Would circumnavigating the Universe allow one way measurement of light's speed?

[A.T.]

I am thinking of the Sagnac effect which measures only rotation and does not measure anisotropy of the one way speed of light.

I would say the Sagnac effect measures the anisotropy of the one way speed of light in a rotating reference frame.

 

[Dale]

I would say the Sagnac effect measures the anisotropy of the one way speed of light in a rotating reference frame.

Consider a rotating reference frame and a family of Sagnac interferometers throughout the rotating reference frame and all at rest with respect to the reference frame. For convenience we will use cylindrical $(r, \theta, z)$ coordinates with $\hat r$, $\hat \theta$, and $\hat z$ being the radial, tangential, and longitudinal unit vectors respectively, and of course the rotation is about the $r=0$ axis with an angular velocity $\omega \hat z$ and the interferometers are limited to the region $r<c/\omega$.

Now, at each point the one way speed of light in this frame is anisotropic about the $\hat \theta$ direction, and the degree of anisotropy increases with increasing $r$. In contrast, at each point the rotation is about the $\hat z$ direction and the rate of rotation is $\omega$ independent of $r$.

Now, consider the measurements of the Sagnac interferometers. The Sagnac interferometers detect no signal when they are oriented in the $\hat r$ or $\hat \theta$ directions, they only detect signal when oriented in the $\hat z$ direction. So the Sagnac signal is anisotropic about the $\hat z$ direction. Furthermore, the magnitude of the Sagnac signal is independent of $r$ and is proportional to $\omega$. Therefore, the Sagnac signal is strictly a measure of rotation and definitively not a measurement of the anisotropy of the one way speed of light in the rotating frame.

This is an occasional source of confusion because the Sagnac effect can be derived in the rotating frame using the one-way speed of light in that frame, but despite the derivation, it does not possesses the right signal behavior to be considered a measurement of the one way speed of light. This should not be terribly surprising since the Sagnac effect can also be derived in an inertial frame with no anisotropy in the one way speed of light. So it must be measuring something else, and that something is the rotation which is present in both derivations and which fits the observed signal behavior.

 

[Sagittarius A-Star]

This should not be terribly surprising since the Sagnac effect can also be derived in an inertial frame with no anisotropy in the one way speed of light. So it must be measuring something else, and that something is the rotation which is present in both derivations and which fits the observed signal behavior.

Also, the relativistic calculation shows, that the $\Delta t$ of the Sagnac effect is independent of the signal speed. It is a relativistic effect, related to the Born coordinates.

Rather remarkably, this is independent of the signal velocity! A beam of light sent around the disk each way, or two people strolling slowly around the disk each way, will encounter the same difference in the time it takes to go all the way around.

Source:
http://www.physicsinsights.org/sagnac_1.html

 

[Sagittarius A-Star]

Let's say one these observers sends 2 light signals simultaneously in opposite directions, and checks if they reach him again simultaneously. Is this not a test for isotropic propagation speed?

I would say no.

Suppose you have an observer $A$ with a clock $A$ in the light sphere and a second observer $B$ with a clock $B$ on the opposite side of the black hole in the light sphere.

Clock synchonization:
Observer $A$ sends a light pulse at 01:00 A.M. according to clock $A$ via the light sphere to observer $B$, who sets his clock $B$ to 01:00 A.M. just as the light pulse arrives and reflects with a mirror the light pulse back along the same semi circle.

Then the light moved instantaniously from $A$ to $B$ and back with c/2.

The same experiment is repeated via the other semi circle. Also then the light moved from $A$ to $B$ instantaniously and back with c/2.

If you define a related sphere-coordinate system, then $A$ can send a light pulse around the whole circle, which moves instantaniously in the 1st semi circle and with c/2 in the 2nd semi circle. But $A$ measures then only the 2-way speed of light c.

 

[Vanadium 50]

Let's go back to basics for a moment.

SR says that c is c. A different speed in different directions would mean SR is wrong. GR builds on SR, and cosmology builds on GR. So it's not clear that this question even makes sense.

