On May 12, 10:55 pm, Koobee Wublee wrote:
On May 11, 11:43 pm, JanPB wrote:



On May 11, 9:53 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.

I meant I computed only R_theta,theta and didn't bother with the rest.

Remember dO^2 consists of r, theta, and
phi as the Ricci and the Einstein tensors are only valid in four
dimensions. Since you have not computed the Ricci tensor, you have no
right to call the above solution not a null result of the Ricci and
the Einstein tensors. shrug There is no point to continue.

If Ricci tensor were zero, then R_theta,theta would be zero. It isn't,
hence Ricci is not zero.

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.

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.

OTOH if the radial coordinates in these formulas are related by
substitutions, then they are the same solution, only written in
different bases. Just like dx equals cos(theta)dr - r sin(theta)dtheta
- just written in different bases.

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?

--
Jan Bielawski

On May 12, 11:05 pm, Mike wrote:
On May 13, 1:55 am, Koobee Wublee wrote:
[...]
There are an infinite number of solutions found. 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.

Of course there are infinite solutions. But they will not accept this.
What they do is select the solution they want to get as a result. Then
work the math to get that solutions. Then claim they got the solution.

Nice, eh?

Nice but it's baloney. Good for a Hollywood movie. It's much easier to
believe in silly conspiracies than to do honest work.

--
Jan Bielawski

On May 12, 11:24 pm, Koobee Wublee wrote:
On May 12, 11:16 pm, Eric Gisse wrote:

On May 12, 9:55 pm, Koobee Wublee wrote:
You continue to say "infinite number" like that's supposed to have any
meaning.

So, you don’t understand what infinity represents.

There is an infinite number of coordinate systems

Yes. shrug

that the Schwarzschild solution can be projected on,

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 same North-Star-pointing vector has different component triplets
in different coordinates. Do you agree? Tensors of rank higher than 1
are exactly the same way.

So, you do not understand
the Schwarzschild metric, either.

like any other metric.

Other metric as a solution to the Einstein field equations also only
applies to the spherically symmetric polar coordinate system and
nothing else,

That's false. Would you say that a vector field (say, velocity field
of a fluid) "only applies to" a particular system?

--
Jan Bielawski

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

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. Closing your
eyes and yelling they are not does not make them not. shrug It is
all in the mathematics. shrug

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

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

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

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

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?

On May 13, 1:24*am, Koobee Wublee wrote:
On May 12, 11:16 pm, Eric Gisse wrote:

On May 12, 9:55 pm, Koobee Wublee wrote:
You continue to say "infinite number" like that's supposed to have any
meaning.

So, you don’t understand what infinity represents.

There is an infinite number of coordinate systems

Yes. *shrug

that the Schwarzschild solution can be projected on,

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.

Likewise, the hydrogen atom wavefunctions are most conveniently
written in spherically polar coordinates, so that one can take
algebraic advantage of the properties of Legendre polynomials and
Bessel radial functions, and because the spherically symmetric
potential allows for the quick finding of these solutions by virtue of
the algebraic method called separation of variables. But if you were
to say that it is simply impossible to write the hydrogen atom
wavefunctions in anything other than spherically polar coordinates,
you'd be daft.

To take it one step further just to hammer in the point, writing the
equation of a sphere of radius R is easiest in spherical polar
coordinates, but it is certainly possible to write the equation for
the same sphere in rectangular coordinates, or cylindrical
coordinates, or any other coordinate system you choose.


like any other metric.

Other metric as a solution to the Einstein field equations also only
applies to the spherically symmetric polar coordinate system and
nothing else, and you have never understood the field equations.
shrug

What makes me wonder is why you think this is meaningful or even
relevant to the uniqueness of the solution.

You remain a multi-year super-senior. *shrug

You write down multiple representations of the same thing and expect
us to agree with you that they are different. It doesn't work that way.

You were expecting to graduate this spring. *Did it happen? *No.
shrug

On May 12, 10:24*pm, Koobee Wublee wrote:
On May 12, 11:16 pm, Eric Gisse wrote:

On May 12, 9:55 pm, Koobee Wublee wrote:
You continue to say "infinite number" like that's supposed to have any
meaning.

So, you don’t understand what infinity represents.

I usually take it to mean "lots", or in a more general sense, "more
than one".

You chose the coordinate form so there is only one solution for that
particular coordinate form. Ever read anything about ODE uniqueness
theorems?

The other solutions are clearly in different coordinate systems.


There is an infinite number of coordinate systems

Yes. *shrug

that the Schwarzschild solution can be projected on,

The Schwarzschild metric only applies to the spherically symmetric
polar coordinate system and nothing else. *So, you do not understand
the Schwarzschild metric, either.

There are still an infinite number of spherically symmetric coordinate
representations of Schwarzschild - you have shown me at least 3 of
them.

Who says that the metric is only true in a spherically symmetric
coordinate system? I can write the line element in Cartesian, prolate
spherical, or cylindrical coordinates. Or can I not do that for some
reason?


like any other metric.

Other metric as a solution to the Einstein field equations also only
applies to the spherically symmetric polar coordinate system and
nothing else, and you have never understood the field equations.
shrug

Since when does the coordinate system determine the physics? Spherical
symmetry is a property of the manifold, not the coordinate system.
Writing something in spherical coordinates doesn't make it spherically
symmetric - or did you not know what?

Glass houses, kooby.


What makes me wonder is why you think this is meaningful or even
relevant to the uniqueness of the solution.

You remain a multi-year super-senior. *shrug

You write down multiple representations of the same thing and expect
us to agree with you that they are different. It doesn't work that way.

You were expecting to graduate this spring. *Did it happen? *No.
shrug

I was also expecting a pony. Where's my pony?

I'll say this once and no mo my personal life is exactly none of
your business.

On May 13, 12:37 am, Eric Gisse wrote:

You chose the coordinate form so there is only one solution for that
particular coordinate form. [The rest of crap is mercifully
snipped.]

You continue to whine and distort what I have said. Go back to sit on
your piles of books.

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. 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 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. 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. 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.

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