Definition of "Spatial X Direction" in Spacetime Context

[cianfa72]

TL;DR 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 ? Thank you.

 

[Ibix]

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

Suppose that at event A all of them have their axes aligned in the same common direction.

What do you mean by "common direction" here? Because, as you note, they point in different 4d directions. I think what you want to say is that all of them will agree that their rotation axes are pointing in the same spatial direction, but you can't do that without defining space first.

The point is that what I call the $x-t$ plane is also what you call the $x'-t'$ plane, although we use different pairs of vectors to define it. Thus if your gyroscope's axis lies in the $x'$ direction then its projection perpendicular to my four velocity lies in my $x$ direction.

 

[cianfa72]

I think what you want to say is that all of them will agree that their rotation axes are pointing in the same spatial direction, but you can't do that without defining space first.

Yes, if at event A their rotation axes are aligned then by definition they are pointing in the same spatial direction. In a sense we can say 'at event A the equivalence class of those gyroscopes having their rotation axes aligned' actually define a spatial direction at event A.

Thus if your gyroscope's axis lies in the $x'$ direction then its projection perpendicular to my four velocity lies in my $x$ direction.

Actually $x'$ and $x$ directions are spacetime directions. So the perpendicular projection you were talking about is a projection in spacetime, I believe.

 

[PeroK]

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 ? Thank you.

There are no absolute x-y-z directions at a point in space. Note that in flat spacetime a straight line in one IRF maps to a straight line in any other IRF. But, angles are not preserved. So, mutually orthogonal coordinate axes in one IRF are not necessarily mutually orthogonal in another IRF (but only in the special case where the relative motion is along one of the coordinate axes). That means, of course, for any two IRF's you can always choose coordinate axes so that the relative motion is along one of these and hence they can share Cartesian coordinate axes (in a sense).

You cannot have a common set of Cartesian axes across any three IRF's (only in special cases). And, in fact, if IRFs $S$ and $S'$ share coordinate axes with relative motion along the x-axis; and, $S'$ and $S''$ share coordinate axes with relative motion along the y-axis; then, perhaps surprisingly, $S$ and $S''$ do not share coordinate axes, but each IRF's coordiante axes are rotated, as measured in the other IRF. This is called the Wigner Rotation:

https://en.wikipedia.org/wiki/Wigner_rotation

This issue came up on another recent thread:

https://www.physicsforums.com/threads/time-dilation-along-multiple-axes.1012568/

 

[Ibix]

Yes, if at event A their rotation axes are aligned then by definition they are pointing in the same spatial direction. In a sense we can say 'at event A the equivalence class of those gyroscopes having their rotation axes aligned' actually define a spatial direction at event A.

I don't think so. Consider the set of all gyroscopes moving in the x direction. Any pair overlap in a region, but I think that the only place all pairs mutually overlap (i.e. the intersection of all worldsheets in this set) is the event A.

 

[cianfa72]

I don't think so. Consider the set of all gyroscopes moving in the x direction. Any pair overlap in a region, but I think that the only place all pairs mutually overlap (i.e. the intersection of all worldsheets in this set) is the event A.

Sorry, I missed your point. Which is your definition of 'x direction' shared from the set of all gyroscopes ?

 

[Ibix]

Sorry, I missed your point. Which is your definition of 'x direction' shared from the set of all gyroscopes ?

Pick one gyroscope's definition. It doesn't matter which, as far as I can see. The requirement is that all their four velocities and rotation axes be coplanar - i.e., you can draw them all on a 2d Minkowski diagram.

 

[cianfa72]

Pick one gyroscope's definition. It doesn't matter which, as far as I can see. The requirement is that all their four velocities and rotation axes be coplanar - i.e., you can draw them all on a 2d Minkowski diagram.

Sorry for the stupid/basic question: as far as I understand, you are saying to consider the four velocities and the rotation axes of all the gyroscopes in the set. Since all of them are actually directions in spacetime then we can draw any of them on a 2d Minkowski diagram, right ?

