Sagnac effect for matter beams

[alexandrinushka]

TL;DR Summary

Proper time identical for any beams sent in opposite directions (Sagnac style experiment)

Hello,
I have recently come across this article by Rizzi and Ruggiero, called "The Relativistic Sagnac Effect: Two Derivations": https://arxiv.org/pdf/gr-qc/0305084.pdf
In section 3, the authors derive the Sagnac proper time difference for all beams (light beams and matter beams, including muons, neutrinos, electrons, acoustic waves, etc).

I feel like I have misread or not understood something from the paper, because it seems to me that:
Whatever the angular velocity of the matter beams (be it 0.001c or 0.1c or 0.8c) when we read the time lapse between the counter-propagating beams, we'll register it as being a constant on the rotating clock.

They say this condition holds only if the rotating clock is Einstein synchronized. So, can this be actually tested? Or has it? Do we actually have the possibility to Einstein synchronize the rotating clock and, in this case, will we indeed ALWAYS obtain the same time lapse for matter beams, no matter how slow (or fast) these are?

The math in their derivation adds up nicely, but I find this statement counterintuitive.
Can someone explain it to me in a simple way and suggest the experimental setup for testing it?
Thank you in advance.

 

[Sagittarius A-Star]

Summary: Proper time identical for any beams sent in opposite directions (Sagnac style experiment)

In the linked text they speak of the elapsed proper time of a clock at rest on the rotating disk at the start- and endpoint of the beams. This does not depend on the velocity, both matter beams have relative to the rim of the disk. There is a time difference according to this clock, although the elapsed proper times of both matter beams are the same. This is explained in the following text.

we realize at once that the time difference between clockwise and counterclockwise traversals must be independent of the signal velocity.
...
What About Proper Time? What does the Signal See?
So to find out how much proper time elapses for the signal as it goes around the disk, we can just look at the case of a straight cable, and that's trivial.

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

 

[alexandrinushka]

In the linked text they speak of the elapsed proper time of a clock at rest on the rotating disk at the start- and endpoint of the beams. This does not depend on the velocity, both matter beams have relative to the rim of the disk. There is a time difference according to this clock, although the elapsed proper times of both matter beams are the same. This is explained in the following text.Source:
http://www.physicsinsights.org/sagnac_1.html

That is a great link, thanks.
I liked that it mentions that acceleration is irrelevant. The opposite claim is something we find a lot here and there, but it never made sense to me: only the instantaneous velocity at a certain moment defines the relativistic phenomena.

So just to make sure I understood it correctly:
Two people running along the rim in opposite directions (to borrow your example), each carrying a watch, when they meet and compare their clocks, will register the same difference as two electrons running at 0.5 around the rim (if electrons carried watches). Right?

Yet this will not be the case of the clock placed on the rim, which cannot be properly synchronized. Actually, I think my confusion came from the fact that I thought "proper time" refers to the time that this clock, turning around on the rim of the turntable, registers...

 

[Sagittarius A-Star]

So just to make sure I understood it correctly:
Two people running along the rim in opposite directions (to borrow your example), each carrying a watch, when they meet and compare their clocks, will register the same difference as two electrons running at 0.5 around the rim (if electrons carried watches). Right?

The two people will find no difference between their clock measurements of travel time when each arrives at the starting point. The same is valid for the elapsed times of the two electrons.

Yet this will not be the case of the clock placed on the rim, which cannot be properly synchronized.

It makes no sense to speak of the synchronization of only one clock. What they say is that the two beams have the same one-way-speed (in opposite direction) relative to the rim of the disk, if they reference to an Einstein-synchronized LCIF (local co-moving inertial frame). See in the paper, linked by you, under equation (24).

Actually, I think my confusion came from the fact that I thought "proper time" refers to the time that this clock, turning around on the rim of the turntable, registers...

No, you thought this correctly. They mean the elapsed proper time of the clock, that is at rest relative to the rim of the disk. As I already wrote, it's elapsed proper time between the arrivals of the two beams (which were sent-out at the same time) does not depend on the velocity, both matter beams have each locally relative to the rim of the disk.

