# The Sagnac Effect: Exploring Length Contraction with a Thought Experiment

• MattRob
In summary, according to this book, if you spin a disk quickly, it will seem to have a circumference that is less than what you would expect. This is because the radius of the disk decreases as you move away from the center.
MattRob
So, I've been reading on GR and I've come across this.

Awhile ago I read a smaller book on relativity that ended with a neat thought experiment: take a disk, lay out the circumference with many small rods, and start spinning it at relativistic speeds. Because of SR length contraction, each of the rods seems shorter, therefore the circumference of the circle as measured by you, the inertial observer, would seem to be less than 2πr. As odd as it is, I've been able to visualize it pretty well thanks to embedding diagrams like what I have linked below, where in (b) the radial distance is greater than what the circumference would imply for flat euclidean space.

So, overall, it seems as though this should make sense, except the equation the book's offered is throwing me off. It's:

$C=2πr(1+\frac{ω^{2}r^{2}}{2c^{2}}) > 2πr$

However, instead of implying that the circumference is less than what you'd expect with a given radial distance, this equation seems to imply that the circumference is more. Is that the case, or am I understanding this equation incorrectly? Is

$\frac{ω^{2}r^{2}}{2c^{2}}$

A negative term?

What book is that? If you work with consistent units then the circumference is always 2πr. However if you measure the different distances with differently contracted rods, then you can get such strange looking results. And I don't think it has anything to do with event horizons.

In real life the disc will explode if you rotate it very fast, but if it doesn't then it will be expanded due to inertial (centrifugal) forces, and this expansion will be slightly offset by a shrinkage due to length contraction. Lorentz calculated in 1921 the shrinkage of a solid disc due to Lorentz contraction, not accounting for the expansion which depends on material properties. I can look up that equation if you like.

Anyway, you should find that you can place slightly more than 3.14159 contracted meter bands* on the circumference of a very fast rotating disc that has a diameter of 1 m while rotating.

*edited: Einstein spoke of rods, but you should be able to bend them along the circumference

PS: on second thoughts, the equation to use for stress-free metering "rods" is just the Lorentz contraction factor, applied to this specific case. Surely you can derive that yourself. No need to rely on a book!

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MattRob
Well, that can surely be no a negative term because everything is squared and hence is always positive.

A little bit of rearrangement would give: C/d=π + πω^2r^2/2c^2 >π

d=diameter of the disk

this equation says that the measured value of pi is greater than usual.

Well,this conclusion is correct. I would like you to read this chapter :http://www.bartleby.com/173/23.html

This also says a similar conclusion, but no equations are given. And this thus makes sense.

Well, you are right, your measuring rods shrink.But when you try to measure the circumference using the shrunken measuring rod,measured circumference would be larger,since your measuring rod shrunk.

MattRob
First of all, this sounds more like:
than:
http://en.wikipedia.org/wiki/Sagnac_effect

MattRob said:
Awhile ago I read a smaller book on relativity that ended with a neat thought experiment: take a disk, lay out the circumference with many small rods, and start spinning it at relativistic speeds. Because of SR length contraction, each of the rods seems shorter, therefore the circumference of the circle as measured by you, the inertial observer, would seem to be less than 2πr.
No, the inertial frame uses non rotating rulers and measures the Euclidean ratio. The co-rotating frame uses co-rotating rulers and measures more than 2πr for the circumference.

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MattRob
One can derive that equation.

Let me take a rod of meter length as the standard rod of measurement.

In a non-rotating case, the number of times the standard rod would be equal to the circumference of the disc measured. let that be $n_1$

Similarly, let the length of the diameter be $n_2$

then ${\frac{n_1}{n_2}}=π$

Now,let the disc rotate uniformly in an inertial frame of reference.Now,with respect to the disc,disc is at rest but a gravitational field exist.

A rod tangentially placed will suffer a length contraction equal to $(1-{\frac{r^2ω^2}{c^2}})^{1/2}$ as seen from inertial frame of reference.

