Bob Knodel and the laser ring gyroscope

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Re: Bob Knodel and the laser ring gyroscope
« Reply #30 on: February 19, 2019, 07:19:17 AM »
In 2007, R. Sungenis and R. Bennett published a 1147 page treatise, Galileo Was Wrong. They included copious amounts of information which was very well presented, having outdone any previous work on the subject.

Yet, on page 745 they state:

The Sagnac time difference is (for the Michelson-Gale experiment):

Δt = 4Aω (sinφ)/c2

As such, they practically stated to their readers that Galileo was right.

For the MGX, the figures for the area of the path, latitude (41deg. 46'), wavelength of the light, speed of light, and the expected fringe shifts are well known.

Expected fringe shift: 0.2364

Measured fringe shift: 0.230 +/- 0.005

Then, the angular velocity of the Earth can be easily computed.

By having made the outrageous claim that the above formula is the SAGNAC EFFECT formula, Albert Michelson put an end to any debate on geocentrism vs. heliocentrism.

Modern day ring laser interferometers also feature the same formula, while the physicists running the experiment are claiming that it is the Sagnac formula.

That is why the author of the video finds himself facing the same conundrum.

Sungenis and Bennett, each geocentrist, the author of the video, CANNOT claim that it is the ether which is rotating above the surface of the Earth, since THEY ACCEPT that the formula published by Michelson is the SAGNAC EFFECT formula (exactly as Sungenis and Bennett did on page 745 of their treatise).

Encasing one arm of the interferometer in lead will result in Hammar's experiment, which then will have the heliocentrists claim that there is no ether entrapment.

Few scientists, especially including all of the geocentrists, understand that the MGX put an end in 1925 to any debate on heliocentrism vs. geocentrism.

If the heliocentrists are told that the formula published by Michelson is wrong, as a last attempt to provide any kind of an explanation for the MGX, all they have to do is respond: "show us the correct formula then", "go ahead and derive the right formula". They do this because they know that if Einstein, Lorentz, Michelson, Pauli, Langevin, Post were not able to derive the SAGNAC EFFECT formula, no one else will.

Unless the geocentrists come up with the SAGNAC EFFECT formula, there isn't any debate at all: the heliocentrists win hands down, since they claim that the formula published by Michelson is the SAGNAC EFFECT formula which measures rotation.

Every single day thousands of papers are published that say you are wrong. We don't have to go back to 1947 or whenever to find supporting evidence. But again the scientists who wrote the paper you quoted were they flat earth believers........I think not. In fact none of the scientists you are so fond of quoting believed in a flat earth or any of the other things you believe in.

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zork

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Re: Bob Knodel and the laser ring gyroscope
« Reply #31 on: February 19, 2019, 12:41:03 PM »
Sungenis is christian apologetic and Bennet is speculative fiction writer. Good to know from who sandokhan gets his information.
Rowbotham had bad eyesight
-
http://thulescientific.com/Lynch%20Curvature%202008.pdf - Visually discerning the curvature of the Earth
http://thulescientific.com/TurbulentShipWakes_Lynch_AO_2005.pdf - Turbulent ship wakes:further evidence that the Earth is round.

Re: Bob Knodel and the laser ring gyroscope
« Reply #32 on: February 19, 2019, 02:12:48 PM »
Regardless of the discussion on the differing interpretations of the math the fact is that the measured drift is 15°/hr.  The exact value it should get because of the earths rotation.  Bob Knodel (the person doing the experiment) find it unnerving and tries it again with the same results.   Bob was so sure it would disprove rotation and wants to keep it under wraps because it blows their whole theory out of the water.

Mike
Since it costs 1.82˘ to produce a penny, putting in your 2˘ if really worth 3.64˘.

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rabinoz

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Re: Bob Knodel and the laser ring gyroscope
« Reply #33 on: February 19, 2019, 02:26:14 PM »
This post deserved a fuller answer so here is the first part about Silberstein's paper.

You do not have the necessary mathematical understanding to quote from Zendri's paper.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
CAN YOU READ ENGLISH RABINOZ?
Yes!
Quote from: sandokhan
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.

Dr. Silberstein:
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/c^2
No, he does not! Here read again exactly what he wrote and not what you read into it:

And read very carefully his last paragraph!

Quote from: sandokhan
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.
Agreed, "effect of the Coriolis force upon the interferometer will be to create a convex and a concave shape of the areas: σ1 and σ".
But he then goes on to show that the lengths of these paths are virtually the same and the Coriolis effect makes no significant change to the Sagnac shift.

Quote from: sandokhan
The sum of these two areas is replaced by 2A and this is how the final formula achieves its final form:
dt = 4ωA/c^2
A = σ1 + σ2

That is, the CORIOLIS EFFECT upon the light beams is totally related to the closed contour area.
No! It is not the Coriolis effect.
Silberstein takes great pains to show that while the Coriolis effect does cause the light paths to curve very slightly it can have no effect on the Sagnac delay.