 

[alexandrinushka]

The two people will find no difference between their clocks when each arrives at the starting point. The same is valid for the elapsed times of the two electrons.

It makes no sense to speak of the synchronization of only one clock. What they say is that the two beams have the same one-way-speed (in opposite direction) relative to the rim of the disk, if they reference to an Einstein-synchronized LCIF (local co-moving inertial frame). See in the paper, linked by you, under equation (24).

No, you thought this correctly. They mean the elapsed proper time of the clock, that is at rest relative to the rim of the disk. As I already wrote, it's elapsed proper time between the arrivals of the two beams (which were sent-out at the same time) does not depend on the velocity, both matter beams have each locally relative to the rim of the disk.

Thanks a lot.

 

[alexandrinushka]

@Sagittarius A-Star in the link you provided, why is u_- = (k - v)/(1 - kv) (5) and u₊ = (k + v)/(1 + kv) (7) ?
I understand the upper part, but not the denominator.
Thanks.

 

[Sagittarius A-Star]

@Sagittarius A-Star in the link you provided, why is u_- = (k - v)/(1 - kv) (5) and u₊ = (k + v)/(1 + kv) (7) ?
I understand the upper part, but not the denominator.
Thanks.

They use the formula for "relativistic velocity addition" and calculate with a unit system, in which $c=1$.

https://en.wikipedia.org/wiki/Velocity-addition_formula#Special_relativity

A derivation of the formula for "relativistic velocity addition" from inverse Lorentz transformation:

$x = \gamma (x' + vt'), \ \ \ \ \ t = \gamma (t' + \frac{v}{c^2}x')$

with

$x' := v_1 \ t'$​

$v := v_2$​

$V = x/t$​

=>
$$V = \frac{v_1 + v_2} {1+ v_1 v_2/c^2}$$
Here I transformed $v_1$ from the primed frame to the unprimed frame.

 

[alexandrinushka]

@Sagittarius A-Star
Ok, I think I got the relativistic velocity addition.
Actually, this video helped me a lot:

If I get it right, this is what the velocity of each beam would be for me (expressions (5) and (7)), the guy watching the beams from the lab, right?

What I still don't get is why in both cases we use 2pi*r : expressions (6) and (8). To me, in the lab, I'd probably say that one beam made a longer journey than 2pi*r and the other a shorter one, no?

 

[Sagittarius A-Star]

If I get it right, this is what the velocity of each beam would be for me (expressions (5) and (7)), the guy watching the beams from the lab, right?

Yes.

What I still don't get is why in both cases we use 2pi*r : expressions (6) and (8). To me, in the lab, I'd probably say that one beam made a longer journey than 2pi*r and the other a shorter one, no?

There they calculate with the closing speed $u+v$, as you can see in the dominator. That is the speed, by which the distance between the sender/receiver and the signal becomes smaller, described in the lab frame. This distance goes down from $2\pi r$ to $0$.

 

[alexandrinushka]

@Sagittarius A-Star Thanks a lot for your patience. It takes some time for me to understand it, to assemble all the pieces together.
So... if I am the guy watching the disk from the lab, don't I need to calculate (6) by applying the Lorentz contraction factor to 2Πr?
On the denominator side, I get that we add u_- (which has been calculated by the relativistic addition of velocities) to the disk angular velocity, but don't we need to perform the same procedure when we add v to it? Is it legit to just add u_- and v directly?
Side note: are you the author of the site? You just find it useful for beginners?
Another side note: how do you get all the nice formatting for your math? Mine is awful...
Thanks.

 

[Sagittarius A-Star]

So... if I am the guy watching the disk from the lab, don't I need to calculate (6) by applying the Lorentz contraction factor to 2Πr?