A person in the disc won't notice this(he thinks he is at rest) and he would measure the circumference of the disc. This time he gets length $n'_1$

But a length perpendicular to the rotation of the disc won't suffer any length contraction as seen from inertial frame of reference and hence length will remain the same as that of non-rotating case. i.e even in rotation, length of diameter measured would be $n_2 =d$

Now $C=πd$ And the no of times the observer in the rotating frame lays to measure the circumference is i.e $n'_1={\frac{πd}{(1-{\frac{r^2ω^2}{c^2}})^{1/2}}}$

Therefore he measures ${\frac{n'_1}{n_2}}={\frac{π}{(1-{\frac{r^2ω^2}{c^2}})^{1/2}}}$

Circumference measured $C={\frac{πd}{(1-{\frac{r^2ω^2}{c^2}})^{1/2}}}$

which is equal to $C=2πr(1-{\frac{r^2ω^2}{c^2}})^{-1/2}$
If $v<<c$ then the above expression equals

$C=2πr(1+{\frac{r^2ω^2}{2c^2}})$

MattRob
ash64449 said:
One can derive that equation.

Let me take a rod of meter length as the standard rod of measurement.

In a non-rotating case, the number of times the standard rod would be equal to the circumference of the disc measured. let that be $n_1$

Similarly, let the length of the diameter be $n_2$

then ${\frac{n_1}{n_2}}=π$

Now,let the disc rotate uniformly in an inertial frame of reference.Now,with respect to the disc,disc is at rest but a gravitational field exist.

If a gravitational field exists then this will affect clocks and rulers which you have not taken into account.

In flat spacetime the 4-velocity of a circular path is ##u= \gamma \partial_t + \omega r \gamma \partial_\phi##, ##\gamma=1/\sqrt{1-\omega^2 r^2}##.

We have

##\frac{dt}{d\tau}=\gamma## and ##\frac{d\phi}{d\tau}=\omega r \gamma##.

So for one revolution, with ##t_p=2\pi/\omega##

##\tau_p=\int_0^{t_p} \gamma^{-1} dt = \int_0^{t_p} \sqrt{1-\omega^2 r^2} dt=\frac{2\,\pi\,\sqrt{1-{r}^{2}\,{w}^{2}}}{w}##

The distance covered is ##\int_0^{\tau_p} \frac{r\,w}{\sqrt{1-{r}^{2}\,{w}^{2}}} d\tau= 2\pi r##

Caution : I have just worked this out so it could be wrong. But it is invariant if right because it is based on proper time which is frame independent.

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ash64449 said:
One can derive that equation.

[..] .Now,with respect to the disc,disc is at rest but a gravitational field exist.[..]
Clarification in addition to Mentz114: Your derivation is pure SR, and at a quick glance it looks correct - except of course for the cited phrase which should simply be omitted. No gravitational field effect should be introduced in such a derivation!

Mentz114 said:
If a gravitational field exists then this will affect clocks and rulers which you have not taken into account.

Sorry, I actually meant what Einstein meant, That in Non-inertial Frame of reference, Events takes place in the same way as that a Gravitational Field exist. I meant that Gravitational Field,didn't mean that already a Gravitational Field existed before.

MattRob
harrylin said:
Clarification in addition to Mentz114: Your derivation is pure SR, and at a quick glance it looks correct - except of course for the cited phrase which should simply be omitted. No gravitational field effect should be introduced in such a derivation!

No Gravitational Field exist before. But if the system is rotating as given in the problem being discussed,events takes place with respect to an observer in that rotating system in such a manner that Gravitational Field exists-This cannot be ignored.

ash64449 said:
No Gravitational Field exist before. But if the system is rotating as given in the problem being discussed,events takes place with respect to an observer in that rotating system in such a manner that Gravitational Field exists-This cannot be ignored.
Not so; and in view of the first post, this merits elaboration.