 

[Ibix]

I'm only considering the set that can be drawn on a Minkowski diagram. There certainly exist ones that cannot be drawn on it, but I don't think we need to consider them.

 

[cianfa72]

The requirement is that all their four velocities and rotation axes be coplanar - i.e., you can draw them all on a 2d Minkowski diagram.

I take it as if the four velocities and rotation axes of all gyroscopes in the set are coplanar, then that set of gyroscopes actually point to the same spatial direction. Right ?

 

[Ibix]

I take it as if the four velocities and rotation axes of all gyroscopes in the set are coplanar, then that set of gyroscopes actually point to the same spatial direction. Right ?

Yes. But just one gyroscope, with its four velocity and the direction along its rotation axis, is enough to pick out that plane. Having more of them doesn't pick out a particular direction in that plane, is what I was attempting to point out.

 

[cianfa72]

Yes. But just one gyroscope, with its four velocity and the direction along its rotation axis, is enough to pick out that plane.

The plane you were talking about is a plane in 4d Minkowski spacetime, I believe.

Having more of them doesn't pick out a particular direction in that plane, is what I was attempting to point out.

So all the gyroscopes in that set (namely the set of gyroscopes with their rotation axes aligned) actually pick the same plane in spacetime just with a different pair of basis vectors spanning it.

 

[vanhees71]

I don't think so. Consider the set of all gyroscopes moving in the x direction. Any pair overlap in a region, but I think that the only place all pairs mutually overlap (i.e. the intersection of all worldsheets in this set) is the event A.

The axes of a gyroscope are Fermi-Walker transported along the time-like worldline of the gyroscope. If this worldline is not a straight line the axes nevertheless precess wrt. a momentaneous local restframe defined by a rotation-free boost wrt. a fixed inertial frame. That's the Thomas precession. For details see Sect. 1.8 in

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

 

[cianfa72]

The axes of a gyroscope are Fermi-Walker transported along the time-like worldline of the gyroscope.

Sorry for the basis question: at first glance the axes of a gyroscope seems to be interpreted simply as a spatial direction. Instead in this context it is actually understood as a direction in spacetime.

 

[cianfa72]

You will excuse me if I try to recap my understanding.

Consider the set of all events in the 4D Minkowski spacetime. Take a free-falling gyroscope with 3 axes mutually orthogonal in a given state of motion.

Consider the inertial frame in which the gyroscope is 'at rest'. The set of events with coordinate time $t=0$ spatially 'aligned' say along the x-axis (one of the 3 gyroscope axes) map to a straight line in spacetime mathematical model representing the x-axis in the give inertial frame.

In this sense the gyroscope spatial x-axis actually maps to a spacelike direction is spacetime.

 

[vanhees71]

If the gyroscope is not in uniform motion wrt. an inertial frame then there is no single inertial frame where the gyroscpe is at rest but only instantaneous ones. The spin of the gyroscope obeys the Fermi-Walker transport equations along the worldline of the gyroscope, if there is no torque acting on it. This, however, implies that if the gyroscope is accelerated and moving not on a straight line, there is the Thomas precession, i.e., the spin precesses wrt. its instantaneous rest frames which are defined by a rotation free Lorentz boost from a fixed inertial frame (so to say the "lab frame" of an inertial observer who observes the motions of the gyroscope). As I said already above:
For details see Sect. 1.8 in

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

 

[cianfa72]

To explain how the gyroscope spatial axis is represented in spacetime.

The blue arrow represents the gyroscope axis in its rest inertial frame. Its worldsheet (light blue region) intersects the primed inertial frame's simultaneity plane (x' axis in the picture). So basically the primed frame describes the gyroscope axis as the projection of the blue arrow onto its simultaneity plane.

 

[PeroK]

To explain how the gyroscope spatial axis is represented in spacetime.

The blue arrow represents the gyroscope axis in its rest inertial frame. Its worldsheet (light blue region) intersects the primed inertial frame's simultaneity plane (x' axis in the picture). So basically the primed frame describes the gyroscope axis as the projection of the blue arrow onto its simultaneity plane.

