On May 13, 5:46*pm, Koobee Wublee wrote:
On May 13, 12:21 am, PD wrote:

On May 13, 1:24 am, Koobee Wublee wrote:
The Schwarzschild metric only applies to the spherically symmetric
polar coordinate system and nothing else. *So, you do not understand
the Schwarzschild metric, either.

No, that's incorrect. The Schwarzchild metric is a *shape*. It is most
conveniently described in one set of coordinates, but it can certainly
be described in more than one coordinate system.

The spacetime described by the Schwarzschild metric only applied to
the spherically symmetric polar coordinate. *You are welcome to
transform to another coordinate system with a different metric. *

No, the Schwarzschild metric IS the metric. There is no such thing as
Schwarzschild metric in a coordinate system with a metric.

I think you are pounding on a non-issue. The real issue is a more
important one: in the limit the Schwarzschild metric reduc es to
Minkowski spacetime, which is gravity free. Yet, in physical reality
there is gravity is such limit.

Thus, the solution is empirically falsified.

Mike



The
segment of spacetime remains the same and thus invariant while the
coordinate is your choice as well as the metric associated with this
coordinate to describe the invariant segment of spacetime.

This concept is so basic, but you have a lot of trouble understanding
that. *There is no need to go through the rest of your babbling.
shrug