On May 19, 5:10*am, Mike wrote:
On May 18, 9:40*pm, Eric Gisse wrote:



On May 18, 3:22*am, Mike wrote:

On May 15, 8:04*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).

What KW is saying, which is obvious to a 10 years old buy NOT TO YOU,
it's that the metric, without the corresponding set, does not specify
a geometry.

YOU ASSUME a priori --- somethign that he does not --- that if someone
tells you the metric is:

dr^2 + r^2 dtheta^2

then you know that:

x = r cos(theta) and y = r sin(theta)

Why don't you show us how you would transform the metric in Cartesian
coordinates to polar coordinates. I mean do the actual calculation.

Although this is a trivial example you presented and in most cases
someone will assume that, he is saying that in general, without the
coordinate transformation to (x,y), given the above metric, you cannot
deduce it is equivalent to dx^2 + dy^2.

Yes, you can. He did it explicitly - calculate the length of a vector.
Scalar quantities are independent of coordinates.

You can also brute force it. Can you guess how one might do that?

When you understand you are a troll, when you understand you ate
nothing but an incompetent individual who fails common sense tests,
you may become a better person.

I know I'm a troll - I troll idiots like you. Usually its' with
insults combined a liberal dose of pointing out stupidity. Doesn't
make me incompetent or stupid, just an asshole.


You stupid troll, the calculated the length of the vector after
assuming the transformation. Given the metric dr^2 + r^2 dtheta^2
without the coordinate transformation it is impossible to prove it is
equivalent to dx^2 + dy^2.

Want to bet? Both manifolds have the exact same curvature - zero.
Establishing that is enough to ensure that they are the same.


In other words, you stupid totally insane troll, without knowing the
geometry, it is impossible to deduce metric equivalence.

Unless, of course, you calculate some coordinate-independent
quantities with the two supposedly different metrics and find that
there is equivalence every time.


Now, get a life.

Mike



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On May 20, 1:53*pm, Eric Gisse wrote:

I know I'm a troll - I troll idiots like you. Usually its' with
insults combined a liberal dose of pointing out stupidity. Doesn't
make me incompetent or stupid, just an asshole.

When you do this you become more and more like them, thus slowly it's
next to impossible to tell the difference.

Now what was first your post or theirs?

Raghar wrote on Tue, 20 May 2008 06:11:08 -0700:

On May 20, 1:53Â*pm, Eric Gisse wrote:

I know I'm a troll - I troll idiots like you. Usually its' with insults
combined a liberal dose of pointing out stupidity. Doesn't make me
incompetent or stupid, just an asshole.

When you do this you become more and more like them, thus slowly it's
next to impossible to tell the difference.

One main difference is that one poster does not know dimensional analysis

http://groups.google.com/group/sci.p...b49966853da7e4

On May 20, 7:53*am, Eric Gisse wrote:
On May 19, 5:10*am, Mike wrote:





On May 18, 9:40*pm, Eric Gisse wrote:

On May 18, 3:22*am, Mike wrote:

On May 15, 8:04*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).

What KW is saying, which is obvious to a 10 years old buy NOT TO YOU,
it's that the metric, without the corresponding set, does not specify
a geometry.

YOU ASSUME a priori --- somethign that he does not --- that if someone
tells you the metric is:

dr^2 + r^2 dtheta^2

then you know that:

x = r cos(theta) and y = r sin(theta)

Why don't you show us how you would transform the metric in Cartesian
coordinates to polar coordinates. I mean do the actual calculation.

Although this is a trivial example you presented and in most cases
someone will assume that, he is saying that in general, without the
coordinate transformation to (x,y), given the above metric, you cannot
deduce it is equivalent to dx^2 + dy^2.

Yes, you can. He did it explicitly - calculate the length of a vector.
Scalar quantities are independent of coordinates.

You can also brute force it. Can you guess how one might do that?

When you understand you are a troll, when you understand you ate
nothing but an incompetent individual who fails common sense tests,
you may become a better person.

I know I'm a troll - I troll idiots like you. Usually its' with
insults combined a liberal dose of pointing out stupidity. Doesn't
make me incompetent or stupid, just an asshole.



You stupid troll, the calculated the length of the vector after
assuming the transformation. Given the metric dr^2 + r^2 dtheta^2
without the coordinate transformation it is impossible to prove it is
equivalent to dx^2 + dy^2.

Want to bet? Both manifolds have the exact same curvature - zero.
Establishing that is enough to ensure that they are the same.

Every day you prove that you are an even bigger troll than anyone
thought.


In other words, you stupid totally insane troll, without knowing the
geometry, it is impossible to deduce metric equivalence.

Unless, of course, you calculate some coordinate-independent
quantities with the two supposedly different metrics and find that
there is equivalence every time.

You do not know whay you are talking about. You have no clue.

Dig hard troll....


Mike









Now, get a life.

Mike

[snip whining]- Hide quoted text -

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On May 20, 9:11*am, Raghar wrote:
On May 20, 1:53*pm, Eric Gisse wrote:

I know I'm a troll - I troll idiots like you. Usually its' with
insults combined a liberal dose of pointing out stupidity. Doesn't
make me incompetent or stupid, just an asshole.

When you do this you become more and more like them, thus slowly it's
next to impossible to tell the difference.

Now what was first your post or theirs?

