On Sandokhan definitions of the Sagnac and Coriolis Effects

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rabinoz

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On Sandokhan definitions of the Sagnac and Coriolis Effects
« on: August 26, 2019, 09:38:06 PM »
This is simply a reply to a post by Sandokhan in the thread: Flat Earth General   Re: how to know the altitude of a DirecTV satelite where it is quite off-topic.

Georges Sagnac derived the CORIOLIS EFFECT formula, which features the AREA.
Look, YOU cannot arbitrarily define what is Coriolis Effect and what is Sagnac Effect. That has been decided long before YOU came on the scene!

No! Georges Sagnac derived the SAGNAC EFFECT formula, which bears HIS name and features the AREA.
Go and read:
Quote
Regarding the Proof for the Existence of a Luminiferous Ether Using a Rotating Inteferometer Experiment by Georges Sagnac
Abstract: This is English translation of Georges Sagnac’s second paper, which presents his “rotating interferometer experiment” where
the phenomenon known as the Sagnac effect manifests itself. This paper was originally published, in French, as: Sur la preuve de la realite de l’ether lumineux par l’experience de l’interferographe tournant. Note de G. Sagnac, presentee par E. Bouty. Comptes rendus, 1913, tome 157, pages 1410 –1413. Translated from the French in 2008 by William Lonc, Canada. The Editor of The Abraham Zelmanov Journal thanks William Lonc for this effort, and also Ioannis Haranas, Canada, for assistance.
Special thanks go to the National Library of France and Nadege Danet in person for the permission to reproduce the originally Sagnac paper in English.

Quote from: sandokhan
The definition of the Sagnac effect is applied to a closed loop (either circular or a uniform path).
Loop = a structure, series, or process, the end of which is connected to the beginning.
Thus, from a mathematical point of view, Michelson did not derive the Sagnac effect formula at all, since he compared two open segments, and not two loops.
No! Whatever YOU CLAIM, Michelson derived the Sagnac effect formula!

How can you ever claim that this is "two open segments, and not two loops"?


Quote from: sandokhan
The definition of the Sagnac effect.
Using the correct definition, we recover not only the error-free formula, but also the precise velocity addition terms.
For the Coriolis effect, one has a formula which is proportional to the area; only the phase differences of EACH SIDE are being compared, and not the continuous paths.
No! The Coriolis effect is completely different and is not related to area in the slightest!

The theory was originally developed by Gaspard-Gustave de Coriolis and applied to water-wheels. His original paper is the second in:


Quote from: sandokhan
For the Sagnac effect, one has a formula which is proportional to the velocity of the light beam; the entire continuous clockwise path is being compared to the other continuous counterclockwise path exactly as required by the definition of the Sagnac effect.

Experimentally, the Michelson-Gale test was a closed loop, but not mathematically. Michelson treated mathematically each of the longer sides/arms of the interferometer as a separate entity: no closed loop was formed at all. Therefore the mathematical description put forth by Michelson has nothing to do with the correct definition of the Sagnac effect (two pulses of light are sent in opposite direction around a closed loop) (either circular or a single uniform path). By treating each side/arm separately, Michelson was describing and analyzing the Coriolis effect, not the Sagnac effect.

Loop = a structure, series, or process, the end of which is connected to the beginning.
No! The Sagnac loop can be any shape at all! And, of course, you would analyse any straight edged loop one edge at a time! Whyever NOT?

Quote from: sandokhan
Connecting the two sides through a single mathematical description closes the loop; treating each side separately does not. The Sagnac effect requires, by definition, a structure, the end of which is connected to the beginning.
No! The Sagnac loop can be any shape at all! And, of course, you would analyse any straight edged loop one edge at a time! Whyever NOT?

Quote from: sandokhan
HERE IS THE DEFINITION OF THE SAGNAC EFFECT:

Two pulses of light sent in opposite direction around a closed loop (either circular or a single uniform path), while the interferometer is being rotated.
So?
Quote from: sandokhan
Loop = a structure, series, or process, the end of which is connected to the beginning.

A single continuous pulse A > B > C > D > A, while the other one, A > D > C > B > A is in the opposite direction, and has the negative sign.



Let's ignore your derivation and use the result from what appears to be the source of your diagram: The Michelson-Gale Experiment by Doug Marett (2010)

Look at this from what appears to be the source of your diagram:
Quote from: Doug Marett
Conspiracy of Light, The Michelson-Gale Experiment
In refining his argument, he proposed that it was not necessary for the light to go all the way around the globe - since there should be a velocity difference for any closed path rotating on the surface of the earth. He presented the following equation to calculate the time difference expected, using the shift in the interference fringes when the two beams overlap at the detector as a measure of the time difference:
Fig.1:
where:  Vo = the tangential velocity of the earth's rotation at the equator (465m/s)
              A = the area of the circular path
              R = the radius of the earth (6371000 m)
              c = speed of light (3E8 m/s)
              f = the latitude in degrees where the experiment is conducted.
              l = wavelength of the light
And those 2's should be 4's because even Michelson didn't initially get it quite right and it was corrected by Silberstein:
Quote from: Doug Marett
   The experiment remained in abeyance for several years, until Silberstein published a paper in 1921 on the theory of light propagation in rotating systems. In this article, Silberstein discusses Michelson's proposed experiment and through calculations of his own demonstrated that the time difference expected in such an experiment would be double what Michelson suggested.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
After taking all these factors into account, the expected fringe shift becomes:

Quote from: sandokhan
Dr. Ludwik Silberstein, a physicist on the same level with Einstein and Michelson, partially inspired and supported the Michelson-Gale experiment.

In 1921, Dr. Silberstein proposed that the Sagnac effect, as it relates to the rotation of the Earth or to the effect of the ether drift, must be explained in terms of the Coriolis effect: the direct action of Coriolis forces on counterpropagating waves.
<< Let's ignore all that for the moment - I simply do not have the time! >>

He proved that the real cause of the phenomenon measured by Georges Sagnac was the CORIOLIS FORCE EFFECT.
No! Dr. Ludwik Silberstein did not "prove that the real cause of the phenomenon measured by Georges Sagnac was the CORIOLIS FORCE EFFECT."
He showed that the light paths deviated from straight lines due to the Coriolis effect but by a negligible amount and so that was completely left out of his final result!

Malykin and Pozdnyakova do say:
Quote from: Grigorii B. Malykin and Vera I. Pozdnyakova
Siberstein [798, 799] suggested an explanation of the Sagnac effect based on the direct consideration of effect of the Sagnac forces on the counterpropagating waves. . . . . The areas of the triangles are different."
Now, while "the areas of the triangles are different" Dr Siberstein had previously shown "Thus, even for a ≈ 10 or 20 km the difference would certainly be too small to be measured directly.
So Dr Silberstein certainly does not "derive the Coriolis effect" and on the contrary, he shows that its effect is "certainly be too small to be measured directly."

And again: Malykin and Pozdnyakova say:
Quote from: Grigorii B. Malykin and Vera I. Pozdnyakova
Siberstein [798, 799] suggested an explanation of the Sagnac effect based on the direct consideration of effect of the Sagnac forces on the counterpropagating waves. . . . . The areas of the triangles are different."
Now, while "the areas of the triangles are different" Dr Siberstein had previously shown "Thus, even for a ≈ 10 or 20 km the difference would certainly be too small to be measured directly.
So Dr Silberstein certainly does not "derive the Coriolis effect" and on the contrary, he shows that its effect is "certainly be too small to be measured directly."

Quote from: sandokhan
http://www.conspiracyoflight.com/Michelson-Gale/Silberstein.pdf

The formula derived by Dr. Silberstein, peer reviewed in the IOP article, and described by the author as the "effect of the Coriolis forces" is this:
dt = 4ωA/c^2
Excuse me but is it described by the author as the "effect of the Coriolis forces"? Silberstein explicitly neglects the "effect of the Coriolis forces"!

So dt = 4ωA/c^2 is simply the Sagnac Delay, nothing more!

Quote from: sandokhan
Here is the Maraner-Zendri formula:


What Maraner and Zendri did is to derive the CORIOLIS EFFECT formula with relativistic corrections which are dependent on the center of rotation, and NOT the SAGNAC EFFECT.
No, they did not "derive the CORIOLIS EFFECT formula" but they did "derive the Sagnac Effect formula with relativistic corrections which are dependent on the center of rotation".

And that "dependence on the center of rotation" is only part of the relativistic corrects and in any case he showed that in practical cases the relativistic corrections are negligible.

If those corrections are neglected it reduces to the usual Sagnac expression which is independent of the shape of the loop and the centre of rotation.

Quote from: sandokhan
They used the SAME derivation as did Michelson based on a comparison of two sides, AND NOT THE TWO LOOPS as required by the definition of the Sagnac error, a huge error on their part.
Nope, that's just your incorrect interpretation.

Quote from: sandokhan
For the uninformed RE: here is the correct definition of the Sagnac effect.

https://www.mathpages.com/rr/s2-07/2-07.htm
You mean the one that goes on to say:
Quote
2.7  The Sagnac Effect
This phenomenon applies to any closed loop, not necessarily circular. For example, suppose a beam of light is split by a half-silvered mirror into two beams, and those beams are directed in a square path around a set of mirrors in opposite directions as shown below.
Just as in the case of the circular loop, if the apparatus is unaccelerated, the two beams will travel equal distances around the loop, and arrive at the detector simultaneously and in phase. However, if the entire device (including source and detector) is rotating, the beam traveling around the loop in the direction of rotation will have farther to go than the beam traveling counter to the direction of rotation, because during the period of travel the mirrors and detector will all move (slightly) toward the counter-rotating beam and away from the co-rotating beam. Consequently the beams will reach the detector at slightly different times, and slightly out of phase, producing optical interference "fringes" that can be observed and measured.

Quote from: sandokhan
THE SAGNAC EFFECT DOES NOT REQUIRE AN AREA, only the CORIOLIS EFFECT is proportional to an area.
No, you have never shown that "THE SAGNAC EFFECT DOES NOT REQUIRE AN AREA" and the CORIOLIS EFFECT is quite unrelated to an area.

Go and read what the Coriolis Force is! Here is the expression for it

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sandokhan

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Re: On Sandokhan definitions of the Sagnac and Coriolis Effects
« Reply #1 on: August 26, 2019, 11:07:53 PM »
The rotational and the orbital CORIOLIS EFFECT formulas were derived for the first time in 1904 by A. Michelson, and they feature AN AREA:

http://www.conspiracyoflight.com/Michelson-Gale/Michelson_1904.pdf

Georges Sagnac derived THE SAME FORMULA, featuring an area, the CORIOLIS EFFECT formula.

The original papers published by G. Sagnac (The Luminiferous Ether is Detected as a Wind Effect Relative to the Ether Using a Uniformly Rotating Interferometer):

http://zelmanov.ptep-online.com/papers/zj-2008-07.pdf

http://zelmanov.ptep-online.com/papers/zj-2008-08.pdf

In 1913, Georges Sagnac measured ONLY the Coriolis effect, and not the true Sagnac effect (proportional to the linear velocity and radius of rotation).

Here is the shape of the interferometer used by Sagnac:




Is there a way to clearly distinguish between the formulas required for an irregularly shaped interferometer and a symmetrically shaped interferometer?

Yes, there is: using graduate level differential geometry.

Please read:


https://link.springer.com/article/10.1023/A:1023972214666

https://arxiv.org/pdf/gr-qc/0103091.pdf

Coriolis Force and Sagnac Effect

Because of acting of gravity-like Coriolis force the trajectories of co- and anti-rotating photons have different radii in the rotating reference frame, while in the case of the equal radius the effective gravitational potentials for the photons have to be different.




An interferometer with DIFFERENT RADII (located away from the center of rotation) will manifest the Coriolis force in the form of a phase shift 4AΩ/c^2.

Different sets of radii and the center of rotation do not coincide with the geometrical center of the interferometer.

That is why Sagnac had to use the formula which features the area and the angular velocity: he only measured the CORIOLIS EFFECT.

Even if the shape of the interferometer is made to look more symmetrical, there are still two different radii to deal with:




This is the formula derived by G. Sagnac:

4AΩ/c^2.



THIS IS THE CORIOLIS EFFECT FORMULA.

Here is the precise proof, peer-reviewed in an IOP article.

THIS IS AN IOP ARTICLE, one of the most comprehensive papers on the Sagnac effect ever published.






Here is reference #27:






http://www.conspiracyoflight.com/Michelson-Gale/Silberstein.pdf

The formula derived by Dr. Silberstein, peer reviewed in the IOP article, and described by the author as the "effect of the Coriolis forces" is this:

dt = 4ωA/c^2


Here is a direct derivation of the same formula using only the Coriolis force:

https://www.ias.ac.in/article/fulltext/pram/087/05/0071

The derivation has NO LOOPS at all.

Just a comparison of two sides.



Here is how Dr. Silberstein's CORIOLIS EFFECT formula was derived:

The propagation of light in rotating systems, Journal of the Optical Society of America, vol. V, number 4, 1921

He proved that the real cause of the phenomenon measured by Georges Sagnac was the CORIOLIS FORCE EFFECT.



Dr. Silberstein proved that the effect measured by Sagnac is A PHYSICAL EFFECT, a deflection/inflection of the light beams due to the CORIOLIS FORCE.

In 1922, he extended the definition used in his 1921 paper on the nature of the rays arriving from the collimator:

http://gsjournal.net/Science-Journals/Historical%20Papers-Mechanics%20/%20Electrodynamics/Download/2645

Dr. Ludwik Silberstein, a physicist on the same level with Einstein and Michelson, partially inspired and supported the Michelson-Gale experiment.

In 1921, Dr. Silberstein proposed that the Sagnac effect, as it relates to the rotation of the Earth or to the effect of the ether drift, must be explained in terms of the Coriolis effect: the direct action of Coriolis forces on counterpropagating waves.

http://www.conspiracyoflight.com/Michelson-Gale/Silberstein.pdf

The propagation of light in rotating systems, Journal of the Optical Society of America, vol. V, number 4, 1921

Dr. Silberstein developed the formula published by A. Michelson using very precise details, not to be found anywhere else.

He uses the expression kω for the angular velocity, where k is the aether drag factor.

