sandokhan lies regarding the Sagnac effect

You are embarrassing yourself.

This phase difference has been demonstrated in a rotating interferometer as well as in a geostationary one.
In the latter case, the Sagnac effect is due solely to earth’s rotation with ωl = ωE. The earth rotation rate ωE is about 2π/(86400 s) and the corresponding maximum phase difference is as large as 2 rad, when the wavelength is 0.6 µm and the loop area S = 0.2 km2. Thus, a loop interferometer can be utilized as a precise means to detect earth’s rotation rate.
Moreover, according to the local-ether model, earth’s orbital motion around the sun or others does not contribute to the Sagnac effect in an earthbound propagation loop. In as early as 1904 Michelson supposed that the Sagnac effect due to the orbital motion of the earth around the sun might be detectable, although the angular speed of the orbital motion is about 1/365 times that of the rotation .

Again, you cannot read.

We've been through this before.

MICHELSON SUPPOSED THAT THE ORBITAL SAGNAC MIGHT BE DETECTED, ALTOUGH HE THOUGHT THAT THE ANGULAR SPEED OF THE ORBITAL MOTION IS 1/365 TIMES THAT OF THE ROTATION.

Here is what was published though:

C.C. Su, "A Local-ether model of propagation of electromagnetic wave," in Bull. Am. Phys. Soc., vol. 45, no. 1, p. 637, Mar. 2000 (Minneapolis, Minnesota).

http://www.ee.nthu.edu.tw/ccsu/










Both the rotational and the orbital motions of the earth together with the orbital
motion of the target planet contribute to the Sagnac
effect. But the orbital motion of the sun has no effects
on the interplanetary propagation. On the other hand, as
the unique propagation frame in GPS and intercontinental
links is a geocentric inertial frame, the rotational motion
of the earth contributes to the Sagnac effect. But the orbital
motion of the earth around the sun and that of the
sun have no effects on the earthbound propagation. By
comparing GPS with interplanetary radar, it is seen that
there is a common Sagnac effect due to earth’s rotation
and a common null effect of the orbital motion of the sun
on wave propagation. However, there is a discrepancy in
the Sagnac effect due to earth’s orbital motion. Moreover,
by comparing GPS with the widely accepted interpretation
of the Michelson–Morley experiment, it is seen that
there is a common null effect of the orbital motions on
wave propagation, whereas there is a discrepancy in the
Sagnac effect due to earth’s rotation.


Based on this characteristic of uniqueness and switchability of the propagation frame,
we propose in the following section the local-ether model
of wave propagation to solve the discrepancies in the in-
fluences of earth’s rotational and orbital motions on the
Sagnac effect and to account for a wide variety of propagation
phenomena.


Anyway, the interplanetary Sagnac effect is due to
earth’s orbital motion around the sun as well as earth’s
rotation. Further, for the interstellar propagation where
the source is located beyond the solar system, the orbital
motion of the sun contributes to the interstellar Sagnac
effect as well.

Evidently, as expected, the proposed local-ether model
accounts for the Sagnac effect due to earth’s rotation and
the null effect of earth’s orbital motion in the earthbound
propagations in GPS and intercontinental microwave link
experiments. Meanwhile, in the interplanetary radar, it accounts
for the Sagnac effect due both to earth’s rotation
and to earth’s orbital motion around the sun.


Based on the local-ether model, the propagation is entirely
independent of the earth’s orbital motion around
the sun or whatever and the velocity v for such an earthbound
experiment is referred to an ECI frame and hence
is due to earth’s rotation alone. In the original proposal,
the velocity v was supposed to incorporate earth’s orbital
motion around the sun. Thus, at least, v²/c²
=~ 10^-8. Then the amplitude of the phase-difference variation
could be as large as π/3, when the wavelength is
0.6 µm and the path length is 10 m. However, as the velocity
v is the linear velocity due to earth’s rotation alone,
the round-trip Sagnac effect is as small as v²/c²∼ 10^-12 which is merely 10^-4 times that due to the orbital motion.


The Sagnac effect is a FIRST ORDER effect in v/c.

Even in the round-trip nature of the Sagnac effect, as it was applied in the Michelson-Morley experiment, thus becoming a second order effect within that context, we can see that the ORBITAL SAGNAC IS 10,000 TIMES GREATER than the rotational Sagnac effect.


What exactly is wrong with you rabinoz?

Leaving not the slightest doubt that the earth rotates and orbits the sun - two this that you vehemently deny and that the orbital Sagnac delay would be 1/365 times that of the rotational Sagnac delay.

Dr. Su quoted the opinion of A. Michelson, that's all.

Then, he went on to write:

Based on the local-ether model, the propagation is entirely
independent of the earth’s orbital motion around
the sun or whatever and the velocity v for such an earthbound
experiment is referred to an ECI frame and hence
is due to earth’s rotation alone. In the original proposal,
the velocity v was supposed to incorporate earth’s orbital
motion around the sun. Thus, at least, v²/c²
=~ 10^-8. Then the amplitude of the phase-difference variation
could be as large as π/3, when the wavelength is
0.6 µm and the path length is 10 m. However, as the velocity
v is the linear velocity due to earth’s rotation alone,
the round-trip Sagnac effect is as small as v²/c²∼ 10^-12 which is merely 10^-4 times that due to the orbital motion.

10,000 is not the same as 1/365, is it now?

What I like best about sandokhan is how every single scientist he references and quotes believes the earth is a globe.

show explicitly what is wrong with my derivation
It is my derivation which is irrefutable
can you show anything wrong with the derivation?

In fact jackblack never disclosed at all HOW he arrived at his calculations, the actual methodology.

That is, until today.

The fact that he used the same area and the same radius for two different Sagnac effects, and thus was off by a scale of 3,650,000 was just a consequence.

Having been forced to explain WHY he substracted the two values, finally jack revealed what he so shrewdly was hiding from the view of his readers:

I explained why you MUST subtract the 2 values.
If you have beam a and beam b, where they are counterpropogating, then in the arc with radius R2, say beam a is moving forwards and beam b is moving backwards (relative to the orbital motion).
This means the path length for b is much shorter, and thus the shift (measured as extra time taken for beam a relative to beam b) will be positive.

The figure used is this:

https://i.imgur.com/NEacN3P.png

This means the path length for b is much shorter, and thus the shift (measured as extra time taken for beam a relative to beam b) will be positive.

And you actually had the audacity to challenge me with this piece of shit?

A terrible, catastrophic error made by jackblack.

Dr. A.G. Kelly explains:



The shift IS DUE TO THE DIFFERENT SPEEDS RECORDED, AND NOT DUE TO THE DISTANCE.

You get a shift because of the different speeds recorded, c + v and c - v.

Basic theory of the Sagnac effect:

http://www.naturalphilosophy.org/pdf/ebooks/Kelly-TimeandtheSpeedofLight.pdf

The shift is due to the DIFFERENT SPEEDS RECORDED, C + V AND C - V, AND NOT DUE TO THE LENGTHS.


The Dufour-Prunier experiment proves beyond a shadow of a doubt that the speeds recorded are in fact c + v and c - v:

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1978311#msg1978311


There will never be a "the path length for b is much shorter".

The entire derivation is based upon a significant error which speaks volumes about jack's poor scientific background.

Only someone who does not understand the theory of the Sagnac effect can state that the path will be shorter.

There will be a difference in speed, c + v and c - v, and not a difference in the lengths of the paths.


In order to make his contraption work, he made another mistake:

One beam has the rotation make the path longer

Then, you no longer have the Sagnac effect.

Sagnac means you have different SPEEDS, and not different lengths for the paths.

while the other has it shrink

You cannot.

The paths must be the same.

It is the difference in speed, c + v and c - v which counts.

with the light travelling at a speed of c.

It cannot be c.

“In his final essay on the subject in 1937, Langevin proposed that the results published that year by Dufour and Prunier showed that one had to assume either (a) the light speed varied to c + wr in one direction and c – wr in the other direction, or (b) the time aboard the spinning apparatus had to change by a factor of +/-2wA/c2 in either direction. Indeed, Langevin went as far as to say that assuming (a), “we find, by a very simple and very general reasoning, the formula for the difference of the times of the path of the two light beams in the Sagnac experiment.” .

The proposition (b) though is untenable because if this were true then when the light beam passed back to the moving detector, the local time from each direction would be out of synchronization, meaning that the clocks cannot be counting real time and that the effective time dilation is meaningless. This was also pointed out by Herbert Ives in his 1938 paper criticizing Langevin. Ives says about the absurdity of Langevin’s proposition (b):

” There are of course not merely two clocks, but an infinity of clocks, where we include those that could be transported at finite speeds, and around other paths. As emphasized previously, the idea of “local time” is untenable, what we have are clock readings. Any number of clock readings at the same place are physically possible, depending on the behaviour and history of the  clocks used. More than one “time” at one place is a physical absurdity. “

The only explanation left, is Langevin’s proposition a) that the light speed varies by C+/-wr in one or the other direction around the disk, consistent with Dufour and Prunier’s experimental results."


If you have beam a and beam b, where they are counterpropogating, then in the arc with radius R2, say beam a is moving forwards and beam b is moving backwards (relative to the orbital motion).
This means the path length for b is much shorter, and thus the shift (measured as extra time taken for beam a relative to beam b) will be positive.

Dr. A.G. Kelly explains jack's monumental error:



This is how jack's piece of shit derivation got started, but he never revealed his actual warped logic until today, making it very difficult to find the actual cause of his catastrophic result (1/365 the incorrect value vs. 10,000 the correct value).

Here is how to best define your mental condition jack.
Your derivation reaches a figure of 1/365.

Yes, and until you show a problem with that derivation, it stands correct.
Now stop with your BS appeals to authority and deal with the derivation.

Is this not a sign of sure idiocy?

Yes, it is a sure sign of your idiocy, that you are unable to point out a flaw with my derivation, which clearly means the 1/365 figure is correct, yet you go on acting like it is wrong.
A sure sign of your idiocy.

For your information, the Harvard ads abstract service only published MAJOR ASTRONOMY AND PHYSICS PUBLICATIONS.

They don't publish. The collate abstracts, nothing more.
If a paper is published in a journal they collate the abstracts for, it will appear there.

Not garbage like your derivation.

You mean my completely correct derivation?
If so, no, they wont put that there, because it is nothing noteworthy as it is already well established by others (like post) that the Sagnac effect is dependent upon the area of the loop, not the distance from the centre of the orbit.
But don't worry, they do have some of the papers I have published.

Published in Physics Letters A, a journal primarily for letters rather than strong journal articles.
It has an impact factor of 1.8, making it quite low.
Are we to understand that you jack have such a low degree of ignorance as to try to minimize the PRL journal?

Yes, I have such a low degree of ignorance that I understand that it's impact factor is quite low, making it not a very good journal.

And no acknowledgement that you were completely wrong about where it was published?

I'm done with your ignorance regarding the academic publishing world.

Deal with the derivation or shut up.

derive the calculation of the Sagnac effect (i.e. the phase shift between 2 counterpropagating beams of light around a loop), and show that the orbital Sagnac is much greater,

This is exactly what I have done.

No it isn't. You have repeatedly ignored the requirement that there exists a loop with 2 counterpropagating beams of light.
The closest you have come is a hypothetical case where the sagnac loop is the entire orbit or entire planet.
But then you use 2 different loops.

The calculations for the rotational Sagnac.
And we can calculate the amount of time it would take light to travel once around a non-rotating circumference:

Do it for a small interferometer on the surface of Earth, not one which is the entire planet.

The calculations for the orbital Sagnac:
And we can calculate the amount of time it would take light to travel once around a non-rotating circumference:

And now you make your comparison meaningless by comparing 2 completely different loops.
Do it for the same loop as you had for Earth.
If you don't, then you are showing the Sagnac effect for Earth's orbit in one system, is larger than the Sagnac effect for Earth's rotation in another system.

We can continue dishonest comparisons like this.
Lets assume a hypothetical interferometer with a radius of 1 000 000 000 km, rotating with Earth, now, this 1 light year centred on Earth has a shift given by dt=4Aw/c^2, where A=pi*r^2.
So this gives dt=4 000 000 000 000 000 000*pi km^2*(1/86400 s)/90 000 000 000 (km/s)^2 = 4 000 000 *pi*(1/864*9) s = 1616 s.

But you only got 2 us for your orbit. So it looks like Earth's rotational Sagnac effect is much larger than its orbital one.

See what happens when you use 2 different loops?
The comparison is meaningless.

Now show the derivation FOR THE SAME LOOP!!!!!!
Until you do, your comparison is meaningless.

The rotational and the orbital Sagnac can be compared immediately:

That's right. For a given loop, A=A.
For the rotation we have wr, for the orbit we have wo.

∆t=4*A*w/c^2

Thus ∆to=4A*wo/c^2
and ∆tr=4A*wr/c^2.
Comparing the 2:
∆to/∆tr=(4A*wo/c^2)/(4A*wr/c^2)=wo/wr.
Thus it is quite simple to see that the orbital Sagnac is much less than the rotational one due to the much lower angular velocity.

Once again, if you use different loops, your comparison is meaningless.

I provided a quote from it, where it clearly indicates uniformly moving fibre doesn't produce a phase shift.

You are confusing FOG with FOC.
The authors explained in detail how the phase shift is obtained.

No. I'm not confusing anything. All a FOG is, is an optic fibre with 2 counter propagating beams of light, where this optic fibre is a loop, which when rotated results in a phase shift.
A FOC basically the same, except instead of rotation, you can pull the fibre along a loop and have the same result. You still need some form of loop, and it is the fibres non-uniform motion around this loop that results in the phase shift.

The author is clearly stating that uniform motion itself does not produce a phase shift. Instead it is something else which does it.

So this uniform motion did not produce a phase shift. As such, there is more to it than simple translational motion.
The authors explained:

Stop just quote mining the author. I have shown the problem with that.
Explain it yourself, or shut up.

The uniform motion did not produce a phase shift.
It was only in the loop of the fibre with non-uniform motion that produced a phase shift.

Professor Wang's seminal did prove that the Sagnac applied to linear motion.

No. Not linear motion. A loop. It is impossible for linear motion to be a loop.

If you don't, you will not correctly calculate the fringe shift.
But you do get the correct fringe shift using the correct calculation.

Yes, as I did, showing the orbital Sagnac is much smaller than the rotational one.

Try to imagine the center of the earth as the sun and the earth's orbit as the surface. So, the earth is in a rotational/circular pattern in its orbit. Just like the rotational sagnac, the earth rotates toward where the satellite emitted the signal, and with the orbit, the earth, the unit is orbited toward where the signal was emitted.

STOP JUST REPEATING THE SAME REFUTED CRAP!!!!!!
I already explained the issue. That would require the satellite to be orbiting the sun at a velocity significantly different to Earth.

This means the path length for b is much shorter, and thus the shift (measured as extra time taken for beam a relative to beam b) will be positive.
COMPLETELY WRONG!
Dr. A.G. Kelly explains:

Explains nothing related to that specific issue.
Explain what is wrong with it or shut up.

The shift IS DUE TO THE DIFFERENT SPEEDS RECORDED, AND NOT DUE TO THE DISTANCE.

Only in a non-inertial reference frame, which is not the frame being used.
It requires the loop to be stationary (not rotating), and then the distortion due to the non-inertial reference frame results in an anisotropy of the speed of light resulting in a fringe shift.

In a non-inertial reference frame, the loop rotates (and can translate). As such, this results in a change in distance, while the speed of light remains constant.

The difference in time is due to the different distances.

So no, I am still completely correct.

If you do not know how the Sagnac is derived please study:

I do know how, and I have shown how. You are yet to show a single error.

http://www.naturalphilosophy.org/pdf/ebooks/Kelly-TimeandtheSpeedofLight.pdf

And this still agrees with me. It states A is the area enclosed by the light path. No mention of the orbital distance.
So yet another of your sources indicates you are full of shit.


So once again, my derivation remains unrefuted.

You are embarrassing yourself.
Again, you cannot read.
We've been through this before.

Yes we have. You got completely destroyed with your claims shown to be pure garbage.

10,000 is not the same as 1/365, is it now?

That's right. All derivations of the Sagnac effect show the shift to be proportional to the area of the loop and angular velocity.
As such, for a given loop, the orbital Sagnac is much smaller than the rotational one.
As such, anyone claiming 10 000 rather than 1/365 is full of shit.

In fact jackblack never disclosed at all HOW he arrived at his calculations, the actual methodology.

Yes I did, right from the start, from the very first time we started discussing this and I presented my derivation. It was all there and you were unable to show a single thing wrong with it.

The fact that he used the same area and the same radius for two different Sagnac effects

Not the same radius, just the same area, as that way it is an honest comparison of the 2 effects on a given system.

thus was off by a scale of 3,650,000 was just a consequence.

Nope. I am correct. You are yet to show anything wrong with the derivation, so your claims of me being off is pure garbage.

Having been forced to explain WHY he substracted the two values, finally jack revealed what he so shrewdly was hiding from the view of his readers:

Again, I explained that right from the start.
You can also do it without subtracting the 2 values and instead adding the 2 values for the time taken by each beam and then subtracting them.

Once again, you are unable to show any problem.

And you actually had the audacity to challenge me with this piece of shit?

No. I have the audacity to challenge you with this irrefutable derivation, which you have been unable to show any issue with.

Dr. A.G. Kelly explains:

The Shift is proportional to the area enclosed by the light path, and thus the distance from the centre of rotation is irrelevant.
As such, Dr Kelly agrees with me that the correct formula is:
dt=4*A*w/c^2.
This means the orbital Sagnac (with a much smaller w) is much smaller than the rotational Sagnac.

Basic theory of the Sagnac effect:

No, new attempt at an interpretation.
But I do see you discarding all your old sources to try and come up with an excuse.

Feel free to provide the derivation for the loop in question (not your garbage ones), for both the rotation and orbit, using this methodology. I might get around to it later.