Here is another usenet troll. This is his suggestion:

"Fight. What else should they do? They would surely will not throw
away money for salaries of these persons, just to have a group for
danger assessment with respect to orbital space. "

http://groups.google.com/group/sci.s...9?dmode=source

Another one:

"To quote General Schwarzkopf "The Israelis are a bunch of third rate
swaggerers who wouldn't last five minutes in a European war""
Raghar: Nice quote, could you post a reference for it? "

http://groups.google.com/group/uk.po...9?dmode=source


Raghar, stick your fingure into vazeline and then plug the hole in
your brain.

Mike

On May 19, 9:23 am, PD wrote:
On May 16, 11:48 pm, Koobee Wublee wrote:

Consider the Schwarzschild spacetime as described below.

ds^2 = c^2 (1 – 2 U) dt^2 – dr^ / (1 – 2 U) – r^2 dO^2

Where

** U = G M / c^2 / r

** The geometry is ds^2.

Uh, no. That is the measure. Please proceed to real analysis, do not
pass go, do not collect $20.

Have I said something wrong? Does ds^2 not represent the geometry?

I stand by what I said. You are just not fit to teach anything.
shrug

** The coordinate is (t, r, O) or (t, r, longitude, latitude).

** The metric is

[c^2 (1 – 2 U), 0, 0, 0]
[0, 1 / (1 – 2 U), 0, 0]
[0, 0, r^2 cos^2(latitude), 0]
[0, 0, 0, r^2]

According to the mathematics of the field equations and thus GR, the
metric is not the geometry, and this is very obvious as explained
above.

If you cannot understand this, you are not fit to teach. If you
understand this, your claim is an utter lie. That would make it
(LYING IS TEACHING) on your part. shrug

Either way, you are not fit to teach. How can I be so point blank?

You are under the influence of the following toxic Orwellian
education. shrug

** MYSTICISM IS WISDOM
** PLAGIARISM IS CREATIVITY
** CONJECTURE IS REALITY
** FAITH IS THEORY
** LYING IS TEACHING
** BELIEVING IS LEARNING

On May 21, 3:20*am, Koobee Wublee wrote:
On May 19, 9:23 am, PD wrote:

On May 16, 11:48 pm, Koobee Wublee wrote:
Consider the Schwarzschild spacetime as described below.

ds^2 = c^2 (1 – 2 U) dt^2 – dr^ / (1 – 2 U) – r^2 dO^2

Where

** *U = G M / c^2 / r

** *The geometry is ds^2.

Uh, no. That is the measure. Please proceed to real analysis, do not
pass go, do not collect $20.

Have I said something wrong? *Does ds^2 not represent the geometry?

You now have proof that you are dealing with cons and troll.

When suits them , ds^2 IS the geometry.

When suits them, ds^2 IS the measure.

They have made little autistic troll Erica Gissa so confused, it don't
know what to do...

Mike






I stand by what I said. *You are just not fit to teach anything.
shrug





** *The coordinate is (t, r, O) or (t, r, longitude, latitude).

** *The metric is

[c^2 (1 – 2 U), 0, 0, 0]
[0, 1 / (1 – 2 U), 0, 0]
[0, 0, r^2 cos^2(latitude), 0]
[0, 0, 0, r^2]

According to the mathematics of the field equations and thus GR, the
metric is not the geometry, and this is very obvious as explained
above.

If you cannot understand this, you are not fit to teach. *If you
understand this, your claim is an utter lie. *That would make it
(LYING IS TEACHING) on your part. *shrug

Either way, you are not fit to teach. *How can I be so point blank?

You are under the influence of the following toxic Orwellian
education. *shrug

** *MYSTICISM IS WISDOM
** *PLAGIARISM IS CREATIVITY
** *CONJECTURE IS REALITY
** *FAITH IS THEORY
** *LYING IS TEACHING
** *BELIEVING IS LEARNING- Hide quoted text -

- Show quoted text -

On May 21, 2:20*am, Koobee Wublee wrote:
On May 19, 9:23 am, PD wrote:

On May 16, 11:48 pm, Koobee Wublee wrote:
Consider the Schwarzschild spacetime as described below.

ds^2 = c^2 (1 – 2 U) dt^2 – dr^ / (1 – 2 U) – r^2 dO^2

Where

** *U = G M / c^2 / r

** *The geometry is ds^2.

Uh, no. That is the measure. Please proceed to real analysis, do not
pass go, do not collect $20.

Have I said something wrong? *Does ds^2 not represent the geometry?

It does not. It is the *measure*. The metric tensor (regardless of its
coordinate representation) is the geometry.


I stand by what I said.

I know you do. You've done that for years, without learning much of
anything.

*You are just not fit to teach anything.

It's not my job to teach you on usenet or convince you of anything. If
you want to be taught, engage in a teaching venue.

shrug



** *The coordinate is (t, r, O) or (t, r, longitude, latitude).

** *The metric is

[c^2 (1 – 2 U), 0, 0, 0]
[0, 1 / (1 – 2 U), 0, 0]
[0, 0, r^2 cos^2(latitude), 0]
[0, 0, 0, r^2]

According to the mathematics of the field equations and thus GR, the
metric is not the geometry, and this is very obvious as explained
above.

If you cannot understand this, you are not fit to teach. *If you
understand this, your claim is an utter lie. *That would make it
(LYING IS TEACHING) on your part. *shrug

Either way, you are not fit to teach. *How can I be so point blank?

You are under the influence of the following toxic Orwellian
education. *shrug

** *MYSTICISM IS WISDOM
** *PLAGIARISM IS CREATIVITY
** *CONJECTURE IS REALITY
** *FAITH IS THEORY
** *LYING IS TEACHING
** *BELIEVING IS LEARNING