# Sagnac explanation question

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Here is a typical explanation of the Sagnac effect:

Clearly the pulse traveling in the same direction as the rotation of the loop must travel a slightly greater distance than the pulse traveling in the opposite direction, due to the angular displacement of the loop during the transit. . . [T]he expressions "c+v" and "c-v" appearing in the derivation of the phase shift . . . do not refer to the speed of light, but rather to the sum and difference of the speed of light and the speed of some other object, both with respect to a single inertial coordinate system, which can be as great as 2c according to special relativity . . . http://mathpages.com/rr/s2-07/2-07.htm

I am having difficulty grasping this explanation because all parts of the interferometer in the classic Sagnac experiment are fixed on the rotating disk (including both the light source and the film that receives and records the light waves). So the experiment does not seem to measure light speed plus or minus rotational speed with respect to some inertial coordinate system (such as the lab system in which the disk rotates). The experiment seems to measure light speed entirely with respect to points on the disk.

Can anyone offer suggestions about how to interpret or supplement this explanation? Thanks.

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Do you understand why you get a fringe shift in a rotating Sagnac interferometer but not in a translating one? The light travels a different distance depending on the direction around the interferometer, because the detector has moved during the time the light was traveling. That's not the case in translation (similar to the Michelson-Morley experiment), since you can always analyze it as being at rest — all inertial frames are equivalent.

If you are in the rotating frame of reference, but want to pretend you are in an inertial frame (i.e. you want to say you are at rest), then it's as if light travels at c±v, depending on the direction.

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Here is a typical explanation of the Sagnac effect:

Clearly the pulse traveling in the same direction as the rotation of the loop must travel a slightly greater distance than the pulse traveling in the opposite direction, due to the angular displacement of the loop during the transit. . . [T]he expressions "c+v" and "c-v" appearing in the derivation of the phase shift . . . do not refer to the speed of light, but rather to the sum and difference of the speed of light and the speed of some other object, both with respect to a single inertial coordinate system, which can be as great as 2c according to special relativity . . . http://mathpages.com/rr/s2-07/2-07.htm

I am having difficulty grasping this explanation because all parts of the interferometer in the classic Sagnac experiment are fixed on the rotating disk (including both the light source and the film that receives and records the light waves). So the experiment does not seem to measure light speed plus or minus rotational speed with respect to some inertial coordinate system (such as the lab system in which the disk rotates). The experiment seems to measure light speed entirely with respect to points on the disk.

Can anyone offer suggestions about how to interpret or supplement this explanation? Thanks.

Consider the loop as a circle and consider angular velocity relative to the center. The light moves independently of the setup, while the setup rotates.

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Thanks to both swansont and phyti. The thing that confuses me about the particular quotation is the phrase "the sum and difference of the speed of light and the speed of some other object, both with respect to a single inertial coordinate system . . ." Given that the emitter and detector are both on the disk, the result (fringe shift recorded on film) has nothing to do with any inertial coordinate system.

If you set the interferometer into translational movement rather than rotation, the film will not record any fringe shift (swansont: "you get a fringe shift in a rotating Sagnac interferometer but not in a translating one"). It is only in another inertial coordinate system (say a laboratory relative to which the interferometer is moving) that one would observe light to move relative to the interferometer at c plus or minus v.

Given that the entire experiment (including the photographing of the fringe shift) occurs on the disk, which is not inertial, utilizing the concept of "a single inertial coordinate system" seems irrelevant and confusing.

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Thanks to both swansont and phyti. The thing that confuses me about the particular quotation is the phrase "the sum and difference of the speed of light and the speed of some other object, both with respect to a single inertial coordinate system . . ." Given that the emitter and detector are both on the disk, the result (fringe shift recorded on film) has nothing to do with any inertial coordinate system.

If you set the interferometer into translational movement rather than rotation, the film will not record any fringe shift (swansont: "you get a fringe shift in a rotating Sagnac interferometer but not in a translating one"). It is only in another inertial coordinate system (say a laboratory relative to which the interferometer is moving) that one would observe light to move relative to the interferometer at c plus or minus v.

Given that the entire experiment (including the photographing of the fringe shift) occurs on the disk, which is not inertial, utilizing the concept of "a single inertial coordinate system" seems irrelevant and confusing.

It's there because an inertial frame allows you to use special relativity and you are guaranteed that the physics will work as advertised. If you're in an accelerated frame, you can't a priori be sure what effects are going to creep in.