Sagnac means you have different SPEEDS, and not different lengths for the paths.

Only in a non-inertial reference frame.
The earliest derivations of the Sagnac effect used different lengths, not different speeds.
For a simple case of a rotating loop, one light beam has to rotate all the way around the loop, and then a little bit more to reach the source/detector.
The other one has to rotate a little bit less.
This derivation works without appealing to an anisotropy of the speed of light due to a non-inertial reference frame.

The paths must be the same.

No. As the loop is moving the paths CANNOT be the same.

“In his final essay on the subject in 1937, Langevin proposed

Yes, proposed, not proved.

So once again, my derivation remains correct and unrefuted and you are yet to provide your own derivation for the system in question.

So once again, my derivation remains correct and unrefuted and you are yet to provide your own derivation for the system in question.

Your derivation is no more.

It is totally false.

Explains nothing related to that specific issue.
Explain what is wrong with it or shut up.
Only in a non-inertial reference frame, which is not the frame being used.
It requires the loop to be stationary (not rotating), and then the distortion due to the non-inertial reference frame results in an anisotropy of the speed of light resulting in a fringe shift.

In a non-inertial reference frame, the loop rotates (and can translate). As such, this results in a change in distance, while the speed of light remains constant.

See, this is what was wrong with your piece of shit derivation.

It does not require AT ALL for the loop to be stationary, on the contrary.

And it is not the change in distance at all.

You cannot have the speed of light constant.

The Dufour-Prunier experiment proved as much.

Please read.

“In his final essay on the subject in 1937, Langevin proposed that the results published that year by Dufour and Prunier showed that one had to assume either (a) the light speed varied to c + wr in one direction and c – wr in the other direction, or (b) the time aboard the spinning apparatus had to change by a factor of +/-2wA/c2 in either direction. Indeed, Langevin went as far as to say that assuming (a), “we find, by a very simple and very general reasoning, the formula for the difference of the times of the path of the two light beams in the Sagnac experiment.” .

The proposition (b) though is untenable because if this were true then when the light beam passed back to the moving detector, the local time from each direction would be out of synchronization, meaning that the clocks cannot be counting real time and that the effective time dilation is meaningless. This was also pointed out by Herbert Ives in his 1938 paper criticizing Langevin. Ives says about the absurdity of Langevin’s proposition (b):

” There are of course not merely two clocks, but an infinity of clocks, where we include those that could be transported at finite speeds, and around other paths. As emphasized previously, the idea of “local time” is untenable, what we have are clock readings. Any number of clock readings at the same place are physically possible, depending on the behaviour and history of the  clocks used. More than one “time” at one place is a physical absurdity. “

The only explanation left, is Langevin’s proposition a) that the light speed varies by C+/-wr in one or the other direction around the disk, consistent with Dufour and Prunier’s experimental results."

Langevin did propose, but IVES DID PROVE.

Please read the classic paper signed Herbert Ives:

http://www.conspiracyoflight.com/Ives/Herbert_Ives_Light_Signals_Sent_Around_a_Closed_Path.pdf


Here is what you had the audacity to write:

One beam has the rotation make the path longer

Then, you no longer have the Sagnac effect.

Sagnac means you have different SPEEDS, and not different lengths for the paths.

while the other has it shrink

You cannot.

The paths must be the same.

It is the difference in speed, c + v and c - v which counts.

with the light travelling at a speed of c.

It cannot be c.

As such your derivation is thrashed and is shown for what it is: a useless piece of shit.


Here is what you wrote:

I explained why you MUST subtract the 2 values.
If you have beam a and beam b, where they are counterpropogating, then in the arc with radius R2, say beam a is moving forwards and beam b is moving backwards (relative to the orbital motion).
This means the path length for b is much shorter, and thus the shift (measured as extra time taken for beam a relative to beam b) will be positive.

Your latest excuse is worth nothing, you are not fooling anybody jack.

Only in a non-inertial reference frame.
No. As the loop is moving the paths CANNOT be the same.

But they are the same.

One of the greatest experts on the Sagnac effect of the 20th century, Dr. A.G. Kelly explains:



A clear and concise proof that you are wrong, and thus so is your derivation.

You still need some form of loop, and it is the fibres non-uniform motion around this loop that results in the phase shift.

The author is clearly stating that uniform motion itself does not produce a phase shift. Instead it is something else which does it.

You do not need a loop, that is the entire point of the paper.

It is a new discovery in Sagnac theory, that it applies precisely to uniform/translational motion.

No. Not linear motion. A loop. It is impossible for linear motion to be a loop.


But it is possible.

Professor Wang proved just that.

The authors explained:

The FOG travelled with the mechanical conveyor and its uniform motion did not cause any phase shift because the FOG is only sensitive to the rotational movement.

The experiment was repeated with 24 different
arrangements of conveyor speeds, fiber lengths, and the
three different FOC configurations shown in Fig.1.
The conveyor speeds were between 3 and 9 cm/s. The
loops had perimeters of 2.5, 4.0, 8.0, and 16.0 m; in
each case there were three turns of the fiber wound on
the loop.

As shown in Fig. 3, the phase shift or the traveltime
difference between two counter-propagating light
beams in the moving optic fiber was clearly observed
in all different configurations of FOCs. The phase shift
Δφ, and therefore, the travel-time difference Δt are
proportional to both the total length and the speed of
the moving fiber whether the motion is circular or
uniform. Other tests using smaller end wheels for the
FOC and fiber loops with additional curves also
confirmed the same finding.

Professor Wang's seminal did prove that the Sagnac applied to linear motion.


The crucial aspect of your piece of shit derivation is this:

I explained why you MUST subtract the 2 values.
If you have beam a and beam b, where they are counterpropogating, then in the arc with radius R2, say beam a is moving forwards and beam b is moving backwards (relative to the orbital motion).
This means the path length for b is much shorter, and thus the shift (measured as extra time taken for beam a relative to beam b) will be positive.

The Dufour-Prunier experiment, also proven by Ives, prove that you are wrong.

Dr. Kelly concise explanation shows that you are wrong.

And you were able to hide your actual derivation from your readers, were it not for my investigation which discovered your monumental error.


Please learn, here is how to precisely calculate the orbital Sagnac.

"The motion of the earth's orbit is also a sagnac effect. We should see light path distance differentials caused by the orbit just like we see if for earth's rotation.

The orbital path is simply longer and nothing else.

The earth - sun orbital frame is a sagnac rotating frame.

The Sagnac correction for the earth's rotation is applied because as the light moves toward the receiver, the receiver rotates with the earth changing the distance the signal travels.

In the same light, if the unit had been at the equator at noon, then it should see the full effect of the Sagnac effect of the earth's revolution around the sun.
In other words, assume a satellite is low on the horizon in the east at the equator.

We should measure a sagnac correction for the earth's rotation on its axis and a sagnac correction of the earth's rotation/revolution around the sun.
If sagnac is true for the earth's rotation, then light travels at one speed c. the speed of light cannot be increased by circular motion and presumably not by linear motion either.

If light travels at one speed c, then as the earth moves in it's revolution loop at 30k/s, while light moves c through space, the unit at the equator at noon would move with the earth' rotation and the earth's revolution cutting the distance the signal must travel to meet the unit.
The earth is rotating at 1000 mph. This shows up in GPS as c+v and c-v as you would expect with Sagnac.

All that is fine.

When the satellite emits at c, the earth rotates the receiver at v and so a correction is needed.

This is all OK.

Now, the earth is revolving around the sun at 67000 mph, as we are told by the heliocentrists.

Let's say the unit is at the equator and the satellite is low on the horizon in the east at noon.

That means the unit is traveling at the orbital speed of the earth at 67,000 MPH.

The satellite emits at one speed c in space. While the light travels through space toward the unit at c, the unit moves with the earth at 67,000 MPH. The unit cuts the distance that the light must travel.

This is not being seen by any experiements nor GPS."

Try to imagine the center of the earth as the sun and the earth's orbit as the surface. So, the earth is in a rotational/circular pattern in its orbit. Just like the rotational sagnac, the earth rotates toward where the satellite emitted the signal, and with the orbit, the earth, the unit is orbited toward where the signal was emitted.

Then we simply apply the well-known formula to this situation, where R = 150,000,000 km, v = wR (=30km/s).


This view is supported by mainstream science which acknowledge that the orbital Sagnac is much greater than the rotational Sagnac and that it is missing.

Journal of Electromagnetic Waves and Applications:

http://www.ee.nthu.edu.tw/ccsu/qem/f3c.pdf

For the interplanetary propagation, earth’s orbital
motion contributes to the Sagnac effect as well. This local-ether model
has been adopted to account for the Sagnac effect due to earth’s
motions in a wide variety of propagation phenomena, particularly the
global positioning system (GPS), the intercontinental microwave link,
and the interplanetary radar.


The peer reviewers at the Journal of Electromagnetic Waves and Applications agree that the orbital Sagnac is larger than the rotational Sagnac, that it is missing, and that a local-ether model has to be adopted in order to account for this fact.

So once again, my derivation remains correct and unrefuted and you are yet to provide your own derivation for the system in question.

Your derivation is no more.

It is totally false.

Present your derivation or the Sagnac delay of run away!

By the way, the Sagnac effect does not depend on relativity, nor does it conflict with it.
And depending on the reference frame used for the analysis it can be looked on as differing distances od different velocities.

By the way, in case you have not heard, the earth is a huge Globe that rotates once in roughly 23.9344696 Hours and orbits the sun once in about 365.256363004 mean solar days.

On the Heliocentric Globe part, you'll just have to take word of
who says:

This phase difference has been demonstrated in a rotating interferometer as well as in a geostationary one. In the latter case, the Sagnac effect is due solely to earth’s rotation with ω_l = ω_E. The earth rotation rate ω is about 2π/(86400 s) and the corresponding maximum phase difference is as large as 2 rad, when the wavelength is 0.6 µm
and the loop area S = 0.2 km². Thus, a loop interferometer can be utilized as a precise means to detect earth’s rotation rate.

Moreover, according to the local-ether model, earth’s orbital motion around the sun or others does not contribute to the Sagnac effect in an earthbound propagation loop. In as early as 1904 Michelson supposed that the Sagnac effect due to the orbital motion of the earth around the sun might be detectable, although the angular speed of the orbital motion is about 1/365 times that of the rotation.

Leaving not the slightest doubt that the earth rotates and orbits the sun - two things that you vehemently deny and that the orbital Sagnac delay would be 1/365 times that of the rotational Sagnac delay as we already knew.
So, go and find some flat earth scientists to support your silly ideas!

So once again, my derivation remains correct and unrefuted and you are yet to provide your own derivation for the system in question.
Your derivation is no more.

Nope. Until you show what is wrong with it, it remains correct and unrefuted.
Until you show your "correct" derivation (for a single loop with counterpropagating light beams), mine remains correct and unrefuted.

It is totally false.

Then why are you completely unable to show a single issue with it???

It does not require AT ALL for the loop to be stationary, on the contrary.

Yes it does. It uses a stationary loop in a non-inertial (rotating) reference frame to create an anisotropy in the speed of light around the loop to produce a phase shift.

You cannot have the speed of light constant.

Yes we can. MM proved as much.

Please read.

No. Provide the argument yourself or shut up. Stop with your BS appeals to authority.

Please read the classic paper signed Herbert Ives:

No. Explain the argument yourself.
And if you want me to read any paper, make sure it is published in an accepted scientific journal, not just some shitty website. Provide the DOI.

One beam has the rotation make the path longer
Then, you no longer have the Sagnac effect.

No, that is exactly what you have, and I have explained why. Stop just ignoring what I have said. Stop just repeating the same refuted BS.

You do not need a loop, that is the entire point of the paper.

You do need a loop. The entire point of the paper was to show that it was motion around a loop, not just rotation.

It is a new discovery in Sagnac theory, that it applies precisely to uniform/translational motion.

No, it doesn't. Its proof requires that it doesn't apply to translation.

But it is possible.

No it isn't.
Please describe how you can have linear motion in a loop without any other motion.

The authors explained:
The FOG travelled with the mechanical conveyor and its uniform motion did not cause any phase shift because the FOG is only sensitive to the rotational movement.

That's right, directly contradicting your claims.

The Dufour-Prunier experiment, also proven by Ives, prove that you are wrong.

No they don't.

Dr. Kelly concise explanation shows that you are wrong.

Nope. It agrees with me, the shift is proportional to the area of the loop times the angular velocity, making the orbital Saganc much smaller than the rotational one.
As such, it shows you are wrong.

And you were able to hide your actual derivation from your readers

Nope. I provided it in the prior thread, and provided it again in this thread. You are the one hiding, not me.

Please learn, here is how to precisely calculate the orbital Sagnac.

Stop just quoting crap. Do it yourself, using a common loop for the orbital and rotational sagnac.

The Sagnac correction for the earth's rotation is applied because as the light moves toward the receiver, the receiver rotates with the earth changing the distance the signal travels.

Really?
What were you saying before?
Here, let me remind you:

it is not the change in distance at all.
Sagnac means you have different SPEEDS, and not different lengths for the paths.

You might want to get your story straight before spouting such crap.
Which is it? Distance or speed? Or, does it not matter and they are just 2 ways of looking at the same thing?

Regardless, what you are discussing now is not the Sagnac effect. The Sagnac effect is not a linear effect, it is an effect resulting from 2 counterpropagating beams of light moving around a loop.

Now then, either show an actual problem with my derivation and provide your own derivation for one of the loops in question or SHUT UP!

I'll even be nice and do the derivation with keeping the distance the same and changing the speed.
For those that haven't been following along from the other thread, here is a link to my original derivation (I think):
https://www.theflatearthsociety.org/forum/index.php?topic=70614.msg1915658#msg1915658

And here is the picture that accompanies it (which I made after):


We have 2 beams of light moving around the interferometer loop shown in red.

Now what Sandokhan is objecting to is this part:

The total time for each will be the sum of the a and b part, so now we need to figure out what they are.
Well, in t1a, the light (travelling at speed c) travels a distance alpha*R2+omega*t1a.
Thus t1a*c=alpha*R2+omega*R2*t1a
Thus t1a=alpha*R2/(c-omega*R2).
Similarly, t2b=alpha*R1/(c-omega*R1).
And t1b*c=alpha*R1-omega*t1b
Thus t1b=alpha*R1/(c+omega*R1)
and t2a=alpha*R2/(c+omega*R2).

He is saying that the distance doesn't change. Instead the speed does. The speed becomes c+v or c-v depending on if it goes with or against the motion.
So let's try that:
In t1a, the light (which now travels at a speed of c-v) travels a distance of alpha*R2.
Thus t1a*(c-v)=alpha*R2.
Thus t1a=alpha*R2/(c-v)
But what is v?
Well this is the velocity of the arc, which is given by the angular velocity (omega) times the radius.
Thus v=omega*R2

So we end up with t1a=alpha*R2/(c-omega*R2)
Likewise t2b=alpha*R1/(c-omega*R1)
Then for t1b and t2a you have c+v instead (as it is going the opposite way) giving you:
t1b=alpha*R1/(c+omega*R1)
t2a=alpha*R2/(c+omega*R2)

Notice how this is exactly the same as what Sandokhan is objecting to?
That means it doesn't matter if you treat it as a different distance in an inertial reference frame or a different speed in a non-inertial reference frame; YOU GET THE SAME RESULT!!!

As such, Sandokhan bitching about it allegedly being a difference in speed is just another of its pathetic distractions.

The Sagnac effect remains proportional to the area of the loop, regardless of how far the loop is away from the centre of rotation, and the orbital sagnac remains 1/365 times the rotational sagnac.

jack, these are your own words:

I explained why you MUST subtract the 2 values.
If you have beam a and beam b, where they are counterpropogating, then in the arc with radius R2, say beam a is moving forwards and beam b is moving backwards (relative to the orbital motion).
This means the path length for b is much shorter, and thus the shift (measured as extra time taken for beam a relative to beam b) will be positive.

As such, your piece of shit derivation DOES NOT APPLY TO THE SAGNAC EFFECT.

Your own words betrayed you.

Sagnac effect = SAME DISTANCES/LENGTHS, DIFFERENT SPEEDS


This means the path length for b is much shorter, and thus the shift (measured as extra time taken for beam a relative to beam b) will be positive.

One beam has the rotation make the path longer
while the other has it shrink
with the light travelling at a speed of c.

Then, your piece of shit derivation has NOTHING to do with the Sagnac effect.

By your own admition.


Sagnac effect = SAME DISTANCES/LENGTHS, DIFFERENT SPEEDS.


You do need a loop.

You do not.

Professor Ruyong Wang proved that the Sagnac effect applies to uniform/translational motion.

This is an accepted fact of science.

Here is none other than Dr. A.G. Kelly telling you so:



Here are Professor Wang's seminal papers on the subject:

The experiment was repeated with 24 different
arrangements of conveyor speeds, fiber lengths, and the
three different FOC configurations shown in Fig.1.
The conveyor speeds were between 3 and 9 cm/s. The
loops had perimeters of 2.5, 4.0, 8.0, and 16.0 m; in
each case there were three turns of the fiber wound on
the loop.

As shown in Fig. 3, the phase shift or the traveltime
difference between two counter-propagating light
beams in the moving optic fiber was clearly observed
in all different configurations of FOCs. The phase shift
Δφ, and therefore, the travel-time difference Δt are
proportional to both the total length and the speed of
the moving fiber whether the motion is circular or
uniform. Other tests using smaller end wheels for the
FOC and fiber loops with additional curves also
confirmed the same finding.

Professor Wang's seminal did prove that the Sagnac applied to linear motion.


https://arxiv.org/ftp/physics/papers/0609/0609222.pdf (first experiment conducted by R. Wang)

https://arxiv.org/ftp/physics/papers/0609/0609202.pdf (second experiment carried out by R. Wang)


The first proof that the Sagnac effect applies to linear/translational motion AND to rotational motion was offered by the French physicists Dufour and Prunier.

