A Simple Description of the Sagnac Effect |

In any context, however, the effect is rather strange, and it can be quite confusing. Typical treatments get lost in the math rather quickly and can leave one with the impression that several pages of algebra are required to "prove" that it happens. A little later on this page, I'm going to present a simple graphical argument (here) showing why it must occur, as viewed from the rotating disk itself.

Throughout this page, I'm going to treat the effect using straight-line motion rather than circular motion as much as possible, which greatly simplifies the math. To do that, I'll make heavy use of the fact that

But before we look at the effect as seen by an observer moving with the disk, we need a little background, and we'll also look at a direct calculation of the effect from the point of view of a stationary (non-rotating) observer.

As usual, we will assume

Now, suppose the disk is

Clearly

To get an idea of what's actually going on in our real (relativistic) universe, let's view the disk from a stationary (non-rotating) frame of reference. In a small neighborhood of a single point on the rim of the disk, we observe that the fiber optic cable is traveling at velocity

Clearly (6), the time for the signal to go around the ring clockwise, is

But if we subtract (6) from (8) to obtain the actual difference in the transit times, we obtain:

Rather remarkably, this is

Let's approach this with three diagrams, one trivial, one peculiar, and one (hopefully!) enlightening. We'll start by replacing the problem with a

We can easily find the time difference between clocks A and B as viewed from the stationary frame just by using a Lorentz transform. If the length of the cable in the stationary frame is

Within the cable, the only difference is that it's now accelerating sideways at every point; but that doesn't change its physical properties. Viewed locally, signals still travel at velocity

But what about those clocks? Clocks A and B are now adjacent; they must show the same time. In fact, to get from Figure 4 to Figure 5, we don't need to adjust the clocks at all -- they're necessarily synchronized in the stationary frame in this case, as well. And

Furthermore, since the time difference is caused by the clocks being "out of sync", we realize at once that the time difference between clockwise and counterclockwise traversals must be

where we have made use of the fact that

and so the difference in arrival times between signals going in opposite directions, viewed from the stationary frame, must be

which is exactly what we found in eq (9) by direct calculation of the transit times of the signals in the stationary frame.

This does it for the pictures, and for the basic description of the effect.

There is, however, lots more that one could say about it. A few more aspects of the problem are touched on, below.

This is, in fact, the essence of the Sagnac effect. The clocks on the rotating disk can't be properly synchronized: any attempt to do so leads to a discontinuity somewhere on the disk. Any signal which goes all the way around the disk crosses the discontinuity.

The Earth provides us with a rotating frame, also. In fact, it's impossible to exactly synchronize all Earth clocks such that light pulses sent between pairs of clocks will appear to have the same velocity regardless of the direction they're going in.

The errors due to the rotation of the slowly turning Earth are so small they're not normally of consequence, however.

Acceleration does not affect time. An accelerating clock shows time passing at the same rate as a non-accelerating clock which happens to be moving at the same velocity. So, the effect on time of straight-line motion is the same as the effect of motion in a circle (since the only difference is that circular motion requires centripetal acceleration), and the rate at which time passes for the signal itself will be the same in figure 5 as it is in figure 3 or figure 4. So to find out how much proper time elapses for the signal as it goes around the disk, we can just look at the case of a straight cable, and that's trivial. Proper time is the same in all coordinate systems, so we can take one more step, and just look at the case where the cable is stationary. Assume the proper length of the cable is

and that's true regardless of which direction the signal travels in.

This brings up another question, which is: What happens if a person, carrying a watch, strolls around the edge of the disk? In the low speed limit, τ =

Again, the problem is that you cannot synchronize the clocks in a rotating frame. If all clocks on the disk are set to the same time

As I understand it, it's most often demonstrated using a fiber optic ring around the edge of a disk, and an interferometer which is also on the disk. Light traveling in opposite directions around the disk is combined and produces interference fringes. As the initially stationary disk "spins up", the fringes shift; classically they should not. The amount of the shift as a function of the rotational velocity of the disk matches that predicted by relativity.

The same effect can be used at larger scales (with large rings and interferometers) to detect even very slow rotation, such as that of the Earth.