He proves that the formula for the Coriolis effect on the light beams is:

dt = 2ωσ/c2

Then, Dr. Silberstein analyzes the area σ and proves that it is actually a SUM of two other areas (page 300 of the paper, page 10 of the pdf document).

The effect of the Coriolis force upon the interferometer will be to create a convex and a concave shape of the areas: σ1 and σ2.

The sum of these two areas is replaced by 2A and this is how the final formula achieves its final form:

dt = 4ωA/c2

A = σ1 + σ2

That is, the CORIOLIS EFFECT upon the light beams is totally related to the closed contour area.



Here are the DEFINITIONS USED BY MODERN SCIENCE TO DESCRIBE THE SAGNAC EFFECT:

https://www.mathpages.com/rr/s2-07/2-07.htm

If two pulses of light are sent in opposite directions around a stationary circular loop of radius R, they will travel the same inertial distance at the same speed, so they will arrive at the end point simultaneously.

http://www.cleonis.nl/physics/phys256/sagnac.php

Essential in the Sagnac effect is that a loop is closed.

http://www.einsteins-theory-of-relativity-4engineers.com/sagnac-effect.html

The Sagnac effect is observed when coherent light travels around a closed loop in opposite directions and the phases of the two signals are compared at a detector.



THE SAGNAC EFFECT DOES NOT REQUIRE AN AREA, only the CORIOLIS EFFECT is proportional to an area.



Michelson and Gale COMPARED TWO SIDES ONLY, not any loops at all:


http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1925ApJ....61..137M&amp;data_type=PDF_HIGH&amp;whole_paper=YES&amp;type=PRINTER&amp;filetype=.pdf



The final formula used by Michelson features an AREA: it is the CORIOLIS EFFECT formula.



Using a phase-conjugate mirror, for the first time in 1986, Professor Yeh was able to derive the TRUE SAGNAC FORMULA which is proportional to the velocity of the light beams.




page 152 of the pdf document, section Recent Advances in Photorefractive Nonlinear Optics page 4

The MPPC acts like a normal mirror and Sagnac interferometry is obtained.

Here is the derivation of my formula, using TWO LOOPS:

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg2117351#msg2117351

Here is the final formula:

2(V1L1 + V2L2)/c2

My formula is confirmed at the highest possible scientific level, having been published in the best OPTICS journal in the world, Journal of Optics Letters, and it is used by the US NAVAL RESEARCH OFFICE, Physics Division.

A second reference which confirms my global/generalized Sagnac effect formula.

https://apps.dtic.mil/dtic/tr/fulltext/u2/a206219.pdf

Studies of phase-conjugate optical devices concepts

US OF NAVAL RESEARCH, Physics Division

Dr. P. Yeh
PhD, Caltech, Nonlinear Optics
Principal Scientist of the Optics Department at Rockwell International Science Center
Professor, UCSB
"Engineer of the Year," at Rockwell Science Center
Leonardo da Vinci Award in 1985
Fellow of the Optical Society of America, the Institute of Electrical and Electronics Engineers

page 152 of the pdf document, section Recent Advances in Photorefractive Nonlinear Optics page 4

The MPPC acts like a normal mirror and Sagnac interferometry is obtained.



Phase-Conjugate Multimode Fiber Gyro

Published in the Journal of Optics Letters, vol. 12, page 1023, 1987

page 69 of the pdf document, page 1 of the article


A second confirmation of the fact that my formula is correct.

Here is the first confirmation:



Self-pumped phase-conjugate fiber-optic gyro, I. McMichael, P. Yeh, Optics Letters 11(10):686-8 · November 1986 

http://www.dtic.mil/dtic/tr/fulltext/u2/a170203.pdf (appendix 5.1)


Exactly the formula obtained by Professor Yeh:

φ = -2(φ2 - φ1) = 4π(R1L1 + R2L2)Ω/λc = 4π(V1L1 + V2L2)/λc

Since Δφ = 2πc/λ x Δt, Δt = 2(R1L1 + R2L2)Ω/c2 = 2(V1L1 + V2L2)/c2

CORRECT SAGNAC FORMULA:

2(V1L1 + V2L2)/c2

The very same formula obtained for a Sagnac interferometer which features two different lengths and two different velocities.

http://www.dtic.mil/dtic/tr/fulltext/u2/a170203.pdf

ANNUAL TECHNICAL REPORT PREPARED FOR THE US OF NAVAL RESEARCH.

Page 18 of the pdf document, Section 3.0 Progress:

Our first objective was to demonstrate that the phase-conjugate fiberoptic gyro (PCFOG) described in Section 2.3 is sensitive to rotation. This phase shift plays an important role in the detection of the Sagnac phase shift due to rotation.

Page 38 of the pdf document, page 6 of Appendix 3.1


it does demonstrate the measurement of the Sagnac phase shift Eq. (3)


HERE IS EQUATION (3) OF THE PAPER, PAGE 3 OF APPENDIX 3.1:

φ = -2(φ2 - φ1) = 4π(R1L1 + R2L2)Ω/λc = 4π(V1L1 + V2L2)/λc

Since Δφ = 2πc/λ x Δt, Δt = 2(R1L1 + R2L2)Ω/c2 = 2(V1L1 + V2L2)/c2

CORRECT SAGNAC FORMULA:

2(V1L1 + V2L2)/c2




The Coriolis effect is a physical effect upon the light beams: it is proportional to the area of the interferometer. It is a comparison of two sides.

The Sagnac effect is an electromagnetic effect upon the velocities of the light beams: it is proportional to the radius of rotation. It is a comparison of two loops.

Two different phenomena require two very different formulas.


My SAGNAC EFFECT formula proven and experimentally fully established at the highest possible level of science.



Let us now compare the two derivations, using two loops (Sagnac effect) and two sides (Coriolis effect):



Point A is located at the detector
Point B is in the bottom right corner
Point C is in the upper right corner
Point D is in the upper left corner

l1 is the upper arm.
l2 is the lower arm.

Here is the most important part of the derivation of the full/global Sagnac effect for an interferometer located away from the center of rotation.

A > B > C > D > A is a continuous counterclockwise path, a negative sign -

A > D > C > B > A is a continuous clockwise path, a positive sign +

The Sagnac phase difference for the clockwise path has a positive sign.

The Sagnac phase difference for the counterclockwise has a negative sign.


Sagnac phase components for the A > D > C > B > A path (clockwise path):

l1/(c - v1)

-l2/(c + v2)

Sagnac phase components for the A > B > C > D > A path (counterclockwise path):

l2/(c - v2)

-l1/(c + v1)


For the single continuous clockwise path we add the components:

l1/(c - v1) - l2/(c + v2)

For the single continuous counterclockwise path we add the components:

l2/(c - v2) - l1/(c + v1)


The net phase difference will be (let us remember that the counterclockwise phase difference has a negative sign attached to it, that is why the substraction of the phase differences becomes an addition):

{l1/(c - v1) - l2/(c + v2)} - (-){l2/(c - v2) - l1/(c + v1)} = {l1/(c - v1) - l2/(c + v2)} + {l2/(c - v2) - l1/(c + v1)}

Rearranging terms:

l1/(c - v1) - l1/(c + v1) + {l2/(c - v2) - l2/(c + v2)} =

2(v1l1 + v2l2)/c2

Exactly the formula obtained by Professor Yeh:

φ = -2(φ2 - φ1) = 4π(R1L1 + R2L2)Ω/λc = 4π(V1L1 + V2L2)/λc

Since Δφ = 2πc/λ x Δt, Δt = 2(R1L1 + R2L2)Ω/c2 = 2(V1L1 + V2L2)/c2

CORRECT SAGNAC FORMULA:

2(V1L1 + V2L2)/c2

Self-pumped phase-conjugate fiber-optic gyro, I. McMichael, P. Yeh, Optics Letters 11(10):686-8 · November 1986 

http://www.dtic.mil/dtic/tr/fulltext/u2/a170203.pdf (appendix 5.1)


This is how the correct Sagnac formula is derived: we have single continuous clockwise path, and a single continuous counterclockwise path.

If we desire the Coriolis effect, we simply substract as follows:

dt = l1/(c - v1) - l1/(c + v1) - (l2/(c - v2) - l2/(c + v2))

Of course, by proceeding as in the usual manner for a Sagnac phase shift formula for an interferometer whose center of rotation coincides with its geometrical center, we obtain:

2v1l1/(c2 - v21) - 2v2l2/(c2 - v22)

l = l1 = l2

2l[(v1 - v2)]/c2

2lΩ[(R1 - R2)]/c2

R1 - R2 = h

2lhΩ/c2

By having substracted two different Sagnac phase shifts, valid for the two different segments, we obtain the CORIOLIS EFFECT formula.


However, for the SAGNAC EFFECT, we have a single CONTINUOUS CLOCKWISE PATH, and a single CONTINUOUS COUNTERCLOCKWISE PATH, as the definition of the Sagnac effect entails.


« Last Edit: August 26, 2019, 11:11:49 PM by sandokhan »

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sandokhan

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Re: On Sandokhan definitions of the Sagnac and Coriolis Effects
« Reply #2 on: August 27, 2019, 01:14:39 AM »
I am now going to derive the true/correct SAGNAC EFFECT formula for the interferometer used by G. Sagnac in 1913:



Topologically, it is a square interferometer (in order to avoid dealing with the vx and vy components of v), the two segments M4l and lM1 can be joined into a single segment, M4M1.

Let us remember that Sagnac used ONLY THE AREA to derive the final phase-shift formula: 16πAΩ/λc.




The Sagnac effect formula for a square interferometer which rotates around its own geometrical center.

Let L = r√2 (r = distance from point O to one of the corners)

Time travel along side AB:

dtab = L/(c - v/√2)

(distance from point O to one of the sides is r/√2, and since v = r x ω, velocity for the light beam traveling along a side is v/√2)

dtcounterclockwise = 8r/(√2c + v)

dtclockwise = 8r/(√2c - v)

Δt = 8rv/c2



Now, the much more difficult case for the same square interferometer located away from the center of rotation.

The laboratory at Sorbonne, in France, where Sagnac performed his experiments, ALSO was rotating, according to heliocentrism/RE theory, along with the Earth.

Let us now rotate the square interferometer by 135° in the clockwise direction: point A will be located in the uppermost position (the source of light will be placed at point A as well).

Distance from the center of rotation to point C is k2, while the distance from the center of rotation to point A is k1.

v1 = k1 x ω

v2 = k2 x ω

Proceeding exactly as in the case of the interferometer in the shape of a rectangle, we have two loops, one counterclockwise, one clockwise.

A > B > C > D > A is the clockwise path

A > D > C > B > A is the counterclockwise path

Sagnac phase components for the counterclockwise path (only the vx components of the velocity vector are subject to a different time phase difference in rotation, not the vy components):

L/(c - v1)

-L/(c + v2)

-L/(c + v2)

L/(c - v1)

Sagnac phase components for the clockwise path:

-L/(c + v1)

L/(c - v2)

L/(c - v2)

-L/(c + v1)

For the single continuous counterclockwise path we add the components:

L/(c - v1) - L/(c + v2) - L/(c + v2) + L/(c - v1) = 2L/(c - v1) - 2L/(c + v2)

For the single continuous clockwise path we add the components:

-L/(c + v1) + L/(c - v2) + L/(c - v2) - L/(c + v1) = -2L/(c + v1) + 2L/(c - v2)

The net time phase difference will be (let us remember that the counterclockwise phase difference has a negative sign attached to it, that is why the substraction of the phase differences becomes an addition):

2L/(c - v1) - 2L/(c + v2) -(-)[-2L/(c + v1) + 2L/(c - v2)] = 2L(2v1/c2) + 2L(2v2/c2) = 4L(v1 + v2)/c2

This is the correct global/generalized SAGNAC EFFECT formula for a square shaped  interferometer:

4L(v1 + v2)/c2

For the same interferometer, the CORIOLIS EFFECT formula is:

4Aω/c2


The phase difference for the SAGNAC EFFECT is:

Δφ = Δt x c/λ = [4L(v1 + v2)]/c2 x c/λ = [4L(v1 + v2)]/cλ


Sagnac did not derive a formula using TWO LOOPS, as required in the correct definition of the Sagnac effect: he simply used the AREA of the interferometer to obtain the CORIOLIS EFFECT formula, an equation derived nine years earlier by Michelson.

« Last Edit: August 27, 2019, 02:38:09 AM by sandokhan »

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sandokhan

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Re: On Sandokhan definitions of the Sagnac and Coriolis Effects
« Reply #3 on: August 27, 2019, 02:09:16 AM »
Let us now compare the two formulas: the CORIOLIS EFFECT formula derived by G. Sagnac in 1913 and the SAGNAC EFFECT formula obtained using two loops.


CORIOLIS EFFECT formula (G. Sagnac, 1913)

Time difference

8AΩ1/c2


SAGNAC EFFECT formula

4L(v1 + v2)/c2

Latitude of Sorbonne: 48.85°

Thus, using the latitude of the laboratory in France, the formula becomes:

4Lv(cos2Φ1 + cos2Φ2)/c2

v = R x Ω2, where R = 4,197 km


RATIO

{4Lv(cos2Φ1 + cos2Φ2)/c2}/8AΩ1/c2 = RΩ2((cos2Φ1 + cos2Φ2)/2LΩ1

Φ1 = Φ2 = 48.85°

2((cos2Φ1 + cos2Φ2) = [3634.66 km x 7.29 x 10-5] = 0.265 km

2LΩ1 = 2 x 0.000294 km x 25 = 0.0147 km

0.265 km/0.0147 km = 18


Let us also compare the CORIOLIS EFFECT formula derived by Michelson in 1925 with the true/correct SAGNAC EFFECT formula:

The turning of the MGX area at the hypothetical rotational speed of the Earth takes place a distance of some 4,250 km from the center of the Earth (latitude 41°46').

FULL CORIOLIS EFFECT FOR THE MGX:

4AΩsinΦ/c2

FULL SAGNAC EFFECT FOR THE MGX:

4Lv(cos2Φ1 + cos2Φ2)/c2


Sagnac effect/Coriolis effect ratio:

R((cos2Φ1 + cos2Φ2)/hsinΦ

R = 4,250 km

h = 0.33924 km

The rotational Sagnac effect is much greater than the Coriolis effect for the MGX.

