I Relativistic centrifugal force

[PAllen]

One thing to add to the beautiful posts above, that @pervect mentioned in passing, is how much worse it gets going from a ring to a disc. In the ring case, there is continuous curve orthogonal to all congruence lines - it is unfortunately an infinitely long spacelike spiral, parts of which are in the future of others; that is, it intersects each world line an infinite number of times. However, the anomalies are all global; over reasonable finite lengths, you see no problem. Locally, this spiral is everywhere orthogonal to the rotating ring congruence.

Once you go to a disc, the failure becomes local. You cannot find any finite area slice orthogonal to all the intersected congruence lines. This comes about because while the disc congruence has no expansion or shear, it has vorticity. Each congruence line ‘sees’ nearby world lines rotating around them, at fixed distance. This turns out to make it impossible to find a slice orthogonal to all of a bundle of the congruence world lines.

 

[Ibix]

I can post a zoomed in view of the relevant part of the diagram later if that would be helpful.

The cyan lines are the worldlines of clocks moving on a circular track, with their ticks marked by the red blobs. The green line joins one set of blobs, which is a chain of clocks each Einstein synchronised to its neighbour (in fact, if you slit the cylinder along the white line and flattened it out, the green line would be the x-axis of a frame moving at $v=\omega r$). But this approach inevitably goes wrong somewhere - as can be seen from the failure of the green line to close.

 

[pervect]

Actually, forget the Euclidean part. There is no spacelike slice (at all) through the world tube that corresponds to intuitive notions of "surface of the disk". Specifically, no spacelike slice that is a local time slice for every particle of the disk surface. Mathematically, the stationary rotating congruence has no hypersuface orthogonal spatial slices at all.

There is no actual space-like surface that corresponds to the surface of the disk, but mathematically one can project the worldlines of the disk in the 4 dimensional space-time manifold to a 3-dimensional "quotient space". With the understanding that this is a quotient space, and not any sort of surface, we can then discuss the spatial geoemtry of the quotient space. The technique is most applicable to a disk that spins at a constant angular velocity.

A paper that discusses this approach is "Relative space: space measurements on a rotating platform" by M. L. Ruggerio, https://arxiv.org/abs/gr-qc/0309020.

I like Ruggerio's approach, because it ties the notion of distance in the quotient space to the then-current SI defintion of the meter as the distance light travels in 1/c seconds, c being the defined constant, the speed of light. It's got a reasonable number of citations according to google scholar, though I'm sure there are papers on the rotating disk with higher citation counts.

 

[PAllen]

There is no actual space-like surface that corresponds to the surface of the disk, but mathematically one can project the worldlines of the disk in the 4 dimensional space-time manifold to a 3-dimensional "quotient space". With the understanding that this is a quotient space, and not any sort of surface, we can then discuss the spatial geoemtry of the quotient space. The technique is most applicable to a disk that spins at a constant angular velocity.

A paper that discusses this approach is "Relative space: space measurements on a rotating platform" by M. L. Ruggerio, https://arxiv.org/abs/gr-qc/0309020.

I like Ruggerio's approach, because it ties the notion of distance in the quotient space to the then-current SI defintion of the meter as the distance light travels in 1/c seconds, c being the defined constant, the speed of light. It's got a reasonable number of citations according to google scholar, though I'm sure there are papers on the rotating disk with higher citation counts.

I am well aware of that approach, and agree it is possibly the best notion available, but it does not give you an intuitive notion of the body as a whole. I think the rotating disk is important precisely because it is the simplest example in SR of the general fact that a the whole notion of a body undergoing arbitrary motion in SR breaks down. The general case in SR is the congruence of particle world lines bounded by a world tube with no well defined notion of a body as a whole undergoing motion and rotation.

In particular, finite distance calculated on quotient space don’t correspond to any measurement. Only ultralocal distances correspond to radar distances.

 

[DrGreg]

In particular, finite distance calculated on quotient space don’t correspond to any measurement.

I don't agree. Unless I've completely misunderstood this, it's exactly what you'd measure with a tape measure.