View attachment 297549

If I'm interpreting that correctly, then that is a spacetime diagram for two reference frames in uniform relative motion in the direction of a shared $x-x'$ axis?

 

[cianfa72]

If I'm interpreting that correctly, then that is a spacetime diagram for two reference frames in uniform relative motion in the direction of a shared $x-x'$ axis?

Yes, the two reference frames moves along the same shared spatial direction.

 

[PeterDonis]

The blue arrow represents the gyroscope axis in its rest inertial frame.

Not quite. The blue arrow represents the 4-vector that describes the gyroscope's axis. The gyroscope has a definite state of motion in spacetime; it's 4-velocity is the vertical "t" arrow in your diagram. The 4-vector describing the gyroscope's axis is orthogonal to its 4-velocity and in the spacetime plane in which the axis points. This 4-vector is an invariant; it doesn't change when you change frames. It would only change if the gyroscope changed its state of motion (since the 4-vector describing the axis must remain orthogonal to the gyroscope's 4-velocity).

the primed frame describes the gyroscope axis as the projection of the blue arrow onto its simultaneity plane

No. The primed frame might describe a component of the gyroscope axis this way. But, as above, the gyroscope axis itself is a 4-vector that is invariant; it doesn't change when you change frames. The key point in the primed frame is that the 4-vector of the gyroscope axis is not a purely spatial vector in this frame; it has a $t'$ component as well as an $x'$ component. Or, to put it another way, it is not orthogonal to the $t'$ basis vector.

 

[Ibix]

No. The primed frame might describe a component of the gyroscope axis this way.

I'm not sure I agree - or at least, it depends what one means by "axis". Certainly the blue and red arrows lie along what the two frames would call the axle. I'd have said that was a perfectly acceptable interpretation of "the gyroscope axis". Agreed that this isn't the same as the angular momentum axis, which is an invariant representable by the blue arrow, I think.

 

[PeterDonis]

it depends what one means by "axis".

The meaning I am giving it is the standard one in relativity: that spacelike vector in the tetrad of which the gyroscope's 4-velocity is the timelike vector, that points along the gyroscope's axis. In other words, that particular spacelike vector in the 2-plane spanned by the 4-velocity and the "spatial $x$ direction" that is picked out by the invariant condition of being orthogonal to the 4-velocity.

Agreed that this isn't the same as the angular momentum axis, which is an invariant representable by the blue arrow, I think.

Angular momentum in relativity is not a 4-vector, it's an antisymmetric 2nd-rank tensor. And in 4 dimensions, unlike 3, there is not a one-to-one correspondence between vectors--more precisely pseudovectors--and antisymmetric 2nd rank tensors. The best you can do is pick out a 2-plane orthogonal to the angular momentum, and that plane is the one we have been talking about, the one @cianfa72 drew the diagram in. Which unfortunately is not sufficient by itself to resolve any of the issues in this thread. You need an additional condition to pick out one particular spacelike vector in this 2-plane; I gave the additional condition above.

 

[cianfa72]

In other words, that particular spacelike vector in the 2-plane spanned by the 4-velocity and the "spatial $x$ direction"

I'm stuck in grasping the following: how does a "spatial $x$ direction" define a 4-vector in spacetime ?

 

[vanhees71]

The meaning I am giving it is the standard one in relativity: that spacelike vector in the tetrad of which the gyroscope's 4-velocity is the timelike vector, that points along the gyroscope's axis. In other words, that particular spacelike vector in the 2-plane spanned by the 4-velocity and the "spatial $x$ direction" that is picked out by the invariant condition of being orthogonal to the 4-velocity.Angular momentum in relativity is not a 4-vector, it's an antisymmetric 2nd-rank tensor. And in 4 dimensions, unlike 3, there is not a one-to-one correspondence between vectors--more precisely pseudovectors--and antisymmetric 2nd rank tensors. The best you can do is pick out a 2-plane orthogonal to the angular momentum, and that plane is the one we have been talking about, the one @cianfa72 drew the diagram in. Which unfortunately is not sufficient by itself to resolve any of the issues in this thread. You need an additional condition to pick out one particular spacelike vector in this 2-plane; I gave the additional condition above.