None other than Einstein's most fervent supporter, Langevin, was baffled by the experiments and had to admit:


http://www.conspiracyoflight.com/pdf/Dufour_and_Prunier-On_the_Fringe_Movement_Registered_on_a_Platform_in_Uniform_Motion_%281942%29.pdf

A. Dufour and F. Prunier created Sagnac interferometers that were composites of moving and stationary paths, including stationary sources and stationary detectors. This was essentially to test if the relativistic approach could be distinguished from the classical approach.

"In all cases of this experimental test, the Sagnac effect was the same. This overturned Langevin’s analysis, and in 1937, he had to revise his explanation, as pointed out by Kelly: 

“In his final essay on the subject in 1937, Langevin proposed that the results published that year by Dufour and Prunier showed that one had to assume either (a) the light speed varied to c + wr in one direction and c – wr in the other direction, or (b) the time aboard the spinning apparatus had to change by a factor of +/-2wA/c2 in either direction. Indeed, Langevin went as far as to say that assuming (a), “we find, by a very simple and very general reasoning, the formula for the difference of the times of the path of the two light beams in the Sagnac experiment.” .

The proposition (b) though is untenable because if this were true then when the light beam passed back to the moving detector, the local time from each direction would be out of synchronization, meaning that the clocks cannot be counting real time and that the effective time dilation is meaningless. This was also pointed out by Herbert Ives in his 1938 paper criticizing Langevin. Ives says about the absurdity of Langevin’s proposition (b):

” There are of course not merely two clocks, but an infinity of clocks, where we include those that could be transported at finite speeds, and around other paths. As emphasized previously, the idea of “local time” is untenable, what we have are clock readings. Any number of clock readings at the same place are physically possible, depending on the behaviour and history of the  clocks used. More than one “time” at one place is a physical absurdity. “

The only explanation left, is Langevin’s proposition a) that the light speed varies by C+/-wr in one or the other direction around the disk, consistent with Dufour and Prunier’s experimental results."

In 1939, Dufour and Prunier carried out their
final experiment. They did a test with both the
beginning and end of the light path on the
spinning disc, but with the middle portion of
the path reflected off mirrors fixed in the
laboratory (directly above the disc). In this
test, they had both the light emitter and the
photographic recorder fixed in the laboratory.

The fringe shifts resulting from all the above
Dufour and Prunier tests were the same as in
their original Sagnac-type tests.

In 1942 Dufour and Prunier published a
composite paper reviewing their total
experimental work to date. At the end of this
paper they state that "the relativity theory
seems to be in complete disagreement with the
result which was garnered from the
experiment ".

Herbert Ives, Light Signals Sent Around a Closed Path:

http://www.conspiracyoflight.com/Ives/Herbert_Ives_Light_Signals_Sent_Around_a_Closed_Path.pdf

 Ives was a pioneer in the development of television at Bell Telephone Laboratories.  The following quotations are from his 1938 article.
 
     The experiment was interpreted by its author as positive evidence for the existence of the luminiferous ether…

     It is the purpose of this paper first to show that the Sagnac experiment in its essentials involves no consideration of rotation, and second to investigate the results obtained when transported clocks are used.
 
Ives analyzed the Sagnac experiment using a hexagonal path rather than a circular one.
 
He concluded with this statement:
 
     The net result of this study appears to be to leave the argument of Sagnac as to the significance of his experiment as strong as it ever was.

The Sagnac effect is not due to rotation, but instead is a linear effect due to a true anisotropic light speed in a moving frame.


Dufour and Prunier also proved that the light speed varied to c + wr in one direction and c – wr in the other direction for the Sagnac effect.

Which directly contradicts the main portion of your piece of shit derivation:

with the light travelling at a speed of c.


Your piece of shit derivation has nothing to do with the Sagnac effect.

This means the path length for b is much shorter, and thus the shift (measured as extra time taken for beam a relative to beam b) will be positive.

Dr. A.G. Kelly explains where you went wrong:




The extent of your utter ignorance as it pertains to the Sagnac effect is mind boggling.

Your posted a piece of shit derivation which is not related to the Sagnac effect, by your own admission.

SAGNAC EFFECT = SAME DISTANCES/LENGTHS, DIFFERENT SPEEDS C + V AND C - V

He is saying that the distance doesn't change. Instead the speed does. The speed becomes c+v or c-v depending on if it goes with or against the motion.
So let's try that:
In t1a, the light (which now travels at a speed of c-v) travels a distance of alpha*R2.
Thus t1a*(c-v)=alpha*R2.
Thus t1a=alpha*R2/(c-v)
But what is v?
Well this is the velocity of the arc, which is given by the angular velocity (omega) times the radius.
Thus v=omega*R2

So we end up with t1a=alpha*R2/(c-omega*R2)
Likewise t2b=alpha*R1/(c-omega*R1)
Then for t1b and t2a you have c+v instead (as it is going the opposite way) giving you:
t1b=alpha*R1/(c+omega*R1)
t2a=alpha*R2/(c+omega*R2)


You numskull!

Now you have FOUR light beams instead of the TWO REQUIRED BY THE SAGNAC EFFECT.

That is why I cordially invited you to read up on the Sagnac effect:

http://www.naturalphilosophy.org/pdf/ebooks/Kelly-TimeandtheSpeedofLight.pdf

The fact that a light signal, that is sent both
clockwise and anticlockwise around a path on
a rotating disc, takes different times to return
to the source was discovered by Sagnac over
eighty years ago.


TWO SIGNALS

NOT FOUR

ONE LOOP

Not only your derivation had nothing to do with the Sagnac, your latest attempt also falls within the same category.

SAGNAC EFFECT APPLIES TO BOTH LOOPS AND TRANSLATIONAL MOTION.

ONE LOOP, TWO COUNTER-PROPAGATING BEAMS

YOU HAVE FOUR COUNTER-PROPAGATING BEAMS.

A total failure on your part to understand the Sagnac.

As such, your piece of shit derivation DOES NOT APPLY TO THE SAGNAC EFFECT.

You are yet to justify that.

Your own words betrayed you.

Nope. That would be you, were you can't decide on if it is distance or speed that changes.

Sagnac effect = SAME DISTANCES/LENGTHS, DIFFERENT SPEEDS

No. Sagnac effect = a phase shift of 2 counterpropagating beams of light around a loop.
This can be considered as a change in distance in an inertial reference frame or a change in the speed of light in a non-inertial (rotating) reference frame.
The same result is achieved.

You do not.
Professor Ruyong Wang proved that the Sagnac effect applies to uniform/translational motion.

Stop just repeating the same crap.
That work showed you need a fibre optic conveyer, a loop, where the loop was moving around some path. It was not uniformly moving. To achieve their results they neede uniform translation motion to produce no phase shift.

Now stop lying.

The experiment was repeated with 24 different
arrangements of conveyor speeds, fiber lengths

Nope. It all used the same length of fibre. They rolled up the unused section with the FOG and just had it move uniformly, and NOT CONTRIBTUE!!!!

The loops had perimeters of 2.5, 4.0, 8.0, and 16.0 m

See how it appeals to loops?
Still going to claim you can do it without a loop?

Professor Wang's seminal did prove that the Sagnac applied to linear motion.

Again, stop lying. It proved that it applies to a loop, where the fibre optics is moving around that loop. It did not show it applies to linear motion.

Which directly contradicts the main portion of your piece of shit derivation:

Nope. It produces the same result.

Now you have FOUR light beams instead of the TWO REQUIRED BY THE SAGNAC EFFECT.

Nope. I just have 2.

I have 4 different segments of time. Each of the light beams has to go through 2 of these temporal segments.

One beam of light travels along the arc with radius R2 (in time t1a), then in, then along the arc of radius R1 (in time t1b), then out.
The other beam travels a similar path in an opposite direction with times t2a and t2b.

So I have 2 beams.
One takes a total time of t1a+t1b+tc, where tc is the time taken to go in and out and is the same for both beams.
The other takes a time of t2a+t2b+tc.

So just 2 beams.

TWO SIGNALS
NOT FOUR
ONE LOOP

Yes, just like I have, and completely unlike what you have.

Not only your derivation had nothing to do with the Sagnac, your latest attempt also falls within the same category.

Nope. My derivation is still to do with the Sagnac effect and still falls in the same category, completely disproving your BS with you finding whatever pathetic excuse you can to try and dismiss it.

What will your next pathetic excuse be? Or will you run away again now?

Ever since you disclosed HOW you achieved your piece of shit derivation, and I was able immediately to debunk it, YOUR STORY HAS BEEN CHANGING BY THE MINUTE.

Faced with the proofs that the Sagnac effect means THE SAME DISTANCES/LENGTHS AND DIFFERENT SPEEDS (C + V AND C - V), you finally gave in:

I'll even be nice and do the derivation with keeping the distance the same and changing the speed.

Which means your messages were just a charade meant to fool your readers.


That work showed you need a fibre optic conveyer, a loop, where the loop was moving around some path. It was not uniformly moving. To achieve their results they neede uniform translation motion to produce no phase shift.

No such thing ever happened.

Here is reality starring you in the face:

The experiment was repeated with 24 different
arrangements of conveyor speeds, fiber lengths, and the
three different FOC configurations shown in Fig.1.
The conveyor speeds were between 3 and 9 cm/s. The
loops had perimeters of 2.5, 4.0, 8.0, and 16.0 m; in
each case there were three turns of the fiber wound on
the loop.

As shown in Fig. 3, the phase shift or the traveltime
difference between two counter-propagating light
beams in the moving optic fiber was clearly observed
in all different configurations of FOCs. The phase shift
Δφ, and therefore, the travel-time difference Δt are
proportional to both the total length and the speed of
the moving fiber whether the motion is circular or
uniform. Other tests using smaller end wheels for the
FOC and fiber loops with additional curves also
confirmed the same finding.

Professor Wang's seminal did prove that the Sagnac applied to linear motion.


https://arxiv.org/ftp/physics/papers/0609/0609222.pdf (first experiment conducted by R. Wang)

https://arxiv.org/ftp/physics/papers/0609/0609202.pdf (second experiment carried out by R. Wang)

Dr. A.G. Kelly explains:



Sagnac effect = a phase shift of 2 counterpropagating beams of light around a loop.

Not necessarily a loop.

Dr. Ives and Dr. Wang proved that it can be uniform/translational motion.

But yes, it does involve two counter-propagating beams of light.

It proved that it applies to a loop, where the fibre optics is moving around that loop. It did not show it applies to linear motion.

Take a look jack at your miserable path of least resistance to denying the evident truth.

The loops are a different part of the entire path.

The Sagnac effect was observed on the linear path.

Read the papers.

Nope. I just have 2.

See how you are changing your story line again?

Here is your own description:

He is saying that the distance doesn't change. Instead the speed does. The speed becomes c+v or c-v depending on if it goes with or against the motion.
So let's try that:
In t1a, the light (which now travels at a speed of c-v) travels a distance of alpha*R2.
Thus t1a*(c-v)=alpha*R2.
Thus t1a=alpha*R2/(c-v)
But what is v?
Well this is the velocity of the arc, which is given by the angular velocity (omega) times the radius.
Thus v=omega*R2

So we end up with t1a=alpha*R2/(c-omega*R2)
Likewise t2b=alpha*R1/(c-omega*R1)
Then for t1b and t2a you have c+v instead (as it is going the opposite way) giving you:
t1b=alpha*R1/(c+omega*R1)
t2a=alpha*R2/(c+omega*R2)

I explained why you MUST subtract the 2 values.
If you have beam a and beam b, where they are counterpropogating, then in the arc with radius R2, say beam a is moving forwards and beam b is moving backwards (relative to the orbital motion).
This means the path length for b is much shorter, and thus the shift (measured as extra time taken for beam a relative to beam b) will be positive.
They then continue along the light path and travel along the arc of radius R1. Now beam a is moving backwards and beam b is moving forwards. That means the shift will now be negative.
You add these 2 shifts and get a shift proportional to R2^2-R1^2.


YOUR OWN WORDS.

TWO SHIFTS.

FOUR COUNTER-PROPAGATING BEAMS OF LIGHT.

You have a terrible knowledge of the Sagnac effect.

Your piece of shit derivation, the latest version, ALSO DOES NOT APPLY TO THE SAGNAC EFFECT.

SAGNAC EFFECT = A SINGLE SOURCE OF LIGHT, TWO COUNTER-PROPAGATING BEAMS OF LIGHT (EITHER IN A LOOP OR ON A STRAIGHT LINE), SAME DISTANCES, DIFFERENT SPEEDS (C + V/C - V).

One beam of light travels along the arc with radius R2 (in time t1a), then in, then along the arc of radius R1 (in time t1b), then out.
The other beam travels a similar path in an opposite direction with times t2a and t2b.


THAT IS NOT WHAT YOU STATED YESTERDAY.

YOUR OWN WORDS.

I explained why you MUST subtract the 2 values.
If you have beam a and beam b, where they are counterpropogating, then in the arc with radius R2, say beam a is moving forwards and beam b is moving backwards (relative to the orbital motion).
This means the path length for b is much shorter, and thus the shift (measured as extra time taken for beam a relative to beam b) will be positive.
They then continue along the light path and travel along the arc of radius R1. Now beam a is moving backwards and beam b is moving forwards. That means the shift will now be negative.
You add these 2 shifts and get a shift proportional to R2^2-R1^2.


FOUR COUNTER-PROPAGATING BEAMS OF LIGHT.

A FORWARD, AND B BACKWARDS.

THEN, B FORWARD AND A BACKWARDS.

BUT THIS IS NOT SAGNAC.

IN THE SAGNAC YOU HAVE A SINGLE SOURCE OF LIGHT AND TWO COUNTER-PROPAGATING BEAMS OF LIGHT, IN THIS INSTANCE IN A LOOP.

THAT'S IT.

YOU CAME UP WITH FOUR COUNTER-PROPAGATING BEAMS OF LIGHT.

BUT THIS IS NOT SAGNAC.

So I have 2 beams.

No you don't.

You have four beams.

Remember this?

I'll even be nice and do the derivation with keeping the distance the same and changing the speed.

SINCE NOW YOU HAVE THE SAME DISTANCE, YOU HAVE A SINGLE LOOP.

NO NEED TO CALCULATE TWO DIFFERENT TIMES FOR A SINGLE LOOP FOR EACH BEAM.

A SINGLE LOOP = TWO COUNTER-PROPAGATING BEAMS OF LIGHT.

BUT YOU STATED YOU HAVE FOUR BEAMS:

I explained why you MUST subtract the 2 values.
If you have beam a and beam b, where they are counterpropogating, then in the arc with radius R2, say beam a is moving forwards and beam b is moving backwards (relative to the orbital motion).
This means the path length for b is much shorter, and thus the shift (measured as extra time taken for beam a relative to beam b) will be positive.
They then continue along the light path and travel along the arc of radius R1. Now beam a is moving backwards and beam b is moving forwards. That means the shift will now be negative.
You add these 2 shifts and get a shift proportional to R2^2-R1^2.

WHY DID YOU DO THIS?

BECAUSE YOU THOUGHT YOU HAVE TWO DIFFERENT DISTANCES, THAT'S WHY.

NOW YOU NO LONGER HAVE THE COMFORT OF DIFFERENT DISTANCES.

YOU HAVE A SINGLE LOOP.

AND TWO BEAMS OF LIGHT.

You really have a terrible knowledge of the Sagnac effect, no other conclusion is possible here.

Other than the fact that you are deliberately trying to fool your readers.

The loop interferometer described in the following graphic has nothing to do with the physical description of the orbital Sagnac effect:

https://i.imgur.com/NEacN3P.png

We have 2 beams of light moving around the interferometer loop shown in red.

The only way to build such a device would be to use a PHASE-CONJUGATE GYRO INTERFEROMETER:

https://arxiv.org/ftp/physics/papers/0609/0609202.pdf (page 5, figures 5 and 6)

In the real physical world, YOU CANNOT HAVE THE EARTH MOVE BACKWARDS WITH A 30KM/S SPEED NECESSARY FOR THE R1 ARC.

We can easily conceive of a situation like this, where we have a rectangular interferometer at the surface of the Earth, being rotated with a speed of 30km/s (v=wR, w angular velocity of orbital path, R = 150,000,000):



For the sake of simplicity we can replace the rectangular shape with a circular interferometer.

The graphic proposed in this thread is a THEORETICAL situation, which can be described mathematically, but it has absolutely no connection to the real world, in this case, the orbital Sagnac.


So now lets let alpha be the angle subtended by the arcs, and omega be the angular velocity.
So, one beam of light will propagate in the same direction as the big arc and the opposite direction as the little arc.
What this means is its trip along the big arc will be longer than if it was stationary and shorter along the little arc.

If you have a stationary interferometer in the approximate shape of a rectangle (just like in the graphic) and a source of light located right on the path, then the Sagnac effect would be 2Lv/c², where L = 4a + 4b and the observer, moving at speed v to the right in the laboratory frame, sends two counter propagating light signals when he is at the origin.

What you are describing is a mechanical device, in the shape of a phase-conjugate gyro interferometer, which can go back and forth to create the actual graphic described.

But the Earth cannot go back on its orbit to actually create the interferometer described in the graphic.

Meanwhile, the light going the other way has to travel a distance of alpha*R1+omega*R1*t2b+alpha*R2-omega*R2*t2a, for similar reasons.

YOU CAN ONLY DO THAT IN A LAB WITH A PHASE CONJUGATE GYRO INTERFEROMETER.

If you have an interferometer, in the shape described, two satellites and two receivers/emitters on the surface of the Earth, then you have a very different situation best visualized thusly:

Try to imagine the center of the earth as the sun and the earth's orbit as the surface. So, the earth is in a rotational/circular pattern in its orbit. Just like the rotational sagnac, the earth rotates toward where the satellite emitted the signal, and with the orbit, the earth, the unit is orbited toward where the signal was emitted.

