The first shill has been confined to the AR.
The second shill can no longer use trolling, stalling, spamming to escape the final conclusion: my formula is correct.
But you are the one who is repeatedly trolling, stalling and spamming.
When you first brought up your claims regarding the Sagnac effect you fully accepted that the formula was 4*A*omega/c^2, and accepted that it applied to any interferometer, but claimed that the area used was the area of the "orbit", such that the orbital Sagnac was based on the area of Earth's orbit around the sun, while for Earth's rotation it would be the area or a circle centred on Earth's axis and passing through the interferometer.
But you had that argument completely refuted and I provided a derivation from first principles for an annular interferometer showing it was the area of the interferometer which mattered, not the orbit.
You were completely unable to refute that and spammed a bunch more of baseless assertions before running away.
You then came up with a few false derivations which were also refuted.
For example, you brought in fibre optic conveyors, which while having a similar origin, are fundamentally different.
The apparatus no longer moves as one, it moves relative to itself.
It is irrelevant to the discussion on the Sagnac effect for a simple ring interferometer.
It also doesn't show what you claim.
The shift is not proportional to some fictitious absolute velocity of the interferometer. It is proportional to the velocity of the source/detector relative to the conveyor.
So it doesn't prove what you are claiming at all.
Another important distinction is that it uses the entire length, not just the length of 2 arms.
So no, it doesn't apply.
You have been completely unable to justify your derivation at all. As such it is proof of nothing. It remains a pile of refuted garbage.
Meanwhile, you have been completely unable to refute my derivation at all (either of them).
This formula does not include the area at all, and is proportional to the VELOCITY of the light beams (and thus is proportional to the RADIUS of rotation).
Two different formulas, featuring two different physical descriptions.
This means that the formulas must be describing TWO DIFFERENT PHYSICAL PHENOMENA.
Not quite. It is the velocity, not of the light beams, but of the source/detector relative to the conveyor.
There is no radius here. There is no rotation, except where the conveyor turns a corner.
A fibre optic conveyor, while similar to a Sagnac interferometer, is not one. Its formula is not that for the Sagnac effect.
In order to derive this formula, you compared two sides, not two loops, as required by the definition of the SAGNAC EFFECT.
No, that would be what you have repeatedly done.
I have compared 2 loops (really just the one loop, with 2 beams of light going in different directions).
I added up the time taken for each loop to complete, and then I found the difference between those 2 times, to find the shift.
Instead, you looked at the time taken for a beam of light to traverse each arm and found the difference in those times, not corresponding to anything in reality.
When you have the centre of the interferometer aligned with the centre of rotation, you don't change sign with direction. Instead you just add up the times.
Completely wrong.
No, completely true.
You take the time taken to traverse each tiny section of the arm and add them all up to find the total time taken to traverse the loop.
But now you want to throw that out the window and find some difference in this time which makes no sense at all.
THEY SUBSTRACT THE TIMES.
For the 2 different counterpropogating beams, not for the same beam.
For a single beam they add up the time.
This matches what I did.
We have one beam which traverses the loop in one direction. The time taken to traverse the arms of interest are the following:
l1/(c - v1)l2/(c + v2)Notice the difference in sign of the velocity term as in one case the light is moving with the arm, while in the other it is moving against it.
This means for this beam of light, the total time (that we care about) is:
l1/(c - v1) +
l2/(c + v2).
Likewise we do the same for the other BEAM OF LIGHT and end up with:
l2/(c - v2) +
l1/(c + v1).
Note: these correspond to the time taken for the clockwise path and counterclockwise path for a simple ring interferometer:
2πR(1/(c - v))
2πR(1/(c + v))
Notice that they add up the single light path.
They don't do something like this:
"Well for half of the path it goes in one direction, so it has a Sagnac phase component of πR(1/(c - v)), but then for the other half it goes in the opposite direction so it has a phase component of -πR(1/(c - v)), giving us a total of 0."
