sandokhan lies regarding the Sagnac effect

It is measured with angular velocity, given dt=4*A*w/c^2.

Then, our discussion is over.

You lose.

The Sagnac effect is a function purely of the linear velocity NOT THE ANGULAR VELOCITY of the receiver on the circular path.

The linear velocity of the rotation of the earth is 0.465 km/s
The linear velocity of the orbit of the earth is 30 km/s

Take the linear velocity ratios
30/ 0.465 = 64.5

The ratio of the Sagnacs is based only on the linear speeds of each.

It is not at all based on the ratios of the angular speeds.


If you use the SAME AREA for the annular sector, that means you are effectively saying that R2 = r2 and R1 = r1.

By using the SAME AREA the ratio will become a comparison of the angular velocities of the loops.

But that is not Sagnac.

Sagnac is a comparison of the linear velocities (circumferential tangential speed at the end point).

It is measured with angular velocity, given dt=4*A*w/c^2.
Then, our discussion is over.

Not until you answer the question, as the whole point of this debate has been to determine this very question.

The Sagnac effect is a function purely of the linear velocity NOT THE ANGULAR VELOCITY of the receiver on the circular path.

PURE BULLSHIT!
The Sagnac effect is a function of the angular velocity and the area of the loop.

Do you accept that the area of an annular sector is given by:
A=θ_o(R₂² - R₁²)/2
or equally that the following equation holds:
2A=θ_o(R₂² - R₁²)

YES OR NO?

If you can't answer it, shut up.

The Sagnac effect is a function of the angular velocity and the area of the loop.

Let's put your statement to the test.

The translational/linear Sagnac effect IS A FACT OF SCIENCE.

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."

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

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

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




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

DUFOUR-PRUNIER EXPERIMENT

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

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

In all cases of this experimental test, the Sagnac effect was the same.

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

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

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


You have just been disproven.

The Sagnac effect is a function of the angular velocity and the area of the loop.

Let's put your statement to the test.
The translational/linear Sagnac effect IS A FACT OF SCIENCE.
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.

But, Wang is not testing absolute linear motion.

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

<< I've read all that stuff, no problem, >>

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."

That is not a Fibre Optic Gyroscope and Wang does not call it one, he calls it a Fibre Optic Conveyor.
An FOG has a fixed geometry that rotates as a whole, in the FOC the fibre, the source and the detector all move move relative to the laboratory.

A big difference, which Wang clearly recognises and describes.

The FOC dies not measure absolute linear velocity and Wang does not claim that it does.

You might like to read, General relativistic Sagnac formula revised, Paolo Maraner, Jean-Pierre Zendri, but of course you don't accept GR.

Now, Present your derivation of time difference in a genuine Sagnac gyroscope configuration.

Have fun.

PS If your are raving on with all this stuff in an attempt to disprove the earth's rotation and orbiting the sun, forget it!
      All of these learned authors you quote from are convinced of the heliocentric globe and a universe of cosmic dimensions.
      Even your idol,  Nicola Tesla wrote strongly supporting heliocentric globe. Sure, some differ on relativity and the cause of gravitation.

but of course you don't accept GR.

There is no such thing as GR.

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg769750#msg769750

The original set of Maxwell equations are invariant under Galilean transformations, thus eliminating the GR completely:

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1918701#msg1918701

You haven't done your homework on the Sagnac.

There have been multiple attempts to explain in terms of theories which are much more complex than GR.

J.H. Field (STR, time dilation), R. Klauber (NTO, non-time orthogonal metric in spacetime), A. Tartaglia/M.L. Ruggiero/G. Rizzi (flat spacetime, Aharonov-Bohm effect), P. Maraner/J.P. Zendri (Minkowski spacetime, Aharonov-Bohm effect), S.J.G. Gift (GTR), M.F. Yagan (cumulative Doppler effect using STR), A.G. Kelly (universal relativity), A. Sfarti (STR), F. Amador (Evans field theory), F. Selleri (specific set of spacetime transformations which lead directly to MLET, Modified Lorentz Ether Theory).


The method employed by your friend to actually build the interferometer in the shape of an annular sector makes the entire project UNFEASIBLE.

It cannot be built. The orbital Sagnac and the rotational Sagnac simply cannot be compared at all using this method.

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

Thus dt=2*omega*alpha*R22/c2-2*omega*alpha*R1^2/c2
=2*omega*alpha*(R22-R12)/c2

Now, can we simplify this any more?
I know, lets work out the area.
Note that alpha has always been in radians.
A circle has an area pi*r2.
This circle is a circular sector which subtends an angle of 2*pi.
If it only subtends an angle of alpha, then it will only have an area of alpha/(2*pi) of the circle.
Thus A (for a circular sector) is (alpha/(2*pi))*pi*r2
=alpha*r2/2
This means from the center to the outer arc you have an area of:
A2=alpha*R22/2.
And for the inner one you have an area:
A1=alpha*R12/2.
Thus the area between them is:
A=alpha*R22/2-alpha*R12/2
=alpha*(R22-R12)/2
Thus 2*A=alpha*(R22-R12).

Notice that was in the formula above?
That means we can sub it in.
As a reminder we had:
dt=2*omega*alpha*(R22-R12)/c2
By subbing in the above we get:
dt=2*omega*2*A/c2
And thus:
dt=4*A*omega/c2


One can only SUB IT IN simply to express the formula in a different manner, one which involves the area.

But this is done with many other interferometers in the shape of a rectangle, triangle.

You can express the Sagnac as linear velocity or as angular velocityxarea, BUT YOU CANNOT COMPARE TWO SAGNACS USING ANGULAR VELOCITY, because then you'll end up comparing the angular velocities of the loop, and NOT the circumferential tangential speed of the end point.

This is important to understand.


Without having an intelligent clue as to what his method of constructing an interferometer actually means, our friend continued his derivation:

Thus 4*A/c^2 is constant, thus you get the relation:
dt=k*w, where k=4*A/c^2.
As such, the shift is directly proportional to the angular velocity.

Thus comparing 2:
dto=k*wo
dtr=k*wr
And thus:
dto/dtr=k*wo/k*wr=wo/wo=1/365.

But such in interferometer CANNOT BE BUILT AT ALL.

No comparison can be made between the orbital Sagnac and the rotational Sagnac.

TAKE A LOOK AT WHAT IS ACTUALLY BEING REQUIRED.

1.00007811833281, 1.000078125 (s1 and s2), your data for s1o and s2o.

An interferometer which has to be built to a precision of 0.00000000666719km.

That is, down to the level of actual MICRONS. It is not feasible at all.

If you are off by a couple of digits, say 0.999999666, then you get this:

R1 = 149,999,949.9 km

Your R1 value was 149,999,999

A difference of 49.1 km.


You cannot substitute an area involving r2 and r1 into a formula which features R2 and R1.

θo = angle subtended by the two radii, R2 and R1 = orbital angle

s2 = R2 x θo

s1 = R1 x θo


θr = angle subtended by the two radii, r2 and r1 = rotational angle

s2 = r2 x θr

s1 = r1 x θr


R2 - R1 = r2 - r1

r2 x θr = R2 x θo

r1 x θr = R1 x θo

r2/r1 = R2/R1

(r2 x R1) = (r1 x R2)

Since the two areas must be equal,

r1/R1 = (r2 + r1)/(R2 + R1)



Right away, one runs into huge problems with this scenario.

R2 = r2 - r1 + R1

(r2 x R1) = r1r2 - r12 + (R1 x r1)

r2(R1 - r1) = r1(R1 - r1)

So we end up with: r2 = r1, which is impossible


YOU COMPARE SAGNAC EFFECT USING LINEAR VELOCITY (CIRCUMFERENTIAL TANGENT SPEED AT THE END POINT) AND NOT THE ANGULAR VELOCITY.

Here, let me exemplify how to CORRECTLY calculate the Sagnac for a triangular shaped interferometer.

C will be the center of a circle, where the triangle will be inscribed.

Sides BC and AC are equal (isosceles triangle).

Points A and B do not have to touch the circle.

Side AB is closest to the side of the circle.


dt = (tAC + tCB + tBA) - (tAB + tBC + tCA)

The travel times along AC and BC will not differ.

Then, tAB - tBA = 4ωA/c2, where A the area of the triangle.