Actually, I think it doesn't make sense as asked, although it may be possible to pose a well-defined similar question. Or maybe not? If you don't have spacetime, what does it mean for spacetime to be curved and the universe closed? If you have spacetime without SR, how exactly does this work?

Next, let's examine why you can't measure the one-way speed of light. You need a distance, a time, and a synchronization convention.

Sagredus: We don't need a synchronization convention because the light comes back to you.
Salvatus: Provided you are at the same point. How do you know that?
Sagredus: Um, absolute space?
Salvatus: There is no such thing. How do you know you receive the light at the same place you emitted it? After all, there are two parts to a synchronization convention - space and time.
Sagredus: If you move from A to B, you feel the acceleration as you start and stop.
Salvatus: Not if you are moving at constant velocity.
Sagredus: But you can tell if you are moving!
Salvatus: How?
Sagredus: Relative to the fixed stars! Or, if you like, in a frame where the totality of the rest of the universe has no net momentum with respect to you.
Salvatus: That is picking a synchronization convention. One can do this even in expanding FRW universe, if one could imagine such a crazy thing. The time is measured in comoving frames relative to the big bang. A person at one place shines a light at another person, and tells that person how many seconds it has been since the big bang in his comoving frame. The second person knows the time the pulse was sent, how to synchronize their clocks, the distance between himself and the source, and off he goes to calculate. He will of course get c, because that's a consequence of his synchronization convention.

So, I believe that if it were possible to replace the question asked by a similar question that is well-defined, the answer would be "you are picking a synchronization convention. The most obvious one, so obvious that you don't even know you are picking it, gives c as the result of the proposed experiment."

 

[PeterDonis]

They do not have multiple clocks on the same worldline running at different rates. But they do have two clocks at the same rate with an offset, at the "same" location(s) $\phi = 0$ and $\phi = 2\pi$, assigned to different events.

If $\phi = 0$ and $\phi = 2 \pi$ are the same location (same worldline), then the two clocks with different offsets, even if they run at the same rate, must still be assigning different times to the same event on that worldline. Which is not a valid coordinate chart.

 

[PeterDonis]

Let's say one these observers sends 2 light signals simultaneously in opposite directions, and checks if they reach him again simultaneously. Is this not a test for isotropic propagation speed?

Not unless you make an appropriate coordinate choice. What you are describing is a test for the Sagnac effect, and the only "objective" (i.e., invariant) property that tests for is, heuristically, whether or not your worldline has nonzero angular velocity relative to a particular global symmetry of the spacetime. With an appropriate choice of coordinates, you can interpret this nonzero angular velocity relative to the global symmetry as implying an anisotropic speed of light; but nothing requires you to adopt such coordinates.

 

[PeterDonis]

Wouldn't rotation also imply a proper acceleration?

No, in the sense of the simple presence of proper acceleration being an independent test for rotation. The "rotation" being referred to here is not the same as following a circular path in ordinary Minkowski spacetime.

In the "Minkowski rolled up into a cylinder" case and the Einstein static universe case, none of the observers in question have any proper acceleration, whether they are "rotating" with respect to the global symmetry of the spacetime or not.

In the "photon sphere around the black hole" case, all of the observers have proper acceleration, whether they are "rotating" or not. Their proper acceleration does vary with angular velocity--it is greatest for zero angular velocity, and approaches zero in the limit as the angular velocity approaches the maximum possible, the angular velocity corresponding to a light ray in the photon sphere circular orbit--but the proper acceleration is never exactly zero except for the light ray itself (its circular orbit is a geodesic).

 

[PeterDonis]

You could also pick the family of observers who needed the least tape to circumvent the universe.

There is only one family of observers who can even have a tape at rest relative to them that circumnavigates the universe. That is the family of observers whose worldlines are integral curves of the timelike Killing vector field that I have been calling a "global symmetry" of the spacetime--the one that is orthogonal to a set of closed $S^1$ curves that foliate the $R^1 \times S^1$ submanifold.

