 ## A Simple Description of the Sagnac Effect

If you search the web for "sagnac effect" you're likely to find a number of sites that claim that relativity can't explain the Sagnac effect.  (In fact, when I google'd it just now the second hit claims not only that the Sagnac effect disproves relativity but also suggests that we live in a geocentric universe!)  This is absurd.  In the context of relativity, the Sagnac effect is easy to explain, as we shall see below.  However, from the point of view of Newtonian mechanics, the effect is harder to understand.  If anything, the Sagnac effect is a demonstration of the validity of relativity.

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 acceleration, by itself, does not affect time.  If an accelerating clock is momentarily adjacent to an "inertial" clock which happens to be moving at the same velocity, both clocks will show time is passing at the same rate.  I will also make use of the fact that (small) distances are unaffected by acceleration:  If an accelerating observer is momentarily adjacent to an inertial observer who is moving at the same velocity, both will measure distances between (nearby) points as being the same.  If we keep these two facts in mind, we'll find that the Sagnac effect is really pretty simple to analyze, because, from the point of view of a "stationary" observer, an object moving in a circle at constant speed behaves identically to an object moving in a straight line at constant velocity, save that the object traveling in a circle is constantly accelerating toward the center of the circle.

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.

### Some Basics

Within glass with a refractive index of N, light travels at c/N, where c is the velocity of light in a vacuum.  Typical values for N for real glass are between 1.1 and 3. If we let k = c/N, then the velocity of a signal within a (single-mode) fiber optic cable is k.  (Among other things, this implies that fiber optic links don't get your data there as fast as radio links -- but the difference is minuscule in most cases.)

As usual, we will assume c=1, and we'll explicitly use the symbol c only where we want to emphasize that it's the velocity of light in vacuum that we're referring to.

### The Effect:  A Brief Introduction

Suppose we have a fiber optic ring -- a piece of fiber optic cable wrapped around the edge of a (stationary) disk.  If we send a signal through the cable, around the ring, and measure the time it takes, it will (obviously) take the same amount of time to go around the ring one way as it takes to go around the other way (see figure 1).  The time it takes will depend on the length of cable and the refractive index of the glass.  If the refractive index of the glass is N, then the velocity of the signal is k = c/N.  If the radius of the disk is r, the signal will take 2πr/k to travel around the circle, in either direction.

Figure 1: Now, suppose the disk is spinning.  As Sagnac found in 1913, the transit time is no longer the same in the two directions -- the signal traveling "with" the direction of the spin takes longer to go around the disk than the signal traveling "against" the direction of spin (see figure 2).

Figure 2: This is not obvious!  Consider an observer riding on the rim of the disk.  If he measures the velocity of the signals as they go by, he'll find that they're both going the same speed, regardless of which direction they're going in.  Classically, in the fixed frame of reference of the laboratory, we might expect to find the counterclockwise signal velocity = v + k, and the clockwise signal velocity would be = v - k, and a moment's calculation will show that the time to circle the disk doesn't depend on the direction.  If we let u- = clockwise signal speed, t- = clockwise traversal time, u+ = counterclockwise signal speed, and t+ = counterclockwise traversal time, then classically we would find:  Clearly  t- is equal to  t+ and, in fact, both are equal to the time it takes for the signal to go around a stationary disk.  Using simple Newtonian mechanics to compute the velocity of the signal, we find that the effect is not just strange; it's impossible!  (But note: it's typically called a "classical effect", and there is more than one version of "classical" dynamics.)

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 v.  It's also accelerating centripetally, but we can ignore that -- it has no effect on what our observation of a small neighborhood over a very brief time period.  How fast are the signals in the cable traveling?  Of course, we can just use the composition of velocities formula to find the answer (again, the acceleration has no effect on this).  Analogously to the Newtonian calculations above, we find:  Clearly (6), the time for the signal to go around the ring clockwise, is not equal to (8), the time to go around the ring counterclockwise.  In contrast with the Newtonian case, as the signal velocity approaches c, the effect of the rotation on that velocity becomes smaller; when k=c (=1), u+ = u-, and the difference in transit time is apparently due to the fact that the counterclockwise signal must travel farther than the clockwise signal due to the rotation of the disk.

But if we subtract (6) from (8) to obtain the actual difference in the transit times, we obtain: Rather remarkably, this is independent of the signal velocity!  A beam of light sent around the disk each way, or two people strolling slowly around the disk each way, will encounter the same difference in the time it takes to go all the way around.  This will seem a bit less strange, I think, when we view the problem from the point of view of someone riding on the disk, in the succeeding sections.

### But What About an Observer On the Disk?

Ah, here's the rub.  An observer riding on the disk, stopwatch in hand, who carefully measures the speed of an impulse in a short length of the cable as it passes him by, will find that the physical laws of the universe are working just the way they always do:  The velocity of the signal is k no matter which way it's going.  (If the cable is replaced by a wire he can measure the signal speed with an oscilloscope and some probes.  If it's a fiber optic cable he can cut it at a couple of points, insert electro-optical probes, and once again use his oscilloscope to measure the signal velocity.  He'll find that it's isotropic.)  And yet, if he sends a signal all the way around the disk, it takes longer to go one way than the other way.  How can this be?

