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