 

[PeterDonis]

SR says that c is c.

Actually, that's not what SR says in the modern view. In the modern view, you are perfectly free to do SR in a coordinate chart in which the coordinate speed of light is not always $c$. You don't even have to get into all the weird possibilities; the Rindler coordinate chart is a simple, well-known example.

What SR says in the modern view is that spacetime has an invariant geometric light cone structure and zero curvature (the latter condition is what distinguishes SR from GR, in which the curvature can be nonzero). How that structure "appears" in quantities like the speed of light can depend on your choice of coordinates; only the underlying geometric structure itself is invariant. As you note, the most common choice of coordinates when spacetime is flat is the one that makes the coordinate speed of light invariant and always $c$.

 

[Dale]

There is only one family of observers who can even have a tape at rest relative to them that circumnavigates the universe.

So, here is the issue in geometric terms as I see it. If you have such a tape then that constitutes a cylinder in spacetime. The tape can be given marks, each of which becomes a line along the length of the cylinder, and a pulse of light of the type we are interested in is a helix wrapping around the cylinder.

Now, if you take one of the mark lines and the two events where the light intersects it then there are only two relevant frame invariant measures you can make: the invariant length of the two paths between the two events. One is the proper time, but the other is null, not the distance.

The cylinder has timelike lines on it from the marks, but you can draw rather arbitrary spacelike lines around it. And many different sets of spacelike lines are valid, each representing a different synchronization convention. Those spacelike lines will give different values for the total circumference as well as different values for the local one way speed of light at each point.

So, in the end, whether this is a “one way” measurement or not it still suffers from the same ambiguity as a typical one way measurement. Namely, you still need to define a synchronization convention which changes the result both locally along the trip and overall.

 

[PeterDonis]

If you have such a tape then that constitutes a cylinder in spacetime.

Yes; the "worldsheet" of the tape is precisely the $R^1 \times S^1$ submanifold I have been referring to.

The tape can be given marks, each of which becomes a line along the length of the cylinder

Yes; these are the "at rest" worldlines I have been describing.

a pulse of light of the type we are interested in is a helix wrapping around the cylinder

Yes.

if you take one of the mark lines and the two events where the light intersects it then there are only two relevant frame invariant measures you can make: the invariant length of the two paths between the two events.

Yes, these are frame invariant measures, but not the only ones. As you note, the invariant length along the mark line defines an invariant proper time between the two "light intersection" events. But you can also define an invariant proper distance around the cylinder: just take the invariant length of the closed spacelike circle that intersects the mark line at one of the two"light intersection" events (either one will do--and in fact any event along the mark worldline will do--since the metric of the submanifold is stationary). And dividing that invariant proper distance by the invariant proper time gives a measurement of the speed of light.

many different sets of spacelike lines are valid, each representing a different synchronization convention

No. Only one set of spacelike lines is valid, because it is the only one that is a set of closed circles that each intersect every mark line at precisely one event. Every other set of spacelike lines intersects each mark line at multiple events, because the spacelike lines are helixes, not closed circles. And that means you can't use those spacelike lines to define a valid synchronization convention, because any such convention must lead to a foliation by spacelike curves that each only intersect a timelike curve once.

In other words, the $R^1 \times S^1$ submanifold in question has a unique foliation in terms of worldlines of marks on the tape and closed spacelike circles, picked out by its geometry. And that unique foliation can be used to measure the speed of light that circumnavigates the manifold.

 

[Dale]

But you can also define an invariant proper distance around the cylinder: just take the invariant length of the closed spacelike circle that intersects the mark line at one of the two"light intersection" events (either one will do--and in fact any event along the mark worldline will do--since the metric of the submanifold is stationary)

The problem is that “the closed spacelike circle” is not uniquely required. You can take any number of closed spacelike paths.

No. Only one set of spacelike lines is valid, because it is the only one that is a set of closed circles that each intersect every mark line at precisely one event. Every other set of spacelike lines intersects each mark line at multiple events, because the spacelike lines are helixes, not closed circles

This is not correct. There are also spacelike ellipses which are closed and intersect each mark line at one event. Think of slicing a baguette straight crosswise (one of your circles) or at an angle (one of my ellipses).