4ωA/c2 = (4ωhL/2)/c2 where h is the height of the triangle to point D (midpoint between A and B), and L is the distance from A to B.

Then the final answer:

dt = 2vL/c2 where v = the speed at point D, the midpoint between A and B


dto/dtr=[4*wo*Ao/c2]/[4*wr*Ar/c2]
=wo/wr

Let us now proceed as jack requires.

dto = 2wohL/c2

dtr = 2wrhL/c2

Then, obviously we would be left with a ratio of the angular velocities of the loops, and NOT the circumferential speeds at the end point.

Let us now proceed correctly.

dto = 2voL/c2

where ωR = 30km/s (center of loop located at the Sun, since we are calculating the correct orbital Sagnac effect)

dtr = 2vrL/c2

where ωr = 0.465km/s (calculation for the rotational Sagnac)

Then the final ratio will be: vo/vr = ~60.

The linear velocities (the circumferential tangent speeds at the end point) are being compared and NOT the angular velocities.

It is the circumferential speeds at the end point which are being compared, and not the angular velocities of the loop itself.

The Sagnac effect is a function of the angular velocity and the area of the loop.

Let's put your statement to the test.

Fine, lets do that. BY ANSWERING THE QUESTION AND SEEING IF THE DERIVATION IS CORRECT!!!
Not with you pathetic distractions.

ANSWER THE QUESTION:
Do you accept that the area of an annular sector is given by:
A=θ_o(R₂² - R₁²)/2
or equally that the following equation holds:
2A=θ_o(R₂² - R₁²)

YES OR NO?

If you can't answer it, shut up.

Take a look at my previous message. Your question and much more (so as not waste everybody's time here) was answered.

The method employed by your friend to actually build the interferometer in the shape of an annular sector makes the entire project UNFEASIBLE.

It cannot be built. The orbital Sagnac and the rotational Sagnac simply cannot be compared at all using this method.

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

Thus dt=2*omega*alpha*R22/c2-2*omega*alpha*R1^2/c2
=2*omega*alpha*(R22-R12)/c2

Now, can we simplify this any more?
I know, lets work out the area.
Note that alpha has always been in radians.
A circle has an area pi*r2.
This circle is a circular sector which subtends an angle of 2*pi.
If it only subtends an angle of alpha, then it will only have an area of alpha/(2*pi) of the circle.
Thus A (for a circular sector) is (alpha/(2*pi))*pi*r2
=alpha*r2/2
This means from the center to the outer arc you have an area of:
A2=alpha*R22/2.
And for the inner one you have an area:
A1=alpha*R12/2.
Thus the area between them is:
A=alpha*R22/2-alpha*R12/2
=alpha*(R22-R12)/2
Thus 2*A=alpha*(R22-R12).

Notice that was in the formula above?
That means we can sub it in.
As a reminder we had:
dt=2*omega*alpha*(R22-R12)/c2
By subbing in the above we get:
dt=2*omega*2*A/c2
And thus:
dt=4*A*omega/c2


One can only SUB IT IN simply to express the formula in a different manner, one which involves the area.

But this is done with many other interferometers in the shape of a rectangle, triangle.

You can express the Sagnac as linear velocity or as angular velocityxarea, BUT YOU CANNOT COMPARE TWO SAGNACS USING ANGULAR VELOCITY, because then you'll end up comparing the angular velocities of the loop, and NOT the circumferential tangential speed of the end point.

This is important to understand.


Without having an intelligent clue as to what his method of constructing an interferometer actually means, our friend continued his derivation:

Thus 4*A/c^2 is constant, thus you get the relation:
dt=k*w, where k=4*A/c^2.
As such, the shift is directly proportional to the angular velocity.

Thus comparing 2:
dto=k*wo
dtr=k*wr
And thus:
dto/dtr=k*wo/k*wr=wo/wo=1/365.

But such in interferometer CANNOT BE BUILT AT ALL.

No comparison can be made between the orbital Sagnac and the rotational Sagnac.

TAKE A LOOK AT WHAT IS ACTUALLY BEING REQUIRED.

1.00007811833281, 1.000078125 (s1 and s2), your data for s1o and s2o.

An interferometer which has to be built to a precision of 0.00000000666719km.

That is, down to the level of actual MICRONS. It is not feasible at all.

If you are off by a couple of digits, say 0.999999666, then you get this:

R1 = 149,999,949.9 km

Your R1 value was 149,999,999

A difference of 49.1 km.


You cannot substitute an area involving r2 and r1 into a formula which features R2 and R1.

θo = angle subtended by the two radii, R2 and R1 = orbital angle

s2 = R2 x θo

s1 = R1 x θo


θr = angle subtended by the two radii, r2 and r1 = rotational angle

s2 = r2 x θr

s1 = r1 x θr


R2 - R1 = r2 - r1

r2 x θr = R2 x θo

r1 x θr = R1 x θo

r2/r1 = R2/R1

(r2 x R1) = (r1 x R2)

Since the two areas must be equal,

r1/R1 = (r2 + r1)/(R2 + R1)



Right away, one runs into huge problems with this scenario.

R2 = r2 - r1 + R1

(r2 x R1) = r1r2 - r12 + (R1 x r1)

r2(R1 - r1) = r1(R1 - r1)

So we end up with: r2 = r1, which is impossible

Take a look at my previous message. Your question and much more (so as not waste everybody's time here) was answered.

No. Deal with it one question at a time rather than trying to bury it in bullshit.
Also don't try and cover up the fact that you failed to answer this simple question by attempting to cover it in bullshit.

It is a simple yes or no question.
Now answer the question:
Do you accept that the area of an annular sector is given by:
A=θ_o(R₂² - R₁²)/2
or equally that the following equation holds:
2A=θ_o(R₂² - R₁²)

YES OR NO!
Be clear with your answer, without resorting to loads of bullshit.

but of course you don't accept GR.

There is no such thing as GR.
https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg769750#msg769750

Says the know-it-all, sandokhan, ignoring the numerous times it has been verified right from things like the anomalous precession planetary orbits to the predictions about clock rates as influenced by velocity and gravitational potential.

The original set of Maxwell equations are invariant under Galilean transformations, thus eliminating the GR completely:
https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1918701#msg1918701

Sure, back when the luminiferous aether was generally accepted, but numerous observations including the experiments of:

• Bradley (stellar aberration),
• Michelson and Morley (plus dozens of more recent versions),
• Bessel (stellar parallax),
• Hammar
• Michelson-Gale-Pearson experiment

Certainly seem to disprove the existence of the luminiferous aether and prove that the earth rotates and orbits the sun.

The linear velocities (the circumferential tangent speeds at the end point) are being compared and NOT the angular velocities.

It is the circumferential speeds at the end point which are being compared, and not the angular velocities of the loop itself.

If it is the "circumferential speeds at the end point which are being compared" it is not Sagnac.

If you claim that "it is the circumferential speeds at the end point which are being compared" why didn't that show up in the original Sagnac experiment in 1913, in the Michelson, Gale and Pearson experiment in 1925 or in the numerous ring laser gyroscopes used around the world today.

Please explain which words in the following that you cannot understand!

E. J. POST specifically states,
          "does not depend on the shape of the surface A;"
          "does not depend on the location of the centre of rotation;"
Care to explain in your own words what 
"does not depend on the shape of the surface A" and  "does not depend on the location of the centre of rotation" mean?

I am sure I can find many more references right back to Sagnac and Michelson and Gale asserting exactly the same thing.
In any case,
but did not include a path extending to the sun and back.

What I find so strange is that, as far as I can see, all your sources certainly believe the earth to be a sphere, and probably all strongly support a heliocentric universe.
They certainly differ on the matter of the existence of aether and how much it may or may not be dragged by massive bodies.

You might be interested to read:

Whereas the leading Sagnac time delay does not depend on the position of the center of rotation, relativistic corrections do. On the one side, this has the effect of amplifying the order of the first correction from (ωr/c)² to the generally much larger (ωR/c)².

From: General relativistic Sagnac formula revised, Paolo Maraner · Jean-Pierre Zendri - 1110.1643.pdf

But then GR doesn't exist for you, so that won't help your "orbital Sagnac".

Bottom line,
And from what I can see none of you references claim any experimental evidence to the contrary.