In the graphic, you are actually constructing the interferometer using the angular velocity, in the real world the interferometer exists already, and is being rotated as a whole:

"The motion of the earth's orbit is also a sagnac effect. We should see light path distance differentials caused by the orbit just like we see if for earth's rotation.

The orbital path is simply longer and nothing else.

The earth - sun orbital frame is a sagnac rotating frame.

The Sagnac correction for the earth's rotation is applied because as the light moves toward the receiver, the receiver rotates with the earth changing the distance the signal travels.

In the same light, if the unit had been at the equator at noon, then it should see the full effect of the Sagnac effect of the earth's revolution around the sun.
In other words, assume a satellite is low on the horizon in the east at the equator.

We should measure a sagnac correction for the earth's rotation on its axis and a sagnac correction of the earth's rotation/revolution around the sun.
If sagnac is true for the earth's rotation, then light travels at one speed c. the speed of light cannot be increased by circular motion and presumably not by linear motion either.

If light travels at one speed c, then as the earth moves in it's revolution loop at 30k/s, while light moves c through space, the unit at the equator at noon would move with the earth' rotation and the earth's revolution cutting the distance the signal must travel to meet the unit.
The earth is rotating at 1000 mph. This shows up in GPS as c+v and c-v as you would expect with Sagnac.

All that is fine.

When the satellite emits at c, the earth rotates the receiver at v and so a correction is needed.

This is all OK.

Now, the earth is revolving around the sun at 67000 mph, as we are told by the heliocentrists.

Let's say the unit is at the equator and the satellite is low on the horizon in the east at noon.

That means the unit is traveling at the orbital speed of the earth at 67,000 MPH.

The satellite emits at one speed c in space. While the light travels through space toward the unit at c, the unit moves with the earth at 67,000 MPH. The unit cuts the distance that the light must travel.

This is not being seen by any experiements nor GPS."

The only way to physically have an interferometer as seen in the graphic would be to employ the use of a phase conjugate gyro interferometer. But the Earth cannot go back on its orbital path to physically create the interferometer described.


If you want an interferometer which is already built and fixed to function near the surface of the Earth, then this situation can be described easily as I have done already in terms of the orbital Sagnac effect.

In the graphic, you are actually constructing an interferometer based on the orbital path of the Earth, as such the R1 arc is physically impossible, as the Earth cannot go back on its orbital path with 30km/s.

What you want is an already built interferometer where you have a light source in one of the corners, or from a side, WHICH IS MOVING ALONG WITH THE EARTH ON A ORBITAL PATH.


Moreover, satellites transmit their signal to Earth and that’s that.  There’s no transmission back to the satellites, or between satellites, and none of their signals are made to loop the equator.

("One place where the Sagnac effect would have relevance would be in the synching of ground-monitoring stations.  There are six monitoring stations located roughly along the equator and these would need to be kept in-synch with each other.  Such a process would need to adjust for the Sagnac effect, as the signal from the designated master station was transmitted to the others.")

A SITUATION TOTALLY DIFFERENT THAN THE LOOP INTERFEROMETER DESCRIBED IN THE GRAPHIC ITSELF.

Actually what we have is a uniform/translational motion toward the source of the light, the delay being the Sagnac effect, exactly as described by Ives, Wang and Dufour-Prunier.


So, the graphic itself has nothing to do with the orbital Sagnac.

If you are physically building the actual interferometer, as you have described it, going back and forth with a certain angular velocity, it can only be done with a phase conjugate gyro interferometer.

If the interferometer exists already (say two satellites and two receivers/emitters on the surface of the Earth), then we have the following situation:

Try to imagine the center of the earth as the sun and the earth's orbit as the surface. So, the earth is in a rotational/circular pattern in its orbit. Just like the rotational sagnac, the earth rotates toward where the satellite emitted the signal, and with the orbit, the earth, the unit is orbited toward where the signal was emitted.

A FIXED INTERFEROMETER WHICH IS BEING ROTATED WITH A CERTAIN VELOCITY.

Then we simply apply the well-known formula to this situation, where R = 150,000,000 km, v = wR (=30km/s).

Since speed of light is constant c+v and c-v should still equal c. Although those formulas are not correct.

Here is how to best visualize the orbital Sagnac:



The satellites transmit their signal to Earth and that’s that.  There’s no transmission back to the satellites, or between satellites, and none of their signals are made to loop the equator.

As such we do not have a loop interferometer at all.

What we have is exactly the uniform/translational motion exemplified in the Ives, Wang and Dufour-Prunier experiments.

Here is how modern science calculates the rotational Sagnac:




Try to imagine the center of the earth as the sun and the earth's orbit as the surface. So, the earth is in a rotational/circular pattern in its orbit. Just like the rotational sagnac, the earth rotates toward where the satellite emitted the signal, and with the orbit, the earth, the unit is orbited toward where the signal was emitted.

"The motion of the earth's orbit is also a sagnac effect. We should see light path distance differentials caused by the orbit just like we see if for earth's rotation.

The orbital path is simply longer and nothing else.

The earth - sun orbital frame is a sagnac rotating frame.

The Sagnac correction for the earth's rotation is applied because as the light moves toward the receiver, the receiver rotates with the earth changing the distance the signal travels.

In the same light, if the unit had been at the equator at noon, then it should see the full effect of the Sagnac effect of the earth's revolution around the sun.
In other words, assume a satellite is low on the horizon in the east at the equator.

We should measure a sagnac correction for the earth's rotation on its axis and a sagnac correction of the earth's rotation/revolution around the sun.

If light travels at one speed c, then as the earth moves in it's revolution loop at 30k/s, while light moves c through space, the unit at the equator at noon would move with the earth' rotation and the earth's revolution cutting the distance the signal must travel to meet the unit.

The earth is rotating at 1000 mph. This shows up in GPS as c+v and c-v as you would expect with Sagnac.

All that is fine.

When the satellite emits at c, the earth rotates the receiver at v and so a correction is needed.

This is all OK.

Now, the earth is revolving around the sun at 67000 mph, as we are told by the heliocentrists.

Let's say the unit is at the equator and the satellite is low on the horizon in the east at noon.

That means the unit is traveling at the orbital speed of the earth at 67,000 MPH.

The satellite emits at one speed c in space. While the light travels through space toward the unit at c, the unit moves with the earth at 67,000 MPH. The unit cuts the distance that the light must travel.

This is not being seen by any experiements nor GPS."

Now, the final piece of the puzzle.

In a remarkable achievement, Dr. Paul Marmet has shown that the velocity of light is c±v with respect to an observer, moving in straight line at velocity v, as it applies to the situation being discussed here, GPS satellites and the rotational/orbital motion of the Earth (heliocentric version).

Beginning in 1967 Marmet served as director of the laboratory for Atomic and Molecular Physics at Laval University in Quebec City, Canada, serving in that role until 1982. From 1983 to 1990, Marmet was a senior researcher at the Herzberg Institute of Astrophysics of the National Research Council of Canada in Ottawa. In 1990 Marmet was an Assistant Professor of Physics at the University of Ottawa.

A past president of the Canadian Association of Physicists (1981-1982), he also served as a member of the executive committee of the Atomic Energy Control Board of Canada from 1979 to 1984.  Marmet was elected Fellow of the Royal Society of Canada in 1973 and was made an Officer of the Order of Canada in 1981.  The Order of Canada is the highest decoration bestowed by the Canadian government.


Experimental Confirmation of the Discordant Einstein's Synchronization Method with the GPS
             There are direct measurements proving that the velocity of light in one direction is c±v with respect to the moving observer. This discordant synchronization given in equation 13 has been measured in the world system of clock synchronization with the Global Positioning System. It is then observed experimentally that the Einstein's method of synchronization using the "half time interval" taken by a reflected beam of light is inadequate to determine the correct time. A correction (which is the Sagnac effect) has to be added.
            As an example, let us assume that clock a (from figure 2) is in New York (N.Y.), and clock ß is in San Francisco (S.F.) as illustrated on figure 3. The velocity v is the velocity of rotation of the Earth around the pole axis, at the location where the experiment is done. The distance l is the distance between New York and San Francisco (dotted line on figure 3).



After the initial synchronization of clock a with a mobile atomic clock called µ, that clock is moved from New York to San Francisco at a constant altitude and slow velocity e (see figure 3). The constant altitude (at sea level) avoids other corrections due to the change of gravitational energy, which is irrelevant in this paper. The equivalent of such an experiment has been done by Sadeh [3] using a truck containing a number of accurate atomic clocks, previously synchronized with a primary standard of time. In the truck, moving clocks were sent down across USA. This experiment is reported in Science [4]. Using the GPS correction (which is mathematically identical to equation 13, the correct time is set up between clock a in New York and clock ß in San Francisco.
            The reader must be aware of the fundamental principles of physics involved in the GPS. The standards for the synchronization of clocks stations used by the Global Positioning System have been published in 1990 by the International Radio Consultative Committee: International Telecommunication Union CCIR [5] which uses similar rules as the 1980 publication of the CCDS (Comité Consultatif pour la définition de la Seconde: Bureau International des Poids et Mesures) [6].
            The Global Positioning System (GPS) determines that after clock µ moves away from clock a in New York, toward clock ß in San Francisco, its display accumulates an extra 14 ns (approximately) with respect to clock ß.  We know that due to the Earth rotation, between N.Y. and S.F. clock µ moves at velocity (v-e), which is the velocity of rotation of the Earth "v" minus the velocity of the truck "e". Therefore 14 ns are subtracted to its display at its arrival in order to give a correct synchronization of time on clock ß in S.F.. This correction is identical to equation 13. This correction is the same as the one programmed automatically in the GPS.
            Experimentally, an equivalent experiment has also been done carrying a clock between Washington and Tokyo by Saburi et al. [7]. It is then an experimental fact that the two clocks (a and ß) are not naturally synchronized at the same value, as a result of the discordant Einstein's synchronization method explained above.
            There is another well-known way to synchronize the clocks between these two stations (a and ß). It is done sending radio signals transmitted simultaneously (east-west and west-east) between these two cities. Again, it is observed that a simultaneous transmission of radio signals between New York and San Francisco does not give "directly" the same correct clock display (time) in both cities. There is a difference of about 14 ns that must be subtracted to the clock in San Francisco in order to get the correct GPS time. This correction is identical to the one when we are carrying clocks. This correction corresponds to a change of velocity c±v between stations.
            This GPS synchronization has been verified in numerous experiments. It is identical to the calculations presented in this paper and also to the Sagnac's effect, (which is included in the GPS). Among the GPS list of corrections, there is a correction involving a parameter taking into account how many Earth meridians are crossed by light or by the moving clock µ, between the two locations. Kelly [8] explains that the correction used by the GPS is:



where w is the angular velocity of rotation of the Earth, A_E is the projected area on the Earth equator plane of the path used by light (or by a slowly moving clock) between the two stations. We define l as the distance between the two stations, both moving at velocity v. The circumference of the Earth is called "circ". Therefore the area A_E is



The angular velocity  w is equal to v/r. The circumference of the Earth is 2pr. Equation 23 in equation 22 gives:



We see that the GPS correction of clocks (equation 24)  is identical to the Sagnac effect.

One must conclude that there exists no space-time distortion of any kind. It is no longer necessary to fascinate people with the magic of relativity. Unless we accept the absurd solution that the distance between N.Y. to S.F. is smaller than the distance between S.F. and N.Y., we have to accept that in a moving frame, the velocity of light is different in each direction. As mentioned above, this difference is even programmed in the GPS computer in order to get the correct Global Positioning. This proves that the experimental velocity of light with respect to a moving observer is c±v.

 Let us show now that the velocity of light is c±v with respect to an observer, moving in straight line at velocity v.  Let us consider two flashes of light emitted from our Earth moving in a direction parallel to its motion, during its orbit around the Sun.  The two light beams are emitted simultaneously in opposite directions along the Earth’s orbit.  Around the Sun, there are some stationary stations reflecting light, which are distributed so that light emitted from Earth can circle the Sun simultaneously in both directions.  For example, we can assume eight stationary mirrors reflecting light and forming an octagon around the sun.  For the beam moving in the same direction as the Earth, the moving observer will measure a longer time interval (than in opposite direction) before light completes the rotation around the Sun and reaches Earth again, because the Earth has moved a small distance, during that travel time interval of light.  The velocity of light calculated is then c-v (for the Earth observer).  Also, for the observer on the moving Earth, the velocity will be calculated as c+v for the beam of light moving in the opposite direction.
        Between each of the eight pairs of mirrors, in the forward direction, when the observer uses his proper units in his frame and Einstein’s synchronization, the moving observer will “believe” that the velocity of light is c.  This is due to the erroneous clock synchronization explained above.  Also, we could install a GPS around the Sun (as the one on Earth) and we would find, just as on Earth, those clocks are slowing down and standard meters dilated due to the kinetic energy of the observer.  The distance between a pair of mirrors corresponds to the distance between New York and San Francisco on Earth, as in the example above.  Again, the velocity of light moving along a straight line moves at velocity c-v with respect to the observer on Earth, if he uses a correct synchronization method (i.e. from the Sun's pole).

The most remarkable thought experiment is when the Sun is completely removed from the center of the system above.  We mean, an experiment done very far in empty outer space.  Then again, the observer moves at velocity v, in straight line, along one side of the octagon.  Due to the observer’s velocity, light will take more time to complete the complete octagonal path when he moves in the same direction as light, than if light and the observer are moving in the opposite direction.  In fact, any mass, either the Earth or the Sun is irrelevant.  Again, this is the classical Sagnac effect.


Try to imagine the center of the earth as the sun and the earth's orbit as the surface. So, the earth is in a rotational/circular pattern in its orbit. Just like the rotational sagnac, the earth rotates toward where the satellite emitted the signal, and with the orbit, the earth, the unit is orbited toward where the signal was emitted.

Ever since you disclosed HOW you achieved your piece of shit derivation, and I was able immediately to debunk it, YOUR STORY HAS BEEN CHANGING BY THE MINUTE.

Nope. I disclosed how right from the start and you were completely unable to debunk you. Instead you just bitched about the conclusion.
Now you seem to finally understand that you can't just bitch about the conclusion and isntead need to focus on the derivation so you contradict yourself and come up with whatever crap you can to try and reject (not refute) the derivation.

Faced with the proofs that the Sagnac effect means THE SAME DISTANCES/LENGTHS AND DIFFERENT SPEEDS (C + V AND C - V), you finally gave in:

Nope. Not faced with proof.
The Sagnac effect's simplest explanation is one that uses an inertial frame of reference in which there are different distances with the same speed.
An alternative explanation is that it is the same distance but different speeds.

I'll even be nice and do the derivation with keeping the distance the same and changing the speed.
Which means your messages were just a charade meant to fool your readers.

No, which meant your objection was just a pathetic distraction to try and escape your defeat.

Like I said, you can do it based upon either method. You get the same result.

Now quit your whinging and focus on the derivation itself.

That work showed you need a fibre optic conveyer, a loop, where the loop was moving around some path. It was not uniformly moving. To achieve their results they neede uniform translation motion to produce no phase shift.
No such thing ever happened.
Here is reality starring you in the face:
The experiment was repeated with 24 different
arrangements of conveyor speeds, fiber lengths, and the
three different FOC configurations shown in Fig.1.
The conveyor speeds were between 3 and 9 cm/s. The
loops had perimeters of 2.5, 4.0, 8.0, and 16.0 m; in
each case there were three turns of the fiber wound on
the loop.

No. Here is reality staring you in the face. It clearly states it is a LOOP!!!!
Do you understand that?
No loop, no Sagnac effect.

Also, if you bother reading the paper you will see the same actual length of fibre was used the entire time.
They had the FOG with an additional 50m length of optic fibre attached.

A portion (2.5, 4.0, 8.0 or 16.0 m) was used to construct the loop, with the rest bundled up with the FOG, which was moved uniformly yet developed no phase shift.
If uniform motion develops a phase shift the dependence upon length is voided as they used the same length each time.

Sagnac effect = a phase shift of 2 counterpropagating beams of light around a loop.
Not necessarily a loop.
Dr. Ives and Dr. Wang proved that it can be uniform/translational motion.
But yes, it does involve two counter-propagating beams of light.

No. They didn't. It was motion around a loop, not uniform/translational motion.
You need 2 counterpropagating beams of light to be sent from one location and then be received at one location.
It is impossible to do this unless you have a loop.

You need 2 beams of light travelling in opposite directions around a loop.
No loop, no Sagnac effect.
And that means it can't be purely translational motion.


Now then, quit your bitching about the papers and focus on my derivation (or provide your own for the loop(s) in question.
If you can't then SHUT UP!!!

Nope. I just have 2.
See how you are changing your story line again?

No. I see how you are grasping at whatever straws you can to try and blatantly misrepresent my derivation so you can try and dismiss it rather than have to refute it as you are completely unable to refute it.

At no point in time did I ever indicate there were 4 beams of light.

What I showed were the travel time for 2 sections where a phase shift develops for 2 beams of light.
The time for 2 sections * 2 beams of light gives 4 time periods.

Here is your own description:
YOUR OWN WORDS.
TWO SHIFTS.
FOUR COUNTER-PROPAGATING BEAMS OF LIGHT.

Nope. 2 shifts, one from each of 2 sections of the loop.
Still just 2 counter-propagating beams.

This was already explained and was made clear by my own words previously.
Now stop trying to blatantly lie about the derivation and either admit it is correct, shut up or refute it.

Your piece of shit derivation, the latest version, ALSO DOES NOT APPLY TO THE SAGNAC EFFECT.

You mean my still correct derivation? That still applies to the Sagnac effect.

