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Cosmological space-times are for the most part not static, so this trick won't work for them.

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- #76

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Cosmological space-times are for the most part not static, so this trick won't work for them.

- #77

JDoolin

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Flatlanders living on a sphere:

"Which way is your home?

"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."

"Which way is your home?

"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."

Last edited:

- #78

JDoolin

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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.

Cosmological space-times are for the most part not static, so this trick won't work for them.

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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]

whichisa closed 1-form, but it isnota 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.

https://www.physicsforums.com/showpost.php?p=3436925&postcount=21

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'velied. 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.

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