So your claim is a total distortion of what Silberstein wrote. Sure, "A = σ1 + σ2" but Silberstein carefully points out that "σ1 + σ2" is twice "the area (σ) of the rectilinear (dotted) triangle, and similarly in the case of any polygons.

Silberstein knows he is deriving the Sagnac effect and he clearly says so and he goes on the derive the relativistic case:
                 

Exactly as before but without the drag coefficient and it is quite obvious that Silberstein is calculating the Sagnac effect!

In the meantime, please carefully check your derivation of your "for the first time, the correct Sagnac formula for an interferometer located away from the center of rotation has been derived in a precise manner."

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sandokhan

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Re: Bob Knodel and the laser ring gyroscope
« Reply #34 on: February 19, 2019, 02:49:22 PM »
You are destroying whatever credibility you had left here, in the upper forums.

You are not able to even read a scientific paper properly.

You are confusing two very different derivations which speaks volumes of your training as a researcher into scientific matters.

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

One of them, the first one, ends on page 298, and Dr. Silberstein is deriving the equation of the light path in relation to Fermat's principle.

It is there where he states that: "the difference would certainly be too small to be measured directly".

Next, he takes up another matter, the derivation of the CORIOLIS EFFECT.

Silberstein takes great pains to show that while the Coriolis effect does cause the light paths to curve very slightly it can have no effect on the Sagnac delay.

He does?



Dr. Silberstein:

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/c^2

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/c^2

A = σ1 + σ2

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


Dr. Silberstein is describing the Coriolis effect, whether the lines are straight or not, NOT the electromagnetic effect (the Sagnac effect).

HERE IS THE PROOF THAT DR. SILBERSTEIN DERIVED THE CORIOLIS EFFECT:

One of the most in-depth treaties on the ring laser interferometers.

https://books.google.ro/books?id=8c_mBQAAQBAJ&pg=PA15&lpg=PA15&dq=malykin+silberstein+coriolis&source=bl&ots=JrMqF2vmto&sig=xCnMB4hL_J_ESg9Xdfhye1ahVjA&hl=en&sa=X&ved=2ahUKEwiE0ZDWxeXeAhXwkYsKHYxwBMYQ6AEwCXoECAUQAQ#v=onepage&q=malykin%20silberstein%20coriolis&f=false

CAN YOU READ ENGLISH RABINOZ?

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.


You confused TWO VERY DIFFERENT DERIVATIONS, deviously assigning the conclusion for the first derivation to the second derivation, which really means you are getting very desperate.


I understand your predicament.



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


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



The promise made by A. Michelson, "the difference in time required for the two pencils to return to the starting point will be...", never materialized mathematically.

Instead of applying the correct definition of the Sagnac effect, Michelson compared TWO OPEN SEGMENTS/ARMS of the interferometer, and not the TWO LOOPS, as required by the exact meaning of the Sagnac experiment.

As such, his formula captured the Coriolis effect upon the light beams.

Not even the formal derivation of the Sagnac effect formula is not entirely correct.





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 -

The net phase difference will be (let us remember that the counterclockwise phase difference has a negative sign attached to it):

2πR(1/(c - v)) - (-){-2πR(1/(c + v))} = 2πR(1/(c - v)) - (+)2πR(1/(c + v)) = 2πR(1/(c - v)) - 2πR(1/(c + v)) = 2vL/c2


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.

Using the correct definition, we recover not only the error-free formula, but also the precise velocity addition terms.



Practically, A. Michelson received the Nobel prize (1907) for the wrong formula (published in 1904 and 1887; E.J. Post proved in 1999 that the Michelson-Morley interferometer is actually a Sagnac interferometer).

No other physicist has been able to derive the correct Sagnac formula: for the past 100 years they have been using the wrong formula (the Coriolis effect equation) to describe a very different physical situation.

Here, for the first time, the correct Sagnac formula for an interferometer located away from the center of rotation has been derived in a precise manner.




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.

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.


We can see at a glance each and every important detail.


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.

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.

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.


The derivation used by Dr. Silberstein leads to the same formula derived by Michelson where THERE IS NO LOOP WHATSOEVER in the analysis.

The SAGNAC EFFECT requires two loops, neither Silberstein nor Michelson ever offered at least one loop.

Again, this speaks volumes of your training as a physicist, and of your true motives for trolling the upper forums.



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rabinoz

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Re: Bob Knodel and the laser ring gyroscope
« Reply #35 on: February 19, 2019, 03:03:38 PM »
Sungenis is christian apologetic and Bennet is speculative fiction writer. Good to know from who sandokhan gets his information.
While Robert Sungenis is a strong Geocentrist but he is very much against the very idea of a Flat Earth.
Have a look at: Flat Earth Geography: The Flat Earth Frenzy, Unscientific and Unbiblical

There's a great deal of good anti-flat earth material in his "The Flat Earth Frenzy" yet Sandokhan dares to refer to Robert Sungenis for support - Strange bed-fellows!