 

[PeterDonis]

Unless I've completely misunderstood this, it's exactly what you'd measure with a tape measure.

If this were true, then the distance would have to correspond to a particular spacelike slice taken out of the congruence of worldlines describing the tape measure. Which in turn would imply that the overall spatial geometry would have to correspond to a spacelike slice taken out of the congruence of worldlines describing the disk as a whole. But there is no such spacelike slice.

What is true is that if you take a large number of "ultralocal" measurements using tape measures, you can "assemble" them together into a global 3-geometry described by the quotient space. But there is no way to relate distances in that geometry to measurements made by tape measures of finite (i.e., non-ultralocal) extent. (Similar remarks apply to radar distances.)

 

[DrGreg]

If this were true, then the distance would have to correspond to a particular spacelike slice taken out of the congruence of worldlines describing the tape measure.

Modelling the tape measure as a one-dimensional curve in space, i.e. a two-dimensional surface in spacetime, yes. The "spacelike slice" is a one-dimensional line within this two-dimensional sub-manifold of 4D spacetime. And the ultralocal lengths within this 1D slice are the same as the corresponding lengths in the quotient manifold. Or to put it another way, the 1D "spacelike slice" in 4D spacetime can be identified (isometrically) with a 1D line within the 3D quotient manifold.

Which in turn would imply that the overall spatial geometry would have to correspond to a spacelike slice taken out of the congruence of worldlines describing the disk as a whole.

I don't see why that would follow. You can't glue all possible 1D "spacelike slices" together in 4D spacetime to get a spacelike 3-surface, but so what? That's irrelevant.

 

[PeterDonis]

The "spacelike slice" is a one-dimensional line within this two-dimensional sub-manifold of 4D spacetime.

Not if you want it to be orthogonal to each of the worldlines that form the sub-manifold. There is no such one-dimensional line, because the congruence of worldlines is not hypersurface orthogonal. And if the line is not orthogonal to each of the worldlines, then it is not a valid realization of the intuitive concept of "the tape measure at an instant of time".

 

[DrGreg]

There is no such one-dimensional line, because the congruence of worldlines is not hypersurface orthogonal.

I don't follow your logic. Surely the green line in Ibix's diagram in post #32 is an example of such a line. It's orthogonal to all the cyan lines.

 

[PeterDonis]

Surely the green line in Ibix's diagram in post #32 is an example of such a line.

No, it isn't. It's not a closed curve and does not form a spacelike submanifold, since some points on it are null or timelike separated.

 

[PeterDonis]

Surely the green line in Ibix's diagram in post #32 is an example of such a line.

Also, that line does not correspond to points on a tape measure laid out to measure a spacelike geodesic distance over a finite (non-ultralocal) region of the disk. All of its points are at the same radius, but no spacelike geodesic over a finite region has that property.

 

[DrGreg]

No, it isn't. It's not a closed curve and does not form a spacelike submanifold, since some points on it are null or timelike separated.

But you're now not talking about the 1D helical arc, you're talking about the non-existence of a 2D surface enclosed by it (and orthogonal to the congruence), which, in my view, is irrelevant. The null or timelike separations you refer to are in the embedded 4D spacetime, not within the 2D cylindrincal surface.

In this example we are using a tape measure to measure the circumference of a circle. We do this by integrating ds along the green curve, and each "ds" segment is orthogonal to the local worldline (of the rotating disk) that it intersects with. And each "ds" in the green curve in 4D spacetime corresponds to an identical-length "ds" in the 3D quotient space (by the definition of the quotient space's metric).

Also, that line does not correspond to points on a tape measure laid out to measure a spacelike geodesic distance over a finite (non-ultralocal) region of the disk. All of its points are at the same radius, but no spacelike geodesic over a finite region has that property.

By referring to a tape measure (rather than a ruler), I'm not restricting myself to geodesics; it applies to any curve in space.

I then used a circular arc as an example since Ibix had already drawn a diagram to illustrate it. But I can't see why the same method wouldn't apply to any other curve, including geodesics. (If there is a flaw in my argument, then it's likely this is where it is, but I can't see one.)

