If we have two lasers based at the equator and we direct one of them east and the other west on to two screens each 10 km away, will light emitted synchronously at the lasers arrive at the same time as each other on thier respective screens?

My interest is that one laser moves with the rotation of the earth and the other one the opposite.

From what I know the light shouldn't be affected by the rotation of the earth so the laser on the west side will arrive faster than the east side since the screen on the west side will come to the laser reference point. Correct me if I'm wrong. (with mathematics if you can)

  • 7
    $\begingroup$ See also the Michaelson-Morley Experiment, which sort of was a test of this same idea, the results of which influenced Einstein's thinking on Relativity. $\endgroup$ Oct 19, 2018 at 20:40
  • 2
    $\begingroup$ Yes, this is one of those things where you don't need to take a theoretical approach - literally we tried it and the result is that they arrive at the same time, and theory had to be built to accomodate this result. $\endgroup$
    – pjc50
    Oct 19, 2018 at 21:15
  • 2
    $\begingroup$ Related: Sagnac Effect and Ring Laser Gyroscope. $\endgroup$ Oct 19, 2018 at 21:42
  • 1
    $\begingroup$ Yet another related experiment is that by Hafele & Keating, who flew an atomic clock around the world, once eastward and once westward, and compared the elapsed proper times with a ground-based clock. $\endgroup$ Oct 24, 2018 at 1:15

2 Answers 2


It depends on your frame of reference.

If you are positioned by the laser on the Earth's surface you would observe the two laser beams to have the same speed and your would observe the two targets to be stationary and you would observe the light to take the same amount of time to reach each target.

On the other hand if you were not at rest compared to the laser and screens, say for example you are observing from the moon, you would still observe the laser beams having the same speed but you would see the screen as moving at about 460 m/s. From this frame of reference you would see the light hit the screen to the west first.

What is simultaneous in one frame of reference is not necessarily simultaneous in a different frame of reference.

  • 1
    $\begingroup$ Would a Coriolis force come into effect in the first picture? Light moves at constant speed, but if it takes a longer path... $\endgroup$ Oct 18, 2018 at 20:07
  • 6
    $\begingroup$ @John No. This is exclusively a relativistic phenomenon - the beams' arrivals at the screens are simultaneous in the rest frame of the screen, but they are not simultaneous as observed from the Moon. Simultaneity, like space and time, is relative. $\endgroup$ Oct 18, 2018 at 20:15
  • 3
    $\begingroup$ There's also another component to this: even if you're at rest, if you're standing closer to one of the screens than the other, you will see the light scattered from that screen first. You will only observe simultaneity if you're at rest with respect to the screens and equidistant from both of them. $\endgroup$ Oct 18, 2018 at 20:16
  • 3
    $\begingroup$ Actually in a frame of reference where the targets and and the air are moving the speed of light is different in the two directions is different by about 0.5 m/s. $\endgroup$
    – M. Enns
    Oct 18, 2018 at 20:16
  • 3
    $\begingroup$ @EmilioPisanty: There is Coriolis force at the equator; it just happens to be vertical! $\endgroup$ Oct 18, 2018 at 23:55

The one-way speed of light relatively to Earth's surface is anisotropic, i.e. it is different in different directions. We must remember that the equality of the one-way speed of light to constant c is a convention.

So as to measure the one-way speed of light we need a pair of synchronized clocks, and the outcome of the measurement depends on the convention of how to synchronize them.

Special relativity employs the Einstein synchrony convention for all inertial frames of reference.

In rotating frames, even in special relativity, the non-transitivity of Einstein synchronisation diminishes its usefulness. If clock 1 and clock 2 are not synchronised directly, but by using a chain of intermediate clocks, the synchronisation depends on the path chosen. Synchronisation around the circumference of a rotating disk gives a non-vanishing time difference that depends on the direction used. This is important in the Sagnac effect and the Ehrenfest paradox. The Global Positioning System accounts for this effect.

Einstein synchronisation

Let us imagine that a short-wave radar is stationed close to the city of Quito, sending a narrow-angle signal in the east direction. Let us also imagine that all over the equator line a great number of reflectors are stationed in such a way that any of the adjacent reflectors is within the field of vision from another. Let the reflectors deflect the radar signal emitted in Quito in such a way that it, propagating zigzag-wise near the Earth’s surface, circles the Earth along the equator, coming back to the radar in Quito from the west.

