Integrating Polygonal Sagnac Paths

 

As discussed in the note on Sagnac and Fizeau, if two signal 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 upper-right 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 right-hand expression by cos(θ)2 to give the lower right-hand expression. This shows that the expression in the square brackets on the left side is the anti-derivative 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)R2/4, this is

 

 

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

 

If we repeat this analysis for a fiber optic path in the shape of any regular n-gon, 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 n-gon:

 

 

Noting that the area A of the n-gon is given by 2A = nR2sin(2π/n), the power series for this expression can be written as

 

 

where T = tan(π/n). The non-zero coefficients aj of (ωR/c)j in the series can be written as

 

 

where C = cos(π/n). Thus we have

 

 

and in general, beginning with a0 = 1, we have the recursive relation

 

 

The coefficient a2 of the second-order 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 second-order 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 n-gon are different, according to

 

 

Therefore, the highest-order 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.

 

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