Once more, you did an SR analysis using inertial frames for the physics. That has nothing to do with gravitational fields, not even fictitious ones. One could unnecessarily complicate matters by using a rotating system as if it is not a rotating but a rest system, and then use the equivalence principle to calculate what according to that principle (and thus GR) would be measured if the felt forces were caused by a gravitational field instead of inertia from rotation. The predictions for such a fictitious gravitational field are based on the SR prediction for rotation without gravitation, which you just calculated.

harrylin said:
The predictions for such a fictitious gravitational field are based on the SR prediction for rotation without gravitation, which you just calculated.

Yes. I agree with this. I made SR prediction without gravitation.

"A person in that rotating system which he considers as being rest in his frame would get the measurement of the circumference as the one i derived.

That conclusion doesn't disturb him.. Because he knows that a Gravitational Field exist and it is similar to the case as that of Non-Euclidean continuum where Euclid's axioms won't work." - this is going to be his conclusion as he knows nothing about his rotation of the disc in an inertial frame.

I agree that every gravitational field in a particular system cannot be seen to vanish when viewed in a different frame as Einstein pointed it out.

But a person in the accelerated frame(w.r.t to an inertial frame) would try to explain events involving the use of the existence of gravitational field.

But the person in the rotating disc won't get the formula since he knows nothing about it. The calculation for the circumference was made by a person in the inertial frame which sees the disc rotating uniformly and he considers that the value he predicted must be the one that a person in the rotating system should get.

Edit: Here in my derivation where i have using SR prediction helped to find that the value of the pi is greater than the usual as measured by a person in the disc. Then after using equivalence principle,one can find that this is the property of a system in which gravitational field is present.

So here we can see that a SR prediction helped describe the property of gravitational field.

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MattRob
ash64449 said:
Then after using equivalence principle,one can find that this is the property of a system in which gravitational field is present.

You're reading the equivalence principle backwards. It says that gravitational forces are members of the class of inertial pseudo-forces (those forces that can be made to disappear by appropriate choice of coordinate system), not that all inertial pseudo-forces are gravitational.

There is no gravity involved in Ehrenfest's rotating disk; no matter what frame of reference you choose the spacetime is flat and the Riemann tensor is zero. In fact, there is no configuration of masses that will produce gravitational effects that are exactly (as opposed to locally and and approximately) the same as the pseudo-force seen inside an accelerating spaceship or along the edge of a rotating disk.

This confusion about the equivalence principle is fairly common, in part because Einstein's own writing on the subject is less than completely clear, in part because a lot of physicists wrote a lot of generally reasonable stuff during the decade or so that it took to come to a complete and clear understanding of the theory, in part because the mathematical methods that are optionally used to solve flat-spacetime SR problems involving acceleration are those that are required to solve curved-spacetime GR problems.

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PeterDonis
Nugatory said:
You're reading the equivalence principle backwards

Do you mean that i was considering that all pseudo-forces were gravitational? Well, you are right. I was considering like that. Thank you for pointing out my mistake.

Nugatory said:
There is no gravity involved in Ehrenfest's rotating disk; no matter what frame of reference you choose the spacetime is flat

If i take the rotating disk as the reference frame,doesn't a pseudo force exist in this frame? Did you consider that this pseudo force cannot be corresponded to any gravitational field?

Nugatory said:
This confusion about the equivalence principle is fairly common, in part because Einstein's own writing on the subject is less than completely clear
Do you consider that i should use a different book to study GR to get complete understanding of it?

ash64449 said:
If i take the rotating disk as the reference frame,doesn't a pseudo force exist in this frame? Did you consider that this pseudo force cannot be corresponded to any gravitational field?
If we choose to use coordinates in which the disk is not rotating, there will be a pseudo-force, which we usually call "centrifugal force". No matter what coordinates we use all points rotating as part of the disk will experience proper acceleration due to a real force, which we usually call "centripetal force". There is no way of arranging masses around the disk to create a gravitational field that will produce the same effects.

Do you consider that i should use a different book to study GR to get complete understanding of it?
I learned my general relativity from MTW. If you want to get to the fun stuff quickly and don't care if some corners are cut on the way there, Hartle is a good start.