PS Read:

HOW COSMIC FORCES SHAPE OUR DESTINIES, ("Did the War Cause the Italian Earthquake") by Nikola Tesla also at — How Cosmic Forces Shape Our Destinies — ("Did the War Cause the Italian Earthquake"), New York American, February 7, 1915  in which he states: Quote from: Nicola Tesla NATURAL FORCES INFLUENCE US . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Accepting all this as true let us consider some of the forces and influences which act on such a wonderfully complex automatic engine with organs inconceivably sensitive and delicate, as it is carried by the spinning terrestrial globe in lightning flight through space. For the sake of simplicity we may assume that the earth's axis is perpendicular to the ecliptic and that the human automaton is at the equator. Let his weight be one hundred and sixty pounds then, at the rotational velocity of about 1,520 feet per second with which he is whirled around, the mechanical energy stored in his body will be nearly 5,780,000 foot pounds, which is about the energy of a hundred-pound cannon ball. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The sun, having a mass 332,000 times that of the earth, but being 23,000 times farther, will attract the automaton with a force of about one-tenth of one pound, alternately increasing and diminishing his normal weight by that amount Though not conscious of these periodic changes, he is surely affected by them. The earth in its rotation around the sun carries him with the prodigious speed of nineteen miles per second . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . From the above address. Sure, Nicola Tesla had a lot of "different ideas", but he most certainly did not believe in a flat stationary earth.

Well, it seems Sandy is still here, lying on other posts in the liars only section, yet he is unable to rationally defend his claims here.
So I will just continue without him.
Sandy, if you want to comment, start with the question you have refused to answer and take it one step at a time. If you try bringing any other garbage in, I will be ignoring it until you answer the question or flee yet again.

Now lets just continue where he left off, he has admitted this is the correct formula to determine the shift due to the Sagnac effect:
dt=2θ_oω_o(R₂² - R₁²)/c²

So we can go from there, I will discuss his lies later.
He has repeatedly refused to answer this simple qusetion:

Do you accept that the area of an annular sector is given by:
A=θ_o(R₂² - R₁²)/2
or equally that the following equation holds:
2A=θ_o(R₂² - R₁²)

This is because he must either lie and thus be humiliated by not even being able to get the area of an annular sector correct, or tells the truth and gets defeated.

But rather than just accept the answer, I will prove it.
The area of a circle is given by pi*r².
The area of a fraction (or sector) of a circle, is given by multiplying the area of the circle by the fraction the sector is, which can be determined by the angle it subtends (θ)
Using radians as angles (where 2pi radians=360 degrees), the total angle is 2pi, and thus the fraction is θ/2pi.
Thus a sector will have the area=(θ/2pi)*pi*r²=(θ*r²)/2

For an annular sector, this is simply the difference of 2 circular sectors which have 2 radii, R2 being the larger and R1 being the smaller. Thus we simply subtract the 2 and get:
A=(θ_o*R₂²)/2-(θ_o*R₁²)/2
A=θ_o*(R₂²-R₁²)/2
And thus:
θ_o(R₂²-R₁²)=2A

This is an irrefutable fact, which is why Sandokhan needs to avoid it like the plague as it destroys his claim.

We can easily sub this equation into the above formula for the shift:
dt=2θ_oω_o(R₂² - R₁²)/c²
dt=2θ_o(R₂² - R₁²)ω_o/c²
dt=2(2A)ω_o/c²
dt=4Aω_o/c²


He even accepts this when you have a triangular interferometer, such as here:
https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1982635#msg1982635

He accepts that the shift is given by:
dt=4Aω/c²

This shows that for a given interferometer (which has the same area), the shift is proportional to the angular velocity and thus the orbital Sagnac will be much smaller than the rotational one.

And this is where Sandy fails.
We notice that the shift is based purely upon the area of the interferometer and the angular velocity.
There is no dependence upon the linear velocity nor the distance from the centre of rotation.
This means Earths rotational Sagnac effect will be much larger than that due to its orbit (and so on for the galactic orbit, etc).
This means there is no missing orbital Sagnac effect, it is just tiny.

This is another irrefutable fact. But as it is slightly more complex and requires some math to show it, Sandy is happy to lie about it, even trying to manipulate the equation to pretend it is false.

He repeatedly wants to try and manipulate it to a form like:
dt=2Lv/c², pretending it still works fine.
But this requires the above triangular interferometer to go to the centre of the circle.
As such, it means you have completely different interferometers for the orbit and rotation.
It would require an interferometer reaching the sun.

Alternatively he will manipulate it to a form like:
dt=2ωS₂(R₂ - R₁S₁/S₂)/c²

where he can then proceed to ignore the fact that the S value is like the R value and is dependent upon the interferometer used. It will be different between the orbital and rotational Sagnac interferometer approximations.

Alternatively, he will appeal to pure nonsense regarding the accuracy required (like saying 1 mm off will make it 50 km off, note: numbers may be slightly off, but this conveys the point). We don't need it to be constructed that accurately. A slight imperfection wont do much. We are approximating it an an annular sector. The point is the slight differences wont change it, it just changes the math making it a lot more complex.

He also resorts to lies about a rectangular interferometer, like this:

It is effectively a fibre optic conveyer. The 4 mirrors need to remain stationary while the observer moves.
This is not the Sagnac effect.
But it is the Sagnac effect.
The entire interferometer moves with speed v.

Yet, the source he is quoting shows this quite clearly:

Notice how the mirrors remain stationary and it is just the source and detector which are moving?
This means he is lying yet again, all because he knows if he analyses it honestly he will be shown to be wrong.

Sandy, as you seem quite intent on bringing up your ignorance and lies regarding the Sagnac effect, perhaps you care to come back and try and address your repeated failings.

I believe we left off with you being unable to answer this simple question:
Do you accept that the area of an annular sector is given by:
A=θ_o(R₂² - R₁²)/2
or equally that the following equation holds:
2A=θ_o(R₂² - R₁²)

YES OR NO!
Be clear with your answer, without resorting to loads of bullshit.

Only a bot could substitute the area, which uses the R1 and R2 radii, for the area which is calculated using r1 and r2.

A total disaster.

But for a bot thereis no problem.

In the real world, however, it really is a problem.

You cannot substitute an area involving r2 and r1 into a formula which features R2 and R1.

θo = angle subtended by the two radii, R2 and R1 = orbital angle

s2 = R2 x θo

s1 = R1 x θo


θr = angle subtended by the two radii, r2 and r1 = rotational angle

s2 = r2 x θr

s1 = r1 x θr


R2 - R1 = r2 - r1

r2 x θr = R2 x θo

r1 x θr = R1 x θo

r2/r1 = R2/R1

(r2 x R1) = (r1 x R2)

Since the two areas must be equal,

r1/R1 = (r2 + r1)/(R2 + R1)



Right away, one runs into huge problems with this scenario.

R2 = r2 - r1 + R1

(r2 x R1) = r1r2 - r12 + (R1 x r1)

r2(R1 - r1) = r1(R1 - r1)

So we end up with: r2 = r1, which is impossible.


The best astrophysicists from NASA and ESA proved that the orbital Sagnac is much larger than the rotational Sagnac.

Here is the demonstration.



LISA: Light Interferometer Space Antenna

https://lisa.nasa.gov/

https://www.elisascience.org/

http://sci.esa.int/lisa/

The orbital calculations for the LISA project were performed by some of the very best astrophysicists in the world.




http://tycho.usno.navy.mil/ptti/2003papers/paper34.pdf

Dr. Massimo Tinto, Jet Propulsion Laboratory, Principal Scientist

In the SSB frame, the differences between back-forth delay times are very much larger than has been previously recognized. The reason is in the aberration due to motion and changes of orientation in the SSB frame. With a velocity V=30 km/s, the light-transit times of light signals in opposing directions (Li, and L’i) will differ by as much as 2VL (a few thousands km).

SSB = solar system barycenter

Published in the Physical Review D

http://tycho.usno.navy.mil/ is the U.S. Naval Observatory website


https://arxiv.org/pdf/gr-qc/0310017.pdf

Within this frame, which we can assume to be Solar System Barycentric (SSB), the differences between back-forth delay times that occur are in fact thousands of kilometers, very much larger than has been previously recognized by us or others. The problem is not rotation per se, but rather aberration due to motion and changes of orientation in the SSB frame.