There are also more complicated slicings which are permissible, but more difficult to describe (like Pringles potato chips or waffle fries)

Every other set of spacelike lines intersects each mark line at multiple events, because the spacelike lines are helixes, not closed circles.

Are you thinking of rotating frames here?

 

[A.T.]

Sagredus: But you can tell if you are moving!
Salvatus: How?
Sagredus: Relative to the fixed stars! Or, if you like, in a frame where the totality of the rest of the universe has no net momentum with respect to you.

My understanding is that in a closed (cylindrical universe) there is a preferred rest frame:

https://www.tandfonline.com/doi/abs/10.1080/00029890.2001.11919789

We can find it by sending clocks around the universe, and comparing the proper times between two meetings. No need to refer to fixed stars or other references.

 

[PeterDonis]

There are also spacelike ellipses which are closed and intersect each mark line at one event.

Yes, you're right, this is true; my statement as I gave it was too strong.

I think these ellipses will not work as a synchronization convention, but I'll have to think some more to see if I can come up with a rigorous way to express why not.

Are you thinking of rotating frames here?

No. I'm thinking of the natural way a Lorentz transformation acts. Suppose we have a frame with the mark lines and closed circles (not ellipses--the fact that they are circles and not ellipses does geometrically pick out this particular foliation) as its timelike and spacelike axes. If we now transform to a "moving frame" whose time axis is a helical timelike worldline that wraps around the cylinder, with some velocity $v$ relative to the mark lines of the original frame, then the corresponding spacelike axis of this "frame" is a helical spacelike curve like the ones I described. (To see why this must be the case, unroll the cylinder.) And of course that means this "frame" is not actually a valid coordinate chart globally; it can only be used on a local patch that does not wrap all the way around the cylinder.

 

[PeterDonis]

My understanding is that in a closed (cylindrical universe) there is a preferred rest frame

Yes. It is the frame whose "at rest" timelike worldlines go "straight up the cylinder" and don't wrap around it at all, and whose spacelike slices of constant time are circles.

The question is what other frames it is possible to define.

 

[A.T.]

There is only one family of observers who can even have a tape at rest relative to them that circumnavigates the universe.

Can this one family of observers (O_A), which has tape (T_A) at rest relative to them that circumnavigates the universe, synchronize their clocks using the Einstein convention without any discontinuity?

If yes, O_A could do this:

1) Build a second tape (T_B) around the universe, side by side with the first one

2) Cut it in small segments

3) Accelerate the T_B segments (simultaneously according to their frame) parallel to T_A.

What will O_A observe after the acceleration? A moving tape T_B with some gaps, because the T_B segments are contracted in their frame?

Let's say they accelerate a second family of observers (O_B) along with the tape segments. Will O_B not observe their local T_B segments at rest?

If the gaps in T_B are a problem, O_A could build two new tapes T_B1 and T_B2. Accelerate T_B1 segments until they are half their proper length, then accelerate T_B2 to fill the gaps in T_B1, to form a continuous moving T_B.

Will O_B not observe their local T_B at rest, just like O_A observe their local T_A at rest?

 

[PeterDonis]

Can this one family of observers (OA), which has tape (TA) at rest relative to them that circumnavigates the universe, synchronize their clocks using the Einstein convention without any discontinuity?

They can if they make sure to not use any light pulses that circumnavigate the universe while doing the synchronization.

O_A could do this

No, he can't. It has nothing to do with clock synchronization; it has to do with the fact that it's impossible to accelerate all the tape segments (or observers that go along with them) so that they remain at rest relative to each other, because of the $S^1$ topology along the "spatial" dimension of the submanifold. (To see why, unroll the cylinder and compare what would have to happen in the cylinder case with what is possible in the ordinary Minkowski $R^2$ submanifold case, i.e., with a family of Rindler observers.)

 

[A.T.]

