= Sagnac fringe shift worked example
{c}
{title2=$N\simeq4\boldsymbol\Omega\cdot\mathbf A/(c\lambda_0)$}
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 $\Delta\phi$ of the $+$ contribution and the corresponding displacement $N$ in units of one <interference fringe> by
$$
\Delta\phi=2\pi\nu_0\Delta\tau,\qquad
N=\frac{\Delta\phi}{2\pi}\simeq\frac{4\boldsymbol\Omega\cdot\mathbf A}{c\lambda_0}.
$$
The 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 $I(x)=I_{\max}[1+\cos(Kx-\Delta\phi)]/2$, where $K$ sets the chosen spatial <interference fringe spacing> $2\pi/K$. Moving $\Delta\phi$ 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 $\Omega=+10\,\mathrm{rad\,s^{-1}}$, 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 $c=299792458\,\mathrm{m\,s^{-1}}$, the leading predictions are
$$
\begin{aligned}
\Delta t_{\rm square}&\simeq4.006\times10^{-17}\,\mathrm s,&N_{\rm square}&\simeq0.0189705,\\
\Delta t_{\rm rectangle}&\simeq3.560\times10^{-17}\,\mathrm s,&N_{\rm rectangle}&\simeq0.0168626.
\end{aligned}
$$
These 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.
\Image[/sagnac-area-law-fringe-model.png]
{title=Independent path-segment calculation of the predicted fringe displacement for two equal-perimeter loops}
{id=image-sagnac-area-law-fringe-model}
{description=The generator sums $(\boldsymbol\Omega\times\mathbf r)\cdot d\mathbf r$ along each straight segment, rather than inserting the enclosed <area> into the answer. Both loops are oriented counterclockwise. Lines and marked values are leading-order vacuum model predictions; no historical measurements are plotted.}
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