Φ1 = Φ = 41°46' = 41.76667°

Φ2 = 41°45' = 41.75°

R((cos2Φ1 + cos2Φ2) = 4729.885

hsinΦ = 0.225967

4729.885/0.225967 = 20,931.72

THE ROTATIONAL SAGNAC EFFECT IS 21,000 TIMES GREATER THAN THE CORIOLIS EFFECT.


G. Sagnac recorded ONLY the CORIOLIS EFFECT (proportional to the AREA) and not the ROTATIONAL SAGNAC EFFECT (proportional to the velocity/radius of rotation).

« Last Edit: August 27, 2019, 02:47:41 AM by sandokhan »

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rabinoz

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Re: On Sandokhan definitions of the Sagnac and Coriolis Effects
« Reply #4 on: August 27, 2019, 05:46:31 AM »
The rotational and the orbital CORIOLIS EFFECT formulas were derived for the first time in 1904 by A. Michelson, and they feature AN AREA:
So YOU say, but I'd believe Sagnac and Michelson before you any day!

Quote from: sandokhan
http://www.conspiracyoflight.com/Michelson-Gale/Michelson_1904.pdf
Georges Sagnac derived THE SAME FORMULA, featuring an area, the CORIOLIS EFFECT formula.

The original papers published by G. Sagnac (The Luminiferous Ether is Detected as a Wind Effect Relative to the Ether Using a Uniformly Rotating Interferometer):
http://zelmanov.ptep-online.com/papers/zj-2008-07.pdf
http://zelmanov.ptep-online.com/papers/zj-2008-08.pdf
And I hope that YOU read the first paper and found this:


If Georges Sagnac first observed the effect I will call it the Sagnac Effect as does everybody else that I've read (except YOU)!

Quote from: sandokhan
In 1913, Georges Sagnac measured ONLY the Coriolis effect, and not the true Sagnac effect (proportional to the linear velocity and radius of rotation).
No Georges Sagnac did NOT measure the Coriolis effect!
The Coriolis effect depends on the velocity of an object not on any area!

Here is the expression for the Coriolis Acceleration
Where
aC isthe Coriolis acceleration.
Omega is the angular velocity and  v is the velocity of the object normal to the axis of rotation.

Quote from: sandokhan
Here is the shape of the interferometer used by Sagnac:

Please read:
https://link.springer.com/article/10.1023/A:1023972214666

https://arxiv.org/pdf/gr-qc/0103091.pdf

Coriolis Force and Sagnac Effect

Because of acting of gravity-like Coriolis force the trajectories of co- and anti-rotating photons have different radii in the rotating reference frame, while in the case of the equal radius the effective gravitational potentials for the photons have to be different.

You did read the abstract?
Quote
Abstract
We consider the optical Sagnac effect, when the fictitious gravitational field simulates the reflections from the mirrors. It is shown that no contradiction exists between the conclusions of the laboratory and rotated observers. Because of acting of gravity-like Coriolis force the trajectories of
co- and anti-rotating photons have different radii in the rotating reference frame, while in the case of the equal radius the effective gravitational potentials for the photons have to be different.

Nowhere does he say he is deriving the Coriolis effect. He is simply using it to derive the Sagnac effect.

Quote from: sandokhan
That is why Sagnac had to use the formula which features the area and the angular velocity: he only measured the CORIOLIS EFFECT.

So, no! Sagnac had to use the formula which features the area and the angular velocity and he did measure the Sagnac effect.

Again, I repeat! The Coriolis effect is completely different LOOK:  the Coriolis Acceleration

Quote from: sandokhan
. . . .

Here is the precise proof, peer-reviewed in an IOP article.

THIS IS AN IOP ARTICLE, one of the most comprehensive papers on the Sagnac effect ever published.

No, it is NOT "precise proof". I have the utmost respect for Grigorii B. Malykin and Vera I. Pozdnyakova but I doubt you interpreted what they wrote correctly! Read this:
Quote from: Grigorii B. Malykin and Vera I. Pozdnyakova
Siberstein [798, 799] suggested an explanation of the Sagnac effect based on the direct consideration of effect of the Sagnac forces on the counterpropagating waves. . . . . The areas of the triangles are different."
Now, while "the areas of the triangles are different" Dr Siberstein had previously shown "Thus, even for a ≈ 10 or 20 km the difference would certainly be too small to be measured directly.
So Dr Silberstein certainly does not "derive the Coriolis effect" and on the contrary, he shows that its effect is "certainly be too small to be measured directly."

Quote from: sandokhan

The formula derived by Dr. Silberstein, peer reviewed in the IOP article, and described by the author as the "effect of the Coriolis forces" is this:

dt = 4ωA/c^2
Where is it "described by the author as the 'effect of the Coriolis forces' "? Silberstein showed that the Coriolis forces had negligible effect!

Read again what Grigorii B. Malykin and Vera I. Pozdnyakova wrote!
Quote from: Grigorii B. Malykin and Vera I. Pozdnyakova
Siberstein [798, 799] suggested an explanation of the Sagnac effect based on the direct consideration of effect of the Sagnac forces on the counterpropagating waves. . . . . The areas of the triangles are different."
Now, while "the areas of the triangles are different" Dr Siberstein had previously shown "Thus, even for a ≈ 10 or 20 km the difference would certainly be too small to be measured directly.

Dr. Ludwik Silberstein did not ever "prove that the real cause of the phenomenon measured by Georges Sagnac was the CORIOLIS FORCE EFFECT."
He showed that the light paths deviated from straight lines due to the Coriolis effect but by a negligible amount and so that was completely left out of his final result!

So Dr Silberstein certainly does not "derive the Coriolis effect" and on the contrary, he shows that its effect is "certainly be too small to be measured directly."

Quote from: sandokhan
<< I do not have the time to wade through all of this especially as much is simply repeated and the rest has been posted and answered before! >>

My SAGNAC EFFECT formula proven and experimentally fully established at the highest possible level of science.
No, it certainly has not been "proven and experimentally fully established at the highest possible level of science." because it is wrong!
Quote from: sandokhan
Let us now compare the two derivations, using two loops (Sagnac effect):
No, let's not bother with your so-called "Coriolis effect" because it is nothing like any real Coriolis effect!

Quote from: sandokhan


Point A is located at the detector
Point B is in the bottom right corner
Point C is in the upper right corner
Point D is in the upper left corner

l1 is the upper arm.
l2 is the lower arm.

Here is the most important part of the derivation of the full/global Sagnac effect for an interferometer located away from the center of rotation.
A > B > C > D > A is a continuous counterclockwise path, a negative sign -
A > D > C > B > A is a continuous clockwise path, a positive sign +

The Sagnac phase difference for the clockwise path has a positive sign.

The Sagnac phase difference for the counterclockwise has a negative sign.

Sagnac phase components for the A > D > C > B > A path (clockwise path):
l1/(c - v1)
-l2/(c + v2)
Look your so-called "Sagnac phase components" are NOT "phase components" but are "time delay components".
Just look at the dimensions - they are all (length)/(velocity) which is the travel time along each segment.
And travel times cannot possibly be negative, so please correct your signs! You need to remove the negative signs between the terms

I do not have the time to go through all your working to correct it.

Quote from: sandokhan
Sagnac phase components for the A > B > C > D > A path (counterclockwise path):
l2/(c - v2)
-l1/(c + v1)

For the single continuous clockwise path we add the components:
l1/(c - v1) - l2/(c + v2)

For the single continuous counterclockwise path we add the components:
l2/(c - v2) - l1/(c + v1)

The net phase difference will be (let us remember that the counterclockwise phase difference has a negative sign attached to it, that is why the substraction of the phase differences becomes an addition):

{l1/(c - v1) - l2/(c + v2)} - (-){l2/(c - v2) - l1/(c + v1)} = {l1/(c - v1) - l2/(c + v2)} + {l2/(c - v2) - l1/(c + v1)}

Rearranging terms:

l1/(c - v1) - l1/(c + v1) + {l2/(c - v2) - l2/(c + v2)} =
2(v1l1 + v2l2)/c2

Exactly the formula obtained by Professor Yeh:
But it should NOT be "Exactly the formula obtained by Professor Yeh" because his formula was for a Phase Conjugate Gyro!

Quote from: sandokhan

φ = -2(φ2 - φ1) = 4π(R1L1 + R2L2)Ω/λc = 4π(V1L1 + V2L2)/λc

Since Δφ = 2πc/λ x Δt, Δt = 2(R1L1 + R2L2)Ω/c2 = 2(V1L1 + V2L2)/c2

CORRECT SAGNAC FORMULA:
No, this is NOT the "CORRECT SAGNAC FORMULA"!

Quote from: sandokhan
2(V1L1 + V2L2)/c2

Self-pumped phase-conjugate fiber-optic gyro, I. McMichael, P. Yeh, Optics Letters 11(10):686-8 · November 1986 

http://www.dtic.mil/dtic/tr/fulltext/u2/a170203.pdf (appendix 5.1)

This is how the correct Sagnac formula is derived: we have single continuous clockwise path, and a single continuous counterclockwise path.

If we desire the Coriolis effect, we simply substract as follows:

dt = l1/(c - v1) - l1/(c + v1) - (l2/(c - v2) - l2/(c + v2))

Of course, by proceeding as in the usual manner for a Sagnac phase shift formula for an interferometer whose center of rotation coincides with its geometrical center, we obtain:

2v1l1/(c2 - v21) - 2v2l2/(c2 - v22)

l = l1 = l2

2l[(v1 - v2)]/c2

2lΩ[(R1 - R2)]/c2

R1 - R2 = h

2lhΩ/c2
And that is the Sagnac effect NOT Coriolis! As I noted above, you got half the signs incorrect in your so-called Sagnac time-delay formula!

Quote from: sandokhan
By having substracted two different Sagnac phase shifts, valid for the two different segments, we obtain the CORIOLIS EFFECT formula.
No it is NOT the "CORIOLIS EFFECT formula"! This is the formula for the Coriolis acceleration:

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sandokhan

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Re: On Sandokhan definitions of the Sagnac and Coriolis Effects
« Reply #5 on: August 27, 2019, 06:43:10 AM »
The CORIOLIS EFFECT applied to light beams certainly involves an AREA.

Here is a complete demonstration.

https://www.ias.ac.in/article/fulltext/pram/087/05/0071

Spinning Earth and its Coriolis effect on the circuital light beams

A very direct, undergraduate level derivation.

The final formula is this:

dt = 4ωA/c^2

Therefore, contrary to the desperate attempts to suggest otherwise, the CORIOLIS EFFECT formula for light beams does involves the area of the interferometer.


https://link.springer.com/article/10.1023/A:1023972214666

https://arxiv.org/pdf/gr-qc/0103091.pdf

Coriolis Force and Sagnac Effect

Because of acting of gravity-like Coriolis force the trajectories of co- and anti-rotating photons have different radii in the rotating reference frame, while in the case of the equal radius the effective gravitational potentials for the photons have to be different.

The author describes the fact that if the interferometer has sides which are not equal, or which have an irregular shape, as in the case referenced to G. Sagnac's 1913 experiment, then the CORIOLIS FORCE upon the light beams will be recorded.

Very simple.

Here is the demonstration using differential geometry.




Therefore, contrary to the desperate attempts to suggest otherwise, the CORIOLIS FORCE effect upon the light beams will be recorded in Sagnac's irregularly shaped interferometer.


And the desperate tactics continue unabated on the part of the RE.

Here read this from p 298:

On page 298 there are two DIFFERENT derivations, a fact quite obvious for anyone but yourself.

You are quoting from the previous derivation, the equation of the light path in relation to Fermat's principle.

he shows that the deviations from straight are so slight that it cannot affect the result.

Yes, for the light path in terms of Fermat's principle.

Then, he starts to derive the CORIOLIS EFFECT on the same page, a totally different derivation.

The quote refers to FERMAT'S PRINCIPLE, not to the next derivation which takes place on the same page:



You must be pretty desperate to use these kinds of tactics in a debate.


http://www.conspiracyoflight.com/Michelson-Gale/Silberstein.pdf


The derivation for the light path in terms of FERMAT'S PRINCIPLE starts on page 293 and ends on page 298.

Now, while "the areas of the triangles are different" Dr Siberstein had previously shown "Thus, even for a ≈ 10 or 20 km the difference would certainly be too small to be measured directly."

For FERMAT'S PRINCIPLE, yes, NOT for the next derivation.

So, your entire argument amounts to nothing at all, with the exception of your devious and miserable tactics you are using to satisfy your cognitive dissonance.

The derivation for the CORIOLIS EFFECT starts quite obviously with these words, right on the same page 298:

The experimental possibilities with regard to the optical effect of the rotation of the Earth lie in another direction...

Yet, you quoted from the PREVIOUS derivation, based on FERMAT'S PRINCIPLE, which has nothing to do with the NEXT derivation, right on the same page, which is the CORIOLIS EFFECT.

This means that you are UNABLE to read a scientific paper.


The analysis in terms of FERMAT'S PRINCIPLE starts on page 293 and ends on page 298: the two formulas are of different orders!


Here is the final formula:

4akw/c x (...)

THIS FORMULA IS OF THE ORDER OF 1/C: O(1/c).

By constrast, the CORIOLIS EFFECT formula, whose derivation also begins on page 298, is of the order O(1/c2).


Two different formulas, yet the RE seem to be unable to differentiate between them.
Silberstein (798, 799) suggested an explanation for the Sagnac effect based on the direct consideration of the effect of the Coriolis force on the counterpropagating waves.

Those two references, 798 and 799 are EXACTLY the ones I provided in my messages.
Make no mistake about it: Dr. Silberstein derives the Coriolis effect, which is directly related to the area of the interferometer.


Therefore, the fact that Dr. Silberstein did derive the CORIOLIS EFFECT formula, which features an area, remains undisputed.

THIS IS AN IOP ARTICLE, one of the most comprehensive papers on the Sagnac effect ever published.