 

[PeterDonis]

you're now not talking about the 1D helical arc

Yes, I am. As a 1-D submanifold, this arc is not spacelike (there are points on it that are timelike or null separated) and is not a closed curve.

The null or timelike separations you refer to are in the embedded 4D spacetime, not within the 2D cylindrincal surface.

No, those separations are within the 2D cylindrical surface. That is, assuming you mean the 2D cylindrical surface formed by the congruence of worldlines that the 1-D arc is orthogonal to.

We do this by integrating ds along the green curve, and each "ds" segment is orthogonal to the local worldline (of the rotating disk) that it intersects with. And each "ds" in the green curve in 4D spacetime corresponds to an identical-length "ds" in the 3D quotient space (by the definition of the quotient space's metric).

I agree that this process gives the circumference of the circle in the quotient space. I do not agree that this process is equivalent to laying an actual, physical tape measure around the circumference of the circle. Instead, it's equivalent to having a huge number of infinitesimal tape measures, each of which measures an infinitesimal circumferential distance that corresponds to an infinitesimal spacelike curve; but those individual spacelike curves cannot be "assembled" into a single closed spacelike curve that describes "the circumference of the disk at an instant of time".

By referring to a tape measure (rather than a ruler), I'm not restricting myself to geodesics

Yes, agreed. But my comment above still applies.

 

[pervect]

I don't agree. Unless I've completely misunderstood this, it's exactly what you'd measure with a tape measure.

In particular, finite distance calculated on quotient space don’t correspond to any measurement. Only ultralocal distances correspond to radar distances.

I do agree that one needs to take an appropriate limit to recover the geometry - for a distance of finite length, one does not directly use "radar methods", the radar method gives the length in the limit of small distance and is not directly used to measure longer distances. But I don't see this as an obstacle.

Informally, I would describe the issues as follows. A numerical description of the rigidity of a body can be tied to the speed of sound in that body, with higher speeds implying a more rigid body. In special relativity, the speed of sound cannot exceed "c", so a ruler made out of light is as rigid as it is possible to construct.

When we attempt to measure lengths on the rotating disk, "Centrifugal forces" in the rotating frame warp the shape of any ruler we can construct. And all rulers capable of physical construction are non-rigid.

The way we get around this issue is to say that we can use rulers to measure infinitesimal lengths, and sub-divide a longer length into smaller lengths. As we sub-divide the long ruler into shorter and shorter length, the distortion of the rulers (which can typically be modeled as the deflection of a beam) due to the centrifugal forces becomes lower and lower. So by taking the limit, since the beam deflection increases faster than lineararly (wiki gives cubic deflection with length for an end-loaded beam), we can get rid of the distortion of the rulers from centrifugal force by taking the limit, reducing the ruler length until it's small enough to not be appreciably distorted.

We can use the old-fashioned, lower-performance, "prototype meter" rulers with this technique, or we can use the modern SI rulers, based on light, that don't rely on any physical artifact. It doesn't really matter, except the modern techniques are both more accurate and MUCH more rigid.

One then does need a strategy for defining how to orient the infinitesimally small ruler segements so they form a "straight line". We can use the infinitesimal rulers to find the length of any curve, now we need to find the specific curve connecting two points that are a finite distance apart that we want to measure the length of.

We could make this pretty complicated, but basically the curve we are after is the shortest curve connecting the two points that are a finite distance apart. And we can use pretty standard methods to find this curve. The tricky part is that we do this in the quotient space. But once we realize we want to use a quotient space, having infinitesimal rulers available is sufficient for us to define length.

 

[PeterDonis]

I don't see this as an obstacle.

It's not an obstacle to constructing the quotient space and a reasonable physical interpretation of what the metric of that quotient space means. I'm just pointing out that any interpretation in this case has limitations because the congruence of worldlines that describes the object in question (the rotating disk) is not hypersurface orthogonal.

 

[Ibix]

No, it isn't. It's not a closed curve and does not form a spacelike submanifold, since some points on it are null or timelike separated.

