sandokhan lies regarding the Sagnac effect

sandokhan, considering you seem intent on repeatedly bringing it up, here is a chance for you to try and prove your BS once and for all (or be shown to be full of shit once and for all).

You claim Earth's orbital Sagnac effect is much greater than Earth's rotational Sagnac effect, but all the evidence contradicts you.

The Sagnac effect is observed for any loop which contains 2 counterpropagating beams of light, which is rotating.
before you start claiming otherwise, no one has ever built an interferometer which spans the entirety of Earth, nor the entirety of Earth's orbit. As such, claiming the loop needs to be centred on the centre of rotation is pure garbage as it would mean no Sagnac effect could have ever been observed. However, you can treat a non-centred interferometer as parts of multiple centered interferometers to determine the result.

Assuming the rotational speed is not massive (i.e. it is significant less than the speed of light), then the formula to determine the Sagnac effect is quite simple:
Δt=4Aω/c²
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.

So now then, here is your chance.
Explicitly derive the Sagnac effect due to Earth's orbit and Earth's rotation on a ring interferometer constructed as in the following (very much not to scale) diagram:


The Earth is shown in blue.
The sun is shown in red.
This has a square loop interferometer, of side length l, which is sitting on the surface of Earth (which has a radius of r, and is following a circular orbit of radius R), at the equator.
This is taken on the equinox, simplified to a perfectly circular orbit of 365 days for simplicity (so it will just be an approximation).
The lengths of the interferometer are straight, but are sufficiently short enough to be approximated as axial spokes or as sections of a circular arc centred on either the centre of Earth or the centre of Earth's orbit.
Also note that the interferometer stands vertically, that is one arm is on the ground, and another is l above the ground.

Now then, derive the expected Sagnac effect for this loop based upon Earth's orbital motion, and separately derive it for Earth's rotation, and then compare the 2.
Be explicit in your derivation, don't skip any steps and explain why things must be the case.

Using the simple formula above is also acceptable.
The area of the loop is l², as it is a square of side length l.
That means Δt=4l²ω/c²
And as Earth's rotational motion (measured as angular velocity or ω) is 365 times as fast as Earth's orbital motion, that means the rotational sagnac effect will be 365 times that of the orbital sagnac effect.

Once you have tried to show a derivation I can do the more complex derivation (which I already had before).

You do realise you have just performed the Invocation Of Sandokhan? Following this he will be conjured forth, and as written in the Great Scrolls, will perform the following actions:
Start by saying he will address your question, then go off into a tangential ramble never to return.
Post 565 lines of copypasta.
Claim that there was no human history prior to the reign of Henry VIII.
Post a further 312 lines of copypasta.
Print some indecipherable maths.
Post that picture of a man skiing towards an eclipse.

INB4 Giant wall of CTRL-V gobbledygook!

You do realise you have just performed the Invocation Of Sandokhan? Following this he will be conjured forth, and as written in the Great Scrolls, will perform the following actions:
Start by saying he will address your question, then go off into a tangential ramble never to return.

This is incorrect. He will start by saying, "You haven't done your homework."

It is really quite simple, and Sandokhan repeats this same ridiculous nonsense over and over:

Alright, let's make this easy and see how much different an orbit and rotation would be.

Sagnac Effect: dt = 4Aw/c^2
dt= Difference in time
A= Area enclose by light path
w= angular velocity of the spin in radians per second
c= speed of light

I don't even have to calculate to know your "the sagnac effect for the earth's orbit is greater than that of the rotation" is wrong. The orbital vs rotational sagnac has the same speed of light and same area enclosed by the path, the only think left is the angular velocity of the rotation and you get dt. Angular velocity is the rate of change in angular displacement with respect to time, which is far more in a rotation of Earth than an orbit. The angular velocity of Earth's rotation is 7.2921159 × 10^-5, and the earth's orbit would be 2.0x10^-7 rad/s.

So, no, the orbital sagnac is not greater but less than the rotational, because the angular velocity is more for the rotational, which is why I say the rotational is 1/365th of the orbital, which shows the angular velocity difference in a day. You mistakenly use linear velocity which does not hold in the sagnac effect, but rather angular velocity.

Great read and explanation of the Sagnac Effect: http://www.space-lab.ru/files/pages/PIRT_VII-XII/pages/text/PIRT_IX/Kelly_2.pdf

And he responds with this:

I don't even have to calculate to know your "the sagnac effect for the earth's orbit is greater than that of the rotation" is wrong. The orbital vs rotational sagnac has the same speed of light and same area enclosed by the path, the only think left is the angular velocity of the rotation and you get dt.

Then you have a poor understanding of physics, in particular you do not understand the Sagnac effect.

Let me explain.

The orbital Sagnac and the rotational Sagnac DO NOT and CANNOT have the same area enclosed by the path.

Absolute BS and a lie^

What is the center of rotation for the orbit of the earth?

Here is the equation.

∆t = 4πRv / ( c² - v²) = 4Aω / ( c² - v²)

Where A = πR² and v = ωR

So, it is easy to calculate the orbital sagnac is more than 60 times that of the rotational.

More absolute BS^

But, A is based on R and according to mathpages, "circular loop of radius R".

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

Mathpages says one must use the center of rotation which is the sun.

Exactly^

It is a loop and the earth is moving along the loop in its orbit around the sun.

If light travels at one speed c, then as the earth supposedly moves in it's revolution loop at 30k/s, while light moves c through space, the unit at the equator at noon would move with the earth' rotation and the earth's revolution cutting the distance the signal must travel to meet the unit.

"Let's say the unit is at the equator and the satellite is low on the horizon in the east at noon.

That means the unit is traveling at the orbital speed of the earth at 67,000 MPH.

How is this relevant to the sagnac effect?^

The satellite emits at one speed c in space. While the light travels through space toward the unit at c, the unit moves with the earth at 67,000 MPH. The unit cuts the distance that the light must travel.

This is not being seen by any experiements nor GPS."

Yet, this same logic applies and works with the earth's supposed rotation.


Published by the BULLETIN OF THE AMERICAN PHYSICAL SOCIETY, one of the most prestigious journals in the world today.

C.C. Su, "A Local-ether model of propagation of electromagnetic wave," in Bull. Am. Phys. Soc., vol. 45, no. 1, p. 637, Mar. 2000 (Minneapolis, Minnesota).

http://www.ee.nthu.edu.tw/ccsu/










Both the rotational and the orbital motions of the earth together with the orbital
motion of the target planet contribute to the Sagnac
effect. But the orbital motion of the sun has no effects
on the interplanetary propagation. On the other hand, as
the unique propagation frame in GPS and intercontinental
links is a geocentric inertial frame, the rotational motion
of the earth contributes to the Sagnac effect. But the orbital
motion of the earth around the sun and that of the
sun have no effects on the earthbound propagation. By
comparing GPS with interplanetary radar, it is seen that
there is a common Sagnac effect due to earth’s rotation
and a common null effect of the orbital motion of the sun
on wave propagation. However, there is a discrepancy in
the Sagnac effect due to earth’s orbital motion. Moreover,
by comparing GPS with the widely accepted interpretation
of the Michelson–Morley experiment, it is seen that
there is a common null effect of the orbital motions on
wave propagation, whereas there is a discrepancy in the
Sagnac effect due to earth’s rotation.


Based on this characteristic of uniqueness and switchability of the propagation frame,
we propose in the following section the local-ether model
of wave propagation to solve the discrepancies in the in-
fluences of earth’s rotational and orbital motions on the
Sagnac effect and to account for a wide variety of propagation
phenomena.


Anyway, the interplanetary Sagnac effect is due to
earth’s orbital motion around the sun as well as earth’s
rotation. Further, for the interstellar propagation where
the source is located beyond the solar system, the orbital
motion of the sun contributes to the interstellar Sagnac
effect as well.

Evidently, as expected, the proposed local-ether model
accounts for the Sagnac effect due to earth’s rotation and
the null effect of earth’s orbital motion in the earthbound
propagations in GPS and intercontinental microwave link
experiments. Meanwhile, in the interplanetary radar, it accounts
for the Sagnac effect due both to earth’s rotation
and to earth’s orbital motion around the sun.


Based on the local-ether model, the propagation is entirely
independent of the earth’s orbital motion around
the sun or whatever and the velocity v for such an earthbound
experiment is referred to an ECI frame and hence
is due to earth’s rotation alone. In the original proposal,
the velocity v was supposed to incorporate earth’s orbital
motion around the sun. Thus, at least, v²/c²
=~ 10^-8. Then the amplitude of the phase-difference variation
could be as large as π/3, when the wavelength is
0.6 µm and the path length is 10 m. However, as the velocity
v is the linear velocity due to earth’s rotation alone,
the round-trip Sagnac effect is as small as v²/c²∼ 10^-12 which is merely 10^-4 times that due to the orbital motion.


The Sagnac effect is a FIRST ORDER effect in v/c.

Even in the round-trip nature of the Sagnac effect, as it was applied in the Michelson-Morley experiment, thus becoming a second order effect within that context, we can see that the ORBITAL SAGNAC IS 10,000 TIMES GREATER than the rotational Sagnac effect.

ABSOLUTE BULLOCKS! That is either a flat out lie or you have no idea what the sagnac effect is^

Here is how to correctly calculate the orbital Sagnac effect:

Earth's radius = 6357 km; r² = 40411449

Earth's orbital radius = 150,000,000 km r² = 22500000000000000

∆t = 4πR²ω/(c²-v²)
or

I use the linear velocity.

NO YOU DON'T, YOU USE ANGULAR VELOCITY!^

∆t = 4πRv/( c² - v² ), where v is the linear velocity.

For the earth's rotation, it is 0.4638333 km/ sec and the orbit v = 30km/sec.

∆t = 0.62831852628 for the earth's orbit.
Total path of the orbit is 2πr=2π(150,000,000 km) = 942,477,780km

Hence, the sagnac effect for a 1 km path, that means light source in the center and two receivers placed at .5km is:
0.62831852628 / 942,477,780km = 6.6666667 e-10 sec / km

Now, for the earth's rotation.
∆t = 4.1170061 e-7 seconds
Total path of the rotation is 2πr=2π(6357 km) = 39942.21 km


4.1170061 e-7 seconds / 39942.21 km = 1.0307407 e-11 sec / km


The sagnac effect for the earth's orbit is greater than that of the rotation.

Failed to show this, you ignore angular velocity which is the main variable here, the angular velocity of the Earth's orbit would be far less and so the the Sagnac effect far less, the speed of light and area enclosed by the path are the same^

The orbital Sagnac, though much larger than the rotational Sagnac, is not being registered by GPS satellites.

IT IS NOT DETECTABLE REGARDLESS OF WHETHER THE EARTH ORBITS THE SUN!^

The lunar laser ranging experiment is an astronomical version of the Sagnac experiment.

However, G. Sagnac used the fringe-shift method to measure indirectly light travel time;
while Dr. Daniel Gezari uses clocks to measure directly light travel time in both directions.

Shooting light to the moon has to do with the behavior of light like GPS.

The arrival time of light to a receptor is influenced by the motion of
the receptor relative to the earth: this is the basic discovery of G. Sagnac.

This fact has to be incorporated into the lunar laser ranging calculations.

Here is a basic reference which confirms this fact:

Ring-laser tests of fundamental physics and geophysics, G.E. Steadman, 1997, pg 15



One needs both the orbital and rotational Sagnac to calculate the correct timing, there is no way around that.


Dr. Daniel Gezari emitted a pulse of photons from a point on earth, bounced those photons off a reflector on the moon, and then recorded the photons’ arrival time at that same point on earth.


Please note the theoretical orbital sagnac shows up in these calculations, but is not picked up/registered/recorded by GPS satellites.

Motion of the Earth-Moon system in orbit around the Sun would average out in a two-way measurement, and only appear as a small (∼3 m/s) second-order residual.

Because of the two-way averaging, the orbital Sagnac effect registered is smaller than usual, however it is not 1/365 of the rotational Sagnac effect, in fact even in the diluted form permitted by the two-way averaging calculation, it represents a significant percentage of the rotational Sagnac effect.


THE SMALL (~3M/S) SECOND ORDER RESIDUAL IS THE ORBITAL SAGNAC.


For instance, the Earth’s full 30 km/s orbital velocity along the line-of-sight would produce a second-order residual velocity of only ~3 m/s, so we cannot preclude the possibility that some part of the 8.4 m /s difference between co and c measured here is a real second-order residual due to motion of the Earth-Moon system relative to an absolute frame.

THE 8.4 M/S DIFFERENCE IS THE ROTATIONAL SAGNAC.


Dr. Daniel Gezari:


For instance, the Earth’s full 30 km/s orbital velocity along the line-of-sight would produce a second-order residual velocity of only ~3 m/s, so we cannot preclude the possibility that some part of the 8.4 m /s difference


3/8.4 = 0.357

1/365 = 0.00274

0.357/0.00274 = 130.3

Now, because of the vast distance, if the RE were correct, we should see 1/365 of the rotational sagnac in the measurements and that will show up on this vast distance.

So, if they are correct, then we should see the 1/365 conclusions in the measurements. Guess what. We do not.

Dr. Daniel Gezari's calculations prove otherwise: even in the diluted two way averaging form, the orbital Sagnac amounts for a 3/8.4 = 0.357 (35.7%) percentage of the rotational Sagnac.


It is also of interest to note that the missing orbital Sagnac effect proves that the lunar missions never occurred in reality, that the lunar laser ranging is actually a small mirror (in the form of a minuscule satellite) orbiting above the flat surface of the Earth right in front of the Moon, using the Biefeld-Brown effect to stay in orbit.


More information on Dr. C.C. Su's paper on the orbital Sagnac effect.

His paper was also published by HARVARD UNIVERSITY:

http://adsabs.harvard.edu/cgi-bin/nph-bib_query?2001EPJC...21..701S

See the headline at the top:

NASA ADS Physics/Geophysics Abstract Service



So far, Dr. C.C. Su's papers, which include the correct orbital Sagnac calculations, based on a circular loop with the center of rotation located at the Sun, have been published by:

HARVARD UNIVERSITY

BULLETIN OF THE AMERICAN PHYSICAL SOCIETY

EUROPEAN PHYSICAL JOURNAL

EUROPHYSICS LETTERS JOURNAL

JOURNAL OF ELECTROMAGNETIC WAVES AND APPLICATIONS

Further information here:

http://www.ee.nthu.edu.tw/ccsu/



Journal of Electromagnetic Waves and Applications:

http://www.ee.nthu.edu.tw/ccsu/qem/f3c.pdf

For the interplanetary propagation, earth’s orbital
motion contributes to the Sagnac effect as well. This local-ether model
has been adopted to account for the Sagnac effect due to earth’s
motions in a wide variety of propagation phenomena, particularly the
global positioning system (GPS), the intercontinental microwave link,
and the interplanetary radar.


The peer reviewers at the Journal of Electromagnetic Waves and Applications agree that the orbital Sagnac is larger than the rotational Sagnac, that it is missing, and that a local-ether model has to be adopted in order to account for this fact.



You do understand English, do you not?

Based on the local-ether model, the propagation is entirely
independent of the earth’s orbital motion around
the sun or whatever and the velocity v for such an earthbound
experiment is referred to an ECI frame and hence
is due to earth’s rotation alone. In the original proposal,
the velocity v was supposed to incorporate earth’s orbital
motion around the sun. Thus, at least, v²/c²
=~ 10^-8. Then the amplitude of the phase-difference variation
could be as large as π/3, when the wavelength is
0.6 µm and the path length is 10 m. However, as the velocity
v is the linear velocity due to earth’s rotation alone,
the round-trip Sagnac effect is as small as v²/c²∼ 10^-12 which is merely 10^-4 times that due to the orbital motion.


The Sagnac effect is a FIRST ORDER effect in v/c.

Even in the round-trip nature of the Sagnac effect, as it was applied in the Michelson-Morley experiment, thus becoming a second order effect within that context, we can see that the ORBITAL SAGNAC IS 10,000 TIMES GREATER than the rotational Sagnac effect.