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Makes sense, thanks.

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Thanks to both swansont and phyti. The thing that confuses me about the particular quotation is the phrase "the sum and difference of the speed of light and the speed of some other object, both with respect to a single inertial coordinate system . . ." Given that the emitter and detector are both on the disk, the result (fringe shift recorded on film) has nothing to do with any inertial coordinate system.

From the POV of an inertial observer situated in the center of the disc (non-rotating), the equations of motion are:

1. In the clockwork direction (see the mathpages drawing):

$ct_{CW}=2 \pi R+\omega R t_{CW}$ (the wavefront "chases" the detector)

2. In the counter clockwork direction:

$ct_{CCW}+\omega R t_{CCW}=2 \pi R$ (the wavefront "runs into" the detector)

The above resolve to:

1a. $t_{CW}=\frac{2 \pi R}{c-\omega R}$

2a. $t_{CCW}=\frac{2 \pi R}{c+\omega R}$

Hence, the $c \pm \omega R$

Edited by xyzt
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Thanks to both swansont and phyti. The thing that confuses me about the particular quotation is the phrase "the sum and difference of the speed of light and the speed of some other object, both with respect to a single inertial coordinate system . . ." Given that the emitter and detector are both on the disk, the result (fringe shift recorded on film) has nothing to do with any inertial coordinate system.

I don't know anything about the Sagnac explanation.

Variables v and c are speeds, not velocities; there is no object in the rotating ring that has a constant velocity. Could you treat the ring as a set of individual objects, and compare the speed of light to the nearest object, considering only the object's "momentarily comoving inertial frame" (MCIF)? Could the explanation refer to each infinitesimal section of the ring as the "some other object"? Each section of the ring would have its own MCIF, but since they're all the same except for their direction, if you're only dealing with speed and ignore direction, you can deal with all the sections as if they're the same (object).

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I don't know anything about the Sagnac explanation.

Variables v and c are speeds, not velocities; there is no object in the rotating ring that has a constant velocity. Could you treat the ring as a set of individual objects, and compare the speed of light to the nearest object, considering only the object's "momentarily comoving inertial frame" (MCIF)? Could the explanation refer to each infinitesimal section of the ring as the "some other object"? Each section of the ring would have its own MCIF, but since they're all the same except for their direction, if you're only dealing with speed and ignore direction, you can deal with all the sections as if they're the same (object).

I suspect that you can do this. You might not even need more than one section. Say that you have (1) a stationary, hollow cylinder with its inside surface mirrored, and (2) a single light emitter/light receiver unit. You set the unit into motion circling around the interior of the cylinder like a motorcycle in the classic "wall of death" carnival sideshow, see http://en.wikipedia.org/wiki/Wall_of_death_(carnival_sideshow). Then the unit sends a flash of light in each direction. I suspect that the same result would occur (the unit would record a fringe shift for the returning flashes when they arrive at the unit). Dufour and Prunier ran the Sagnac experiment with different parts of the device fixed in the lab rather than on the disk (such as fixing some of the mirrors in the lab), and their experiments got the same approximate fringe shift results as the one with all parts fixed on the disk. See http://www.conspiracyoflight.com/pdf/Dufour_and_Prunier-On_the_Fringe_Movement_Registered_on_a_Platform_in_Uniform_Motion_(1942).pdf

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I suspect that you can do this. You might not even need more than one section. Say that you have (1) a stationary, hollow cylinder with its inside surface mirrored, and (2) a single light emitter/light receiver unit. You set the unit into motion circling around the interior of the cylinder like a motorcycle in the classic "wall of death" carnival sideshow, see http://en.wikipedia.org/wiki/Wall_of_death_(carnival_sideshow). Then the unit sends a flash of light in each direction. I suspect that the same result would occur (the unit would record a fringe shift for the returning flashes when they arrive at the unit). Dufour and Prunier ran the Sagnac experiment with different parts of the device fixed in the lab rather than on the disk (such as fixing some of the mirrors in the lab), and their experiments got the same approximate fringe shift results as the one with all parts fixed on the disk. See http://www.conspiracyoflight.com/pdf/Dufour_and_Prunier-On_the_Fringe_Movement_Registered_on_a_Platform_in_Uniform_Motion_(1942).pdf

I meant only to describe the original experiment with some simplification of the concepts, not to actually describe a modified experiment.