A Geocentric Solar System makes much more sense than any flat earth but still cannot be supported if Newtons Laws are valid.

Re: Bob Knodel and the laser ring gyroscope
« Reply #36 on: February 19, 2019, 03:20:44 PM »
The formula used in laser optics is the CORIOLIS EFFECT equation; however, the counterpropagating beams of light in an interferometers is a SAGNAC EFFECT experiment.

The author of the video does not understand that such an experiment, where an arm of the interferometer was encased in lead, has already been done a long time ago:
No, you are the one who does not understand. The 2 effects are the same.
You have been completely unable to refute this in any way.

A laser ring gyroscope is fully equipped to register/record TWO TYPES OF ROTATIONAL MOTIONS: either the ether drift rotation
As the aether doesn't exist, it is completely incapable of doing so.

What it is capable of doing is recording rotation. That is what it does. This is the rotation of Earth.
However in some fantasy models that will be represented by the rotation of a magical aether around Earth.

has registered ONLY the CORIOLIS EFFECT, nothing else.
As that is all there is to detect, as that Coriolis effect is the Sagnac effect.

You wish to claim the 2 are magically different yet have been completely unable to substantiate it in any way.

You are yet to show any magical derivation for such a magically different formula, not have you been able to refute my refutation in any way.

Again, if you want to start, start from the basics. Rather than focus on a rotating/moving loop, start with a stationary one.
No copying and pasting massive walls of text, a lot of which is completely irrelevant as the experiments discussed don't use a phase conjugate mirror.

The Sagnac phase difference for the clockwise path has a positive sign.
No it doesn't. There is no phase difference for the clockwise path. Just a time taken.
There are 2 options to calculate the overall phase difference.
1 - Calculate the time taken for the clockwise path, and the counterclockwise path and find the difference.
2 - Calculate the phase difference for each segment and add them.


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

l1/(c - v1)

-l2/(c + v2)
Again, these should be times, there is no negative here. You need to add the 2 times.
That is where your massive error is.

This is why you need to start from the basics.

How long will it take the light to propagate around the loop if it is completely stationary?
If you like we can even shrink arm 2 and 4 to be 0, i.e. so it is effectively just a light path back and forth. According to your nonsense it takes no time at all.

So again, start from the basics, a simple stationary loop. How long does it take the light to propagate around it in a clockwise direction?

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sandokhan

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Re: Bob Knodel and the laser ring gyroscope
« Reply #37 on: February 19, 2019, 10:17:14 PM »
The RE are dodging the very definition of the SAGNAC EFFECT.

Michelson did the same thing.

The SAGNAC EFFECT is a phenomenon encountered in interferometry that is elicited by rotation

A stationary loop does not reveal 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.

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

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.


Let us take a look at the humongous error perpetrated by Michelson and also by the RE.

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.

Re: Bob Knodel and the laser ring gyroscope
« Reply #38 on: February 19, 2019, 11:23:54 PM »
The RE are dodging the very definition of the SAGNAC EFFECT.
No, that would be you, repeatedly.
In fact, lets look at your definition and your claimed formula:

The SAGNAC EFFECT is a phenomenon encountered in interferometry that is elicited by rotation
Notice the key part? ROTATION!

Now your alleged formula:
CORRECT SAGNAC FORMULA:
2(V1L1 + V2L2)/c2
Notice how there is nothing in there to do with rotation?
You have lengths and linear velocities. Nothing to do with rotation at all.
In fact, we can even setup the interferometer to be a square such that L=L1=L2, and have it undergo linear motion such that V=V1=V2, with no rotation at all.
What do we get with your "correct" formula?
2(V*L + V*L)/c2
=4*V*L/c2
So in a system which is undergoing no rotation at all, your formula still claims you should measure a shift due to the Sagnac effect, which is purely for rotation.

So who is the one ignoring the definition?
It sure doesn't seem to be us REers.

A stationary loop does not reveal the SAGNAC EFFECT.
I know. It provides a baseless to start with.
It shows your derivation is pure nonsense.
So like I said, start with a stationary loop. Once you have figured out that, you can move on to more complex systems.

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.
Yes, and as we are dealing with times, i.e. the time taken for the light to propagate, the sign will be positive.
Or as we are focusing on the difference between the time taken for each beam, we can focus on the light going one way vs the other.

This guy has the same sign for opposite directions:
l1/(c - v1)+l2/(c + v2)
No, that would be you, where you also have the opposite sign for the same direction.
In the equation you quoted, the direction is the same, they are going the one direction around the loop.
The change in direction of motion of the individual beams is expressed entirely in the c-v or c+v component, based upon the light going with or against the motion.

For the single continuous clockwise path we add the components:
l1/(c - v1) - l2/(c + v2)
That isn't adding. That is subtracting. Do you know the difference?
Adding goes A+B.
Subtracting goes A-B.
You are subtracting 2 times from one another to find the difference in time taken for one beam to go along one arm vs going along the other arm.
It equates to nothing physical.