The kinematics of the LISA  orbit brings in the effects of motion at several orders of magnitude larger than any previous papers on TDI have addressed. The instantaneous rotation axis of LISA swings about the Sun at 30 km/sec, and on any leg the transit times of light signals in opposing directions can differ by as much as 1000 km.

Aberration due to LISA’s orbit about the Sun dominates its instantaneous rotation.

The formula is 2VL/c.

V = RΩ




THIS IS THE ORBITAL SAGNAC EFFECT.

Its magnitude is much larger than the rotational Sagnac effect.


Actually, there is no difference in the path lengths: the light signals take different times around the path, amounting to two different speeds c + v and c - v, which of course would be equivalent to admitting that STR is false.

Therefore, the papers have to mention the difference in path lengths to avoid admitting that STR is false.



The formula for the difference in path lengths is:

dp = 2ΩA/c (p = path length)

Then, the difference in time will be:

dt = 2dp/c


The ORBITAL SAGNAC calculated at the Jet Propulsion Laboratory amounts to an admitted difference in path lengths of 1,000 kilometers.

The difference in path lengths for the rotational Sagnac is 14.4 kilometers:

https://arxiv.org/pdf/gr-qc/0306125.pdf (Dr. Daniel Shaddock, Jet Propulsion Laboratory)

https://gwic.ligo.org/thesisprize/2011/yu_thesis.pdf (pg. 63)

Therefore the difference in path lengths for the ORBITAL SAGNAC is some 60 times greater than the difference in path lengths for the rotational Sagnac, according to these calculations.

The formula used for the ORBITAL SAGNAC (difference in path lengths) is 2VL/c, V = RΩ.

R = Earth - Sun distance

Ω = orbital angular velocity


Algebraic approach to time-delay data analysis: orbiting case
K Rajesh Nayak and J-Y Vinet

https://www.cosmos.esa.int/documents/946106/1027345/TDI_FOR_.PDF/2bb32fba-1b8a-438d-9e95-bc40c32debbe

This is an IOP article, published by the prestigious journal Classic and Quantum Gravity:

http://iopscience.iop.org/article/10.1088/0264-9381/22/10/040/meta





In this work, we estimate the effects due to the Sagnac phase by taking the realistic model for LISA orbital motion.

This work is organized as follows: in section 2, we make an estimate of Sagnac phase
for individual laser beams of LISA by taking realistic orbital motion. Here we show that, in general, the residual laser noise because of Sagnac phase is much larger than earlier estimates.

For the LISA geometry, R⊙/L is of the order 30 and the orbital contribution to the Sagnac phase is larger by this factor.

The computations carried out by Dr. R.K. Nayak (over ten papers published on the subject) and Dr. J.Y. Vinet (Member of the LISA International Science Team), and published by prestigious scientific journals and by ESA, show that the orbital Sagnac is 30 times greater than the rotational Sagnac for LISA.



"In reality the motion of LISA is much more complex and our study shows that the main term for Sagnac effect comes from orbital motion."

ORBITAL SAGNAC/ROTATIONAL SAGNAC =~ R/L = 30

This fact is true for each and every satellite orbiting above the surface of the Earth, especially the GPS satellites.


But the orbital Sagnac is not being recorded/registered by the GPS satellites even though is much larger than the rotational Sagnac effect.

As has been proven above, the calculations of Dr. R.K. Nayak (over ten papers published on the subject) and Dr. J.Y. Vinet (Member of the LISA International Science Team), published by prestigious scientific journals and by ESA, show that the orbital Sagnac is 30 times greater than the rotational Sagnac for LISA.


Algebraic approach to time-delay data analysis: orbiting case
K Rajesh Nayak and J-Y Vinet

https://www.cosmos.esa.int/documents/946106/1027345/TDI_FOR_.PDF/2bb32fba-1b8a-438d-9e95-bc40c32debbe

This is an IOP article, published by the prestigious journal Classic and Quantum Gravity:

http://iopscience.iop.org/article/10.1088/0264-9381/22/10/040/meta

See also: Algebraic approach to time-delay data analysis for orbiting LISA

https://journals.aps.org/prd/abstract/10.1103/PhysRevD.70.102003

"In this work, we estimate the effects due to the Sagnac phase by taking the realistic model for LISA orbital motion."

"Earlier results assume a simple module in which LISA rotates only about its own axis!!

In reality the motion of LISA is much more complex and our study shows that the main term for Sagnac effect comes from orbital motion."



Conclusions:

The contribution from the Sagnac effect is much larger than earlier predicted.

Full calculations comparing the rotational Sagnac with the orbital Sagnac lead to the final result:



The original arm length for LISA: 5,000,000 km (L)

Earth - Sun radius: 150,000,000 km (R)

ORBITAL SAGNAC/ROTATIONAL SAGNAC =~ R/L = 30




http://qem.ee.nthu.edu.tw/f1b.pdf

This is an IOP article.

The author recognizes the earth's orbital Sagnac is missing whereas the earth's rotational Sagnac is not.

He uses GPS and a link between Japan and the US to prove this.

In GPS the actual magnitude of the Sagnac correction
due to earth’s rotation depends on the positions of
satellites and receiver and a typical value is 30 m, as the
propagation time is about 0.1s and the linear speed due
to earth’s rotation is about 464 m/s at the equator. The
GPS provides an accuracy of about 10 m or better in positioning.
Thus the precision of GPS will be degraded significantly,
if the Sagnac correction due to earth’s rotation
is not taken into account. On the other hand, the orbital
motion of the earth around the sun has a linear speed of
about 30 km/s which is about 100 times that of earth’s
rotation. Thus the present high-precision GPS would be
entirely impossible if the omitted correction due to orbital
motion is really necessary.


In an intercontinental microwave link between Japan and
the USA via a geostationary satellite as relay, the influence
of earth’s rotation is also demonstrated in a high-precision
time comparison between the atomic clocks at two remote
ground stations.
In this transpacific-link experiment, a synchronization
error of as large as about 0.3 µs was observed unexpectedly.


Meanwhile, as in GPS, no effects of earth’s orbital motion
are reported in these links, although they would be
easier to observe if they are in existence. Thereby, it is evident
that the wave propagation in GPS or the intercontinental
microwave link depends on the earth’s rotation, but
is entirely independent of earth’s orbital motion around
the sun or whatever. As a consequence, the propagation
mechanism in GPS or intercontinental link can be viewed
as classical in conjunction with an ECI frame, rather than
the ECEF or any other frame, being selected as the unique
propagation frame. In other words, the wave in GPS or the
intercontinental microwave link can be viewed as propagating
via a classical medium stationary in a geocentric
inertial frame.


The GPS on your phone works because the MISSING ORBITAL SAGNAC EFFECT does not show up.


But according to the calculations referenced above by some of the very best astrophysicists in the world, this orbital effect is tens of times larger than the rotational effect.

None of that answers the question.
We left off going through this step by step as you demonstrated that is the only way you are capable of advancing.

Now answer the question:
Do you accept that the area of an annular sector is given by:
A=θ_o(R₂² - R₁²)/2
or equally that the following equation holds:
2A=θ_o(R₂² - R₁²)

YES OR NO!
Be clear with your answer, without resorting to loads of bullshit.

If you start posting a bunch of garbage I will continue to ignore it.
Answer the question. It is a very simple one.

Could an intelligent person write something like this? No. But a bot could.


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

Thus dt=2*omega*alpha*R22/c2-2*omega*alpha*R1^2/c2
=2*omega*alpha*(R22-R12)/c2

Now, can we simplify this any more?
I know, lets work out the area.
Note that alpha has always been in radians.
A circle has an area pi*r2.
This circle is a circular sector which subtends an angle of 2*pi.
If it only subtends an angle of alpha, then it will only have an area of alpha/(2*pi) of the circle.
Thus A (for a circular sector) is (alpha/(2*pi))*pi*r2
=alpha*r2/2
This means from the center to the outer arc you have an area of:
A2=alpha*R22/2.
And for the inner one you have an area:
A1=alpha*R12/2.
Thus the area between them is:
A=alpha*R22/2-alpha*R12/2
=alpha*(R22-R12)/2
Thus 2*A=alpha*(R22-R12).

