I still have to read more. Maybe when you put enough static schwarzschild metrics together, and rub them against each other, it becomes a nonstatic space-time.There's a well defined notion of a static observer in a static space-time like the Scwarzschild space-time and there is an unambiguous notion of the speed of any object relative to the static observer that's co-located with him.
Cosmological space-times are for the most part not static, so this trick won't work for them.
If the GR metric is valid only in a path dependent situation. I would request Ben Niehoff to go through the following posting:By holding the other coordinates constant, you are able to do the integral, sure. But when you take the exterior derivative of both sides, you obtain
[tex]dT = \frac{\partial}{\partial x^\alpha} \Big( \int^{x^0} \sqrt{g_{00}(t',x^1,x^2,x^3)} \; dt' \Big) dx^\alpha[/tex]
which is a closed 1-form, but it is not a 1-form that fits nicely into your metric! It has a bunch of [itex]dx^1,\ dx^2,\ dx^3[/itex] terms in addition to [itex]dx^0[/itex]. You will run into a similar problem with the rest of your integrals.
You should see now why the above claim is false. The point is that you have a system of differential equations (i.e., a bunch of 1-forms which you assume can be written as [itex]dx^\alpha[/itex] for some coordinate functions [itex]x^\alpha[/itex]), and this system of differential equations is not integrable in any finite-sized open region, in general.
The Science Advisor considers it essential to use impolite language to express himself---possibly he feels that it should provide a special parameter to the forumRight here you've lied. By calling it [itex]dT[/itex], you are claiming that the one-form [itex]\sqrt{g_{00}} dt[/itex] is closed (i.e., locally exact). But as others have pointed out, this is not true unless [itex]g_{00}[/itex] happens to be a function of [itex]t[/itex] alone.