Edit: I think most of the following, and my previous post, are irrelevant. In the inertial lab frame, any part of the ring is at a constant speed, though not constant velocity. From this frame, you can measure the speed of light and the speed of an object on the ring, along the non-straight-line distance each travels. It doesn't matter that the object is not inertial.

-------

I don't think I've quite got it so I'll try to restate it: In explaining the effect, you can consider each point around the ring at the moment light passes through. Here, the point acts the same as an object sharing that point's MCIF. Do this all around the ring and you get a bunch of different objects, each moving in a different direction. But, since you're only dealing with speed, you don't care about direction. The ring has radial symmetry, so each point around the ring is identical with respect to speeds.

Ignoring direction, an object that shares an MCIF with any point around the ring is equivalent to one that shares an MCIF with some other point. You could integrate using velocities, and the effects of direction would cancel out, or you could ignore direction and just look at speed, where all the points around the ring are treated the same. That's what I think the "some other object" refers to in the explanation: an object that shares an MCIF with whatever part of the ring you want to consider at the moment, and is inertial at that point. Ignoring direction, the properties of this imagined object are the same all around the ring. Also inconsequential in the explanation is that a single such object would be physically impossible (it wouldn't be inertial and it would have to move around the ring at the speed of light.)

Yet another way to put it is, the "some other object" is a measuring device that touches the ring tangentially but is inertial instead of rotating with the ring. You would need them all around the ring, ideally infinitely many of them, but they all measure the same thing, so you can calculate and explain it as if it was one object.

I think I'm explaining this in such detail more to see if I get it myself.

Edit: Better yet, the explanation might be "Measure the speed of light relative to a momentarily comoving inertial object at one point in the ring. The measurement everywhere else around the ring is the same." Whether it's one object whose single measurement is the used everywhere, or an abstract object that is reused everywhere and produces the same measurement, the imaginary object is simply called "some other object".

... but I think I've gone off track. It doesn't matter that the "some other object" is momentarily inertial??? With respect to a suitable inertial frame of reference, any point on the ring has a constant speed, even without constant velocity (not inertial).

Edited by md65536
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There is also the Hafele-Keating experiment with clocks moving with and against earth rotation, one gains, one loses.

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I suspect that you can do this. You might not even need more than one section. Say that you have (1) a stationary, hollow cylinder with its inside surface mirrored, and (2) a single light emitter/light receiver unit. You set the unit into motion circling around the interior of the cylinder like a motorcycle in the classic "wall of death" carnival sideshow, see http://en.wikipedia.org/wiki/Wall_of_death_(carnival_sideshow). Then the unit sends a flash of light in each direction. I suspect that the same result would occur (the unit would record a fringe shift for the returning flashes when they arrive at the unit). Dufour and Prunier ran the Sagnac experiment with different parts of the device fixed in the lab rather than on the disk (such as fixing some of the mirrors in the lab), and their experiments got the same approximate fringe shift results as the one with all parts fixed on the disk. See http://www.conspiracyoflight.com/pdf/Dufour_and_Prunier-On_the_Fringe_Movement_Registered_on_a_Platform_in_Uniform_Motion_(1942).pdf

The modern re-enactments of the Sagnac experiments use Fiber Optics Gyroscopes (FOGs) or Ring Laser Gyroscopes (RLGs). The precision of the experiments is much increased over the original experiments and the principle is exactly the same:

$t_{CW}-t_{CCW}=\frac{2 \pi R}{c-\omega R}-\frac{2 \pi R}{c+\omega R}=\frac{4 \pi R^2 \omega}{c^2-\omega^2R^2}$

There is also the Hafele-Keating experiment with clocks moving with and against earth rotation, one gains, one loses.

The principle is different from Sagnac (Sagnac has a Newtonian/Galilean explanation), Hafele-Keating has a GR explanation.

Edited by xyzt
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JVNY initially questioned why the setup is considered an inertial frame.

Isn't this because Einstein considered the rotating disk as an inertial frame with a g-field?

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JVNY initially questioned why the setup is considered an inertial frame.

It isn't if you are looking at the periphery of the rotating disc, as explained. The inertial frame is the one attached to the center of the disc. When the calculations are done from the perspective of that frame, everything is very clear, not to mention easy to calculate.

Isn't this because Einstein considered the rotating disk as an inertial frame with a g-field?

No.

