Integrating Polygonal Sagnac Paths 

As discussed in the note on Sagnac and Fizeau, if two signals move in opposite directions around a guided closed loop path on a rotating platform, always maintaining a constant speed relative to the local point on the platform, the time difference (in terms of the time coordinate for an inertial system in which the center of rotation is at rest) for the signals to complete their journeys is given exactly by the integral around the path 



where ω is the angular rotation speed of the platform, θ is the angle about the point of rotation, and r is the distance (in units of time) from the center of rotation to a point on the path. We illustrated this by applying it to elliptical paths, rotating about the center or focus. What about a equilateral triangle path, rotating about the center of the triangle? (This is not the same as having three mirrors at the vertices of a triangle, because we require the light to follow the guided path, such as through a fiber optic line. Also note that the fiber optic lines would be situated such that while rotating at speed ω the path in terms of the inertial coordinates of the center of rotation at any instant of coordinate time is an equilateral triangle.) The path is illustrated in the figure below, showing an equilateral triangle inscribed in a circle of radius R centered on the origin. 



The edges of the triangle represent fiber optic cables, and the triangle rotates about the origin with angular speed ω. For any point on the triangle we have the relations 



Also, the equation of the line corresponding to the upperright edge is 



so for the point on that edge we have 



We substitute this into equation (1), and recognize that the integral for each edge will give the same result, so we can multiple the integral for a single edge by 3. Thus we get 



Note that v = ωR is just the normalized velocity of each vertex of the triangle in geometrical units. (In conventional units this would be written as v = ωR/c where c is the speed of light in vacuum.) To evaluate the integral, we notice the following derivative 



where we multiplied through the numerator and denominator of the upper righthand expression by cos(θ)^{2} to give the lower righthand expression. This shows that the expression in the square brackets on the left side is the antiderivative of the integrand in the previous equation. So, since tan(0) = 0 and tan(2π/3) = –√3, we can evaluate the integral to give 



Making use of the addition formula for the hyperbolic arctangent 



we can simplify the result to 



Ordinarily v is quite small, so only the highest order term is significant. Expanding the expression into a power series in v, the result can be written as 



In terms of conventional units, so that v = ωR/c, and recalling that the area of the equilateral triangle is A = (3√3)R^{2}/4, this is 



As expected, the first term is just the familiar firstorder approximation for the Sagnac effect. 

If we repeat this analysis for a fiber optic path in the shape of any regular ngon, rotating about its center, equation (1) gives 



Again putting v = ωR (which is ωR/c in conventional units), we note the derivative 



where 


Making use of this identity to evaluate the integral for θ = 0 to 2π/n, and simplifying, we arrive at the general result for the time difference of a regular ngon: 



Noting that the area A of the ngon is given by 2A = nR^{2}sin(2π/n), the power series for this expression can be written as 



where T = tan(π/n). The nonzero coefficients a_{j} of (ωR/c)^{j} in the series can be written as 



where C = cos(π/n). Thus we have 



and in general, beginning with a_{0} = 1, we have the recursive relation 



The coefficient a_{2} of the secondorder correction term has the values 1/2, 2/3, and 5/6 for n = 3, 4, and 6 respectively. All the coefficients approach 1 as n increases to infinity, consistent with the fact that the secondorder coefficient for a circle is 1, recalling that the exact Sagnac effect for a circle is 



where v = ωR/c. Of course, for equal areas, the values of “R” for the circle and ngon are different, according to 



Therefore, the highestorder difference between the Sagnac time effect is 



Thus a regular polygonal path has a (very) slightly greater Sagnac effect than a circular path of the same area. For triangular, square, and hexagonal paths, this gives the specific results 



The factors in square brackets are approximately 0.209, 0.047, and 0.008 respectively. 

Incidentally, since Sagnac loops with equal areas but different shapes have slightly different time delays, we can obviously construct a loop with two such shapes, routing the light in opposite directions around each shape (as in a figure8 pattern), to give a Sagnac loop with zero net area but a nonzero time delay. Perhaps the simplest example consists of two circular arcs of different radii as shown below. 


The light in one direction propagates along the path 1, 2, 3, 4, 5, 6, 1, with circular arcs from 3 to 4 and from 6 to 1. The light in the opposite direction propagates along the reverse path, and the entire configuration rotates about the hub at point 2 (and 5). Now, equation (1) applied to these two circular arcs with opposite signs (noting that radial segments contribute nothing to the integral, since dθ = 0 along those segments) immediately gives 



The areas will be equal (with opposite sign) just if 



in which case the equation reduces to 



where we have put v_{1} = ωr_{1} and v_{2} = ωr_{2}, representing the speeds of points on the circular arcs. 
