I Physical Interpretation of Frame Field

[cianfa72]

Moderator's note: Spin-off from another thread due to topic change.

Regarding the general topic of Fermi-Walker transport and "precessions", the following Insights articles might be of interest:

https://www.physicsforums.com/insights/precession-in-special-and-general-relativity/

https://www.physicsforums.com/insights/how-to-study-fermi-walker-transport-in-minkowski-spacetime/

https://www.physicsforums.com/insights/fermi-walker-transport-in-schwarzschild-spacetime/

https://www.physicsforums.com/insights/fermi-walker-transport-in-kerr-spacetime/

In the second link referenced, there is a claim about a physical interpretation of frame field. Consider a family of observers whose worldlines fill a region of spacetime. Each of them carries a clock and a set of mutually orthogonal rulers.

Each observer points in the (timelike) direction defined by its worldline's tangent at any given event along it. What about the rulers each of them carries ? My interpretation: each observer's ruler defines a worldtube. Then take the "local orthogonal spacelike hypersurface" at that event and consider the intersection with the worldtube. Such intersection is a curve pointing in a spacelike direction from that given event. That (spacetime) direction is the spacelike direction the ruler is pointing to (at that event).

 

[PeterDonis]

My interpretation: each observer's ruler defines a worldtube. Then take the "local orthogonal spacelike hypersurface" at that event and consider the intersection with the worldtube. Such intersection is a curve pointing in a spacelike direction from that given event. That (spacetime) direction is the spacelike direction the ruler is pointing to (at that event).

Not quite. But if you imagine basically the same process in the tangent space at an event on the observer's worldline, you'll be more or less there. The "rulers" actually define spacelike unit tangent vectors that are mutually orthogonal to each other and to the observer's 4-velocity. The "tangent vector" part defines a direction in spacetime for each ruler, and the "unit" part defines the length scale along the ruler. In the tangent space, the "unit" part can be thought of as defining the size of the world tube you describe.

 

[cianfa72]

Not quite. But if you imagine basically the same process in the tangent space at an event on the observer's worldline, you'll be more or less there. The "rulers" actually define spacelike unit tangent vectors that are mutually orthogonal to each other and to the observer's 4-velocity. The "tangent vector" part defines a direction in spacetime for each ruler, and the "unit" part defines the length scale along the ruler. In the tangent space, the "unit" part can be thought of as defining the size of the world tube you describe.

Yes, a physical ruler has a finite extension. As far as I can tell, by using exponentiation of one of the mutually orthogonal unit spacelike vectors in the tangent space at an event, one gets a piece of geodesics with "unit" lenght on the "local spacelike section" intersecting that rulers's worldtube.

 

[PeterDonis]

by using exponentiation of one of the mutually orthogonal unit spacelike vectors in the tangent space at an event, one gets a piece of geodesics with "unit" lenght on the "local spacelike section" intersecting that rulers's worldtube.

As long as it's a sufficiently small length compared with the scale of the spacetime curvature.

 

[cianfa72]

Ok. Let me use a more heuristic/intuitive description. Let's fix an event A along the observer's (timelike) worldline. Consider the worldtube's worldlines of one of the observer's carried rulers in a "small" neighborhood of A. Take the (unique) local spacelike hyperplane orthogonal to the observer's worldline at A and consider its intersections with the ruler worldtube's worldlines in that "small" neighborhood. Basically they define what should be understood as the spacelike direction the ruler points to in the tangent space at point A.

If the above makes sense, then I can understand as follows your claim

As long as it's a sufficiently small length compared with the scale of the spacetime curvature.

In a curved spacetime (nonzero geodesic deviation), the exponentiation of the ruler defined's spacelike direction in the tangent space at A may give a spacelike geodesic that "leaves" the previously defined local spacelike hyperplane and no longer stays/remains orthogonal to the ruler worldtube's worldlines.