So far, Dr. C.C. Su's papers, which include the correct orbital Sagnac calculations, based on a circular loop with the center of rotation located at the Sun, have been published by:

HARVARD UNIVERSITY

BULLETIN OF THE AMERICAN PHYSICAL SOCIETY

EUROPEAN PHYSICAL JOURNAL

EUROPHYSICS LETTERS JOURNAL

JOURNAL OF ELECTROMAGNETIC WAVES AND APPLICATIONS

Further information here:

http://www.ee.nthu.edu.tw/ccsu/



Journal of Electromagnetic Waves and Applications:

http://www.ee.nthu.edu.tw/ccsu/qem/f3c.pdf

For the interplanetary propagation, earth’s orbital
motion contributes to the Sagnac effect as well. This local-ether model
has been adopted to account for the Sagnac effect due to earth’s
motions in a wide variety of propagation phenomena, particularly the
global positioning system (GPS), the intercontinental microwave link,
and the interplanetary radar.


The peer reviewers at the Journal of Electromagnetic Waves and Applications agree that the orbital Sagnac is larger than the rotational Sagnac, that it is missing, and that a local-ether model has to be adopted in order to account for this fact.

You responded in the wrong forum topic, but showed interest, come here and do what the OP asks of you or at least respond in this thread.

The peer reviewers at the Journal of Electromagnetic Waves and Applications agree that the orbital Sagnac is larger than the rotational Sagnac, that it is missing, and that a local-ether model has to be adopted in order to account for this fact.

I don't care, prove it to me, do it by following the steps above by JackBlack and I will believe you. Walk me and others through it.

altspace, are you scientifically illiterate?

For a circular path of radius R, the formula: ∆t = 2vl/c², where v = ω R is the speed of the circular motion and l=2πR is the circumference of the circle.

And you said you were using linear velocity, do you know what linear velocity is and how it differs from angular velocity? I mean, seriously?

You responded in the wrong forum topic, but showed interest, come here and do what the OP asks of you or at least respond in this thread.

Quote from: sandokhan on October 31, 2017, 12:59:56 AM

The peer reviewers at the Journal of Electromagnetic Waves and Applications agree that the orbital Sagnac is larger than the rotational Sagnac, that it is missing, and that a local-ether model has to be adopted in order to account for this fact.

I don't care, prove it to me, do it by following the steps above by JackBlack and I will believe you. Walk me and others through it.

Considering he wants to appeal to journals so much, perhaps he should accept the statements in a journal, like this one Rab repeatedly pointed out (and I believe I have pointed it out as well).
The easiest way to find the citation is with the DOI, which in this case is 10.1103/RevModPhys.39.475  but it can also just have a citation either in full, such as saying it is in the journal "Reviews of Modern Physics", volume 39, issue 2, page 475 by E. J. Post, published 1 April 1967. Or the short version, Post, E. J., Rev. Mod. Phys., 39, 2 (1967)

Either way, it has this nice little fact:

A is the area enclosed by the light path.

Not the area of the orbit, the light path. Big difference.

It explains what it should be on a rotating Earth for an interferometer which only covers a small section of Earth's surface.

So it shows quite conclusively that it is dependent upon the area of the loop, and thus for a circular loop, the radius of the loop, not the orbit.

Sandy, this is the thread for you to spout your ignorance about the Sagnac effect.
No need to pollute other thread with your BS.

No need to pollute other thread


You were given ample space, some nine pages, to present your derivation. The debate was over with your complete defeat.


For the sake of the readers I will now offer even more details regarding your catastrophic mistake inherent in your piece of shit derivation.

Let us go back to the derivation itself.

YOU USED A SINGLE INTERFEROMETER, INSTEAD OF THE TWO REQUIRED: THE ROTATIONAL SAGNAC AND THE ORBITAL SAGNAC.

But we are not dealing with an interferometer that goes the entire length of Earth's orbit. We are dealing with a small one on Earth. Thus we are only able to use a small section of the circle.

The orbital Sagnac is the size of Earth's orbit and centered on the Sun.





Notice the 30km/s speed in the figure.

It is a loop and the earth is moving along the loop in its orbit around the sun.

If light travels at one speed c, then as the earth supposedly moves in it's revolution loop at 30k/s, while light moves c through space, the unit at the equator at noon would move with the earth' rotation and the earth's revolution cutting the distance the signal must travel to meet the unit.


"Let's say the unit is at the equator and the satellite is low on the horizon in the east at noon.

That means the unit is traveling at the orbital speed of the earth at 67,000 MPH.

The satellite emits at one speed c in space. While the light travels through space toward the unit at c, the unit moves with the earth at 67,000 MPH. The unit cuts the distance that the light must travel.

This is not being seen by any experiements nor GPS."

Yet, this same logic applies and works with the earth's supposed rotation.


Dr. Herbert Ives, in the one of seminal papers of the 20th century on the Sagnac effect, proved that the path is linear (he used a hexagonal path):



A path using even more linear segments, forming a circle:




https://www.google.com.pg/patents/US20060145063


As the GPS satellites together with the Earth move in the orbital path around the Sun, YOU HAVE A INTERFEROMETER CENTERED ON THE SUN, just like proven by Dr. Herbert Ives.

http://www.conspiracyoflight.com/Ives/Herbert_Ives_Light_Signals_Sent_Around_a_Closed_Path.pdf


Using your SINGLE INTERFEROMETER, you proceeded to do calculations, as follows.

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.


YOU USED THE SAME AREA AND THE SAME RADIUS IN YOUR CALCULATIONS!!!

So to continue:
Δto/Δt r=[4Aiωo/( c² - vo²)] / [4Aiωr/( c² - vr²)]
Obviously, ( c² - vo²) and ( c² - vr²) are very close to the same number, so let's lave them off.
=4Aiωo/ 4Aiωr
Then to simplify:
=ωo/ ωr


AGAIN, YOU USED THE SAME AREA AND RADIUS TO SIMPLIFY THE EQUATIONS!!!


A terrible mistake.


That is not how the calculations are done.

Let Δt _o= the sagnac correction for the earth's orbital path

Let Δt _r= the sagnac correction for the earth's rotational path

RE claim: Δt_o/Δt _r= 1/365

[4A_oω_o/( c² - v_o²)] / [4A_rω_r/( c² - v_r²)]
[A_oω_o/( c² - v_o²)] / [A_rω_r/( c² - v_r²)]

Obviously, ( c² - v_o²) and ( c² - v_r²) are very close to the same number, so let's lave them off.

A_oω_o/ A_rω_r
A_o = πR_o²
A_r = πR_r²

So, πR_o² ω_o / πR_r² ω_r

R_o² ω_o / R_r² ω_r.

Earth's radius = 6357 km; r² = 40411449

Earth's orbital radius = 150,000,000 km r² = 22500000000000000


Make sure you understand that the Sagnac effect applies to translational/linear paths.

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."




So there you have it jack.

A terrible error on your part, to assume a single interferometer.

But there are two of them, just as shown above.

No need to pollute other thread

Since you clearly are incapable of showing where JackBlack's derivation of Sagnac delay is incorrect, at least believe this:


See where Mathpages implies it with, "(8πAcw/l)/(c² – v²)" and E. J. POST very specifically states that,

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
You referred to Mathpages and 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 = πR² is the area enclosed by the loop".
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Since E. J. POST quite unambiguously states that the Sagnac delay:
          "does not depend on the shape of the surface A;"
          "does not depend on the location of the centre of rotation;"

I assume that you now claim to be smarter than both E. J. POST and the authors of Mathpages.

The Sagnac delay is, whether you agree or not, independent of the shape of the loop and its centre of rotation.

For the case where the Earth would orbit the Sun, YOU ALREADY HAVE A SHAPE TO START WITH, you cannot modify it, unless of course you'd care to tell your readers that the nearly circular shape of the orbit has a different geometrical shape.


Even in the official version of heliocentricity, the Earth's orbit around the Sun is assumed to be nearly circular.

The radius of the earth very nearly circular orbit around the sun is 1.50⋅10^11 m.

THE GPS SATELLITES' ORBIT AROUND THE EARTH IS ALSO NEARLY CIRCULAR, YET THE SAGNAC EFFECT IS CALCULATED PRECISELY USING THE KNOWN FORMULA:

4Aω/( c² - v²)

https://www.faa.gov/about/office_org/headquarters_offices/ato/service_units/techops/navservices/gnss/faq/gps/

The orbits are nearly circular.

http://www.navipedia.net/index.php/GPS_Space_Segment

Orbits are nearly circular.


Eccentricity of Earth's orbit around the Sun (official science information):

Earth's orbit has an eccentricity of 0.0167.

Eccentricity of GPS satellites orbit around the Earth:

Orbits are nearly circular, with eccentricity less than 0.02.

The orbital eccentricity of an astronomical object is a parameter that determines the amount by which its orbit around another body deviates from a perfect circle. A value of 0 is a circular orbit, values between 0 and 1 form an elliptical orbit.


Once you are stuck with a certain geometrical shape, in this case the orbital path of the Earth, there's nothing else that you can do.


The basic mistake made by jack was to use a single interferometer.

Then, he used the same area/radius in a calculation where two different areas were required.

Please read my previous message.

It is irrefutable.

You were given ample space, some nine pages, to present your derivation. The debate was over with your complete defeat.

You mean in the thread we discussed it in previously, where I presented my derivation for a similar system and you were unable to point out any flaws in it?
And then you ran away like a coward because you were unable to answer a simple question?

I would say that was the debate being over with your complete defeat and total failure to defend your claims in any rational manner.

For the sake of the readers I will now offer even more details regarding your catastrophic mistake inherent in your piece of shit derivation.

A mistake in my derivation, or you bitching and moaning about the conclusion? It sure seems like the latter.

YOU USED A SINGLE INTERFEROMETER, INSTEAD OF THE TWO REQUIRED: THE ROTATIONAL SAGNAC AND THE ORBITAL SAGNAC.

Nope. A single interferometer is fine for both. If you actually needed separate interferometers then anything which would be affected by the rotational sagnac effect would be unable to be affected by the orbital sagnac effect, making your objections complete crap.

Instead, the only way to honestly compare the 2, to determine the magnitude of their effects on a specific system, is to compare them using the same interferometer.

What you are doing is akin to saying a feather is much more dense than lead because 1 ton of feathers weighs more than 1 g of lead.

So, on the same interferometer, like the one in the OP, can you derive the Sagnac effect for Earth's rotational and orbital motions?

But we are not dealing with an interferometer that goes the entire length of Earth's orbit. We are dealing with a small one on Earth. Thus we are only able to use a small section of the circle.
The orbital Sagnac is the size of Earth's orbit and centered on the Sun.

No. Earth's orbit is the size of Earth's orbit and centred in the sun.
The Sagnac effect as we have been discussing is a time difference and thus saying the size is in units of distance makes no sense.

The size of the Sagnac effect is dependent upon the area of the interferometer loop. The larger the loop, the larger the shift.
The position of the centre of the loop is irrelevant. It doesn't matter if the loop is centred on the centre of rotation, or off someplace far away.

Notice the 30km/s speed in the figure.
It is a loop and the earth is moving along the loop in its orbit around the sun.

Yes, Earth is moving along it, in one direction. It is not 2 counter-propagating beams of light which travel around the loop. It also isn't the interferometer in the OP, as such it is irrelevant.

We are discussing the Sagnac effect, not the failure to detect the anisotropy of light in a linear fashion.

Yet, this same logic applies and works with the earth's supposed rotation.

No, it doesn't.
They note the relative velocities of the satellite w.r.t. Earth.
Doing that for the orbital motion results in 0.

So no, they do something fundamentally different. Something which works fine  with relativity but completely fails in aether based models, yet GPS works. Further evidence
that the aether is a load of crap.

Dr. Herbert Ives, in the one of seminal papers of the 20th century on the Sagnac effect, proved that the path is linear (he used a hexagonal path):

You are aware a hexagon is not linear?
A hexagon is a 6 sided closed polygon, and still with 2 beams of light traversing the loop, not just one beam of light traversing a tiny portion of it.

As the GPS satellites together with the Earth move in the orbital path around the Sun, YOU HAVE A INTERFEROMETER CENTERED ON THE SUN, just like proven by Dr. Herbert Ives.

No. You don't.
If you think you do, draw up a picture, like his, showing the sun, showing Earth, showing the satellites and most importantly, showing the path of the light.
Be very clear in indicating the 2 counterpropagating beams of light which are travelling in opposite directions all the way around Earth's orbit.

Unless you can do that, you are full of shit.

Using your SINGLE INTERFEROMETER, you proceeded to do calculations, as follows.

Yes, I know the calculations I did and how they showed you were full of shit and that the orbital Sagnac effect is much smaller than the rotation Sagnac effect due to the significantly lower angular velocity of the orbit.

YOU USED THE SAME AREA AND THE SAME RADIUS IN YOUR CALCULATIONS!!!

Yes, because I was being honest and comparing the effects in the same interferometer loop, rather than being extremely dishonest and using a hypothetical interferometer which sends light beams all the way around Earth's orbit.

AGAIN, YOU USED THE SAME AREA AND RADIUS TO SIMPLIFY THE EQUATIONS!!!
A terrible mistake.

No. Not to simplify the equations and not a mistake. It was to make an honest comparison, something you are unable to do as it shows you are full of shit.

That is not how the calculations are done.

No, it is.
And your sources agree with me. You use the area for the interferometer loop, i.e. the physical loop you are using to measure the Sagnac effect.
You do not use the area of the orbit or how far you are from the centre of rotation.
You use the area of the loop.

I proved that with my derivation which you are yet to show a single error with.

Make sure you understand that the Sagnac effect applies to translational/linear paths.

No, it doesn't. It applies to rotating paths.
It also applies in a more complex manner to closed loops where the loop itself remains but your position along it changes.

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.

No, he didn't, and MM disproved that.

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

No, not a uniformly moving fibre. A fibre where some sections where moving in one direction, other sections were moving in another direction, some sections were going around a curve and so on.

So no, it was more akin to increasing the area of the loop, and changing how the rotation works.
It was not a linear effect, it still required the loop. It still required 2 counterpropagating beams of light following the loop in different directions and being affected by the motion differently.

If you take a uniformly moving fibre, you find no effect.

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.

Again, there was no experiment carried out with uniformly moving fibre.

And there you have a nice picture of your failure.
The fibre isn't uniformly moving.
You have F and R fixed in space.

So you have a terrible error on your part.

A terrible error on your part, to assume a single interferometer.
But there are two of them, just as shown above.

Nope. Just the one. Even if you want to use GPS as an interferometer, it is still a single interferometer. We aren't sending their signals around the sun in a counterpropagating manner.

For the case where the Earth would orbit the Sun, YOU ALREADY HAVE A SHAPE TO START WITH, you cannot modify it, unless of course you'd care to tell your readers that the nearly circular shape of the orbit has a different geometrical shape.

You have the shape of the orbit. That is irrelevant to the shape of the interferometer.

Do you understand that?
Earth's orbit is not the interferometer.
We are not sending 2 beams of light in opposite directions around Earth's orbit and measuring a shift.
Instead, we are sending 2 beams of light around a small loop (as shown in the OP) and measuring a shift.

That small loop is the interferometer, not Earth's orbit.

Once you are stuck with a certain geometrical shape, in this case the orbital path of the Earth, there's nothing else that you can do.

Sure there is, you can construct an interferometer that doesn't follow that shape.

Please read my previous message.
It is irrefutable.

If it was irrefutable, how come I was able to refute it so easily?

It is my derivation which is irrefutable, as shown by your complete inability to refute it.
And no, bitching about the conclusion is not refuting it.
I provided a derivation showing that you use the area of the loop. You need to show an error in that, such as an assumption that was wrong or a math error. Using the same area was a conclusion, not an assumption.

jack, shut the f*ck up.

It's over and you know it.

Yes, Earth is moving along it, in one direction. It is not 2 counter-propagating beams of light which travel around the loop.

The calculations for the rotational Sagnac.

