Sure, for the case being discussed here we have rotation (MGX).

Sagnac effect while in uniform/translational/linear motion is a proven fact.

Professor Ruyong Wang, in two well-designed experiments showed unambiguously that an identical Sagnac effect appearing in uniform radial motion occurs in linear inertial motion.

He tested the travel-time difference between two counter-propagating light beams in uniformly moving fiber.

The travel-time difference was found to be:

Δt = 2vΔL/c^2

where ΔL is the length of the fiber segment moving with the source and detector at a v, whether the segment was moving uniformly or circularly.

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

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

"For a circular path of radius R, the difference between the different time intervals can also be represented as Δt = 2vl/c^2, where v = ΩR is the speed of the circular motion and l = 2πR is the circumference of the circle.

The travel-time difference of two counterpropagating light beams in moving fiber is proportional to both the total length and the speed of the fiber, regardless of whether the motion is circular or uniform.

In a segment of uniformly moving fiber with a speed of v and a length of Δl, the travel-time difference is 2vΔl/c^2."

Here is another demonstration.

*To assign a negative time you would be saying that the light is travelling backwards in time as it propagates along the loop.*You don't have a loop and there is no negative time.

You simply do not understand the SAGNAC EFFECT.

The RE standard for the Sagnac effect:

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

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

l

_{1} is the upper arm.

l

_{2} 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):

l

_{1}/(c - v

_{1})

-l

_{2}/(c + v

_{2})

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

l

_{2}/(c - v

_{2})

-l

_{1}/(c + v

_{1})

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

l

_{1}/(c - v

_{1}) - l

_{2}/(c + v

_{2})

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

l

_{2}/(c - v

_{2}) - l

_{1}/(c + v

_{1})

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:

l

_{1}/(c - v

_{1})

l

_{2}/(c + v

_{2})

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):

{l

_{1}/(c - v

_{1}) - l

_{2}/(c + v

_{2})} - (-){l

_{2}/(c - v

_{2}) - l

_{1}/(c + v

_{1})} = {l

_{1}/(c - v

_{1}) - l

_{2}/(c + v

_{2})}

+ {l

_{2}/(c - v

_{2}) - l

_{1}/(c + v

_{1})}

Rearranging terms:

l

_{1}/(c - v

_{1}) - l

_{1}/(c + v

_{1})

+ {l

_{2}/(c - v

_{2}) - l

_{2}/(c + v

_{2})} =

2(v

_{1}l

_{1} + v

_{2}l

_{2})/c

^{2}Exactly the formula obtained by Professor Yeh:

φ = -2(φ

_{2} - φ

_{1}) = 4π(R

_{1}L

_{1} **+** R

_{2}L

_{2})Ω/λc = 4π(V

_{1}L

_{1} + V

_{2}L

_{2})/λc

Since Δφ = 2πc/λ x Δt, Δt = 2(R

_{1}L

_{1} **+** R

_{2}L

_{2})Ω/c

^{2} = 2(V

_{1}L

_{1} + V

_{2}L

_{2})/c

^{2}CORRECT SAGNAC FORMULA:

**2(V**_{1}L_{1} + V_{2}L_{2})/c^{2}**A stationary loop does not involve the Sagnac effect. For the Sagnac effect only arms l1 and l2 come into play, as simple as this. Only the arms involved in the rotation analysis are used, and the Sagnac effect has nothing to do with a stationary interferometer. You are trolling the upper forums, desperately trying to avoid the final conclusion: I am right, you are wrong.**

The analysis for the Sagnac effect is completely different than for a stationary interferometer, yet this very simple fact seems to escape your attention.*According to your "reasoning" if the loop is stationary, it takes no time for the light to travel arm 1 and arm 3 as the light magically moves backwards in arm 3.*We are not concerned here AT ALL with the stationary case: my analysis only applies to the rotational case. In the SAGNAC EFFECT, only arms l1 and l2 come into play to provide the time differences: we are now rotating the interferometer. What you are stating to your readers is that you do not understand the workings of the Sagnac effect.

*With my derivation you have 2 light paths, one path goes around A>D>C>B>A, with each component adding time taken (but with the vertical arms ignored).*

The other path goes A>B>C>D>A.

This produces the 2 times provided:

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

l1/(c + v1)+l2/(c - v2)You are lying through your teeth jack, and that is not nice.

What is you have is this:

Path 1 - A>B, D>C.

Path 2 - C>D, B>A

YOU ARE DISREGARDING THE CORRECT DEFINITION OF THE SAGNAC EFFECT IN FULL VIEW.

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

You are comparing two sides, NOT TWO LOOPS.

You admitted that those beams have opposite directions.

*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 you have opposite directions, you must use different signs.

A humongous error on your part.

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.

*This is meant to be a time difference, so it amounts to one path of light, with a time of:*

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

and a second path with a time of:

l1/(c + v1) + l2/(c + v2)THOSE ARE NOT TWO LOOPS, BUT TWO SEGMENTS.

I am not comparing two segments at all, you are.

The correct way is to compare two loops.

The definition of the Sagnac effect involves two loops.

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 +

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):

{l

_{1}/(c - v

_{1}) - l

_{2}/(c + v

_{2})} - (-){l

_{2}/(c - v

_{2}) - l

_{1}/(c + v

_{1})} = {l

_{1}/(c - v

_{1}) - l

_{2}/(c + v

_{2})}

+ {l

_{2}/(c - v

_{2}) - l

_{1}/(c + v

_{1})}

Rearranging terms:

l

_{1}/(c - v

_{1}) - l

_{1}/(c + v

_{1})

+ {l

_{2}/(c - v

_{2}) - l

_{2}/(c + v

_{2})} =

2(v

_{1}l

_{1} + v

_{2}l

_{2})/c

^{2}Exactly the formula obtained by Professor Yeh:

φ = -2(φ

_{2} - φ

_{1}) = 4π(R

_{1}L

_{1} **+** R

_{2}L

_{2})Ω/λc = 4π(V

_{1}L

_{1} + V

_{2}L

_{2})/λc

Since Δφ = 2πc/λ x Δt, Δt = 2(R

_{1}L

_{1} **+** R

_{2}L

_{2})Ω/c

^{2} = 2(V

_{1}L

_{1} + V

_{2}L

_{2})/c

^{2}CORRECT SAGNAC FORMULA:

**2(V**_{1}L_{1} + V_{2}L_{2})/c^{2}My formula is in complete agreement with the formula published by Professor Yeh in the Journal of Optics Letters.

Your formula is in complete agreement with the published CORIOLIS EFFECT formula:

Spinning Earth and its Coriolis effect on the circuital light beams

http://www.ias.ac.in/article/fulltext/pram/087/05/0071The final formula is this:

4AΩsinΦ/c^2

This is the CORIOLIS EFFECT formula, not the SAGNAC EFFECT formula.

You are about to become a flat earth believer, no doubt about that.