 

[PeterDonis]

Take the (unique) local spacelike hyperplane orthogonal to the observer's worldline at A and consider its intersections with the ruler worldtube's worldlines in that "small" neighborhood. Basically they define what should be understood as the spacelike direction the ruler points to in the tangent space at point A.

No, they don't, because the spacelike hyperplane you describe isn't in the tangent space. It's in spacetime. What you're describing is a small displacement in spacetime, which in a curved spacetime can't even be described as a vector. The tangent space is an abstraction. It contains tangent vectors (which are directional derivatives), not displacements.

In a curved spacetime (nonzero geodesic deviation), the exponentiation of the ruler defined's spacelike direction in the tangent space at A may give a spacelike geodesic that "leaves" the previously defined local spacelike hyperplane and no longer stays/remains orthogonal to the ruler worldtube's worldlines.

This is the same confusion: the tangent space doesn't contain displacements.

 

[PeterDonis]

the exponentiation of the ruler defined's spacelike direction

Note that this "exponentiation" operation isn't really well-defined, rigorously speaking, because it treats tangent vectors as infinitesimal displacements, which they aren't. Physicists know what they mean when they talk this way, but it's the sort of physicist talk that gives mathematicians fits because of its lack of rigor.

A more rigorous treatment would have to make use of the theorems that establish a one-to-one mapping between tangent vectors in the tangent space at an event in spacetime, and geodesics in the spacetime that pass through that event.

 

[cianfa72]

No, they don't, because the spacelike hyperplane you describe isn't in the tangent space. It's in spacetime. What you're describing is a small displacement in spacetime, which in a curved spacetime can't even be described as a vector. The tangent space is an abstraction. It contains tangent vectors (which are directional derivatives), not displacements.

Ok, that's my point though: a ruler is a physical thing described by its worldtube in the spacetime model. My difficulty in understanding this clearly is: how can a physical thing define/locate/identify an element in the tangent space at event A (the spacelike direction the ruler points to)? My intuition tells me that such identification takes place through a limiting process occurring in the neighborhood of that event.

Yes, from a formal viewpoint, a displacement in spacetime is different matter from a vector that "lives" in the tangent space at event A. However there is one-to-one mapping that assign a vector in the tangent space at A to a displacement in spacetime starting from A.

 

[PeterDonis]

how can a physical thing define/locate/identify an element in the tangent space at event A (the spacelike direction the ruler points to)?

The ruler and the observer's 4-velocity at the given event span a 2-dimensional subspace of the tangent space at that event; that constraint plus orthogonality to the 4-velocity is sufficient to pick out a unique spacelike direction in the tangent space. There will be a unique spacelike unit vector pointing in that direction.

 

[cianfa72]

Sorry to insist. An element (vector) of the tangent space at point P is a function derivative operator (derivation). Take the set of smooth curves C passing though P and consider the composite map for any smooth function $f$ and curve C
$$f(x(t))$$ where $x(t)$ is the representative of C in a chart containing P. The set of such smooth curves is partioned in equivalence classes based on the value returned from the standard derivatives calculated at point P (note that such partitioning is well-defined since it doesn't depend on the chart being used). Each equivalence class defines a unique function derivative operator at P, i.e. an element of the tangent space at P. Well, the derivative operator we get is a different object from any curve member of the relevant equivalence class, yet any of them uniquely defines the same derivative operator at P.

The same is true for the spacelike direction defined by a ruler carried by the observer. Namely the curve we get as intersection in spacetime of the ruler worldtube's worldlines with the local spacelike hyperplane at point P (orthogonal to the observer's worldline at P) in a neighborhood of it, uniquely defines/locates the unit spacelike vector the ruler points to in the tangent space at P.

 

[PeterDonis]

An element (vector) of the tangent space at point P is a function derivative operator (derivation).

I'm not sure what you mean by this. It's a tangent vector, which is equivalent to a directional derivative. Is that what you mean?

the curve we get as intersection in spacetime of the ruler worldtube's worldlines with the local spacelike hyperplane at point P (orthogonal to the observer's worldline at P) in a neighborhood of it, uniquely defines/locates the unit spacelike vector the ruler points to in the tangent space at P.

