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?