# I Relativistic centrifugal force

#### m4r35n357

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.
In the Rindler spacetime I am used to the proper time of an extended accelerated object running at different rates along the object (i.e. slower at the rear where the acceleration is greatest). Are you just saying that proper time passes at different rates with radius? It sounds like more than that to be honest but I had to ask.

#### PAllen

In the Rindler spacetime I am used to the proper time of an extended accelerated object running at different rates along the object (i.e. slower at the rear where the acceleration is greatest). Are you just saying that proper time passes at different rates with radius? It sounds like more than that to be honest but I had to ask.
I am saying more. Nearby clocks slowly diverging form each other is not the issue. Each world line of a congruence has a local notion of space around it evolving in time. That space is the 3 surface 4 orthogonal to their world line. For the Rindler congruence, you can knit these together to represent the whole body at a moment of time. In the Rindler case, different particles will label the time of consecutive moments of the body as a whole differently, but all agree on the set of events that make up a moment for the body as a whole, by knitting together their local space.

For the stationary rotating congruence, there is no way to do this at all. Any set of events you try to say is a moment of the disk as a whole will fail to represent a locally defined momentary space for some particles of the disk.

This is what is captured in the notion of hypersurface othogonality.

#### Dale

Mentor
So the global metric for the spinning disk is Minkowski?
Yes. The metric is a tensor, so it is by design independent of the coordinate system. However, you typically represent a tensor by its components in some basis (usually a coordinate basis). As you change your coordinates you change your coordinate basis vectors and therefore you change the components of the tensor.

So the metric tensor is the Minkowski metric, but the components will not be the same as the components in a standard inertial frame.

#### pervect

Staff Emeritus
As far as hypersurface orthagonality goes, look at the diagram below. I didn't draw it, the original was drawin by another PF poster, Ibix.

Ignoring the fine points of Lorentzian geometry for now, try to imagine a line that is everywhere orthogonal to the slanted, green, vertical lines on the diagram. You can find segments of such a curve, but it isn't a closed curve. The vertical lines represent the notion of "time" on a 2 space + 1 time space-time diagram on a rotating hoop, and the curve that we asked for, one that is orthogonal to all the lines, would represent the "hypersurface" that's orthogonal to time. The problem is that the curves are not closed.

The problem becomes more acute if you consider a cylinder rather than a hoop, but it's hard to draw the needed lines clearly of this situation on a 2d piece of paper.

That's basically enough to get the idea. The hard part really is getting people to draw space-time diagrams, or to get them to look at and interpret ones that others have already drawn. While I recognize the difficutly, I don't know what to do about it .

#### Ibix

imagine a line that is everywhere orthogonal to the slanted, green, vertical lines on the diagram
I was going to repost this myself last night, but got interrupted and forgot. The slanted lines are actually cyan. In the left hand image their bottom ends are joined by a green line, which is the line you are talking about. It is everywhere orthogonal to the cyan lines, but does not close because it forms a helix.

I can post a zoomed in view of the relevant part of the diagram later if that would be helpful. Currently on my phone and I don't have the necessary software or files.

#### 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

Staff Emeritus
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

Gold Member
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

Mentor
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

Gold Member
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

Mentor
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

Gold Member
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

Mentor
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

Mentor
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

Gold Member
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

Mentor
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

Staff Emeritus
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

Mentor
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

Mentor
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.

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#### pervect

Staff Emeritus
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.

"Relativistic centrifugal force"

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