The speed of light is:
c = 299,792 km/s

The circumference of the earth at the equator is:
C = 40,075 km

The duration of one day is:
d = 24 hr = 1440 min = 86400 sec

So now we can calculate the instantaneous tangential velocity of the earth at the equator to be:
v = (C / d) = 40,075 km / 86400 sec = 0.4638310185 km/sec

And we can calculate the amount of time it would take light to travel once around a non-rotating circumference:
t = (C / c) = 40,075 km / 299,792.458 km/s = 0.1336758111s

But, in the heliocentric hypothesis, the earth circumference is rotating, so there are actually going to be two different times in two different directions:
t1 = (C / (c - v)) = 40,075 km / (299,792.458 - 0.4638310185) km/s = 0.1336760179s
t2 = (C / (c + v)) = 40,075 km / (299,792.458 + 0.4638310185) km/s = 0.1336756043s

The difference between the above times is:
(t - t1) = 0.1336758111s - 0.1336760179s = -.0000002068s = -207 nanoseconds
(t - t2) = 0.1336758111s - 0.1336756043s = +.0000002068s = +207 nanoseconds


The calculations for the orbital Sagnac:

The speed of light is:
c = 299,792.458 km/s

The circumference of the earth orbit:
O = 939,951,145 km

The duration of one year is:
Y = d*(365.25) = 86400*(365.25) = 31,557,600s

So now we can calculate the instantaneous tangential velocity of the earth orbit to be:
V = (O / Y) = 939,951,145 km / 31,557,600s = 29.785254 km/s

And we can calculate the amount of time it would take light to travel once around a non-rotating circumference:
T = (O / c) = 939,951,145 km / 299,792.458 km/s = 3135.339532s

But, under the heliocentric hypothesis, the earth orbit is revolving around the sun, so there are actually going to be two different times in two different directions:
T1 = (O / (c - V)) = 939,951,145 km / (299,792.458 - 29.785254) km/s = 3135.651068s
T2 = (O / (c + V)) = 939,951,145 km / (299,792.458 + 29.785254) km/s = 3135.028057s

The difference between the above times is:
(T - T1) = 3135.339532s - 3135.651068s = -0.311536 seconds
(T - T2) = 3135.339532s - 3135.028057s = +0.311536 seconds

The percentage of orbital path that the earth occupies at any given time:
P = (C / (2πO)) = 40,075 km / (6.283185307)(939,951,145 km) = .0006785601973%

Correcting the time differences for the above percentage:
(-.311536s)(.0006785601973%) = -.000002113959296s = -2,114 nanoseconds
(+.311536s)(.0006785601973%) = +.000002113959296s = +2,114 nanoseconds


The mistake committed by you, jack, is amateurish to say the least.

Luckily for you, I spared you the embarrassment of sending your piece of shit analysis to any of the journals listed in my previous messages.


"By applying Stokes’ rule the derivation of Sagnac effect can be changed from an
integration over a surface to an integration along a line, which is correct in relation to
where the light really is. This demonstrates that Sagnac effect is translational, and that an ether-wind has been detected by Sagnac and by the GPS system.

Since Sagnac effect is an effect in light that is enclosed inside an optical fiber we can
conclude that Sagnac effect is distributed along a line and not over an area. No light and
no rotation exists in the enclosed area. Sagnac detected therefore an effect of translation
although he had to rotate the equipment to produce the effect inside the fiber. By this
solution Sagnac found a method to circumvent Einstein's clock synchronization problem. The Sagnac effect is distributed in every small part of the line.

The fact that Sagnac effect is caused by translation means that the same effect as in a rotating circle also must exist in a translating straight line.

The most important error regarding Sagnac effect is the classification of the effect as rotational. The effect is translational since it is distributed along a line. This means that the same effect must exist along a straight line. In the global positioning system (GPS) a compensation for this translational effect is done.  The high precision in the GPS system demands this correction, when time stations on our planet are compared.

The GPS system cannot afford to ignore the ether wind."


"If the Sagnac effect can be produced in linear uniform motion, then the claim that it
is a characteristic of radial motion is simply incorrect. Because the rules of SR apply to linear uniform motion, the only conclusion is that SR is incorrect."


In 1938 Ives showed by analysis that the measured Sagnac effect would be unchanged if the Sagnac phase detector were moved along a cord of a hexagon-shaped light path rather than rotating the entire structure. Thus, he showed the effect could be induced without rotation or acceleration.


No, not a uniformly moving fibre.

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)

The title of the first paper is:

Modified Sagnac experiment for measuring travel-time difference
between counter-propagating light beams in a uniformly moving fiber

Can you read English jack?

A fiber optic conveyor has been developed for investigating the travel-time difference between two counter-propagating light beams in uniformly moving fiber. Our finding is that there is a travel-time difference Δt = 2vΔl/c^2 in a fiber segment of length Δl moving with the source and detector at a speed v, whether the segment is moving uniformly or circularly.


From the second paper:

We can test the isotropy of the one-way speed of light in a system moving uniformly in a straight line and conduct the one-way Sagnac experiment.


A quote from the paper signed Tartaglia/Ruggiero which was used a bibliographical reference by me:

A “generalized Sagnac effect” arises in a uniformly moving fiber.


You have been shown to be a poor researcher in physics jack.

You do not stand a chance with me on the orbital Sagnac.

Europhysics Letters Journal

http://iopscience.iop.org/journal/0295-5075/page/For%20Authors

EPL publishes original, high-quality Letters in all areas of physics, ranging from condensed matter topics and interdisciplinary research to astrophysics, geophysics, plasma and fusion sciences, including those with application potential. Articles must contain sufficient argument and supporting information to satisfy workers in the field, and must also be of interest and relevance to wider sections of the physics community. In order to comply with general interest, special care should be directed to the introduction and conclusion sections of the articles. Both should be clearly written in a style comprehensible to the general physics community.

By publishing your work with EPL you can benefit from:

High standards of peer review – articles are sent to external reviewers (usually two) selected by EPL's co-editors, a truly international team of active research scientists in your field


PUBLISHED IN THE EUROPHYSICS LETTERS JOURNAL

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 author actually present a local-ether model (MLET, Modified Lorentz Ether Theory) in order to account for the MISSING ORBITAL SAGNAC EFFECT.


I repeat: this is an IOP article, the highest standard of mainstream science.

The paper was peer reviewed and published.

Each and every scientist working at that journal understood the meaning of these words:

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.

DR. C.C. SU'S PAPER WAS PUBLISHED AND PEER REVIEWED.

jack, please provide ANY mainstream paper which agrees with your derivation.

You cannot.

I have just listed mainstream bibliographical references which agree with me.

You have nothing going for you at all.

Your derivation accompanied by the same graphic has already been addressed.

You made a huge error from the very start.

YOU HAVE PRACTICALLY ELIMINATED THE SUN-EARTH DISTANCE, THE RADIUS IN THE SAGNAC FORMULA.

You wanted a round earth without curvature, now you want an orbital Sagnac without the Earth orbiting the Sun.

For a circular path of radius R, the formula: ∆t = 2vl/c^2, where v = ω R is the speed of the circular motion and l=2πR is the circumference of the circle.

Rotational Sagnac, r = 6378 km

ORBITAL SAGNAC, R = 150,000,000 KM

This is what you are missing.


Both Dr. Daniel Gezari and Dr. C.C. Su have published papers, in the most respected scientific journals, which include the correct orbital Sagnac calculation, based on a circular loop with the center of rotation located at the Sun.


You, jackblack, have eliminated the Sun-Earth distance with the stroke of a pen, and you ask innocently, "where did I go wrong?".


Let us see where your error creeps up in your calculations.

And thus:
dt_o/dt_r=k*w_o/k*w_r=w_o/w_o=1/365.

Just like I said.
I also backed up this formula with my own derivation.

jack claims that the orbital Sagnac is 1/365 of the rotational Sagnac, and that, read carefully: "I also backed up this formula with my own derivation."

Using his own very words, jack is telling us that his derivation leads directly to this figure: 1/365.


Here is another quote now:

This is the correct calculation:
Δt_o/Δt _r=[4A_iω_o/( c² - v_o²)] / [4A_iω_r/( c² - v_r²)]

Note, the area here has nothing to do with the area of Earth's orbit or radius of it or the radius of Earth. It is the area of the interferometer, as my derivation.

No where in any derivation did the area of Earth's orbit come into it.

So to continue:
Δt_o/Δt _r=[4A_iω_o/( c² - v_o²)] / [4A_iω_r/( c² - v_r²)]
Obviously, ( c² - v_o²) and ( c² - v_r²) are very close to the same number, so let's lave them off.
=4A_iω_o/ 4A_iω_r
Then to simplify:
=ω_o/ ω_r

And would you look at that? It ends up being just like what we claim.
You have ω_o/ ω_r.
As Earth rotates roughly 365 times for each orbit, ω_r=365*ω_o.
Thus we get:
Δt_o/Δt _r=ω_o/ ω_r
=ω_o/ (365*ω_o)
=1/365

Just like we claim.

A very clear claim based on jack's derivation.

These are his own very words.


He claims that the derivation, as shown above, leads to the figure 1/365.


And he says that he bases his entire derivation on this: Note, the area here has nothing to do with the area of Earth's orbit or radius of it or the radius of Earth.

This hypothesis, in turn, jack claims leads to this conclusion:

=1/365

Just like we claim.


jack simply threw out the basic requirement that the orbital Sagnac is a circular loop, and reached the conclusion that the orbital Sagnac is 1/365 of the rotational Sagnac.


HERE IS YOUR ERROR OF JUDGEMENT.


Note, the area here has nothing to do with the area of Earth's orbit or radius of it or the radius of Earth. It is the area of the interferometer, as my derivation.

No where in any derivation did the area of Earth's orbit come into it.


For a circular path of radius R, the formula: ∆t = 2vl/c^2, where v = ω R is the speed of the circular motion and l=2πR is the circumference of the circle.

Rotational Sagnac, r = 6378 km

ORBITAL SAGNAC, R = 150,000,000 KM

This is what you are missing.


You threw out the R = 150,000,000 km requirement of the orbital Sagnac effect.


No where in any derivation did the area of Earth's orbit come into it.

Then, what you are saying is that the Earth is stationary and does not orbit the Sun, since you eliminated the very Sun-Earth distance with the stroke of a pen. Take a look at your unbelievable delusion jack. You threw out the 150,000,000 km distance. That is why your piece of shit analysis is just that, a piece of shit.


Take a look at how a CORRECT CALCULATION IS PERFORMED.




Can you read English, jack?

Based on the local-ether model, the propagation is entirely
independent of the earth’s orbital motion around
the sun or whatever and the velocity v for such an earthbound
experiment is referred to an ECI frame and hence
is due to earth’s rotation alone. In the original proposal,
the velocity v was supposed to incorporate earth’s orbital
motion around the sun. Thus, at least, v^2/c^2
=~ 10^-8. Then the amplitude of the phase-difference variation
could be as large as π/3, when the wavelength is
0.6 µm and the path length is 10 m. However, as the velocity
v is the linear velocity due to earth’s rotation alone,
the round-trip Sagnac effect is as small as v^2/c^2∼ 10^-12 which is merely 10^-4 times that due to the orbital motion.


Your assumption lead to this figure: 1/365.

The calculations done by Dr. C.C. Su lead to this figure: 10,000


You are off by a scale of 3,650,000.

But it seems you are happier to resume your cognitive dissonance condition than to accept defeat.

Well, as he seems unable to provide a derivation, I guess I will provide my derivation.

So there are 2 light beams counterpropagating along the loop.
There may be a shift in the time taken for the 2 light beams to complete the loop due to the motion of the object.
The path's going out and in (in purple) are equivalent and result in no shift. That is to say, each light beam has a section where it would go straight out, and each has a section where it goes straight in.

So the only shift will be a result of the black lines.

As I said before, these can be approximated as 2 circular arcs.

So let's deal with Earth's rotation first, it has a radius of r, and is turning at a rate of w_r.
So along the inside section you have 2 beams of light going opposite directions.
One beam has the rotation make the path longer, where it moves l+w_r*r*t_i1 in time t_i1, while the other has it shrink and it travels l-w_r*r*t_i2 in time t_i2, with the light travelling at a speed of c.

Thus t_i1*c=l+w_r*r*t_i1=>t_i1*(c-w_r*r)=l=>t_i1=l/(c-w_r*r) and similarly t_i2=l/(c+w_r*r)

On the outer section, the light beams now travel the other way, so the one which used to have the path longer now has it shorter, also to complete the approximation, we need to make the path a little longer (if you don't you are out by a factor of 2). Rather than being l, it will be (l/r)*(r+l)=l*(1+l/r), and it thus has to travel l*(1+l/r)-w_r*(r+l)*t_o1 in time t_o1, while the other beam travels l*(1+l/r)+w_r*(r+l)*t_o2 in time t_o2.

Thus c*t_o1=l*(1+l/r)-w_r*(r+l)*t_o1=>t_o1*(c+w_r*(r+l))=l*(1+l/r)=>t_o1=l*(1+l/r)/(c+w_r*(r+l)), and similarly t_o2=l*(1+l/r)/(c-w_r*(r+l))

So the time shift will simply be the difference between the sum (if you did it the other way you just get a negative value, but the same one):
dt=t_o2+t_i2-t_o1-t_i1
=l*((1+l/r)/(c-w_r*(r+l))+1/(c+w_r*r)-(1+l/r)/(c+w_r*(r+l))-1/(c-w_r*r))

If we were to simplify straight away we will just end up with 0 (l*((1+l/r)/c+1/c-(1+l/r)/c-1/c)), so we need to do something else first to find out what the small shift will be.
We also need to make sure we get a component for the outside and for the inside as each will contribute a small difference.

=l*((1+l/r)/(c-w_r*(r+l))-(1+l/r)/(c+w_r*(r+l))+1/(c+w_r*r)-1/(c-w_r*r))
=l*((1+l/r)*(c+w_r*(r+l)-c+w_r*(r+l))/(c²-(w_r*(r+l))²)+(c-w_r*r-c-w_r*r)/(c²-(w_r*r)²))
=2*l*((1+l/r)*w_r*(r+l)/(c²-(w_r*(r+l))²)-w_r*r/(c²-(w_r*r)²))

So now we can start the simplification.
w_r*(r+l)<<c, thus (w_r*(r+l))²)<<c², thus c²-(w_r*(r+l))²~=c², and likewise c²-(w_r*r)²~=c²
Thus:
dt=2*l*((1+l/r)*w_r*(r+l)/c²-w_r*r/c²)
=2*l*((1+l/r)*w_r*(r+l)-w_r*r)/c²
=2*l*(w_r*(r+l)+(l/r)*w_r*(r+l)-w_r*r)/c²
=2*l*(w_r*r+w_r*l+(l/r)*w_r*r+(l/r)*w_r*l-w_r*r)/c²
=2*l*(w_r*l+l*w_r+l*l*w_r/r)/c²
=2*l*l*w_r*(2+l/r)/c²

To simplify one step more, l<<r, thus l/r<<1<2, thus 2+l/r~=2
Thus:
dt=2*l*l*w_r*(2)/c²
=4*l*l*w_r/c²


dt=4*A*w_r/c²

Just like the simple formula, with no dependence on r at all, only w_r

So once again, we have shown the area is dependent on the area of loop, not the "orbit".

We can follow the above procedure for the Sagnac effect due to Earth's orbit, either jumping straight to the end, or going through each step where we substitue w_r with w_o, and we substitute r with R-r-l.
Either way, we end up with the equivalent result:
dt=4*A*w_o/c²

And thus by comparing them, we get the orbital sagnac as 1/365 times the rotational one.

Now then Sandy, rather than bitching about the conclusion, can you show anything wrong with the derivation?

jack, you are committing the very same error. Again.

You are using a single interferometer.

So, you are using A SINGLE AREA AND A SINGLE RADIUS.

That is why you get a final result which is worthless.

So the only shift will be a result of the black lines.

You followed the same steps before.

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.


YOU USED THE SAME AREA AND THE SAME RADIUS IN YOUR CALCULATIONS!!!

So to continue:
Δto/Δt r=[4Aiωo/( c² - vo²)] / [4Aiωr/( c² - vr²)]
Obviously, ( c² - vo²) and ( c² - vr²) are very close to the same number, so let's lave them off.
=4Aiωo/ 4Aiωr
Then to simplify:
=ωo/ ωr


AGAIN, YOU USED THE SAME AREA AND RADIUS TO SIMPLIFY THE EQUATIONS!!!


You cannot fool anybody here jack.


This is the CORRECT WAY TO PROCEED WITH THE CALCULATIONS:

The calculations for the rotational Sagnac.