In the lab frame, the circumference of the rotating disk is $2\pi r$. But there is a Lorentz contraction. Therefore, the circumference must be $\gamma 2 \pi r$ in the disk frame:

The paradox has been deepened further by Albert Einstein, who showed that since measuring rods aligned along the periphery and moving with it should appear contracted, more would fit around the circumference, which would thus measure greater than 2πR. This indicates that geometry is non-Euclidean for rotating observers, and was important for Einstein's development of general relativity.[4]

Source:
https://en.wikipedia.org/wiki/Ehrenfest_paradox

On the denominator side, I get that we add u_- (which has been calculated by the relativistic addition of velocities) to the disk angular velocity, but don't we need to perform the same procedure when we add v to it? Is it legit to just add u_- and v directly?´

Yes, that's legit.

• You need to calculate with the formula for "relativistic velocity addition", if you transform a velocity from one frame to another. The maximum possible result is $c$.
• You need to just add, if you want to get the closing speed within one frame. The maximum possible result for this is $2c$.

Side note: are you the author of the site? You just find it useful for beginners?

I am not the author. The Sagnac effect itself is not easy to understand for beginners. I like this site, because it shows as simple as it possible for such a topic, that the $\Delta t$ is independent on the signal speed $k$ and because it discusses the description in both frames, the lab frame and the (more complicated to understand) rotating frame.

Another side note: how do you get all the nice formatting for your math? Mine is awful...
Thanks.

Follow the link "LaTeX Guide" on the left side under your text input field.

Here's an example, first with the delimiters disabled so you can see the raw code:

A quadratic equation is of the form ##ax^2 + bx + c = 0##.
To solve for x, use the quadratic formula $$x = \frac {-b \pm \sqrt{b^2 -4ac}} {2a},$$
substituting the values of the coefficients a, b and c.

If you want to see examples for the raw code between the delimiters, you can right-click on a formula in my above postings, then click on "Show Math As" and then click on "TeX Commands".

You will see the nicely formatted result after click on "Preview" or "Post reply".

 

[alexandrinushka]

In the lab frame, the circumference of the rotating disk is $2\pi r$. But there is a Lorentz contraction.

Yes, that's legit.

• You need to calculate with the formula for "relativistic velocity addition", if you transform a velocity from one frame to another. The maximum possible result is $c$.
• You need to just add, if you want to get the closing speed within one frame. The maximum possible result for this is $2c$.

I am not the author. The Sagnac effect itself is not easy to understand for beginners.

@Sagittarius A-Star
So. Taking into account the Lorentz time dilation and length, buuuut having the Ehrenfest paradox interfere... makes the disk itself keep its original circumference length to our eyes and measuring devices ("our" meaning in the lab).

A clock riding on the disk seen from the lab will appear to run slower. A meter riding on the disk will appear to us shorter. That justifies the relativistic velocity. But the disk will still be $2 \pi r$, which we legitimately use calculating the closing speed.

What I like about the link you shared is that it provides an explanation for the phenomenon as seen from the lab first (well on the rim too, but that I already get in Rizzi's paper), whereas in the paper I cited initially I only get this result looking from the rim. And isn't the point of SR to get the same result, weather we're looking from the lab or riding the disk?

Another question. Rizzi and Ruggiero do actually obtain the same results with a clock, synced with the lab. But they claim the result is only made possible using an ugly formula (31). What do you think about it? To me, we should be able to obtain this result using any frame of reference. No?

 

[Sagittarius A-Star]

@Sagittarius A-Star
So. Taking into account the Lorentz time dilation and length, buuuut having the Ehrenfest paradox interfere... makes the disk itself keep its original circumference length to our eyes and measuring devices ("our" meaning in the lab).

Yes.

A clock riding on the disk seen from the lab will appear to run slower. A meter riding on the disk will appear to us shorter. That justifies the relativistic velocity. But the disk will still be $2 \pi r$, which we legitimately use calculating the closing speed.

Yes.

And isn't the point of SR to get the same result, weather we're looking from the lab or riding the disk?

Yes.

Another question. Rizzi and Ruggiero do actually obtain the same results with a clock, synced with the lab. But they claim the result is only made possible using an ugly formula (31). What do you think about it? To me, we should be able to obtain this result using any frame of reference. No?