Just to check I'm following you - all points on the helix as drawn (a line segment of finite length) are spacelike separated. However, as such it isn't a 1d manifold because it's not open (it has end points). If you fix that by connecting the end points somehow, the new piece of path is necessarily timelike or null somewhere.

 

[PeterDonis]

all points on the helix as drawn (a line segment of finite length) are spacelike separated.

Not if it goes enough of the way around the cylinder. Take a point on the helix and construct its future light cone. Points far enough around the helix, well before you reach the point where it has gone one complete turn around the cylinder, will be within that future light cone.

It is true that the helix itself is a "spacelike curve" in the sense that its tangent vector is everywhere spacelike. But that is not a sufficient condition for all pairs of points on the helix to be spacelike separated in the sense that no timelike or null curve exists between them. (And this is true even if you restrict to curves that lie completely within the 2-D cylinder.)

 

[Ibix]

Take a point on the helix and construct its future light cone. Points far enough around the helix, well before you reach the point where it has gone one complete turn around the cylinder, will be within that future light cone.

Of course - thanks.

 

[PAllen]

On further thought about idealized tape measure and geometry of the quotient manifold of the rotating disc, I am going to half agree with Dr. Greg.

Half? Well I can think of at least two ways to use a tape measure on rotating disc. I believe one of them might agree with quotient space geometry.

Note that the congruence orthogonal spiral (which is only quasilocally spacelike, as normally defined) of @Ibix is irrelevant to this discussion for two reasons:

1) it does not reflect the geometry of the quotient manifold
2) it is certainly not a geodesic of the quotient manifold.

With that out of the way, here are two ways to use a tape measure on a rotating disc surface:

1) Local to one observer, mark lines on tape, then extend it to some other disc observer and have them pull it taught. Have them mark where they are, and communicate the result to the other end. I claim this is what would normally be thought of as using a tape measure, and this will not match the quotient space metric.

2) First extend the tape measure between two disc surface observers, pulling it taught. Then have a string of observers along it put markings on, local to them, starting from the what the next nearest observer did. Then communicate the result from one end to the other. I believe this, in an appropriate limit, for an ideal tape measure, will match length of a quotient space geodesic between the end observers. However, if the tape measure is reeled back to either end, it will appear unevenly marked.

So I will agree there is some physical procedure that corresponds, over finite distances. to quotient space metric computations.

 

[pervect]

On further thought about idealized tape measure and geometry of the quotient manifold of the rotating disc, I am going to half agree with Dr. Greg.

Half? Well I can think of at least two ways to use a tape measure on rotating disc. I believe one of them might agree with quotient space geometry.

Note that the congruence orthogonal spiral (which is only quasilocally spacelike, as normally defined) of @Ibix is irrelevant to this discussion for two reasons:

1) it does not reflect the geometry of the quotient manifold
2) it is certainly not a geodesic of the quotient manifold.

With that out of the way, here are two ways to use a tape measure on a rotating disc surface:

1) Local to one observer, mark lines on tape, then extend it to some other disc observer and have them pull it taught. Have them mark where they are, and communicate the result to the other end. I claim this is what would normally be thought of as using a tape measure, and this will not match the quotient space metric.

2) First extent the tape measure between two disc surface observers, pulling it taught. Then have a string of observers along it put markings on, local to them, starting from the what the next nearest observer did. Then communicate the result from one end to the other. I believe this, in an appropriate limit, for an ideal tape measure, will match length of a quotient space geodesic between the end observers. However, if the tape measure is reeled back to either end, it will appear unevenly marked.

So I will agree there is some physical procedure that corresponds, over finite distances. to quotient space metric computations.

I'm not quite sure I understand procedure 2, I'd suggest the following simple one for measuring the circumference of the disk.

Make up a large number of short rods. You could replace them with taunt tape measures, as well, or little radar sets. It's simpler with rods, though. For more simplicity, make all of the rods of uniform, length.

Count how many of these little rods you can fit around the circumference of the disk.

Multiply the length of each rods by the number of rods. In the limit as the rods become shorter and shorter, the result is the circumference of the disk.