If the curve is a geodesic, yes, there is a one-to-one mapping between geodesics through P and tangent vectors in the tangent space at P. I already said that in post #34.

I'm not sure what the rest of your post has to do with this, though.

 

[cianfa72]

It's a tangent vector, which is equivalent to a directional derivative. Is that what you mean?

Yes, exactly (directional derivative aka derivation at P).

If the curve is a geodesic, yes, there is a one-to-one mapping between geodesics through P and tangent vectors in the tangent space at P. I already said that in post #34.

Yes, indeed. My point was on the other way around: any curve (even a non-geodesic one) through P defines there an unique directional derivative/derivation (i.e. an element of the tangent space at P).

I'm not sure what the rest of your post has to do with this, though.

My point was related how the ruler worldtube's in spacetime actually defines the spacelike direction the ruler points to in the tangent space at event/point P.

 

[PeterDonis]

any curve (even a non-geodesic one) through P defines there an unique directional derivative/derivation (i.e. an element of the tangent space at P).

No, this is not correct. The one-to-one mapping I referred to no longer holds if you include non-geodesic curves. You have to restrict to geodesics.

To see why this makes sense physically, consider that, given a tangent vector at P, there are an infinite number of non-geodesic curves that all have that tangent vector--because there are an infinite number of possible proper accelerations that an object with that 4-velocity at P could have, and each one leads to a different curve passing through P.

My point was related how the ruler worldtube's in spacetime actually defines the spacelike direction the ruler points to in the tangent space at event/point P.

The ruler's world tube alone is not sufficient to define a unique spacelike direction at P; it can only pick out a 2-dimensional subspace of the tangent space at P (the 2-plane spanned by the ruler and the observer's 4-velocity). You need the orthogonality condition to pick out a unique spacelike tangent vector within that 2-dimensional subspace.

 

[cianfa72]

No, this is not correct. The one-to-one mapping I referred to no longer holds if you include non-geodesic curves. You have to restrict to geodesics.

I meant the way/implication curve -> directional derivative not the other way around (i.e. a curve defines a unique directional derivative at point P).

The ruler's world tube alone is not sufficient to define a unique spacelike direction at P; it can only pick out a 2-dimensional subspace of the tangent space at P (the 2-plane spanned by the ruler and the observer's 4-velocity). You need the orthogonality condition to pick out a unique spacelike tangent vector within that 2-dimensional subspace.

Consider a ruler with zero tickness. Every point on it describes a timelike worldline in spacetime. The set of ruler worldtube's worldlines is one-parameter (which worldline in the set). What you get intersecting it with the local three-dimensional spacelike hyperplane orthogonal at point P to the observer's timelike worldline passing through it?

P.s. perhaps the better term to use for the set of zero-tickness ruler's worldlines is worldsheet.

 

[PeterDonis]

The set of ruler worldtube's worldlines is one-parameter (which worldline in the set). What you get intersecting it with the local three-dimensional spacelike hyperplane orthogonal at point P to the observer's timelike worldline passing through it?

A set of points forming a one-dimensional curve. But that's because you specified the orthogonality condition. If you don't specify the orthogonality condition, then you don't have a unique curve describing the ruler; there will be an infinite number of spacelike curves through point P that are contained in the worldsheet of the ruler, and no way of picking out any one of them.

P.s. perhaps the better term to use for the set of zero-tickness ruler's worldlines is worldsheet.

Yes. See above.

 

[cianfa72]

A set of points forming a one-dimensional curve. But that's because you specified the orthogonality condition. If you don't specify the orthogonality condition, then you don't have a unique curve describing the ruler;

Are you referring to the condition that the local 3d spacelike hyperplane picked is orthogonal to the observer's 4-velocity ?

 

[PeterDonis]

Are you referring to the condition that the local 3d spacelike hyperplane picked is orthogonal to the observer's 4-velocity ?