SAGNAC EFFECT = A SINGLE SOURCE OF LIGHT, TWO COUNTER-PROPAGATING BEAMS OF LIGHT (EITHER IN A LOOP OR ON A STRAIGHT LINE), SAME DISTANCES, DIFFERENT SPEEDS (C + V/C - V).

No, it MUST be a loop. A straight line produces no Sagnac effect.

If you wish to claim it does then provide a derivation showing that it produces a shift. If you can't, shut up.
Here is one for you:

SD---------------------R

SD is the source and detector. R is the reflector.
The 2 beams of light travel the exact same path in the exact same time. As such, they can be considered 2 co-propagating beams of light.

Instead, you would need something more like this:
R-------------SD------------------RThis way the light starts out travelling in opposite directions and thus can be deemed counterpopogating.

But it doesn't matter where SD is place along the path, you get the same result. As such, this can be reduced to the first case, which has the same light paths. Thus there is no phase shift, thus there is no Sagnac effect.

This can also be considered the limit of something more like this:
R--------------------R
|                    |
|                    |
|                    |
R--------------------R
|
SD
Where the width (or height) tends to 0.
The Sagnac effect is based upon the area of the loop, and thus as the width approaches 0, so does the shift.

Thus you need a loop, not a line.

If you want to use one based upon length, then you need a FOC.
In this the "top" moves one way, while the "bottom" moves the other.
Thus it needs to be a loop or it tears itself apart.

One beam of light travels along the arc with radius R2 (in time t1a), then in, then along the arc of radius R1 (in time t1b), then out.
The other beam travels a similar path in an opposite direction with times t2a and t2b.

THAT IS NOT WHAT YOU STATED YESTERDAY.

It is what I stated quite some time ago.

So now lets let alpha be the angle subtended by the arcs, and omega be the angular velocity.
So, one beam of light will propagate in the same direction as the big arc and the opposite direction as the little arc.

where t1a is the time taken to go along the big arc and t1b is the time taken to go along the little arc

Sure seems like I am indicating the same thing.

But even recently, from what you quoted:

If you have beam a and beam b, where they are counterpropogating, then in the arc with radius R2, say beam a is moving forwards and beam b is moving backwards (relative to the orbital motion).
...
They then continue along the light path and travel along the arc of radius R1. Now beam a is moving backwards and beam b is moving forwards.

Sure seems like I am discussing 2 counterpropagating beams, beam a and beam b and discussing the time it takes for them to move along the arcs with radii R1 and R2.

So 2 beams, 2 arcs, 4 time periods.

So stop lying.

IN THE SAGNAC YOU HAVE A SINGLE SOURCE OF LIGHT AND TWO COUNTER-PROPAGATING BEAMS OF LIGHT, IN THIS INSTANCE IN A LOOP.

Which is what I had. So you lose, yet again.

NO NEED TO CALCULATE TWO DIFFERENT TIMES FOR A SINGLE LOOP FOR EACH BEAM.

Sure I do. One time for moving along the big arc and one time for moving along the little arc.

How would you calculate it?
Remember, the one beam of light is moving through both of these arcs. In one case it is moving with the rotation in the other it is moving against the rotation.
If you think you can do it with a single time for each beam (which isn't just broken into 2 more), feel free to provide such a derivation.

No. Because the simpler explanation uses 2 different distances.

Quote from: sandokhan on November 03, 2017, 05:31:53 AM

You really have a terrible knowledge of the Sagnac effect, no other conclusion is possible here.
Other than the fact that you are deliberately trying to fool your readers.

No. The obvious conclusion is that I have a decent knowledge of the Sagnac effect, far more than you were hoping for, which allows me to show you to be full of shit.
The real question is you. Are you just deliberately trying to fool the readers? Did you honestly think you were correct and just don't want to admit you were wrong, or are you genuinely stupid enough to believe the bullshit you are spouting?

Quote from: sandokhan on November 03, 2017, 08:17:05 AM

The loop interferometer described in the following graphic has nothing to do with the physical description of the orbital Sagnac effect:
In the real physical world, YOU CANNOT HAVE THE EARTH MOVE BACKWARDS WITH A 30KM/S SPEED NECESSARY FOR THE R1 ARC.

No, this is an image to derive the orbital Sagnac effect.
It is a ring interferometer on Earth.
It doesn't need Earth to move backwards 30 km/s.
You are the one that needs that to try and make the orbital Sagnac much larger than it really is.
In reality, all that R1 arc is, is the light beams completing the loop, which cancels out the majority of the phase shift if the larger loop.

Quote from: sandokhan on November 03, 2017, 08:17:05 AM

being rotated with a speed of 30km/s

That is a linear speed, not a rotational speed. Try again.

Quote from: sandokhan on November 03, 2017, 08:17:05 AM

If you have a stationary interferometer in the approximate shape of a rectangle (just like in the graphic) and a source of light located right on the path, then the Sagnac effect would be 2Lv/c², where L = 4a + 4b and the observer, moving at speed v to the right in the laboratory frame, sends two counter propagating light signals when he is at the origin.

Nope. If you have a ring interferometer like that, the shift will be 4*A*w/c^2.

In order to have it be 2Lv/c^2 you need a FOC type setup, where the top moves one way and the bottom moves the other, like this:

Or, like the "paper" you linked, figure 7.

So you know how you said this:

Quote from: sandokhan on November 03, 2017, 08:17:05 AM

In the real physical world, YOU CANNOT HAVE THE EARTH MOVE BACKWARDS WITH A 30KM/S SPEED NECESSARY FOR THE R1 ARC.

This applies to your derivation.
You need the bottom section to move forward at 30 km/s, while the top section moves backwards at 30 km/s.

Quote from: sandokhan on November 03, 2017, 08:17:05 AM

What you are describing is a mechanical device, in the shape of a phase-conjugate gyro interferometer, which can go back and forth to create the actual graphic described.
But the Earth cannot go back on its orbit to actually create the interferometer described in the graphic.

Nope. That would be you.
What I am describing is a simple loop which has 2 counterpropagating beams of light proceeding around it while it orbits the sun (or any object).
Earth doesn't need to go backwards.

Quote from: sandokhan on November 03, 2017, 08:17:05 AM

If you have an interferometer, in the shape described, two satellites and two receivers/emitters on the surface of the Earth, then you have a very different situation best visualized thusly

Or visualised like I showed.
The 2 receives on Earth can be the 2 ends of the big arc. The 2 satellites can be the 2 ends of the little arc.

Stop trying to change it to a different setup. Deal with this one first.

Quote from: sandokhan on November 03, 2017, 08:17:05 AM

changing the distance the signal travels.

Remember how you said the distance doesn't change?
Are you planning on making up your mind any time soon rather than repeatedly contradicting yourself?

Quote from: sandokhan on November 03, 2017, 08:17:05 AM

A SITUATION TOTALLY DIFFERENT THAN THE LOOP INTERFEROMETER DESCRIBED IN THE GRAPHIC ITSELF.

Thanks for admitting your crap is irrelevant. Now deal with the loop interferometer described in the graphic before moving on.

Quote from: sandokhan on November 03, 2017, 08:17:05 AM

going back and forth with a certain angular velocity

No, it is only going forward with a certain angular velocity, the angular velocity of the orbit.

Quote from: sandokhan on November 03, 2017, 08:17:05 AM

If the interferometer exists already (say two satellites and two receivers/emitters on the surface of the Earth), then we have the following situation:

The situation described in my graphic, where the phase shift of the orbital sagnac effect is much smaller than the rotational Sagnac.

Quote from: sandokhan on November 03, 2017, 08:17:05 AM

Then we simply apply the well-known formula to this situation, where R = 150,000,000 km, v = wR (=30km/s).

And A is the area of the interferometer, with v being insignificant compared to c. Thus dt=4*A*w/c^2

And I'm just skipping the rest of your BS.


Deal with the interferometers already discussed rather than trying to invoke a new one.

There are no loops involved in the orbital Sagnac.

There are no loops involved in the rotational Sagnac.

Just a one-way communication between the satellite and the receiver.

Modern science assumes that the Earth rotates, thus causing the light signal delay.

However, if ether can be proven to exist, and there multiple ways to do this, it means the delay is caused by the rotating field of ether over a stationary earth.

Take a look at these graphics, taken straight from the manuals on GPS/GNSS communication:





GPS time and frequency comparison is a one-way technique, since signals are only transmitted by the satellites and received on the ground.


Your graphic is a physical impossibility.

If we have an interferometer on the surface of the Earth, in the shape of a parallelogram or of a circle, we can readily apply the Sagnac, whether it be rotational or orbital.

In reality, all that R1 arc is, is the light beams completing the loop, which cancels out the majority of the phase shift if the larger loop.

BUT THEN YOU ARE DEALING WITH AN ALREADY BUILT INTERFEROMETER.

NOT WITH THE ACTUAL CONSTRUCTION OF AN INTERFEROMETER USING ANGULAR VELOCITIES TO CREATE THE ARCS.

YOU CAN ONLY DO THAT IN A LAB USING PHASE CONJUGATE GYRO INTERFEROMETERS.

As such, your contraption is worthless in analyzing the orbital Sagnac.

You cannot have the Earth move back along the lower arc portion of the path to accomodate for your graphic.

You can only have a stationary interferometer on the surface of the Earth which is then subjected to a rotation or to a orbital rotation.

Nope. If you have a ring interferometer like that, the shift will be 4*A*w/c^2.

No, the shift will be 2Lv/c2, where L = 4a + 4b and the observer, moving at speed v to the right in the laboratory frame, sends two counter propagating light signals when he is at the origin.

That is the Sagnac in TRANSLATIONAL MOTION.

Proven by the very trajectory of the GPS signal to the ground, a one-way light signal which is being delayed by the ether field.

No loop, no Sagnac effect.

Professor Wang proved otherwise.

The linear/uniform part of the path produces the delay.

Not the loops.

It is an accepted fact of science.



As shown in Fig. 3, the phase shift or the traveltime
difference between two counter-propagating light
beams in the moving optic fiber was clearly observed
in all different configurations of FOCs. The phase shift
Δφ, and therefore, the travel-time difference Δt are
proportional to both the total length and the speed of
the moving fiber whether the motion is circular or
uniform. Other tests using smaller end wheels for the
FOC and fiber loops with additional curves also
confirmed the same finding.

Professor Wang's seminal did prove that the Sagnac applied to linear motion.


https://arxiv.org/ftp/physics/papers/0609/0609222.pdf (first experiment conducted by R. Wang)

https://arxiv.org/ftp/physics/papers/0609/0609202.pdf (second experiment carried out by R. Wang)


It was motion around a loop, not uniform/translational motion.
You need 2 counterpropagating beams of light to be sent from one location and then be received at one location.
It is impossible to do this unless you have a loop.

Professor Wang did manage to do the impossible by actually sending two counterpropagating beams on a straight line, and measuring the uniform Sagnac effect.

Moreover, it is easy to prove it mathematically that you do not need a loop using the Stokes formula.

The difference in propagation time Δt for two opposite directions of light is described as

Δt = 4ΏA/c²

Where A is enclosed area. Δt is derived based on an integration of Ώ over A.

According to Stokes' rule an integration of angular velocity Ώ over an area A can be substituted by an integration of tangential component of translational velocity v along the closed line of length L limiting the given area. This interpretation gives

Δt = 2vL/c²

producing the same value as the earlier expression. This can also be demonstrated by geometrical relations. These two integrations have different physical implications. So the decision is whether the effect is caused by a rotating area or by a translating line. Since Sagnac effect is an effect in light that is enclosed inside an optical fiber we can conclude that Sagnac effect is distributed along a line and not over an area. No light and no rotation exists in the enclosed area. Sagnac detected therefore an effect of translation although he had to rotate the equipment to produce the effect inside the fiber.

The 2 receives on Earth can be the 2 ends of the big arc. The 2 satellites can be the 2 ends of the little arc.

Stop trying to change it to a different setup. Deal with this one first.

Then, your entire derivation is wrong.

Any interferometer on the surface of the Earth, subjected to rotation, will reveal the Sagnac effect, be it rotational or orbital.

What you did is to actually CONSTRUCT THE INTERFEROMETER AS YOU GO, BY USING THE ANGULAR VELOCITY IN A POSITIVE WAY, ALONG R2 AND THEN THE SAME ANGULAR VELOCITY IN A NEGATIVE DIRECTION, TO ACCOUNT FOR R1.

Therefore, your contraption can be produced only in a lab with a phase conjugate gyro interferometer, NOT IN SPACE.

You simply had no idea what you were doing.


No matter what you say, the uniform/translational Sagnac is a proven fact of science.

I have at my disposal multiple sources, peer reviewed, while you have nothing.

Professor Wang's seminal papers are an accepted fact of science.

Dufour and Prunier PROVED THE SAME THING, USING AN INGENIOUS PATH, PART CIRCULAR, PART UNIFORM.


http://www.conspiracyoflight.com/pdf/Dufour_and_Prunier-On_the_Fringe_Movement_Registered_on_a_Platform_in_Uniform_Motion_%281942%29.pdf

A. Dufour and F. Prunier created Sagnac interferometers that were composites of moving and stationary paths, including stationary sources and stationary detectors. This was essentially to test if the relativistic approach could be distinguished from the classical approach.

"In all cases of this experimental test, the Sagnac effect was the same. This overturned Langevin’s analysis, and in 1937, he had to revise his explanation, as pointed out by Kelly: 

“In his final essay on the subject in 1937, Langevin proposed that the results published that year by Dufour and Prunier showed that one had to assume either (a) the light speed varied to c + wr in one direction and c – wr in the other direction, or (b) the time aboard the spinning apparatus had to change by a factor of +/-2wA/c2 in either direction. Indeed, Langevin went as far as to say that assuming (a), “we find, by a very simple and very general reasoning, the formula for the difference of the times of the path of the two light beams in the Sagnac experiment.” .

The proposition (b) though is untenable because if this were true then when the light beam passed back to the moving detector, the local time from each direction would be out of synchronization, meaning that the clocks cannot be counting real time and that the effective time dilation is meaningless. This was also pointed out by Herbert Ives in his 1938 paper criticizing Langevin. Ives says about the absurdity of Langevin’s proposition (b):

” There are of course not merely two clocks, but an infinity of clocks, where we include those that could be transported at finite speeds, and around other paths. As emphasized previously, the idea of “local time” is untenable, what we have are clock readings. Any number of clock readings at the same place are physically possible, depending on the behaviour and history of the  clocks used. More than one “time” at one place is a physical absurdity. “

The only explanation left, is Langevin’s proposition a) that the light speed varies by C+/-wr in one or the other direction around the disk, consistent with Dufour and Prunier’s experimental results."

In 1939, Dufour and Prunier carried out their
final experiment. They did a test with both the
beginning and end of the light path on the
spinning disc, but with the middle portion of
the path reflected off mirrors fixed in the
laboratory (directly above the disc). In this
test, they had both the light emitter and the
photographic recorder fixed in the laboratory.

The fringe shifts resulting from all the above
Dufour and Prunier tests were the same as in
their original Sagnac-type tests.

In 1942 Dufour and Prunier published a
composite paper reviewing their total
experimental work to date. At the end of this
paper they state that "the relativity theory
seems to be in complete disagreement with the
result which was garnered from the
experiment ".

Herbert Ives, Light Signals Sent Around a Closed Path:

http://www.conspiracyoflight.com/Ives/Herbert_Ives_Light_Signals_Sent_Around_a_Closed_Path.pdf

 Ives was a pioneer in the development of television at Bell Telephone Laboratories.  The following quotations are from his 1938 article.
 
     The experiment was interpreted by its author as positive evidence for the existence of the luminiferous ether…

     It is the purpose of this paper first to show that the Sagnac experiment in its essentials involves no consideration of rotation, and second to investigate the results obtained when transported clocks are used.
 
Ives analyzed the Sagnac experiment using a hexagonal path rather than a circular one.
 
He concluded with this statement:
 
     The net result of this study appears to be to leave the argument of Sagnac as to the significance of his experiment as strong as it ever was.

The Sagnac effect is not due to rotation, but instead is a linear effect due to a true anisotropic light speed in a moving frame.


The fringe shifts resulting from all the above
Dufour and Prunier tests were the same as in
their original Sagnac-type tests.

They used BOTH A SPINNING DISK AND A FIXED PORTION OF THE PATH.

THEY OBTAINED THE SAME RESULTS.

THIS, ALONG WITH THE WANG EXPERIMENT, PROVES THE EXISTENCE OF UNIFORM/TRANSLATIONAL SAGNAC BEYOND A SHADOW OF A DOUBT.

Now, the final piece of the puzzle.

In a remarkable achievement, Dr. Paul Marmet has shown that the velocity of light is c±v with respect to an observer, moving in straight line at velocity v, as it applies to the situation being discussed here, GPS satellites and the rotational/orbital motion of the Earth (heliocentric version).

Beginning in 1967 Marmet served as director of the laboratory for Atomic and Molecular Physics at Laval University in Quebec City, Canada, serving in that role until 1982. From 1983 to 1990, Marmet was a senior researcher at the Herzberg Institute of Astrophysics of the National Research Council of Canada in Ottawa. In 1990 Marmet was an Assistant Professor of Physics at the University of Ottawa.

A past president of the Canadian Association of Physicists (1981-1982), he also served as a member of the executive committee of the Atomic Energy Control Board of Canada from 1979 to 1984.  Marmet was elected Fellow of the Royal Society of Canada in 1973 and was made an Officer of the Order of Canada in 1981.  The Order of Canada is the highest decoration bestowed by the Canadian government.

So you claim that he cannot be wrong?

<< I'll accept all that >>
The Global Positioning System (GPS) determines that after clock µ moves away from clock a in New York, toward clock ß in San Francisco, its display accumulates an extra 14 ns (approximately) with respect to clock ß.  We know that due to the Earth rotation, between N.Y. and S.F. clock µ moves at velocity (v-e), which is the velocity of rotation of the Earth "v" minus the velocity of the truck "e". Therefore 14 ns are subtracted to its display at its arrival in order to give a correct synchronization of time on clock ß in S.F.. This correction is identical to equation 13. This correction is the same as the one programmed automatically in the GPS.

