# Sagnac Effect and clocks synchronization

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

...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 $$R \omega\over c$$, the $$ds^2$$ (relativistic invariant) takes the form $$ds^2=c^2t^2-dr^2-r^2d\theta^2 -2r^2\omega\ d\theta\ dt$$ Following Langevin, this expression is called the metric of the rotating disk. It will be noted the presence of a cross term in $$d\theta\ dt$$, 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 $$d\theta\ dt$$ makes the clocks synchronization impossible?

• (For clarity, the author of that article is called Gianni Pascoli.) I rather suspect that sentence (or part of sentence) "makes the clocks synchronization impossible" must be due to some translation error or so. It comes out of the blue, and there is no follow-up. Possibly you can try and find other authors who describe Langevin's treatment, and compare that with Pascoli. – Cleonis Aug 15 '19 at 16:38

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

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

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

$$\Delta t = t_1-t_2= 2 \pi r (\frac {1}{c-r\omega}-\frac {1}{c+r\omega})= \frac {2\pi r2r\omega}{c^2-r^2\omega^2}=\gamma^2 \frac {4A\omega}{c^2}$$

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

In rotating frame:

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

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

Let $$\dot \theta= \frac {d\theta} {dt}$$

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

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

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:

$$\Delta t= \frac {2 \pi r}{c-r\omega} - \frac {2 \pi r}{c+r\omega} = \gamma^2 \frac {4A\omega}{c^2}$$

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

• I am confused by your last equation: in the rotating frame, the perimeter of the circle is contracted and is not $2\pi r$. On the other hand, writing $T_\pm=\int dt=\int (\dot\theta)^{-1}d\theta$ and paying attention to the fact that one photon goes from $\theta=0$ to $\theta=2\pi$ while the other goes in the other way, one gets $\Delta T=\gamma^2{4A\omega\over c^2}$ as expected. – Christophe May 23 '20 at 14:00