Nugatory said:
There is no way of arranging masses around the disk to create a gravitational field that will produce the same effects.

yes. you are right and i agree with this. But still Einstein used this frame of reference to arrive at the result of behavior of clocks and rods in gravitational field even though the gravitational field(the pseudo-force present in rotating system) cannot correspond to any real gravitational field(which obey Newton's law). Still we can see considerable measure of truth since the same effect is produced with ordinary fields.( fields which obey Newton's law)

EDIT: Einstein pointed this out in chapter 23,note 1:http://www.bartleby.com/173/23.html

You know you've asked an interesting question when it spurs this kind of discussion.

Great replies, everyone, and especially thanks for the links to other articles on the subject. The effect seems pretty straightforward in one sense - SR length contraction leading to a rotating frame of reference measuring a larger value of pi than in flat, Euclidean space. And in another sense, leads to some profound implications about gravity; it seems like a fascinating problem that, with the equivalence principle, to some extent bridges the gap in-between SR and GR.

This does make me wonder, though; in that rotating frame of reference, wouldn't space have a negative curvature?

If we use the action of parallel lines as a test of the curvature of a space, then this rotating frame of reference will have the separation of two parallel lines increase with an increase in radial distance.* So wouldn't this be a negative curvature, similar to what would be caused by exotic matter? Could it somehow be used as a substitute for exotic matter in GR solutions that require it? (Wormholes, Alcubierre metric, etc.)

*In the inertial frame of reference, let's say we have two light beams to act as our parallel lines. The two lasers are facing the same direction, but laterally separated by .5 m. In the inertial frame of reference, if you set measuring rods in-between the beams, they will continue to measure a .5 m separation out to infinity. But in a rotating frame of reference with co-moving 1m measuring rods, one placed at a small radius from the axis of rotation will be capable of blocking both beams at once. At an appropriate distance, however, length contraction would shorten the rod so the beams would graze both ends, and at an even greater distance, the rod could be placed in-between the beams, and due to even greater length contraction (greater velocity with greater radial distance from center), it would be smaller than the separation in-between the two. Hence the separation of the two beams in the rotating frame of reference - although they're parallel - will measure less than 1m, 1m, and greater than 1m, respectively, as you measure the separation of the beams at greater distances from the axis of rotation.

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MattRob said:
This does make me wonder, though; in that rotating frame of reference, wouldn't space have a negative curvature?
If you define "spatial geometry" as "what rods at rest measure" then yes. See also:
http://www.projects.science.uu.nl/igg/dieks/rotation.pdf

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MattRob
This thread will remain open to clear up misconceptions.

MattRob said:
wouldn't this be a negative curvature, similar to what would be caused by exotic matter?

No. It's a negative spatial curvature, but not a negative spacetime curvature; spacetime is flat in this scenario. Exotic matter would cause a negative spacetime curvature.

PeterDonis said:
No. It's a negative spatial curvature, but not a negative spacetime curvature; spacetime is flat in this scenario. Exotic matter would cause a negative spacetime curvature.

Exactly. Furthermore, a coordinate change will not turn a flat space-time into a curved one. You need a non-zero stress-energy tensor to do that.

MattRob
A.T. said:
If you define "spatial geometry" as "what rods at rest measure" then yes. See also:
http://www.projects.science.uu.nl/igg/dieks/rotation.pdf
I'm glad you cited Dieks because my own attempt at a covariant solution (my earlier post) was clearly wrong but I could not see why. Dieks' method is arcane but covariant.

The error in my earlier post was using ##\omega## when I should have used ##\omega/\sqrt{1-\omega^2 r^2}##. I also calculated the spatial projection tensor as described here
http://en.wikipedia.org/wiki/Congruence_(general_relativity)[/PLAIN]
and find the same ##d\phi## coeffiecient as Dieks.

Phew. I was worried there.

Thanks, Evo.

