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It was my understanding that the speed of light is supposed to be constant for every observer, as in the classical mental experiment of the guy in the train with the flashlights which explain the relativity of simultaneity.

(i.e. observer A outside of the train, will see one beam hit the back wall before the other beam hit the front wall, because for him the “front beam” has to go for a longer way. Observer B in the moving train, instead, will see the beams hit the two walls simultaneously. But both the observers see the two beams move at c).

Now: with the Sagnac experiment, I would expect the same thing: I would expect observer A (still) to see the two beams arriving at the detector at different times because one beam has gone for a longer way. I would expect Observer B, rotating with the device (including source and detector), to see the beams arrive simultaneously.

However, that is not the case, as observer B actually see one beam arrive after the other: how is that? What am I missing?

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    $\begingroup$ If observer B is rotating then they are accelerating, and therefore not in an inertial frame. Thus special relativity does not require them to measure the speed of light tangential to the Sagnac loop to be $c$. This point is explained briefly in the Wiki page. $\endgroup$ – Mark Mitchison May 18 '15 at 20:21
  • $\begingroup$ Special or General Relativity or both? $\endgroup$ – Geremia Jan 22 '16 at 0:38
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@JohnDuffield is indeed correct. Let me try to quantitatively show the anisotropy of the speed of light for an observer on the rotating platform. I think the following reasoning is to be credited to Langevin (I don't remember the reference, sorry, and anyway it would be in French!).

So let's denote by $\omega$ the angular speed of the platform with respect to the still observer. Clocks and rods on the rotating platform will be affected by their motion with respect to the still observer. What order in $\omega$ shall we expect? If there is a term proportional to $\omega$, then the clocks and rods will behave differently depending on whether the platform rotate in one or the other direction. This is silly, as this amount to simply having the still observer look from the top instead of from the bottom for example. So the first term has to be proportional to $\omega^2$. But then, that means that if we limit ourselves to an approximation at order $\omega$, we can completely neglect the fact that clocks and rods on the rotating platform will not measure time and lengths as for the still observer. What that means is that we can use a good old Galilean transform!

So let's denote by $(x, y, z, t)$ and $(x',y',z',t')$ the spacetime coordinates on the platform and of the still observer respectively, then

$$\begin{align} x'&=x\cos\omega t-y\sin\omega t\\ y'&=x\sin\omega t + y\cos\omega t\\ z'&=z\\ t'&=t \end{align}$$

In differential form, this reads

$$\begin{align} dx'&=dx\cos\omega t-dy\sin\omega t - \omega (x\sin\omega t+y\cos\omega t)dt\\ dy'&=dx\sin\omega t + dy\cos\omega t + \omega(x\cos\omega t-y\sin\omega t)dt\\ dz'&=dz\\ dt'&=dt \end{align}$$

Then we look at the spacetime interval $ds$. The still observer being inertial,

$$ds^2 = dt'^2-dx'^2-dy'^2-dz'^2.$$

I have used units where the speed of light $c=1$. Then substituting the above change of coordinates, and a bit of trigonometry, gives

$$ds^2 = dt^2 -2\omega(xdy-ydx)dt\underbrace{-dx^2-dy^2-dz^2}_{-dl^2}.$$

The propagation of a light signal corresponds to $ds^2=0$. The presence of cross terms $dxdt$ and $dydt$ results in an anisotropic speed of light. Let's investigate that precisely.

We note that $dA=\frac{1}{2}(xdy-ydx)$ is the area of the infinitesimal triangle whose vertices are the origin of the coordinates, the point $(x,y,z)$ and the point $(x+dx, y+dy, z+dz)$, that is to say the area swept by the vector from the origin to the position of the light signal during the duration $dt$. This is an area with a sign: positive if the sweeping is anti-clockwise and negative otherwise. So we get

$$dt^2 - 4\omega dAdt -dl^2=0.$$

We can solve this 2nd order equation in $dt$:

$$dt = 2\omega dA + \sqrt{dl^2+4(\omega dA)^2}.$$

But since we neglect terms of order $\omega^2$,

$$dt = dl + 2\omega dA,$$

i.e.

$$\frac{dl}{dt} = 1 - 2\omega \frac{dA}{dt}.$$

This is the speed of light for an observer on the platform: not only it is not equal to 1 but it depends on the direction of propagation since it depends on the sign of $dA/dt$.

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What you're missing is that the speed of light is not constant. There's this modern-day myth that says "Einstein told us that the speed of light is constant". But search the Einstein digital papers on "speed of light" or "velocity of light" for examples like this:

enter image description here

The speed of light is spatially variable. And that isn't some discarded idea from 1911, see Shapiro's 4th test of General Relativity: "the speed of a light wave depends on the strength of the gravitational potential along its path". Also see The Deflection and Delay of Light by Ned Wright and this PhysicsFAQ article by Don Koks:

"Einstein talked about the speed of light changing in his new theory. In the English translation of his 1920 book "Relativity: the special and general theory" he wrote: "according to the general theory of relativity, the law of the constancy of the velocity [Einstein clearly means speed here, since velocity (a vector) is not in keeping with the rest of his sentence] of light in vacuo, which constitutes one of the two fundamental assumptions in the special theory of relativity [...] cannot claim any unlimited validity. A curvature of rays of light can only take place when the velocity [speed] of propagation of light varies with position." This difference in speeds is precisely that referred to above by ceiling and floor observers."

