# Is acceleration absolute or relative - follow up

• I
• cianfa72
In summary: Minkowski spacetime are moving inertially, but the frame as a whole is not inertial), and the pseudorandom KVFs (corresponding to objects that are moving but have a definite velocity, but the velocity is not constant).Any static congruence is hypersurface orthogonal. That's part of the definition of "static".Most static spacetimes have only one timelike KVF (as, for example, Schwarzschild spacetime), but not all. Minkowski spacetime has two infinite families of them: the inertial KVFs (corresponding to... objects at rest in Minkowski spacetime are moving inertially, but the frame
cianfa72 said:
any (an entire family of)
The parenthetical qualifier here is crucial. Without it, "any" just means one single geodesic, not a family of them.

We said that in an arbitrary spacetime there is no guarantee to find a global coordinate chart in which every timelike geodesic is "at rest" in it.

Does exist an example of curved spacetime in which there is a timelike geodesic congruence filling it all ?

In that case however, since the spacetime curvature, the geodesics deviation (i.e. tidal gravity) must be not null.

cianfa72 said:
We said that in an arbitrary spacetime there is no guarantee to find a global coordinate chart in which every timelike geodesic is "at rest" in it.
"Every timelike geodesic" is way too strong. You need to say "every one of some particular family of timelike geodesics". And to really have it be meaningful, the family of timelike geodesics needs to fill the entire spacetime.

For example, consider flat Minkowski spacetime. There are an infinite number of possible inertial coordinate charts on this spacetime. Each one has one family of timelike geodesics that fills the entire spacetime at rest in it; but only that one family. There are also an infinite number of other families of timelike geodesics that are not at rest in any given inertial chart. It's impossible to find a single chart in which all timelike geodesics in the entire spacetime are at rest.

Also, since we know there are global coordinate charts in flat spacetime in which a family of timelike geodesics that fill the entire spacetime is at rest, saying that there is no guarantee of this in an "arbitrary" spacetime obviously means an arbitrary curved spacetime.

cianfa72 said:
Does exist an example of curved spacetime in which there is a timelike geodesic congruence filling it all ?
Of course: FRW spacetime.

cianfa72 said:
In that case however, since the spacetime curvature, the geodesics deviation (i.e. tidal gravity) must be not null.
Yes.

PeterDonis said:
"Every timelike geodesic" is way too strong. You need to say "every one of some particular family of timelike geodesics". And to really have it be meaningful, the family of timelike geodesics needs to fill the entire spacetime.
Yes that's right, I was inaccurate in describing it. As you said the main requirement is that 'that some particular family of timelike geodesics' needs actually to fill the entire spacetime.

Thanks for pointing that out.

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