Like I said, start from a stationary loop. See if you can get it then.
How long does it take the light to propagate around the stationary loop?

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sandokhan

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Re: Bob Knodel and the laser ring gyroscope
« Reply #39 on: February 20, 2019, 12:05:04 AM »
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.

« Last Edit: February 20, 2019, 02:13:52 AM by sandokhan »

Re: Bob Knodel and the laser ring gyroscope
« Reply #40 on: February 20, 2019, 03:30:56 AM »
I am going to explain the entire phenomenon using even more details
You mean spam with even more nonsense.
The problem is you are just repeating the same refuted nonsense and magically subtracting times for a single path, completely ignoring the massive problems, like your formula producing a shift for an interferometer which is just moving in a straight line without rotation, clearly indicating it isn't the Sagnac effect.

Like I said, go back to the basics.
What is the time taken for light to propagate around a stationary loop?
Can you answer that? If not, you shouldn't even attempt to deal with the Sagnac effect.

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sandokhan

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Re: Bob Knodel and the laser ring gyroscope
« Reply #41 on: February 20, 2019, 03:40:34 AM »
You are trolling the upper forums.

Do you understand the definition of the SAGNAC EFFECT?

It involves both rotation and two loops.

A stationary interferometer has nothing to do the SAGNAC EFFECT, which comes into play once the table is being rotated.

We are dealing here with the SAGNAC EFFECT.

You need TWO LOOPS, you have neither.

I have the TWO LOOPS, and the correct signs, and the correct counterclockwise/clockwise paths, everything is in plain sight.

Moreover, my formula agrees with the experimental proof provided by Professor Yeh, a paper published in the Journal of Optics Letters.

A beautiful and striking generalization of the Sagnac effect formula:



Everything you want to know about the universe in just one formula: the most important formula in physics, by far.

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rabinoz

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Re: Bob Knodel and the laser ring gyroscope
« Reply #42 on: February 20, 2019, 04:18:01 AM »
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.
We understand perfectly, thank you, but I'm afraid that your derivation of the "CORRECT SAGNAC FORMULA" is obviously incorrect.

Quote from: sandokhan
The RE standard for the Sagnac effect:
https://www.mathpages.com/rr/s2-07/2-07.htm
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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 subtracted?
We are not dealing with "TWO LOOPS". Mathpages simply draws the light paths separate for clarity.
And of course we can but calling it "the differences in time travel" that "have to be subtracted" seems misleading.
Each of the delay times are positive values and it would be better to simply say "that the time travels have to be subtracted?".

And before going on I must stress that all through Dr Silberstein is deriving nothing other than the Sagnac effect, though he presents fringes not time - either is OK.

He makes this quite obvious at the end of the preamble about the shape of light paths where he says:
Quote from: Dr L. Silberstein
The experimental possibilities with regard to the optical effects of the rotation of the Earth lie in another direction, to wit in the phase retardation in an optical circuit (i.e., closed light path) as in the well-known laboratory experiment of Sagnac with a small spinning interferometer as our system S.
The corresponding formula used by Sagnac and before him by Michelson (Phil. Mag. vol. 2, 1904, p. 716-719) who actually proposed but never carried
out a terrestrial experiment of the kind here aimed at, can be most simply deduced in the following way.
And he proceeds to derive the Sagnac effect!

Quote from: sandokhan
Sagnac phase component for the clockwise path: 2πR(1/(c - v))
Sagnac phase component for the counterclockwise path: -2πR(1/(c + v))
What on earth are those expressions? You call them "phase components" but the dimensions are of time delays.

Quote from: sandokhan
This is the correct way to derive the Sagnac formula:
So what is wrong with the Mathpages derivation? Leading to .
What is wrong with Dr Silberstein's Sagnac effect allowing for aether dragging?
What is wrong with Dr Silberstein's relativistic derivation?

Quote from: sandokhan
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.
Sure, you have (c - v) and (c + v) but both are still positive time delays that you are subtracting.

Quote from: sandokhan
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.


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.
Why TWO LOOPS? It is simply light propagated in two directions around the one loop.

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.

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.
No you need the total time around the clockwise path, (A > D > C > B > A) so you must add the time delay in each leg.

Quote from: sandokhan
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.
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)
You are igoring the time delays in the "vertical, or north south, segments" but that is probably acceptable as they should be the same if they follow meridians.
Quote from: sandokhan
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)
But your making that second term in each negative is quite incorrect.

Even though the propagation is against the velocity of the segment all of the delays are positive,
       so the total delay for the A > D > C > B > A path (clockwise) should be: l1/(c - v1) + l2/(c + v2)
       and the total delay for the the A > B > C > D > A path (counterclockwise ) should be: l2/(c - v2) + l1/(c + v1)

So everything from here on is incorrect.
Quote from: sandokhan
CORRECT SAGNAC FORMULA:

2(V1L1 + V2L2)/c2
No, it cannot be correct.
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 .

