I Definition of "Spatial X Direction" in Spacetime Context

[PeterDonis]

the set of events representing the 'spatial direction' in the tangent space at event A is the blue area in the following diagram

The points in the diagram, since it's strictly speaking a diagram of the tangent space, represent vectors, not events; the vector represented by a given point is, heuristically, the arrow from the origin to that point, which gives a magnitude and a direction. The blue area thus represents the set of spacelike vectors in the plane of the diagram, i.e., the vectors that point in the "spatial $x$ direction".

The red arrow represents the gyroscope 4-velocity at event A

Yes.

The inertial coordinates are chosen in such way that the gyroscope 4-velocity at event A is the axis and the axis is the spacelike vector orthogonal to the 4-velocity at event A (hence orthogonal to the axis).

Yes.

 

[cianfa72]

The points in the diagram, since it's strictly speaking a diagram of the tangent space, represent vectors, not events; the vector represented by a given point is, heuristically, the arrow from the origin to that point, which gives a magnitude and a direction. The blue area thus represents the set of spacelike vectors in the plane of the diagram, i.e., the vectors that point in the "spatial $x$ direction".

ok, so the gyroscope axis actually picks a 2-plane in the tangent space at each event along its worldline.
In a neighborhood of event A the set of spacelike vectors in that 2-plane (arrows from event A) let me say pick events spacelike separated from the event A in 'spatial $x$ direction'.

 

[PeterDonis]

so the gyroscope axis actually picks a 2-plane in the tangent space at each event along its worldline.

No.

The actual gyroscope axis is represented by the blue arrow: the spacelike vector in the 2-plane that is orthogonal to the gyroscope's 4-velocity.

The other spacelike vectors in the 2-plane represent other possible gyroscope axes pointing in the "spatial $x$ direction", for gyroscopes whose 4-velocities are different at the event whose tangent space the diagram represents. The fact that they all lie in the same 2-plane in the tangent space, the $x$, $t$ 2-plane, is what justifies us saying that all of those different gyroscopes, with different 4-velocities, all have axes that point in the "spatial $x$ direction".

 

[cianfa72]

The fact that they all lie in the same 2-plane in the tangent space, the $x$, $t$ 2-plane, is what justifies us saying that all of those different gyroscopes, with different 4-velocities, all have axes that point in the "spatial $x$ direction".

So let me say the arrows in blue region in my diagram actually represent in spacetime the 'class of equivalence of gyroscopes having axes pointing in the same 'spatial $x$ direction' at event A'.

 

[PeterDonis]

So let me say the arrows in blue region in my diagram actually represent in spacetime the 'class of equivalence of gyroscopes having axes pointing in the same 'spatial $x$ direction' at event A'.

Strictly speaking, the arrows in the blue region represent the class of axes of such gyroscopes. The gyroscopes themselves have a 4-velocity as well as an axis so the axis itself does not completely represent the gyroscope.

However, since the axis and the 4-velocity are orthogonal and the 4-velocity is timelike, if you know the 4-vector representing the axis, you also know the 4-vector representing the 4-velocity. So in that sense, yes, the arrows in the blue region are sufficient to represent this class of gyroscopes.

 

[cianfa72]

The points in the diagram, since it's strictly speaking a diagram of the tangent space, represent vectors, not events; the vector represented by a given point is, heuristically, the arrow from the origin to that point, which gives a magnitude and a direction.

ok so, technically, we get the set of 'events spacelike separated from event A in spatial $x$ direction in a neighborhood of A' by means of the exponential map of those spacelike vectors lying in that 2-plane (the 2-plane that each gyroscope's spatial axis in the class pointing in $x$ spatial direction picks/selects in the 4D tangent space at event A).

 

[PeterDonis]

we get the set of 'events spacelike separated from event A in spatial $x$ direction in a neighborhood of A' by means of the exponential map of those spacelike vectors lying in that 2-plane

Yes.

 

[pervect]

Summary:: Clarification about the definition of 'spatial x direction' in the context of flat or curved spacetime.

Hi,
although there is a lot of discussion here in PF, I'd like to ask for a clarification about the definition of 'spatial x direction' in the context of flat or curved spacetime.

Consider a set of free-falling gyroscopes (zero proper acceleration) passing through an event A with different relative velocities. Suppose that at event A all of them have their axes aligned in the same common direction.

I believe that set of gyroscopes -- having their axes aligned in that same common direction-- actually defines the notion of 'spatial x direction'.

The difference between each of them is that the axes of each gyroscope defines a different spacelike direction through spacetime since each gyroscope is actually at rest in different inertial frames at event A.

Does it make sense ? Thanks

This response is a bit rushed, I'll try and get back to it and read the whole thread. But here are my initial thoughts.

Gyroscopes Fermi-walker transport vectors, but since you have specified the gyroscopes are in free fall, Fermi-Walker transport is equivalent to parallel transport. So your gyroscopes are parallel transporting vectors.

You will have the issue that parallel transporting a vector around a loop in curved space-time will generally result in rotating the vector being transported, in this case by your gyroscope. Another way of saying this - if you transport a vector along two different paths, the results will not necessarily be the same in curved space-timer.

At the particular event in question, you already have the notion of a tangent vector pointing in the same direction. Vectors can be transported from one point to another via a connection. GR uses the Levi-Civita connection. But in curved spacetime you'll in general have the issue that I mentioned, that transporting a vector along two different paths won't necessarily have the same result.

 

[cianfa72]

Gyroscopes Fermi-walker transport vectors, but since you have specified the gyroscopes are in free fall, Fermi-Walker transport is equivalent to parallel transport. So your gyroscopes are parallel transporting vectors.

Yes definitely. The topic of the thread was about the representation of gyroscope axis in spacetime.

 

[pervect]

Yes definitely. The topic of the thread was about the representation of gyroscope axis in spacetime.

Since you have specified a geodesic path, your approach will give a notion of "the same direction" in the region around a point where the geodesics do not cross, but in general this region won't cover all of space-time.

The region where geodesics don't cross is called the "geodesically convex region" or sometimes just the "convex region", see https://en.wikipedia.org/wiki/Geodesic_convexity. I don't know a lot about it really, but I do know the name :).

Orbits in the Schwarzschild space-time are geodesics, so any region of space-time large enough for two orbits to cross is not convex. For instance an "orbit" going radially outwards and falling back in will cross a circular orbit.

Fermi normal coordinates use a similar construction, and share the property that they only cover a subset of the manifold.