Sagnac Effect and clocks synchronization

Question

I read this from "The Sagnac effect and its interpretation by Paul Langevin", in cylindrical coordinates 
$$ ds^2=c^2t^2-dr^2-r^2dtheta^2 $$

...This transformation means that the observer O (inertial) now uses a coordinate system that accompanies the disk in its rotation. From the inertial observer O point of view, the source, the detector and the non-inertial observer  (assumed to be at the same place) are located on a radius with a fixed orientation, for instance $theta=0$ . Neglecting a small second-order term in $ R omegaover c $, the $ ds^2$ (relativistic invariant) takes the form
  $$ ds^2=c^2t^2-dr^2-r^2dtheta^2 -2r^2omega dtheta dt $$
  Following Langevin, this expression is called the metric of the rotating disk. It will be noted the presence of a cross term in $dtheta dt $, source of the impossibility of synchronizing clocks, uniformly distributed around the periphery of the disk and connected thereto (a relativistic phenomenon which is the essence of the Sagnac effect).

Could anyone clarify why the term $dtheta dt $ makes the clocks synchronization impossible?

Albert · Answer

I believe that it means, that is not possible to synchronize clocks on the rim of rotating disc by Einstein technique; because it gives a non vanishing time difference that depends on the direction used. Sagnac effect demonstrates anisotropy of the one - way speed of light when measured in rotating reference frame.

To be exact it is possible to synchronize clocks on the rim or rotating ring by emitted from the center of the circumference beams of light; but, again, these clock will be not Einstein - synchronized; measured by these clocks one - way speed of light will be different from c.

However, measured back-and forth speed of light on any segment of rotating ring (say on a segment AB), for example on the path  A-B-A or B-A-B will be isotropic and equal precisely to c (Michelson - Morley experiment)

The article in Wikipedia concludes:

"The Einstein synchronisation looks this natural only in inertial frames. One can easily forget that it is only a convention. In rotating frames, even in special relativity, the non-transitivity of Einstein synchronisation diminishes its usefulness. If clock 1 and clock 2 are not synchronised directly, but by using a chain of intermediate clocks, the synchronisation depends on the path chosen. Synchronisation around the circumference of a rotating disk gives a non vanishing time difference that depends on the direction used. This is important in the Sagnac effect and the Ehrenfest paradox. The Global Positioning System accounts for this effect."

It should look like this:

In inertial frame of reference:

Let $t_1$ be the travel time of photons which move with the rotation, then

$$2 pi r + r omega t_1=ct_1$$
$$t_1 = frac {2 pi r}{c-romega}$$

Let $t_2$ be the travel time for photon moving against the rotation of the disc. The difference of travel time is:

$$Delta t = t_1-t_2= 2 pi r (frac {1}{c-romega}-frac {1}{c+romega})= frac {2pi r2romega}{c^2-r^2omega^2}=gamma^2 frac {4Aomega}{c^2}$$

$A$ is the area enclosed by the photon path or orbit

In rotating frame:

($ds^2 = 0$ along the
world line of a
photon )

$$ds^2 = - (1 - frac {r^2omega^2}{c^2}) c^2dt + r^2d theta^2+2r^2omega dtheta dt$$

Let $dot theta= frac {dtheta} {dt}$

$$r^2 dot theta^2 + 2r^2 omega dot theta –(c^2-r^2 omega^2)=0$$

$$ dot theta = frac {-r^2 omega pm sqrt {r^4omega^2+r^2c^2-r^4 omega^2}}{r^2}$$
$$dottheta = - omega pm frac {rc}{r^2}=-omega pm c/r$$

The speed of light: 
$$v_pm=r dot theta = -r omega pm c$$

We see that in the rotating frame  the measured (coordinate) velocity of light is not isotropic. The difference in the travel time of the two beams is:

$$Delta t= frac {2 pi r}{c-romega} - frac {2 pi r}{c+romega} = gamma^2 frac {4Aomega}{c^2}$$

Does Sagnac effect imply anisotropy of speed of light in this inertial frame of reference?

Sagnac Effect and clocks synchronization

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