Calculation of the Sagnac effect in restricted relativity

After having seen the Effet Sagnac it is advisable to make of it a detailed and rigorous analysis within the framework of the restricted Relativité.

Sagnac effect

We will derive the Sagnac effect within the framework of restricted relativity. That will assure us the fact that restricted relativity predicts well the anisotropy speed of light in the turning reference mark, anisotropy speed characteristic of the effect Sagnac (shift of the moments of arrival on a transmitter-receiver E of two light signals emitted at the same time by E being propagated in opposite direction along the circumference of a turning disc). It could also be useful for the reader studied the Géométrie of the space time in the revolving reference marks in order to have complementary details.

The reference mark R will indicate an inertial reference mark with its origin in the center of the revolving disc of $R$ . The transmitter-receiver E is located in a point (noted E) of the periphery of the disc. It is rotated with the disc. The turning reference mark, attached to the disc, will be noted R'.

We will call R1 the inertial reference mark (local) of origin located in E and moving at the same speed $V$ as E. In the vicinity of E, R' goes at the same speed as R1 during an infinitesimal time. It is in the inertial reference frame tangent R1 that the relative speed of the light is isotropic (in accordance with Restricted Relativity) and not in the reference mark turning R'. Confusion between relative speed of the light in the reference mark turning R' (relative speed whose definition rests on the relativistic synchronization of the distant clocks having course in the inertial reference frame R) and relative speed of the light in the inertial reference frame tangent R1 (relative speed whose definition rests on relativistic synchronization having courses in the inertial reference frame tangent R1) is at the origin of errors of reasoning which have, in the past, conduit to wrongly interpret the Sagnac effect like a setting at fault of Restricted Relativity.

R 1, R 2, R 3… will indicate tangent inertial reference marks going at the speed of R' in an infinitesimal vicinity of a point of the edge of the revolving disc (and for one negligible length of time). The transformations of Lorentz between inertial reference marks can be used whereas to treat the case of R' in its globality the transformation of Lorentz is inapplicable.

The disc is in rotation with the angular velocity $\ omega$ . The disc being flat, with two dimensions, we choose to work in polar coordinates $\ left \ {x^ \ driven \ right \} = \ left (T, R, \ theta \ right)$ .

The Cartesian and polar coordinates are connected simply by:

$\ begin {matrix} x=r \ cos \ left (\ theta \ right) \ \ y=r \ sin \ left (\ theta \ right) \ end {matrix}$

The relativistic interval is given by

$ds^2=g_ {\ driven \ naked} dx^ \ driven dx^ \ nu=-c^2dt^2+dr^2+r^2d \ theta^2$

(here we noted the " tensor métrique" $g_ {\ driven \ naked}$ because, after the change of coordinates, it is not equal to the tensor of Minkowski $\ eta_ {\ driven \ naked}$ ). As one easily checks it by using the bond between polar and Cartesian coordinates.

The line of universe of E is given by the coordinates:

$E= \ left (ct, R, \ theta \ right) = \ left (ct, R, \ Omega T \ right)$

I.e.

$E= \ left (\ frac {C} {\ Omega} \ theta, R, \ theta \ right)$

E emits two signals $S_1$ and $S_2$ (not necessarily of the light) of which the angular velocities are $\ omega_1$ (signal in the same direction as E) and $\ omega_2$ (signal in the opposite direction). Their line of universe is given, just like above, by

$\ begin {matrix} S_1= \ left (\ frac {C} {\ omega_1} \ theta, R, \ theta \ right) \ \ S_2= \ left (\ frac {C} {\ omega_2} \ theta, R, \ theta \ right) \ end {matrix}$

What interests us is the reception of the signals $S_1$ and $S_2$ by E after a full rotation, i.e. moment when the lines of universe are recut. These events take place at the moments

$\ begin {matrix} \ frac {\ theta_1} {\ Omega} = \ frac {1} {\ omega_1} \ left (\ theta_1+2 \ pi \ right) \ \ \ frac {\ theta_2} {\ Omega} = \ frac {1} {\ omega_2} \ left (\ theta_2-2 \ pi \ right) \ end {matrix}$