... it's impossible to accelerate all the tape segments (or observers that go along with them) so that they remain at rest relative to each other,...

Before jumping into what O_B will observe in their rest frame, let's clarify what O_A will observe after the acceleration of the T_B segments. Will all O_A observe the segments of tape T_B moving at the same constant speed? This is basically what the initial rest frame in Bell's spaceship paradox observes after the acceleration.

 

[anuttarasammyak]

If we know radius of universe R, may we say light speed c is
$$c=\frac{2\pi R}{t}$$
where t is measured from the start and the goal ?

 

[Dale]

No. I'm thinking of the natural way a Lorentz transformation acts.

The purpose of the whole thread is to explore if the described experiment leads to an assumption-free measurement of the one way speed of light. Since the Lorentz transform has that assumption built in we must not limit ourselves to the Lorentz transform. All valid coordinate transforms must be admitted.

If we now transform to a "moving frame" whose time axis is a helical timelike worldline that wraps around the cylinder, with some velocity v relative to the mark lines of the original frame, then the corresponding spacelike axis of this "frame" is a helical spacelike curve like the ones I described.

By “corresponding” here I think you are inadvertently using the very assumption we are trying to avoid.

Before changing the time axis, let’s see what can be done keeping the time axis straight along the length of the cylinder. Now, imagine cutting the cylinder lengthwise along the time axis and unrolling it. In order to represent the fact that it is a cylinder we simply treat it as periodic in the unrolled view. Then the spacelike lines forming the circular surfaces of simultaneity are straight horizontal lines.

The ellipses I describe are sinusoidal lines. As long as they are periodic in the unrolled view then there is no problem when we roll the cylinder back together. Any smooth periodic function whose slope never exceeds 45 deg can be used. And choosing that function determines both the overall length of the spacelike path and also the local one-way speed of light at each point along the null path.

the corresponding spacelike axis of this "frame" is a helical spacelike curve like the ones I described

I agree, those straight lines are not periodic. The horizontal straight lines are the only straight lines that are periodic. I am not certain if that periodicity is actually mandatory, but it seems reasonable to expect that it is, so I would enforce it for now.

 

[Dale]

If we know radius of universe R, may we say light speed c is
$$c=\frac{2\pi R}{t}$$
where t is measured from the start and the goal ?

The issue is that we don’t know the radius. Or rather, defining the radius requires an assumption and a different assumption can be made. This is the same problem as with a traditional one-way measurement. It requires an assumption.

 

[Vanadium 50]

The issue is that we don’t know the radius. Or rather, defining the radius requires an assumption and a different assumption can be made.

Suppose you could see the same galaxy multiple times. The first image is at x, the second at x + R, the third at x + 2R and so on. You could use Cepheids to measure the distances.

I think the bigger issue is how to tell you are at rest.

 

[PeterDonis]

This is basically what the initial rest frame in Bell's spaceship paradox observes after the acceleration.

Ah, I see, you were thinking of a Bell spaceship paradox acceleration, not a Rindler acceleration.

For the Bell acceleration case, yes, you could realize it in the cylinder submanifold, and your observers $O_A$ would observe the tape segments after acceleration to be all moving with the same speed $v$ relative to them and to be length contracted (so there would be gaps between them).

Your observers $O_B$, also moving at speed $v$ relative to the $O_A$ observers, would observe the tape segments to be at rest relative to them. They could also use Einstein clock synchronization to synchronize their clocks "locally", i.e., with near neighbor $O_B$ observers--as long as they made sure not to use light signals that circumnavigated the universe, just as with the $O_A$ observers--but they would find that they could not fit together their individual "local" Einstein synchronizations into a single global Einstein synchronization that worked for all of them; any such attempt would end up assigning the same clock time to multiple events on at least one of the $O_B$ observer worldlines (because the spacelike curves of constant time in any such synchronization are helixes, not closed circles, and will end up intersecting the same timelike worldline at multiple events).

So being able to set up a single global Einstein clock synchronization is a property that only the $O_A$ family of observers has.

