It is generally accepted that the fringe shift produced in a fiber optic Sagnac interferometer is independent of the refractive index of the fiber. In fact, the waveguide aspect of the fiber is not of any consequence, the counter propagating beams may propagate in any dielectric medium. Does anybody know a physical explanation for this effect, other than the special relativistic (On the Experiment of F. Harress) or general relativistic (Sagnac Effect) explanations? It has been reported that(Modified Sagnac experiment for measuring traveltime difference between counter-propagating light beams in a uniformly moving fiber) the Sagnac effect is produced even when the moving path is effectively a straight line. This later configuration approximates an inertial frame, and with in an inertial frame one should not be able to measure one's state of motion!
1 Answer
In order for the refractive index to be of no concequence it must be the case that the refractive index has at least two effects, in such a way that they drop away against each other. As we know, the higher the refractive index, the slower the propagation of the light, and the wavelength is correspondingly smaller. That is: the frequency of the light is independent of the refractive index of the medium.
As we know, the frequency of the light comes into effect in the following way. The Sagnac device executes comparison of the amount of cycles of the counter-propagating light and the amount of cycles of the co-propagating light. if the counter-propagating light goes through half a cycle more than the co-propgating light the shift of the interference pattern will be half the distance between two lines of the interference pattern
I'm guessing that the refraction index effects drop away against each other in such a way that the difference in amount of cycles remains the same, whatever the refractive index of the medium.
Further reading:
The discussions of the Sagnac effect by Kevin Brown are very good, and quite accessible:
Sagnac and Fizeau
The Sagnac effect
For a deeper discussion than that I recommend the following article by Olaf Wucknitz:
The Sagnac effect, Twin Paradox, and space-time topology
Olaf Wucknitz also discusses the fiber optic conveyer experiments
In my opinion the article by Olaf Wucknitz is by far the most insightful article about the Sagnac effect, way ahead of anyone else.
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$\begingroup$ Thank you. After going through the suggested references and others, I guess it may be some thing along these lines. If we ignore the so called Fresnel drag for a moment, in a dielectric medium light travels with a velocity of c/n. Compared to a free space interferometer, this would have produced a larger fringe shift (by n^2). But with the increased velocity produced by the Fresnel drag, this amplification effect of the refractive index is reduced. In practical situations, where (c n)^2 >> (r \omega)^2, the observed fringe shift is essentially equal to that of a free space interferometer. $\endgroup$– JKrslCommented May 31, 2022 at 1:07
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$\begingroup$ @JKrsl Well, the concept of Fresnel dragging is rooted in pre-relativistic physics. If it is granted that special relativity is valid then Fresnel dragging is not applicable. I don't know whether the independence of index of refraction is explained. Max von Laue did not address it, in my opinion. I concur with Olaf Wucknitz that GR isn't necessary to account for the Sagnac effect. Kevin Brown (mathpages) asserts specifically that the index of refraction drops out of the equation. Sagnac and Fizeau $\endgroup$– CleonisCommented May 31, 2022 at 19:49
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$\begingroup$ We may call it SR velocity addition or Fresnel drag (FD). With FD, the time required for the propagation of the beams are: $\frac{2n^2\pi r}{cn-r\omega}$ and $\frac{2n^2\pi r}{cn+r\omega}$. The difference of the propagation times is $\frac{4n^2\pi r^2\omega}{c^2n^2-r^2\omega^2}$. When $c^2n^2>>r^2\omega^2$, the $n^2$ term cancels out, giving the familiar expression. $\endgroup$– JKrslCommented Jun 1, 2022 at 7:53