Notice that was in the formula above?
That means we can sub it in.
As a reminder we had:
dt=2*omega*alpha*(R22-R12)/c2
By subbing in the above we get:
dt=2*omega*2*A/c2
And thus:
dt=4*A*omega/c2


The bot continued along the same lines:

Thus 4*A/c^2 is constant, thus you get the relation:
dt=k*w, where k=4*A/c^2.
As such, the shift is directly proportional to the angular velocity.

Thus comparing 2:
dto=k*wo
dtr=k*wr
And thus:
dto/dtr=k*wo/k*wr=wo/wo=1/365.


The bot simply substituted the formula for the area for the rotational Sagnac (which uses r1 and r2) FOR THE AREA assigned to the orbital Sagnac (which features R1 and R2).

A total disaster.

Again, in the real world one cannot do that.

You cannot substitute an area involving r2 and r1 into a formula which features R2 and R1.

θo = angle subtended by the two radii, R2 and R1 = orbital angle

s2 = R2 x θo

s1 = R1 x θo


θr = angle subtended by the two radii, r2 and r1 = rotational angle

s2 = r2 x θr

s1 = r1 x θr


R2 - R1 = r2 - r1

r2 x θr = R2 x θo

r1 x θr = R1 x θo

r2/r1 = R2/R1

(r2 x R1) = (r1 x R2)

Since the two areas must be equal,

r1/R1 = (r2 + r1)/(R2 + R1)



Right away, one runs into huge problems with this scenario.

R2 = r2 - r1 + R1

(r2 x R1) = r1r2 - r12 + (R1 x r1)

r2(R1 - r1) = r1(R1 - r1)

So we end up with: r2 = r1, which is impossible.


The best astrophysicists in the world from NASA and ESA have shown that the orbital Sagnac is much larger than the rotational Sagnac, A SURE PROOF THAT THE CALCULATIONS RUN BY THE BOT IN THIS THREAD ARE HORRENDOUSLY WRONG.

Algebraic approach to time-delay data analysis: orbiting case
K Rajesh Nayak and J-Y Vinet

https://www.cosmos.esa.int/documents/946106/1027345/TDI_FOR_.PDF/2bb32fba-1b8a-438d-9e95-bc40c32debbe

This is an IOP article, published by the prestigious journal Classic and Quantum Gravity:

http://iopscience.iop.org/article/10.1088/0264-9381/22/10/040/meta



In this work, we estimate the effects due to the Sagnac phase by taking the realistic model for LISA orbital motion.

This work is organized as follows: in section 2, we make an estimate of Sagnac phase
for individual laser beams of LISA by taking realistic orbital motion. Here we show that, in general, the residual laser noise because of Sagnac phase is much larger than earlier estimates.

For the LISA geometry, R⊙/L is of the order 30 and the orbital contribution to the Sagnac phase is larger by this factor.

The computations carried out by Dr. R.K. Nayak (over ten papers published on the subject) and Dr. J.Y. Vinet (Member of the LISA International Science Team), and published by prestigious scientific journals and by ESA, show that the orbital Sagnac is 30 times greater than the rotational Sagnac for LISA.

Conclusions:

The contribution from the Sagnac effect is much larger than earlier predicted.

Full calculations comparing the rotational Sagnac with the orbital Sagnac lead to the final result:

Could an intelligent person write something like this? No. But a bot could.

Is this a confession?
Are you a bot?

If so, stop posting, you clearly aren't helping your case.

If not, perhaps you can try to answer the question?

Do you accept that the area of an annular sector is given by:
A=θ_o(R₂² - R₁²)/2
or equally that the following equation holds:
2A=θ_o(R₂² - R₁²)

Yes or no???

It is a very simple question.

You continually refusing to answer it shows you are either incapable of rational thought or know you are full of shit.

Lets see if you can answer it by tomorrow.

Only a bot could substitute the area, which uses the R1 and R2 radii, for the area which is calculated using r1 and r2.

Only a bot could substitute the area, which uses the R1 and R2 radii, for the area which is calculated using r1 and r2.

Sorry if I'm confused here, but as a spectator, I'm a little lost.  Was that a yes or a no?

Only a bot could substitute the area, which uses the R1 and R2 radii, for the area which is calculated using r1 and r2.

And you still failed to answer.

Is the formula I provided the formula for an annular sector? Yes or no?
Or equivalently, is the relation I provided the formula for an annular sector? Yes or no?

Chatbots are prone to make hilarious mistakes when they get involved in scientific discussions since they lack the ability to reason.

Thus 4*A/c^2 is constant, thus you get the relation:
dt=k*w, where k=4*A/c^2.
As such, the shift is directly proportional to the angular velocity.

Thus comparing 2:
dto=k*wo
dtr=k*wr
And thus:
dto/dtr=k*wo/k*wr=wo/wo=1/365.


But comparing the two means one has to substitute the formula used the area of the rotational Sagnac (which uses r1 and r2) FOR THE AREA assigned to the orbital Sagnac (which features R1 and R2).

But for a bot this represents no problem at all.



The computations carried out by Dr. R.K. Nayak (over ten papers published on the subject) and Dr. J.Y. Vinet (Member of the LISA International Science Team), and published by prestigious scientific journals and by ESA, show that the orbital Sagnac is 30 times greater than the rotational Sagnac for LISA.

Conclusions:

The contribution from the Sagnac effect is much larger than earlier predicted.

For the LISA geometry, R⊙/L is of the order 30 and the orbital contribution to the Sagnac phase is larger by this factor.

Full calculations comparing the rotational Sagnac with the orbital Sagnac lead to the final result:



30/(1/365) = 30 x 365 = 10,950

The scale of the error made by the chatbot is 10,950.

Chatbots are prone to make hilarious mistakes when they get involved in scientific discussions since they lack the ability to reason.

Is that why you continually screw up? Because you are a pathetic chatbot that is incapable of rational thought?

Or are you trying to suggest that I am wrong, and that the area of an annular sector is not given by:
A=θ_o(R₂² - R₁²)/2?

It is a very simple question.
All it takes is a simple yes or no answer.

So how about you try answering it:
Is the area of an annular sector given by:
A=θ_o(R₂² - R₁²)/2?

Yes or no?

Thus 4*A/c^2 is constant, thus you get the relation:
dt=k*w, where k=4*A/c^2.
As such, the shift is directly proportional to the angular velocity.

Thus comparing 2:
dto=k*wo
dtr=k*wr
And thus:
dto/dtr=k*wo/k*wr=wo/wo=1/365.

The comparison involves a substitution where the area which features the radii R1 and R2 (on the order of 150,000,000 km) is put in place in the formula which includes two different radii, r1 and r2 (on the order of 6,400 km).

Calculations performed at JPL and ESA reveal that the orbital Sagnac for the LISA space antenna is at least 30 times larger than the rotational Sagnac.

The substitution made by the chatbot leads to a value of 1/365.

Now, the chatbot will deny even the calculations published in the best scientific journals in the world.

Do you accept that the area of an annular sector is given by:
A=θ_o(R₂² - R₁²)/2
or equally that the following equation holds:
2A=θ_o(R₂² - R₁²)

Yes or no???

It is a very simple question.

I will continue to ask this question until you have answered it.
I will be ignoring whatever other BS you provide.

Answer the question. It is quite a simple one.
I know you likely don't want to as it will just move you one step closer to your inevitable defeat.
But if you aren't going to answer it, continually spamming crap will just make you look worse.
If you want to save face without being refuted your only option is to run away and never bring up the Sagnac effect again.

But don't worry, I shall continue from a prior point and go through (noting the φ will have subscripts, and the area of the 2 loops are the same):

dto = 2φoωoR22/c2 - 2φoωoR12/c2
=2φoωo*(R22 - R12)/c2
=4*wo*A/c2

dtr = 2φrωrr22/c2 - 2φrωrr12/c2
=2φrω*(rr22- r12)/c2
=4*wr*A/c2

Now the comparison:
dto/dtr=[4*wo*Ao/c2]/[4*wr*Ar/c2]
=wo/wr

So no worries for the chatbot even though it substituted an area with features radii R1 and R2 into a formula where we have an area which includes r1 and r2.

