# Does Sagnac effect imply anisotropy of speed of light in this inertial frame of reference? There seems to be a consensus that the one - way speed of light is anisotropic in a rotating frame of reference (Sagnac Effect).

According to this article Einstein synchronization "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 synchronization diminishes its usefulness. If clock 1 and clock 2 are not synchronized directly, but by using a chain of intermediate clocks, the synchronization depends on the path chosen. Synchronization around the circumference of a rotating disk gives a non vanishing time difference that depends on the direction used.

Imagine a rotating ring of arbitrarily large diameter. In accordance with the foregoing the one - way speed of light along the ring clockwise and counterclockwise will be different, because simultaneously emitted in opposite directions beams of light that go along the ring will return to the starting point at different times. Hence, it is reasonable to assume that it is anisotropic on any segment of a ring, large or small, say on a segment AB.

Of course, taking into account the Lorentz contraction, the measured round - trip speed of light on any segment of the ring will be exactly equal to c.

Suppose that, a purely inertial laboratory S’ for a very long time moves tangentially to the circumference on which the ring lies, very near to the AB segment.

How does the anisotropic one – way speed of light on the AB segment can magically turn into isotropic one - way speed of light in the co-moving inertial laboratory S’, as the Einstein’s relativity teaches us?

• I'll do you one better: how can rotation magically turn into translation just by using an infinitely large ring? – John Dvorak May 20 at 11:16
• @JohnDvorak, I would ask a different question, namely whether rotation around an infinitely large ring has any true meaning? If you accept the existence of such a ring, then rotation around its axis (which is at an infinitely large radius from the ring) would perhaps be indistinguishable from translation. – Steve May 20 at 11:32
• For any rotating ring $\omega=v/r$ and since $v<c$ then $\lim_{r\to\infty} \omega = 0$ so the rotating reference frame is no longer rotating. – Dale May 22 at 15:57

## 1 Answer

There is nothing "magical" about this. For a rotating ring $$\omega=v/R$$ where $$v$$ is the tangential velocity of the ring and $$R$$ is the radius of the ring. Since $$v then $$\lim_{R\to\infty}\omega=0$$. So then the rotation is 0 and the speed of light is isotropic for any $$v$$.

This should not be surprising at all. The whole reason that you can approximate a large rotating ring as nearly inertial (to first order) is precisely because as the ring becomes large the angular velocity becomes small. This eliminates both the centrifugal force and the Coriolis force, as well as the Sagnac effect and any other first order non-inertial effects.

• Thank you very much for your answer! Sad to say, but your explanation seems to me to be very incomprehensible and unusual, completely contradicting elementary common sense, hence absolutely wrong. It only convinces me more and more that the Lorentz theory is correct, in which the speed of light in one direction is isotropic only in a „preferred“ frame of reference. Of course, in a moving frame of reference, the one way speed of light is anisotropic, while the „round – trip “ speed of light is isotropic. – Albert Jun 2 at 11:49
• @Albert, can you point out what specifically is "very incomprehensible and unusual, completely contradicting elementary common sense, hence absolutely wrong". Are you unaware that $\omega = v/R$? Are you unaware that $v<c$? Do you not understand how to evaluate the limit of $\omega$ as $R$ approaches infinity? – Dale Jun 2 at 12:37
• It is clear that if the rotation stops, the laboratory at the edge will stop and the speed of light will become isotropic. And why have you decided that it would stop? Infinitely tend to zero doesn’t mean equal zero. We can keep linear velocity of the edge infinitely close to c and yes, $\omega$ will infinitely tend to zero if the radius tends to infinity. Is it not clear? If this is hard for you to understand, you can reduce the radius slightly. – Albert Jun 3 at 19:46
• “if the rotation stops, the laboratory at the edge will stop”, nope, this is incorrect. You can fix v at any desired value <c. As R increases $\omega$ decreases even if v is constant. – Dale Jun 4 at 1:37
• $\lim_{R\to \infty} v/R = \omega = 0$ does not imply $lim_{R\to\infty}v=0$. Don't blame SR for your inability to correctly take a limit. – Dale Jun 5 at 13:10