The solution of this system of equations is

$\ begin {matrix} \ theta_1= \ frac {2 \ pi \ Omega} {\ omega_1- \ Omega} \ \ \ theta_2=- \ frac {2 \ pi \ Omega} {\ omega_2- \ Omega} \ end {matrix}$

To reduce the writing, we will use the fallback speeds $B= \ frac {\ Omega R} {C}$ , $B_1= \ frac {\ omega_ 1R} {C}$ and $B_2= \ frac {\ omega_ 2R} {C}$ . The preceding relations are written then

$\ begin {matrix} \ theta_1= \ frac {2 \ pi B} {B_1-B} \ \ \ theta_2=- \ frac {2 \ pi B} {B_2-B} \ end {matrix}$ .

Clean time is given by the interval $cd \ tau=-ds$ . Consequently, to obtain the clean time of E, i.e. the time read by E on the clock at rest in R', it is necessary to integrate the interval on its line of universe (along the circumference of the disc, R' is locally identified during time with the various noted tangent inertial reference marks R 1, R 2… in the vicinity of their origin).

$\ tau=- \ frac {1} {C} \ int ds= \ frac {1} {C} \ int \ sqrt {c^2dt^2-R^2d \ theta^2} = \ frac {1} {\ Omega} \ sqrt {1-B^2} \ int D \ theta$

With times calculated higher, we now have the time indicated by the clock of E at the time of the events of reception of the two signals emitted by E.

$\ begin {matrix} \ tau_1= \ frac {2 \ pi B} {\ Omega} \ frac {\ sqrt {1-B^2}} {B_1-B} \ \ \ tau_2=- \ frac {2 \ pi B} {\ Omega} \ frac {\ sqrt {1-B^2}} {B_2-B} \ end {matrix}$

And the difference is equal to

$\ Delta \ tau= \ tau_1- \ tau_2= \ frac {2 \ pi B} {\ Omega} \ sqrt {1-B^2} \ frac {B_2-2B+B_1} {\ left (B_1-B \ right) \ left (B_2-B \ right)}$

One is interested if the transmitter-receiver E uses signals of which speed is isotropic in R1 (like light signals for example). If E emits particles, it is enough that it emits them same manner in the two directions. For light being propagated in a medium where its speed is lower than $c$ (of water, a fiberoptic), one needs that the medium moves with E, i.e. is interdependent of the disc. The same if one uses sound waves the air must thus move with the disc (it is compared to the propagation medium that the speed of sound is constant).

To impose isotropic speeds in R1, we need to know speeds $B'_1$ and $B'_2$ signals in R1. R and R1 being inertial reference marks, we can use the addition speeds and we have

$\begin{matrix}B_1=\frac{B'_1+B}{1+B'_1B}\\B_2=\frac{B'_2+B}{1+B'_2B}\end{matrix}$

While replacing in the difference in time, we obtain

$\ Delta \ tau= \ frac {4 \ pi B^2} {\ Omega} \ frac {1} {\ sqrt {1-B^2}} + \ frac {2 \ pi B} {\ Omega} \ frac {1} {\ sqrt {1-B^2}} \ left (\ frac {1} {B'_1} + \ frac {1} {B'_2} \ right)$

We can now take account of the constraint of emission with an isotropic speed

$B'_1=-B'_2$

The preceding relation becomes then

$\ Delta \ tau= \ frac {4 \ pi B^2} {\ Omega} \ frac {1} {\ sqrt {1-B^2}} = \ frac {4 \ pi R^2 \ Omega} {c^2} \ left (1 \ frac {\ omega^ 2R ^2} {c^2} \ right) ^ {- 1/2}$

It is the relativistic Sagnac effect.

The calculation of the difference in time of arrival of the signals in E in R' is given in the technical part. This result is derived in any general information then by supposing that the speed of the signals is isotropic in R1.

Some remarks are essential.