Thus 4*A/c^2 is constant, thus you get the relation:
dt=k*w, where k=4*A/c^2.
As such, the shift is directly proportional to the angular velocity.

Thus comparing 2:
dto=k*wo
dtr=k*wr
And thus:
dto/dtr=k*wo/k*wr=wo/wo=1/365.

However, this kind of substitution will also lead directly to this result:

θo = angle subtended by the two radii, R2 and R1 = orbital angle

s2 = R2 x θo

s1 = R1 x θo


θr = angle subtended by the two radii, r2 and r1 = rotational angle

s2 = r2 x θr

s1 = r1 x θr


R2 - R1 = r2 - r1

r2 x θr = R2 x θo

r1 x θr = R1 x θo

r2/r1 = R2/R1

(r2 x R1) = (r1 x R2)

Since the two areas must be equal,

r1/R1 = (r2 + r1)/(R2 + R1)



Right away, one runs into huge problems with this scenario.

R2 = r2 - r1 + R1

(r2 x R1) = r1r2 - r12 + (R1 x r1)

r2(R1 - r1) = r1(R1 - r1)

So we end up with: r2 = r1, which is impossible.


The chatbot equates r2 with r1, and obtains a final result of 1/365.


The best astrophysicists in the world obtain a value of 30.

For the LISA geometry, R⊙/L is of the order 30 and the orbital contribution to the Sagnac phase is larger by this factor.

Algebraic approach to time-delay data analysis: orbiting case
K Rajesh Nayak and J-Y Vinet

https://www.cosmos.esa.int/documents/946106/1027345/TDI_FOR_.PDF/2bb32fba-1b8a-438d-9e95-bc40c32debbe

This is an IOP article, published by the prestigious journal Classic and Quantum Gravity:

http://iopscience.iop.org/article/10.1088/0264-9381/22/10/040/meta


Thus, the chatbot is forced to deny the RE calculations which invalidate its erroneous substitution.

Question: why is this kind of trolling allowed in the upper forums?

But don't worry, I shall continue from a prior point and go through (noting the φ will have subscripts, and the area of the 2 loops are the same):

dto = 2φoωoR22/c2 - 2φoωoR12/c2
=2φoωo*(R22 - R12)/c2
=4*wo*A/c2

dtr = 2φrωrr22/c2 - 2φrωrr12/c2
=2φrω*(rr22- r12)/c2
=4*wr*A/c2

Now the comparison:
dto/dtr=[4*wo*Ao/c2]/[4*wr*Ar/c2]
=wo/wr

<< xxxxxxxxxxx >>

Time to remind you again that the Sagnac delay is independent of the shape of the loop and  independent of the centre of rotation.

If that were not true neither of the experiments of Sagnac nor Michelson-Gale-Pearson would have given the correct results.

Quote from: sandokhan on July 17, 2017, 10:42:26 PM

But one could say that the three experiments taken together provide strong evidence for the rotating earth an the non-existence of aether with the properties then assumed..

The missing orbital Sagnac effect shatters any hypotheses about a rotating earth.

Your opinions, on the other hand, are based on false evidence, proven to be so in my messages.

Funny that only you seem to be able to find it.

Every reference I can find explicitly states that the Sagnac delay is independent of the shape of the loop and the centre of rotation.
Sorry, there's one, an obscure one that harps on the "missing orbital Sagnac", Sandokhan I think.
 
I wrote this before, but if you can use copy-pasta, so can I.

You keep saying that we cannot calculate he Sagnac effect correctly, well [b]YOU show us how it should be done.[/b]

No more of you ridiculous copy-pasta! You show your own calculations.
You referred to Mathpages, it says:

Quote from: Mathpages,  2.7  The Sagnac Effect

where A = πR² is the area enclosed by the loop. The corresponding phase difference for light of frequency n radians/second (in the rest frame of the center of rotation) is simply Df = nDt, and since n = 2πc/l, the phase difference can be written as (8πAcw/l)/(c² – v²).

Just note, "where A = πR² is the area enclosed by the loop".

And again in Sagnac Effect, E. J. POST, Rev. Mod. Phys. 39, 475 (1967) – Published 1 April 1967 we have "in which A is the area enclosed by the loop" and
further on in Section III. General Aspects of the Theory, near end p. 478

Quote

Summarizing, the experiments of Sagnac, Pogany and Michelson-Gale and the results of Harress, as re-interpreted by Harzer, demonstrate beyond doubt the following features  of the Sagnac effect. The observed fringe shift
a) obeys formula (1);
b) does not depend on the shape of the surface A;
c) does not depend on the location of the centre of rotation;
d) does not depend on the presence of a comoving refracting medium in the path of the beam.

Please note that E. J. POST specifically states,
          "does not depend on the shape of the surface A;"
          "does not depend on the location of the centre of rotation;"
Care to explain in your own words what 
"does not depend on the shape of the surface A" and  "does not depend on the location of the centre of rotation" mean?

So quit your miles of wasted copying and show your analysis proving that Mathpages, E. J. POST and we are wrong.

I don't want references or copies, only your analysis showing that the

Sagnac delay is not ∆t = (4.A.ω)/(c² – v²), "where A = πR²is the area enclosed by the loop".

I'll just finish with a bit from the Mathpages paper you keep referring us to

Quote

Michelson was not enthusiastic, since classical optics on the assumption of a stationary ether predicted exactly the same shift does special relativity (as explained above). He said,
         "We will undertake this, although my conviction is strong that we shall prove only that the earth rotates on its axis,
           a conclusion which I think we may be said to be sure of already."
. . . . . . . .
Michelson himself wrote that "this result may be considered as an additional evidence in favor of relativity - or equally as evidence of a stationary ether".

Michelson certainly was sure that the "earth rotates on its axis".

But somehow, the great and mighty, Sandokhan, manages to disagree with Michelson.
  I suppose Michelson was hiding his true feelings and deep inside really thought that the earth was flat and stationary.

And the great and mighty Sandokhan thinks he is also smarter than  Sagnac, Gale and Pearson.

You have still failed to answer this simple question:

Do you accept that the area of an annular sector is given by:
A=θ_o(R₂² - R₁²)/2
or equally that the following equation holds:
2A=θ_o(R₂² - R₁²)

Yes or no???

ANSWER IT!
We can't move on until you do.

A summary for all of the readers.



Here is the original presentation made by the chatbot:

Now lets try 2 arcs, one at R1 and one at R2, where the sections connecting the 2 arcs are along the radii.
What this means is that going between them is the same for both directions.
In each one you have it going to the inner arc along a radii, and going to the outer arc along a radii, so the path length and time taken will be equal for those sections.

But this analysis applies ONLY to the orbital Sagnac.

For the rotational Sagnac, the graphic showing the annular sector of a circle would have to be rotated by 180 degrees, using the r1 and r2 radii.

R1 and R2 are estimated to have about 150,000,000 km (the orbital Sagnac).

r1 and r2 are estimated to have about 6,400 km (the rotational Sagnac).

So, even before we even start to decipher this failed analysis, we need TWO INTERFEROMETERS, one for the orbital Sagnac, and one for the rotational Sagnac.

The analysis cannot be applied to a single interferometer.

And yet, this is exactly what the chatbot did.

Thus the area between them is:
A=alpha*R22/2-alpha*R12/2
=alpha*(R22-R12)/2
Thus 2*A=alpha*(R22-R12).

Notice that was in the formula above?
That means we can sub it in.
As a reminder we had:
dt=2*omega*alpha*(R22-R12)/c2
By subbing in the above we get:
dt=2*omega*2*A/c2
And thus:
dt=4*A*omega/c2

Here is the final fulminant posting that announced to the world the hare brained analysis of the chatbot:

Thus 4*A/c^2 is constant, thus you get the relation:
dt=k*w, where k=4*A/c^2.
As such, the shift is directly proportional to the angular velocity.

Thus comparing 2:
dto=k*wo
dtr=k*wr
And thus:
dto/dtr=k*wo/k*wr=wo/wo=1/365.

THE CHATBOT USED THE VERY SAME INTERFEROMETER FOR BOTH SITUATIONS: dto uses the R1 and R2 radii while dtr has to use the r1 and r2 radii.


This is the crucial, catastrophic, monumental error made by the chatbot.


Only a chatbot could have made such a humongous mistake.