Here is reference #27:




http://www.conspiracyoflight.com/Michelson-Gale/Silberstein.pdf

The formula derived by Dr. Silberstein, peer reviewed in the IOP article, and described by the author as the "effect of the Coriolis forces" is this:

dt = 4ωA/c^2

« Last Edit: August 27, 2019, 07:23:55 AM by sandokhan »

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sandokhan

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Re: On Sandokhan definitions of the Sagnac and Coriolis Effects
« Reply #6 on: August 27, 2019, 06:44:21 AM »
There are no "negative times", a concept most laughable.

I am going to explain the entire phenomenon using even more details, so that everyone here will understand, once and for all, the correct description of the SAGNAC EFFECT.

The RE standard for the Sagnac effect:

https://www.mathpages.com/rr/s2-07/2-07.htm

If two pulses of light are sent in opposite directions around a stationary circular loop of radius R, they will travel the same inertial distance at the same speed, so they will arrive at the end point simultaneously. This is illustrated in the left-hand figure below.



If the interferometer is being rotated, both pulses begin with an initial separation of 2piR from the end point, so the difference between the travel times is:



Can everyone understand the mechanism?

Opposite directions, therefore WE SUBSTRACT THE DIFFERENCE IN TIME TRAVEL.

Moreover, we are dealing with TWO LOOPS.

Can everyone understand that the differences in time travel have to be substracted?

This is the correct way to derive the Sagnac formula:

Sagnac phase component for the clockwise path:

2πR(1/(c - v))

Sagnac phase component for the counterclockwise path:

-2πR(1/(c + v))

The continuous clockwise loop has a positive sign +

The continuous counterclockwise loop has a negative sign -

Good.

That is, if we want to find out the difference in travel times (opposite directions) we must substract them.


For an interferometer which is now located AWAY FROM THE CENTER OF ROTATION, the situation is a bit more complicated, but the same principle applies.



Point A is located at the detector
Point B is in the bottom right corner
Point C is in the upper right corner
Point D is in the upper left corner

l1 is the upper arm.
l2 is the lower arm.

Let us remember that now we are dealing with DIFFERENT VELOCITIES for each arm, and DIFFERENT LENGTHS of each arm, a situation a bit more complex than the previous case analyzed here.


We need to designate the TWO LOOPS, as required by the definition of the Sagnac effect.

HERE IS THE DEFINITION OF THE SAGNAC EFFECT:

Two pulses of light sent in opposite direction around a closed loop (either circular or a single uniform path), while the interferometer is being rotated.

Loop = a structure, series, or process, the end of which is connected to the beginning.

A single continuous pulse A > B > C > D > A, while the other one, A > D > C > B > A is in the opposite direction, and has the negative sign.


So, for the first loop, the clockwise path, the A > D > C > B > A path, we have to deal with beams which are traveling IN OPPOSITE DIRECTIONS, that is, in order to find out the total time travel we need to substract the time differences, just like we did the first time: in effect we are adding two transit times, one of which is traveling in a opposite direction to the first, hence the opposite signs.

We substracted the time differences the first time around for the interferometer whose center of rotation coincides with its geometric center.

Now, we have a loop consisting of two different paths, which travel in opposite directions.

Therefore, to get the TOTAL TIME DIFFERENCE FOR THE CLOCKWISE PATH, we substract the time differences: again, in effect we are adding the transit times, but since one of them has an opposite direction, it will have a different sign than the first transit time, just like in the first example of the Sagnac interferometer.

Very simple, and at the same time we are dealing with a LOOP, as required by the defintion of the Sagnac effect.

Sagnac phase components for the A > D > C > B > A path (clockwise path):

l1/(c - v1)

-l2/(c + v2)

Now, we do the same thing for the counterclockwise path, the A > B > C > D > A path:

l2/(c - v2)

-l1/(c + v1)


For the single continuous clockwise path we now have the total time difference:

l1/(c - v1) - l2/(c + v2)


For the single continuous counterclockwise path we have the total difference:

l2/(c - v2) - l1/(c + v1)


TWO LOOPS as required by the definition of the Sagnac effect.

If we change the sign of the second term/phase component to +, that is:

l1/(c - v1)

l2/(c + v2)

then, we no longer have a LOOP, and moreover we are using the wrong sign for the direction of the second transit time; each transit time has a different direction, hence we must use opposite signs to correctly designate them in our analysis.

Let us remember the very defintion of the Sagnac effect: two loops are required to properly derive the formula.


Now, to obtain the final answer, WE SUBSTRACT THE TOTAL TIME DIFFERENCES FOR EACH PATH, since we are dealing with a counterclockwise path and a clockwise path, if we want the time phase, we need to substract the total time differences for each LOOP. Each loop has a different direction, as such it must have a different sign assigned to it.

The net phase difference will be (let us remember that the counterclockwise phase difference has a negative sign attached to it, that is why the substraction of the phase differences becomes an addition):

{l1/(c - v1) - l2/(c + v2)} - (-){l2/(c - v2) - l1/(c + v1)} = {l1/(c - v1) - l2/(c + v2)} + {l2/(c - v2) - l1/(c + v1)}

Rearranging terms:

l1/(c - v1) - l1/(c + v1) + {l2/(c - v2) - l2/(c + v2)} =

2(v1l1 + v2l2)/c2

Exactly the formula obtained by Professor Yeh:

φ = -2(φ2 - φ1) = 4π(R1L1 + R2L2)Ω/λc = 4π(V1L1 + V2L2)/λc

Since Δφ = 2πc/λ x Δt, Δt = 2(R1L1 + R2L2)Ω/c2 = 2(V1L1 + V2L2)/c2

CORRECT SAGNAC FORMULA:

2(V1L1 + V2L2)/c2


By contrast, what Michelson did is to remove the SIGN from each loop, in effect nullifying the very definition of the Sagnac effect: he compared two different sides, not the two loops, thus he obtained the CORIOLIS EFFECT formula.

The CORIOLIS EFFECT and the SAGNAC EFFECT are two very different phenomena, one is a physical effect while the other one is an electromagnetic effect: two different phenomena require two different formulas.




page 152 of the pdf document, section Recent Advances in Photorefractive Nonlinear Optics page 4

The MPPC acts like a normal mirror and Sagnac interferometry is obtained.

Here is the derivation of my formula, using TWO LOOPS:

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg2117351#msg2117351

Here is the final formula:

2(V1L1 + V2L2)/c2

My formula is confirmed at the highest possible scientific level, having been published in the best OPTICS journal in the world, Journal of Optics Letters, and it is used by the US NAVAL RESEARCH OFFICE, Physics Division.

A second reference which confirms my global/generalized Sagnac effect formula.

https://apps.dtic.mil/dtic/tr/fulltext/u2/a206219.pdf

Studies of phase-conjugate optical devices concepts

US OF NAVAL RESEARCH, Physics Division

Dr. P. Yeh
PhD, Caltech, Nonlinear Optics
Principal Scientist of the Optics Department at Rockwell International Science Center
Professor, UCSB
"Engineer of the Year," at Rockwell Science Center
Leonardo da Vinci Award in 1985
Fellow of the Optical Society of America, the Institute of Electrical and Electronics Engineers

page 152 of the pdf document, section Recent Advances in Photorefractive Nonlinear Optics page 4

The MPPC acts like a normal mirror and Sagnac interferometry is obtained.



Phase-Conjugate Multimode Fiber Gyro

Published in the Journal of Optics Letters, vol. 12, page 1023, 1987

page 69 of the pdf document, page 1 of the article


A second confirmation of the fact that my formula is correct.

Here is the first confirmation:



Self-pumped phase-conjugate fiber-optic gyro, I. McMichael, P. Yeh, Optics Letters 11(10):686-8 · November 1986 

http://www.dtic.mil/dtic/tr/fulltext/u2/a170203.pdf (appendix 5.1)


Exactly the formula obtained by Professor Yeh:

φ = -2(φ2 - φ1) = 4π(R1L1 + R2L2)Ω/λc = 4π(V1L1 + V2L2)/λc

Since Δφ = 2πc/λ x Δt, Δt = 2(R1L1 + R2L2)Ω/c2 = 2(V1L1 + V2L2)/c2

CORRECT SAGNAC FORMULA:

2(V1L1 + V2L2)/c2

The very same formula obtained for a Sagnac interferometer which features two different lengths and two different velocities.

http://www.dtic.mil/dtic/tr/fulltext/u2/a170203.pdf

ANNUAL TECHNICAL REPORT PREPARED FOR THE US OF NAVAL RESEARCH.

Page 18 of the pdf document, Section 3.0 Progress:

Our first objective was to demonstrate that the phase-conjugate fiberoptic gyro (PCFOG) described in Section 2.3 is sensitive to rotation. This phase shift plays an important role in the detection of the Sagnac phase shift due to rotation.

Page 38 of the pdf document, page 6 of Appendix 3.1


it does demonstrate the measurement of the Sagnac phase shift Eq. (3)


HERE IS EQUATION (3) OF THE PAPER, PAGE 3 OF APPENDIX 3.1:

φ = -2(φ2 - φ1) = 4π(R1L1 + R2L2)Ω/λc = 4π(V1L1 + V2L2)/λc

Since Δφ = 2πc/λ x Δt, Δt = 2(R1L1 + R2L2)Ω/c2 = 2(V1L1 + V2L2)/c2

CORRECT SAGNAC FORMULA:

2(V1L1 + V2L2)/c2




The Coriolis effect is a physical effect upon the light beams: it is proportional to the area of the interferometer. It is a comparison of two sides.

The Sagnac effect is an electromagnetic effect upon the velocities of the light beams: it is proportional to the radius of rotation. It is a comparison of two loops.

Two different phenomena require two very different formulas.


My SAGNAC EFFECT formula proven and experimentally fully established at the highest possible level of science.



The desperate tactics used by the RE cannot succeed at all.




page 152 of the pdf document, section Recent Advances in Photorefractive Nonlinear Optics page 4

The MPPC acts like a normal mirror and Sagnac interferometry is obtained.


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rabinoz

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Re: On Sandokhan definitions of the Sagnac and Coriolis Effects
« Reply #7 on: August 27, 2019, 02:47:54 PM »
There are no "negative times", a concept most laughable.
That is exactly what I said, had you bothered to read it. But your signs do give "negative times" for some of the components!

Sagnac phase components for the A > D > C > B > A path (clockwise path):
l1/(c - v1)
-l2/(c + v2)
Look your so-called "Sagnac phase components" are NOT "phase components" but are "time delay components".
Just look at the dimensions - they are all (length)/(velocity) which is the travel time along each segment.
And travel times cannot possibly be negative, so please correct your signs! You need to remove the negative signs between the terms

I do not have the time to go through all your working to correct it.

Look at this term that you call a "Sagnac phase component", "-l2/(c + v2)".
          "l2" is a length and presumably positive,
          "c" and "v2" are velocities and c must be positive and much greater than "v2".

So your "Sagnac time delay component", -l2/(c + v2), must be negative.

And you, yourself said, "There are no "negative times", a concept most laughable."  QED!

Now fixed up the signs in your "Sagnac time delay" and then I might look at the rest of your copy-pasta!

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rabinoz

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Re: On Sandokhan definitions of the Sagnac and Coriolis Effects
« Reply #8 on: August 27, 2019, 03:08:44 PM »
The Coriolis effect is a physical effect upon the light beams: it is proportional to the area of the interferometer. It is a comparison of two sides.
This is the formula derived by G. Sagnac: 4AΩ/c^2.

THIS IS THE CORIOLIS EFFECT FORMULA.

Sure, but your expression, "4AΩ/c^2 . . . . derived by G. Sagnac" and almost everybody else except YOU is NOT the Coriolis effect!

This shows the correct Coriolis effect, with a bit of explanation!
Quote
UNDERSTANDING THE CORIOLIS FORCE
And:
It is proportional to the velocity of the object and has no connection with any area!

Now FIX your own Sagnac Time Delay!

You might even look at the
Quote
The Michelson-Gale Experiment by Doug Marett (2010)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
    In refining his argument, he proposed that it was not necessary for the light to go all the way around the globe - since there should be a velocity difference for any closed path rotating on the surface of the earth. He presented the following equation to calculate the time difference expected, using the shift in the interference fringes when the two beams overlap at the detector as a measure of the time difference:

where:  Vo = the tangential velocity of the earth's rotation at the equator (465m/s)
              A = the area of the circular path
              R = the radius of the earth (6371000 m)
              c = speed of light (3E8 m/s)
              f = the latitude in degrees where the experiment is conducted.
              l = wavelength of the light
It's funny how everybody seems to agree that the "4AΩ/c^2 . . . . derived by G. Sagnac" is the Sagnac effect.

*

sandokhan

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Re: On Sandokhan definitions of the Sagnac and Coriolis Effects
« Reply #9 on: August 27, 2019, 10:12:20 PM »
The CORIOLIS EFFECT applied to light beams certainly involves an AREA.

Here is a complete demonstration.

https://www.ias.ac.in/article/fulltext/pram/087/05/0071

Spinning Earth and its Coriolis effect on the circuital light beams

A very direct, undergraduate level derivation.

The final formula is this:

dt = 4ωA/c^2


The topological aspects of the SAGNAC EFFECT have been researched only recently, this being the main reason why the CORIOLIS FORMULA, which features the AREA, has been substituted for the true SAGNAC EFFECT formula.

"Sagnac effect is a change in propagation time for light going in a closed path. The time delay Δt appears when a test equipment is rotated with an angular velocity Ώ. Sagnac effect is frequently used in rate gyros in navigational systems. Fiber optics is used with light-speed c inside the fiber in a circular light path. The difference in propagation time Δt for two opposite directions of light is described as

Δt = 4AΩ/c2

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

According to Stokes' rule can an integration of angular velocity Ω over an area A 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 = 4vL/c2

producing the same value as the earlier expression. This can also be demonstrated by geometrical relations. These two integrations have different physical implications. We must therefore decide which one is correct from a physical aspect. Mathematics can not tell us that. 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.

We conclude that the later expression

Δt = 4vL/c2

is the correct interpretation."

http://www.gsjournal.net/Science-Journals/Research%20Papers-Astrophysics/Download/2159

"Sagnac effect is distributed along a line and not over a surface. The assumption that starts from an integration over a surface (2Aw; rotation) is mathematically correct (due to Stokes' rule) but equal to a line integral (vL; translation). We must decide if the reason is a translating line or a rotating surface from a physical point of view. The rotation theory is correct only mathematically. Since the effect is locked inside an optical fiber the translating line is the correct interpretation. Classification as a rotational effect is wrong."