The spin $s_{\mu \nu}$ is defined as the angular momentum of a system wrt. its center of momentum. Let the four-velicity of the center of momentm be $u^{\mu}$ with $u^{\mu} u_{\nu}=1$. Now by definition $s_{\mu\nu}u^{\nu}=0$. Due to this constraint the spin tensor is uniquely represented by th spin vector
$$S^{\mu}=\epsilon^{\mu \nu \rho \sigma} s_{\nu\rho} u_{\sigma}/2.$$
Then
$$s_{\mu \nu}=-\epsilon_{\mu \nu\rho \sigma}S^{\rho}u^{\sigma}.$$

 

[PeterDonis]

how does a "spatial direction" define a 4-vector in spacetime ?

It doesn't. The "spatial $x$ direction" is defined by a 2-plane in spacetime (more precisely, in the tangent space at some particular event in spacetime), not a single 4-vector. All spacelike vectors in that 2-plane could be said to be pointing in the "spatial $x$ direction". What picks out one particular one of those spacelike vectors--the blue arrow in your diagram--is that it's the one that's orthogonal to the 4-velocity of the gyroscope--the "t" arrow in your diagram.

 

[cianfa72]

It doesn't. The "spatial $x$ direction" is defined by a 2-plane in spacetime (more precisely, in the tangent space at some particular event in spacetime), not a single 4-vector.

Ah ok, indeed the 'spatial $x$ direction' is actually the set of events parametrized by 2 coordinates (for example w.r.t a fixed inertial frame at a given coordinate time $t$ it is the set of events having $y,z$ coordinates equal to a constant times the $x$ coordinate). Hence that set is a spacelike 2-plane in spacetime.

 

[PeterDonis]

Ah ok, indeed the 'spatial $x$ direction' is actually the set of events parametrized by 2 coordinates

Yes, $x$ and $t$.

(for example w.r.t a fixed inertial frame at a given coordinate time $t$ it is the set of events having $y,z$ coordinates equal to a constant times the $x$ coordinate).

No. The $y$ and $z$ coordinates will be fixed at $0$, and the $t$ coordinate will not be constant.

Hence that set is a spacelike 2-plane in spacetime.

No, the $x$, $t$ plane is not a spacelike 2-plane. But the "spatial $x$ direction" is the set of spacelike vectors that lie within that plane.

 

[cianfa72]

Yes, $x$ and $t$.

Ok yes, to me 'x' was a generic direction however it makes sense to associate 'x' with the direction of ##x# coordinate.

The $y$ and $z$ coordinates will be fixed at $0$, and the $t$ coordinate will not be constant.

ok, yes.

No, the $x$, $t$ plane is not a spacelike 2-plane. But the "spatial $x$ direction" is the set of spacelike vectors that lie within that plane.

So, the 'spatial $x$ direction' picks the $(x,t)$ 2-plane in spacetime, however the set of events corresponding to it is actually a subset of the 2-plane -- namely the subset of events that in the tangent space at some particular event A in spacetime are spacelike separated from it.

 

[PeterDonis]

Ok yes, to me 'x' was a generic direction however it makes sense to associate 'x' with the direction of ##x# coordinate.

You can always choose coordinates so that the $x$, $t$ plane includes whichever spatial direction you like. That's the standard convention when drawing spacetime diagrams.

So, the 'spatial $x$ direction' picks the $(x,t)$ 2-plane in spacetime, however the set of events corresponding to it is actually a subset of the 2-plane -- namely the subset of events that in the tangent space at some particular event A in spacetime are spacelike separated from it.

Yes.

 

[cianfa72]

So the set of events representing the 'spatial $x$ direction' in the tangent space at event A is the blue area in the following diagram (region outside the light cone at event A). The red arrow represents the gyroscope 4-velocity at event A (the gyroscope has a given state of motion represented at each event along its worldline by the timelike tangent vector that lies in the tangent space at that point).

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

 

[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.