The modern-day myth says the speed of light is absolutely constant, and any challenge to this overturns relativity. Nothing could be further from the truth. What's constant is the locally measured speed of light. Because of a tautology wherein we use the local motion of light to define our second and our metre, which we then use... to measure the local motion of light. Duh! See http://arxiv.org/abs/0705.4507 where Magueijo and Moffat talked about it a few years back.

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  • $\begingroup$ If I read you correctly (which I may not be), you are saying that light is constant to each observer, but not constant among observers. I think this is the generally accepted meaning of constant. The inconsistency among observers is due to space-time dilation, but the light itself is traveling at a constant speed relative to any observer. $\endgroup$ – SigSeg Oct 11 '17 at 17:21
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enter image description here

Let's look at it another way. ( Note: Length of train track loop above is 150,000 km long )

Imagine that the length of the train, while it was at rest at the train station, was 300,000 km long. Also, synchronized clocks, 1 and 2, are located at opposite ends of the train. If a burst of light was sent from either end of the train at 12:00 AM through a flexible vacuum tube to the opposite end, and the burst of light lit up the face of the clock at this opposite end, this lit up clock face would indicate that the light burst took 1 second to cross the total length of the train.

If the train was to travel at 260,000 km/s, external observers at rest relative to the tracks would see that the train has shrunk to half length, 150,000 km, the clocks are ticking at half speed, and that clock #1 at the rear end of the train was 0.866 of a second ahead of the clock #2 at the front end.

Because of these changes, via the clocks it will still be indicated to those on board the train that it always takes 1 second for light to go from one end of the train to the other end, and does so in either direction.

Now, while the train is still on the move at 260,000 km/s, picture the train as it is on the loop at the very moment that clocks #1 and #2 are located back to back at the clock #3/#4 position. Here, the external observers now notice that clocks #1 and #2 are once again synchronized, and from that moment onward they also see that these clocks are still ticking at half speed, and they see that the train is still only 150,000 km long and thus can and does fit on the 150,000 km long track loop.

If while at this location, or position, the time is 12:00 AM, and bursts of light are released from both ends at this 12:00 AM time, by the time these two bursts of light have reached the opposite ends, both clocks will have been repositioned. Thus the distance that the light has to travel in the flexible hollow vacuum tube in one direction, is longer, while the distance the light has to travel in the opposite direction, has become shorter.

Since clocks #1 and #2 are no longer offset from each other by 0.866 of a second as they were beforehand, different time periods will be measured for light to go in one direction compared to the other. Therefore this difference in light travel length is no longer being clock compensated for as it was beforehand.

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  • $\begingroup$ extremetech.com/computing/… If this fiber optic were used in the above train, there would only be a 0.3 % error to take into account. $\endgroup$ – Sean Oct 17 '17 at 2:26
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    $\begingroup$ I find it mesmerising how much of a hit and miss this site can be: we both write thorough answers, making effort to be understood, and we both get 0 votes!! $\endgroup$ – user154997 Oct 20 '17 at 12:00
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The Sagnac effect is a proof of SR contrary to what many thought at the time of its discovery. The effect is independent of the 'area' of the loop and is the same even if the loop area is zero. It is also independent of the 'refractive index' - as the term that has such dependency cancels out in the subtraction. The time difference is the integral of the velocity of the transmitter-receiver combo(T/R) and the length of its travel in going and coming back. See https://arxiv.org/ftp/physics/papers/0609/0609222.pdf The derivation of the fromula above is shown in https://arxiv.org/ftp/physics/papers/0609/0609222.pdf using the Lorenz transfom.

The simple explanation is that the T/R moves at a certain velocity but the light it is sending has only a constant velocity hence the difference in time.. due to the sign change of the T/R velocity direction and not due to a change in the velocity of light. See also this quote from the last ref. ; '' Remarkably, the time difference is independent of the refraction index {\displaystyle n}n and the velocity of light in the fiber. ''

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After all, the observer A and B and the light source C here are the same object. A and B are watching C. They are at rest with respect to each other. The problem is more complex. Reflections in mirrors as they move are the source of the new EM wave. Mirrors should be treated as objects D and E that do not form an inertial frame with A, B, C.

However, there is no such thing as an inertial frame.

We simply cannot know our dynamic vector of motion. SR has postulates that look good on a piece of paper, but this model cannot be used for physical studies because there is no such thing as an inertial observer.

The absolute rotation is detected by the Sagnac experience and the current laser gyroscopes. However, with this instrument you cannot detect transnational motion of a rectilinear observer.

The fact is that the postulates of the SR theory concerning the observer are impossible to fulfill. Such a tool is useless and unreliable.We can't figure out what the Doppler effect observers are under.You know nothing. You simply assume that the Doppler effect was not dynamic during the course of the experiment and was acting on all subjects simultaneously.

You also need to know that the change in Doppler effect propagates with a finite velocity in space. This propagation is limited in terms of the speed of light. It follows from this observation that changes in the energies of matter particles do not take place in parallel. They are not synchronous. Therefore, there is no such thing as an inertial frame, which would fulfill the assumptions of the definition.

This is only my free hypothesis, but I will express it:

The only chance for an inertial system to exist is to prove that the ether exists and is static. That is, the motion vector of the ether is 0, or the energy level of the ether at rest is known.

The fact is that Einstein proposed the zero point energy model and did not put it in the SR. This model turned out to be confirmed by the results of today's research. The ZPE was simply a model of the ether as a medium. This is the obvious conclusion from the analysis of this model.

Rather, Sagnac's experiment showed that object movement is important and that if it can be detected, its effects can be compensated.

Sorry for My English.

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