Professor Ruyong Wang has proven the Sagnac effect applies to uniform/translational/linear motion:

https://arxiv.org/ftp/physics/papers/0609/0609222.pdf

https://arxiv.org/ftp/physics/papers/0609/0609202.pdf

http://web.stcloudstate.edu/ruwang/ION58PROCEEDINGS.pdf


Using very advanced concepts from topology, T.W. Barrett proves that the Sagnac effect can only be described by the original set of the equations published by J.C. Maxwell.

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1919728#msg1919728

https://www.researchgate.net/publication/288491190_SAGNAC_EFFECT_A_consequence_of_conservation_of_action_due_to_gauge_field_global_conformal_invariance_in_a_multiply-joined_topology_of_coherent_fields

Moreover, in the Sagnac effect there are two vector potential components with respect to clockwise and counterclockwise beams. The measured quantity, as will be explained more fully below, is then the phase factor or the integral of the potential difference between those beams and related to the angular velocity difference between the two beams. Therefore, as the vector potential measures the momentum gain and the scalar potential measures the kinetic energy gain, the photon will acquire “mass.”














https://books.google.ro/books?id=qsOBhKVM1qYC&pg=PA6&lpg=PA6&dq=electromagnetic+phenomena+not+explained+by+maxwell%27s+equations&source=bl&ots=Hurq5SQ-EG&sig=iMhWIxjuFrg9Co763une7Dnpmf0&hl=ro&sa=X&ved=0ahUKEwi1m-6DmfDZAhUiSJoKHR7RCikQ6AEIOzAC#v=onepage&q=electromagnetic%20phenomena%20not%20explained%20by%20maxwell's%20equations&f=false

T.W. Barrett, "Electromagnetic Phenomena Not Explained by Maxwell's Equations" pg 6 - 85

From a topological point of view, the Heaviside-Lorentz equations are a LINEAR THEORY, U(1).

When extended to SU(2) or higher symmetry forms, Maxwell's theory possesses non-Abelian commutation relations, and addresses global, i.e., nonlocal in space, as well as local phenomena with the potentials used as local-to-global operators.

Dr. Terence W. Barrett (Stanford Univ., Princeton Univ., U.S. Naval Research Laboratory, Univ. of Edinburgh, author of over 200 papers on advanced electromagnetism)


The different velocities of clockwise- and counter-clockwise-rotating light beams in the Sagnac interferometer are due to the motion of the ether.


The observed interference effect is clearly the optical whirling effect due to the movement of the system in relation to the ether and directly manifests the existence of the ether.

G. Sagnac

This ether is a dynamic, and not an inert, ether.


However, the use of the phase-conjugate mirror has permitted physicists to obtain the true SAGNAC EFFECT formula, which is proportional to the VELOCITY of the light beams.



page 152 of the pdf document, section Recent Advances in Photorefractive Nonlinear Optics page 4

The MPPC acts like a normal mirror and Sagnac interferometry is obtained.

Here is the derivation of my formula, using TWO LOOPS:

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg2117351#msg2117351

Here is the final formula:

2(V1L1 + V2L2)/c2

My formula is confirmed at the highest possible scientific level, having been published in the best OPTICS journal in the world, Journal of Optics Letters, and it is used by the US NAVAL RESEARCH OFFICE, Physics Division.

A second reference which confirms my global/generalized Sagnac effect formula.

https://apps.dtic.mil/dtic/tr/fulltext/u2/a206219.pdf

Studies of phase-conjugate optical devices concepts

US OF NAVAL RESEARCH, Physics Division

Dr. P. Yeh
PhD, Caltech, Nonlinear Optics
Principal Scientist of the Optics Department at Rockwell International Science Center
Professor, UCSB
"Engineer of the Year," at Rockwell Science Center
Leonardo da Vinci Award in 1985
Fellow of the Optical Society of America, the Institute of Electrical and Electronics Engineers

Here is the first confirmation:



Self-pumped phase-conjugate fiber-optic gyro, I. McMichael, P. Yeh, Optics Letters 11(10):686-8 · November 1986 

http://www.dtic.mil/dtic/tr/fulltext/u2/a170203.pdf (appendix 5.1)


Exactly the formula obtained by Professor Yeh:

φ = -2(φ2 - φ1) = 4π(R1L1 + R2L2)Ω/λc = 4π(V1L1 + V2L2)/λc

Since Δφ = 2πc/λ x Δt, Δt = 2(R1L1 + R2L2)Ω/c2 = 2(V1L1 + V2L2)/c2

CORRECT SAGNAC FORMULA:

2(V1L1 + V2L2)/c2

The very same formula obtained for a Sagnac interferometer which features two different lengths and two different velocities.


The Coriolis effect is a physical effect upon the light beams: it is proportional to the area of the interferometer. It is a comparison of two sides.

The Sagnac effect is an electromagnetic effect upon the velocities of the light beams: it is proportional to the radius of rotation. It is a comparison of two loops.

Two different phenomena require two very different formulas.

*

sandokhan

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Re: On Sandokhan definitions of the Sagnac and Coriolis Effects
« Reply #10 on: August 27, 2019, 10:18:50 PM »
Each time transit has a DEFINITE DIRECTION.

Opposite directions have opposite signs.

Let us go back to the original derivation of the Sagnac effect.



To get the time transits IN OPPOSITE DIRECTIONS, you must assign a NEGATIVE SIGN to one of them.


If the interferometer is being rotated, both pulses begin with an initial separation of 2piR from the end point, so the difference between the travel times is:



Opposite directions, therefore WE SUBSTRACT THE DIFFERENCE IN TIME TRAVEL.

Moreover, we are dealing with TWO LOOPS.

This is the correct way to derive the Sagnac formula:

Sagnac phase component for the clockwise path:

2πR(1/(c - v))

Sagnac phase component for the counterclockwise path:

-2πR(1/(c + v))

The continuous clockwise loop has a positive sign +

The continuous counterclockwise loop has a negative sign -

That is, if we want to find out the difference in travel times (opposite directions) we must substract them.


For an interferometer which is now located AWAY FROM THE CENTER OF ROTATION, the situation is a bit more complicated, but the same principle applies.



Point A is located at the detector
Point B is in the bottom right corner
Point C is in the upper right corner
Point D is in the upper left corner

l1 is the upper arm.
l2 is the lower arm.

Let us remember that now we are dealing with DIFFERENT VELOCITIES for each arm, and DIFFERENT LENGTHS of each arm, a situation a bit more complex than the previous case analyzed here.


We need to designate the TWO LOOPS, as required by the definition of the Sagnac effect.

HERE IS THE DEFINITION OF THE SAGNAC EFFECT:

Two pulses of light sent in opposite direction around a closed loop (either circular or a single uniform path), while the interferometer is being rotated.

Loop = a structure, series, or process, the end of which is connected to the beginning.

A single continuous pulse A > B > C > D > A, while the other one, A > D > C > B > A is in the opposite direction, and has the negative sign.


So, for the first loop, the clockwise path, the A > D > C > B > A path, we have to deal with beams which are traveling IN OPPOSITE DIRECTIONS, that is, in order to find out the total time travel we need to substract the time differences, just like we did the first time: in effect we are adding two transit times, one of which is traveling in a opposite direction to the first, hence the opposite signs.

We substracted the time differences the first time around for the interferometer whose center of rotation coincides with its geometric center.

Now, we have a loop consisting of two different paths, which travel in opposite directions.

Therefore, to get the TOTAL TIME DIFFERENCE FOR THE CLOCKWISE PATH, we substract the time differences: again, in effect we are adding the transit times, but since one of them has an opposite direction, it will have a different sign than the first transit time, just like in the first example of the Sagnac interferometer.

Very simple, and at the same time we are dealing with a LOOP, as required by the defintion of the Sagnac effect.

Sagnac phase components for the A > D > C > B > A path (clockwise path):

l1/(c - v1)

-l2/(c + v2)

Now, we do the same thing for the counterclockwise path, the A > B > C > D > A path:

l2/(c - v2)

-l1/(c + v1)


For the single continuous clockwise path we now have the total time difference:

l1/(c - v1) - l2/(c + v2)


For the single continuous counterclockwise path we have the total difference:

l2/(c - v2) - l1/(c + v1)


TWO LOOPS as required by the definition of the Sagnac effect.

If we change the sign of the second term/phase component to +, that is:

l1/(c - v1)

l2/(c + v2)

then, we no longer have a LOOP, and moreover we are using the wrong sign for the direction of the second transit time; each transit time has a different direction, hence we must use opposite signs to correctly designate them in our analysis.

Let us remember the very defintion of the Sagnac effect: two loops are required to properly derive the formula.


Now, to obtain the final answer, WE SUBSTRACT THE TOTAL TIME DIFFERENCES FOR EACH PATH, since we are dealing with a counterclockwise path and a clockwise path, if we want the time phase, we need to substract the total time differences for each LOOP. Each loop has a different direction, as such it must have a different sign assigned to it.

The net phase difference will be (let us remember that the counterclockwise phase difference has a negative sign attached to it, that is why the substraction of the phase differences becomes an addition):

{l1/(c - v1) - l2/(c + v2)} - (-){l2/(c - v2) - l1/(c + v1)} = {l1/(c - v1) - l2/(c + v2)} + {l2/(c - v2) - l1/(c + v1)}

Rearranging terms:

l1/(c - v1) - l1/(c + v1) + {l2/(c - v2) - l2/(c + v2)} =

2(v1l1 + v2l2)/c2

Exactly the formula obtained by Professor Yeh:

φ = -2(φ2 - φ1) = 4π(R1L1 + R2L2)Ω/λc = 4π(V1L1 + V2L2)/λc

Since Δφ = 2πc/λ x Δt, Δt = 2(R1L1 + R2L2)Ω/c2 = 2(V1L1 + V2L2)/c2

CORRECT SAGNAC FORMULA:

2(V1L1 + V2L2)/c2


By contrast, what Michelson did is to remove the SIGN from each loop, in effect nullifying the very definition of the Sagnac effect: he compared two different sides, not the two loops, thus he obtained the CORIOLIS EFFECT formula.

The CORIOLIS EFFECT and the SAGNAC EFFECT are two very different phenomena, one is a physical effect while the other one is an electromagnetic effect: two different phenomena require two different formulas.

*

rabinoz

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Re: On Sandokhan definitions of the Sagnac and Coriolis Effects
« Reply #11 on: August 28, 2019, 01:17:05 AM »
Each time transit has a DEFINITE DIRECTION.

Opposite directions have opposite signs.
But you end up with negative times! Which you then subtract again and
       so if V1 has the magnitude as V2 and if L1 has the magnitude as L2
You total transit time comes out as zero, which cannot be correct.

Quote from: sandokhan
Let us go back to the original derivation of the Sagnac effect.
Let's not!

Quote from: sandokhan
The continuous clockwise loop has a positive sign +

The continuous counterclockwise loop has a negative sign -
Agreed.
The total time around the "clockwise loop has a positive sign +" and the total time around the "counterclockwise loop has a negative sign -".

Quote from: sandokhan
That is, if we want to find out the difference in travel times (opposite directions) we must substract them.

For an interferometer which is now located AWAY FROM THE CENTER OF ROTATION, the situation is a bit more complicated, but the same principle applies.



Point A is located at the detector
Point B is in the bottom right corner
Point C is in the upper right corner
Point D is in the upper left corner

l1 is the upper arm.
l2 is the lower arm.

Let us remember that now we are dealing with DIFFERENT VELOCITIES for each arm, and DIFFERENT LENGTHS of each arm, a situation a bit more complex than the previous case analyzed here.

We need to designate the TWO LOOPS, as required by the definition of the Sagnac effect.
. . . . . . .
A single continuous pulse A > B > C > D > A, while the other one, A > D > C > B > A is in the opposite direction, and has the negative sign.

So, for the first loop, the clockwise path, the A > D > C > B > A path, we have to deal with beams which are traveling IN OPPOSITE DIRECTIONS,
No. There is a single "beam" is travelling around the "A > D > C > B > A path" and the transit times in every segment must be added.

Quote from: sandokhan
that is, in order to find out the total time travel we need to substract the time differences, just like we did the first time: in effect we are adding two transit times, one of which is traveling in a opposite direction to the first, hence the opposite signs.

We substracted the time differences the first time around for the interferometer whose center of rotation coincides with its geometric center.

Now, we have a loop consisting of two different paths, which travel in opposite directions.

No.  The loop does not consist of two different paths. It is a single loop through the "A > D > C > B > A path" in that order.

Quote from: sandokhan
Therefore, to get the TOTAL TIME DIFFERENCE FOR THE CLOCKWISE PATH, we substract the time differences: again, in effect we are adding the transit times, but since one of them has an opposite direction, it will have a different sign than the first transit time, just like in the first example of the Sagnac interferometer.

This is where we really disagree! There is no "TOTAL TIME DIFFERENCE FOR THE CLOCKWISE PATH". 
The times must all be added to get the total time around a path. No transit time can be negative - can't you see that is obvious?

Quote from: sandokhan
Very simple, and at the same time we are dealing with a LOOP, as required by the defintion of the Sagnac effect.

Sagnac phase components for the A > D > C > B > A path (clockwise path):
l1/(c - v1)
-l2/(c + v2)

That is the crux of the problem.
Can't you see that if l1 = l2 AND v1 = v2 you would end up with zero transit time around that loop - impossible?

It could easily happen that l2/(c + v2) > l1/(c - v1) and that would end up with negative transit time "for the A > D > C > B > A path".

And we both agree that a negative transit time around any loop is quite impossible.

So there is no point in carrying on till this sticking point is resolved!

And also there is no point in repeating the same old claims that you keep posting!

*

sandokhan

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Re: On Sandokhan definitions of the Sagnac and Coriolis Effects
« Reply #12 on: August 28, 2019, 02:01:47 AM »
Only someone who has nothing left to say, and is pretty desperate, can make a statement such as this:

Can't you see that if l1 = l2 AND v1 = v2 you would end up with zero transit time around that loop - impossible?

Which you then subtract again and  so if V1 has the magnitude as V2 and if L1 has the magnitude as L2
You total transit time comes out as zero, which cannot be correct.


