On May 6, 9:55*pm, Koobee Wublee wrote:
On May 6, 8:28 pm, JanPB wrote:

On May 6, 2:24 pm, Koobee Wublee wrote:
Another solution that is static, spherically symmetric, and (this
time) asymptotically flat is the following.

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^4 dO^2 / K^2

It's the same solution as Schwarzschild's, you merely changed the
numbers labelling the spheres of symmetry.

The Schwarzschild metric has the following form.

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

For the queer of England to continue living in that fat castle in the
air, he/she/it needs to talk himself/herself/itself into believing
both of the above equations are exactly the same.

You did not specify whether you wrote both formulas with respect to
the same coordinate system or not.

Assuming you meant the same coordinates in both formulas, then you are
WRONG: the _first_ formula is NOT a solution. You can verify (it's
quite tedious) that in fact it doesn't satisfy Ricci=0. The second one
does.

On the other hand, if you intended the first formula to be obtained by
substituing r in the second, then you are also WRONG: in that case
both formulas describe the same dot product (i.e., the same solution).

It's exactly the same with dr in polar coordinates and x/
sqrt(x^2+y^2)dx + y/sqrt(x^2+y^2)dy in Cartesian -- they are equal
covectors. Different formulas summing up to the same value. Same thing
when these covectors sit inside a formula for ds^2.

Well, both are solutions to the field equations using the spherically
symmetric polar coordinate system and nothing else.

So you say "the" spherically symmetric polar, so it looks like you
indeed intend both expressions for ds^2 to be written in the same
coordinate system. In this case your first formula is not a solution.

There are no
coordinate transformations involved. *They are just independent
solutions to the field equations. *shrug

Only the second one is a solution in this case.

As I said before, for a mere $99,999, I will design the universe of
your choice from the field equations.

You've just failed.

[...]

--
Jan Bielawski