 

[PeterDonis]

imagine cutting the cylinder lengthwise along the time axis and unrolling it. In order to represent the fact that it is a cylinder we simply treat it as periodic in the unrolled view. Then the spacelike lines forming the circular surfaces of simultaneity are straight horizontal lines.

Yes, this is what I mean by "unrolling the cylinder".

The ellipses I describe are sinusoidal lines.

Yes, agreed.

However, the ellipses are not what you get when you do a standard Lorentz transformation on the unrolled cylinder. When you do that, you get a set of non-vertical timelike lines and a set of non-horizontal spacelike lines; when the cylinder is rolled back up, those become a set of timelike helixes winding around the cylinder, and a set of spacelike helixes winding around the cylinder, and there are multiple intersections between any single timelike helix and any single spacelike helix, which means this construction does not define a valid coordinate chart on the cylinder.

I agree that the above means we can't use the "standard Lorentz transformation" construction to obtain a different coordinate chart from the "vertical-horizontal" one for use in a one-way speed of light calculation. I also agree that your ellipses can be used to obtain such a different coordinate chart. It just won't be a coordinate chart that looks like an "inertial frame". I also don't think the surfaces of constant coordinate time in such a chart will correspond to any realizable clock synchronization among observers at rest in the chart, i.e., whose worldlines have unchanging spatial coordinates.

In fact, it's not clear to me exactly how the "time axis" of such a chart (which would define the set of worldlines with unchanging spatial coordinates) would be defined. Would it be the "vertical" axis (i.e, the same as the original "vertical-horizontal" chart)? Or would the timelike curves be "tilted" (so they are helixes winding around the cylinder)? If so, by how much? Since the slope of the ellipses varies from point to point (because in the unrolled cylinder they are sinusoids), it is impossible to pick a single "tilt" for the timelike curves that would be orthogonal to the ellipses everywhere. Or would the timelike curves have to have varying "tilt" as well (so they would also be sinusoids in the unrolled cylinder) to try to make them always orthogonal to the spacelike curves (if that is even possible)?

 

[Dale]

However, the ellipses are not what you get when you do a standard Lorentz transformation on the unrolled cylinder.

Agreed, but the whole point is to not limit ourselves to the simultaneity convention of the Lorentz transform.

It just won't be a coordinate chart that looks like an "inertial frame".

Agreed. Inertial frames include the isotropic one way speed of light assumption by definition, so we should be explicitly clear that these anisotropic c frames are not inertial.

I also don't think the surfaces of constant coordinate time in such a chart will correspond to any realizable clock synchronization among observers at rest in the chart, i.e., whose worldlines have unchanging spatial coordinates.

I disagree here. Any of these conventions are perfectly realizable. Given anyone of these conventions you know the distance to your neighboring clock and the one way speed of light from them to you. So when you receive a time stamped signal from your neighbor you can correct it for the known distance and speed to synchronize your clock.

Would it be the "vertical" axis (i.e, the same as the original "vertical-horizontal" chart)?

I would leave the time axis unchanged. I don’t know if it is possible to use a tilted axis in this sort of spacetime. I would need to see a proof before being willing to try it with a tilted axis.

it is impossible to pick a single "tilt" for the timelike curves that would be orthogonal to the ellipses everywhere

Yes, clearly. But with anisotropic one way speed of light discussions that is expected.

Or would the timelike curves have to have varying "tilt" as well (so they would also be sinusoids in the unrolled cylinder) to try to make them always orthogonal to the spacelike curves (if that is even possible)?

That is interesting. I had not considered that.

 

[PeterDonis]

I would leave the time axis unchanged.

Ok. That's the simplest way to do it, although I think it would also be possible to use a "tilted" time axis--it would just complicate the construction by clock offset that I describe below. (The "tilt" would also have to be limited, since we can't have more than one intersection between any timelike "grid line" and any spacelike "grid line"; I think that means the "tilt" of the timelike lines, relative to the "vertical" ones, would have to be no greater than the "tilt" of the spacelike ellipses relative to the spacelike circles.)