This result is quite identical, except for the first order (i.e. by considering small angular velocities), with the result obtained in the simple study of the Effet Sagnac obtained within a traditional framework. It is also the result obtained in experiments.
This result shows that at high speed there is a typical relativistic correction.
Même if the speed of emission is isotropic locally in R1, after a full rotation of the signals, E notes Time Gap well given by the relation in the technical part. This confirms the aptitude of relativity restricted to model the anisotropy speed of light in the revolving reference frame.
the speed of the signals does not appear in the expression obtained. The Sagnac effect is independent the speed of the signals (provided that they are isotropic) and depends only on the rotation of the disc. The Sagnac effect is universal. It is predicted by restricted relativity and is confirmed in experiments.

Some remarks on detection by interferometry of the Sagnac effect

Notice 1

Time Gap which we have just calculated also applies to the components of Fourier (decomposition in various frequencies) of a package of wave associated with two matter beams (in quantum Mécanique, the particles are also waves) being propagated in the two directions along the circumference with same speed.

Of course, only matter flows are physical entities, while the components of Fourier are right mathematical entities which do not transport energy. From the point of view of detection by interferometry of the Sagnac effect, the crucial point is the following. In spite of the absence of a direct physical significance and transfer of energy, the speed of phase of these components of Fourier obeys the law of Lorentz of composition speeds and is shared by the components of Fourier in the two directions.

Moreover, detection by interferometry of the Sagnac effect requires that the package of wave associated with the matter beam is sufficiently concentrated in the space of the frequencies to allow the appearance, in the area of interferometry, of a network of observable fringes. It can be useful to recall that:

the displacement of the interference rings $\ Delta z$ depends on the speed of phase of the component of Fourier of the package of wave.
In an inertial reference mark synchronized with the method of Einstein, the speed of any component of Fourier of a package of wave associated with a matter beam, moving with speed (in absolute value) $v=c \ left|B'_1 \ right|$ or $v=c \ left|B'_2 \ right|$ , is given by the expression of Broglie $v_f=v^2/c$ (mechanical quantum).

Consequently, the shift of the fringes in the Sagnac effect due to the difference in time relativistic calculated above is:

$\ Delta \ Phi=2 \ pi \ Delta z=2 \ pi \ left (\ frac {v_f} {\ lambda} \ Delta \ tau \ right) = \ frac {8 \ pi^ 2R ^2 \ Omega} {\ lambda v} \ left (1 \ frac {\ omega^ 2R ^2} {c^2} \ right) ^ {- 1/2}$

Notice 2

Note that Anderson, Stedman and Bilger found with the approximation with the first order the following difference in time:

$\ Delta t= \ frac {4 \ Omega \ pi R^2} {v^2}$

and the displacement of the fringes according to:

$\ Delta \ Phi= \ frac {8 \ pi^2 \ Omega R^2} {\ lambda v}$

where $v$ is speed " not entraînée" beams (speed in R).

Of course, this difference in time is not in agreement with the approximation with the first order (compared to $B= \ Omega R/c$ ) of the value which we calculated. However, it is consistent with the approximation with the first order of the general difference in time (with speeds not necessarily isotropic) provided that $B_1=-B_2 \ equiv v/c$ : that represents a physical situation completion different where the two beams are injected along the circumference in directions opposed with identical speeds in inertial reference mark R. speed being lower than $c$ , it is not invariant and the signals will not have an isotropic speed in R1.

On another side, the difference in phase that they give and which is the only observable quantity in a device of interferometry, is not in agreement with the physical situation considered by these authors. But surprisingly, it is perfectly in agreement with the approximation with the first order of the value which we calculated. If the condition of isotropy in R1 is imposed, the value that they give for the time lag is incorrect but the value of the shift of phase is correct.

Calculation of Time Gap

It is interesting to derive Time Gap in a another way, at the same time more intuitive and simpler. Although it is less rigorous (the only really rigorous method is that presented higher).

The idea is to take as a starting point the noted shift when one tries to synchronize the clocks on the disc.

One uses a series of observers and their reference marks inertial local R 1, R 2, R 3, etc separate one of the other of the infinitesimal distance which we note $dx=Rd \ theta$ (where $d \ theta$ is the infinitesimal angle which separates two small successive fields).

R is as usual the inertial reference mark.

Each observer is framed by a pair of events: R1 by has and B, R2 by B and C, etc the events are also separate (in R) by the distance $dx$ . And speed (longitudinal compared to $dx$ ) is $V=R \ omega$ as usual.

