= Sagnac area law
{c}
{title2=$\Delta t\simeq(2/c^2)\oint_C\mathbf v\cdot d\mathbf r=4\boldsymbol\Omega\cdot\mathbf A/c^2$}
For a rigidly rotating prescribed vacuum path, use spatial coordinates attached to the apparatus and laboratory time $t$. At a path <point> $\mathbf r$, the local apparatus <velocity> is $\mathbf v=\boldsymbol\Omega\times\mathbf r$. A <light> displacement has instantaneous laboratory displacement $d\mathbf r+\mathbf v\,dt$, up to the common rotation of the spatial axes. The <null condition> is therefore
$$
c^2dt^2=|d\mathbf r+\mathbf v\,dt|^2.
$$
Write $d\ell=|d\mathbf r|$. To first order in $|\mathbf v|/c$, the positive root is
$$
dt\simeq\frac{d\ell}{c}+\frac{\mathbf v\cdot d\mathbf r}{c^2}.
$$
Reversing the path reverses the <dot product> term but leaves the leading path-length term unchanged. Subtract the two return times and use the <scalar triple product> identity:
$$
\Delta t\simeq\frac{2}{c^2}\oint_C(\boldsymbol\Omega\times\mathbf r)\cdot d\mathbf r
=\frac{2}{c^2}\boldsymbol\Omega\cdot\oint_C\mathbf r\times d\mathbf r
=\frac{4\boldsymbol\Omega\cdot\mathbf A}{c^2}.
$$
This is a <line integral> version of the <Sagnac area law>; the <vector area> identity also follows from <Stokes theorem>. For a mirror-defined polygon the actual ray paths move slightly as the <mirrors> rotate; the stationary optical path at zero <rotation> gives the same first-order result. The prescribed-path calculation is an idealization, not an exact finite-speed formula for every moving-mirror layout.
The <dot product> explains the geometry: only the projection of the oriented <vector area> along the <angular velocity> contributes. A planar loop whose <normal vector> is perpendicular to $\boldsymbol\Omega$ has zero leading delay. For repeated traversals with the same orientation, the signed <vector areas> add; oppositely oriented lobes can cancel even when the total unsigned <area> is large. Equal perimeter alone does not imply equal response. A constant translational <velocity> contributes nothing to this leading closed-path <line integral>, since $\oint_Cd\mathbf r=0$.
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