The speed of light is:
c = 299,792 km/s

The circumference of the earth at the equator is:
C = 40,075 km

The duration of one day is:
d = 24 hr = 1440 min = 86400 sec

So now we can calculate the instantaneous tangential velocity of the earth at the equator to be:
v = (C / d) = 40,075 km / 86400 sec = 0.4638310185 km/sec

And we can calculate the amount of time it would take light to travel once around a non-rotating circumference:
t = (C / c) = 40,075 km / 299,792.458 km/s = 0.1336758111s

But, in the heliocentric hypothesis, the earth circumference is rotating, so there are actually going to be two different times in two different directions:
t1 = (C / (c - v)) = 40,075 km / (299,792.458 - 0.4638310185) km/s = 0.1336760179s
t2 = (C / (c + v)) = 40,075 km / (299,792.458 + 0.4638310185) km/s = 0.1336756043s

The difference between the above times is:
(t - t1) = 0.1336758111s - 0.1336760179s = -.0000002068s = -207 nanoseconds
(t - t2) = 0.1336758111s - 0.1336756043s = +.0000002068s = +207 nanoseconds


The calculations for the orbital Sagnac:

The speed of light is:
c = 299,792.458 km/s

The circumference of the earth orbit:
O = 939,951,145 km

The duration of one year is:
Y = d*(365.25) = 86400*(365.25) = 31,557,600s

So now we can calculate the instantaneous tangential velocity of the earth orbit to be:
V = (O / Y) = 939,951,145 km / 31,557,600s = 29.785254 km/s

And we can calculate the amount of time it would take light to travel once around a non-rotating circumference:
T = (O / c) = 939,951,145 km / 299,792.458 km/s = 3135.339532s

But, under the heliocentric hypothesis, the earth orbit is revolving around the sun, so there are actually going to be two different times in two different directions:
T1 = (O / (c - V)) = 939,951,145 km / (299,792.458 - 29.785254) km/s = 3135.651068s
T2 = (O / (c + V)) = 939,951,145 km / (299,792.458 + 29.785254) km/s = 3135.028057s

The difference between the above times is:
(T - T1) = 3135.339532s - 3135.651068s = -0.311536 seconds
(T - T2) = 3135.339532s - 3135.028057s = +0.311536 seconds

The percentage of orbital path that the earth occupies at any given time:
P = (C / (2πO)) = 40,075 km / (6.283185307)(939,951,145 km) = .0006785601973%

Correcting the time differences for the above percentage:
(-.311536s)(.0006785601973%) = -.000002113959296s = -2,114 nanoseconds
(+.311536s)(.0006785601973%) = +.000002113959296s = +2,114 nanoseconds


It is a loop and the earth is moving along the loop in its orbit around the sun.

If light travels at one speed c, then as the earth supposedly moves in it's revolution loop at 30k/s, while light moves c through space, the unit at the equator at noon would move with the earth' rotation and the earth's revolution cutting the distance the signal must travel to meet the unit.


"Let's say the unit is at the equator and the satellite is low on the horizon in the east at noon.

That means the unit is traveling at the orbital speed of the earth at 67,000 MPH.

The satellite emits at one speed c in space. While the light travels through space toward the unit at c, the unit moves with the earth at 67,000 MPH. The unit cuts the distance that the light must travel.

This is not being seen by any experiements nor GPS."

Yet, this same logic applies and works with the earth's supposed rotation.

jack, shut the f*ck up.

No. I will keep talking and showing you to be full of shit.
If you want me to shut up, show the problem with my derivation and show the correct one.

It's over and you know it.

Yes, to some extent. Now I am just beating a dead horse. You are completely unable to show a problem with my derivation and thus it is over for you. Until you can show a problem, your BS stands refuted.

The circumference of the earth at the equator is:
And we can calculate the amount of time it would take light to travel once around a non-rotating circumference:
The circumference of the earth orbit:
And we can calculate the amount of time it would take light to travel once around a non-rotating circumference:

Irrelevant.
We are not sending beams of light all the way around Earth. We are sending it around this small loop.

What is the expected sagnac effect on this loop?


If you wish to use those calculations to dishonestly compare the 2, then go and construct these interferometers and see what results you get.
Have fun constructing an interferometer which follows Earth's orbit.

The mistake committed by you, jack, is amateurish to say the least.

What mistake? You are yet to show any.
All you can do is continually repeat the same dishonest crap.

Luckily for you, I spared you the embarrassment of sending your piece of shit analysis to any of the journals listed in my previous messages.

You mean your embarrassment, where you continually get your ass handed to you and make dishonest comparisons?

By applying Stokes’ rule the derivation of Sagnac effect can be changed from an
integration over a surface to an integration along a line, which is correct in relation to
where the light really is. This demonstrates that Sagnac effect is translational, and that an ether-wind has been detected by Sagnac and by the GPS system.

Nope. If it was the result of aether wind you would not need a loop and 2 counterpropogating beams of light. You would be able to do it with an interferometer like that used by MM.

It is not simply a result of translation.

Since Sagnac effect is an effect in light that is enclosed inside an optical fiber we can
conclude that Sagnac effect is distributed along a line and not over an area.

But since we get the same result for different shapes, like squares, with it based upon the area, we can't.

The fact that Sagnac effect is caused by translation means that the same effect as in a rotating circle also must exist in a translating straight line.

You mean the baseless assertion that it is caused by translation would lead to that.
The fact we don't observe this and instead need a loop shows this is not the case.

"If the Sagnac effect can be produced in linear uniform motion

Irrelevant as it is yet to be produced in linear uniform motion.

In 1938 Ives showed by analysis that the measured Sagnac effect would be unchanged if the Sagnac phase detector were moved along a cord of a hexagon-shaped light path

i.e. rotating it in a more convoluted way.
To show it is an effect of translation he would need to translate the entire apparatus equally, rather than differently for each part.

No, not a uniformly moving fibre.
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)

Neither of which have a uniformly moving fibre.

The title of the first paper is:

I don't give a damn. The simple fact is the fibre is not uniformly moving.
Look at the experimental and results, not the title.

By "uniformly moving" they meant moving at the same speed, but the velocity is quite different.

Can you read English jack?

Yes. Can you? If so, GO AND READ WHAT I SAID ABOVE!!

You have been shown to be a poor researcher in physics jack.

Nope. That would be you, continually focusing on pathetic quotes rather than what was actually shown.

You do not stand a chance with me on the orbital Sagnac.

I have already wiped the floor repeatedly with you. You are the one that doesn't stand a chance.

Skipping your repeated, refuted BS regarding EPL

Your derivation accompanied by the same graphic has already been addressed.

You mean accompanied by a different graphic.
You have repeatedly commenting regarding it, but you are yet to point out any problem.

You made a huge error from the very start.
YOU HAVE PRACTICALLY ELIMINATED THE SUN-EARTH DISTANCE, THE RADIUS IN THE SAGNAC FORMULA.

Nope. I started with that distance in it.
I then followed through with the math, and it eliminated itself.
The math shows it is not dependent on the sun-Earth distance. It is only dependent upon the area of the loop.

You wanted a round earth without curvature, now you want an orbital Sagnac without the Earth orbiting the Sun.

Nope. I have never wanted a round Earth without curvature, nor have I wanted an orbital sagnac without an orbit.
On the other hand, you want a sagnac effect without 2 counterpropagating beams of light.

For a circular path of radius R, the formula: ∆t = 2vl/c^2, where v = ω R is the speed of the circular motion and l=2πR is the circumference of the circle.

Yes, the circumference of the circular path that the light propagates around in both directions.
That is the path of the loop, not the path of the orbit.
The 2 are very different. Stop pretending they are the same.

Rotational Sagnac, r = 6378 km
ORBITAL SAGNAC, R = 150,000,000 KM
This is what you are missing.

I am missing nothing.
It is the same loop, thus it would have the same radius.
That is what you are repeatedly ignoring.
Rotational Sagnac, r = x
ORBITAL SAGNAC, R = x

They are the same.

Both Dr. Daniel Gezari and Dr. C.C. Su have published papers, in the most respected scientific journals, which include the correct orbital Sagnac calculation, based on a circular loop with the center of rotation located at the Sun.

And that is not what we are discussing.
We are discussing a small interferometer on Earth, with a small loop which has its centre no where near the sun.
So that calculation is irrelevant.

You, jackblack, have eliminated the Sun-Earth distance with the stroke of a pen, and you ask innocently, "where did I go wrong?".

Nope. Not the stroke of a pen, the math that shows it isn't needed.
You need to show the problem with that math. Until you do, you have no basis to say I have gone wrong.

Let us see where your error creeps up in your calculations.

By that do you mean the calculations in the derivation or just the conclusion?
If the conclusion, showing it doesn't match what you want isn't showing a problem.

You need to show an error in the math used for the derivation.

jack claims that the orbital Sagnac is 1/365 of the rotational Sagnac, and that, read carefully: "I also backed up this formula with my own derivation."

So just more bitching about the conclusion, not showing any problem with my derivation at all.

No where in any derivation did the area of Earth's orbit come into it.

That's right, because it doesn't need to.
But the radius of the orbit did. But the math then eliminated it as it isn't what the Sagnac effect is dependent upon.

A very clear claim based on jack's derivation.

No. A conclusion based upon the derivation, a derivation you are yet to show an issue with. Instead you just repeatedly bitch and moan about it not matching what you want.

jack simply threw out the basic requirement that the orbital Sagnac is a circular loop, and reached the conclusion that the orbital Sagnac is 1/365 of the rotational Sagnac.

Again, that isn't the basic requirement. The basic requirement is that you have 2 beams of light travelling along the same loop in opposite directions. I have that.

HERE IS YOUR ERROR OF JUDGEMENT.

Again, focusing on conclusions rather than the derivation. That is not showing any error in my derivation. It just shows you don't like the conclusion.

Take a look at your unbelievable delusion jack. You threw out the 150,000,000 km distance.

No I didn't. It was used, but was illiminated.
The R is in there in the form of
dt=2*omega*alpha*(R₂²-R₁²)/c2
R₂ and R₁ and the radii of the 2 circular arcs. These are the orbital distances from the sun (or Earth or whatever).
So they were used in the derivation.
But the area of the loop is alpha*(R₂²-R₁²)/2, so they are eliminated from the final result.

Take a look at how a CORRECT CALCULATION IS PERFORMED.

Which doesn't focus on the original problem.
You have already provided links to other calculations which clearly state A IS THE AREA OF THE LOOP!!! No mention of the orbital radius at all.

Can you read English, jack?

Again, can you?
If you can, why are you continually bringing up irrelevant crap?

Your assumption lead to this figure: 1/365.

No. My correct calculation leads to this.

The calculations done by Dr. C.C. Su lead to this figure: 10,000

The calculations for something completely different, as such  you cannot simply compare the two.

But it seems you are happier to resume your cognitive dissonance condition than to accept defeat.

Nope. That would be you, happier to continue bitching about the conclusions not matching what you want while being completely unable to show any error in the derivation.

jack, you are committing the very same error. Again.

No. I am using the same correct approach again, and you are bitching about it again.

You are using a single interferometer.

Again, to compare the Sagnac effect from Earth's orbit and Earth's rotation you need to use the same system, the same interferometer.
It is a dishonest comparison to compare 2 different interferometer results.

That is why you get a final result which is worthless.

No. This is why I get an honest result from which a meaningful comparison can be made.
Using different interferometers produces a meaningless result which makes comparisons impossible.

it would be akin to using an hourglass to measure a period of time, where to do it honestly, you use the same one, while your comparison has one where it takes 1 second for the grains of sand to pass through while you measure another where it takes 5 minutes to pass through.

That is not an honest comparison at all.

You followed the same steps before.

Not quite the same, but I reached the same result where you end up with the Sagnac effect being proportional to the area of the loop, not the area of the orbit.

AGAIN, YOU USED THE SAME AREA AND RADIUS TO SIMPLIFY THE EQUATIONS!!!

No, I used different radii which were eliminated by the math.
I used the same area to produce a meaningful comparison.

It is a loop and the earth is moving along the loop in its orbit around the sun.

Again, EARTH IS MOVING ALONG THE ORBIT!!!
The light is moving around the interferometer loop. This loop is much smaller than the orbit of Earth.

If light travels at one speed c, then as the earth supposedly moves in it's revolution loop at 30k/s, while light moves c through space, the unit at the equator at noon would move with the earth' rotation and the earth's revolution cutting the distance the signal must travel to meet the unit.

This would be aether drift, not Sagnac. Try again.
Again, the basic requirement for Sagnac is a loop which has 2 counterpropagating beams of light.
Where is your loop?
Where are you counterpropagating beams of light?

This is not being seen by any experiements nor GPS.

Yes, because the aether doesn't exist.

Yet, this same logic applies and works with the earth's supposed rotation.

Again, it isn't. Completely different logic applies and works for Earth's actual rotation, which is based upon rotation and relative velocity.

Now then, quit with all the bullshit and focus on the derivation itself. This will be the last time (at least for this thread) where I discuss other irrelavent BS.

Deal with the Sagnac effect on the interferometer in the OP, nothing else.

Show a derivation for the Sagnac effect IN THIS INTERFEROMETER or one very much like it, due to both Earth's orbit and rotation, or show an error IN MY DERIVATION (not the conclusion, the derivation itself).

If you are unable to then shut up.

jack, there is no one in the mainstream scientific community agreeing with you.

You have no other bibliographical references to back up your piece of shit derivation.

I, on the other hand, have brought before the readers multiple sources, IOP articles, which agree with me.

You are on your own with this mindless approach where you are using a single interferometer.

We are not sending beams of light all the way around Earth.

You still don't get it.

Where is your loop?
Where are you counterpropagating beams of light?

This is the CORRECT WAY TO PROCEED WITH THE CALCULATIONS:

The calculations for the rotational Sagnac.

The speed of light is:
c = 299,792 km/s

The circumference of the earth at the equator is:
C = 40,075 km

The duration of one day is:
d = 24 hr = 1440 min = 86400 sec

So now we can calculate the instantaneous tangential velocity of the earth at the equator to be:
v = (C / d) = 40,075 km / 86400 sec = 0.4638310185 km/sec

And we can calculate the amount of time it would take light to travel once around a non-rotating circumference:
t = (C / c) = 40,075 km / 299,792.458 km/s = 0.1336758111s

But, in the heliocentric hypothesis, the earth circumference is rotating, so there are actually going to be two different times in two different directions:
t1 = (C / (c - v)) = 40,075 km / (299,792.458 - 0.4638310185) km/s = 0.1336760179s
t2 = (C / (c + v)) = 40,075 km / (299,792.458 + 0.4638310185) km/s = 0.1336756043s

The difference between the above times is:
(t - t1) = 0.1336758111s - 0.1336760179s = -.0000002068s = -207 nanoseconds
(t - t2) = 0.1336758111s - 0.1336756043s = +.0000002068s = +207 nanoseconds


The calculations for the orbital Sagnac:

The speed of light is:
c = 299,792.458 km/s

The circumference of the earth orbit:
O = 939,951,145 km

The duration of one year is:
Y = d*(365.25) = 86400*(365.25) = 31,557,600s

So now we can calculate the instantaneous tangential velocity of the earth orbit to be:
V = (O / Y) = 939,951,145 km / 31,557,600s = 29.785254 km/s

And we can calculate the amount of time it would take light to travel once around a non-rotating circumference:
T = (O / c) = 939,951,145 km / 299,792.458 km/s = 3135.339532s

But, under the heliocentric hypothesis, the earth orbit is revolving around the sun, so there are actually going to be two different times in two different directions:
T1 = (O / (c - V)) = 939,951,145 km / (299,792.458 - 29.785254) km/s = 3135.651068s
T2 = (O / (c + V)) = 939,951,145 km / (299,792.458 + 29.785254) km/s = 3135.028057s

The difference between the above times is:
(T - T1) = 3135.339532s - 3135.651068s = -0.311536 seconds
(T - T2) = 3135.339532s - 3135.028057s = +0.311536 seconds

The percentage of orbital path that the earth occupies at any given time:
P = (C / (2πO)) = 40,075 km / (6.283185307)(939,951,145 km) = .0006785601973%

Correcting the time differences for the above percentage:
(-.311536s)(.0006785601973%) = -.000002113959296s = -2,114 nanoseconds
(+.311536s)(.0006785601973%) = +.000002113959296s = +2,114 nanoseconds


It is a loop and the earth is moving along the loop in its orbit around the sun.


Your result is 1/365.