Let's approach this with three diagrams, one trivial, one peculiar, and one (hopefully!) enlightening.  We'll start by replacing the problem with a straight length of fiber optic cable traveling in a straight line, with a clock at each end.

### The Trivial Case:  Clocks Properly Synchronized

We have a length of cable traveling left to right, with clock "A" at the front and clock "B" at the back.  (See figure 3.)  The clocks are properly synchronized in the frame of reference of the cable, and an observer who is comoving with the cable and who times signals using clocks A and B will see that the signals travel at the same speed in both directions.

Figure 3: 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 LS and γ = 1/sqrt(1-v2), and if we assume clock A is at the origin, then at time 0 on clock A, on clock B we'll have: ### The Peculiar Case:   Clocks Not Properly Synchronized

Now we take the same cable, and the same clocks, and we change the clocks so that they're synchronized in the stationary frame.  They're no longer properly synchronized in the frame of the cable.  (See figure 4.)  Now, an observer moving with the cable who uses clocks A and B to time signals will find that the signals appear to move faster when they're going to the left than when they're going to the right.  (Of course, a stationary observer who uses clocks A and B to time the signal will also find that it goes from A to B faster than it goes from B to A, but that's as we would expect.)

Figure 4: ### The Final Case:  Wrap the Cable Around the Rim of a Disk

Curl up the cable.  It's still moving at velocity v parallel to its own length, but now it's going in a circle.  (See figure 5.)

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 k regardless of the direction they're going.

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 that is the root of the effect, as viewed by a rider on the disk:  When the signal travels all the way around the disk and comes back, it finds that the clock at the "end point" of its journey is incorrect.  It should be set as in Figure 3, not Figure 4!  But since clocks A and B are really just one clock, we're stuck, and a rider on the disk will see exactly what someone traveling with the cable who used the clocks in Figure 4 to time signals would see:  The time to go all the way around will be measured as different, depending on the direction.

Figure 5: 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 independent of the signal velocity.  It obviously depends only on the velocity of rotation of the disk!  In one direction, the "sync error" is added to the apparent time to go around the disk; in the other direction, it's subtracted.  Thus, the difference in the time it takes between the two directions must be just twice the difference between the settings on clock B in figure 3 and figure 4.  Or, in other words, it must be twice the value given in eq. (10): where we have made use of the fact that LS = 2πr.  When we view the  rotating cable from the stationary frame, if t is time in the stationary frame and τ is time in the rotating frame, then we know 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.

### Note that You Can't Synchronize the Clocks in a Rotating Frame

When clocks are "synchronized" in an inertial frame, and we use them to time the velocity of beams of light, we'll find that it's the same no matter what direction the beams are traveling in.  Indeed, that's the point of the synchronization procedure outlined by Einstein in his 1905 paper on electrodynamics.  But in a rotating frame, it doesn't work.  If there are closely spaced clocks all around the rim of the disk, and we synchronize adjacent pairs of clocks (using pulses of light to carry the time back and forth), then we'll find, when we work our way all the way around the disk, that the time on the first clock doesn't match the time on the last clock.  Since the "unrolled" cable in figure 4 is (more or less) equivalent to the disk, if you could synchronize the clocks all the way around the disk, you could do it for the moving and stationary frames, too, and time would be absolute!

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.

### What About Proper Time?  What does the Signal See?

I've said the signal travels at velocity k relative to the cable, and k < c.  So, time will pass for the signal, and it makes sense to ask how long it appears to take to go around the disk in the frame of the signal.

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 L.  Then, if the signal starts at location 0 at time 0 and travels to location L at time L/k, and if we let γk = 1/sqrt(1-k2), then we can use the Lorentz transform to see that the arrival time in the signal's frame will be 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 = L/k, and it seems like the same amount of time should elapse for the person walking around the disk's edge and someone sitting stationary on the disk.  Yet, a glance at figures 3 and 4 is enough to show that the walker's clock surely can't agree with the clock of the person who did not move when the walker returns to his starting point!  How can this be?

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 in the stationary frame then the person who walks slowly around the disk will find that they are farther and farther out of sync with his watch, until he arrives back at his starting point and finds that they're off by 4πrvγ, as given in eq (11).   And if he then turns around, and walks back around the disk the other way, when he finally arrives back at his starting point (for the third time) he'll find that his watch is back in sync with a clock that remained fixed to the disk!  If instead of walking around the disk and back, we imagine him strolling from one end of a straight cable to the other and back in figure 4, this is reasonable and in fact inevitable; it's only the fact that the cable has had its ends tied together to make a ring that makes it confusing. 