On May 13, 7:20 pm, PD wrote:
On May 13, 4:46 pm, Koobee Wublee wrote:

On May 13, 12:21 am, PD wrote:

On May 13, 1:24 am, Koobee Wublee wrote:
The Schwarzschild metric only applies to the spherically symmetric
polar coordinate system and nothing else. So, you do not understand
the Schwarzschild metric, either.

No, that's incorrect. The Schwarzchild metric is a *shape*. It is most
conveniently described in one set of coordinates, but it can certainly
be described in more than one coordinate system.

The spacetime described by the Schwarzschild metric only applied to
the spherically symmetric polar coordinate.

Repeating the same erroneous statement does not inch it any further
toward being right.



You are welcome to
transform to another coordinate system with a different metric. The
segment of spacetime remains the same and thus invariant while the
coordinate is your choice as well as the metric associated with this
coordinate to describe the invariant segment of spacetime.

This concept is so basic, but you have a lot of trouble understanding
that. There is no need to go through the rest of your babbling.
shrug- Hide quoted text -

- Show quoted text -

On May 13, 2:48 pm, Koobee Wublee wrote:
On May 12, 11:51 pm, JanPB wrote:

On May 12, 11:24 pm, Koobee Wublee wrote:
The Schwarzschild metric only applies to the spherically symmetric
polar coordinate system and nothing else.

No, that's false. It's exactly like saying "the vector pointing to the
North Star applies only to the spherically symmetric polar coordinate
system and nothing else".

The spacetime described by the Schwarzschild metric only applied to
the spherically symmetric polar coordinate.

Not _spacetime_, the _coordinate representation of the metric_. There
is no such thing as "spacetime" that can be "applied" to a coordinate
system. The concept doesn't exist.

I notice that you didn't comment on the North Star vector example. You
always evade points that disprove your claims and then you post semi-
responses on a slightly altered topic. (A standard strategy around
these parts - it won't work with me as I've been posting to this NG
since its inception and have seen it all (by definition :-) ).

You are welcome to
transform to another coordinate system with a different metric.

Metric is like that vector pointing to the North Star - a coordinate-
independent entity. Here I'm standing, here is the North Star, here is
my finger pointing to it. This is a vector. No coordinate system
involved. [No answer, right?] Metric is the same way except it needs
more data to describe it than just magnitude and direction as in the
vector case.

The
segment of spacetime remains the same and thus invariant while the
coordinate is your choice as well as the metric associated with this
coordinate

....there is no such thing as "metric associated with this coordinate"
just like there is no "vector pointing to the North Star associated
with this coordinate".

to describe the invariant segment of spacetime. shrug
This concept is so basic, but you have a lot of trouble understanding
that. There is no point to go through the north star nonsense.

But it's the crux of the matter. For months now you have been avoiding
answering this question: why is vector coordinate independent and
metric suddenly _isn't_?

--
Jan Bielawski

On May 12, 11:54 pm, Koobee Wublee wrote:
On May 12, 11:32 pm, JanPB wrote:



On May 12, 10:55 pm, Koobee Wublee wrote:
ds^2 = G c^2 dt^2 / (1 + K^2 / r^2) – 4 r^2 (1 + K^2 / r^2) dr^2 / K^2
– (r^2 + K^2)^2 dO^2 / K^2

Its Ricci and Einstein tensors both vanish. In doing so, it is also
sharing the same coordinate system with the Schwarzschild metric.

Looking for a generic solution that is static and spherically
symmetric, we employ the spherically symmetric coordinate system that
describes a segment of spacetime in general of the following.

ds^2 = c^2 T(r) dt^2 – P(r) dr^2 – Q(r) dO^2

Where

** T(r), P(r), Q(r) = Functions of r only
** dO^2 = cos^2(Phi) dTheta^2 + dPhi^2

Yes. But the formula you presented:

ds^2 = G c^2 dt^2 / (1 + K^2 / r^2) – 4 r^2 (1 + K^2 / r^2) dr^2 / K^2
– (r^2 + K^2)^2 dO^2 / K^2

...is not a solution if you take r to be the same as in
Schwarzschild's formula.

Mathematically, what you are saying is just wrong. shrug

It's very easy to see who is wrong: I say the Ricci tensor of that is
nonzero. You are saying it is zero. So just calculate the thing and
see for yourself! What are you waiting for?

Then, the null Einstein tensor (not Ricci) tensor (in free space) is
represented by the following 3 differential equations. You are
supposed to get 4, but 2 of these are identical.

