On May 16, 11:26*am, Eric Gisse wrote:
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.- Hide quoted text -

It's a good idea when you insult people, but like all psychos think,
it's not a good idea when others insult them.

No get off my case otherwise I will make a bot to reply each and every
post you make with the ODE post you made.

Do you know what a bot is? Something liuke you bozzo. Spending 24/7 in
usenet.

Mike







- Show quoted text -

On May 16, 7:58*am, Mike wrote:
On May 16, 11:26*am, Eric Gisse wrote:



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.- Hide quoted text -

It's a good idea when you insult people, but like all psychos think,
it's not a good idea when others insult them.

No get off my case otherwise I will make a bot to reply each and every
post you make with the ODE post you made.

If you want me 'off your case', try shutting the **** up for a change.
Nothing gets me in snark mode like self righteous idiots.


Do you know what a bot is? Something liuke you bozzo. Spending 24/7 in
usenet.

Mike



- Show quoted text -

On May 16, 4:38*pm, Eric Gisse wrote:
On May 16, 7:58*am, Mike wrote:





On May 16, 11:26*am, Eric Gisse wrote:

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.- Hide quoted text -

It's a good idea when you insult people, but like all psychos think,
it's not a good idea when others insult them.

No get off my case otherwise I will make a bot to reply each and every
post you make with the ODE post you made.

If you want me 'off your case', try shutting the **** up for a change.
Nothing gets me in snark mode like self righteous idiots.


Yes, I know a self righteous idiot. Is an stupid moron who thinhs a =
GM/r^2 is an Euler ODE:

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

He also thinks that he can solve it in no time:

"Your problem bores me - walking to the toilet and peeing takes
longer,
and required the same amount of mental horsepower."

Youy must be peeing for a long time moron.

http://www.albinoblacksheep.com/flash/youare

Mike











Do you know what a bot is? Something liuke you bozzo. Spending 24/7 in
usenet.

Mike

- Show quoted text -- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -

On May 16, 1:30*pm, Mike wrote:
[snip]

What a boring little person you are.

On May 16, 5:45 am, PD wrote:
On May 16, 1:24 am, Koobee Wublee wrote:

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.

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.

** 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?

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, 8:48*pm, Koobee Wublee wrote:
[snip]

Arrogant and stupid, what a wonderful combination to watch.

On May 16, 9:59 pm, Eric Gisse wrote:

Arrogant and stupid, what a wonderful combination to watch.

Did you stand on top of the piles of your books and make that claim of
yourself? Hey, whacko behavior like that goes with why you remain a
multi-year super-senior. shrug

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)

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.


I think it is obvious that you manipulate words in a way to suit you.
You distort other peoples arguments and you present trivial examples
that do not correspond to the original problem posed.

You, PD, Roberts, Carlipp and other not worth mentioning all exhibit
the same exact pattern of behavior.

I wonder if behind all these nicknames there is just one person, or
several persons instructred by the same person on how to act.

Remember, a 10 year old understands you are wrong. How long are you
going to play this game?

Mike






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.

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 18, 4:22 am, Mike wrote:
On May 15, 8:04 pm, JanPB wrote:

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.

Yes, a metric without a set of coordinate system cannot possibly
describe the geometry. Thank you. This is indeed something a 10-year
old kid learning geometry would have no problem understanding this.
These matheMagicians practice the dark side of mathematics.

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)

Once again, you are correct. Thanks for pointing this one out. It is
his priori that is confusing himself.

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.

I am jumping for joy. Someone finally understands what I have been
saying in the past two years.

I think it is obvious that you manipulate words in a way to suit you.
You distort other peoples arguments and you present trivial examples
that do not correspond to the original problem posed.

What do you expect someone who worships the nonsense of GR and SR.
shrug

Hilbert, Klein, and the rest all knew the field equations yield an
infinite number of solutions. Being very smart, they just walked away
from it. That left Einstein the nitwit, the plagiarist, and the liar
to carry the banner to accept all credit that is not anything
representing his work.

You, PD, Roberts, Carlipp and other not worth mentioning all exhibit
the same exact pattern of behavior.

From the posts I have seen among all these professors. The following
are the profiles I have compiled of them.

** Professor Carlip is the type that he’d rather cease to exist
before accepting what he and others have been taught was total garbage
right from the very start. Don’t expect him to offer any comments.

** Professor Roberts on the other hand will listen to arguments. He
is beginning to show signs of defection from the camp of Einstein
Dingleberries. However, he must be very careful doing that because of
his new job. I have no doubt he will publish something posthumously
to clear up his position just like Copernicus or go public in the last
moments of his productive life just like Constantine accepted
Christianity at his deathbed.

** Professor Drape is very mentally challenged. I don’t think he
knows what he is teaching is lying.

I wonder if behind all these nicknames there is just one person, or
several persons instructred by the same person on how to act.

Remember, a 10 year old understands you are wrong. How long are you
going to play this game?

Mr. Bielawski will retreat back in his fat castle in the air until the
sheriff to come to kick him out. shrug

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?

[snip whining]