For a Sagnac interferometer which is located AWAY FROM THE CENTER OF ROTATION, v1 and v2 can NEVER be the same, since l1 and l2 are located on different latitudes.

This much was assumed from the very start by Michelson, here is his derivation:




Therefore, to make such a mindless statement, to actually assume that v1 = v2, where, by definition, they can NEVER be the same, is beyond belief.


There is A DEFINITE direction of spin of rotation of the Earth, according to heliocentrism. That is why this fact is readily incorporated into the formula itself (by Michelson, by the ring laser gyroscopes physicists, by myself).


There is a single "beam" is travelling around the "A > D > C > B > A path" and the transit times in every segment must be added.

Exactly.

Both the counterclockwise and clockwise loops consist of TWO DIFFERENT PATHS/SIDES, l1 and l2.


Let us go back to how the original SAGNAC formula was derived.

The RE standard for the Sagnac effect:

https://www.mathpages.com/rr/s2-07/2-07.htm

If two pulses of light are sent in opposite directions around a stationary circular loop of radius R, they will travel the same inertial distance at the same speed, so they will arrive at the end point simultaneously. This is illustrated in the left-hand figure below.



If the interferometer is being rotated, both pulses begin with an initial separation of 2piR from the end point, so the difference between the travel times is:



There are NO "negative times": we have OPPOSITE DIRECTIONS, so we simply SUBSTRACT the time travels.


Very simple to understand.


This is the correct way to derive the Sagnac formula:

Sagnac phase component for the clockwise path:

2πR(1/(c - v))

Sagnac phase component for the counterclockwise path:

-2πR(1/(c + v))

The continuous clockwise loop has a positive sign +

The continuous counterclockwise loop has a negative sign -

That is, if we want to find out the difference in travel times (opposite directions) we must substract them.


For an interferometer which is now located AWAY FROM THE CENTER OF ROTATION, the situation is a bit more complicated, but the same principle applies.



Point A is located at the detector
Point B is in the bottom right corner
Point C is in the upper right corner
Point D is in the upper left corner

l1 is the upper arm.
l2 is the lower arm.

Let us remember that now we are dealing with DIFFERENT VELOCITIES for each arm, and DIFFERENT LENGTHS of each arm, a situation a bit more complex than the previous case analyzed here.


We need to designate the TWO LOOPS, as required by the definition of the Sagnac effect.

HERE IS THE DEFINITION OF THE SAGNAC EFFECT:

Two pulses of light sent in opposite direction around a closed loop (either circular or a single uniform path), while the interferometer is being rotated.

Loop = a structure, series, or process, the end of which is connected to the beginning.

A single continuous pulse A > B > C > D > A, while the other one, A > D > C > B > A is in the opposite direction, and has the negative sign.


So, for the first loop, the clockwise path, the A > D > C > B > A path, we have to deal with beams which are traveling IN OPPOSITE DIRECTIONS, that is, in order to find out the total time travel we need to substract the time differences, just like we did the first time: in effect we are adding two transit times, one of which is traveling in a opposite direction to the first, hence the opposite signs.

We substracted the time differences the first time around for the interferometer whose center of rotation coincides with its geometric center.

Now, we have a loop consisting of two different paths, which travel in opposite directions.

Therefore, to get the TOTAL TIME DIFFERENCE FOR THE CLOCKWISE PATH, we substract the time differences: again, in effect we are adding the transit times, but since one of them has an opposite direction, it will have a different sign than the first transit time, just like in the first example of the Sagnac interferometer.

Very simple, and at the same time we are dealing with a LOOP, as required by the defintion of the Sagnac effect.

Sagnac phase components for the A > D > C > B > A path (clockwise path):

l1/(c - v1)

-l2/(c + v2)

Now, we do the same thing for the counterclockwise path, the A > B > C > D > A path:

l2/(c - v2)

-l1/(c + v1)


For the single continuous clockwise path we now have the total time difference:

l1/(c - v1) - l2/(c + v2)


For the single continuous counterclockwise path we have the total difference:

l2/(c - v2) - l1/(c + v1)


TWO LOOPS as required by the definition of the Sagnac effect.

If we change the sign of the second term/phase component to +, that is:

l1/(c - v1)

l2/(c + v2)

then, we no longer have a LOOP, and moreover we are using the wrong sign for the direction of the second transit time; each transit time has a different direction, hence we must use opposite signs to correctly designate them in our analysis.

Let us remember the very defintion of the Sagnac effect: two loops are required to properly derive the formula.


Now, to obtain the final answer, WE SUBSTRACT THE TOTAL TIME DIFFERENCES FOR EACH PATH, since we are dealing with a counterclockwise path and a clockwise path, if we want the time phase, we need to substract the total time differences for each LOOP. Each loop has a different direction, as such it must have a different sign assigned to it.

The net phase difference will be (let us remember that the counterclockwise phase difference has a negative sign attached to it, that is why the substraction of the phase differences becomes an addition):

{l1/(c - v1) - l2/(c + v2)} - (-){l2/(c - v2) - l1/(c + v1)} = {l1/(c - v1) - l2/(c + v2)} + {l2/(c - v2) - l1/(c + v1)}

Rearranging terms:

l1/(c - v1) - l1/(c + v1) + {l2/(c - v2) - l2/(c + v2)} =

2(v1l1 + v2l2)/c2

Exactly the formula obtained by Professor Yeh:

φ = -2(φ2 - φ1) = 4π(R1L1 + R2L2)Ω/λc = 4π(V1L1 + V2L2)/λc

Since Δφ = 2πc/λ x Δt, Δt = 2(R1L1 + R2L2)Ω/c2 = 2(V1L1 + V2L2)/c2

CORRECT SAGNAC FORMULA:

2(V1L1 + V2L2)/c2


By contrast, what Michelson did is to remove the SIGN from each loop, in effect nullifying the very definition of the Sagnac effect: he compared two different sides, not the two loops, thus he obtained the CORIOLIS EFFECT formula.

The CORIOLIS EFFECT and the SAGNAC EFFECT are two very different phenomena, one is a physical effect while the other one is an electromagnetic effect: two different phenomena require two different formulas.


Each time transit has a DEFINITE DIRECTION.

Opposite directions have opposite signs.

Let us go back to the original derivation of the Sagnac effect.



To get the time transits IN OPPOSITE DIRECTIONS, you must assign a NEGATIVE SIGN to one of them.




*

rabinoz

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Re: On Sandokhan definitions of the Sagnac and Coriolis Effects
« Reply #13 on: August 28, 2019, 03:18:03 AM »
Only someone who has nothing left to say, and is pretty desperate, can make a statement such as this:

Can't you see that if l1 = l2 AND v1 = v2 you would end up with zero transit time around that loop - impossible?

Which you then subtract again and  so if V1 has the magnitude as V2 and if L1 has the magnitude as L2
You total transit time comes out as zero, which cannot be correct.


For a Sagnac interferometer which is located AWAY FROM THE CENTER OF ROTATION, v1 and v2 can NEVER be the same, since l1 and l2 are located on different latitudes.
Yes, they can be the same! The two different latitudes could be equally spaced either side of the equator.
So a huge loop could  hypothetically have been at 1°N and 1°S and so V1 would have the same magnitude as V2 and L1 can easily be made the same magnitude as L2.

If your calculations are correct then they should show zero Sagnac delay in that situation but they do not.

And I've found that a good technique for checking a calculation like that is to test it with extreme situations where the results are obvious.

And, of course, your expression can only be accurate when the centre of rotation is far outside the loop
so is not a general expression as is Maraner and Zendri's:

But you claim this:
Quote from: sandokhan

CORRECT SAGNAC FORMULA:

2(V1L1 + V2L2)/c2
And for V1 = V2 ≠ 0 and L1 = L2 ≠ 0 your expression will give a non-zero answer but the Sagnac delay should be zero for that case.

Look at the result the "Conspiracy of Light" site (the source of your diagram,  I believe) gives for it
Quote
The Michelson-Gale Experiment by Doug Marett (2010)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
    In refining his argument, he proposed that it was not necessary for the light to go all the way around the globe - since there should be a velocity difference for any closed path rotating on the surface of the earth. He presented the following equation to calculate the time difference expected, using the shift in the interference fringes when the two beams overlap at the detector as a measure of the time difference:

Fig.1:
where:  Vo = the tangential velocity of the earth's rotation at the equator (465m/s)
              A = the area of the circular path
              R = the radius of the earth (6371000 m)
              c = speed of light (3E8 m/s)
              Φ = the latitude in degrees where the experiment is conducted.
              l = wavelength of the light
It's funny how everybody seems to agree that the "4AΩ/c^2 . . . . derived by G. Sagnac" is the Sagnac effect.

And you seems totally confused about the real Coriolis effect!

This is the formula derived by G. Sagnac: 4AΩ/c^2.

THIS IS THE CORIOLIS EFFECT FORMULA.
Sure, but your expression, "4AΩ/c^2 . . . . derived by G. Sagnac" is NOT the Coriolis effect as everybody else seems to know!

This shows the correct Coriolis effect, with a bit of explanation!
Quote
UNDERSTANDING THE CORIOLIS FORCE
And:
It is proportional to the velocity of the object and has no connection with any area!


*

sandokhan

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Re: On Sandokhan definitions of the Sagnac and Coriolis Effects
« Reply #14 on: August 28, 2019, 03:59:39 AM »
Your tricks don't work with me.

v1 and v2 cannot, EVER, be the same, since l1 and l2 are located on different latitudes: the situation where you'd have the interferometer located right on the equator is a special case of my formula which, again, provides the correct answer.

It can easily be seen to be incorrect by simply centring the loop over the equator when: V1 = V2 = V and L1 = L2 = L.

In that situation there should obviously be no Sagnac delay, but your expression gives a delay of: 4(V L)/c2 .

But there is a Sagnac delay, right on the equator!


https://www.researchgate.net/publication/260796097_Light_Transmission_and_the_Sagnac_Effect_on_the_Rotating_Earth

"Kelly [25]  also  noted that  measurements  using  the GPS  reveal that  a light signal takes  414 nanoseconds  longer  to  circumnavigate  the  Earth  eastward  at  the  equator  than  in the westward direction around the same path. This is as predicted by GPS equations (11) and (12)."

Quote from: sandokhan
https://www.researchgate.net/publication/260796097_Light_Transmission_and_the_Sagnac_Effect_on_the_Rotating_Earth

"Kelly [25]  also noted that measurements using the GPS  reveal that a light signal takes  414 nanoseconds longer to circumnavigate the Earth eastward at the equator than in the westward direction around the same path. This is as predicted by GPS equations (11) and (12)."
Agreed, but that is around the whole equator of the rotating earth and is quite irrelevant to your loop.

The global SAGNAC EFFECT formula applies to the interferometer whose center of rotation is located away from its geometrical center.

It works perfectly.

If the center of rotation coincides with the geometrical center, then you use the local SAGNAC EFFECT formula:  if v1 = v2 and l1 = l2, then quite simply, my formula becomes dt = 4VL/c^2, which is the local Sagnac effect formula.

Agreed, but that is around the whole equator of the rotating earth and is quite irrelevant to your loop.

There is no agreement with your previous statement:

In that situation there should obviously be no Sagnac delay,

But THERE IS a Sagnac delay, right on the line of the equator:

https://www.researchgate.net/publication/260796097_Light_Transmission_and_the_Sagnac_Effect_on_the_Rotating_Earth

"Kelly [25]  also  noted that  measurements  using  the GPS  reveal that  a light signal takes  414 nanoseconds  longer  to  circumnavigate  the  Earth  eastward  at  the  equator  than  in the westward direction around the same path. This is as predicted by GPS equations (11) and (12)."

If you now have equal radii and equal velocities, the local Sagnac formula comes into play at once.

You stated that there is no delay at the equator, yet you were proven to be quite wrong: there is a delay right on the equator, which is picked up by the local Sagnac effect formula, a successful trial for my global formula.

Once again, you seem to be very confused, you have no idea of what you are talking about.

You are simply trolling the lower forums.

A perfect "trial" for the global Sagnac effect formula: if you have equal lengths/velocities, then you use the local Sagnac effect formula, and you do have a delay just like proven by the above reference.


Now, if the center of rotation coincides with the geometrical center of the interferometer,  the instance where the interferometer is placed right on the equator, OBVIOUSLY the derivation will be modified for that situation: we use the classic Sagnac formula for two equal sides and two equal velocities:

dt = 4vL/c2


Proof (using my formula):

Sagnac components for the first loop:

l/(c - v)

-l/(c + v)

Sagnac components for the second loop:

l/(c - v)

-l/(c + v)


For the first loop:

l/(c - v) - l/(c + v) = 2vl/c2

For the second loop:

l/(c - v) - l/(c + v) = 2vl/c2

The net phase difference will be (let us remember that the counterclockwise phase difference has a negative sign attached to it, that is why the substraction of the phase differences becomes an addition):

{2vl/c2} -(-){2vl/c2} = 4vl/c2

My formula:

2(v1l1 + v2l2)/c2

If we now let l1 = l2 and v1 = v2, we get 4vl/c2, in perfect agreement.


And I've found that a good technique for checking a calculation like that is to test it with extreme situations where the results are obvious.

My formula works perfectly: therefore you must accept it.


It is proportional to the velocity of the object and has no connection with any area!

The Coriolis FORCE applied to light beams becomes the CORIOLIS EFFECT which actually is directly proportional to the AREA:

Spinning Earth and its Coriolis effect on the circuital light beams

http://www.ias.ac.in/article/fulltext/pram/087/05/0071

« Last Edit: August 28, 2019, 04:09:36 AM by sandokhan »

*

rabinoz

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Re: On Sandokhan definitions of the Sagnac and Coriolis Effects
« Reply #15 on: August 28, 2019, 04:26:58 AM »
Your tricks don't work with me.

v1 and v2 cannot, EVER, be the same, since l1 and l2 are located on different latitudes: the situation where you'd have the interferometer located right on the equator is a special case of my formula which, again, provides the correct answer.

It can easily be seen to be incorrect by simply centring the loop over the equator when: V1 = V2 = V and L1 = L2 = L.

In that situation there should obviously be no Sagnac delay, but your expression gives a delay of: 4(V L)/c2 .