 

[PAllen]

I'm not quite sure I understand procedure 2, I'd suggest the following simple one for measuring the circumference of the disk.

Make up a large number of short rods. You could replace them with taunt tape measures, as well, or little radar sets. It's simpler with rods, though. For more simplicity, make all of the rods of uniform, length.

Count how many of these little rods you can fit around the circumference of the disk.

Multiply the length of each rods by the number of rods. In the limit as the rods become shorter and shorter, the result is the circumference of the disk.

I am not concerned with measuring the circumference. I am talking about measuring the distance between two arbitrary particles of the disc via a taught tape measure (which should ideally be a geodesic of the quotient manifold geometry . I thought that was clear, since I never mentioned circumference.

 

[A.T.]

... but those individual spacelike curves cannot be "assembled" into a single closed spacelike curve that describes "the circumference of the disk at an instant of time"...

If the geometry of the disc doesn't change over time, that "at an instant of time" part might be irrelevant for practical purposes.

So I will agree there is some physical procedure that corresponds, over finite distances. to quotient space metric computations.

If you wanted to build an already rotating truss structure from pre-fabricated short beams (not spin one up), the quotient space metric would tell you how many of these short beams you need to order.

 

[PeterDonis]

If the geometry of the disc doesn't change over time, that "at an instant of time" part might be irrelevant for practical purposes.

This is what the quotient space construction does: it makes use of the fact that the congruence of worldlines describing the disk is stationary (which is what "doesn't change over time" means in the math) to derive an abstract 3-dimensional space in which each point corresponds to a worldline. This abstract space has a geometry and a metric that can be viewed as "the geometry of the disk". However, this abstract space does not correspond to any 3-dimensional subspace of the 4-dimensional spacetime, and that means it violates some intuitions about what "the geometry of the disk" means physically.

 

[pervect]

I am not concerned with measuring the circumference. I am talking about measuring the distance between two arbitrary particles of the disc via a taught tape measure (which should ideally be a geodesic of the quotient manifold geometry . I thought that was clear, since I never mentioned circumference.

OK, let's consider that case. I would say that we have a bucket of rods of uniform, very small, length, and we ask the question "what's the least number of rods can we use to go from one mark on the surface of the disk to another". Technically, we are taking the limit as the length of the rods grows shorter and shorter.

Do you think your idea of pulling a tape measure taut gives the same answer as this approach, or a different answer? If you think it's different, we might have to delve into the mathematical representation of the tape measure some more.

You claim that there is some technique that gives a different answer for the distance, but I don't quite follow what it is. Let's focus on that, the technique you claim gives a different answer.

1) Local to one observer, mark lines on tape, then extend it to some other disc observer and have them pull it taught. Have them mark where they are, and communicate the result to the other end. I claim this is what would normally be thought of as using a tape measure, and this will not match the quotient space metric.

I've been trying to imagine what you are saying here, an failing. I think of a "mark on the tape", and also "a mark on the disk", that I used earlier, as necessarily being some worldline in the congruence of worldlines that represents the spinning disk.

As long as the tape is static on the disk (not vibrating), marks on the disk should be the same as marks on the tape. And both are represented by worldlines that are in the congruence of worldlines that represent the spinning disk.

You seem to be claiming that we cannot use the quotient manifold to intepret the idea of the distance between worldlines as a distance between points in the quotient space. But I don't see why you are claiming this. Basically, to my mind, marks on the disk, marks on the tape (when the tape is at rest on the disk), and points in the quotient manifold all represent the same thing in different words.

 

[PeterDonis]

As long as the tape is static on the disk

But the tape can't stay static on the disk during the process of it being pulled taut after marks have been made on it. So if the marks are made on the tape before it is pulled taut, while the tape is restricted to a single local region of the disk, as @PAllen describes, then the relationship between marks on the tape and marks on the disk will change when the tape is pulled taut between points that are not restricted to a single local region of the disk.