There, Rizzi and Ruggiero make use of the fact, that a one-way-speed is only isotropic (the same speed from A to B as from B to A), if the speed is measured with two Einstein-synchronized clocks for each pair of sender and receiver.

You find a good explanation for these two possibilities of clock synchronization in section "But What About an Observer On the Disk?" with Figures 3, 4 and 5.

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

Also, there you can see from equation (10), that the formula for the Sagnac $\Delta t$ ist nothing else than the term for "relativity of simultaneity" in the Lorentz transformation, multiplied with $2$:

$\require {color} \Delta t = \gamma (\Delta t' + \color{red} {\frac {v} {c^2} \Delta x'} \color{black})$

 

[alexandrinushka]

@Sagittarius A-Star yeah, I have read it several times. Well, that's the thing, I find the website you've sent much more intuitive in this regard and, like I said in the previous comment, things "add up" both ways here (lab clock and rotating clock).

I'm familiar with the Lorentz factor and the relativity of simultaneity.

I get confused by R&R's approach though. Their math is beyond my level of proficiency and I am a bit disappointed to NOT see what I saw in your link on their side. It seems to me they claim that their approach is the only one giving the right time gap for subluminal beams (I mean using clocks Einstein synchronized in the rotating frame). Whereas in your link the whole analysis starts with the observers in the lab and gets to prove the independence of the Sagnac time difference on the signal speed. So I've managed to follow this approach on the site and I (thought I) had my aha moment.

But once I jump back to the R&R paper and look at their formula 31... I am just lost. Clearly they obtained it somehow, but I can't picture the math behind it. I don't know if you get my "pain point"... I must be missing something essential. Or maybe I just have not tried to derive it all myself to arrive to their (31).

What I want to see and am not seeing in their paper is: here are two beams of electrons, we look from the lab, we launch them on a rotating disk, we use relativistic velocity addition (like in the link you've cited) and voilà, we get the time gap independent of the beams' speed. Because otherwise they proove what? That on a moving disk you cannot mix Einstein synchronization with lab-frame synchronization? Well, that is not something anyone attempted to do in the first place, no?

Here I'm just quoting them:
However, any non relativistic explanation completely fails for subluminally traveling entities (such as matter waves, sound waves, electromagnetic waves in an homogeneous co-moving medium, and so on). In fact, for subluminally traveling entities the vital condition is: ”equal relative velocity in opposite directions”. If this condition is expressed by eq. (24) (which explicitly requires the local Einstein synchronization) the Sagnac proper time difference (25) arises, as we carefully showed before. On the contrary, if the condition ”equal relative velocity in opposite directions” is expressed by an analogous relation, in which the local Einstein synchronization is replaced by a synchronization borrowed from the global synchronization of the central IF (see Subsection 3.5), no time difference arises. Let us prove this claim.

By now I am really embarrassed to insist so much. I must be missing something or maybe I need to reread all of this on a fresh morning.

Anyway, you've been really kind and patient to me. I'm sorry I have taken so much of your time already. Thanks a lot.

 

[alexandrinushka]

@Sagittarius A-Star Ok, I think I know where it all "went wrong" in the second derivation. Equations 26, 27 and 28 are all non-relativistic (therefore wrong) and thus will lead to a complete cancellation of the effect, as in eq. 1, 2, 3, 4 of your link.

 

[Sagittarius A-Star]

@Sagittarius A-Star Ok, I think I know where it all "went wrong" in the second derivation. Equations 26, 27 and 28 are all non-relativistic (therefore wrong) and thus will lead to a complete cancellation of the effect, as in eq. 1, 2, 3, 4 of your link.

No.

In my link, equations (1) to (4), a non-relativistic calculation is done, that's right. It is based on the Galileo-Transformation, from which follows the ordinary velocity-addition.

But in your link, equations (26) to (28), they do a relativistic calculation with non-Einstein-synchronization locally along the rim of the rotating disk:

On the contrary, if the condition ”equal relative velocity in opposite directions” is expressed by an analogous relation, in which the local Einstein synchronization is replaced by a synchronization borrowed from the global synchronization of the central IF (see Subsection 3.5), no time difference arises. Let us prove this claim.