Dr. C.C. Su's result is 10,000.

Published and peer reviewed in an IOP article.

You are off by a factor of 3,650,000.

Because you used a single interferometer.


This would be aether drift, not Sagnac.

But it is Sagnac.

Try to imagine the center of the earth as the sun and the earth's orbit as the surface. So, the earth is in a rotationa/circular pattern in its orbit. Just like the rotational sagnac, the earth rotates toward where the satellite emitted the signal, and with the orbit, the earth, the unit is orbited toward where the signal was emitted.


You are going against mainstream science which reveals your cognitive dissonance condition.

Rotational Sagnac, r = x
ORBITAL SAGNAC, R = x

They are the same.

This is madness.

r = 6,378 km

R = 150,000,000 km

How can they be the same?

Neither of which have a uniformly moving fibre.

A fiber optic conveyor has been developed for investigating the travel-time difference between two counter-propagating light beams in uniformly moving fiber. Our finding is that there is a travel-time difference Δt = 2vΔl/c^2 in a fiber segment of length Δl moving with the source and detector at a speed v, whether the segment is moving uniformly or circularly.

We conclude that a segment of uniformly moving fiber with a speed of v and a length of Δl contributes Δφ = 4πvΔl/cλ or Δt = 2vΔl/c^2 , like a segment of circularly moving fiber does.

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

PUBLISHED IN THE PRESTIGIOUS JOURNAL: PHYSICS LETTER A

PEER REVIEWED

You are on your own on this one jack.

Make no mistake about it, you are totally defeated here.

Short answer:

The Sagnac delay is, whether you agree or not, independent of the shape of the loop and its centre of rotation.

For the case where the Earth would orbit the Sun, YOU ALREADY HAVE A SHAPE TO START WITH, you cannot modify it, unless of course, you'd care to tell your readers that the nearly circular shape of the orbit has a different geometrical shape.

Irrelevant because the signal/light path does not go to the sun and back, end of story!

Please explain which words in the following that you cannot understand!
If you are having difficulty understanding plain English there are numerous online dictionaries,

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 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.

From what I can see though, the experiments of:

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

Certaily prove that the earth rotates and orbits the sun.

Notice the 30km/s speed.

This is Sagnac.

Try to imagine the center of the earth as the sun and the earth's orbit as the surface. So, the earth is in a rotational/circular pattern in its orbit. Just like the rotational sagnac, the earth rotates toward where the satellite emitted the signal, and with the orbit, the earth, the unit is orbited toward where the signal was emitted.

Then we simply apply the well-known formula to this situation, where R = 150,000,000 km, v = wR (=30km/s).


What jack has done is to use the wrong radius for the orbital sagnac and then claim the distance to the sun is the same as the radius of the earth and he comes up with a sagnac based only on angular velocity.

But the sagnac is a function of the product of the area swept out by the path and the angular velocity. That path is based on the distance from the center of rotation to the path.

This correct calculation is supported by mainstream papers.

Which are peer reviewed.

The v of the moving frame is what matters in the calculations for the sagnac effect.

And v = 30km/s in the heliocentric hypothesis.

The radius for such a speed is 150,000,000.


But jack has no problem at all to sweep all this away.

Rotational Sagnac, r = x
ORBITAL SAGNAC, R = x

They are the same.

This is madness.

r = 6,378 km

R = 150,000,000 km

How can they be the same?


This is why the relativitsts HAD to invent a local aether model which hides the orbital Sagnac.

This was published in a mainstream respected journal.

Journal of Electromagnetic Waves and Applications:

http://www.ee.nthu.edu.tw/ccsu/qem/f3c.pdf

For the interplanetary propagation, earth’s orbital
motion contributes to the Sagnac effect as well. This local-ether model
has been adopted to account for the Sagnac effect due to earth’s
motions in a wide variety of propagation phenomena, particularly the
global positioning system (GPS), the intercontinental microwave link,
and the interplanetary radar.


The peer reviewers at the Journal of Electromagnetic Waves and Applications agree that the orbital Sagnac is larger than the rotational Sagnac, that it is missing, and that a local-ether model has to be adopted in order to account for this fact.


What counts is the distance of the light emission point to the hand held unit at the time of emission. The unit moves toward that position and so sagnac must be applied.

The speed is 30km/s.

The radius is 150,000,000 km.


jack has failed to provide papers that support his position.


Here is Dr. C.C. Su explaining how the orbital Sagnac is much greater than the rotational Sagnac:



Based on the local-ether model, the propagation is entirely
independent of the earth’s orbital motion around
the sun or whatever and the velocity v for such an earthbound
experiment is referred to an ECI frame and hence
is due to earth’s rotation alone. In the original proposal,
the velocity v was supposed to incorporate earth’s orbital
motion around the sun. Thus, at least, v2/c2
=~ 10-8. Then the amplitude of the phase-difference variation
could be as large as π/3, when the wavelength is
0.6 µm and the path length is 10 m. However, as the velocity
v is the linear velocity due to earth’s rotation alone,
the round-trip Sagnac effect is as small as v2/c2∼ 10-12 which is merely 10-4 times that due to the orbital motion.


The Sagnac effect is a FIRST ORDER effect in v/c.

Even in the round-trip nature of the Sagnac effect, as it was applied in the Michelson-Morley experiment, thus becoming a second order effect within that context, we can see that the ORBITAL SAGNAC IS 10,000 TIMES GREATER than the rotational Sagnac effect.


"does not depend on the shape of the surface A;"
"does not depend on the location of the centre of rotation;"

This provision was made for situations where this kind of calculations might be made, BUT NOT HERE in the case of the orbital path of the Earth.

Dr. C.C. Su's was published so far by these journals:


HARVARD UNIVERSITY

BULLETIN OF THE AMERICAN PHYSICAL SOCIETY

EUROPEAN PHYSICAL JOURNAL

EUROPHYSICS LETTERS JOURNAL

JOURNAL OF ELECTROMAGNETIC WAVES AND APPLICATIONS


Do you really think that the peer reviewers over at the Journal of Electromagnetic Waves and Applications could not have reminded Dr. Su of the quote from E.J. Post's 1967 paper?

But they could not, as they know very well that the orbital Sagnac is much larger than the rotational Sagnac.

http://www.ee.nthu.edu.tw/ccsu/qem/f3c.pdf

For the interplanetary propagation, earth’s orbital
motion contributes to the Sagnac effect as well. This local-ether model
has been adopted to account for the Sagnac effect due to earth’s
motions in a wide variety of propagation phenomena, particularly the
global positioning system (GPS), the intercontinental microwave link,
and the interplanetary radar.


Here is how to correctly calculate the rotational and the orbital Sagnac:

The calculations for the rotational Sagnac.

The speed of light is:
c = 299,792 km/s

The circumference of the earth at the equator is:
C = 40,075 km

The duration of one day is:
d = 24 hr = 1440 min = 86400 sec

So now we can calculate the instantaneous tangential velocity of the earth at the equator to be:
v = (C / d) = 40,075 km / 86400 sec = 0.4638310185 km/sec

And we can calculate the amount of time it would take light to travel once around a non-rotating circumference:
t = (C / c) = 40,075 km / 299,792.458 km/s = 0.1336758111s

But, in the heliocentric hypothesis, the earth circumference is rotating, so there are actually going to be two different times in two different directions:
t1 = (C / (c - v)) = 40,075 km / (299,792.458 - 0.4638310185) km/s = 0.1336760179s
t2 = (C / (c + v)) = 40,075 km / (299,792.458 + 0.4638310185) km/s = 0.1336756043s

The difference between the above times is:
(t - t1) = 0.1336758111s - 0.1336760179s = -.0000002068s = -207 nanoseconds
(t - t2) = 0.1336758111s - 0.1336756043s = +.0000002068s = +207 nanoseconds


The calculations for the orbital Sagnac:

The speed of light is:
c = 299,792.458 km/s

The circumference of the earth orbit:
O = 939,951,145 km

The duration of one year is:
Y = d*(365.25) = 86400*(365.25) = 31,557,600s

So now we can calculate the instantaneous tangential velocity of the earth orbit to be:
V = (O / Y) = 939,951,145 km / 31,557,600s = 29.785254 km/s

And we can calculate the amount of time it would take light to travel once around a non-rotating circumference:
T = (O / c) = 939,951,145 km / 299,792.458 km/s = 3135.339532s

But, under the heliocentric hypothesis, the earth orbit is revolving around the sun, so there are actually going to be two different times in two different directions:
T1 = (O / (c - V)) = 939,951,145 km / (299,792.458 - 29.785254) km/s = 3135.651068s
T2 = (O / (c + V)) = 939,951,145 km / (299,792.458 + 29.785254) km/s = 3135.028057s

The difference between the above times is:
(T - T1) = 3135.339532s - 3135.651068s = -0.311536 seconds
(T - T2) = 3135.339532s - 3135.028057s = +0.311536 seconds

The percentage of orbital path that the earth occupies at any given time:
P = (C / (2πO)) = 40,075 km / (6.283185307)(939,951,145 km) = .0006785601973%

Correcting the time differences for the above percentage:
(-.311536s)(.0006785601973%) = -.000002113959296s = -2,114 nanoseconds
(+.311536s)(.0006785601973%) = +.000002113959296s = +2,114 nanoseconds


If light travels at one speed c, then as the earth supposedly moves in it's revolution loop at 30k/s, while light moves c through space, the unit at the equator at noon would move with the earth' rotation and the earth's revolution cutting the distance the signal must travel to meet the unit.


"Let's say the unit is at the equator and the satellite is low on the horizon in the east at noon.

That means the unit is traveling at the orbital speed of the earth at 67,000 MPH.

The satellite emits at one speed c in space. While the light travels through space toward the unit at c, the unit moves with the earth at 67,000 MPH. The unit cuts the distance that the light must travel.

This is not being seen by any experiements nor GPS."

Yet, this same logic applies and works with the earth's supposed rotation.

That path is based on the distance from the center of rotation to the path.

No.
Have you lost the ability to read?


If you claim that your papers prove otherwise, show a link to those papers, so they can be read in their entirety.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Funny that YOU refuse to analyse the off-centre Sagnac Loop yourself, but everyone else seem able to show this sort of thing!

Quote from: Eyal Schwartz, Published 30 November 2016 • © 2016 IOP Publishing Ltd , European Journal of Physics, Volume 38, Number 1

Sagnac effect in an off-center rotating ring frame of reference

Abstract
Interference resides deeply in our understanding of the wave properties of light. In this paper, the century famous Sagnac effect is demonstrated to be independent of the rotation axis position, using a rotating ring optical fiber in a straightforward laboratory experiment. A simple theoretical explanation for this result is given for any arbitrary closed loop interferometer. The level of this discussion should be suitable for undergraduate physics or engineering courses where electromagnetic theory and optics are discussed. The experiment described utilizes basic and important aspects in modern optics which every science student should acquire.

And

Quote from: A.G.Kelly

Sagnac showed experimentally that the centre of rotation can be away from the geometric centre of the apparatus, without affecting the above result.

From: TIME and the SPEED of LIGHT - a NEW INTERPRETATION, A.G.Kelly

Which really puts the kibosh on you interpretation of the "Orbital Sagnac Effect", so nice to know.
Bye bye any massive orbital Sagnac Effect.
And
 

Quote

This phase difference has been demonstrated in a rotating interferometer as well as in a geostationary one.
In the latter case, the Sagnac effect is due solely to earth’s rotation with ω_l = ω_E. The earth rotation rate ω_E is about 2π/(86400 s) and the corresponding maximum phase difference is as large as 2 rad, when the wavelength is 0.6 µm and the loop area S = 0.2 km². Thus, a loop interferometer can be utilized as a precise means to detect earth’s rotation rate.
Moreover, according to the local-ether model, earth’s orbital motion around the sun or others does not con-
tribute to the Sagnac effect in an earthbound propagation loop. In as early as 1904 Michelson supposed that
the Sagnac effect due to the orbital motion of the earth around the sun might be detectable, although the angular
speed of the orbital motion is about 1/365 times that of the rotation .

Plenty more where they came from!

The Sagnac delay is independent of the shape of the loop and on its centre of rotation!
Only on the area projected on the plane of rotation and the angular velocity.

And

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
You might also read:
          Sagnac Effect, E. J. POST, Rev. Mod. Phys. 39, 475 (1967) – Published 1 April 1967
Where Post reviews some of the Sagnac experiments and states:

Note that this too states, "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 he specifically states, "does not depend on the location of the centre of rotation;"

Now stop re-posting the same old complete crap.  Your pages and pages of repeated garbage don't impress anyone.

Just get this straight,
1) I do not trust your interpretation of the most learned papers and
2) Not all IOP papers are necessarily accepted theory. "Modern science" does allow dissenting views to be published.

rabinoz, you do not have experience in dealing with scientific papers.

You are omitting a crucial fact.

You have been reminded of this thing over and over again, but to you it is not a problem.

"Sagnac showed experimentally that the centre of rotation can be away from the geometric centre of the apparatus, without affecting the above result."


CAN BE! COULD BE!


Not always. Not for each and every experiment.


This is what you are missing.


This phase difference has been demonstrated in a rotating interferometer as well as in a geostationary one.
In the latter case, the Sagnac effect is due solely to earth’s rotation with ωl = ωE. The earth rotation rate ωE is about 2π/(86400 s) and the corresponding maximum phase difference is as large as 2 rad, when the wavelength is 0.6 µm and the loop area S = 0.2 km2. Thus, a loop interferometer can be utilized as a precise means to detect earth’s rotation rate.
Moreover, according to the local-ether model, earth’s orbital motion around the sun or others does not contribute to the Sagnac effect in an earthbound propagation loop. In as early as 1904 Michelson supposed that the Sagnac effect due to the orbital motion of the earth around the sun might be detectable, although the angular speed of the orbital motion is about 1/365 times that of the rotation .

Again, you cannot read.

We've been through this before.

MICHELSON SUPPOSED THAT THE ORBITAL SAGNAC MIGHT BE DETECTED, ALTOUGH HE THOUGHT THAT THE ANGULAR SPEED OF THE ORBITAL MOTION IS 1/365 TIMES THAT OF THE ROTATION.

Here is what was published though:

C.C. Su, "A Local-ether model of propagation of electromagnetic wave," in Bull. Am. Phys. Soc., vol. 45, no. 1, p. 637, Mar. 2000 (Minneapolis, Minnesota).

http://www.ee.nthu.edu.tw/ccsu/










Both the rotational and the orbital motions of the earth together with the orbital
motion of the target planet contribute to the Sagnac
effect. But the orbital motion of the sun has no effects
on the interplanetary propagation. On the other hand, as
the unique propagation frame in GPS and intercontinental
links is a geocentric inertial frame, the rotational motion
of the earth contributes to the Sagnac effect. But the orbital
motion of the earth around the sun and that of the
sun have no effects on the earthbound propagation. By
comparing GPS with interplanetary radar, it is seen that
there is a common Sagnac effect due to earth’s rotation
and a common null effect of the orbital motion of the sun
on wave propagation. However, there is a discrepancy in
the Sagnac effect due to earth’s orbital motion. Moreover,
by comparing GPS with the widely accepted interpretation
of the Michelson–Morley experiment, it is seen that
there is a common null effect of the orbital motions on
wave propagation, whereas there is a discrepancy in the
Sagnac effect due to earth’s rotation.


Based on this characteristic of uniqueness and switchability of the propagation frame,
we propose in the following section the local-ether model
of wave propagation to solve the discrepancies in the in-
fluences of earth’s rotational and orbital motions on the
Sagnac effect and to account for a wide variety of propagation
phenomena.


Anyway, the interplanetary Sagnac effect is due to
earth’s orbital motion around the sun as well as earth’s
rotation. Further, for the interstellar propagation where
the source is located beyond the solar system, the orbital
motion of the sun contributes to the interstellar Sagnac
effect as well.

Evidently, as expected, the proposed local-ether model
accounts for the Sagnac effect due to earth’s rotation and
the null effect of earth’s orbital motion in the earthbound
propagations in GPS and intercontinental microwave link
experiments. Meanwhile, in the interplanetary radar, it accounts
for the Sagnac effect due both to earth’s rotation
and to earth’s orbital motion around the sun.