Quote from: sandokhan
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.
Nope! Sagnac, Michelson, Silberstein and Mathpages all got it right and derived the Sagnac effect but you made a simple mistake and ended with an impossible result.

Quote from: sandokhan
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.
Agreed!

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sandokhan

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Re: Bob Knodel and the laser ring gyroscope
« Reply #43 on: February 20, 2019, 05:16:18 AM »
You are trolling the upper forums.

And before going on I must stress that all through Dr Silberstein is deriving nothing other than the Sagnac effect, though he presents fringes not time - either is OK.

It is a well known fact of science that Dr. Silberstein derived the CORIOLIS EFFECT formula.

One of the most in-depth treaties on the ring laser interferometers.

https://books.google.ro/books?id=8c_mBQAAQBAJ&pg=PA15&lpg=PA15&dq=malykin+silberstein+coriolis&source=bl&ots=JrMqF2vmto&sig=xCnMB4hL_J_ESg9Xdfhye1ahVjA&hl=en&sa=X&ved=2ahUKEwiE0ZDWxeXeAhXwkYsKHYxwBMYQ6AEwCXoECAUQAQ#v=onepage&q=malykin%20silberstein%20coriolis&f=false

CAN YOU READ ENGLISH RABINOZ?

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.

Here is another reference, explicitly deriving the CORIOLIS EFFECT, same formula as that published by Dr. Silberstein:

Spinning Earth and its Coriolis effect on the circuital light beams

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


Why TWO LOOPS? It is simply light propagated in two directions around the one loop.

You are continuing your trolling series unabated.

You have ONE interferometer, with two loops: a counterclockwise loop and a clockwise loop.

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.


Make sure you understand the definition of the term LOOP.

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

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 you need the total time around the clockwise path, (A > D > C > B > A) so you must add the time delay in each leg.


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.




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.

so the total delay for the A > D > C > B > A path (clockwise) should be: l1/(c - v1) + l2/(c + v2)

It can't be.

Now, you have TWO OPPOSING DIRECTIONS, while you assign the SAME SIGN, a catastrophic mistake if you are seeking the SAGNAC EFFECT formula.

Moreover, where are the loops ??

What in effect you are saying is this:

A > D > C combined with A > B.

NO LOOP AT ALL.

A tremendous error committed by Michelson.

The SAGNAC EFFECT requires TWO LOOPS. Now we are dealing with the clockwise loop.

What you want is this:

A > D > C > B > A path (clockwise path)

Now we have a loop.

This loop includes two opposite directions, two opposite terms/components.

That is why one must a positive sign, while the other must have a negative sign.

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.


It can easily be seen to be incorrect by simply centring the loop over the equator

Is this supposed to be a joke?

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

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)."

« Last Edit: February 20, 2019, 05:20:32 AM by sandokhan »

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markjo

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Re: Bob Knodel and the laser ring gyroscope
« Reply #44 on: February 20, 2019, 06:51:28 AM »
A beautiful and striking generalization of the Sagnac effect formula:



Everything you want to know about the universe in just one formula: the most important formula in physics, by far.
Where is the term for rotation in your formula?
Science is what happens when preconception meets verification.
Quote from: Robosteve
Besides, perhaps FET is a conspiracy too.
Quote from: bullhorn
It is just the way it is, you understanding it doesn't concern me.

Re: Bob Knodel and the laser ring gyroscope
« Reply #45 on: February 20, 2019, 09:04:54 AM »

The global/generalized Sagnac effect formula, the most important in all of physics:

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


The most important? In all of physics? All of it?

I'd like to hear comments. From anyone except sandokhan.

*

sandokhan

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Re: Bob Knodel and the laser ring gyroscope
« Reply #46 on: February 20, 2019, 09:40:08 AM »
You better believe it.

Michelson was promptly awarded a Nobel prize (1907) for the wrong formula.

He used the same formula in 1925 to claim that his interferometer in Clearing, Illinois measured the angular velocity of the Earth.

The GLOBAL SAGNAC EFFECT formula answers directly the most important questions we might have about the universe, the ones most physicists are seeking at the present time.

?

JCM

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Re: Bob Knodel and the laser ring gyroscope
« Reply #47 on: February 20, 2019, 09:58:12 AM »
My favorite part of all of these arguments from the Bob Knodels of the world is they don’t address the consequences, they just pick some aspects of science they like and then misrepresent it. After which they then claim seemingly all of science to be fake news, while using cherry picked 100 plus year old science experiments to push their agenda. It boggles the mind..

Now another issue is if the Earth is still then the entirety of the Universe is orbiting us.  Every galaxy, every nebula, everything.  That makes more sense to Mr. Knodel then simply the Earth is spinning just like every single object in space we can see. 

Mr. Knodel then claims the laser gyroscope is so sensitive he changes the units from degrees rotation per day to sensing the mph of the rotation and then infers it also would pick up the movement around the Sun and the rotational movement of the Milky Way and finally it would pick up the linear velocity of the galaxy.  How utterly underhanded.  The uniform linear motion cannot be directly detected while inside Earth’s reference frame.  As an “engineer” he would know this.  Otherwise the gyroscope wouldn’t be able to work at all on the airplane.   

