Is acceleration absolute or relative - follow up

[jbriggs444]

You can always find a coordinate chart in which a single timelike geodesic has constant spatial coordinates. Just construct Fermi normal coordinates centered on that geodesic. The chart might not cover the entire spacetime, but it will always cover the entire range of that single geodesic.

Right. If you fix the geodesic you can pick a chart that gives it constant spatial coordinates.

But if you fix a chart, you may find that there is no timelike geodesic which has constant spatial coordinates in that chart. (for instance Rindler coordinates).

 

[PeterDonis]

if you fix a chart

My understanding is that the OP is allowing free choice of coordinate charts, not trying to fix any particular chart.

 

[jbriggs444]

My understanding is that the OP is allowing free choice of coordinate charts, not trying to fix any particular chart.

Right. If he allows free choice of charts, I am allowed to fix a chart that violates his claims about it.

 

[PeterDonis]

If he allows free choice of charts, I am allowed to fix a chart that violates his claims about it.

I don't think he's claiming that every chart must have certain properties. I think he's just trying to figure out whether any chart can have certain proprties.

 

[cianfa72]

What you cannot always do is find a coordinate chart in which an entire family of timelike geodesics has constant spatial coordinates, since the geodesics might cross.

No, that's not what I said in post #25. See above.

My point was to say that there is no guarantee to be able to find a global coordinate chart in which any (an entire family of) timelike geodesic is actually "at rest" in it.

 

[PeterDonis]

any (an entire family of)

The parenthetical qualifier here is crucial. Without it, "any" just means one single geodesic, not a family of them.

 

[cianfa72]

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.

 

[PeterDonis]

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.

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

Of course: FRW spacetime.

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

Yes.

 

[cianfa72]

"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.