Knowing the length of the zigzag line along which the radar signal propagates and the time needed for signal to circle the Earth, an operator of a radio relay station can calculate the propagation speed of the signal circling the Earth from east to west or reverse. That these speeds will not be identical to and differ from the constant C can be upheld by the following:

Let us ideally place a non-rotating detached onlooker at a distance from the Earth point of the imaginary Earth rotation axis. Let this onlooker be immobile in relation to the centre of the Earth’s mass and be watching the northern hemisphere rotating counter clockwise below us, mentally following the signal propagation.

Within the reference frame of the detached onlooker the speed of light propagating in space zigzag-wise is equal to the fundamental constant C. If the Earth were not rotating, then the signal in order to circle the hypothetically non-rotating Earth would need the time equal to the length of the zigzag line encompassing the Earth along the equator, divided by the constant C.

However, the Earth does rotate!

When the signal arrives at the initial point in the space of the detached onlooker, the radar of Quito will move approximately 62 meters east, and the signal arriving from the west would need extra time equal to 0.2 microseconds to return to the radar.

If the operator turned the antenna 180 degrees and directed the signal westward, the signal would need two microseconds less to circle the Earth and return to the radar because during the signal’s travel around the Earth the radar would move 62 meters eastward and the signal coming from the east would have no need to cover these 62 meters. The signal delay is a first order effect in relation to the value of v/С, where v is the linear speed of the rotating Earth’s surface, and this delay is large enough compared to relativistic effects of the second order of smallness.

Considering the Lorentz contraction of the equator and the slowdown of the rate of the clock moving together with the surface of the rotating Earth, the average speed of light on the way there and back would have been precisely equal to the constant C.

The clock synchronization considering the inequality of the speeds of light from west to east and from east to west will indeed give the same result as the clock synchronization through a synchronizing signal transmitted by a detached onlooker from a point on the imaginary Earth’s axis of rotation to all point on the equator. The readings of the clocks synchronized with regard to the inequality of the speeds there and back are conceived by a detached onlooker as the same.

The issue of synchronization becomes even more entertaining, if we ideally replace the Earth with a giant ring of an arbitrarily large diameter, accommodating a transmitter/receiver and a system of reflectors there. In this case, at a given linear speed v of the ring and an arbitrarily small angular speed of the ring rotation, the deviation of the signal propagation speed in one of the directions from the constant C will, as a first approximation, equal v.

If we imagine that an inertial laboratory is tangentially flying to the ring at a speed equal to the linear speed of this ring, and that this laboratory, given an arbitrarily large diameter of the ring, during an arbitrarily long period of time finds itself beside a nearby sector of the ring, then during this period of time this sector of the ring and the laboratory will find themselves practically immobile one to another. If the speed of signal propagation in one direction in relation to a sector of the ring is different from C, then why should the speed of that same signal (in the same direction) in relation to the inertial laboratory unquestionably be thought of as equal to constant C?

A short note with mathematics in this book, p. 42, The Sagnac effect:

Lecture Notes On General Relativity. Øyvind Grøn

  • 2
    $\begingroup$ TL;DR, but right that the answer is in the Sagnac effect (cf. laser ring gyroscope). $\endgroup$
    – amI
    Oct 19, 2018 at 3:55
  • $\begingroup$ You are ignoring the effect of the earth's atmosphere, I presume? Refraction, speed of light in the air, frequency-dependent absorption/attenuation, scattering, etc. $\endgroup$ Oct 20, 2018 at 21:29
  • 1
    $\begingroup$ Thank you for your edits! I think this is an idealized model. It did not even occur to me to take into account the air and weather conditions (wind, rain) as well as giraffes and hippos walking along the equator, which can block the beam with their bodies. Then the beam will not come at all to the target. So I think that this is not about air, but about synchronization of clocks on a rotating ring. Einstein synchronization (from clock to clock) on the entire rotating ring is IMPOSSIBLE. Accordingly, the speed of light is anisotropic on any single segment. $\endgroup$
    – Albert
    Oct 21, 2018 at 7:31
  • $\begingroup$ If we synchronize all clocks on the rim by beam from the center (it is equidistant from all clocks on the rim), as I have already noted in the answer, measured one way speeds of light will be anisotropic, but measured speed of light "back and forth" at any single segment will be equal precisely to c. Good to note, that actually these clocks will appear to be synchronized fully in accordance with Reichenbach synchronization ($\epsilon$ has value between 0 and 1), which keeps one way speeds of light anisotropic while speed of light "back and forth" isotropic. $\endgroup$
    – Albert
    Oct 21, 2018 at 7:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.