Re: Bob Knodel and the laser ring gyroscope
« Reply #48 on: February 20, 2019, 12:07:54 PM »
Do you understand the definition of the SAGNAC EFFECT?
Yes. Do you?
It involves rotation, as you have already admitted, yet your nonsense formula claims you will experience a shift without rotation, with just linear motion.
Do you understand what that means? It means your formula is wrong.
Meanwhile, my formula doesn't predict such nonsense.

So either you aren't trying to calculate the Sagnac effect, or your formula/derivation is completely wrong.

Your formula only displays lengths and linear velocities. Another sign that it is completely wrong.

Yet you refuse to address any of this and instead just keep making the same baseless assertions.

Now, as I actually understand the derivation, I also understand where the big error is in your analysis. It is how you are horribly mistreating times. Yet you cannot argue it and instead just repeatedly assert that you are correct.


This is why I am telling you to start with a stationary loop.
It is the most basic (at least the most basic loop).
If you can't figure out the stationary loop you have no hope of figuring out what happens to a loop in motion.

Once you have figured it out for a stationary loop you can then see what effects motion will cause.

And no, you don't have 2 loops. You have light propagating around a single loop in 2 directions.

So I will ask again, how long does it take for the light to propagate around a stationary loop?

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rabinoz

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Re: Bob Knodel and the laser ring gyroscope
« Reply #49 on: February 20, 2019, 02:18:34 PM »
You are trolling the upper forums.

And before going on I must stress that all through Dr Silberstein is deriving nothing other than the Sagnac effect, though he presents fringes not time - either is OK.

It is a well known fact of science that Dr. Silberstein derived the CORIOLIS EFFECT formula.
No it is not and Grigorii B. Malykin, Vera I. Pozdnyakova do not ever claim that he does.

Quote from: sandokhan
One of the most in-depth treaties on the ring laser interferometers.
https://books.google.ro/books?id=8c_mBQAAQBAJ&pg=PA15&lpg=PA15&dq=malykin+silberstein+coriolis&source=bl&ots=JrMqF2vmto&sig=xCnMB4hL_J_ESg9Xdfhye1ahVjA&hl=en&sa=X&ved=2ahUKEwiE0ZDWxeXeAhXwkYsKHYxwBMYQ6AEwCXoECAUQAQ#v=onepage&q=malykin%20silberstein%20coriolis&f=false

CAN YOU READ ENGLISH RABINOZ?
Perfectly well, thank you!

Quote from: sandokhan
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.

No, Dr. Silberstein derives the curved light paths without reference to or even mentioning the Coriolis effect though it is, in effect, the Coriolis effect.
But, rather than being "directly related to the area of the interferometer" he shows that the deviations from straight are so slight that it cannot affect the result.

Here read this from p 298:


Now read exactly what Grigorii B. Malykin and Vera I. Pozdnyakova say in:[/i]
       

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, "There are recent studies by Bashkov [74, 75, 78] where the erroneous statement of Silberstein [798, 799] is repeated."
In my opinion that "erroneous statement of Silberstein" is simply a case of Grigorii B. Malykin and Vera I. Pozdnyakova quite innocently mistaking Silberstein's words.

Grigorii B. Malykin and Vera I. Pozdnyakova might interpret what Dr Silberstein wrote a little differently but nowhere suggest that Dr Silberstein is deriving the Coriolis effect.

Quote from: sandokhan
Here is another reference, explicitly deriving the CORIOLIS EFFECT, same formula as that published by Dr. Silberstein:

Spinning Earth and its Coriolis effect on the circuital light beams

I'm perfectly aware of that paper but I'll not go into it now.

And I'll ignore the repeated incorrect derivation of your famous "global/generalized Sagnac effect formula"!

Quote from: sandokhan
It can easily be seen to be incorrect by simply centring the loop over the equator

Is this supposed to be a joke?
Sure it's a joke, ON YOU! YOU claimed that "The global/generalized Sagnac effect formula, the most important in all of physics" but it doesn't work!
The global/generalized Sagnac effect formula, the most important in all of physics:
https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg2117351#msg2117351

Quote from: sandokhan
If you now have equal radii and equal velocities, the old Sagnac formula comes into play at once.
Really? Your "Sagnac effect formula, the most important in all of physics" needs an exception in the first trial!

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.

But, I'd agree with Michelson, Silberstein, MathPages, E.J. Post, Maraner and Zendri long before your result.
« Last Edit: August 26, 2019, 03:37:12 PM by rabinoz »

Re: Bob Knodel and the laser ring gyroscope
« Reply #50 on: February 20, 2019, 02:36:22 PM »
Anyone who claims "Everyone who ever worked on X in physics or math was wrong except me!" deserves special scrutiny.

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sandokhan

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Re: Bob Knodel and the laser ring gyroscope
« Reply #51 on: February 20, 2019, 09:46:21 PM »
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.