On a disk that was not rotating, it would be possible to pull the tape taut in a way that minimized distortion of the tape (and in the limit eliminated it) so that there would be no change in the relationship between marks on the tape and marks on the disk. But this is not possible for a rotating disk. This is one of the counterintuitive consequences of the congruence of worldlines that describes the disk not being hypersurface orthogonal.

You seem to be claiming that we cannot use the quotient manifold to intepret the idea of the distance between worldlines as a distance between points in the quotient space.

That's not the issue; distances in the quotient space can indeed be interpreted as one possible meaning of "distance between worldlines". But this meaning of "distance between worldlines" does not correspond to other possible meanings (in terms of various possible physical realizations) in the way that it would if the disk were not rotating. Basically, the rotation of the disk makes different interpretations of "distance" give different answers, which would give the same answers for a non-rotating disk.

 

[pervect]

That's not the issue; distances in the quotient space can indeed be interpreted as one possible meaning of "distance between worldlines". But this meaning of "distance between worldlines" does not correspond to other possible meanings (in terms of various possible physical realizations) in the way that it would if the disk were not rotating. Basically, the rotation of the disk makes different interpretations of "distance" give different answers, which would give the same answers for a non-rotating disk.

It sounds like we agree that the quotient manifold represents the space of the disk.

Do we agree that in the limit as the distance between two points on the disk (two worldlines in 4d spacetime) approaches zero, i.e. for very close points / worldlines, that physically significant notion of distance must agree with the radar notion of distance?

If so, then it follows we agree that the quotient space represents the "space" of the disk, and we agree on the metric associated with the quotient space. Now, while I could imagine using some connection other than the Levi-Civita connection of the metric to define distances (by measuring distances along geodesics defined by this alternative connection), I can't see any motivation for doing such a thing.

 

[PeterDonis]

It sounds like we agree that the quotient manifold represents the space of the disk.

For a suitable interpretation of "the space of the disk", yes. One of the points I'm trying to make is that the ordinary language phrase "the space of the disk" does not have a single unique interpretation. It just so happens that in simpler cases like a non-rotating disk, the different possible interpretations end up leading to the same result. But in the case of the rotating disk, they don't.

Do we agree that in the limit as the distance between two points on the disk (two worldlines in 4d spacetime) approaches zero, i.e. for very close points / worldlines, that physically significant notion of distance must agree with the radar notion of distance?

In the limit of close enough points/worldlines, all of the different possible notions of distance agree.

If so, then it follows we agree that the quotient space represents the "space" of the disk

No, it doesn't, because "the space of the disk" is a global concept, not a local concept. As long as we're clear about exactly which global concept we're talking about, there's no problem. But just saying "the space of the disk" doesn't pick out a single unique global concept. That's why terms like "quotient space", which is unambiguous, are preferred.

 

[A.T.]

However, this abstract space does not correspond to any 3-dimensional subspace of the 4-dimensional spacetime, and that means it violates some intuitions about what "the geometry of the disk" means physically.

To me personally, measuring a stationary geometry by laying out rulers on it seems quite intuitive and physical, while a 3-dimensional subspace of the 4-dimensional spacetime is a more abstract idea.

 

[PeterDonis]

measuring a stationary geometry by laying out rulers on it seems quite intuitive and physical

"Laying out rulers on it" is ambiguous. If it means the particular procedure that has been described, where each ruler only covers an infinitesimal distance and you have to do a careful calculation to combine all the infinitesimal ruler measurements into a total distance between two points that are not infinitesimally close, then I'm not sure that would seem "quite intuitive and physical" to everyone. What I think is "quite intuitive in physical" is the idea of just laying a single ruler between two points, but on the rotating disk you simply can't do that the way you can on a non-rotating disk.

 

[A.T.]

If it means the particular procedure that has been described, where each ruler only covers an infinitesimal distance and you have to do a careful calculation to combine all the infinitesimal ruler measurements into a total distance between two points that are not infinitesimally close, then I'm not sure that would seem "quite intuitive and physical" to everyone.

This is basically how humans have been surveying the curved surface of the Earth for centuries. To me it seems less abstract than slices through space-time.