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PeterDonis said:
No. It's a negative spatial curvature, but not a negative spacetime curvature; spacetime is flat in this scenario. Exotic matter would cause a negative spacetime curvature.
Exactly. Furthermore, a coordinate change will not turn a flat space-time into a curved one. You need a non-zero stress-energy tensor to do that.

Ah, that clears it up rather nicely, thanks.

rude man said:
The Sagnac effect is based on General Relativity

No, it isn't; it's present in flat spacetime.

rude man said:
the reasoning is entirely fallacious.

Please show your work. The derivation of the Sagnac effect in flat spacetime is straightforward; even the Wikipedia page gets it right (and references a number of textbooks and papers that give more details).

PeterDonis said:
No, it isn't; it's present in flat spacetime.
Please show your work. The derivation of the Sagnac effect in flat spacetime is straightforward; even the Wikipedia page gets it right (and references a number of textbooks and papers that give more details).
The Sagnac effect is a phase shift observed between two beams of light traversing in opposite directions the same closed path around a rotating object. A description of this experiment is obtained within the context of general relativity. In this context the effect provides an operational definition of rotation. An expression for the magnitude of the phase shift is derived under fairly general conditions. The general definition of rotation provided by this experiment is shown to reduce, in certain particular cases, to the usual definitions available. It is observed that the Sagnac effect represents a gravitational analog of the Aharanov−Bohm effect in electrodynamics.

© 1975 American Institute of Physics

rude man said:
A description of this experiment is obtained within the context of general relativity.

You're not claiming that the experiment can be described using general relativity. You're claiming that the experiment can only be described using general relativity, which is claiming that the effect is not present in flat spacetime. That is a much stronger claim which you have not justified.

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rude man said:
The Sagnac effect is a phase shift observed between two beams of light traversing in opposite directions the same closed path around a rotating object.

This description, as it stands, is ambiguous. Take the example of a ring laser gyro. By "closed path around a rotating object", do you mean that light beams travel in opposite directions around the ring, which is itself rotating? (In this case, no other objects of significance would be present.) Or do you mean that light beams travel in opposite directions around the ring, which is not itself rotating, but which has a rotating massive object at its center?

PeterDonis said:
You're not claiming that the experiment can be described using general relativity. You're claiming that the experiment can only be described using general relativity, which is claiming that the effect is not present in flat spacetime. That is a much stronger claim which you have not justified.
PeterDonis said:

Oh, thank you.

From the French Wikipedie: http://www.ecoresults.info/search.p...000000;GFNT:0000FF;GIMP:0000FF;FORID:11&hl=en

"… On the contrary, since this emitter-receptor is in rotation, its reference is no longer inertial and therefore special relativity does not permit the direct determination of the observed shift."

And:

In employing general relativity one finds a shift ….

I worked in Honeywell's world-leading FOG program for over 20 years and can promise you I read that fact in English also, just don't have access to those papers any more.

So it would seem that the French wikipedie is somewhat better informed than the English wikipedia.

PeterDonis said:
This description, as it stands, is ambiguous. Take the example of a ring laser gyro. By "closed path around a rotating object", do you mean that light beams travel in opposite directions around the ring, which is itself rotating? (In this case, no other objects of significance would be present.) Or do you mean that light beams travel in opposite directions around the ring, which is not itself rotating, but which has a rotating massive object at its center?
If you need more info on the FOG there are lots of papers available on the Web. And I assume you know that the ring laser gyro operates on the Sagnac principle also.

rude man said:
I assume you know that the ring laser gyro operates on the Sagnac principle also.

Yes, of course. Nobody is questioning that the Sagnac effect exists or that ring laser gyros use it. What I am questioning is your claim that GR is required to explain it. The sources you give only show that GR can be used to explain it--and even here your claim is not based on the actual content of the sources you have given, but simply on the fact that they use the words "general relativity" in various places. That is not a valid argument.