Edited by xyzt
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It isn't if you are looking at the periphery of the rotating disc, as explained. The inertial frame is the one attached to the center of the disc. When the calculations are done from the perspective of that frame, everything is very clear, not to mention easy to calculate.

No.

The process of elimination still works! Thanks

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I am trying to get at the point of view of the rim, not the axis. If a rim is rotating in an inertial laboratory, it makes sense that signals sent in opposite directions around the rim travel in the lab frame relative to their starting point on the rim at c+v or c-v. However, the result of +v or -v applying is that the difference in lab time between the arrival of counter- and co-rotating signals back at their starting point on the rim will depend on the speed of the signals.

On the rim, however, the difference in arrival times does not depend on the speed of the signals. Any pair of opposite signals sent around the rim (light flashes, slow traveling portable clocks, etc.) will have the same difference in arrival time due to the Sagnac effect as measured by an observer fixed on the rotating rim. See, for example, Ashby and Allan on slow clocks carried around the earth (http://tf.nist.gov/timefreq/general/pdf/133.pdf , page 652), and Ashby on clocks around the earth that are synchronized using the Einstein convention with light signals sent around the earth (http://areeweb.polito.it/ricerca/relgrav/solciclos/ashby_d.pdf , page 15) (excerpts attached).

Because the difference in arrival time as measured on the rim (by an emitter and receptor fixed on the rim) does not depend on signal velocity, it cannot be right to explain the Sagnac effect as being due to light traveling at c+v or c-v as measured by a rim observer.

The standard explanation seems to confuse what is happening in the inertial laboratory frame with what is happening on the rim itself.

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I am trying to get at the point of view of the rim, not the axis.

I'm trying to think my way through your post, and perhaps making progress that might be useful to discuss?

To figure out what's going on, it might be useful to place imaginary measuring devices all over the place. So imagine you have detectors placed all around the ring, and they have clocks and can indicate when the beam passes in either direction.

One thing to consider is the synchronization of the clocks. Considering from the lab (center of ring) frame, it must be that all of the ring clocks tick at the same (slowed) rate. Imagine a lab clock controlling the clocks on the ring, by sending out a sync pulse from the center of the ring. Then from the lab frame only, the ring's clocks are all set the same.

From the perspective of a ring observer, the clocks on the ring cannot be synchronized. A single pulse from the center will reach different clocks at different times. Yet the ring clocks must all tick at the same rate, so there must be a "shift" in simultaneity, where --- I think --- the clocks in front of an observer (with respect to the ring's rotation in the lab frame) are behind ahead in time, and ones behind are ahead behind in time. (Is this right??? My grasp on the reasoning is feeble and I think I got it backwards first attempt. ... and second, dammit!!) At some point near halfway around the ring, one of the clocks can be in sync with the observer's.

So immediately you have a problem now of how you're going to measure the speed of light as the beam goes around the ring. You can use each local clock and the local speed will equal c. But you can't use only the one clock at the observer location, because as you make your way around the ring there will be a shift in simultaneity. If you do this while ignoring relative simultaneity, then you should get the "c+/-v" values that don't match the locally measured speed, but you're not properly measuring speed this way.

That's as far as I got but can this line of reasoning be carried further?

Edited by md65536
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. . . Considering from the lab (center of ring) frame, it must be that all of the ring clocks tick at the same (slowed) rate. . .

From the perspective of a ring observer, the clocks on the ring cannot be synchronized. A single pulse from the center will reach different clocks at different times. Yet the ring clocks must all tick at the same rate, so there must be a "shift" in simultaneity, where --- I think --- the clocks in front of an observer (with respect to the ring's rotation in the lab frame) are behind in time, and ones behind are ahead in time. (Is this right???) At some point near halfway around the ring, one of the clocks can be in sync with the observer's. . .

A couple of thoughts.

First, I think that even on the rim (not just in the lab) each clock will tick at the same rate. Say the interior of the rim is mirrored, so that a light signal can travel around the interior. If a clock sends a series of signals around the ring (in either direction), one every second of its proper time, then each other clock on the rim should receive the signals one every second of its proper time (after the initial delay for the first signal of the series to arrive).

Second, you are on to something with the simultaneity shift suggestion -- some analyze the Sagnac effect as a shift in simultaneity, like this: http://www.physicsinsights.org/sagnac_1.html

However, I don't think that the shift in simultaneity view follows the c+v and c-v analysis of how light travels on the rim (to a rim observer), because the Sagnac effect is the same in absolute value for both the counter- and co-rotating signal (plus or minus 207.4 ns for a signal sent around the earth per the Ashby article linked above). It is definitely worth thinking through.