Agreed!

            Experimentally, an equivalent experiment has also been done carrying a clock between Washington and Tokyo by Saburi et al. [7]. It is then an experimental fact that the two clocks (a and ß) are not naturally synchronized at the same value, as a result of the discordant Einstein's synchronization method explained above.
            There is another well-known way to synchronize the clocks between these two stations (a and ß). It is done sending radio signals transmitted simultaneously (east-west and west-east) between these two cities. Again, it is observed that a simultaneous transmission of radio signals between New York and San Francisco does not give "directly" the same correct clock display (time) in both cities. There is a difference of about 14 ns that must be subtracted to the clock in San Francisco in order to get the correct GPS time. This correction is identical to the one when we are carrying clocks. This correction corresponds to a change of velocity c±v between stations.
            This GPS synchronization has been verified in numerous experiments. It is identical to the calculations presented in this paper and also to the Sagnac's effect, (which is included in the GPS). Among the GPS list of corrections, there is a correction involving a parameter taking into account how many Earth meridians are crossed by light or by the moving clock µ, between the two locations. Kelly [8] explains that the correction used by the GPS is:



where w is the angular velocity of rotation of the Earth, A_E is the projected area on the Earth equator plane of the path used by light (or by a slowly moving clock) between the two stations. We define l as the distance between the two stations, both moving at velocity v. The circumference of the Earth is called "circ". Therefore the area A_E is



The angular velocity  w is equal to v/r. The circumference of the Earth is 2πr. Equation 23 in equation 22 gives:



We see that the GPS correction of clocks (equation 24)  is identical to the Sagnac effect.

Agreed!

One must conclude that there exists no space-time distortion of any kind.

No, one need not conclude that at all!

It is no longer necessary to fascinate people with the magic of relativity.

The "magic of relativity" is simply not needed in the explanation of the Sagnac delay provided the observation is done from an inertial frame of reference.

Unless we accept the absurd solution that the distance between N.Y. to S.F. is smaller than the distance between S.F. and N.Y., we have to accept that in a moving frame, the velocity of light is different in each direction.

Quite true, Mr Sandokhan, quite true!
Well, at least the light path between "between N.Y. to S.F. is smaller than the distance between S.F. and N.Y."
Think about it!

The earth is rotating so
         light travelling from San Francisco to New York has to travel about 4,115+0.0049 km and
         light travelling from New York to San Francisco has to travel about 4,115-0.0049 km.
See simple, just the Sagnac effect analysed from an inertial reference frame and different distances for each direction.

So, when looked at in inertial reference frame, it is the distances that are different, not the velocities.

As mentioned above, this difference is even programmed in the GPS computer in order to get the correct Global Positioning.

Sure, "this difference is even programmed in the GPS computer in order to get the correct Global Positioning".
Obviously, those that designed to GPS were smarter than you.

This proves that the experimental velocity of light with respect to a moving observer is c±v.

No, it doesn't prove anything of the sort, just that your understanding of both Sagnac and relativity is crap!
The Sagnac effect is not a relativistic effect and it neither proves nor disproves relativity.

There are no loops involved in the orbital Sagnac.
There are no loops involved in the rotational Sagnac.

By definition, the Sagnac effect is a phase shift between 2 counter-propagating beams of light around a loop.
No loop, no Sagnac.

Deal with the simple case of the loop before moving on to anything more complex.

Your graphic is a physical impossibility.
If we have an interferometer on the surface of the Earth, in the shape of a parallelogram or of a circle, we can readily apply the Sagnac, whether it be rotational or orbital.

We can do it with any shape, as the Sagnac effect is proportional to the area and angular velocity.
But you didn't accept that, so I had to derive it and prove it.

My graphic is one such example.
There is nothing impossible about it.

BUT THEN YOU ARE DEALING WITH AN ALREADY BUILT INTERFEROMETER.

The loop shown in red is the interferometer.

NOT WITH THE ACTUAL CONSTRUCTION OF AN INTERFEROMETER USING ANGULAR VELOCITIES TO CREATE THE ARCS.

I'm not doing that.
The interferometer stretches over an angle of alpha. It is orbiting the centre with an angular velocity of omega.

As such, your contraption is worthless in analyzing the orbital Sagnac.

Nope. What is worthless is your pathetic attempts to avoid the fact that you are wrong.

You cannot have the Earth move back along the lower arc portion of the path to accommodate for your graphic.

Like I said before, IT DOESN"T MOVE BACK!!!!
That lower arc is a path which is moving with Earth.
All that travels "back" is the beam of light.
One beam of light moves forwards along the big arc then backwards along the small arc.
The other moves backwards along the big arc and then forwards along the small arc.

It is your setup and your analysis which needs Earth to magically move backwards.

Now stop repeating the same lies which have already been exposed as lies.

You can only have a stationary interferometer on the surface of the Earth which is then subjected to a rotation or to a orbital rotation.

As I did.

Nope. If you have a ring interferometer like that, the shift will be 4*A*w/c^2.
No, the shift will be 2Lv/c2, where L = 4a + 4b and the observer, moving at speed v to the right in the laboratory frame, sends two counter propagating light signals when he is at the origin.
That is the Sagnac in TRANSLATIONAL MOTION.

Again, the Sagnac effect does not exist for translational motion. It exists for light moving around a loop. With purely translational motion, the 2 phase shifts (one from each arm moving in the direction of the light) cancel out and you are left with no net phase shift.

In order to have it given by 2Lv/c^2 you need to have the loop moving as a conveyer, requiring one side of the loop to move forward at speed v and the other side to move backwards at speed v.

Professor Wang proved otherwise.

No they didn't.
Stop it with your pathetic appeals to autority and deal with the OP.
If you can't, take it elsewhere or shut up.

Moreover, it is easy to prove it mathematically that you do not need a loop using the Stokes formula.
The difference in propagation time Δt for two opposite directions of light is described as
Δt = 4ΏA/c²
Where A is enclosed area. Δt is derived based on an integration of Ώ over A.

And the enclosed area only makes sense for a loop.

What is the enclosed area of a line segment?

These two integrations have different physical implications.

That's right, so you can't just pretend it works after doing the math to manipulate it like that.

Since Sagnac effect is an effect in light that is enclosed inside an optical fiber we can conclude that Sagnac effect is distributed along a line and not over an area.

Nope. Over a loop.

Then, your entire derivation is wrong.

No it isn't.

Any interferometer on the surface of the Earth, subjected to rotation, will reveal the Sagnac effect, be it rotational or orbital.

Unlike your prior claims which indicate the interferometer would need to be centred on the sun or Earth. So thanks for once again showing yourself to be full of shit.

What you did is to actually CONSTRUCT THE INTERFEROMETER AS YOU GO, BY USING THE ANGULAR VELOCITY IN A POSITIVE WAY, ALONG R2 AND THEN THE SAME ANGULAR VELOCITY IN A NEGATIVE DIRECTION, TO ACCOUNT FOR R1.

No I didn't.
Omega was the angular velocity. The interferometer was constructed with the angle alpha.

The reason the sign is reversed is because the direction of light is reversed.

You simply had no idea what you were doing.

No, it seems you are trying to make up whatever excuse you can to pretend my derivation is wrong while being completely unable to refute it.

No matter what you say, the uniform/translational Sagnac is a proven fact of science.
I have at my disposal multiple sources, peer reviewed, while you have nothing.

You have sources which you blatantly misrepresent.
I have the derivations to show it is a load of crap, and your own sources which refute you.

And a large portion of the crap you link to isn't peer reviewed.

And I'm skipping over the rest of your already refuted crap.

The earth is rotating so
         light travelling from San Francisco to New York has to travel about 4,115+0.0049 km and
         light travelling from New York to San Francisco has to travel about 4,115-0.0049 km.

But the distance did not change at all.

It was the speed of light that changed.

I "accept the absurd solution that the distance between N.Y. to S.F. is smaller than the distance between S.F. and N.Y.".

You cannot.

Here is the Dufour Prunier experiment which proved that in a Sagnac experiment the speed of light will be c + v and c - v.

It is an accepted fact of science.


http://www.conspiracyoflight.com/pdf/Dufour_and_Prunier-On_the_Fringe_Movement_Registered_on_a_Platform_in_Uniform_Motion_%281942%29.pdf

A. Dufour and F. Prunier created Sagnac interferometers that were composites of moving and stationary paths, including stationary sources and stationary detectors. This was essentially to test if the relativistic approach could be distinguished from the classical approach.

"In all cases of this experimental test, the Sagnac effect was the same. This overturned Langevin’s analysis, and in 1937, he had to revise his explanation, as pointed out by Kelly: 

“In his final essay on the subject in 1937, Langevin proposed that the results published that year by Dufour and Prunier showed that one had to assume either (a) the light speed varied to c + wr in one direction and c – wr in the other direction, or (b) the time aboard the spinning apparatus had to change by a factor of +/-2wA/c2 in either direction. Indeed, Langevin went as far as to say that assuming (a), “we find, by a very simple and very general reasoning, the formula for the difference of the times of the path of the two light beams in the Sagnac experiment.” .

The proposition (b) though is untenable because if this were true then when the light beam passed back to the moving detector, the local time from each direction would be out of synchronization, meaning that the clocks cannot be counting real time and that the effective time dilation is meaningless. This was also pointed out by Herbert Ives in his 1938 paper criticizing Langevin. Ives says about the absurdity of Langevin’s proposition (b):

” There are of course not merely two clocks, but an infinity of clocks, where we include those that could be transported at finite speeds, and around other paths. As emphasized previously, the idea of “local time” is untenable, what we have are clock readings. Any number of clock readings at the same place are physically possible, depending on the behaviour and history of the  clocks used. More than one “time” at one place is a physical absurdity. “

The only explanation left, is Langevin’s proposition a) that the light speed varies by C+/-wr in one or the other direction around the disk, consistent with Dufour and Prunier’s experimental results."

In 1939, Dufour and Prunier carried out their
final experiment. They did a test with both the
beginning and end of the light path on the
spinning disc, but with the middle portion of
the path reflected off mirrors fixed in the
laboratory (directly above the disc). In this
test, they had both the light emitter and the
photographic recorder fixed in the laboratory.

The fringe shifts resulting from all the above
Dufour and Prunier tests were the same as in
their original Sagnac-type tests.

In 1942 Dufour and Prunier published a
composite paper reviewing their total
experimental work to date. At the end of this
paper they state that "the relativity theory
seems to be in complete disagreement with the
result which was garnered from the
experiment ".

Herbert Ives, Light Signals Sent Around a Closed Path:

http://www.conspiracyoflight.com/Ives/Herbert_Ives_Light_Signals_Sent_Around_a_Closed_Path.pdf

 Ives was a pioneer in the development of television at Bell Telephone Laboratories.  The following quotations are from his 1938 article.
 
     The experiment was interpreted by its author as positive evidence for the existence of the luminiferous ether…

     It is the purpose of this paper first to show that the Sagnac experiment in its essentials involves no consideration of rotation, and second to investigate the results obtained when transported clocks are used.
 
Ives analyzed the Sagnac experiment using a hexagonal path rather than a circular one.
 
He concluded with this statement:
 
     The net result of this study appears to be to leave the argument of Sagnac as to the significance of his experiment as strong as it ever was.

The Sagnac effect is not due to rotation, but instead is a linear effect due to a true anisotropic light speed in a moving frame.


The fringe shifts resulting from all the above
Dufour and Prunier tests were the same as in
their original Sagnac-type tests.


Remember, Langevin was the most ardent supporter of Einstein and he had to accept that the light speed varied to c + wr in one direction and c – wr in the other direction.

Again, the Sagnac effect does not exist for translational motion. It exists for light moving around a loop. With purely translational motion, the 2 phase shifts (one from each arm moving in the direction of the light) cancel out and you are left with no net phase shift.

But they do not.

Here is the celebrated experiment carried out by Professor Ruyong Wang, published in the Physical Review, peer reviewed.

Physical Review Letters (PRL), established in 1958, is a peer-reviewed, scientific journal that is published 52 times per year by the American Physical Society. As also confirmed by various measurement standards, which include the Journal Citation Reports impact factor and the journal h-index proposed by Google Scholar, many physicists and other scientists consider Physical Review Letters one of the most prestigious journals in the field of physics.

As shown in Fig. 3, the phase shift or the traveltime
difference between two counter-propagating light
beams in the moving optic fiber was clearly observed
in all different configurations of FOCs. The phase shift
Δφ, and therefore, the travel-time difference Δt are
proportional to both the total length and the speed of
the moving fiber whether the motion is circular or
uniform. Other tests using smaller end wheels for the
FOC and fiber loops with additional curves also
confirmed the same finding.

Professor Wang's seminal did prove that the Sagnac applied to linear motion.


https://arxiv.org/ftp/physics/papers/0609/0609222.pdf (first experiment conducted by R. Wang)

https://arxiv.org/ftp/physics/papers/0609/0609202.pdf (second experiment carried out by R. Wang)


Can you read English jack?

As shown in Fig. 3, the phase shift or the traveltime
difference between two counter-propagating light
beams in the moving optic fiber was clearly observed
in all different configurations of FOCs.


HERE IS THE CONCLUSION OF THE PAPER, HAVING BEEN PEER REVIEWED AND PUBLISHED IN ONE OF THE MOST RESPECTED JOURNAL OF PHYSICS IN THE WORLD:

Conclusion

The travel-time difference of two counterpropagating
light beams in moving fiber is proportional
to both the total length and the speed of the fiber,
regardless of whether the motion is circular or uniform.
In a segment of uniformly moving fiber with a speed of
v and a length of Δl, the travel-time difference is
2vΔl/c^2.

Professor Wang proved a linear sagnac effect that is effected by linear cable length in then same way that additional circumference affects time delay between signals.



Ives used a hexagonal path, made up of straight lines.

He got the same results.

Herbert Ives, Light Signals Sent Around a Closed Path:

http://www.conspiracyoflight.com/Ives/Herbert_Ives_Light_Signals_Sent_Around_a_Closed_Path.pdf

These are classic results from mainstream science.

You cannot deny them.


Professor Wang's experiment is being used as a bibliographical reference for the uniform Sagnac all over the scientific world.

It is an accepted fact of science.


In order to have it given by 2Lv/c^2 you need to have the loop moving as a conveyer, requiring one side of the loop to move forward at speed v and the other side to move backwards at speed v.

You do not.

All you have to do is move the entire interferometer and it will record the Sagnac effect.


What is the enclosed area of a line segment?

You do not need an area, only a segment.

Moreover, it is easy to prove it mathematically that you do not need a loop using the Stokes formula.

The difference in propagation time Δt for two opposite directions of light is described as

Δt = 4ΏA/c2

Where A is enclosed area. Δt is derived based on an integration of Ώ over A.

According to Stokes' rule an integration of angular velocity Ώ over an area A can be substituted by an integration of tangential component of translational velocity v along the closed line of length L limiting the given area. This interpretation gives

Δt = 2vL/c2

producing the same value as the earlier expression. This can also be demonstrated by geometrical relations. These two integrations have different physical implications. So the decision is whether the effect is caused by a rotating area or by a translating line. Since Sagnac effect is an effect in light that is enclosed inside an optical fiber we can conclude that Sagnac effect is distributed along a line and not over an area. No light and no rotation exists in the enclosed area. Sagnac detected therefore an effect of translation although he had to rotate the equipment to produce the effect inside the fiber.


Professor Wang proved that you do not need an area:

As shown in Fig. 3, the phase shift or the traveltime
difference between two counter-propagating light
beams in the moving optic fiber was clearly observed
in all different configurations of FOCs. The phase shift
Δφ, and therefore, the travel-time difference Δt are
proportional to both the total length and the speed of
the moving fiber whether the motion is circular or
uniform. Other tests using smaller end wheels for the
FOC and fiber loops with additional curves also
confirmed the same finding.

Professor Wang's seminal did prove that the Sagnac applied to linear motion.


https://arxiv.org/ftp/physics/papers/0609/0609222.pdf (first experiment conducted by R. Wang)

https://arxiv.org/ftp/physics/papers/0609/0609202.pdf (second experiment carried out by R. Wang)


SHOW US THE LOOP IN THE FOLLOWING GRAPHICS, TAKEN DIRECTLY FROM THE GPS/GNSS MANUALS:





The satellites transmit their signal to Earth and that’s that.  There’s no transmission back to the satellites, or between satellites, and none of their signals are made to loop the equator.

As such we do not have a loop interferometer at all.

What we have is exactly the uniform/translational motion exemplified in the Ives, Wang and Dufour-Prunier experiments.



Nope. Over a loop.

Dufour and Prunier used BOTH A LOOP AND STRAIGHT LINE SEGMENT AND GOT THE SAME RESULT.

YOU HAVE JUST BEEN DISPROVEN.


Unlike your prior claims which indicate the interferometer would need to be centred on the sun or Earth.

I made both claims, just like in the correct definition of the Sagnac.

You can have an interferometer centered on the Sun, and AN ALREADY BUILT INTERFEROMETER SOMEWHERE ON THE SURFACE OF THE EARTH. I have TWO options at my disposal, you only have one.
Try to imagine the center of the earth as the sun and the earth's orbit as the surface. So, the earth is in a rotational/circular pattern in its orbit. Just like the rotational sagnac, the earth rotates toward where the satellite emitted the signal, and with the orbit, the earth, the unit is orbited toward where the signal was emitted.


YOU ARE BUILDING YOUR INTERFEROMETER USING BOTH THE ANGULAR VELOCITY AND THE OMEGA ANGLE AS YOU GO ALONG, RIGHT IN FRONT OF YOUR READERS.