But there is a Sagnac delay, right on the equator!

https://www.researchgate.net/publication/260796097_Light_Transmission_and_the_Sagnac_Effect_on_the_Rotating_Earth

That is nothing like your loop. That is a loop around the equator and so has the centre of rotation right in the centre of the loop.

A loop like yours or that in the Michelson-Gale-Pearson experiment that is flat on the earth's surface should have a sin(latitude) term in it.

Quote
The Michelson-Gale Experiment by Doug Marett (2010)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
    In refining his argument, he proposed that it was not necessary for the light to go all the way around the globe - since there should be a velocity difference for any closed path rotating on the surface of the earth. He presented the following equation to calculate the time difference expected, using the shift in the interference fringes when the two beams overlap at the detector as a measure of the time difference:

Fig.1:
where:  Vo = the tangential velocity of the earth's rotation at the equator (465m/s)
              A = the area of the circular path
              R = the radius of the earth (6371000 m)
              c = speed of light (3E8 m/s)
              Φ = the latitude in degrees where the experiment is conducted.
              λ = wavelength of the light
It's funny how everybody else seems to agree that the "4AΩ/c2 . . . . derived by G. Sagnac" is the Sagnac effect.

*

Bullwinkle

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Re: On Sandokhan definitions of the Sagnac and Coriolis Effects
« Reply #16 on: August 28, 2019, 04:58:43 AM »
I took some time to quote these monkeys.

Turns out there is a 20,000 word limit.   >:(

*

rabinoz

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Re: On Sandokhan definitions of the Sagnac and Coriolis Effects
« Reply #17 on: August 28, 2019, 05:06:18 AM »
I took some time to quote these monkeys.

Turns out there is a 20,000 word limit.   >:(
Trying to reply to a whole novel succinctly is a bit hard so I delete most of the oft-repeated fiction.

Re: On Sandokhan definitions of the Sagnac and Coriolis Effects
« Reply #18 on: September 12, 2019, 04:26:53 PM »
Sandy, I see you are yet to figure out the time taken for light to propagate around a stationary loop yet.

You are still using the same garbage derivation, producing the same garbage result.
Garbage in, garbage out.

You are still confusing times with time differences, and conflating phase conjugate mirrors with normal mirrors.
Even then, you still aren't doing it properly.

There are 2 valid ways of approaching the simple loop, and neither uses this garbage:
Quote
l1/(c - v1)

-l2/(c + v2)

You have 2 options, you can focus on the arms (equivalent to focusing on the loops in the PCM setup), or you can focus on the light paths.

Note: for the following derivation, all velocities will be measured as fractions of c, so c=1.

When focusing on the arms, you calculate the time differences/sagnac shift.
So arm 1, with length 1 has the following shift:
l1/(c - v1)-l1/(c + v1)=2l1v1.
For arm 2, and the combination we have 2 options.
We can find the shift the same way, and then find the difference, or we can note that the light is going in the opposite direction and thus use the below and then add the 2 shifts:
l2/(c + v2)-l2/(c + v2)=-2l2v2/c^2

Thus the total shift is (2l1v1-2l2v2)/c^2
which under the approximation of a annular sector = 4Aw/c^2

i.e. the same formula that every sane person produces.

Treating it as 2 light beams around a loop we instead focus on the times first:
Notice, the time for the 2 arms are added because it is a time and we are seeing how long it takes to go around the loop, ignoring the 2 arms where there is no shift.
The time for the first beam to go around the loop is given by:
l1/(c - v1)+l2/(c + v2)
Then for the beam going in the opposite directions, we change the sign of the velocity terms:
l1/(c + v1)+l2/(c - v2)

Then to find the shift, we find the difference of these times:
l1/(c - v1)+l2/(c + v2) - [l1/(c + v1)+l2/(c - v2)]
=l1/(c - v1)-l1/(c + v1) + l2/(c + v2)-l2/(c + v2)
Just like before.

It is nothing like the formula you are providing.

Your derivation is nothing more than garbage which has no connection to reality.
When trying to find a time taken to traverse the loop you instead find the time difference for a single beam of light propagating along the 2 arms, which doesn't equate to anything in the experiment.

*

sandokhan

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Re: On Sandokhan definitions of the Sagnac and Coriolis Effects
« Reply #19 on: September 12, 2019, 10:05:06 PM »
Let's put your word to the test.

Now instead of adding and subtracting based upon direction, we will add the terms of the same colour, corresponding to the one beam rotating around the interferometer and then find the difference.
dt=l1/(c - v1)+l2/(c + v2)-l1/(c + v1)-l2/(c - v2)
=l1/(c - v1)-l1/(c + v1)+l2/(c + v2)-l2/(c - v2)
=l1(c + v1-c + v1)/(c2 - v12)+l2(c - v2-c - v2)/(c2 - v22)
=2*l1v1/(c2 - v12)-2*l2v2/(c2 - v22)

Now, what the frell is this?

The author of this unscientific piece of garbage cannot distinguish between two opposite directions.

We no longer have a Sagnac interferometer whose center of rotation coincides with its geometrical center: the interferometer is located away from the center of rotation, as such each and every direction MUST HAVE THE CORRECT SIGN.

This guy has the same sign for opposite directions:

l1/(c - v1)+l2/(c + v2)

and

-l1/(c + v1)-l2/(c - v2)

Catastrophically wrong!!!

Here is the correct analysis:

Sagnac phase components for the A > D > C > B > A path (clockwise path):

l1/(c - v1)

-l2/(c + v2)

Sagnac phase components for the A > B > C > D > A path (counterclockwise path):

l2/(c - v2)

-l1/(c + v1)


For the single continuous clockwise path we add the components:

l1/(c - v1) - l2/(c + v2)

For the single continuous counterclockwise path we add the components:

l2/(c - v2) - l1/(c + v1)


The proper signs, in accordance with the direction, are in place.

What jackblack did is to substract the phase differences for TWO SEPARATE OPEN SEGMENTS, and not for the TWO LOOPS (as required by the defintion of the Sagnac effect).

He assigned the wrong signs, moreover, he did not complete the counterclockwise and the clockwise addition of the components of the phase differences.

We have the following terms, both have the same direction, that means one of them corresponds to the red in the inner segment and one to orange on the outer segment. I will colour code them for clarity:
l1/(c - v1)
l2/(c - v2)

Then, we have the remaining terms, in the opposite direction, likewise meaning one is for orange one is for red, noting that red travelled along l1 in the previous one so now it must travel along l2 in this one:
l1/(c + v1)
l2/(c + v2)

Then, if they ARE in opposite direction, they must have the OPPOSITE SIGN.

Here is the correct analysis:

Sagnac phase components for the A > D > C > B > A path (clockwise path):

l1/(c - v1)

-l2/(c + v2)

Sagnac phase components for the A > B > C > D > A path (counterclockwise path):

l2/(c - v2)

-l1/(c + v1)


For the single continuous clockwise path we add the components:

l1/(c - v1) - l2/(c + v2)

For the single continuous counterclockwise path we add the components:

l2/(c - v2) - l1/(c + v1)


jackblack assigned the SAME SIGN, even though he just said a few lines earlier, that they are in fact in opposite direction.


Where are your loops???

You are still comparing two OPEN SEGMENTS: defying the very definition of the Sagnac effect.

Path 1 - A>B, D>C.
Path 2 - C>D, B>A


Completely wrong!

The paths are very clear:

A > B > C > D > A is a continuous counterclockwise path, a negative sign -
A > D > C > B > A is a continuous clockwise path, a positive sign +

Yes, ignoring the sign which I don't particular care about at this time

You CANNOT ignore the sign, since by your own admission you have light beams travelling in opposite directions.

You are literally saying it takes negative time to do something.

No negative times at all.

Just two loops, continuous paths, as required by the definition of the Sagnac effect.



Point A is located at the detector
Point B is in the bottom right corner
Point C is in the upper right corner
Point D is in the upper left corner

Here is the most important part of the derivation of the full/global Sagnac effect for an interferometer located away from the center of rotation.

A > B > C > D > A is a continuous counterclockwise path, a negative sign -

A > D > C > B > A is a continuous clockwise path, a positive sign +

The Sagnac phase difference for the clockwise path has a positive sign.

The Sagnac phase difference for the counterclockwise has a negative sign.


Sagnac phase components for the A > D > C > B > A path (clockwise path):

l1/(c - v1)

-l2/(c + v2)

Sagnac phase components for the A > B > C > D > A path (counterclockwise path):

l2/(c - v2)

-l1/(c + v1)


For the single continuous clockwise path we add the components:

l1/(c - v1) - l2/(c + v2)

For the single continuous counterclockwise path we add the components:

l2/(c - v2) - l1/(c + v1)


The net phase difference will be (let us remember that the counterclockwise phase difference has a negative sign attached to it, that is why the substraction of the phase differences becomes an addition):

{l1/(c - v1) - l2/(c + v2)} - (-){l2/(c - v2) - l1/(c + v1)} = {l1/(c - v1) - l2/(c + v2)} + {l2/(c - v2) - l1/(c + v1)}

Rearranging terms:

l1/(c - v1) - l1/(c + v1) + {l2/(c - v2) - l2/(c + v2)} =

2(v1l1 + v2l2)/c2


BY CONTRAST, here is what you did:

Now instead of adding and subtracting based upon direction, we will add the terms of the same colour, corresponding to the one beam rotating around the interferometer and then find the difference.
dt=l1/(c - v1)+l2/(c + v2)-l1/(c + v1)-l2/(c - v2)
=l1/(c - v1)-l1/(c + v1)+l2/(c + v2)-l2/(c - v2)
=l1(c + v1-c + v1)/(c2 - v12)+l2(c - v2-c - v2)/(c2 - v22)
=2*l1v1/(c2 - v12)-2*l2v2/(c2 - v22)

We have the following terms, both have the same direction, that means one of them corresponds to the red in the inner segment and one to orange on the outer segment. I will colour code them for clarity:
l1/(c - v1)
l2/(c - v2)

Then, we have the remaining terms, in the opposite direction, likewise meaning one is for orange one is for red, noting that red travelled along l1 in the previous one so now it must travel along l2 in this one:
l1/(c + v1)
l2/(c + v2)

Then, if they ARE in opposite direction, they must have the OPPOSITE SIGN.

You used the SAME sign for opposite directions.

Moreover, you compared two open segments, and not the two loops of the Sagnac interferometer.

l1/(c - v1)
l2/(c + v2)

Again, there are 4 legs, not 2. This means you should actually have 4 components.
If you assume arm 2 and 4 to be insignificant (which is technically wrong for a rectangle, as they need to be radial to have no effect, but then again you don't even have a constant v for a rectangle either), then you end up with arm 1, where the light is propagating with the motion of the apparatus, a time of (again, just accepting the formula you provided rather than double checking it):
l1/(c - v1)
which is larger than if it is at rest.

Then for the time in arm 3 you get:
l3/(c + v3)
which is smaller than if the arm is at rest.
You need to add these 2 POSITIVE times to get a reference time for the loop (as well as 2 lots for arm 2 and 4).


You seem to need medical attention jackblack.

Of course the times will be larger and smaller, since you are dealing with DIFFERENT VELOCITIES, c - v1 - v2 and c + v1 + v2.

Positive times? Everyone is laughing at you.

You used the wrong signs.

You compared two open segments, in full defiance of the definition of the Sagnac effect.

I added correctly the terms for the two loops.

Do you understand the definition of the Sagnac effect?

Let me remind you of it:

https://www.mathpages.com/rr/s2-07/2-07.htm

Two pulses of light are sent in opposite directions around a loop.

Loop = a structure, series, or process, the end of which is connected to the beginning.

What you, jackblack, have done, is to compare two open segments of the interferometer, and not the two loops as required by the definition of the Sagnac effect.

l1/(c - v1) + l2/(c + v2)

You have the wrong sign!!!

These beams are in opposite direction: one has a positive sign l1/(c - v1), the other has a negative sign -l2/(c + v2).

But again, we don't use your nonsense negative times.

There are NO negative signs.

Just TWO LOOPS: one counterclockwise, one clockwise.

Exactly as required by the defintion of the Sagnac effect.


EXPERIMENTAL PROOF THAT MY FORMULA IS ABSOLUTELY CORRECT:



The most ingenious experiment performed by Professor Yeh: light from a laser is split into two separate fibers, F1 and F2 which are coiled such that light travels clockwise in F1 and counterclockwise in F2.

https://www.researchgate.net/publication/26797550_Self-pumped_phase-conjugate_fiber-optic_gyro

Self-pumped phase-conjugate fiber-optic gyro, I. McMichael, P. Yeh, Optics Letters 11(10):686-8 · November 1986 

http://www.dtic.mil/dtic/tr/fulltext/u2/a170203.pdf (appendix 5.1)

The first phase-conjugate Sagnac experiment on a segment light path with a self-pumped configuration.

The Sagnac phase shift for the first fiber F1:

+2πR1L1Ω/λc

The Sagnac phase shift for the second fiber F2:

-2πR2L2Ω/λc

These are two separate Sagnac effects, each valid for the two fibers, F1 and F2.

The use of the phase conjugate mirror permits the revealing of the final formula, the total phase difference:

φ = -2(φ2 - φ1) = 4π(R1L1 + R2L2)Ω/λc

2(v1l1 + v2l2)/c2

Exactly the formula obtained by Professor Yeh:

φ = -2(φ2 - φ1) = 4π(R1L1 + R2L2)Ω/λc = 4π(V1L1 + V2L2)/λc

Since Δφ = 2πc/λ x Δt, Δt = 2(R1L1 + R2L2)Ω/c2 = 2(V1L1 + V2L2)/c2

CORRECT SAGNAC FORMULA:

2(V1L1 + V2L2)/c2



YOU ARE NOT USING THE DEFINITION OF THE SAGNAC EFFECT: TWO COUNTERPROPAGATING LOOPS.

You are comparing two sides, WITHOUT ANY LOOPS.

As such, your analysis is the CORIOLIS EFFECT formula, and not at all the SAGNAC EFFECT.