Source:
https://arxiv.org/pdf/gr-qc/0305084.pdf

The same synchronization scheme is described in my link in section "The Peculiar Case: Clocks Not Properly Synchronized", figure #4.

The Lorentz transformation is only valid, if in both frames the time coordinates are defined via Einstein-synchronization.

If the clocks on the rotating disk are synchronized to each other with reference to the non-rotating inertial frame, this leads to a modification of the Lorentz transformation. In this case, it's term for "relativity of simultaneity" $v x'/c^2$ would be deleted (but still would contain the $\gamma$ for time dilation). As you can see in my posting #7, the formula for "relativistic velocity addition" would then become the normal classical velocity addition formula, in your link equations (26) to (28).

In your link in equation (30) they define, that the particles locally move formally with equal speed on opposite directions. But because that equation is valid under the condition of non-Einstein-synchronized clocks, they have modified the experiment. In "reality", they sent the particles with different speeds, to cancel-out the Sagnac-effect.

In equation (31) they show, how much the speeds in opposite directions must formally differ (based on non-Einstein-synchronized clocks) for the case, in which the particle are sent-out in "reality" with equal speeds.

 

[alexandrinushka]

@Sagittarius A-Star thanks a lot. So given equation (31) relating these opposite velocities is "extremely ad-hoc" (citing the authors here) and, according to R&R, has no obvious physical meaning, why not use this as the ultimate proof of Einstein synchronization being the only legitimate and correct one and thus end the debate on the conventionality of synchronization once and for all?
It is not wrong mathematically, but it is ad-hoc, as they say.

 

[Sagittarius A-Star]

why not use this as the ultimate proof of Einstein synchronization being the only legitimate and correct one and thus end the debate on the conventionality of synchronization once and for all?

If you use the Einstein synchronization, then the math gets not complicated enough

 

[alexandrinushka]

@Sagittarius A-Star actually... equations 5-8 in your link should apply to translational motion as well, no? Just replace the circumference with L, no?

 

[Sagittarius A-Star]

@Sagittarius A-Star actually... equations 5-8 in your link should apply to translational motion as well, no? Just replace the circumference with L, no?

Yes, they apply also to translational motion, like shown in figure 3.

 

[alexandrinushka]

@Sagittarius A-Star ok, so what if I set multiple clocks around a square, not a disk, and move the square like a serpent. I mean, I do not turn it around, I just move the fiber in a closed square.
The clocks that travel on my moving square will be able to send signals to one another and will stay synced, right?
But when I check one of the clocks against my wristwatch (me being the guy in the lab, not on the moving fiber optic serpent), then my clock and the clock that has just aligned with me on the fiber optic serpent will be out of sync. Or will it? Do we actually know the answer from some sort of experiment?

The guy says in the link:
When clocks are "synchronized" in an inertial frame, and we use them to time the velocity of beams of light, we'll find that it's the same no matter what direction the beams are traveling in. Indeed, that's the point of the synchronization procedure outlined by Einstein in his 1905 paper on electrodynamics. But in a rotating frame, it doesn't work. If there are closely spaced clocks all around the rim of the disk, and we synchronize adjacent pairs of clocks (using pulses of light to carry the time back and forth), then we'll find, when we work our way all the way around the disk, that the time on the first clock doesn't match the time on the last clock.

(Maybe my questions is silly, sorry, I needed some time to let it sink *haha* I don't really see a huge difference between rotational and translational motion, other than that centripetal acceleration the author of the site asks us to ignore)

 

[Sagittarius A-Star]

The clocks that travel on my moving square will be able to send signals to one another and will stay synced, right?

Yes and no.

Yes: Locally (that means compared to their neighbors) and along each of the four edges of the square they will stay in sync. Also, two neighboring clocks stay in sync, after they have both turned around an edge.

