. *If you are not
malicious as Eric Gisse is, you need to understand my point of view.
shrug- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -

On May 16, 10:53*am, PD wrote:
On May 16, 9:10*am, Mike wrote:





On May 16, 8:45*am, PD wrote:

On May 16, 1:24*am, Koobee Wublee wrote:

On May 15, 5:04 pm, JanPB wrote:

On May 14, 10:08 pm, Koobee Wublee wrote:
Mathematically, you are just wrong. *For example, describing flat
spacetime using the linear rectangular (Euclidean) and using the
spherically symmetric polar coordinate systems require you to supply
different metric for each choice of coordinate system. *shrug

No, it's the same metric in both cases, e.g. in the plane the
following are equal:

* * dx^2 + dy^2

and:

* * dr^2 + r^2 dtheta^2

...where x = r cos(theta) and y = r sin(theta) (polar coordinates).

The above are two different coordinate decompositions of the same
metric. To see that they are equal, just evaluate them on an arbitrary
vector and see you get the same value in both cases.

You are making the same mistake again. *What you refer to is the
geometry itself not the metric.

Then you are suffering from a basic misalignment on terminology. The
metric *is* the geometry. That's the point.

You are wrong once more along with your wannabees friends.

You confuse metrics with metric spaces. A metric is just a function
that defines distance between elements of a set. A metric space is a
set WITH a metric defined.

A metric induces a topology on a set but it may not be sufficient to
generate a topology.

But the topology doesn't tell you (much of) anything about the *shape*
of the space as is relevant for physics, where the relevant feeds from
shape pertain to *interval* and, for a particular observer, perhaps
space and time coordinates, and these have everything to do with the
metric. Moreover, the notion of a geodesic, which is a path of
extremal measure between two points in the manifold, has everything to
do with the metric. Likewise, acceleration due to non-gravitational
forces is defined in terms of the metric.



Listen you morons: KW is right, you have no freaken idea of what you
are talking about. My suspicion that you are just cranks has been
confirmed many times.

Hey stupidos, I can have a Manhattan metric in Euclidean space. But
given just the Manhattan metric, how would anyone know the geometry is
Euclidean?

Because it is a constant metric with a zero curvature. That's what
flat means.

N, a metric by itself tells you nothing about the UNDERLINE geometry,
what KW I think calls a coordinate system.

I can use that metric in any geometry, flat or non-flat.

A metric space is a set with a metric defined.

Mike









Hey stupid, I tell you my metric is abs(x1-x2)+abs(y1-y2)

what tha fok do you know from that? Pther than how I measure
distance.

The metric, especially a strange one like the ones imagined in GR,
does not tell you much about the topology. OFTEN THE UNDERLINE
TOPOLOGY IS ASSUMED, AND THIS AMOUNTS TO SCIENTIFIC CON.

Mike

*The equations above represent the
same geometry, yes. *They are equivalent. *However, the coordinates
are different, and the metrics are different. *The metric cannot
adequately describe the geometry despite your voodoo conjectures of
dot products, and the coordinates itself cannot adequately describe
the geometry. *It takes both well specified coordinate systems and the
metrics to fully describe the geometries. *shrug

We cannot go on without you understand my point of view, and I have
understood yours and pointed the errors in your logic. *If you are not
malicious as Eric Gisse is, you need to understand my point of view.
shrug- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -

On May 16, 6:51*am, Mike wrote:
On May 16, 10:40*am, Eric Gisse wrote:

On May 16, 6:10*am, Mike wrote:

[snip]

Wow, the first person who actually agrees with ol' kooby is an idiot.
Didn't see that one coming.

YOu stupid dropout. What do you know about metric spaces when you
think that a = GM.r^2 is an Euler ODE?

http://groups.google.gr/group/sci.ph...g/01b8793b2b1f...

How do you have the nerve to discuss such issues when you javen't gone
past high school algebra?

NOw, send email to your friend DIrt to come and throw his Dirt. Cry
stupid to Dirt, cry.

"Dirt, Dirt please wake up, please, wake up, come and post, Mike is
again showing that ODE thing... Please Dirt, when I come to Belgium I
buy you a six pack... please Dirt. You can do to me anything you
want...."

Moron....

Mike

I wonder what's wrong with your head that makes you think writing
stuff like this is a good idea.