What you need to do is show, based on the actual content of these papers, that they use curved spacetime in their models, since the use of curved spacetime is the key content that distinguishes GR from SR. You have not done this. In fact, the quotes you do give from your sources indicate that they, like you, are confused on this point. For example:

rude man said:
"… On the contrary, since this emitter-receptor is in rotation, its reference is no longer inertial and therefore special relativity does not permit the direct determination of the observed shift."

This claim is simply false. Using a non-inertial frame in flat spacetime is still SR. A rotating frame in flat spacetime is still SR. And the Sagnac effect can be predicted using exactly these tools (though they are not needed to make the prediction--see below).

What's more, the apparent implication of the above quote that the Sagnac effect cannot be derived using an inertial frame in flat spacetime (which would unproblematically be "special relativity" even by their definition) is also false. Deriving the prediction of a phase shift using an inertial frame is simple; as I said in a previous post, even the Wikipedia page on the Sagnac effect does it. (It's true that the motion of the emitter-receptor in the ring laser is not inertial, but non-inertial motion of objects does not require a non-inertial frame to describe it.) This clearly shows that "special relativity", even by the strictest possible definition of that term, can explain the Sagnac effect, contrary to your claim that GR is required to do so.

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Mentz114
<h2>1. What is the Sagnac Effect?</h2><p>The Sagnac Effect is a phenomenon in which there is a difference in the time it takes for light to travel in opposite directions along a rotating platform. This effect is caused by the rotation of the platform and the resulting change in the length of the path that the light travels.</p><h2>2. How does the Sagnac Effect relate to length contraction?</h2><p>The Sagnac Effect is often used as a thought experiment to explore the concept of length contraction, which is a phenomenon in which an object appears shorter when it is moving at high speeds. In this thought experiment, the rotating platform represents the moving object and the difference in the time it takes for light to travel in opposite directions represents the perceived difference in length.</p><h2>3. What is the significance of the Sagnac Effect?</h2><p>The Sagnac Effect is significant because it provides evidence for the theory of relativity, specifically the concept of length contraction. This effect has been observed in various experiments and has helped to further our understanding of the nature of space and time.</p><h2>4. Can the Sagnac Effect be observed in everyday life?</h2><p>The Sagnac Effect is not typically observable in everyday life as it requires a rotating platform and precise measurements of light travel time. However, it has been observed in various experiments, such as the Sagnac Interferometer, which is used in navigation systems and gyroscopes.</p><h2>5. How does the Sagnac Effect impact our understanding of the universe?</h2><p>The Sagnac Effect, along with other phenomena explored by the theory of relativity, has greatly impacted our understanding of the universe. It has helped us to develop a more accurate model of space and time, and has opened up new possibilities for space travel and other scientific advancements.</p>

## 1. What is the Sagnac Effect?

The Sagnac Effect is a phenomenon in which there is a difference in the time it takes for light to travel in opposite directions along a rotating platform. This effect is caused by the rotation of the platform and the resulting change in the length of the path that the light travels.

## 2. How does the Sagnac Effect relate to length contraction?

The Sagnac Effect is often used as a thought experiment to explore the concept of length contraction, which is a phenomenon in which an object appears shorter when it is moving at high speeds. In this thought experiment, the rotating platform represents the moving object and the difference in the time it takes for light to travel in opposite directions represents the perceived difference in length.

## 3. What is the significance of the Sagnac Effect?

The Sagnac Effect is significant because it provides evidence for the theory of relativity, specifically the concept of length contraction. This effect has been observed in various experiments and has helped to further our understanding of the nature of space and time.

## 4. Can the Sagnac Effect be observed in everyday life?

The Sagnac Effect is not typically observable in everyday life as it requires a rotating platform and precise measurements of light travel time. However, it has been observed in various experiments, such as the Sagnac Interferometer, which is used in navigation systems and gyroscopes.

## 5. How does the Sagnac Effect impact our understanding of the universe?

The Sagnac Effect, along with other phenomena explored by the theory of relativity, has greatly impacted our understanding of the universe. It has helped us to develop a more accurate model of space and time, and has opened up new possibilities for space travel and other scientific advancements.

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