Using the angular velocity of the Earth and the angle subtended, you are going forward and backwards, just like you described.

But the Earth cannot move backwards.

That is why your interferometer is a physical impossibility in outer space.

You could built one in a lab, using a phase conjugate gyro interferometer.

But that is something else altogether.


That lower arc is a path which is moving with Earth.
All that travels "back" is the beam of light.
One beam of light moves forwards along the big arc then backwards along the small arc.
The other moves backwards along the big arc and then forwards along the small arc.

Then, what you are describing is an ALREADY BUILT/CONSTRUCTED INTERFEROMETER ON THE SURFACE OF THE EARTH.

No problem at all.

We simply apply the orbital Sagnac to it, using R = 150,000,000 km and v = 30 km/s.


But what you want is to actually construct your interferometer as you go along, USING THE ANGULAR VELOCITY OF THE EARTH IN BOTH DIRECTIONS, but the Earth cannot move backwards.

It is either/or.

If you want an already built interferometer, we can apply the Sagnac immediately and I win.

If you want to actually build the interferometer using the angular velocity/omega angle, you can only do that in a lab using a phase conjugate gyro interferometer.

Here are your own words:

So now lets let alpha be the angle subtended by the arcs, and omega be the angular velocity.
So, one beam of light will propagate in the same direction as the big arc and the opposite direction as the little arc.
What this means is its trip along the big arc will be longer than if it was stationary and shorter along the little arc.

You have a contribution from the arc with radius R2. This is based upon the area swept out by this arc, alpha*R2^2/2 (it would be 4*A*w/c^2).
You then have a contribution from the arc with radius R1. This is based upon the area swept out by this arc, and as the light beams are now travelling in an opposite direction and thus is of opposite sign and thus would be proportional to -alpha*R1^2/2.

What you are describing is a mechanical device, in the shape of a phase-conjugate gyro interferometer, which can go back and forth to create the actual graphic described.

But the Earth cannot go back on its orbit to actually create the interferometer described in the graphic.

If the interferometer IS ALREADY BUILT, we do not need a debate, I win hands down.

Where is the source of light located on your interferometer? You have none. That source of light is the source of the counter-propagating beams. But you have none.

You are building the interferometer as you go along, trying to use the angular velocity and the subtended angle: the Earth cannot go backwards for you.

You need a source of light on that interferometer.

BUT ONCE YOU DO INDICATE THAT SOURCE OF LIGHT, YOU ARE OUT OF LUCK, SINCE THEN YOU ARE ACCEPTING THAT THE INTERFEROMETER IS ALREADY BUILT, AND ALL YOU HAVE TO DO IS TO ROTATE IN ANY WAY TO OBTAIN THE SAGNAC.

You can't fool anybody here jack.

The earth is rotating so
         light travelling from San Francisco to New York has to travel about 4,115+0.0049 km and
         light travelling from New York to San Francisco has to travel about 4,115-0.0049 km.
But the distance did not change at all.
It was the speed of light that changed.

No. The distance between the cities didn't change, the cities moved.
As such, the distance that light has to take is different.

As an example, consider a train 100 m long, travelling at 10 m/s.
You need to move either from the front of the train to the back, or from the back of the train to the front, and can travel at 50 m/s, but need to do so outside the train.
How long does it take you to move towards the front of the train and what distance do you cover?
How long does it take you to move to the back of the train and what distance do you cover?
Does this mean the train changes length? NO!

Here is the Dufour Prunier experiment which proved that in a Sagnac experiment the speed of light will be c + v and c - v.

No it didn't.

It is an accepted fact of science.

No, it is accepted that you can use a non-inertial reference frame and have the speeds change. That doesn't mean that is the only way.

The experiment was interpreted by its author as positive evidence for the existence of the luminiferous ether…

That doesn't matter. The simple fact is the aether has been debunked long ago.

The Sagnac effect is not due to rotation, but instead is a linear effect due to a true anisotropic light speed in a moving frame.

Not in a moving frame, in a non-inertial frame. That is the key. The key is the light path is bent.

Again, the Sagnac effect does not exist for translational motion. It exists for light moving around a loop. With purely translational motion, the 2 phase shifts (one from each arm moving in the direction of the light) cancel out and you are left with no net phase shift.
But they do not.
Here is the celebrated experiment carried out by Professor Ruyong Wang, published in the Physical Review, peer reviewed.

Which used a loop of optic fibre where each side of the loop moved in an opposite direction. It was not translational motion. Try again.

Can you read English jack?

Yes. Can you?
Can you read and comprehend the entire paper, rather than just cherry picking key words while ignoring the meaning?

You cannot deny them.

I'm not denying them. I am denying your bullshit misrepresentations of them.

In order to have it given by 2Lv/c^2 you need to have the loop moving as a conveyer, requiring one side of the loop to move forward at speed v and the other side to move backwards at speed v.
You do not.
All you have to do is move the entire interferometer and it will record the Sagnac effect.

No it wont.
I already explained why.
If you think it will, THEN DERIVE IT!!
If you can't, stop repeating the same lie.

Moreover, it is easy to prove it mathematically that you do not need a loop using the Stokes formula.

Yet you are completely unable to.
Don't just appeal to a formula, explain it/derive it. Show how it works.

Where A is enclosed area. Δt is derived based on an integration of Ώ over A.

See, you need an area.

According to Stokes' rule an integration of angular velocity Ώ over an area A can be substituted by an integration of tangential component of translational velocity v along the closed line of length L limiting the given area. This interpretation gives

Once again, appealing to an area. Regardless, PROVE IT!!

Professor Wang proved that you do not need an area:

Instead you need a FOC, which isn't what you had, and isn't what a straight line is.

SHOW US THE LOOP IN THE FOLLOWING GRAPHICS, TAKEN DIRECTLY FROM THE GPS/GNSS MANUALS:

No. Deal with the loop in the OP first, then we can move on.

YOU HAVE JUST BEEN DISPROVEN.

Nope. My derivation remains unrefuted. You are yet to disprove anything.

Unlike your prior claims which indicate the interferometer would need to be centred on the sun or Earth.
I made both claims, just like in the correct definition of the Sagnac.

You are aware those claims are contradictory right?
If it needs to be centred on the sun, you can't have a small one on Earth.
Make up your mind.

I have TWO options at my disposal, you only have one.

No, I have plenty, but I only needed that one to disprove you.

And I'm skipping over your repeated garbage. I have already said why it is wrong.
I have a physical loop indicated in red which subtends an angle alpha which is moving along Earth's orbit at an angular velocity of omega. The only thing going back is the light, not the arc.

Then, what you are describing is an ALREADY BUILT/CONSTRUCTED INTERFEROMETER ON THE SURFACE OF THE EARTH.
No problem at all.

That's right. No problem at all, we apply the simple Sagnac formula to it:
dt=4*A*w/c^2.
And we do the same for Earth and find the orbital Sagnac is much smaller.

We simply apply the orbital Sagnac to it, using R = 150,000,000 km and v = 30 km/s.

And just where is R meant to fit in?
It has A, not R.
Why should R be that?
There are 2 circular arcs. Which R should you use?

If you remember, I showed you the derivation. You are yet to show a problem with it or do a derivation yourself.
It was dt=2*alpha*(R2^2-R1^2)*w/c^2.
So if you insist on using the distance from the sun, you need to use both R values, and they mostly cancel.

in fact, you can represent R2 as R1+b, where b is the length of the axial component of the loop, and get this:
dt=2*alpha*((R1+b)^2-R1^2)*w/c^2.
=2*alpha*(R1^2+2*b*R+b^2-R1^2)*w/c^2.
=2*alpha*(2*b*R1+b^2)*w/c^2.
Noting that w is tiny compared to R, this approximates:
dt=2*alpha*(2*R1*b)*w/c^2.
=4*alpha*R1*b*w/c^2.
and b*R roughly equals the area of the loop (it is off by b^2).

And before you try and suggest this shows it is dependent on R, so comparing the orbtial and rotational you would end up with
dto/dtr=150 000 000 / (365 * 6400), it is also dependent upon alpha, the angle subtended by the arc, and that angle is much smaller for the orbit.

In fact, we can do some more tweaking (especially as the loop is basically a rectangle anyway).
If the arc of radius R1 has a length of l, then alpha*R1=l and thus alpha=l/R1

This means dt=4*(l/R1)*R1*b*w/c^2.
=4*l*b*w/c^2.

And yet again, it ends up as the area of the loop.

But what you want is to actually construct your interferometer as you go along, USING THE ANGULAR VELOCITY OF THE EARTH IN BOTH DIRECTIONS, but the Earth cannot move backwards.

Again, the only thing moving back in mine is the light. It is you that needs Earth to magically move backwards.

If you want an already built interferometer, we can apply the Sagnac immediately and I win.

Nope. We apply the correct Sagnac, dt=4*A*w/c^2 and you are refuted.
Or I can show the derivation, which you are yet to refute, and you are refuted.

The only way for you to win is if you can do a derivation yourself for the loop in question, explaining each step and not have it be refuted.
But that would require you to change physics.

Here are your own words:
So now lets let alpha be the angle subtended by the arcs, and omega be the angular velocity.

Yes, notice the key part? Alpha is the angle subtended by the arc and omega is the angular velocity. I don't have the arc go backwards due to the angular momentum.
Can you tell the difference between alpha and omega?

If the interferometer IS ALREADY BUILT, we do not need a debate, I win hands down.

Nope. I win, hands down.

Where is the source of light located on your interferometer? You have none. That source of light is the source of the counter-propagating beams. But you have none.

It can be placed anywhere along the loop, or off to the side of the loop somewhere with a common light path approaching the loop.
It doesn't matter where it is, so this is just another pathetic distraction from you.

You need a source of light on that interferometer.

Happy?
I did this just for you. Do you feel special?
https://i.imgur.com/BrbOgFL.png
I'll let you figure out what all the extra bits are.

BUT ONCE YOU DO INDICATE THAT SOURCE OF LIGHT, YOU ARE OUT OF LUCK, SINCE THEN YOU ARE ACCEPTING THAT THE INTERFEROMETER IS ALREADY BUILT, AND ALL YOU HAVE TO DO IS TO ROTATE IN ANY WAY TO OBTAIN THE SAGNAC.

Nope. You are the one that is out of luck.
When you rotate it the simple Sagnac formula applies of dt=4*A*w/c^2, which is what my derivation is showing.
So no, you get refuted, yet again.

If you wish to claim otherwise, you need to show the error with my derivation or provide your own.

And you seem capable of neither.

You can't fool anybody here jack.

I don't have to. The question is if you are managing to fool anyone and if you think you are.

No. The distance between the cities didn't change, the cities moved.

You cannot compare trains with the entire Earth.

Your analogy fails.

The distance between the cities did not change at all.

What did change was the speed of light being registered.

And you cannot deny that.

No it didn't.

Let us put your word to the test.

Dufour and Prunier PROVED THE SAME THING, USING AN INGENIOUS PATH, PART CIRCULAR, PART UNIFORM.


http://www.conspiracyoflight.com/pdf/Dufour_and_Prunier-On_the_Fringe_Movement_Registered_on_a_Platform_in_Uniform_Motion_%281942%29.pdf

A. Dufour and F. Prunier created Sagnac interferometers that were composites of moving and stationary paths, including stationary sources and stationary detectors. This was essentially to test if the relativistic approach could be distinguished from the classical approach.

"In all cases of this experimental test, the Sagnac effect was the same. This overturned Langevin’s analysis, and in 1937, he had to revise his explanation, as pointed out by Kelly: 

“In his final essay on the subject in 1937, Langevin proposed that the results published that year by Dufour and Prunier showed that one had to assume either (a) the light speed varied to c + wr in one direction and c – wr in the other direction, or (b) the time aboard the spinning apparatus had to change by a factor of +/-2wA/c2 in either direction. Indeed, Langevin went as far as to say that assuming (a), “we find, by a very simple and very general reasoning, the formula for the difference of the times of the path of the two light beams in the Sagnac experiment.” .

The proposition (b) though is untenable because if this were true then when the light beam passed back to the moving detector, the local time from each direction would be out of synchronization, meaning that the clocks cannot be counting real time and that the effective time dilation is meaningless. This was also pointed out by Herbert Ives in his 1938 paper criticizing Langevin. Ives says about the absurdity of Langevin’s proposition (b):

” There are of course not merely two clocks, but an infinity of clocks, where we include those that could be transported at finite speeds, and around other paths. As emphasized previously, the idea of “local time” is untenable, what we have are clock readings. Any number of clock readings at the same place are physically possible, depending on the behaviour and history of the  clocks used. More than one “time” at one place is a physical absurdity. “

The only explanation left, is Langevin’s proposition a) that the light speed varies by C+/-wr in one or the other direction around the disk, consistent with Dufour and Prunier’s experimental results."

In 1939, Dufour and Prunier carried out their
final experiment. They did a test with both the
beginning and end of the light path on the
spinning disc, but with the middle portion of
the path reflected off mirrors fixed in the
laboratory (directly above the disc). In this
test, they had both the light emitter and the
photographic recorder fixed in the laboratory.

The fringe shifts resulting from all the above
Dufour and Prunier tests were the same as in
their original Sagnac-type tests.

In 1942 Dufour and Prunier published a
composite paper reviewing their total
experimental work to date. At the end of this
paper they state that "the relativity theory
seems to be in complete disagreement with the
result which was garnered from the
experiment ".

Herbert Ives, Light Signals Sent Around a Closed Path:

http://www.conspiracyoflight.com/Ives/Herbert_Ives_Light_Signals_Sent_Around_a_Closed_Path.pdf

 Ives was a pioneer in the development of television at Bell Telephone Laboratories.  The following quotations are from his 1938 article.
 
     The experiment was interpreted by its author as positive evidence for the existence of the luminiferous ether…

     It is the purpose of this paper first to show that the Sagnac experiment in its essentials involves no consideration of rotation, and second to investigate the results obtained when transported clocks are used.
 
Ives analyzed the Sagnac experiment using a hexagonal path rather than a circular one.
 
He concluded with this statement:
 
     The net result of this study appears to be to leave the argument of Sagnac as to the significance of his experiment as strong as it ever was.

The Sagnac effect is not due to rotation, but instead is a linear effect due to a true anisotropic light speed in a moving frame.


The fringe shifts resulting from all the above
Dufour and Prunier tests were the same as in
their original Sagnac-type tests.

They used BOTH A SPINNING DISK AND A FIXED PORTION OF THE PATH.

THEY OBTAINED THE SAME RESULTS.

THIS, ALONG WITH THE WANG EXPERIMENT, PROVES THE EXISTENCE OF UNIFORM/TRANSLATIONAL SAGNAC BEYOND A SHADOW OF A DOUBT.


CAN YOU READ ENGLISH JACK?

In 1939, Dufour and Prunier carried out their
final experiment. They did a test with both the
beginning and end of the light path on the
spinning disc, but with the middle portion of
the path reflected off mirrors fixed in the
laboratory (directly above the disc). In this
test, they had both the light emitter and the
photographic recorder fixed in the laboratory.

The fringe shifts resulting from all the above
Dufour and Prunier tests were the same as in
their original Sagnac-type tests.



Which used a loop of optic fibre where each side of the loop moved in an opposite direction. It was not translational motion. Try again.

No such thing happened.

If it did, the paper would have been rejected.

Read the paper.


https://arxiv.org/ftp/physics/papers/0609/0609222.pdf (first experiment conducted by R. Wang)

https://arxiv.org/ftp/physics/papers/0609/0609202.pdf (second experiment carried out by R. Wang)


Can you read English jack?

As shown in Fig. 3, the phase shift or the traveltime
difference between two counter-propagating light
beams in the moving optic fiber was clearly observed
in all different configurations of FOCs.


HERE IS THE CONCLUSION OF THE PAPER, HAVING BEEN PEER REVIEWED AND PUBLISHED IN ONE OF THE MOST RESPECTED JOURNAL OF PHYSICS IN THE WORLD:

Conclusion

The travel-time difference of two counterpropagating
light beams in moving fiber is proportional
to both the total length and the speed of the fiber,
regardless of whether the motion is circular or uniform.
In a segment of uniformly moving fiber with a speed of
v and a length of Δl, the travel-time difference is
2vΔl/c^2.

Professor Wang proved a linear sagnac effect that is effected by linear cable length in then same way that additional circumference affects time delay between signals.


Instead you need a FOC, which isn't what you had, and isn't what a straight line is.

As shown in Fig. 3, the phase shift or the traveltime
difference between two counter-propagating light
beams in the moving optic fiber was clearly observed
in all different configurations of FOCs. The phase shift
Δφ, and therefore, the travel-time difference Δt are
proportional to both the total length and the speed of
the moving fiber whether the motion is circular or
uniform. Other tests using smaller end wheels for the
FOC and fiber loops with additional curves also
confirmed the same finding.



You really need medical help jack.


I have a physical loop indicated in red which subtends an angle alpha which is moving along Earth's orbit at an angular velocity of omega. The only thing going back is the light, not the arc.

But that is not what you are doing in your derivation.

YOU ARE SAYING THAT THE EARTH HAS TO GO BACK ON ITS ORBIT TO ACCOMODATE YOUR GRAPHIC, THE R1 ARC.

IF YOUR INTERFEROMETER IS ALREADY BUILT YOU DO NOT NEED THE DERIVATION AT ALL.

YOU SIMPLY PLUG IN THE FORMULA.

YOU CHEATED OUTRIGHTLY JACK, THAT IS WHY YOU GOT THE dt=4*A*w/c^2 RESULT.


Happy?
I did this just for you. Do you feel special?
https://i.imgur.com/BrbOgFL.png
I'll let you figure out what all the extra bits are.

YOU ARE NOW OUT OF LUCK JACK!!!

You have just indicated that your interferometer IS ALREADY BUILT IN.

