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Relativistic derivation of Sagnac formula
Consider a flexible optical fiber of arbitrary shape, moving arbitrarily in space, without stretching, see Figure 4. Consider a small segment of the fiber, whose length in an instantaneously inertial frame attached to it is $$d\ell'$$. The time intervals, $$dt'_\pm$$, it takes the left and right moving light rays to traverse the interval, are the same in this frame and are given by$$dt'_\pm={n \over c} d \ell'$$where $$n$$ is the index of refraction of the fiber. Let $d\ell=|d\mathbf{x}|$ be the length of this interval in the lab frame. By the relativistic length contraction formula, $d\ell'=\gamma d\ell\approx d\ell$ correct to first order in the velocity $$\mathbf{v}$$ of the segment. The time intervals $$dt_\pm$$ for traversing the segment in the lab frame are given by Lorentz transformation as:$$dt_\pm=\gamma \left(dt' \pm \frac{\mathbf{v}\cdot d\mathbf{x}'}{c^2}\right)\approx \frac n c d\ell \pm \frac{\mathbf{v}}{c^2}\cdot d\mathbf{x}$$correct to first order in the velocity $$\mathbf{v}$$. (The two beams visit the interval at slightly different times, but, in the absence of stretching $d\ell$  is the same.) It follows that the cycle times  of the two rays are different and the delay is given by$$\Delta T= \int \left(dt_+-dt_-\right)\approx \frac 2 {c^2} \oint  \mathbf{v}\cdot d\mathbf{x}$$Remarkably, the delay is independent of the refraction index $$n$$ and the velocity of light in the fiber. The phase difference at the detector is given by the generalized Sagnac formula, called Sagnac-Wang in $$\Delta \Phi\approx \frac {4\pi} {\lambda c} \oint \mathbf{v}\cdot d\mathbf{x}$$In the special case that the fiber moves like a rigid body with angular frequency $$\boldsymbol{\omega}$$, the velocity is $ \mathbf{v}=\boldsymbol{\omega}\times\mathbf{x}$  and the line integral can be computed in terms of the area of the lop:$$\oint\mathbf{v}\cdot d\mathbf{x}=\oint\boldsymbol{\omega}\times\mathbf{x}\cdot d\mathbf{x}= \oint\boldsymbol{\omega}\cdot\mathbf{x}\times d\mathbf{x}=2\oint\boldsymbol{\omega}\cdot d\mathbf{A}= 2\boldsymbol{\omega}\cdot \mathbf{A}$$This recovers Sagnac formula$$\Delta \Phi\approx \frac {8 \pi} {\lambda c}\boldsymbol{\omega}\cdot \mathbf{A} $$