We consider in R that all the events are simultaneous so as to synchronize the clocks in R.

In R1, because of the relativity of simultaneity, has and B will be shifted.

If R' were inertial, there would be a shift uniform and proportional to the distance (see the transformations of Lorentz), C would be shifted a little more than B compared to has, D, still a little more, and by making a full rotation one would have an increasing shift then decreasing (the horizontal coordinate of R' in the figure above would be increasing while going towards the line of the decreasing figure then while returning towards the left, by exceeding the position of origin, then again increasing while returning towards the line) and a final shift no one between has and has! , as it should be if synchronization did not pose a problem. Synchronization could be done on the disc.

But on the disc total synchronization is not possible. Each observer will note a light shift between the events which frames it and in a full rotation one obtains a total shift not no one, it is Time Gap around the disc.

The transformations of Lorentz give the shift of has and B for R1:

$dt'= \ gamma \ left (dt- \ frac {Vdx} {c^2} \ right)$

Since has and B are simultaneous in R, there is $dt=0$ . Therefore,

$dt'= \ gamma \ frac {driveways and various services \ theta} {c^2}$

While integrating on a full rotation ( $\ theta$ varies from 0 with $2 \ pi$ ), there is Time Gap

$\ Delta t'= \ gamma \ frac {2 \ pi RV} {c^2}$

Maybe, by using the fallback speed $B=V/c$ ,

$\ Delta t'= \ frac {2 \ pi B^2} {\ Omega \ sqrt {1-B^2}}$

If one employs the light (in the vacuum), the speed of the signals is $c$ , the difference in length traversed by the two signals is thus

$\ Delta l'= \ frac {4 \ pi B^2c} {\ Omega \ sqrt {1-B^2}}$

That one can qualify " Length Gap".

If one sends a signal of each with dimensions, one will undergo Time Gap and the other same Time Gap with the opposite sign. The variation of time between the reception of the signals will be the double, and one finds the formula which we calculated.

By using the speed of the signals, one can also calculate the length which they traverse for O' when they make a turn.

One can qualify the difference length of " Length Gap". It is the difference in length according to whether one makes the turn in a direction or the other. Note that shows well that the geometry is not Euclidean in the R' reference mark. Worse still, even with usual curved geometries (for example the surface of the sphere) which you make the turn of a circle in a direction or the other, that gives the same length. Cutting in " slice spatiale" as suggested in the Effect Sagnac before risk thus not only to give us a curved space but in more risk to raise difficulties to us!

Let us note that this Length Gap is not " universel" because it depends on the speed of the signal. In particular, if a O' observer placed on the disc sends a " arpenteur" to measure the length of the circumference at low speed (compared to the number of revolutions and with $c$ ), according to the direction it will measure a different length but with an extremely weak difference.

Length Gap calculated above is the difference in length " maximale".

A small numerical example will give an idea. Either $R=1$ meter and $\ omega=100000$ (that made 16000 turns a second, or almost a turn million per minute, which is already much, hundred times faster than the inflated engines of formula 1, and even unrealizable with real discs which would break under the action of the centrifugal force). Let us suppose that our land-surveyor moves to 1 meter a second (on the disc in rotation).

Then the Sagnac effect will be of a hundredth of billionth of second! It is weak but perfectly measurable with interferometers or even with atomic clocks. Maximum Length Gap will be of 4 millimetres. What is measurable, in particular by interferometry. But the land-surveyor will measure Length Gap only one hundredth of nanometer (smaller than the size of an atom)! Useless to make the experiment, it is impossible to measure. Even if the land-surveyor, equipped with a clock, notes same Time Gap. Small causes, great effects, great causes, small effects!

See too

the Effect Sagnac
the Paradox of Ehrenfest
the Paradox of Selleri
the Synchronization in the revolving reference marks
the Geometry of the space time in the revolving reference marks
the Paradox of the twins and the effect Sagnac
the compact spaces
the Paradox of the twins in compact spaces
the restricted Relativity
the General relativity
the Principle of relativity
the Paradox of the twins

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