No understanding whatsoever of the forces involved.


Since the chatbot used the VERY SAME INTERFEROMETER for both the orbital and the rotational Sagnac, while substituting the area from one to the other, without paying attention to the fact that we have TWO DIFFERENT SETS OF RADII to deal with, I immediately used these facts to provide this very simple proof:

θo = angle subtended by the two radii, R2 and R1 = orbital angle

s2 = R2 x θo

s1 = R1 x θo


θr = angle subtended by the two radii, r2 and r1 = rotational angle

s2 = r2 x θr

s1 = r1 x θr


R2 - R1 = r2 - r1

r2 x θr = R2 x θo

r1 x θr = R1 x θo

r2/r1 = R2/R1

(r2 x R1) = (r1 x R2)

Since the two areas must be equal,

r1/R1 = (r2 + r1)/(R2 + R1)



Right away, one runs into huge problems with this scenario.

R2 = r2 - r1 + R1

(r2 x R1) = r1r2 - r12 + (R1 x r1)

r2(R1 - r1) = r1(R1 - r1)

So we end up with: r2 = r1, which is impossible.


That is, the scenario provided by the chatbot will lead to a glaring contradiction from the very start.

Of course, a chatbot will not be stopped by rational arguments.

It will repeat ad nauseam the very same stupid analysis, having no understanding of the facts.


The conclusion reached by the chatbot, based on its horrendous "analysis", was the figure 1/365.


Here are the very best scientists from NASA and ESA providing the CORRECT calculations for both the orbital and the rotational Sagnac effects.


http://tycho.usno.navy.mil/ptti/2003papers/paper34.pdf

Dr. Massimo Tinto, Jet Propulsion Laboratory, Principal Scientist


https://arxiv.org/pdf/gr-qc/0310017.pdf

Within this frame, which we can assume to be Solar System Barycentric (SSB), the differences between back-forth delay times that occur are in fact thousands of kilometers, very much larger than has been previously recognized by us or others. The problem is not rotation per se, but rather aberration due to motion and changes of orientation in the SSB frame.

The kinematics of the LISA  orbit brings in the effects of motion at several orders of magnitude larger than any previous papers on TDI have addressed. The instantaneous rotation axis of LISA swings about the Sun at 30 km/sec, and on any leg the transit times of light signals in opposing directions can differ by as much as 1000 km.

Aberration due to LISA’s orbit about the Sun dominates its instantaneous rotation.

The formula is 2VL/c.

V = RΩ



The ORBITAL SAGNAC calculated at the Jet Propulsion Laboratory amounts to an admitted difference in path lengths of 1,000 kilometers.

The difference in path lengths for the rotational Sagnac is 14.4 kilometers:

https://arxiv.org/pdf/gr-qc/0306125.pdf (Dr. Daniel Shaddock, Jet Propulsion Laboratory)

https://gwic.ligo.org/thesisprize/2011/yu_thesis.pdf (pg. 63)

Therefore the difference in path lengths for the ORBITAL SAGNAC is some 60 times greater than the difference in path lengths for the rotational Sagnac, according to these calculations.


These are the calculations for the difference in lightpaths.


Now, the calculation for the ratio of the orbital/rotational Sagnac effects.



Algebraic approach to time-delay data analysis: orbiting case
K Rajesh Nayak and J-Y Vinet

https://www.cosmos.esa.int/documents/946106/1027345/TDI_FOR_.PDF/2bb32fba-1b8a-438d-9e95-bc40c32debbe

This is an IOP article, published by the prestigious journal Classic and Quantum Gravity:

http://iopscience.iop.org/article/10.1088/0264-9381/22/10/040/meta




In this work, we estimate the effects due to the Sagnac phase by taking the realistic model for LISA orbital motion.

This work is organized as follows: in section 2, we make an estimate of Sagnac phase
for individual laser beams of LISA by taking realistic orbital motion. Here we show that, in general, the residual laser noise because of Sagnac phase is much larger than earlier estimates.

For the LISA geometry, R⊙/L is of the order 30 and the orbital contribution to the Sagnac phase is larger by this factor.

The computations carried out by Dr. R.K. Nayak (over ten papers published on the subject) and Dr. J.Y. Vinet (Member of the LISA International Science Team), and published by prestigious scientific journals and by ESA, show that the orbital Sagnac is 30 times greater than the rotational Sagnac for LISA.





A total victory for the FE.


Yet, the chatbot will not give up on its silly arguments which lead to a catastrophic final result.


The figure published by the official team of the LISA project, NASA and ESA, is 60.

The chatbot arrived at a value of 1/365.

That is by how much the chatbot went wrong, by a magnitude to 10,950.


QUESTION FOR THE MODERATORS: how much longer will this be allowed to go on?

The chatbot is denying now the NASA and the ESA calculations made for the biggest project ever, THE LISA SPACE ANTENNA.

It is denying reality.

This is called TROLLING.

I even went to the lengths of showing the chatbot where it went wrong.

To no avail.

This no longer just a disagreement.

The chatbot has been provided with the calculations made by NASA and ESA scientists, which show the monumental errors it made in its failed analysis.

A clear victory for FE.

The chatbot comes back ad nauseam using the very same failed arguments to argue its failed case.

It does not realize it has lost the debate, so it continues to troll the upper forums.

So, it is the moderators' job, not to mention the admin's responsability, to step in and inform the chatbot of the reality of the situation.

A summary for all of the readers.

An actual summary for the readers:
I have provided a complete derivation of the Sagnac effect for an annular interferometer (which other shapes can be approximated as, the other shapes don't have as nice a solution). The shape isn't really important as what matters is the area, as all of Sandy's references regarding the Sagnac effect state.
This shows that Sandy is wrong so he refuses to accept it.
Rather than try to honestly or rationally debate, he complete ignores or lies about the derivation to pretend there are issues when there are in fact none.

I have been trying to get him to state what is wrong with the derivation and he can just go off on the same tangents again and again.

So I have now been stepping through the derivation piece by piece.

He has accepted that the correct formula for the Sagnac shift for an annular interferometer is given by:
dt = t₁ - t₂ = 2θ_oω_o(R₂² - R₁²)/c²

But he refuses to honestly take it any further as it shows he is wrong.
He has been completely unable to answer a simple question which has been repeatedly asked to him:
Do you accept that the area of an annular sector is given by:
A=θ_o(R₂² - R₁²)/2
or equally that the following equation holds:
2A=θ_o(R₂² - R₁²)

He knows that if he does this, the shift simplifies and is proportional to area and angular velocity, not the linear velocity as he has repeatedly falsely claimed.

So instead of answering, he continues trolling by the spamming the same refuted BS again and again rather than answering this simple question.

So Sandy, I ask again:
Do you accept that the area of an annular sector is given by:
A=θ_o(R₂² - R₁²)/2
or equally that the following equation holds:
2A=θ_o(R₂² - R₁²)

Yes or no?
Or don't you have any idea?

The issues are very simple.

We have a chatbot which refuses to accept reality, and thus defeat.

Thus 4*A/c^2 is constant, thus you get the relation:
dt=k*w, where k=4*A/c^2.
As such, the shift is directly proportional to the angular velocity.

Thus comparing 2:
dto=k*wo
dtr=k*wr
And thus:
dto/dtr=k*wo/k*wr=wo/wo=1/365.

THE CHATBOT USED THE VERY SAME INTERFEROMETER FOR BOTH SITUATIONS: dto uses the R1 and R2 radii while dtr has to use the r1 and r2 radii.


This is the crucial, catastrophic, monumental error made by the chatbot.


Only a chatbot could have made such a humongous mistake.


No understanding whatsoever of the forces involved.


Since the chatbot used the VERY SAME INTERFEROMETER for both the orbital and the rotational Sagnac, while substituting the area from one to the other, without paying attention to the fact that we have TWO DIFFERENT SETS OF RADII to deal with, I immediately used these facts to provide this very simple proof:

θo = angle subtended by the two radii, R2 and R1 = orbital angle

s2 = R2 x θo

s1 = R1 x θo


θr = angle subtended by the two radii, r2 and r1 = rotational angle

s2 = r2 x θr

s1 = r1 x θr


R2 - R1 = r2 - r1

r2 x θr = R2 x θo

r1 x θr = R1 x θo

r2/r1 = R2/R1

(r2 x R1) = (r1 x R2)

Since the two areas must be equal,

r1/R1 = (r2 + r1)/(R2 + R1)



Right away, one runs into huge problems with this scenario.