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.

Here is a second reference which explicitly derives the CORIOLIS EFFECT formula:

Spinning Earth and its Coriolis effect on the circuital light beams

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


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.

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: if v1 = v2 and l1 = l2, then quite simply, my formula becomes dt = 4VL/c2, which is the local Sagnac effect formula.

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 upper 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.



I have the loops, you have nothing.

I use the signs correctly, you use the same sign for beams traveling in opposite directions.


You have ONE interferometer, with two loops: a counterclockwise loop and a clockwise loop.

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.


Make sure you understand the definition of the term LOOP.

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

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.

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.




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.

so the total delay for the A > D > C > B > A path (clockwise) should be: l1/(c - v1) + l2/(c + v2)

It can't be.

Now, you have TWO OPPOSING DIRECTIONS, while you assign the SAME SIGN, a catastrophic mistake if you are seeking the SAGNAC EFFECT formula.

Moreover, where are the loops ??

What in effect you are saying is this:

A > D > C combined with A > B.

NO LOOP AT ALL.

A tremendous error committed by Michelson.

The SAGNAC EFFECT requires TWO LOOPS. Now we are dealing with the clockwise loop.

What you want is this:

A > D > C > B > A path (clockwise path)

Now we have a loop.

This loop includes two opposite directions, two opposite terms/components.

That is why one must a positive sign, while the other must have a negative sign.

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, using topology it can be proven that the SAGNAC EFFECT formula must include the velocity and the path of the arms of the interferometer, while the CORIOLIS EFFECT formula must include the area:

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg2023979#msg2023979

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg2044039#msg2044039 (the SAGNAC EFFECT can only be described by the SU(2) group of transformations)



Only when the interferometer is being rotated does the TOPOLOGICAL PATCH condition/MULTIPLY-CONNECTED domain come into play.

No rotation, no multiply-connected domain.

Two very different topological situations.



« Last Edit: February 20, 2019, 10:27:54 PM by sandokhan »

Re: Bob Knodel and the laser ring gyroscope
« Reply #52 on: February 20, 2019, 11:53:24 PM »
On page 298 there are two DIFFERENT derivations, a fact quite obvious for anyone but yourself.
Which in no way makes them different things. You can have many derivations for the same thing.

The time or phase shift created by the coriolis effect for 2 beams of light counter-propagating around a ring interferometer IS the Sagnac effect.
You are yet to show any cause to think they are not.
Instead all you have provided is an incorrect derivation producing an incorrect formula.

You have repeatedly avoiding/ignored the massive problems with this and just kept asserting the same nonsense.
You have also refused to do very simple things to show your understanding by starting from the basics.

Your formula is wrong. It has clearly been shown to be wrong. It has clearly be shown to not be linked to rotation.

Now like I said, start from a simple stationary loop.
How long does it take light to propagate around a stationary loop?
Once you have that we can move on to how long it takes for light to propagate around a rotating loop, for each of with and against the rotation.
Then we can focus on the difference, i.e. the Sagnac effect.

So can you start from the basics and show how long it takes for the light to propagate around a stationary loop?

*

sandokhan

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Re: Bob Knodel and the laser ring gyroscope
« Reply #53 on: February 21, 2019, 01:39:01 AM »
For a stationary loop there is no difference in transit times.

NO SAGNAC EFFECT AT ALL.

Sagnac effect = the rotation of the interferometer

Therefore, the RE are avoiding the very definition of the Sagnac effect: rotation of the interferometer, two counterpropagating beams of light, two loops.

No multiply-connected domain for a stationary interferometer, that is why I included the facts from topology in my previous message so that these issues are very well understood.


"It doesn't matter how beautiful your theory is, it doesn't matter how smart you are. If it doesn't agree with experiment, it's wrong."

Richard P. Feynman

Here is the experiment performed by Professor Yeh, one of the top experts in the world in laser optics, which confirms the CORRECT SAGNAC EFFECT formula:



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

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


The formula derived by Silberstein, Post and Michelson has nothing to do with the Sagnac effect.

Very easy to prove.



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.


so the total delay for the A > D > C > B > A path (clockwise) should be: l1/(c - v1) + l2/(c + v2)

It can't be.

Now, you have TWO OPPOSING DIRECTIONS, while you assign the SAME SIGN, a catastrophic mistake if you are seeking the SAGNAC EFFECT formula.

Moreover, where are the loops ??

What in effect you are saying is this:

A > D > C combined with A > B.

NO LOOP AT ALL.

A tremendous error committed by Michelson.

The SAGNAC EFFECT requires TWO LOOPS. Now we are dealing with the clockwise loop.

What you want is this:

A > D > C > B > A path (clockwise path)

Now we have a loop.

This loop includes two opposite directions, two opposite terms/components.

That is why one must a positive sign, while the other must have a negative sign.

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.

I have the loops, you have nothing.

I use the signs correctly, you use the same sign for beams traveling in opposite directions.