Now, we find the difference in time between the light beams to find the time shift at the detector. This is where the difference comes in. It is the difference in time for the 2 beams of light:
dt=
l1/(c - v1) +
l2/(c + v2) - (
l2/(c - v2) +
l1/(c + v1))
=
l1/(c - v1) -
l1/(c + v1) +
l2/(c + v2) -
l2/(c - v2)=l
1(c + v
1 - c + v
1)/(c
2 - v
12) + l
2(c - v
2 - c + v
2)/(c
2 - v
22)
=2 l
1 v
1/(c
2 - v
12) - 2 l
2 v
2/(c
2 - v
22)
Then when you note v is tiny compared to c, this simplifies to:
dt=2 l
1 v
1/c
2 - 2 l
2 v
2/c
22(l
1 v
1 - l
2 v
2)/c
2Just like I have proven countless times.
And when you note that this needs to be an annular interferometer for the other arms to not matter, this results in the same old formula:
dt=4Aw/c^2
Meanwhile you are subtracting the times for a single beam which makes no sense and corresponds to nothing in reality.
Opposite directions, therefore WE SUBSTRACT THE DIFFERENCE IN TIME TRAVEL.
No, 2 different beams which produce an interference pattern, so we find the difference in time taken for the beams. We don't find a difference in time taken for a single beam to traverse the different arms as that corresponds to nothing in reality.
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
Again, THIS MAKES NO SENSE!
Again, if you have someone who can run at a speed of c, who wants to run back and forth down a track of length l, is the time required:
t1=l/c+l/c, where we don't change the sign for opposite spatial directions, as both paths move forwards in time, or is it:
t2=l/c-l/c, where we change the sign for opposite spatial directions, meaning regardless of how long the path is it takes no time to run back and forth down it because you magically go backwards in time when you go backwards down the path?
According to your nonsense, you find the difference and can end up with no time taken at all.
According to almost everyone, you add the times together to find the total time.
Do you understand what a total is? It is what you get when you add up the components. It is not a difference.
For any individual light path you add the time taken for it to traverse the individual components to find the total time taken.
The only time you would subtract is if the light beam magically travels backwards in time.
This is why your derivation amounts to nothing more than pure bovine excrement.
Because you are trying to find a total time by finding a difference in time taken.
Do you not realise that they are vastly different things?
I was able, for the first time in history, to derive the corresponding SECOND formula for the Michelson-Gale experiment.
And have you considered that that the reason for that is because you have made a massive mistake? A mistake which has been pointed out to you countless times, which you have repeatedly ignored?
And again, your formula can be shown to be pure nonsense by also considering what happens with a rectangular interferometer travelling in uniform linear motion (i.e. without any rotation at all).
According to simple symmetry the 2 light paths will be equivalent and thus there will be no shift as neither beam of light can get ahead of the other.
i.e. the light travelling clockwise along arm 1 will be the same as the light travelling counterclockwise along arm 3, and so on for all the other arms. This means the total shift must be 0.
My formula indicates the following:
dt=2(l v - l v)/c
2 = 0.
Meanwhile your nonsense indicates the following:
2(l v + l v)/c
2 = 2 l v / c
2This is a massive problem for you.
Your formula produces an incorrect result in one of the simplest cases.
This shows your formula is nonsense.
So no, you haven't made some historic break through. You have just made a massive mistake.
This is what you actually need to deal with. To summarise:
You need to refute my derivation which you have been unable to show any problem with.
You need to explain why you are finding the difference in time taken for a single beam of light, which corresponds to nothing in reality, and pretending it is the total time taken.
You need to explain why an interferometer which isn't rotating at all and instead is moving with uniform linear motion has a Sagnac shift, when symmetry demands it can't (and my formula says it can't, and the formula produced by so many people says it can't).
And these are all problems which have been pointed out to you before which you have chosen to ignore.