THEN, THE DERIVATION FOR THE LOOP IS QUITE SIMPLE.

YOU HAVE TWO COUNTER-PROPAGATING BEAMS OF LIGHT, AND YOU GET THE SHIFT.


What you have done is to actually construct a interferometer using the angular velocity and the omega angle, back and forth.

Those were your own words.

You simply substracted the two arcs BUILDING YOUR INTERFEROMETER AS YOU WENT ALONG.

But you can't do that.


NOW, YOU HAVE INDICATED THAT THE INTERFEROMETER ALREADY HAS A LIGHT SOURCE.

No need then to substract the two arcs using the angular velocity and the angle subtended of the orbit of the Earth.

You simply cheated jack!!!

Did you really think for a second that nobody here would be able to detect your piece of shit derivation?

Think again.

If you wish to claim otherwise, you need to show the error with my derivation.

Here is the error.

Your own words.

So now lets let alpha be the angle subtended by the arcs, and omega be the angular velocity.
So, one beam of light will propagate in the same direction as the big arc and the opposite direction as the little arc.
What this means is its trip along the big arc will be longer than if it was stationary and shorter along the little arc.

You have a contribution from the arc with radius R2. This is based upon the area swept out by this arc, alpha*R2^2/2 (it would be 4*A*w/c^2).
You then have a contribution from the arc with radius R1. This is based upon the area swept out by this arc, and as the light beams are now travelling in an opposite direction and thus is of opposite sign and thus would be proportional to -alpha*R1^2/2.



Now lets try 2 arcs, one at R1 and one at R2, where the sections connecting the 2 arcs are along the radii.
What this means is that going between them is the same for both directions.

So now lets let alpha be the angle subtended by the arcs, and omega be the angular velocity.
So, one beam of light will propagate in the same direction as the big arc and the opposite direction as the little arc.
What this means is its trip along the big arc will be longer than if it was stationary and shorter along the little arc.

Meanwhile, the light going the other way has to travel a distance of alpha*R1+omega*R1*t2b+alpha*R2-omega*R2*t2a, for similar reasons.

You are actually constructing an interferometer in outer space, demanding that the Earth stop in its orbit, and then go back with 30km/s to accomodate your graphic.

You can't do that jack.

Moreover, you are using four beams of light back and forth.

That is not Sagnac.


I explained why you MUST subtract the 2 values.
If you have beam a and beam b, where they are counterpropogating, then in the arc with radius R2, say beam a is moving forwards and beam b is moving backwards (relative to the orbital motion).
This means the path length for b is much shorter, and thus the shift (measured as extra time taken for beam a relative to beam b) will be positive.
They then continue along the light path and travel along the arc of radius R1. Now beam a is moving backwards and beam b is moving forwards. That means the shift will now be negative.
You add these 2 shifts and get a shift proportional to R2^2-R1^2.

Here is another error.

Sagnac = one shift, two beams

What you have is TWO SHIFTS and four beams.


You cannot have two shifts.

Not by having the Earth actually go back on its orbit for you.

You have now indicated a source of light for the interferometer.

Then, your interferometer is already built and your derivation amounts to nothing at all.


If you think it will, THEN DERIVE IT!!

Sure thing.

Sagnac effect and pure geomatry, Tartaglia/Ruggiero







Published in the  AMERICAN JOURNAL OF PHYSICS, vol. 83, pp. 427-432. - ISSN 0002-9505


A total refutation of your piece of shit derivation.

The Sagnac effect obtained for a uniform paths/lengths.

No. The distance between the cities didn't change, the cities moved.
You cannot compare trains with the entire Earth.

Why not?
Did I need to say it goes along a circular track?

Your analogy fails.

Why?
Do you have anything other than pathetic dismissal?

The distance between the cities did not change at all.

Just like the distance between the front and back of the train didn't change.

What did change was the speed of light being registered.

Only in a non-inertial reference frame.
A just as valid (if not more valid) analysis is using an inertial reference frame in which the cities moved and thus the distance the light took was different in each case.

Let us put your word to the test.

Yes, lets put MY WORDS to the test. Not all this other crap you are trying to come up with.
Quit with all the relevant BS.
If you want to move onto something else, deal with the OP first.
I'm ignoring anything which doesn't deal with it.

I have a physical loop indicated in red which subtends an angle alpha which is moving along Earth's orbit at an angular velocity of omega. The only thing going back is the light, not the arc.
But that is not what you are doing in your derivation.

No. THAT IS EXACTLY WHAT I AM DOING!!!

Stop repeating the same lie.


Point out exactly where in my derivation I am claiming that.
If you can't, then stop repeating the same lie.

YOU SIMPLY PLUG IN THE FORMULA.

Which I did, and you claimed it was wrong, claiming you need to use an area which doesn't even exist. So I had to derive the formula to show you were wrong.
The formula is dt=4*A*w/c.
This formula shows that for a given loop the orbital Sagnac is much smaller.

YOU CHEATED OUTRIGHTLY JACK, THAT IS WHY YOU GOT THE dt=4*A*w/c^2 RESULT.

No. THAT IS THE FORMULA!!!! You didn't accept it so I had to derive it to show that you were full of shit.

You have just indicated that your interferometer IS ALREADY BUILT IN.
THEN, THE DERIVATION FOR THE LOOP IS QUITE SIMPLE.
YOU HAVE TWO COUNTER-PROPAGATING BEAMS OF LIGHT, AND YOU GET THE SHIFT.

That is what I have done from the start.
But yes the derivation is quite simple.
So why do you keep rejecting it?
You have 2 counterpropogating beams and get a shift dt=4*A*w/c, where A is the AREA OF THE LOOP!!!!!
As such, the orbital Sagnac is much smaller than the rotational one.

What you have done is to actually construct a interferometer using the angular velocity and the omega angle, back and forth.
Those were your own words.

No. They weren't my own words.
The angular velocity was omega. The angle subtended by the interferometer was alpha.
The entire interferometer was moving forwards. At no point did it go backwards. The only thing that went backwards was the light.

If you wish to disagree provide the quote (complete with a link to the post) showing exactly where I said that.

You simply cheated jack!!!

Nope. That would be you.
Blatantly lying about what I have done to try and dismiss it.

Here is the error.
Your own words.

No error there.

You are actually constructing an interferometer in outer space, demanding that the Earth stop in its orbit, and then go back with 30km/s to accomodate your graphic.
Moreover, you are using four beams of light back and forth.
What you have is TWO SHIFTS and four beams.

Again, stop lying. I am not doing that at all.
My words do not indicate that at all.
I have already explained this to you.

So are you too stupid to understand or are you intentionally lying to everyone to try and pedal your BS?

Now can you show the exact error in my derivation.
Don't just bitch and moan like this.
Point out explicitly the key thing that is wrong and what it should be, correcting the math as you go.
If you can't, then shut up.

Then, your interferometer is already built and your derivation amounts to nothing at all.

Again, my derivation is to show that I was correct in using the area of the loop.
That was its sole purpose. If you just accepted the formula, there would be no need. But you refused.

If you think it will, THEN DERIVE IT!!
Sure thing.
Sagnac effect and pure geomatry, Tartaglia/Ruggiero

Posting a bunch of links to crap is not deriving it.
Do it yourself.
If you can't, SHUT UP!
Your inability to derive it yourself shows you do not understand it and thus cannot honestly judge it.

A total refutation of your piece of shit derivation.

It didn't deal with my derivation at all. So it didn't refute it at all.

Do you have anything other than pathetic dismissal?

Sure do.

A precise mathematical demonstration that I am right.







“Any clear sign of a variation in c, the speed of light, as the Earth [revolved]would prove that the aether existed.”
George Smoot, Wrinkles in Time, 1993.

Posting a bunch of links to crap is not deriving it.

Tartaglia and Ruggiero are among the foremost experts on the Sagnac effect in the world today.

Their precise demonstration demolishes your claim that "only the area matters" in the Sagnac effect.

Not at all.

Please read:

Sagnac effect and pure geomatry, Tartaglia/Ruggiero







Published in the  AMERICAN JOURNAL OF PHYSICS, vol. 83, pp. 427-432. - ISSN 0002-9505


A total refutation of your piece of shit derivation.

The Sagnac effect obtained for a uniform paths/lengths.


Point out exactly where in my derivation I am claiming that.

You have 2 counterpropogating beams and get a shift dt=4*A*w/c, where A is the AREA OF THE LOOP!!!!!

That is not what you did jack.

Your own words.

I explained why you MUST subtract the 2 values.
If you have beam a and beam b, where they are counterpropogating, then in the arc with radius R2, say beam a is moving forwards and beam b is moving backwards (relative to the orbital motion).
This means the path length for b is much shorter, and thus the shift (measured as extra time taken for beam a relative to beam b) will be positive.
They then continue along the light path and travel along the arc of radius R1. Now beam a is moving backwards and beam b is moving forwards. That means the shift will now be negative.
You add these 2 shifts and get a shift proportional to R2^2-R1^2.


TWO SHIFTS, FOUR BEAMS OF LIGHT

A FORWARD, B BACKWARDS

B FORWARD, A BACKWARDS

BUT THIS IS NOT SAGNAC.

The entire interferometer was moving forwards. At no point did it go backwards. The only thing that went backwards was the light.

That is not what your calculations reveal at all.

YOU ARE ACTUALLY BUILDING THE INTERFEROMETER USING THE ANGULAR VELOCITY AND THE ALPHA ANGLE AS YOU GO ALONG.

Now lets try 2 arcs, one at R1 and one at R2, where the sections connecting the 2 arcs are along the radii.
What this means is that going between them is the same for both directions.

So now lets let alpha be the angle subtended by the arcs, and omega be the angular velocity.
So, one beam of light will propagate in the same direction as the big arc and the opposite direction as the little arc.
What this means is its trip along the big arc will be longer than if it was stationary and shorter along the little arc.

Meanwhile, the light going the other way has to travel a distance of alpha*R1+omega*R1*t2b+alpha*R2-omega*R2*t2a, for similar reasons.

Are these not your words jack?

Are you so dumb as to not realize what you have done here?

You can't have it both ways.

Now you accept that your interferometer is already built and has a light source.

Let us use a circular path for the sake of simplicity.

t' anticlockwise direction
t'' clockwise direction

t' = 2Πr/(c + v)

t" = 2Πr/(c - v)

TWO BEAMS ONE SHIFT

dt = 4Πrv/(c² - v²)


What you did was to actually construct your interferometer in space, using the angular velocity of the Earth to draw your arcs of light back and forth.

But the Earth cannot go back and forth to accomodate for your derivation.

You DID NOT indicate a light source at first.

Only today you added that feature on your graph.

But now you are out of luck jack.

Since you already have a light source, it means the interferometer was already existing to start with: you cannot go back and forth to actually build the interferometer like you did.

Moreover, the CORRECT derivation does not involve the area at all, or any loop.

You have a path.

Tartaglia and Ruggiero calculated the Sagnac for this uniform/translational path.

Here is the conclusion:



Then, if you want the orbital, you simply plug in the 30km/s speed.

You cheated in your derivation to arrive at the final result.

You used two shifts, and you actually constructed the interferometer (the R2 and the R1 arcs) using the angular velocity and the alpha angle. In space you cannot do that at all. Perhaps in a lab using a phase conjugate gyro interferometer. But not in space.


By the way.

Can you show in this graphic WHERE THE LOOP OF THE ROTATIONAL AND/OR ORBITAL SAGNAC IS LOCATED?

IF YOU CANNOT YOU LOSE.



All you have is a straight beam of light from the satellite to the receiver on the ground.

WHERE IS THE LOOP JACK?

No. The distance between the cities didn't change, the cities moved.
You cannot compare trains with the entire Earth.

Your analogy fails.

The distance between the cities did not change at all.

The cities both moved while the signal was in transit meaning the two distances travelled were slightly different.

I fail to see your problem.

Do you have anything other than pathetic dismissal?
Sure do.
A precise mathematical demonstration that I am right.

How does this relate to using a train as an analogy at all?

And it doesn't even address the OP.

And of course, nothing of your own, just copied and pasted from elsewhere.

DEAL WITH THE LOOP IN THE OP OR THE PRIOR THREAD OR SHUT UP!!!

I'm ignoring anything else.

You have 2 counterpropogating beams and get a shift dt=4*A*w/c, where A is the AREA OF THE LOOP!!!!!
That is not what you did jack.

Yes it is. If you think it isn't you will need more than just copying and pasting my prior statements which do not indicate anything else.

I explained why you MUST subtract the 2 values.
If you have beam a and beam b, where they are counterpropogating, then in the arc with radius R2, say beam a is moving forwards and beam b is moving backwards (relative to the orbital motion).
This means the path length for b is much shorter, and thus the shift (measured as extra time taken for beam a relative to beam b) will be positive.
They then continue along the light path and travel along the arc of radius R1. Now beam a is moving backwards and beam b is moving forwards. That means the shift will now be negative.
You add these 2 shifts and get a shift proportional to R2^2-R1^2.

TWO SHIFTS, FOUR BEAMS OF LIGHT

No, a single shift, composed of 2 parts.
There is the part due to the big arc and the part due to the small arc.
There are 2 beams of light. One beam moves forward in the big arc and backwards in the small arc. The other goes the other way.
Both beams of light travel through the both arcs.

STOP REPEATING THE SAME LIES!!!!

A FORWARD, B BACKWARDS
B FORWARD, A BACKWARDS
BUT THIS IS NOT SAGNAC.

No, that is Sagnac.
You have beam A and beam B.
For a section of the loop A is moving forwards while B is moving backwards. In another section you have the opposite.
So 2 beams counterpropagating around a loop.
That is Sagnac.

The entire interferometer was moving forwards. At no point did it go backwards. The only thing that went backwards was the light.
That is not what your calculations reveal at all.

Where do they suggest anything otherwise?
Show explicitly.

Are these not your words jack?

I have mainly been skipping every time you quote me as I already know what I said.
No where in what I have said do I indicate that there are 4 beams of light or that the apparatus ever moves backwards.
It is only ever the beams of light that goes backwards, as they need to to complete the loop as the loop is travelling significantly below the speed of light (or stationary in a non-inertial reference frame).

Are you so dumb as to not realize what you have done here?

Do you mean am I dumb enough to just accept your lies about what I have done rather than to know what I have actually done?

I know what I have done. I correctly derived the Sagnac effect for a ring interferometer which has the centre of rotation far away fromt the loop.
I have done this correctly and explicitly such that you are unable to refute it and instead need to resort to blatantly lying about it.
You are also completely unable to provide your own derivation.

Let us use a circular path for the sake of simplicity.

But a circular path isn't simple.
A circular path is only simple when you have the circle centred on the centre of rotation. As soon as you move it away it is no longer that simple.

A nice simple question to highlight this is to ask what is v meant to be?
Should it be w*R1, the section of the circle closest to the centre of rotation?
Should it be w*R2, the section of the circle furthest from the centre of rotation?

Go and draw a picture of the system, derive the shifts explicitly, making it clear what each thing is.
Until you do, your words are meaningless.

TWO BEAMS ONE SHIFT

Which is what I had.
dt=4*A*w/c^2.
Or if you like:
ta=alpha*R2/(c-omega*R2)+alpha*R1/(c+omega*R1)+k
tb=alpha*R2/(c+omega*R2)+alpha*R1/(c-omega*R1)+k
dt=alpha*R2/(c-omega*R2)+alpha*R1/(c+omega*R1)+k-alpha*R2/(c+omega*R2)+alpha*R1/(c-omega*R1)+k
=2*alpha*omega*R2^2/(c^2-omega^2*R2^2)-2*alpha*omega*R1^2/(c^2-omega^2*R1^2)
~2*alpha*omega*(R2^2-R1^2)/c^2
=4*A*omega/c^2.

So two beams, one shift, with the shift proportional to the area.

Each beam has multiple time components. There is the time traversing the big arc, the time travelling in from the big arc to the small arc, the time traversing the small arc, and the time moving out to the big arc. The time travelling in and out is the same for both are is represented (collectively) as k.

You DID NOT indicate a light source at first.

Again, that is because it is irrelevant. The shift remains the same regardless of where the light source/detector is.

Since you already have a light source, it means the interferometer was already existing to start with: you cannot go back and forth to actually build the interferometer like you did.

I didn't go back and forth to magically build one. I used a simple loop.

Moreover, the CORRECT derivation does not involve the area at all, or any loop.

Yes it does. If you wish to claim it doesn't, then you need to show the error in mine which clearly involves the area.

Tartaglia and Ruggiero calculated the Sagnac for this uniform/translational path.

DEAL WITH THE OP OR SHUT UP!!
Stop trying to distract from your failures with pathetic BS.

You cheated in your derivation to arrive at the final result.

Then why have you been completely unable to show where I "cheated"?

Proving you wrong is not cheating.
You blatantly lying about what I have done is not me cheating.

Can you show in this graphic WHERE THE LOOP OF THE ROTATIONAL AND/OR ORBITAL SAGNAC IS LOCATED?

Like I said, DEAL WITH THE OP BEFORE MOVING ON!!!

Here is the precise mathematical demonstration that you do have a problem:







“Any clear sign of a variation in c, the speed of light, as the Earth [revolved]would prove that the aether existed.”
George Smoot, Wrinkles in Time, 1993.

The results in equations (7) and (11) confirm the independent claims of Marmet and Kelley: Light travels faster westward than eastward relative to the surface of the Earth.

Specifically the one-way measurement of light speed using GPS data in (6) clearly indicates that a signal sent eastward travels at speed c minus the rotational speed of the Earth v at that latitude giving c - v. The GPS data available in (10) also shows that a signal sent westward travels at speed c plus the rotational speed of the Earth v at that latitude giving c + v.

Here is the precise mathematical demonstration that you do have a problem:

Again, this does not deal with the OP at all, and it is not your own derivation.
Provide your own derivation which deals directly with the OP, and show the problem with mine or shut up.