Sagnac fringe shift worked example ($N\simeq 4\mathbf{\Omega }\cdot \mathbf{A}/(c{\lambda }_0)$)

For a stationary rotating apparatus driven by a source with optical coherence and proper frequency ${\nu }_0$, define ${\lambda }_0=c/{\nu }_0$. The source wave phase advances by $2\pi {\nu }_0\Delta \tau$ between the emission times of the two contributions that meet at one readout spacetime event. Define the signed phase lag $\Delta \varphi$ of the $+$ contribution and the corresponding displacement $N$ in units of one interference fringe byThe sign of a photographed spatial displacement also depends on the imposed fringe orientation; this definition fixes the theoretical sign through the delayed beam. In an equal-amplitude output, a representative light intensity is $I(x)=I_{\max }[1+\cos (Kx-\Delta \varphi )]/2$, where $K$ sets the chosen spatial interference fringe spacing $2\pi /K$. Moving $\Delta \varphi$ by $2\pi$ shifts the pattern by one fringe. A fixed output exactly at a bright or dark fringe has zero first-order intensity slope, so a known phase bias or a spatial fringe readout is useful for determining a small signed delay.

Choose a square of side $0.30\mathrm{m}$, a rectangle $0.40\mathrm{m}$ by $0.20\mathrm{m}$, ${\lambda }_0=633\mathrm{nm}$ and $\Omega =+10\mathrm{rad}{\mathrm{s}}^{-1}$, with angular velocity parallel to the positive normal vectors. Both perimeters are $1.20\mathrm{m}$, but the areas are $0.090{\mathrm{m}}^2$ and $0.080{\mathrm{m}}^2$. Using $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$, the leading predictions areThese are original model choices and predictions, not measured points. The square has $9/8$ times the response despite the equal path lengths. Reversing from $+\Omega$ to $-\Omega$ gives $N(-\Omega )-N(+\Omega )=-2N(+\Omega )$: the displacement magnitude between opposite rotations is twice the displacement magnitude from rest. Confusing these two comparisons introduces a factor-of-two error.