I Congruent worldlines in "static" gravitational field

[cianfa72]

No. You can always draw straight axes on a map of the nastiest twistiest spacetime you can imagine (as long as you avoid coordinate singularities and the like). The axes won't necessarily be straight in the real world!

I would like to point out that when you see a map on a piece of paper you have to know how that map is actually built (i.e. what map it is). In the case of post#1 it is the inertial/Minkowski map for Minkowski/flat spacetime.

 

[Ibix]

would like to point out that when you see a map on a piece of paper you have to know how that map is actually built (i.e. what map it is).

Indeed.

In the case of post#1 it is the inertial/Minkowski map for Minkowski/flat spacetime.

No, it's in an arbitrary static spacetime in coordinates that reflect that symmetry. In context, it's probably Schwarzschild spacetime in Schwarzschild coordinates, but the argument is more general.

 

[cianfa72]

No, it's in an arbitrary static spacetime in coordinates that reflect that symmetry. In context, it's probably Schwarzschild spacetime in Schwarzschild coordinates, but the argument is more general.

Why ? If you assume Schwarzschild spacetime in Schwarzschild coordinates then spacetime is curved and that's actually what that argument is supposed to show (i.e. give a proof).

 

[Ibix]

If you assume Schwarzschild spacetime in Schwarzschild coordinates then spacetime is curved and that's actually what that argument is supposed to show (i.e. give a proof).

I was stating the eventual conclusion there. The only assumption being made is that observers at any fixed $z$ always see the same result of any gravitational experiment (so the lines of constant $z$ are what we will eventually call integral curves of the timelike Killing vector field). That's why the light paths are the same. But we aren't assuming anything about what those paths are, which is why they are drawn as arbitrary squiggles.

If the experiment is being carried out in vacuum over a spherical body this will inevitably turn out to be Schwarzschild spacetime. But we aren't assuming that at this stage.

 

[cianfa72]

The only assumption being made is that observers at any fixed $z$ always see the same result of any gravitational experiment (so the lines of constant $z$ are what we will eventually call integral curves of the timelike Killing vector field). That's why the light paths are the same.

Ok, so the lines of constant $z$ in the map represent (in the map) the integral curves of the timelike KVF -- a 'static' spacetime also requires that the timelike KVF is hypersurface orthogonal. Basically the values of $z$ "label/ identify" each of those timelike KVF's integral curves.

The point I do not catch is that the obsever at fixed $z$ doesn't see the path of the light coming from the bottom ($z=0$)...

 

[cianfa72]

The only assumption being made is that observers at any fixed $z$ always see the same result of any gravitational experiment.

I'm stuck with this point: observers at any fixed $z$ always see the same result only for local gravitational experiment. Is that true?

 

[Ibix]

I'm stuck with this point: observers at any fixed $z$ always see the same result only for local gravitational experiment. Is that true?

I'm not sure what "local" is meant to mean in this context, honestly.

A simple experiment you could do is to drop a ball and time its fall. An observer at fixed $z$ in a static field who always releases the ball from rest and allows it to fall the same distance will always measure the same time. That is not a "local" result - the observer could let the ball fall 1cm or 100,000km and the time for the fall would not vary no matter how many times they repeated that experiment. That's the definition of static, really, that there is a family of observers for which this is true. Contrast with an observer who is changing $z$ slowly (in a lift, for example), who will eventually notice changing results as the experiment probes different regions of the field. Or contrast someone on Earth, who detects changing results due to the gravity of the Sun and Moon.

I suspect the point about locality is that if I drop a small ball through 1cm near my head and near my toes, I'll get the same results from each one (very nearly) on Earth, but very different ones if I'm hovering close to the horizon of a small black hole due to the tidak forces. The argument is that the head-height and toe-height experiments are the same but they give different results. Thus we say "local experiments" and say that near the horizon my head and toes don't count as local to each other. I prefer the view that the head-height and toe-height experiments are different ones always (albeit giving very similar results in weak fields). Thus I don't need the "local" restriction.

 

[cianfa72]

An observer at fixed $z$ in a static field who always releases the ball from rest and allows it to fall the same distance will always measure the same time

How does the observer at fixed $z$ measure the time along the path the object is falling from rest ?

 

[Ibix]

How does the observer at fixed $z$ measure the time along the path the object is falling from rest ?

Bounce a radar pulse off it. The nature of a static spacetime means that the radar pulse takes equal times on both legs of its journey, so you may not be able to measure distance accurately that way without further work, but you can determine at what time (by your clock) the echo happens.

Or you could measure the time from when you release the ball (right in front of your nose) to when you see it strike the floor. We don't care about the result, really, just the repeatability.

 

[cianfa72]

The nature of a static spacetime means that the radar pulse takes equal times on both legs of its journey,

I believe that's true only in the coordinate chart adapted to the hypersurface orthogonal's timelike KVF (i.e. in the coordinate chart in which the coordinate time worldlines are orbits of timelike KVF).

 

[Ibix]

I believe that's true only in the coordinate chart adapted to the hypersurface orthogonal's timelike KVF (i.e. in the coordinate chart in which the coordinate time worldlines are orbits of timelike KVF).

Sure, but that's fine, as long as you use the same coordinates for each run. If you want something coordinate free, go with the "record the time you see it land".

 

[cianfa72]

If you want something coordinate free, go with the "record the time you see it land".

Do you mean record the time on the observer's clock at fixed $z$ when the radar pulse bouncing from the land event come back to the observer at fixed $z$ ?

 

[Ibix]

Yes.