On May 11, 9:53 pm, Koobee Wublee wrote:
On May 10, 11:57 pm, JanPB wrote:

On May 10, 9:42 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.

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

What do you mean ‘probably’?

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.

--
Jan Bielawski