Based on the local-ether model, the propagation is entirely
independent of the earth’s orbital motion around
the sun or whatever and the velocity v for such an earthbound
experiment is referred to an ECI frame and hence
is due to earth’s rotation alone. In the original proposal,
the velocity v was supposed to incorporate earth’s orbital
motion around the sun. Thus, at least, v²/c²
=~ 10^-8. Then the amplitude of the phase-difference variation
could be as large as π/3, when the wavelength is
0.6 µm and the path length is 10 m. However, as the velocity
v is the linear velocity due to earth’s rotation alone,
the round-trip Sagnac effect is as small as v²/c²∼ 10^-12 which is merely 10^-4 times that due to the orbital motion.


The Sagnac effect is a FIRST ORDER effect in v/c.

Even in the round-trip nature of the Sagnac effect, as it was applied in the Michelson-Morley experiment, thus becoming a second order effect within that context, we can see that the ORBITAL SAGNAC IS 10,000 TIMES GREATER than the rotational Sagnac effect.


Just take a look at your superficial and laughable research rabinoz.


Dr. Su's paper was published not only by the IOP, but also by the following journals:

HARVARD UNIVERSITY

BULLETIN OF THE AMERICAN PHYSICAL SOCIETY

EUROPEAN PHYSICAL JOURNAL

EUROPHYSICS LETTERS JOURNAL

JOURNAL OF ELECTROMAGNETIC WAVES AND APPLICATIONS

Do you really think that these reviewers could not have reminded Dr. Su of the paper published by Post in 1967?

Post's paper on the Sagnac effect was the first to be published in some 30 years.

Each and every researcher in the field of GPS technology, optical waves, has a perfect knowledge of this paper.

But they did not mention that quote to him.

Because it cannot be done.


It is very simple to understand the orbital Sagnac.



Notice the 30km/s speed.

This is Sagnac.

Try to imagine the center of the earth as the sun and the earth's orbit as the surface. So, the earth is in a rotational/circular pattern in its orbit. Just like the rotational sagnac, the earth rotates toward where the satellite emitted the signal, and with the orbit, the earth, the unit is orbited toward where the signal was emitted.

Then we simply apply the well-known formula to this situation, where R = 150,000,000 km, v = wR (=30km/s).


What jack has done is to use the wrong radius for the orbital sagnac and then claim the distance to the sun is the same as the radius of the earth and he comes up with a sagnac based only on angular velocity.

But the sagnac is a function of the product of the area swept out by the path and the angular velocity. That path is based on the distance from the center of rotation to the path.

This correct calculation is supported by mainstream papers.

Which are peer reviewed.

The v of the moving frame is what matters in the calculations for the sagnac effect.

And v = 30km/s in the heliocentric hypothesis.

The radius for such a speed is 150,000,000.


According to your warped logic, the rotational Sagnac would not need to be calculated using r = 6,378 km and v = 0.465 km/s. But that is precisely how it is being calculated. That is how Sagnac is calculated. The orbital path is the same thing. The same logic applies. Please read:

If light travels at one speed c, then as the earth supposedly moves in it's revolution loop at 30k/s, while light moves c through space, the unit at the equator at noon would move with the earth' rotation and the earth's revolution cutting the distance the signal must travel to meet the unit.


"Let's say the unit is at the equator and the satellite is low on the horizon in the east at noon.

That means the unit is traveling at the orbital speed of the earth at 67,000 MPH.

The satellite emits at one speed c in space. While the light travels through space toward the unit at c, the unit moves with the earth at 67,000 MPH. The unit cuts the distance that the light must travel.

This is not being seen by any experiements nor GPS."

Yet, this same logic applies and works with the earth's supposed rotation.


Tell you what.

FIND ANY PAPERS PUBLISHED IN THE SAME JOURNALS WHICH SUPPORT JACK'S VIEW THAT r = R (the radius of the Earth = the Radius of the orbital path of the Earth).

You won't find any.

Then, it is time for you to shut up and accept the fact that the orbital Sagnac is larger than the rotational Sagnac, and that it is missing.

jack, there is no one in the mainstream scientific community agreeing with you.

No. Mainstream science agrees with me. The aether was refuted long ago.

You have no other bibliographical references to back up your piece of shit derivation.

That is because unlike you, I am capable of making an argument from first principles instead of needing to rely upon fake authorities to pedal bullshit.
However there are plenty of references which can back me up, but they are irrelevant to the discussion.

If my derivation was a piece of shit you would be able to show what is wrong with it. Not bitch and moan about the conclusion, but show an actual problem.

You are on your own with this mindless approach where you are using a single interferometer.

Nope. All those that want an honest answer are with me.

Skipping your irrelevant (and already refuted) BS

Because you used a single interferometer.

Yes, as is required to show the Sagnac effect on a single system.

Try to imagine the center of the earth as the sun and the earth's orbit as the surface. So, the earth is in a rotationa/circular pattern in its orbit. Just like the rotational sagnac, the earth rotates toward where the satellite emitted the signal, and with the orbit, the earth, the unit is orbited toward where the signal was emitted.

But this would require the Satellite to orbit the sun, not Earth, at a significantly different velocity to the point on the surface.

You are going against mainstream science which reveals your cognitive dissonance condition.

Again, that would be you.
Mainstream science accepts Earth is round.
They accept that the aether does not exist.
They accept relativity.

Meanwhile, you reject all three. You claim Earth is round, the aether exists and relativity is wrong.
So no, you are the one gong against mainstream science. Not me.

Rotational Sagnac, r = x
ORBITAL SAGNAC, R = x
They are the same.
This is madness.
How can they be the same?

Because you are using 2 completely different r values.
We are discussing a single interferometer, in order to determine the effect of Earth's orbital and rotational sagnac effect on a particular system.

How can they be different when it is a single system (interferometer)?
Claiming they are different is madness.

The radius of Earth and Earth's orbit is irrelevant as my math demonstrated.

Neither of which have a uniformly moving fibre.
A fiber optic conveyor has been developed

Stop just repeating the same bullshit. Actually deal with what I have said.
Look at fig. 1. b.
Notice how the lower and upper sections are moving in completely different directions?
That is not uniformly moving.

What they meant by "uniformly moving" was that it was all moving at the same speed, not velocity. As such, it is not merely a translational effect.

And regardless, this is irrelevant to the discussion at hand.

Make no mistake about it, you are totally defeated here.

Once again, that would be you who is defeated.

Notice the 30km/s speed.
This is Sagnac.

No it isn't.
The sagnac effect is a fringe shift between 2 counterpropagating beams of light moving around a loop.

You don't have that here.

What jack has done is to use the wrong radius for the orbital sagnac

No, I used the correct one and had it be removed in the math as it isn't relavent.

then claim the distance to the sun is the same as the radius of the earth

No. I said the radius of the loop is the same as we are dealing with a single interferometer.

he comes up with a sagnac based only on angular velocity.

No, angular velocity and the area of the loop, as your papers agreed.

But the sagnac is a function of the product of the area swept out by the path and the angular velocity. That path is based on the distance from the center of rotation to the path.

And that only applies that simply when you have the centre of rotation on the centre of the loop.
If you don't, then this contributes one amount when moving along one section of the loop, and then contributes an opposite amount (not quite equal), when moving along the other section.

I already explained this to you in the other thread.

Remember this picture (which I repeatedly defeated you with):

You have a contribution from the arc with radius R2. This is based upon the area swept out by this arc, alpha*R2^2/2 (it would be 4*A*w/c^2).
You then have a contribution from the arc with radius R1. This is based upon the area swept out by this arc, and as the light beams are now travelling in an opposite direction and thus is of opposite sign and thus would be proportional to -alpha*R1^2/2.

So what happens when you add them up? You get alpha*R2^2/2-alpha*R1^2/2=(alpha/2)*(R2^2-R1^2), which is the area of the loop (the annular segment).
As such, the distance from the centre of the orbit is irrelevant. What matters is the area of the loop.

But jack has no problem at all to sweep all this away.

Yes, because your claims are pure bullshit, as my derivation showed.
Now stop making pathetic appeals to authority and show an actual error with the derivation or provide your own.

Remember, you need to use the interferometer in the OP (or the prior thread) and derive the Sagnac effect for it.

This is why the relativitsts HAD to invent a local aether model which hides the orbital Sagnac.

No, the "relativists" have completely discarded aether.

The peer reviewers at the Journal of Electromagnetic Waves and Applications agree that the orbital Sagnac is larger than the rotational Sagnac

No they don't. If you wish to claim they do, then PROVE IT!!!

What they actually agree is that an aether which isn't local doesn't work.

What counts is the distance of the light emission point to the hand held unit at the time of emission. The unit moves toward that position and so sagnac must be applied.

No. This is not Sagnac.
Sagnac is 2 counterpropagating beams of light in a loop.

jack has failed to provide papers that support his position.

Because I have a derivation that you are unable to refute.
I don't need to make pathetic appeals to authority.

However, you are wrong, yet again.
I have provided a paper (well, Rab did first, I don't care) that shows it is based upon the area of the loop, not on the distance from the centre of rotation.

Dr. C.C. Su's was published so far by these journals:
HARVARD UNIVERSITY

You are aware that is a university, not a journal?
Do you know the difference?

And I'm just skipping the rest of your repeated, refuted BS.

rabinoz, you do not have experience in dealing with scientific papers.
You are omitting a crucial fact.
You have been reminded of this thing over and over again, but to you it is not a problem.
"Sagnac showed experimentally that the centre of rotation can be away from the geometric centre of the apparatus, without affecting the above result."
CAN BE! COULD BE!

No. That would be you omitting this crucial fact.

The Sagnac effect is based upon the area of the loop, which can have a centre of rotation away from the loop.
This shows there is nothing wrong with my conclusion.

Not always. Not for each and every experiment.
This is what you are missing.

No. We aren't missing anything. The simple fact that you can have the centre of rotation away from the loop and still have the shift based upon the area of the loop and angular velocity shows you are completely wrong.

The fact that you can have the centre of rotation at the centre of the loop doesn't refute that at all.

This phase difference has been demonstrated in a rotating interferometer as well as in a geostationary one.
In the latter case, the Sagnac effect is due solely to earth’s rotation with ωl = ωE. The earth rotation rate ωE is about 2π/(86400 s) and the corresponding maximum phase difference is as large as 2 rad, when the wavelength is 0.6 µm and the loop area S = 0.2 km2. Thus, a loop interferometer can be utilized as a precise means to detect earth’s rotation rate.
Moreover, according to the local-ether model, earth’s orbital motion around the sun or others does not contribute to the Sagnac effect in an earthbound propagation loop. In as early as 1904 Michelson supposed that the Sagnac effect due to the orbital motion of the earth around the sun might be detectable, although the angular speed of the orbital motion is about 1/365 times that of the rotation .
Again, you cannot read.

Nope. That would be you.
It is quite clear.
The Sagnac effect due to Earth's orbital motion is based upon the area of the loop and the angular velocity, which is 1/365 that of Earth's rotation.

We've been through this before.

And you have been unable to refute it, instead just appealing to crap.

FIND ANY PAPERS PUBLISHED IN THE SAME JOURNALS WHICH SUPPORT JACK'S VIEW THAT r = R (the radius of the Earth = the Radius of the orbital path of the Earth).

How about you find anywhere where I have suggested that. You wont be able to.

Or better yet, forget about that and show what is actually wrong with either of my derivations.
If you can't, then SHUT UP and accept the FACT that the orbital Sagnac effect is much smaller than the rotational Sagnac effect.

Considering you like bitching about papers so much, perhaps you will like this quote from the paper you linked:

The remainder of the 50 m of fiber was wound on a coil of 9 cm in diameter which moved uniformly with the FOG so as not to produce any phase shift.

So according to your own paper, uniformly moving fibre does not produce a phase shift.

You are aware that is a university, not a journal?
Do you know the difference?

Is this supposed to be a joke?

Dr. C.C. Su's paper was also published by HARVARD UNIVERSITY:

http://adsabs.harvard.edu/cgi-bin/nph-bib_query?2001EPJC...21..701S

See the headline at the top:

NASA ADS Physics/Geophysics Abstract Service

Here is the title of the article:

A local-ether model of propagation of electromagnetic wave


So you lied again, for this is what you said:

No. Mainstream science agrees with me. The aether was refuted long ago.

How can it be refuted if Harvard University agreed to publish a paper entitled, A local-ether model of propagation of electromagnetic wave?

Why is a local aether model needed?

Dr. C.C. Su explains:



Based on the local-ether model, the propagation is entirely
independent of the earth’s orbital motion around
the sun or whatever and the velocity v for such an earthbound
experiment is referred to an ECI frame and hence
is due to earth’s rotation alone. In the original proposal,
the velocity v was supposed to incorporate earth’s orbital
motion around the sun. Thus, at least, v2/c2
=~ 10-8. Then the amplitude of the phase-difference variation
could be as large as π/3, when the wavelength is
0.6 µm and the path length is 10 m. However, as the velocity
v is the linear velocity due to earth’s rotation alone,
the round-trip Sagnac effect is as small as v2/c2∼ 10-12 which is merely 10-4 times that due to the orbital motion.


Can you read English jack?

The orbital Sagnac is 10,000 greater than the rotational Sagnac.

Published by:

HARVARD UNIVERSITY

BULLETIN OF THE AMERICAN PHYSICAL SOCIETY

EUROPEAN PHYSICAL JOURNAL

EUROPHYSICS LETTERS JOURNAL

JOURNAL OF ELECTROMAGNETIC WAVES AND APPLICATIONS


Journal of Electromagnetic Waves and Applications:

http://www.ee.nthu.edu.tw/ccsu/qem/f3c.pdf

For the interplanetary propagation, earth’s orbital
motion contributes to the Sagnac effect as well. This local-ether model
has been adopted to account for the Sagnac effect due to earth’s
motions in a wide variety of propagation phenomena, particularly the
global positioning system (GPS), the intercontinental microwave link,
and the interplanetary radar.


The peer reviewers at the Journal of Electromagnetic Waves and Applications agree that the orbital Sagnac is larger than the rotational Sagnac, that it is missing, and that a local-ether model has to be adopted in order to account for this fact.


You are on your own jack, and nobody in the scientific community will agree with you, and these scientists are not flat earth believers.


Do you have any paper, anybody, agreeing with your approach? You have no such thing.


So according to your own paper, uniformly moving fibre does not produce a phase shift.

You cannot read even a scientific paper.

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

PUBLISHED BY THE PHYSICAL REVIEW, ONE OF THE BEST JOURNALS

The FOG travelled with the mechanical conveyor and its uniform motion did not cause any phase shift because the FOG is only sensitive to the rotational movement.

The FOG with the extended fiber was calibrated and its output rate is 1162.6 mV per radian of detected phase shift. The phase shift of the FOG is linearly proportional to vL at a constant rate of 0.03200 radians per m2/s.

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.


You seem very confused jack.


No, angular velocity and the area of the loop, as your papers agreed.

But you are using the SAME AREA FOR TWO DIFFERENT SAGNAC EFFECTS.


You have a contribution from the arc with radius R2. This is based upon the area swept out by this arc, alpha*R2^2/2 (it would be 4*A*w/c^2).
You then have a contribution from the arc with radius R1. This is based upon the area swept out by this arc, and as the light beams are now travelling in an opposite direction and thus is of opposite sign and thus would be proportional to -alpha*R1^2/2.

That is not how you apply the Sagnac.

Here is how you tried to explain your diagram (by the way, those 9 pages were a total defeat of your ideas, make no mistake about it):

The R values are the distance from the centre of rotation (the sun in this case) to the section of the interferometer. Both values are roughly 150 000 000 km.

YOU CANNOT SUBSTRACT THE TWO VALUES.

That is not how the Sagnac is calculated.

Please learn.

The calculations for the rotational Sagnac.