No: Globally, clocks on the rim of a rotating disk, as well as clocks moving along the edges of the square, cannot be all in sync. That is the reason for the Sagnac effect, which happens as well on the rotating disk as along the square.

Assume you have the following 10 clocks distributed equally around the disk. The next neighbor of J will then be A:

A, B, C, D, E, F, G, H, I, J.​

Now you do Einstein synchronizations:

Synchronize B to A, then C to B, and so on ... until J to I.​

If you check you will see, that J and it's neighbor A have (in the rotating frame) an offset of $\gamma \ 2\pi r \ v/c^2$. That is half of the Sagnac-$\Delta t$.

Edit: I forgot to write, that you are now on the rotating disk. I wanted to describe here, what happens for an observer on the rim of the disk in the rotating frame.

But when I check one of the clocks against my wrist clock (me being the guy in the lab, not on the moving fiber optic serpent), then my clock and the clock that has just aligned with me on the fiber optic serpent will be out of sync.

Your clock and the moving clocks must be generally be out-of sync, because moving clocks run slow. The important point here is, that the adjacent moving clocks are not in synch to each other, from your viewpoint.

 

[alexandrinushka]

@Sagittarius A-Star hm... wait, let's not sync them. A runs on the disk, B runs, C runs etc... they mind their own life of clocks. And we come back in 1 year. I get in front of A and I see it's lagging behind by x, then B gets just in front of me and its lag is also x, right? And so on. They have not sent any signal to each other ever. So they must lag by about the same amount. But then... if I only use inertial movements, never any rotation, and I get this out-of-sync effect when comparing clocks, when exactly is it appropriate to use the Einstein synchronization method? Can I only use it to check clocks that are stationary compared to me?
Funny thing happened the other day: my mom has not synced her fitbit with her phone for two months and... her watch was lagging by almost 2 minutes! That is huge. I know gravitational effects come in here as well. I was actually struggling to explain to her why her "perfect" fitbit has "lost" two minutes.
Thanks

 

[alexandrinushka]

@Sagittarius A-Star see, that is where it gets ****ed up for me. I'm the lab guy and I lab sync all my 10 clock on the square serpent. Great. They are GPS synced and things are neat for them. The serpent is not moving, it's stationary. A sends a signal to B, B to C, and so on and since the speed of light is c in one direction (assume this) then J is synced with all the them and with A as well. Right?

Now I launch my light serpent train movement. And... it gets ****ed up. A sends a signal to B and B to C and so on and then J goes oops when it meets A.

But wait... A and B were in an inertial frame. They don't know they are moving at all. They send the signal, assume it's c and so they should not get unsynced. Ever. I an not getting involved from my lab, except for the very first sync before launching the light train.

Light is defined as c in any inertial frame so... this cannot happen. What Figure 4 says about the signals being anisotropic is... strange. A and B have been synced just before I launched them. A signals B, the signal has velocity c, B signals A, the signal velocity is c too. Perfect. Clocks A and B should be nicely synced. So when they travel from A to B to C and so on to J, no time gap should arrive. Assuming speed of light is c in any inertial frame. No?

 

[Nugatory]

if I only use inertial movements, never any rotation, and I get this out-of-sync effect when comparing clocks, when exactly is it appropriate to use the Einstein synchronization method? Can I only use it to check clocks that are stationary compared to me?

It might be better to say "there exists an inertial frame in which the clocks are at rest throughout the synchronization procedure". We don't need to introduce you or any other observer at rest relative to the clocks to use that frame and what matters is that the clocks are at rest relative to one another.

Funny thing happened the other day: my mom has not synced her fitbit with her phone for two months and... her watch was lagging by almost 2 minutes! That is huge. I know gravitational effects come in here as well. I was actually struggling to explain to her why her "perfect" fitbit has "lost" two minutes.

This has a very mundane explanation. The fitbit has an internal clock. The time the device displays is the time of the last sync (which is very very accurate) plus the time on the internal clock since that sync. The internal clock is much less accurate (apparently your mom's loses about two seconds every day) so if we don't sync for a long time the displayed time will by off until we do another resync.