R2 = r2 - r1 + R1

(r2 x R1) = r1r2 - r12 + (R1 x r1)

r2(R1 - r1) = r1(R1 - r1)

So we end up with: r2 = r1, which is impossible.


That is, the scenario provided by the chatbot will lead to a glaring contradiction from the very start.

Of course, a chatbot will not be stopped by rational arguments.

It will repeat ad nauseam the very same stupid analysis, having no understanding of the facts.


The conclusion reached by the chatbot, based on its horrendous "analysis", was the figure 1/365.

With this formula directly indicating that the orbital sagnac effect will be 1/365 times that of the rotational one due to the significantly reduced value of ω.

This is the conclusion reached by the chatbot.

Thus, it is denying the calculations provided by NASA and ESA.

Here are the very best scientists from NASA and ESA providing the CORRECT calculations for both the orbital and the rotational Sagnac effects.


http://tycho.usno.navy.mil/ptti/2003papers/paper34.pdf

Dr. Massimo Tinto, Jet Propulsion Laboratory, Principal Scientist


https://arxiv.org/pdf/gr-qc/0310017.pdf

Within this frame, which we can assume to be Solar System Barycentric (SSB), the differences between back-forth delay times that occur are in fact thousands of kilometers, very much larger than has been previously recognized by us or others. The problem is not rotation per se, but rather aberration due to motion and changes of orientation in the SSB frame.

The kinematics of the LISA  orbit brings in the effects of motion at several orders of magnitude larger than any previous papers on TDI have addressed. The instantaneous rotation axis of LISA swings about the Sun at 30 km/sec, and on any leg the transit times of light signals in opposing directions can differ by as much as 1000 km.

Aberration due to LISA’s orbit about the Sun dominates its instantaneous rotation.

The formula is 2VL/c.

V = RΩ



The ORBITAL SAGNAC calculated at the Jet Propulsion Laboratory amounts to an admitted difference in path lengths of 1,000 kilometers.

The difference in path lengths for the rotational Sagnac is 14.4 kilometers:

https://arxiv.org/pdf/gr-qc/0306125.pdf (Dr. Daniel Shaddock, Jet Propulsion Laboratory)

https://gwic.ligo.org/thesisprize/2011/yu_thesis.pdf (pg. 63)

Therefore the difference in path lengths for the ORBITAL SAGNAC is some 60 times greater than the difference in path lengths for the rotational Sagnac, according to these calculations.


These are the calculations for the difference in lightpaths.


Now, the calculation for the ratio of the orbital/rotational Sagnac effects.



Algebraic approach to time-delay data analysis: orbiting case
K Rajesh Nayak and J-Y Vinet

https://www.cosmos.esa.int/documents/946106/1027345/TDI_FOR_.PDF/2bb32fba-1b8a-438d-9e95-bc40c32debbe

This is an IOP article, published by the prestigious journal Classic and Quantum Gravity:

http://iopscience.iop.org/article/10.1088/0264-9381/22/10/040/meta


In this work, we estimate the effects due to the Sagnac phase by taking the realistic model for LISA orbital motion.

This work is organized as follows: in section 2, we make an estimate of Sagnac phase
for individual laser beams of LISA by taking realistic orbital motion. Here we show that, in general, the residual laser noise because of Sagnac phase is much larger than earlier estimates.

For the LISA geometry, R⊙/L is of the order 30 and the orbital contribution to the Sagnac phase is larger by this factor.

The computations carried out by Dr. R.K. Nayak (over ten papers published on the subject) and Dr. J.Y. Vinet (Member of the LISA International Science Team), and published by prestigious scientific journals and by ESA, show that the orbital Sagnac is 30 times greater than the rotational Sagnac for LISA.




The chatbot says the figure is 1/365.

The calculations provided by both NASA and ESA say the figure is 30.

So the chatbot is forced to deny reality, and the NASA calculations.


Question for the moderators: how long will this trolling be allowed to go on here in the upper forums?

This no longer just a disagreement.

The chatbot has been provided with the calculations made by NASA and ESA scientists, which show the monumental errors it made in its failed analysis.

The issues are very simple.

Yes, they are.
I have provided a derivation you have been unable to refute.
I have provided a simple question you have been completely unable to answer.

Perhaps you can answer it this time:
Do you accept that the area of an annular sector is given by:
A=θ_o(R₂² - R₁²)/2
or equally that the following equation holds:
2A=θ_o(R₂² - R₁²)

Yes or no?

Until you answer I will continue asking it.

Indeed, the issues are very simple.

The conclusion reached by the chatbot, based on its horrendous "analysis", was the figure 1/365.

With this formula directly indicating that the orbital sagnac effect will be 1/365 times that of the rotational one due to the significantly reduced value of ω.

This is the conclusion reached by the chatbot.

Thus, it is denying the calculations provided by NASA and ESA.

Here are the very best scientists from NASA and ESA providing the CORRECT calculations for both the orbital and the rotational Sagnac effects.


http://tycho.usno.navy.mil/ptti/2003papers/paper34.pdf

Dr. Massimo Tinto, Jet Propulsion Laboratory, Principal Scientist


https://arxiv.org/pdf/gr-qc/0310017.pdf

Within this frame, which we can assume to be Solar System Barycentric (SSB), the differences between back-forth delay times that occur are in fact thousands of kilometers, very much larger than has been previously recognized by us or others. The problem is not rotation per se, but rather aberration due to motion and changes of orientation in the SSB frame.

The kinematics of the LISA  orbit brings in the effects of motion at several orders of magnitude larger than any previous papers on TDI have addressed. The instantaneous rotation axis of LISA swings about the Sun at 30 km/sec, and on any leg the transit times of light signals in opposing directions can differ by as much as 1000 km.

Aberration due to LISA’s orbit about the Sun dominates its instantaneous rotation.

The formula is 2VL/c.

V = RΩ



The ORBITAL SAGNAC calculated at the Jet Propulsion Laboratory amounts to an admitted difference in path lengths of 1,000 kilometers.

The difference in path lengths for the rotational Sagnac is 14.4 kilometers:

https://arxiv.org/pdf/gr-qc/0306125.pdf (Dr. Daniel Shaddock, Jet Propulsion Laboratory)

https://gwic.ligo.org/thesisprize/2011/yu_thesis.pdf (pg. 63)

Therefore the difference in path lengths for the ORBITAL SAGNAC is some 60 times greater than the difference in path lengths for the rotational Sagnac, according to these calculations.


These are the calculations for the difference in lightpaths.


Now, the calculation for the ratio of the orbital/rotational Sagnac effects.



Algebraic approach to time-delay data analysis: orbiting case
K Rajesh Nayak and J-Y Vinet

https://www.cosmos.esa.int/documents/946106/1027345/TDI_FOR_.PDF/2bb32fba-1b8a-438d-9e95-bc40c32debbe

This is an IOP article, published by the prestigious journal Classic and Quantum Gravity:

http://iopscience.iop.org/article/10.1088/0264-9381/22/10/040/meta


In this work, we estimate the effects due to the Sagnac phase by taking the realistic model for LISA orbital motion.

This work is organized as follows: in section 2, we make an estimate of Sagnac phase
for individual laser beams of LISA by taking realistic orbital motion. Here we show that, in general, the residual laser noise because of Sagnac phase is much larger than earlier estimates.

For the LISA geometry, R⊙/L is of the order 30 and the orbital contribution to the Sagnac phase is larger by this factor.

The computations carried out by Dr. R.K. Nayak (over ten papers published on the subject) and Dr. J.Y. Vinet (Member of the LISA International Science Team), and published by prestigious scientific journals and by ESA, show that the orbital Sagnac is 30 times greater than the rotational Sagnac for LISA.




How long will this kind of trolling be allowed in the upper forums?

The chatbot is denying the RE calculations published by NASA and ESA.

It reached a conclusion using a figure of 1/365.

The NASA calculations provided above say otherwise: the figure is 30.

Any sensible, real person would understand the debate has been lost and would move on to something else.

But this chatbot has been programmed to deny everything.