The speed of light is:
c = 299,792 km/s

The circumference of the earth at the equator is:
C = 40,075 km

The duration of one day is:
d = 24 hr = 1440 min = 86400 sec

So now we can calculate the instantaneous tangential velocity of the earth at the equator to be:
v = (C / d) = 40,075 km / 86400 sec = 0.4638310185 km/sec

And we can calculate the amount of time it would take light to travel once around a non-rotating circumference:
t = (C / c) = 40,075 km / 299,792.458 km/s = 0.1336758111s

But, in the heliocentric hypothesis, the earth circumference is rotating, so there are actually going to be two different times in two different directions:
t1 = (C / (c - v)) = 40,075 km / (299,792.458 - 0.4638310185) km/s = 0.1336760179s
t2 = (C / (c + v)) = 40,075 km / (299,792.458 + 0.4638310185) km/s = 0.1336756043s

The difference between the above times is:
(t - t1) = 0.1336758111s - 0.1336760179s = -.0000002068s = -207 nanoseconds
(t - t2) = 0.1336758111s - 0.1336756043s = +.0000002068s = +207 nanoseconds


The calculations for the orbital Sagnac:

The speed of light is:
c = 299,792.458 km/s

The circumference of the earth orbit:
O = 939,951,145 km

The duration of one year is:
Y = d*(365.25) = 86400*(365.25) = 31,557,600s

So now we can calculate the instantaneous tangential velocity of the earth orbit to be:
V = (O / Y) = 939,951,145 km / 31,557,600s = 29.785254 km/s

And we can calculate the amount of time it would take light to travel once around a non-rotating circumference:
T = (O / c) = 939,951,145 km / 299,792.458 km/s = 3135.339532s

But, under the heliocentric hypothesis, the earth orbit is revolving around the sun, so there are actually going to be two different times in two different directions:
T1 = (O / (c - V)) = 939,951,145 km / (299,792.458 - 29.785254) km/s = 3135.651068s
T2 = (O / (c + V)) = 939,951,145 km / (299,792.458 + 29.785254) km/s = 3135.028057s

The difference between the above times is:
(T - T1) = 3135.339532s - 3135.651068s = -0.311536 seconds
(T - T2) = 3135.339532s - 3135.028057s = +0.311536 seconds

The percentage of orbital path that the earth occupies at any given time:
P = (C / (2πO)) = 40,075 km / (6.283185307)(939,951,145 km) = .0006785601973%

Correcting the time differences for the above percentage:
(-.311536s)(.0006785601973%) = -.000002113959296s = -2,114 nanoseconds
(+.311536s)(.0006785601973%) = +.000002113959296s = +2,114 nanoseconds


One way is c + v, the other way is c - v.

YOU DO NOT SUBSTRACT THE VALUES, AS YOU HAVE WRONGLY DONE.

Send your derivation to any university, any journal, and they will be laughing at you.


What they actually agree is that an aether which isn't local doesn't work.

But it does work, that is why they published the paper.

Journal of Electromagnetic Waves and Applications:

http://www.ee.nthu.edu.tw/ccsu/qem/f3c.pdf

For the interplanetary propagation, earth’s orbital
motion contributes to the Sagnac effect as well. This local-ether model
has been adopted to account for the Sagnac effect due to earth’s
motions in a wide variety of propagation phenomena, particularly the
global positioning system (GPS), the intercontinental microwave link,
and the interplanetary radar.


The peer reviewers at the Journal of Electromagnetic Waves and Applications agree that the orbital Sagnac is larger than the rotational Sagnac, that it is missing, and that a local-ether model has to be adopted in order to account for this fact.



Sagnac is 2 counterpropagating beams of light in a loop.

Exactly.

Please learn.


The calculations for the orbital Sagnac:

The speed of light is:
c = 299,792.458 km/s

The circumference of the earth orbit:
O = 939,951,145 km

The duration of one year is:
Y = d*(365.25) = 86400*(365.25) = 31,557,600s

So now we can calculate the instantaneous tangential velocity of the earth orbit to be:
V = (O / Y) = 939,951,145 km / 31,557,600s = 29.785254 km/s

And we can calculate the amount of time it would take light to travel once around a non-rotating circumference:
T = (O / c) = 939,951,145 km / 299,792.458 km/s = 3135.339532s

But, under the heliocentric hypothesis, the earth orbit is revolving around the sun, so there are actually going to be two different times in two different directions:
T1 = (O / (c - V)) = 939,951,145 km / (299,792.458 - 29.785254) km/s = 3135.651068s
T2 = (O / (c + V)) = 939,951,145 km / (299,792.458 + 29.785254) km/s = 3135.028057s

The difference between the above times is:
(T - T1) = 3135.339532s - 3135.651068s = -0.311536 seconds
(T - T2) = 3135.339532s - 3135.028057s = +0.311536 seconds

The percentage of orbital path that the earth occupies at any given time:
P = (C / (2πO)) = 40,075 km / (6.283185307)(939,951,145 km) = .0006785601973%

Correcting the time differences for the above percentage:
(-.311536s)(.0006785601973%) = -.000002113959296s = -2,114 nanoseconds
(+.311536s)(.0006785601973%) = +.000002113959296s = +2,114 nanoseconds


I don't need to make pathetic appeals to authority.

Why does this remind me of the famous quote:

"We don't need no stinking badges".


The simple fact that you can have the centre of rotation away from the loop and still have the shift based upon the area of the loop and angular velocity shows you are completely wrong.

You are using THE SAME AREA FOR TWO DIFFERENT SAGNAC EFFECTS.

The papers published and listed above contradict you.

No, the "relativists" have completely discarded aether.

But they did not.

Journal of Electromagnetic Waves and Applications:

http://www.ee.nthu.edu.tw/ccsu/qem/f3c.pdf

For the interplanetary propagation, earth’s orbital
motion contributes to the Sagnac effect as well. This local-ether model
has been adopted to account for the Sagnac effect due to earth’s
motions in a wide variety of propagation phenomena, particularly the
global positioning system (GPS), the intercontinental microwave link,
and the interplanetary radar.


The peer reviewers at the Journal of Electromagnetic Waves and Applications agree that the orbital Sagnac is larger than the rotational Sagnac, that it is missing, and that a local-ether model has to be adopted in order to account for this fact.



If the orbital Sagnac was really insignificant, the peer reviewers would not have accepted the paper to be published.

Because it is not insignificant, in fact larger than the rotational Sagnac, a local aether has to be adopted, contrary to your assertion.


I have provided a paper (well, Rab did first, I don't care) that shows it is based upon the area of the loop, not on the distance from the centre of rotation.

What you provided is this, a classic result from Sagnac theory:

"Sagnac showed experimentally that the centre of rotation can be away from the geometric centre of the apparatus, without affecting the above result."


CAN BE! COULD BE!


Not always. Not for each and every experiment. Certainly not for the rotational Sagnac, and certainly not for the orbital Sagnac.


This is what you are missing.


Based on the local-ether model, the propagation is entirely
independent of the earth’s orbital motion around
the sun or whatever and the velocity v for such an earthbound
experiment is referred to an ECI frame and hence
is due to earth’s rotation alone. In the original proposal,
the velocity v was supposed to incorporate earth’s orbital
motion around the sun. Thus, at least, v2/c2
=~ 10-8. Then the amplitude of the phase-difference variation
could be as large as π/3, when the wavelength is
0.6 µm and the path length is 10 m. However, as the velocity
v is the linear velocity due to earth’s rotation alone,
the round-trip Sagnac effect is as small as v2/c2∼ 10-12 which is merely 10-4 times that due to the orbital motion.

PUBLISHED BY FIVE DIFFERENT JOURNALS, COMPLETELY AGREEING WITH ME, AND NOT WITH YOU.

YOU REACHED A VALUE OF 1/365 BASED ON YOUR DERIVATION.

DR. C.C. SU REACHED A VALUE OF 10,000.

HIS PAPER WAS PEER REVIEWED AND PUBLISHED.

YOU ARE OFF BY A SCALE OF 3,650,000.

And yet you have the nerve to argue about who got defeated.

Not only you are totally defeated here, you are done here.



Notice the 30km/s speed.

This is Sagnac.

Try to imagine the center of the earth as the sun and the earth's orbit as the surface. So, the earth is in a rotational/circular pattern in its orbit. Just like the rotational sagnac, the earth rotates toward where the satellite emitted the signal, and with the orbit, the earth, the unit is orbited toward where the signal was emitted.

Then we simply apply the well-known formula to this situation, where R = 150,000,000 km, v = wR (=30km/s).


What jack has done is to use the wrong radius for the orbital sagnac and then claim the distance to the sun is the same as the radius of the earth and he comes up with a sagnac based only on angular velocity.

But the sagnac is a function of the product of the area swept out by the path and the angular velocity. That path is based on the distance from the center of rotation to the path.

This correct calculation is supported by mainstream papers.

Which are peer reviewed.

The v of the moving frame is what matters in the calculations for the sagnac effect.

And v = 30km/s in the heliocentric hypothesis.

The radius for such a speed is 150,000,000.

You are aware that is a university, not a journal?
Do you know the difference?
Is this supposed to be a joke?
Dr. C.C. Su's paper was also published by HARVARD UNIVERSITY:
http://adsabs.harvard.edu/cgi-bin/nph-bib_query?2001EPJC...21..701S
See the headline at the top:

Good question.
Is that supposed to be a joke? Because you sure are playing the part of an idiot very well.

It is an abstract service, run by the university, not a journal, and not a place where journal articles are published.

If you bother reading it, you see it is just the abstract for this paper:
The European Physical Journal C, Volume 21, Issue 4, pp. 701-715 (2001).

So thanks for showing everyone you have no idea what you are talking about in regards to published works.

The orbital Sagnac is 10,000 greater than the rotational Sagnac.

PROVE IT!!!
Don't just link to crap and make pathetic appeals to authority, derive the calculation of the Sagnac effect (i.e. the phase shift between 2 counterpropagating beams of light around a loop), and show that the orbital Sagnac is much greater, and refute my derivation by showing something wrong with it.

Until you do, my derivation stands as correct as does the conclusions from it, i.e. the orbital sagnac effect is tiny compared to the rotational sagnac effect.

Do you have any paper, anybody, agreeing with your approach? You have no such thing.

Yes, the ones already provided to you which indicate that it is based upon the area of the loop, and that the loop can be away from the centre of rotation.
Thus for a given loop, the orbital sagnac effect will be much smaller.

So according to your own paper, uniformly moving fibre does not produce a phase shift.

You cannot read even a scientific paper.

No, I can. I provided a quote from it, where it clearly indicates uniformly moving fibre doesn't produce a phase shift.

Of course, I can read it correctly, in context, and understand what they mean, while you just continually quote mine crap.

PUBLISHED BY THE PHYSICAL REVIEW, ONE OF THE BEST JOURNALS

Pure bullshit.
Published in Physics Letters A, a journal primarily for letters rather than strong journal articles.
It has an impact factor of 1.8, making it quite low.
Some of the best journals, like science and nature, have impact factors of 37 and 40.
It is a quite low standard journal.
Yes, it is vastly superior to some, but nothing like the best.

Shall I continue pointing out your ignorance regarding the academic/scientific publishing world?

The FOG travelled with the mechanical conveyor and its uniform motion did not cause any phase shift because the FOG is only sensitive to the rotational movement.

That's right, along with all the extra fibre they had sitting there.
So this uniform motion did not produce a phase shift. As such, there is more to it than simple translational motion.
As such, you are full of shit.

The phase shift of the FOG is linearly proportional to vL at a constant rate of 0.03200 radians per m2/s.

Where L is the length of a part of the optic fibre which is moving in a specific way, not the entire length. Not even the 50m they added to the FOG.

You seem very confused jack.

Nope. Again, that is you.
You don't seem to understand the paper at all.

But you are using the SAME AREA FOR TWO DIFFERENT SAGNAC EFFECTS.

Yes, too compare the 2 effects.
As I said, doing otherwise makes the comparison meaningless.

It would be like timing 2 processes using completely different methods which count at different rates.

The only way to make an honest comparison between the 2 is using the same interferometer or compensating for the area of the interferometers.
Either (honest) way you do it, you end up with the orbital sagnac being 1/365 times the rotational way.

That is not how you apply the Sagnac.

If you think it is wrong, EXPLAIN EXACTLY WHAT IS WRONG!!!
Stop bitching about it.

Here is how you tried to explain your diagram (by the way, those 9 pages were a total defeat of your ideas, make no mistake about it):

No, it was a complete defeat of your pathetic crap with you running away like a pathetic coward when you couldn't honestly answer a simple question without exposing your bullshit.

YOU CANNOT SUBSTRACT THE TWO VALUES.

Why not?
I explained why you MUST subtract the 2 values.
If you have beam a and beam b, where they are counterpropogating, then in the arc with radius R2, say beam a is moving forwards and beam b is moving backwards (relative to the orbital motion).
This means the path length for b is much shorter, and thus the shift (measured as extra time taken for beam a relative to beam b) will be positive.
They then continue along the light path and travel along the arc of radius R1. Now beam a is moving backwards and beam b is moving forwards. That means the shift will now be negative.
You add these 2 shifts and get a shift proportional to R2^2-R1^2.

So not only can you subtract the 2 values, YOU MUST!!!!
If you don't, you will not correctly calculate the fringe shift.

If you wish to disagree, then show a derivation FOR THAT INTERFEROMETER!!!!

That is not how the Sagnac is calculated.

YES IT IS!!!
If you wish to claim otherwise you need to show how it is, and explain what the issue is with doing it this way.

Make sure you do it for this loop, not for a loop the size of Earth's orbit or the size of Earth.

Send your derivation to any university, any journal, and they will be laughing at you.

Yes, saying WHO GIVES A SHIT!!!
Send your garbage to any decent university or journal and they will laugh at you for your stupidity.

But it does work, that is why they published the paper.

No, it doesn't. If you read the paper it clearly indicates it doesn't work.

Sagnac is 2 counterpropagating beams of light in a loop.
Exactly.
Please learn.

I'm not the one that needs to learn. You are.

The calculations for the orbital Sagnac:

Draw a diagram, clearly showing where your loop is and the 2 beams of light counterpropagating through it.

Until you do, your argument is crap as it suffers from the same problem I have already pointed out.

I don't need to make pathetic appeals to authority.
Why does this remind me of the famous quote:
"We don't need no stinking badges".

Because you are looking for any pathetic excuse to avoid having to rely upon yourself to make an argument rather than just being able to spam crap?

If the orbital Sagnac was really insignificant, the peer reviewers would not have accepted the paper to be published.

For such a shitty journal, I would't be surprised if they would. It probably only had a superficial peer review rather than focusing on key words.
The paper doesn't discuss the Sagnac effect.

What you provided is this, a classic result from Sagnac theory:
"Sagnac showed experimentally that the centre of rotation can be away from the geometric centre of the apparatus, without affecting the above result."
CAN BE! COULD BE!

Like I said before, showing that you don't need to use the distance to the sun and instead can just use the area of the loop.
As such, you can use the same loop to compare the rotational and orbital Sagnac effect.

So no, I'm not missing anything. You are just grasping at straws.


Now stop repeating the same refuted BS and show explicitly what is wrong with my derivation (which shows it is the area of the loop, not the distance to the centre of rotation which is important) and provide your own derivation either using one of the systems I already provided or providing your own with a diagram clearly indicating the common loop used for both the rotational and orbital Saganc effect.

Until you do, you will remain defeated.

rabinoz, you do not have experience in dealing with scientific papers.
You are omitting a crucial fact.
You have been reminded of this thing over and over again, but to you it is not a problem.

I take no notice of someone trying to disprove the heliocentric globe by pretending to refer to papers by numerous researchers who themselves have no doubt that the heliocentric globe is correct.

Then that someone   has to assume quite unrealistic and unproven properties of aether to make his unique flat model work.

"Sagnac showed experimentally that the centre of rotation can be away from the geometric centre of the apparatus, without affecting the above result."

CAN BE! COULD BE!

Sorry, neither I nor anyone else seems to interpret Sagnac's words in that way. Your interpretation makes them meaningless!

You ignore the simple fact that every analysis of the Sagnac effect shows that it is independent of the location of the centre of rotation.

You know, these statements, by people much more knowledgeable than you.

It is not just Sagnac's statement, but many others, as I have said all along:
Funny that YOU refuse to analyse the off-centre Sagnac Loop yourself, but everyone else seem able to show this sort of thing!

Sagnac effect in an off-center rotating ring frame of reference

Abstract
Interference resides deeply in our understanding of the wave properties of light. In this paper, the century famous Sagnac effect is demonstrated to be independent of the rotation axis position, using a rotating ring optical fiber in a straightforward laboratory experiment. A simple theoretical explanation for this result is given for any arbitrary closed loop interferometer. The level of this discussion should be suitable for undergraduate physics or engineering courses where electromagnetic theory and optics are discussed. The experiment described utilizes basic and important aspects in modern optics which every science student should acquire.