 

[jartsa]

But wait... A and B were in an inertial frame. They don't know they are moving at all. They send the signal, assume it's c and so they should not get unsynced. Ever. I an not getting involved from my lab, except for the very first sync before launching the light train.

Light is defined as c in any inertial frame so... this cannot happen. What Figure 4 says about the signals being anisotropic is... strange. A and B have been synced just before I launched them. A signals B, the signal has velocity c, B signals A, the signal velocity is c too. Perfect. Clocks A and B should be nicely synced. So when they travel from A to B to C and so on to J, no time gap should arrive. Assuming speed of light is c in any inertial frame. No?

When one pair of those inertially moving clocks properly Einstein-synchronize themselves, they set the leading clock to a later time than the trailing clock. This is of course according to an observer according to which the pair of clocks is moving.

If afterwards said observer accelerates so that according to her the clocks are not moving anymore, then according to her the clocks are not out of sync anymore. Here we can see acceleration having an effect on something.It's easy to imagine moving clocks synchronizing themselves: they set a camera flash with a timer between themselves, when they see the flash they set their time to zero. Leading clock sees the flash later.

 

[Sagittarius A-Star]

Can I only use it to check clocks that are stationary compared to me?

Sorry, I forgot to write, that you are now on the rotating disk, near each adjacent pair of clocks when synchronizing/checking the clocks A to J. Then from your viewpoint, these clocks are all at rest and the lab makes a coordinate-rotation in the opposite direction.

I wanted to describe here, what happens for an observer on the rim of the disk in the rotating frame.

 

[alexandrinushka]

Sorry, I forgot to write, that you are now on the rotating disk, near each adjacent pair of clocks when synchronizing/checking the clocks A to J. Then from your viewpoint, these clocks are all at rest and the lab makes a coordinate-rotation in the opposite direction.

I wanted to describe here, what happens for an observer on the rim of the disk in the rotating frame.

OK for the clocks on the rim. Danke, @Sagittarius A-Star

But I am an observer in the lab, just seeing one clock at a time when it is facing my clock...
So what about the scenario described in the post 24?

Is this a bit like the twin paradox?

 

[Sagittarius A-Star]

OK for the clocks on the rim. Danke, @Sagittarius A-Star

Yes, and equally important: with reference to the rotating frame.

But I am an observer in the lab, just seeing one clock at a time when it is facing my clock...
So what about the scenario described in the post 24?

I try to understand your scenario from:

1)

@Sagittarius A-Star ok, so what if I set multiple clocks around a square, not a disk, and move the square like a serpent. I mean, I do not turn it around, I just move the fiber in a closed square.

2)

@Sagittarius A-Star see, that is where it gets ****ed up for me. I'm the lab guy and I lab sync all my 10 clock on the square serpent. Great. They are GPS synced and things are neat for them. The serpent is not moving, it's stationary. A sends a signal to B, B to C, and so on and since the speed of light is c in one direction (assume this) then J is synced with all the them and with A as well. Right?

Now I launch my light serpent train movement. And... it gets ****ed up. A sends a signal to B and B to C and so on and then J goes oops when it meets A.

But wait... A and B were in an inertial frame. They don't know they are moving at all. They send the signal, assume it's c and so they should not get unsynced. Ever. I an not getting involved from my lab, except for the very first sync before launching the light train.

Light is defined as c in any inertial frame so... this cannot happen. What Figure 4 says about the signals being anisotropic is... strange. A and B have been synced just before I launched them. A signals B, the signal has velocity c, B signals A, the signal velocity is c too. Perfect. Clocks A and B should be nicely synced. So when they travel from A to B to C and so on to J, no time gap should arrive. Assuming speed of light is c in any inertial frame. No?

Clocks are not synced via "A sends a signal to B, B to C ...", because that alone doesn't adjust clock settings. Instead, you could for example be at rest in the middle between A and B and receive light-pulses from both, when they have 12:00 A.M. If you don't receive both light pulses at the same time, you adjust B and repeat these activities until B is in sync with A. Then you stay in the middle between B and C and do the same procedure to get C in sync with B ...

