Sagnac Effect

Suppose two identical clocks are in motion on Earth’s Equator with constant speed v relative to Earth, one moving east and one moving west around the Equator. Do they tick at the same rate? What do their elapsed times reveal when they meet again?

Answer

No, they do not tick at the same rates. Their tick rates are different because Earth is rotating with respect to an inertial reference frame such as the distant stars. The clock moving eastward has a higher velocity with respect to the inertial frame than the clock moving westward at all moments. According to the STR, the higher the velocity, the slower the clock ticks. That is, a clock ticks fastest when at rest in an STR inertial reference frame.

The difference in the elapsed time for the two clocks can be calculated by considering a light clock following a circular light path around the Equator.

One also could use a regular n-gon of flat mirrors to reflect the light around the Equator and then take the limit as n becomes infinite. The light leaves from point P on the Equator of the rotating Earth and returns to point P in time T. The light going eastward has traveled the distance 2πR + ωRT in the inertial system, where ω is the angular frequency of rotation with respect to the inertial reference frame. The point P has traveled ωRT. The ratio of point speed to light speed is ωR/c = ωRT/(2πR + ωRT), from which T = 2πR/(c – ωR). For the system at rest, T = 2πR/c. Hence, when ω ≠ 0, define δT = T – 2πR/c as the extra time required. Substitution for T gives δT = 2πωR²/[c (c – ωR)]. Upon returning to point P on the Equator after one circuit, the clocks will differ by 2δT for the measured elapsed times.

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