We saw previously that the time delay (and therefore the difference in the optical path lengths) for a circular loop is proportional to the area enclosed by the loop. This interesting fact actually applies to arbitrary closed loops. To prove this, we will derive the difference in arrival times of the two pulses of light for an arbitrary polygonal loop inscribed in a circle. Let the (inertial) coordinates of two consecutive mirrors separated by a ubtended angle q be

where w is the angular velocity of the device. Since light rays travel along null intervals, we have  c²(dt)² = (dx)² + (dy)²,  so the coordinate time T required for a light pulse to travel from one mirror to the next in the forward and reverse directions satisfies the equations

Typically ωT is extremely small, i.e., the polygon doesn't rotate through a very large angle in the time it takes light to go from one mirror to the next, so we can expand these equations in wT (up to second order) and collect powers of T to give the quadratic

The two roots of this polynomial are the values of T, one positive and one negative, for the co-rotating and counter-rotating solutions, so the difference in the absolute times is the sum of these roots. Hence we have

This is the net contribution of this edge to the total time increment. Recalling that the area of a regular n-sided polygon of radius R is nR²sin(2π/n)/2, the area of the triangle formed by the hub and the two mirrors is R²sinθ/2. It follows that each edge of an arbitrary polygonal loop inscribed in a circle contributes 4A_iω/(c² - v²cosθ) to the total time discrepancy, where A_i is the area of the ith triangular slice of the loop and v = ωR is the tangential speed of the mirrors. Therefore, the total discrepancy in travel times for the co-rotating and counter-rotating beams around the entire loop is simply

where A is the total area enclosed in the loop. This applies to polygons with any number of sides, including the limiting case of circular fiber-optic loops with virtually infinitely many edges (where the "mirrors" are simply the inner reflective lining of the fiber-optic cable), in which case q goes to zero and the denominator of the phase difference is simply c² - v². For realistic values of v (i.e., very small compared with c), the phase difference reduces to the well-known result  4Aw/c². It's worth noting that nothing in this derivation is unique to special relativity, because the Sagnac effect is a purely "classical" effect. The apparatus is set up as a differential device, so the relativistic effects apply equally in both directions, and hence the higher-order corrections of special relativity cancel out of the phase difference.

It's also worth noting that the amount of angular travel of the device during the time it takes for one pair of light pulses to circumnavigate a circular loop is directly proportional to the net "anisotropy" in the travel times. To prove this, note that in a circular Sagnac device of radius R the beam of light in the direction of rotation travels a distance of (2π - ωt₁)R and the other beam goes a distance of (2π + ωt₂)R where t₁ and t₂ are the travel times of the two beams, and ω is the angular velocity of the loop. The travel times of the beams are just these distances divided by c, so we have

(更正：两公式的右边都丢了R，这就是连初中生也知道的两个典型的伽利略变换公式)

Solving for the times gives

so the difference in times is

where A = 2πR²  and v = ωR.  The "anisotropic ratio" is the ratio of the travel times, which is

Solving this for ωR gives

Letting q denote the angular travel of the loop during the travel of the two light beams, we have

Substituting for wR this reduces to

Therefore, the amount by which the ratio of travel times differs from 1 is exactly proportional to the angle through which the loop rotates during the transit of light, and this is true independent of R. (Of course, increasing the radius has the effect of increasing the difference between the travel times, but it doesn't alter the ratio.)