Given anyone of these conventions you know the distance to your neighboring clock and the one way speed of light from them to you.

If you need to already know the one-way speed of light in order to do the synchronization, then you can't use the synchronization convention in order to measure the one-way speed of light. I thought the point was to define different synchronizations that could be independently set up, so that they could be used as a basis to measure the one-way speed of light.

That said, I think I was wrong about there being no way to setup such a synchronization (without prior knowledge of the one-way speed of light). The "ellipse" synchronization convention basically amounts to giving a space-dependent offset to each clock; the offset will be $A \cos \left ( x / C \right)$, where $x$ is the spatial coordinate of the clock in the original "rest" frame (the one whose spacelike surfaces of constant time are circles), $C$ is the "circumference of the universe" (which is also the circumference of the spacelike circles), and $A$ is the maximum amplitude of the offset (which corresponds to the "tilt" of the ellipses relative to the circles). So given a set of clocks synchronized in the original "rest" frame, it is easy to adjust them to any desired "ellipse" synchronization. However, note that this construction makes no use of light signals (more precisely, none once the synchronization in the original "rest" frame is complete).

 

[Dale]

If you need to already know the one-way speed of light in order to do the synchronization, then you can't use the synchronization convention in order to measure the one-way speed of light.

Yes, of course. It is just an arbitrary choice of convention. The OP I think intended it to be a direct measurement, but I am convinced that it requires a simultaneity convention.

 

[PeterDonis]

The OP I think intended it to be a direct measurement, but I am convinced that it requires a simultaneity convention.

I agree, but I think it's worth explicitly laying out some of the complexities.

We have an invariant timelike interval: the interval between the emission and reception, by the same observer, of a light ray that "circumnavigates the universe" one time.

In order to obtain a value for "the speed of light" (leaving aside the question, which has been discussed elsewhere in this thread, of whether it is appropriate to call this speed a "one-way" speed, a "two-way" speed, both, or neither) from that, we need to have a distance. That means we need to have some way of specifying a closed spacelike curve (or family of curves all of the same circumference that foliate the $R^1 \times S^1$ submanifold--since the manifold is stationary, that is possible) whose circumference will be the distance we need. This is equivalent to specifying a simultaneity convention.

There are at least two ways of obtaining such a family of curves:

(1) We could impose some kind of constraint on the speed of light; for example, we could impose the constraint that it must be isotropic; or we could impose some kind of precisely specified anisotropy. The "isotropic" specification would give us the circles as our spacelike curves; different specifications of anisotropy would give us different sets of ellipses.

(2) We could impose some other kind of constraint that has nothing to do with the speed of light. For example, we could say that the spacelike curves must all be orthogonal to the timelike worldlines of our family of observers (or the segments of our tape measure that runs around the universe), or that they must have some specified "tilt" with respect to those worldlines. The "orthogonal" specification would give us the circles; different specifications of "tilt" would give us different sets of ellipses.

Method #1 would correspond to what you were describing in post #61: we know what we want the speed of light to look like and we deduce everything else accordingly.

Method #2 would correspond to what I was describing in post #62: we have an independent way of obtaining the simultaneity convention, and we deduce the speed of light from that, however it comes out.

Either way, though, as I said, I agree you need the family of curves, which is equivalent to a simultaneity convention, because that's the only way to obtain the distance you need to convert the invariant timelike interval you have into a speed.

 

[anuttarasammyak]

The issue is that we don’t know the radius. Or rather, defining the radius requires an assumption and a different assumption can be made.

Thank you . So I know we do not have sufficient information to answer the question of OP.

I imagine if we emit light to all the directions as the start, we would observe lights from all the directions at the same time or not as the goal to know anisotropy of universe or speed of light.

 

[A.T.]

So being able to set up a single global Einstein clock synchronization is a property that only the $O_A$ family of observers has.

So the $O_B$ observers can also have a continuous tape $T_B$ at relative rest to them around the universe (build as described below), they just cannot synch their clocks in the standard way?