Yes. That's the only orthogonality condition we've discussed.

 

[cianfa72]

Very well, next step: starting from a tetrad at point P (i.e. from just the set of four vectors given by evaluating the frame field at point P) one defines its Fermi-Walker (FW) transport along the curve C from P. All the additional information needed are the connection coefficients in a neighborhood of the curve (like for the parallel transport).

That means to calculate the FW transport above isn't required to know the vectors off the curve, it suffices to know them at point P alone (like for parallel transport equation).

 

[PeterDonis]

one defines its Fermi-Walker (FW) transport along the curve C from P

What curve C?

 

[cianfa72]

First, what curve C?

the curve C is the timelike worldline of an observer carrying three mutually orthogonal rulers.

 

[PeterDonis]

to calculate the FW transport above isn't required to know the vectors off the curve, it suffices to know them at point P alone (like for parallel transport equation).

No, you have to know them all along the curve, not just at one point on the curve.

But even that assumes that you're only concerned with one curve. If you're dealing with a frame field (which is what you originally asked about), you're dealing with a family of curves, not just one--the integral curves of the 4-velocity field defined by the frame field (by picking out the timelike vector of that field at each point in the open region it covers).

 

[cianfa72]

No, you have to know them all along the curve, not just at one point on the curve.

I'm not sure whether the following argument applies to FW transport as well. Take the parallel transport equation along a curve C from point P and fix a chart. It becomes a first-order differential equation, hence the solution depends only on the components of the vector being parallel transported from point P. Does a similar logic/argument apply to FW transport?

 

[PeterDonis]

Take the parallel transport equation along a curve C from point P and fix a chart. It becomes a first-order differential equation, hence the solution depends only on the components of the vector being parallel transported from point P.

Yes.

Does a similar logic/argument apply to FW transport?

No, because FW transport involves the proper acceleration, and knowing the tangent vector at only one point doesn't tell you the proper acceleration along the curve. You need to know the tangent vector at every point along the curve to know the proper acceleration.

 

[PAllen]

Note that this "exponentiation" operation isn't really well-defined, rigorously speaking, because it treats tangent vectors as infinitesimal displacements, which they aren't. Physicists know what they mean when they talk this way, but it's the sort of physicist talk that gives mathematicians fits because of its lack of rigor.

A more rigorous treatment would have to make use of the theorems that establish a one-to-one mapping between tangent vectors in the tangent space at an event in spacetime, and geodesics in the spacetime that pass through that event.

But that operation is typically called the exponential map (and is a completely rigorous construct):

https://en.wikipedia.org/wiki/Exponential_map_(Riemannian_geometry)

I wonder if this is what @cianfa72 is thinking of.

 

[PeterDonis]

that operation is typically called the exponential map (and is a completely rigorous construct)

Yes, I agree, this operation is rigorous, but the actual rigorous operation, if you look at it, doesn't map a vector in the tangent space to an entire geodesic in the spacetime; it maps a vector in the tangent space to a point on that geodesic (the point that, according to the measure defined by the tangent vector, is a unit distance away from point P along that geodesic). To get a mapping to the geodesic itself, you have to invoke an extra step that maps the point to the curve. Physicists routinely elide all this, but it's still there and needs to be paid attention to if rigor is the objective.

 

[cianfa72]

No, because FW transport involves the proper acceleration, and knowing the tangent vector at only one point doesn't tell you the proper acceleration along the curve.

Ok yes.

You need to know the tangent vector at every point along the curve to know the proper acceleration.

Yes, however the curve C is known (it is the observer's worldline) hence the tangent vector $\mathbf u$ along the curve (4-velocity) is also known and proper acceleration is $\nabla_{\mathbf u} \mathbf u$. Therefore, to perform the FW transport of a tetrad (not a tetrad/frame field) along the observer's worldline (curve C), the tetrad at P and the curve C actually suffice.

Note that the timelike vector of the tetrad at P is the 4-velocity of the observer.