*

rabinoz

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Re: On Sandokhan definitions of the Sagnac and Coriolis Effects
« Reply #20 on: September 13, 2019, 01:18:16 AM »
CORRECT SAGNAC FORMULA:

2(V1L1 + V2L2)/c2
If that is the "CORRECT SAGNAC FORMULA" please explain why all of Michelson, Sagnac, Silberstein, Paolo Maranez and Jean-Pierre Zendri all disagree with your expression!

Quote from: sandokhan
As such, your analysis is the CORIOLIS EFFECT formula, and not at all the SAGNAC EFFECT.
Incorrect! Go and read up on the Coriolis acceleration and it's nothing like what you claim!

Re: On Sandokhan definitions of the Sagnac and Coriolis Effects
« Reply #21 on: September 13, 2019, 02:24:25 AM »
Let's put your word to the test.
We have done that repeatedly.
You have been unable to find a single error with my derivation, nor have you been able to support your nonsense.
You can even figure out how long it takes for light to propagate around a stationary ring.


The author of this unscientific piece of garbage cannot distinguish between two opposite directions.
No, I can, quite easily. That is why I have the c+v or c-v terms.

What you, the author of so much unscientific garbage it isn't funny, seems incapable of understanding, is how time doesn't give a damn what direction you are going in.

If you travel at a speed of 1 m/s, do you think it takes you 100 second to walk 100 m to the right, but -100 second to walk 100 m to the left?

Only a complete moron would think that.
Instead, it takes 100 second to walk 100 m to left or to the right at 100 m/s.
The direction does not effect the time.
So when you try to find out how long it takes for the light to propagate around the ring, you add the 2 times, you don't subtract.
Only a moron or a dishonest scumbag would subtract.

If there was a track with a length of l and you wanted to walk down to the end and back, travelling at a speed of c, would you find the time as:
t=l/c+l/c, or would you have it be t=l/c-l/c?
The former is one which makes sense.
Using 100 m and 1 m/s, that means it takes you 100 s to walk to the end and an additional 100 s to walk back.

But according to your garbage, which has no connection to reality at all, it takes no time at all, because the time spent walking in one direction magically cancels out the time spent walking in the other, as if walking to the right advances you forwards in time while walking to the left advances you backwards.

So when adding times, I will do the only sane thing, and ADD the times, not subtract just because the direction was different.
Again, only a complete moron or a dishonest scumbag would do otherwise.

So no, my derivation is correct.

We have the following terms, both have the same direction, that means one of them corresponds to the red in the inner segment and one to orange on the outer segment. I will colour code them for clarity:
l1/(c - v1)
l2/(c - v2)

Then, we have the remaining terms, in the opposite direction, likewise meaning one is for orange one is for red, noting that red travelled along l1 in the previous one so now it must travel along l2 in this one:
l1/(c + v1)
l2/(c + v2)
Then, if they ARE in opposite direction, they must have the OPPOSITE SIGN.
And what do I have?
One direction has c+v. The other has c-v. Notice how they have a different sign? One is -, one is +.
What has the same sign is the time, as they are both moving forwards in time, i.e. the temporal direction is the same.
If you want to tell us how they magically go backwards in time, feel free. Until you can justify such insanity, your "derivation" remains a pile of garbage completely unconnected to reality.

Sagnac phase components for the A > D > C > B > A path (clockwise path):
Again, this makes no sense.
You can discuss the phase components for the individual arms, and then combine them, or you can find the time for the light paths.
If you are taking times, you need to add the components, not subtract them, because they are times.

What you are doing is finding a difference in time for the same beam of light, which does not represent anything in reality.

This has all been pointed out before, and you were able to do was repeatedly spam the same garbage.
You were completely incapable of rationally defending your claims which is why I repeatedly asked you such a simple question:

How long does it take for the light to travel around a stationary loop? Can you tell us that?
Don't worry, the light is still travelling in a different direction, so you can still make up your BS about signs.

*

sandokhan

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Re: On Sandokhan definitions of the Sagnac and Coriolis Effects
« Reply #22 on: September 13, 2019, 08:19:45 AM »
This is the CORIOLIS EFFECT formula:

Δt = 4AΩ/c^2

Here is a very direct proof:

Spinning Earth and its Coriolis effect on the circuital light beams

http://www.ias.ac.in/article/fulltext/pram/087/05/0071

The CORIOLIS EFFECT formula features an AREA.


The SAGNAC EFFECT formula, on the other hand, does not deal with areas, only with velocities.

Here is a nice proof:





At this point, in a normal debate, the discussion is over.

I have just proven that the formula derived by having compared two sides of the interferometer is actually the CORIOLIS EFFECT formula.

I have posted the proof that the SAGNAC EFFECT does not require an area.


But both these shills are allowed to sabotage this forum, by posting the very same nonsense all over again, as if no counter-arguments had been presented to them.


One direction has c+v. The other has c-v. Notice how they have a different sign? One is -, one is +.

That is the VELOCITY, not the direction of the LIGHT BEAM.

My formula was flawlessly derived and I have arrived at the SAME formula featured in Professor Yeh's papers as well.

There is nothing else to discuss here.


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rabinoz

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Re: On Sandokhan definitions of the Sagnac and Coriolis Effects
« Reply #23 on: September 13, 2019, 03:14:39 PM »
This is the CORIOLIS EFFECT formula:

Δt = 4AΩ/c^2

Here is a very direct proof:

Spinning Earth and its Coriolis effect on the circuital light beams

http://www.ias.ac.in/article/fulltext/pram/087/05/0071

The CORIOLIS EFFECT formula features an AREA.

The SAGNAC EFFECT formula, on the other hand, does not deal with areas, only with velocities.
I have to ask, "Do you even read your own references or take note of their titles? "

Look at the full title of the above reference: "Spinning Earth and its Coriolis effect on the circuital light beams: Verification of the special relativity theory SANKAR HAJRA".

Note first his object "Verification of the special relativity theory" which you do not accept.

Now look at his introduction:


If you claim he's deriving the Coriolis effect please explain:
Quote
In Bilger et al [1], Anderson et al [2], and in Michelson and Gale assisted by Pearson [3], Sagnac effect on the circuital laser/light beams on the spinning Earth has been studied.
The formula for Sagnac effect on the spinning Earth for circuital opposing beams of light first calculated by Silberstein [4] and used in the explanation of the Michelson–Gale experiment was:

Your own author, Sankar Ajra, calls it the Sagnac effect and he says the Michelson and Gale assisted by Pearson experiment the Sagnac effect.

All Sankar Ajra is doing is to derive the Sagnac effect using the Coriolis force and verifying of the special relativity theory.

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sandokhan

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Re: On Sandokhan definitions of the Sagnac and Coriolis Effects
« Reply #24 on: September 13, 2019, 09:47:09 PM »
You have been reported for SPAMMING this forum.

Your 'argument' has been debunked before:

https://www.theflatearthsociety.org/forum/index.php?topic=82434.msg2200962#msg2200962

The Coriolis force is exactly as SANKAR HAJRA shows in (2) and that is not the expression that Michelson, Sagnac and Silberstein derived!

But it is the VERY SAME EXPRESSION.

Again, here is the final formula derived by S. Hajra:

EQUATION 12:

dt = 4ωA/c^2

He even SPECIFIES that it is the VERY SAME equation derived by both Sagnac and Silberstein.

Take a look at the title of the section 4, mentioned on page 3 of 5:

Sagnac effect? No, it is Coriolis effect

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rabinoz

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Re: On Sandokhan definitions of the Sagnac and Coriolis Effects
« Reply #25 on: September 14, 2019, 12:36:53 AM »
You have been reported for SPAMMING this forum.

Your 'argument' has been debunked before:

https://www.theflatearthsociety.org/forum/index.php?topic=82434.msg2200962#msg2200962

The Coriolis force is exactly as SANKAR HAJRA shows in (2) and that is not the expression that Michelson, Sagnac and Silberstein derived!

But it is the VERY SAME EXPRESSION.
No, it is not! This is the Coriolis force in Hajra's equation (2). Read what he writes in the intro.

And Hajra uses that Coriolis force to derive that Sagnac effect.

Quote from: sandokhan
Again, here is the final formula derived by S. Hajra:

EQUATION 12:

dt = 4ωA/c^2

He even SPECIFIES that it is the VERY SAME equation derived by both Sagnac and Silberstein.

Take a look at the title of the section 4, mentioned on page 3 of 5:

Sagnac effect? No, it is Coriolis effect
Yes, S. Hajra calls it the Coriolis effect but:
  • Why does Hajra's calling it the Coriolis effect supercede Michelson, Sagnac  (himself), Silberstein, Langevin, Paolo Maraner, Jean-Pierre Zendri, Malykin and, as far as I can tell, everybody else?

    Because YOU say so?

  • Hajra uses the Coriolis force to derive his final expression:


    But using the Coriolis force does not make the final result the the Coriolis effect.
    Any Coriolis effect involves a linear velocity and an angular velocity not an area but
    the Sagnac effect involves an area and an angular velocity but no linear velocity.

  • If you look at Hajra's paper, you'll find in the introduction:
    Quote
    In the calculation, Silberstein has assumed a relative spinning motion between ether and Earth at and near its surface and has reached the well-known formula of Sagnac effect for the circuital opposing light beams on the surface of the spinning Earth as given above.

    Hajra seened to believe that Silberstein was arguing for an "ether interpretation" of the Sagnac effect and Hajra is writing this paper to show that it can also be a relativistic effect.

    Read the title again, "Spinning Earth and its Coriolis effect on the circuital light beams: Verification of the special relativity theory by SANKAR HAJRA".

    Hajra needn't have done that because Silberstein accepted General Relativity and was urging Michelson do verify this experimentally to verify GR.

    And in the second part of Silberstein's paper he shows that GR gives the same result.
And you might read, The Sagnac effect and its interpretation by Paul Langevin Gianni Pascoli https://doi.org/10.1016/j.crhy.2017.10.010.

They all seem to call it the Sagnac effect!

I don't know  where this leaves  you because you are trying to use Hajra's "Spinning Earth and its Coriolis effect" to support you flat stationary earth.

That strikes me as very odd!

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sandokhan

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Re: On Sandokhan definitions of the Sagnac and Coriolis Effects
« Reply #26 on: September 14, 2019, 01:11:43 AM »
YOU HAVE BEEN REPORTED FOR SPAM AGAIN!

Is there no one among the mods here to take care of this deliberate spamming?

rabinoz has deep problems which are not any of our business to take care of.

He spammed this very thread TWO TIMES TODAY IN A ROW, even though he knows it is against the rules.

What is going on here?

His cognitive dissonance condition cannot allow to function in any other way: how are we supposed here to deal with this?

He is manifesting evident psychological problems, he is unable to face up to reality, to accept reality.

How can we here at the FES deal with this?


Your 'argument' has been debunked before:

https://www.theflatearthsociety.org/forum/index.php?topic=82434.msg2200962#msg2200962

But it is the VERY SAME EXPRESSION.

Again, here is the final formula derived by S. Hajra:

EQUATION 12:

dt = 4ωA/c^2

He even SPECIFIES that it is the VERY SAME equation derived by both Sagnac and Silberstein.

Take a look at the title of the section 4, mentioned on page 3 of 5:

Sagnac effect? No, it is Coriolis effect


PAGE 4 OF 5

The only alternative is: Coriolis effect (not the
Sagnac effect) is responsible for the non-null result of
the Michelson–Gale experiment assisted by Pearson and
the experiment of Bilger et al.



https://www.ias.ac.in/article/fulltext/pram/087/05/0071


Why then is rabinoz allowed to spam this forum on a daily basis?

I have just posted the proof which debunks yet again his statements; this was done months ago as well.

Yet here he is spamming this forum again.


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rabinoz

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Re: On Sandokhan definitions of the Sagnac and Coriolis Effects
« Reply #27 on: September 14, 2019, 01:15:36 AM »
YOU HAVE BEEN REPORTED FOR SPAM AGAIN!
I am under no obligation to accept your "proofs" and have every right to reply,  thank you.

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sandokhan

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Re: On Sandokhan definitions of the Sagnac and Coriolis Effects
« Reply #28 on: September 14, 2019, 01:19:40 AM »
All of your statements were debunked months ago!!!

Here you are SPAMMING YET AGAIN, TWICE TODAY, this thread.

Your cognitive dissonance problem is in full view for all to observe.

If you don't like this forum, if you are very unpleased with the discussion here, then please leave, get out!

If the mods intervene and ban you every time you SPAM this forum, you will have nothing left to say.

Without the constant spamming, you are totally devoid of any arguments.

Re: On Sandokhan definitions of the Sagnac and Coriolis Effects
« Reply #29 on: September 16, 2019, 01:57:43 PM »
This is the CORIOLIS EFFECT formula:
It is the Sagnac effect formula. You not liking that wont change it.

At this point, in a normal debate, the discussion is over.
Yes, you have been refuted.
Your argument has been shown to be completely wrong.
You have been unable to back up your derivation nor refute mine.
Instead all you can do is assert that you are correct and ignore any questions asked.

So yes, in a normal debate, the discussion would be over, you would accept that you were wrong and we would move on.

One direction has c+v. The other has c-v. Notice how they have a different sign? One is -, one is +.
That is the VELOCITY, not the direction of the LIGHT BEAM.
Yes, that is the velocity, the part where direction is important.
In one part the light moves with the motion of the beam and in the other the light moves against the motion.
That is the part where the sign matters.

Adding up the times doesn't change the sign.

Again, if you have someone who can run at a speed of c, who wants to run back and forth down a track of length l, is the time required:
t1=l/c+l/c, where we don't change the sign for opposite spatial directions, as both paths move forwards in time, or is it:
t2=l/c-l/c, where we change the sign for opposite spatial directions, meaning regardless of how long the path is it takes no time to run back and forth down it because you magically go backwards in time when you go backwards down the path?

It makes no sense to change the sign like that.
When you add the components for a single light path, you add the signs as you are adding up the time required to traverse the 2 beams.
You do not find the difference.

As such, your derivation has no connection to reality.

This is also why before I was repeatedly asking you how long it takes for light to traverse a stationary loop.
Using your nonsense you end up with a time of 0, regardless of the size of the loop.


Also note, you are being extremely inconsistent with this nonsense. When you have the centre of the interferometer aligned with the centre of rotation, you don't change sign with direction. Instead you just add up the times. So it is clear that you know you are spouting nonsense.