$O_A$ could build two new tapes $T_{B1}$ and $T_{B2}$. Accelerate $T_{B1}$ segments until they are half their proper length, then accelerate $T_{B2}$ to fill the gaps in $T_{B1}$, to form a continuous moving $T_B$.

If $O_B$ rely only on their resting tape $T_B$ to measure the size of the universe (no clocks needed), they would get twice the value that $O_A$ gets?

But $O_B$ would also see the moving contracted tape $T_A$ with just half the number of meter marks spanning the same distance?

 

[PeterDonis]

So the $O_B$ observers can also have a continuous tape $T_B$ at relative rest to them around the universe (build as described below), they just cannot synch their clocks in the standard way?

If $O_B$ rely only on their resting tape $T_B$ to measure the size of the universe (no clocks needed), they would get twice the value that $O_A$ gets?

But $O_B$ would also see the moving contracted tape $T_A$ with just half the number of meter marks spanning the same distance?

The short answer is: it's more complicated than that.

The worldlines of the $O_A$ observers, and of their segments of tape, are timelike curves that go vertically up the cylinder, and don't wrap around it at all. If they Einstein synchronize their clocks, then a snapshot of their tape at an instant of time is just a closed circle around the cylinder. The length of the tape is then just the circumference of the circle.

The worldlines of the $O_B$ observers, and of their segments of tape, are then timelike helixes that wrap around the cylinder. They can't Einstein synchronize their clocks, as I have already said; but that means they also cannot define any set of surfaces of constant time, which would be needed to define the "length of the tape" according to these observers, that matches up with the local notion of "rest" of any of these observers. That is because, as I have already noted, the local notion of "rest" of each of these observers (which is the one that you are implicitly using when you talk about the $O_B$ tape being "contracted") is the spacelike curve that is orthogonal to their worldlines, and that curve is not closed--it is a spacelike helix that winds around the cylinder.

So there is no closed spacelike surface that defines a global notion of "space" for the $O_B$ observers and has the properties you are assuming. So the only way to define such a global notion of "space" is as a quotient space (similar to what is done with a rotating disk in discussions of how to resolve the Ehrenfest paradox). I think that the "circumference of the universe" in this quotient space will indeed be longer (by a factor of 2 in your specific example) than the circumference of the circle for the $O_A$ observers, but I have not done the math to check. However, this "space" also has weird properties such as the speed of light being anisotropic (it is slower in the "co-rotating" direction than the "counter-rotating" direction), and I believe at least one of those speeds of light will be significantly greater than $c$.

Because the global "space" for the $O_B$ observers is a quotient space, not a spacelike surface in the actual spacetime, it's not clear to me how the $O_A$ tape will look in this space, or if that notion is even well-defined.

 

[A.T.]

So the only way to define such a global notion of "space" is as a quotient space (similar to what is done with a rotating disk in discussions of how to resolve the Ehrenfest paradox). I think that the "circumference of the universe" in this quotient space will indeed be longer (by a factor of 2 in your specific example) than the circumference of the circle for the $O_A$ observers, but I have not done the math to check. However, this "space" also has weird properties such as the speed of light being anisotropic (it is slower in the "co-rotating" direction than the "counter-rotating" direction), and I believe at least one of those speeds of light will be significantly greater than $c$.

I agree that this very similar to the rotating frame, but here you have no proper acceleration. So locally you have perfect symmetry between members of $O_A$ and $O_B$.

It's only when $O_B$ starts counting, when they realize that a moving tape-loop $T_A$, has 1/2 the number of segments, each only 1/2 the length, compared to the resting tape-loop $T_B$, laying side by side to $T_A$.

Weird indeed.

 

[StandardsGuy]

No, but it would be a measurement of the size of the universe in meters.

How would you know that you are at rest when you make this measurement, or does it matter?

 

[PeterDonis]

How would you know that you are at rest when you make this measurement

By the absence of a Sagnac effect: light pulses that you send in opposite directions around the universe come back to you at the same instant.