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First, I think that even on the rim (not just in the lab) each clock will tick at the same rate.

Agreed. With constant radial velocity and radial symmetry, at any time any observer on the ring should see the same thing as any other ring observer at any other time. The ring clocks couldn't tick at different rates without diverging, contradicting what the lab observer sees.

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

However, I don't think that the shift in simultaneity view follows the c+v and c-v analysis of how light travels on the rim (to a rim observer), because the Sagnac effect is the same in absolute value for both the counter- and co-rotating signal (plus or minus 207.4 ns for a signal sent around the earth per the Ashby article linked above).

It makes sense to me that the + and - values would end up the same.

Consider some n clocks spaced evenly around the ring. Suppose that all of the ring clocks are set to the same time in the lab frame. The situation is the same at any of the clocks, as any of the others. Then relative to any given individual clock, if the previous clock is ahead by t seconds, then the next will be behind by the same value t. Do you agree?

If you make your way around the ring in one direction, each next clock is ahead by the same value t, making a sync error of t*n all around the ring. In the opposite direction the sync error is -t*n. Edit: I think your link explains this in a slightly different way.

However, if you fix the observer viewpoint at one clock, you'll find something like, the previous clock is ahead by t, and the one before that is ahead of it by a different amount, and about a quarter around the ring subsequent clocks start being behind the previous. A graph of the sync error relative to the single observer would look something like a sin wave, rather than a linearly increasing sync error. The fact that this observer disagrees with eg. an observer on the other side of the ring, about which of a given pair of clocks is ahead of the other, shows that the ring clocks can't be synchronized.

Edit: Sorry I keep switching 'ahead/behind' when I think I got it backwards. It doesn't change the reasoning tho.

Edited by md65536
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I am trying to get at the point of view of the rim, not the axis.

This is extremely tough and unnnecessary. The standard treatment is done from the POV of the inertial frame attached to the center.

If a rim is rotating in an inertial laboratory, it makes sense that signals sent in opposite directions around the rim travel in the lab frame relative to their starting point on the rim at c+v or c-v.

They don't, $c \pm v$ is the closing speed , not the speed of the signal.

Edited by xyzt
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This is extremely tough and unnnecessary. The standard treatment is done from the POV of the inertial frame attached to the center.

They don't, $c \pm v$ is the closing speed , not the speed of the signal.

I agree that it is tough, but that is part of what makes it interesting. The light emitter and the photographic film that recorded the fringe shift were fixed on Sagnac's rotating disk, so what he recorded was from the POV of the rotating disk, not the inertial frame attached to the center. Trying to understand the effect from the POV of the disk is worthy of effort.

And I agree, the c+v and c-v refer to closing speeds.

Agreed. With constant radial velocity and radial symmetry, at any time any observer on the ring should see the same thing as any other ring observer at any other time. The ring clocks couldn't tick at different rates without diverging, contradicting what the lab observer sees.

It makes sense to me that the + and - values would end up the same.

Consider some n clocks spaced evenly around the ring. Suppose that all of the ring clocks are set to the same time in the lab frame. The situation is the same at any of the clocks, as any of the others. Then relative to any given individual clock, if the previous clock is ahead by t seconds, then the next will be behind by the same value t. Do you agree?

If you make your way around the ring in one direction, each next clock is ahead by the same value t, making a sync error of t*n all around the ring. In the opposite direction the sync error is -t*n. Edit: I think your link explains this in a slightly different way.

However, if you fix the observer viewpoint at one clock, you'll find something like, the previous clock is ahead by t, and the one before that is ahead of it by a different amount, and about a quarter around the ring subsequent clocks start being behind the previous. A graph of the sync error relative to the single observer would look something like a sin wave, rather than a linearly increasing sync error. The fact that this observer disagrees with eg. an observer on the other side of the ring, about which of a given pair of clocks is ahead of the other, shows that the ring clocks can't be synchronized.

Edit: Sorry I keep switching 'ahead/behind' when I think I got it backwards. It doesn't change the reasoning tho.