** 1 + dQ/dr dP/dr / (2 P^2) + (dQ/dr)^2 / (4 P Q) – d^2Q/dr^2 / P =
0

** 1 – dQ/dr dT/dr / (1 T P) – (dQ/dr)^2 / (4 P Q) = 0

** - d^2T/dr^2 / T + dT/dr dP/dr / (2 T P) + (dT/dr)^2 / (2 T^2) – dQ/
dr dT/dr / (2 Q T) + dP/dr dQ/dr / (2 P Q) + (dQ/dr)^2 / (2 Q^2) –
d^2Q/dr^2 / Q = 0

There are an infinite number of solutions found.

No. While your statement is true in the generic case, it turns out
that in those coordinates these equations become ordinary diff.
equations ("ordinary" is a technical term meaning "in one variable").
And it is well-known that ordinary differential equations have unique
solutions - it follows, oddly enough, from a fixed point theorem. It's
a classic result called Picard-Lindeloef theorem.

A few examples are
the following in the order of first discovered. These include the
inverse cubed law one at the beginning of the post.
** Schwarzschild’s original solution

ds^2 = G c^2 dt^2 (1 - K / R) – r^4 dr^2 / R^4 / (1 - K / R) – R^2
dO^2

Where

** R = (r^3 + K^3)^(1/3)

** Schwarzschild (Hilbert’s) solution

ds^2 = G c^2 (1 + K / r) dt^2 – dr^2 / (1 + K / r) – r^2 dO^2

** ?’s solution

ds^2 = G c^2 dt^2 / (1 + K / r) – dr^2 (1 + K / r) – (r + K)^2 dO^2

Same thing. If "r" in these formulas are presumed the same, then the
third one does not satisfy Ricci=0.

Well, all these are solutions to the field equations.

Not the third one. Calculate its Ricci curvature.

Closing your
eyes and yelling they are not does not make them not. shrug

I'm doing no such thing - I'm saying their Ricci curvature is nonzero
(an easy although a bit tedious calculation) which means they can't be
solutions.

It is
all in the mathematics. shrug

Exactly.

OTOH if the radial coordinates in these formulas are related by
substitutions, then they are the same solution, only written in
different bases.

You are full of crap. The coordinate system is the spherically
symmetric polar coordinate system and nothing else. shrug

OK, if you want it this way then none of the formulas you wrote except
Schwarzschild's own is the solution. Calculate their Ricci curvature
and see it's nonzero. Why aren't you doing it? You prefer to live in
your fantasy world?

Just like dx equals cos(theta)dr - r sin(theta)dtheta
- just written in different bases.

Wrong analogy.

Actually it's better than an analogy. The dx, dr, etc. above are
objects of exactly the same type as dr dt etc. in Riemannian metrics.

As a film critic of second-rated films, you are very
subjective, aren’t you?

I am not a film critic. I run (with a friend) a web site devoted to
the Russian filmmaker Andrei Tarkovsky who is about as second-rate as
Dostoyevsky. There are some other things in that department that I'm
involved with. Besides that I don't see how pointing out a result of
definite computation being nonzero can be seen as "subjective".

You need to resolve your own mathematical errors. I cannot help you
on that one. shrug

The bottom line is that the Newtonian solution is not unique under the
Einstein field equations, and you need to get over with that. shrug
It is also time to vacate from that fat castle in the air, your
majesty, the queer of England.

That's nice but have you noticed that your claim that the formula:

ds^2 = G c^2 dt^2 / (1 + K^2 / r^2) – 4 r^2 (1 + K^2 / r^2) dr^2 / K^2
– (r^2 + K^2)^2 dO^2 / K^2

...with Schwarzschild's "r" was a solution has just been disproved?

I don’t know what you mean by my own claim. I have claimed that all
these solutions are mathematically tested out to be the solutions to
the Einstein field equations under the spherically symmetric polar
coordinate system. shrug

I'm just saying the above is not a solution because if you assume "r"
is the same then calculating Ricci curvature of the above yields a
nonzero result.

It is better for you majesty, the queer of England, to retire into
that soon-to-demolished fat castle in the air, for it will not be
there long. You don’t have to be such scared sh*tless. At least, you
can still go back being a critic for all these second-rated films, no?

I'm not scared at all, where did you get this idea? I know this stuff
is as solid as 1+1=2.

--
Jan Bielawski

On May 13, 11:52*pm, Koobee Wublee wrote:
On May 13, 7:20 pm, PD wrote:

On May 13, 4:46 pm, Koobee Wublee wrote:

On May 13, 12:21 am, PD wrote:

On May 13, 1:24 am, Koobee Wublee wrote:
The Schwarzschild metric only applies to the spherically symmetric
polar coordinate system and nothing else. *So, you do not understand
the Schwarzschild metric, either.

No, that's incorrect. The Schwarzchild metric is a *shape*. It is most
conveniently described in one set of coordinates, but it can certainly
be described in more than one coordinate system.