You have ONE interferometer, with two loops: a counterclockwise loop and a clockwise loop.

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.


Make sure you understand the definition of the term LOOP.

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

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.

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.



The CORIOLIS EFFECT and the GLOBAL SAGNAC EFFECT are two different phenomena: they require two different formulas.

Professor Yeh's experiment, peer-reviewed at the highest possible level, in the Journal of Optics Letters proves I am correct.


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


*

sandokhan

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Re: Bob Knodel and the laser ring gyroscope
« Reply #54 on: February 21, 2019, 01:57:16 AM »
The CORIOLIS EFFECT is a physical effect: the modification of the paths of the light beams. It compares two arms of the interferometer, no loops at all.

The SAGNAC EFFECT is an electromagnetic effect: the modification of the velocity of the light beams, it compares two loops.

They are not the same phenomena, they require different formulas.

Here is the very easy derivation of the CORIOLIS EFFECT for light beams:

Spinning Earth and its Coriolis effect on the circuital light beams

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

The final formula is this:

4AΩsinΦ/c^2

This is the CORIOLIS EFFECT formula.

By contrast the formula derived by me and proven to be correct by Professor Yeh is this:

2(V1L1 + V2L2)/c2


No loops for the CORIOLIS EFFECT, just a comparison of the two arms, a physical effect.



The promise made by A. Michelson, "the difference in time required for the two pencils to return to the starting point will be...", never materialized mathematically.

Instead of applying the correct definition of the Sagnac effect, Michelson compared TWO OPEN SEGMENTS/ARMS of the interferometer, and not the TWO LOOPS, as required by the exact meaning of the Sagnac experiment.

As such, his formula captured the Coriolis effect upon the light beams.

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.

Let us take a look at the humongous error perpetrated by Michelson and also by the RE.

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.


My analysis is correct, as also proven by Professor Yeh.


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sandokhan

  • Flat Earth Sultan
  • Flat Earth Scientist
  • 6764
Re: Bob Knodel and the laser ring gyroscope
« Reply #55 on: February 21, 2019, 02:55:34 AM »
Which in no way makes them different things.

Are you telling your readers that you do not understand the difference between the analysis for light paths based on FERMAT'S PRINCIPLE and the CORIOLIS EFFECT?

The analysis in terms of FERMAT'S PRINCIPLE starts on page 293 and ends on page 298:

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

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.

« Last Edit: February 21, 2019, 02:59:41 AM by sandokhan »

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zork

  • 3319
Re: Bob Knodel and the laser ring gyroscope
« Reply #56 on: February 21, 2019, 03:08:52 AM »
 I don't really care if you understand what you copy/paste here but thank you for demonstrating how out of date you are. Some hundred and more years always.
Rowbotham had bad eyesight
-
http://thulescientific.com/Lynch%20Curvature%202008.pdf - Visually discerning the curvature of the Earth
http://thulescientific.com/TurbulentShipWakes_Lynch_AO_2005.pdf - Turbulent ship wakes:further evidence that the Earth is round.

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sandokhan

  • Flat Earth Sultan
  • Flat Earth Scientist
  • 6764
Re: Bob Knodel and the laser ring gyroscope
« Reply #57 on: February 21, 2019, 03:13:02 AM »
My formula is totally up to date.

"It doesn't matter how beautiful your theory is, it doesn't matter how smart you are. If it doesn't agree with experiment, it's wrong."

Richard P. Feynman

Here is the experiment performed by Professor Yeh, one of the top experts in the world in laser optics, which confirms the CORRECT SAGNAC EFFECT formula:



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

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

Re: Bob Knodel and the laser ring gyroscope
« Reply #58 on: February 21, 2019, 03:45:38 AM »
For a stationary loop there is no difference in transit times.
I know. Why do you need to keep repeating this?
This is just a simple starting point. If you can't do the derivation correctly for a stationary loop you have no hope of doing it correctly for a rotating loop.
So start with the basics.
How long does it take for light to propagate around a stationary loop?
Can you answer that.

NO SAGNAC EFFECT AT ALL.
Sagnac effect = the rotation of the interferometer
Yes, that is correct. Your formula is not the Sagnac effect at all.
Sagnac is for a rotating interferometer, but yours works with any motion, even linear motion without any rotation at all.
That is a big indicator that your formula isn't for the Sagnac effect.

Your errors have already been pointed out and you just ignore them.
Repeating the same BS wont magically make you correct.
Ignoring what other people say wont magically make you correct.


So like I said, start with the basics.
How long does it take for light to propagate around a stationary loop?

Can you figure that out? If not, anything more advanced is pointless.
« Last Edit: February 21, 2019, 03:47:58 AM by JackBlack »

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Zaphod

  • 137
Re: Bob Knodel and the laser ring gyroscope
« Reply #59 on: February 21, 2019, 03:49:51 AM »
Go on Sandy, have a go at answering JBs step-by-step derivation. As you’re so sure you’re right, it will be easy to show he’s a fool no?