And

Sagnac showed experimentally that the centre of rotation can be away from the geometric centre of the apparatus, without affecting the above result.

From: TIME and the SPEED of LIGHT - a NEW INTERPRETATION, A.G.Kelly

Which really puts the kibosh on you interpretation of the "Orbital Sagnac Effect", so nice to know.
Bye bye any massive orbital Sagnac Effect.
And
 

This phase difference has been demonstrated in a rotating interferometer as well as in a geostationary one.
In the latter case, the Sagnac effect is due solely to earth’s rotation with ω_l = ω_E. The earth rotation rate ω_E is about 2π/(86400 s) and the corresponding maximum phase difference is as large as 2 rad, when the wavelength is 0.6 µm and the loop area S = 0.2 km². Thus, a loop interferometer can be utilized as a precise means to detect earth’s rotation rate.
Moreover, according to the local-ether model, earth’s orbital motion around the sun or others does not con-
tribute to the Sagnac effect in an earthbound propagation loop. In as early as 1904 Michelson supposed that
the Sagnac effect due to the orbital motion of the earth around the sun might be detectable, although the angular
speed of the orbital motion is about 1/365 times that of the rotation .

Plenty more where they came from!

The Sagnac delay is independent of the shape of the loop and on its centre of rotation!

You might also read:
          Sagnac Effect, E. J. POST, Rev. Mod. Phys. 39, 475 (1967) – Published 1 April 1967
Where Post reviews some of the Sagnac experiments and states:

Note that this too states, "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

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 he specifically states, "does not depend on the location of the centre of rotation;"

Here is how to best define your mental condition jack.

Your derivation reaches a figure of 1/365.

Dr. C.C. Su arrives at a figure of 10,000.

His paper is published in peer reviewed journals, who could have reminded him of the quote from Post's article. They did not.

You are off by a scale of 3,650,000.

But this is no problem to you.

Is this not a sign of sure idiocy?


For your information, the Harvard ads abstract service only published MAJOR ASTRONOMY AND PHYSICS PUBLICATIONS.

Not garbage like your derivation.


Published in Physics Letters A, a journal primarily for letters rather than strong journal articles.
It has an impact factor of 1.8, making it quite low.

Physical Review Letters (PRL), established in 1958, is a peer-reviewed, scientific journal that is published 52 times per year by the American Physical Society. As also confirmed by various measurement standards, which include the Journal Citation Reports impact factor and the journal h-index proposed by Google Scholar, many physicists and other scientists consider Physical Review Letters one of the most prestigious journals in the field of physics.

Are we to understand that you jack have such a low degree of ignorance as to try to minimize the PRL journal?


Dr. Su's paper was also published by the Bulletin of the American Physical Society.

The American Physical Society (APS) is the world's second largest organization of physicists. The Society publishes more than a dozen scientific journals, including the prestigious Physical Review and Physical Review Letters, and organizes more than twenty science meetings each year. APS is a member society of the American Institute of Physics.


Dr. Su's paper was also published by the Journal of Electromagnetic Waves and Applications, also a highly respected scientific journal.

His paper was published by the IOP journals.


Journal of Electromagnetic Waves and Applications:

http://www.ee.nthu.edu.tw/ccsu/qem/f3c.pdf

For the interplanetary propagation, earth’s orbital
motion contributes to the Sagnac effect as well. This local-ether model
has been adopted to account for the Sagnac effect due to earth’s
motions in a wide variety of propagation phenomena, particularly the
global positioning system (GPS), the intercontinental microwave link,
and the interplanetary radar.


The peer reviewers at the Journal of Electromagnetic Waves and Applications agree that the orbital Sagnac is larger than the rotational Sagnac, that it is missing, and that a local-ether model has to be adopted in order to account for this fact.



If the orbital Sagnac was really insignificant, the peer reviewers would not have accepted the paper to be published.

Because it is not insignificant, in fact larger than the rotational Sagnac, a local aether has to be adopted, contrary to your assertion.


Go ahead and mail your piece of shit derivation and you will see how you will be laughed at.


derive the calculation of the Sagnac effect (i.e. the phase shift between 2 counterpropagating beams of light around a loop), and show that the orbital Sagnac is much greater,


This is exactly what I have done.

The calculations for the rotational Sagnac.

The speed of light is:
c = 299,792 km/s

The circumference of the earth at the equator is:
C = 40,075 km

The duration of one day is:
d = 24 hr = 1440 min = 86400 sec

So now we can calculate the instantaneous tangential velocity of the earth at the equator to be:
v = (C / d) = 40,075 km / 86400 sec = 0.4638310185 km/sec

And we can calculate the amount of time it would take light to travel once around a non-rotating circumference:
t = (C / c) = 40,075 km / 299,792.458 km/s = 0.1336758111s

But, in the heliocentric hypothesis, the earth circumference is rotating, so there are actually going to be two different times in two different directions:
t1 = (C / (c - v)) = 40,075 km / (299,792.458 - 0.4638310185) km/s = 0.1336760179s
t2 = (C / (c + v)) = 40,075 km / (299,792.458 + 0.4638310185) km/s = 0.1336756043s

The difference between the above times is:
(t - t1) = 0.1336758111s - 0.1336760179s = -.0000002068s = -207 nanoseconds
(t - t2) = 0.1336758111s - 0.1336756043s = +.0000002068s = +207 nanoseconds


This is the well known figure of 207 nanoseconds accepted by the scientific community around the world.

The calculations for the orbital Sagnac:

The speed of light is:
c = 299,792.458 km/s

The circumference of the earth orbit:
O = 939,951,145 km

The duration of one year is:
Y = d*(365.25) = 86400*(365.25) = 31,557,600s

So now we can calculate the instantaneous tangential velocity of the earth orbit to be:
V = (O / Y) = 939,951,145 km / 31,557,600s = 29.785254 km/s

And we can calculate the amount of time it would take light to travel once around a non-rotating circumference:
T = (O / c) = 939,951,145 km / 299,792.458 km/s = 3135.339532s

But, under the heliocentric hypothesis, the earth orbit is revolving around the sun, so there are actually going to be two different times in two different directions:
T1 = (O / (c - V)) = 939,951,145 km / (299,792.458 - 29.785254) km/s = 3135.651068s
T2 = (O / (c + V)) = 939,951,145 km / (299,792.458 + 29.785254) km/s = 3135.028057s

The difference between the above times is:
(T - T1) = 3135.339532s - 3135.651068s = -0.311536 seconds
(T - T2) = 3135.339532s - 3135.028057s = +0.311536 seconds

The percentage of orbital path that the earth occupies at any given time:
P = (C / (2πO)) = 40,075 km / (6.283185307)(939,951,145 km) = .0006785601973%

Correcting the time differences for the above percentage:
(-.311536s)(.0006785601973%) = -.000002113959296s = -2,114 nanoseconds
(+.311536s)(.0006785601973%) = +.000002113959296s = +2,114 nanoseconds


The rotational and the orbital Sagnac can be compared immediately:

Earth's radius = 6357 km; r² = 40411449

Earth's orbital radius = 150,000,000 km r² = 22500000000000000

∆t = 4πR²ω/(c²-v²)
or

I use the linear velocity.

∆t = 4πRv/( c² - v² ), where v is the linear velocity.

For the earth's rotation, it is 0.4638333 km/ sec and the orbit v = 30km/sec.

∆t = 0.62831852628 for the earth's orbit.
Total path of the orbit is 2πr=2π(150,000,000 km) = 942,477,780km

Hence, the sagnac effect for a 1 km path, that means light source in the center and two receivers placed at .5km is:
0.62831852628 / 942,477,780km = 6.6666667 e-10 sec / km

Now, for the earth's rotation.
∆t = 4.1170061 e-7 seconds
Total path of the rotation is 2πr=2π(6357 km) = 39942.21 km


4.1170061 e-7 seconds / 39942.21 km = 1.0307407 e-11 sec / km


The sagnac effect for the earth's orbit is greater than that of the rotation.



I provided a quote from it, where it clearly indicates uniformly moving fibre doesn't produce a phase shift.



You are confusing FOG with FOC.

The authors explained in detail how the phase shift is obtained.

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.

So this uniform motion did not produce a phase shift. As such, there is more to it than simple translational motion.

The authors explained:

The FOG travelled with the mechanical conveyor and its uniform motion did not cause any phase shift because the FOG is only sensitive to the rotational movement.

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 did prove that the Sagnac applied to linear motion.


If you don't, you will not correctly calculate the fringe shift.


But you do get the correct fringe shift using the correct calculation.

Try to imagine the center of the earth as the sun and the earth's orbit as the surface. So, the earth is in a rotational/circular pattern in its orbit. Just like the rotational sagnac, the earth rotates toward where the satellite emitted the signal, and with the orbit, the earth, the unit is orbited toward where the signal was emitted.

Then we simply apply the well-known formula to this situation, where R = 150,000,000 km, v = wR (=30km/s).


What jack has done is to use the wrong radius for the orbital sagnac and then claim the distance to the sun is the same as the radius of the earth and he comes up with a sagnac based only on angular velocity.

But the sagnac is a function of the product of the area swept out by the path and the angular velocity. That path is based on the distance from the center of rotation to the path.

This correct calculation is supported by mainstream papers.

Which are peer reviewed.

The v of the moving frame is what matters in the calculations for the sagnac effect.

And v = 30km/s in the heliocentric hypothesis.

The radius for such a speed is 150,000,000.


Your conclusion leads to this figure: 1/365.

The papers published by Dr. Su in more than five journals, which are peer reviewed, lead to this figure: 10,000.

What does the fact that you are off by a factor 3,650,000 tell you jack?

Δto/Δt r=[4Aiωo/( c² - vo²)] / [4Aiωr/( c² - vr²)]

Note, the area here has nothing to do with the area of Earth's orbit or radius of it or the radius of Earth. It is the area of the interferometer, as my derivation.

No where in any derivation did the area of Earth's orbit come into it.

So to continue:
Δto/Δt r=[4Aiωo/( c² - vo²)] / [4Aiωr/( c² - vr²)]
Obviously, ( c² - vo²) and ( c² - vr²) are very close to the same number, so let's lave them off.
=4Aiωo/ 4Aiωr
Then to simplify:
=ωo/ ωr

And would you look at that? It ends up being just like what we claim.
You have ωo/ ωr.
As Earth rotates roughly 365 times for each orbit, ωr=365*ωo.
Thus we get:
Δto/Δt r=ωo/ ωr
=ωo/ (365*ωo)
=1/365


You are not using the correct Sagnac formula jack.

If light travels at one speed c, then as the earth supposedly moves in it's revolution loop at 30k/s, while light moves c through space, the unit at the equator at noon would move with the earth' rotation and the earth's revolution cutting the distance the signal must travel to meet the unit.


"Let's say the unit is at the equator and the satellite is low on the horizon in the east at noon.

That means the unit is traveling at the orbital speed of the earth at 67,000 MPH.

The satellite emits at one speed c in space. While the light travels through space toward the unit at c, the unit moves with the earth at 67,000 MPH. The unit cuts the distance that the light must travel.

This is not being seen by any experiements nor GPS."

Yet, this same logic applies and works with the earth's supposed rotation.


Please read again.

Try to imagine the center of the earth as the sun and the earth's orbit as the surface. So, the earth is in a rotational/circular pattern in its orbit. Just like the rotational sagnac, the earth rotates toward where the satellite emitted the signal, and with the orbit, the earth, the unit is orbited toward where the signal was emitted.

Then we simply apply the well-known formula to this situation, where R = 150,000,000 km, v = wR (=30km/s).


But that is not what you did.

Rotational Sagnac, r = x
ORBITAL SAGNAC, R = x

They are the same.

This is madness.

r = 6,378 km

R = 150,000,000 km

How can they be the same?


If you have beam a and beam b, where they are counterpropogating,

Exactly.
Here is the correct calculation for you.

The calculations for the orbital Sagnac:

The speed of light is:
c = 299,792.458 km/s

The circumference of the earth orbit:
O = 939,951,145 km

The duration of one year is:
Y = d*(365.25) = 86400*(365.25) = 31,557,600s

So now we can calculate the instantaneous tangential velocity of the earth orbit to be:
V = (O / Y) = 939,951,145 km / 31,557,600s = 29.785254 km/s

And we can calculate the amount of time it would take light to travel once around a non-rotating circumference:
T = (O / c) = 939,951,145 km / 299,792.458 km/s = 3135.339532s

But, under the heliocentric hypothesis, the earth orbit is revolving around the sun, so there are actually going to be two different times in two different directions:
T1 = (O / (c - V)) = 939,951,145 km / (299,792.458 - 29.785254) km/s = 3135.651068s
T2 = (O / (c + V)) = 939,951,145 km / (299,792.458 + 29.785254) km/s = 3135.028057s

The difference between the above times is:
(T - T1) = 3135.339532s - 3135.651068s = -0.311536 seconds
(T - T2) = 3135.339532s - 3135.028057s = +0.311536 seconds

The percentage of orbital path that the earth occupies at any given time:
P = (C / (2πO)) = 40,075 km / (6.283185307)(939,951,145 km) = .0006785601973%

Correcting the time differences for the above percentage:
(-.311536s)(.0006785601973%) = -.000002113959296s = -2,114 nanoseconds
(+.311536s)(.0006785601973%) = +.000002113959296s = +2,114 nanoseconds


This means the path length for b is much shorter, and thus the shift (measured as extra time taken for beam a relative to beam b) will be positive.

COMPLETELY WRONG!

Dr. A.G. Kelly explains:



The shift IS DUE TO THE DIFFERENT SPEEDS RECORDED, AND NOT DUE TO THE DISTANCE.

This is where you went completely wrong.

As such your derivation is a piece of shit.


You get a shift because of the different speeds recorded, c + v and c - v.

The difference in time (dt) for the light to traverse the path, in opposite directions, was derived by Sagnac as:

dt = 4Aw/c^2

where A is the area enclosed by the light path, wRadians per second is the angular velocity of spin and c the speed of light. By reversing the direction of the spin of the disc, he got double the amount, and this made the result more detectable.

If you do not know how the Sagnac is derived please study:

http://www.naturalphilosophy.org/pdf/ebooks/Kelly-TimeandtheSpeedofLight.pdf


So that is where you went wrong.

This means the path length for b is much shorter, and thus the shift (measured as extra time taken for beam a relative to beam b) will be positive.

I explained why you MUST subtract the 2 values.
If you have beam a and beam b, where they are counterpropogating, then in the arc with radius R2, say beam a is moving forwards and beam b is moving backwards (relative to the orbital motion).
This means the path length for b is much shorter, and thus the shift (measured as extra time taken for beam a relative to beam b) will be positive.

The shift is due to the DIFFERENT SPEEDS RECORDED, C + V AND C - V, AND NOT DUE TO THE LENGTHS.

You have just shown jack, your utter ignorance of the physics behind the Sagnac effect.

No wonder you were off by a scale of 3,650,000.

<< Sandokhan's attempted cover-up of his total inability to derive the Sagnac delay deleted >>

Looking in Su's paper, you'll find this:

Now, to an untrained eye like mine (after all I only spent 38 years at UQ) that looks essentially the same as JackBlack's result and everybody else's and the location of the centre of rotation seems to be quite irrelevant!

He then adds a bit, saying:

This phase difference has been demonstrated in a rotating interferometer as well as in a geostationary one. In the latter case, the Sagnac effect is due solely to earth’s rotation with ω_l = ω_E. The earth rotation rate ω is about 2π/(86400 s) and the corresponding maximum phase difference is as large as 2 rad, when the wavelength is 0.6 µm
and the loop area S = 0.2 km². Thus, a loop interferometer can be utilized as a precise means to detect earth’s rotation rate.

Moreover, according to the local-ether model, earth’s orbital motion around the sun or others does not contribute to the Sagnac effect in an earthbound propagation loop. In as early as 1904 Michelson supposed that the Sagnac effect due to the orbital motion of the earth around the sun might be detectable, although the angular speed of the orbital motion is about 1/365 times that of the rotation.

Leaving not the slightest doubt that the earth rotates and orbits the sun - two things that you vehemently deny and that the orbital Sagnac delay would be 1/365 times that of the rotational Sagnac delay as we already knew.