# The General Relativity Metric and Flat Spacetime

pervect
Staff Emeritus
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.

JDoolin
Gold Member
Flatlanders living on a sphere:

"That way." (pointng in two opposite directions)

"And what direction did you just come from?"

"Sorry, that direction does not exist here. I would require a third degree of freedom to tell you, explicitly."

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JDoolin
Gold Member
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.
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.

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

$$dT = \frac{\partial}{\partial x^\alpha} \Big( \int^{x^0} \sqrt{g_{00}(t',x^1,x^2,x^3)} \; dt' \Big) dx^\alpha$$

which is a closed 1-form, but it is not a 1-form that fits nicely into your metric! It has a bunch of $dx^1,\ dx^2,\ dx^3$ terms in addition to $dx^0$. 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 $dx^\alpha$ for some coordinate functions $x^\alpha$), and this system of differential equations is not integrable in any finite-sized open region, in general.
If the GR metric is valid only in a path dependent situation. I would request Ben Niehoff to go through the following posting:
https://www.physicsforums.com/showpost.php?p=3436925&postcount=21

Right here you've lied. By calling it $dT$, you are claiming that the one-form $\sqrt{g_{00}} dt$ is closed (i.e., locally exact). But as others have pointed out, this is not true unless $g_{00}$ happens to be a function of $t$ alone.
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