If the observers on the rim try to synchronize their clocks using the Einstein convention sending signals in one direction, then yes I think that each clock is ahead of the one behind it (or vice versa, if the signal were sent in the other direction). I don't know, however, what is simultaneous with what if you synchronize clocks with any given convention (a light signal from the center, or some other method). I am not sure what simultaneity convention properly applies to a rotating rim. So I was focusing first on the explanation of the Sagnac effect; if the closing speeds of the signals measured on the disk relative to their source (on the disk) are the same as the closing speeds of the signals in the lab frame relative to their source (on the disk), i.e. c+v or c-v, that might give some information to help with the subsequent question of simultaneity. But it seems as if the closing speeds are different on the disk.

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I agree that it is tough, but that is part of what makes it interesting. The light emitter and the photographic film that recorded the fringe shift were fixed on Sagnac's rotating disk, so what he recorded was from the POV of the rotating disk, not the inertial frame attached to the center. Trying to understand the effect from the POV of the disk is worthy of effort.

And I agree, the c+v and c-v refer to closing speeds.

If the observers on the rim try to synchronize their clocks using the Einstein convention sending signals in one direction, then yes I think that each clock is ahead of the one behind it (or vice versa, if the signal were sent in the other direction). I don't know, however, what is simultaneous with what if you synchronize clocks with any given convention (a light signal from the center, or some other method). I am not sure what simultaneity convention properly applies to a rotating rim. So I was focusing first on the explanation of the Sagnac effect; if the closing speeds of the signals measured on the disk relative to their source (on the disk) are the same as the closing speeds of the signals in the lab frame relative to their source (on the disk), i.e. c+v or c-v, that might give some information to help with the subsequent question of simultaneity. But it seems as if the closing speeds are different on the disk.

But it is neither necessary nor useful. By contrast, the calculation in the inertial frame is useful and necessary, this is how it is done in all the published papers.

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If the observers on the rim try to synchronize their clocks using the Einstein convention sending signals in one direction, then yes I think that each clock is ahead of the one behind it (or vice versa, if the signal were sent in the other direction). I don't know, however, what is simultaneous with what if you synchronize clocks with any given convention (a light signal from the center, or some other method). I am not sure what simultaneity convention properly applies to a rotating rim. So I was focusing first on the explanation of the Sagnac effect; if the closing speeds of the signals measured on the disk relative to their source (on the disk) are the same as the closing speeds of the signals in the lab frame relative to their source (on the disk), i.e. c+v or c-v, that might give some information to help with the subsequent question of simultaneity. But it seems as if the closing speeds are different on the disk.

I don't think c+/-v is a meaningful closing speed of anything from the viewpoint of an observer on the ring. If so, what does v refer to?

I think that from this perspective, c+v would be a coordinate speed of light as measured by a single observer on the ring using only its own clock. As discussed in other threads, coordinate speed is not the same as speed. The calculations work out the same (since the different observer viewpoints are consistent)... but I think it would be counter productive to consider it a speed. I think it is "the average rate of light's propagation when measured using a single clock that is out of sync with other clocks along the light's path (and thus can differ from the locally measured speed of light along that path)."

The clocks on the ring can't be synchronized (to the full definition) by any method. I don't know what simultaneity convention you might use either, but whatever you use, the simultaneity of two specific events will generally depend on the choice of observer frame of reference on the ring. Setting the ring's clocks to be the same in the lab frame is useful conceptually, because then every location on the ring can be treated symmetrically, simplifying some ideas.

To me, the lack of synchronization, ie. the relativity of simultaneity on the ring, is the explanation of why the average(?) coordinate speed of light can differ from c. According to your link: http://www.physicsinsights.org/sagnac_1.html

"Note that You Can't Synchronize the Clocks in a Rotating Frame

[...]
This is, in fact, the essence of the Sagnac effect. The clocks on the rotating disk can't be properly synchronized: any attempt to do so leads to a discontinuity somewhere on the disk. Any signal which goes all the way around the disk crosses the discontinuity."
So I don't know if it's possible to figure out the Sagnac effect from a ring perspective before considering relativity of simultaneity. Unfortunately, I'm content with vague ideas of why it works, without the maths that show how.
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The clocks on the ring can't be synchronized (to the full definition) by any method.

That's because synchronization requires them to have the same phase. They can be set to give a consistent result, however.

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• 1 year later...

...Sagnac has a Newtonian/Galilean explanation...

Really? Maybe when we have rotating mirrors.

What Newtonian/Galilean explanation do you know for the fibre optic conveyor (see Figure 5 in https://en.wikipedia.org/wiki/Sagnac_effect)? Anyone?

Edited by DanMP

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