On May 10, 9:42 pm, Koobee Wublee wrote:
On May 10, 3:09 am, JanPB wrote:

On May 9, 12:32 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 curvature is still nonzero if r denotes the same coordinate
as in Schwarzschild. Try it.

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

Calculate the Ricci tensor. It's not zero. The R_theta,theta is
probably the easiest nonzero component to compute.

I recommend Cartan moving frame method as
it's much faster than Christoffel symbols.

There is no short cut in such type of mathematics. shrug

Cartan's method utilizes the skew symmetry of certain combinations of
Christoffel symbols (rather than symmetry of the symbols themselves).
It allows one to refactor the computations in a more efficient way. In
this regard it's a bit like the Euler-Lagrange equation which also
cleverly refactors the rather ugly brute-force general-coordinate form
of "F=ma".

The following is also a solution where the gravitational force follows
the inverse cubed law.

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

Same thing: if r is the same then Ricci of the above is nonzero.

But mathematically, its Ricci and Einstein tensors also vanish.
shrug

Does "mathematically vanish" mean something different than "vanish"?

All I'm saying is that Ricci curvatures of the two metrics you wrote
are nonzero if you think of (t,r,theta,phi) as the coordinate system
of Schwarzschild.

OTOH
if r denotes sqrt(K * Schwarzschild-r) then Ricci of the above equals
0 but in that case it's the same solution as Schwarzschild - you
simply change the labels you attach to the spheres from "r" to "r^2/
K".

Perhaps, there is a flaw in your Cartan whatever.

Use Christoffel symbols then. You'll see the terms don't cancel.

As I said, anyone
who possesses mathematical software like Eric Gisse can easily verify
it, and I do not see any complaint from that multi-year super-senior.
shrug

By all means, use a computer if you prefer.

--
Jan Bielawski