Assume now, all clocks are in sync in the inertial frame.

Then comes "Now I launch my light serpent train movement." This is an acceleration of the clocks. I assume, that the head of the serpent is glued to it's tail. The 10 clocks are mounted at the serpent with equidistant locations, with clock A located in direction of movement before clock B and so on. The serpent should naturally become contiguously shorter with increasing speed by length-contraction, but the square (for example made of steel) hinders the serpent from becoming shorter, while it moves around the square. This created internal mechanical stress in the serpent. The distance between the clocks remains therefore the same with reference to the inertial frame. Because the clocks have the same acceleration profile (and therefore at each time the same speed), they stay in sync to each other (in the inertial frame), but can't be in sync with a clock at rest, because moving clocks run slow.

Now assume an observer moving with clock D. While acceleration, this observer feels a pseudo-gravitational field, like in an accelerating rocket. For this observer, as long as clock C is at the same edge of the square as clock D, clock C ticks faster than clock D, because of gravitational time-dilation. When the acceleration has finished, clock C is ahead of clock D by a constant offset, in the moving frame. If the observer moving with clock D now synchronizes clocks C and D, by adjusting D to a later time, then C and D become out-of-sync with reference to the inertial frame.

The following is wrong: "What Figure 4 says about the signals being anisotropic is... strange. A and B have been synced just before I launched them."

The text does not say, that A and B were synced when they were at rest in the inertial frame. There is no acceleration described. They are only assumed to be in sync to each other with reference to the inertial frame.

Is this a bit like the twin paradox?

I don't know, what a "twin paradox" is. I can't find any discussion about such a topic in PF

 

[alexandrinushka]

@Sagittarius A-Star thanks for your patience and thorough reply.
As for the Twin Paradox: https://en.m.wikipedia.org/wiki/Twin_paradox

 

[Nugatory]

As for the Twin Paradox: https://en.m.wikipedia.org/wiki/Twin_paradox

Sure, but what’s the connection to the topic of this thread?

 

[alexandrinushka]

@Sagittarius A-Star actually... hum...
The final formula for the time lag due to the Sagnac effect calculated by Rizzi and Ruggiero (their formula 25 with the proper Lorentz factor for the local rotating clock time) is different from the one in the link (formulae 9 and 13). The link takes c² - v² as a denominator, while R&R take c² as a denominator.
Wiki gives one that is the same with the link:

On Wiki they do indeed proceed by later removing the v² (as it is negligible due to the high c²), but I don't think R&R ever apply approximations in their derivation.

 

[alexandrinushka]

Sure, but what’s the connection to the topic of this thread?

Hello @Nugatory we were discussing earlier if the Sagnac effect applies as well to non-circular paths. I was thinking about some clocks, that would move around a square shape (but it can also be a triangle, whatever, but not necessarily a circle). When back in front of the stationary clock in the lab, the clocks that have moved around the square will show a time lag. So I was asking if that is similar to the twin paradox, when the twin that left Earth and then came back is actually younger. @Sagittarius A-Star said he could not find it on PF, so I just linked a brief reference.

 

[jartsa]

Hello @Nugatory we were discussing earlier if the Sagnac effect applies as well to non-circular paths. I was thinking about some clocks, that would move around a square shape (but it can also be a triangle, whatever, but not necessarily a circle). When back in front of the stationary clock in the lab, the clocks that have moved around the square will show a time lag. So I was asking if that is similar to the twin paradox, when the twin that left Earth and then came back is actually younger. @Sagittarius A-Star said he could not find it on PF, so I just linked a brief reference.

If you have one clock going around a small square, you will see a clock that is gaining time slowly, when you look at some point on the path. Because the clock is constantly doing small trips at high speed.

If you add more clocks on the path, what you see does not change very much ... because the square is small. i mean those clocks that have tried to synchronize themselves while going around the square.

 

[Sagittarius A-Star]

@Sagittarius A-Star said he could not find it on PF

That was only meant as a joke.