The spacetime described by the Schwarzschild metric only applied to
the spherically symmetric polar coordinate.

Repeating the same erroneous statement does not inch it any further
toward being right.

*You are welcome to
transform to another coordinate system with a different metric. *The
segment of spacetime remains the same and thus invariant while the
coordinate is your choice as well as the metric associated with this
coordinate to describe the invariant segment of spacetime.

This concept is so basic, but you have a lot of trouble understanding
that. *There is no need to go through the rest of your babbling.
shrug- Hide quoted text -

- Show quoted text -

Not saying anything at all doesn't inch the incorrect statement any
more towards being right either.

Do you think you cannot write down the equation of a sphere in
anything other than spherical polar coordinates?

PD

On May 13, 5:46*pm, Koobee Wublee wrote:
On May 13, 12:21 am, PD wrote:

On May 13, 1:24 am, Koobee Wublee wrote:
The Schwarzschild metric only applies to the spherically symmetric
polar coordinate system and nothing else. *So, you do not understand
the Schwarzschild metric, either.

No, that's incorrect. The Schwarzchild metric is a *shape*. It is most
conveniently described in one set of coordinates, but it can certainly
be described in more than one coordinate system.

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.

I think you are pounding on a non-issue. The real issue is a more
important one: in the limit the Schwarzschild metric reduc es to
Minkowski spacetime, which is gravity free. Yet, in physical reality
there is gravity is such limit.

Thus, the solution is empirically falsified.

Mike



The
segment of spacetime remains the same and thus invariant while the
coordinate is your choice as well as the metric associated with this
coordinate to describe the invariant segment of spacetime.

This concept is so basic, but you have a lot of trouble understanding
that. *There is no need to go through the rest of your babbling.
shrug

On May 14, 6:10*am, Mike wrote:
On May 13, 5:46*pm, Koobee Wublee wrote:

On May 13, 12:21 am, PD wrote:

On May 13, 1:24 am, Koobee Wublee wrote:
The Schwarzschild metric only applies to the spherically symmetric
polar coordinate system and nothing else. *So, you do not understand
the Schwarzschild metric, either.

No, that's incorrect. The Schwarzchild metric is a *shape*. It is most
conveniently described in one set of coordinates, but it can certainly
be described in more than one coordinate system.

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.

I think you are pounding on a non-issue. The *real issue is a more
important one: in the limit the Schwarzschild metric reduc es to
Minkowski spacetime, which is gravity free. Yet, in physical reality
there is gravity is such limit.

They reduce to the same limit, Mike. Expand the metric in a power
series expansion of GM/r for GM / r 1.


Thus, the solution is empirically falsified.

Mike

The
segment of spacetime remains the same and thus invariant while the
coordinate is your choice as well as the metric associated with this
coordinate to describe the invariant segment of spacetime.

This concept is so basic, but you have a lot of trouble understanding
that. *There is no need to go through the rest of your babbling.
shrug

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

I think you are pounding on a non-issue. The real issue is a more
important one: in the limit the Schwarzschild metric reduc es to
Minkowski spacetime, which is gravity free. Yet, in physical reality
there is gravity is such limit.

Hmmm... You are very confused.

Thus, the solution is empirically falsified.

shrug

On May 14, 3:03 am, PD wrote:
On May 13, 11:52 pm, Koobee Wublee wrote:

Not saying anything at all doesn't inch the incorrect statement any
more towards being right either.

Go back to read my last post. You are very confused as usual.

Do you think you cannot write down the equation of a sphere in
anything other than spherical polar coordinates?

Yes, I can, but it does not apply to what we are discussing. shrug

On May 13, 11:51 pm, JanPB wrote:
On May 13, 2:48 pm, Koobee Wublee wrote:

The spacetime described by the Schwarzschild metric only applied to
the spherically symmetric polar coordinate.

Not _spacetime_, the _coordinate representation of the metric_. There
is no such thing as "spacetime" that can be "applied" to a coordinate
system. The concept doesn't exist.

You have grossly misunderstood what I said, and I cannot tell if it is
due to your incompetence or just being malicious. There is no point
to go on.

On May 14, 12:09 am, JanPB wrote:
On May 12, 11:54 pm, Koobee Wublee wrote:

Mathematically, what you are saying is just wrong. shrug

It's very easy to see who is wrong: I say the Ricci tensor of that is
nonzero. You are saying it is zero. So just calculate the thing and
see for yourself! What are you waiting for?

I have given you the complete Einstein tensor in the link below. The
solutions I gave all satisfy the null Einstein tensor and thus the
Ricci tensor. Your application of Cartan stuff is grossly
simplistic. There is no short cut in mathematical grunge. shrug

http://groups.google.com/group/sci.p...9098444d9ab5b7