The Sagnac Effect

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Main Question or Discussion Point

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.
bt2lf1711_a.jpg


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

[itex]C=2πr(1+\frac{ω^{2}r^{2}}{2c^{2}}) > 2πr[/itex]

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

[itex]\frac{ω^{2}r^{2}}{2c^{2}}[/itex]

A negative term?
 

Answers and Replies

  • #2
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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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  • #3
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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.
 
  • #4
A.T.
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First of all, this sounds more like:
http://en.wikipedia.org/wiki/Ehrenfest_paradox
than:
http://en.wikipedia.org/wiki/Sagnac_effect

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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  • #5
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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 [itex] n_1 [/itex]

Similarly, let the length of the diameter be [itex] n_2 [/itex]

then [itex]{\frac{n_1}{n_2}}=π[/itex]

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 [itex](1-{\frac{r^2ω^2}{c^2}})^{1/2}[/itex] 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 [itex] n'_1 [/itex]

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 [itex] n_2 =d[/itex]

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

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

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

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

[itex] C=2πr(1+{\frac{r^2ω^2}{2c^2}})[/itex]
 
  • #6
Mentz114
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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 [itex] n_1 [/itex]

Similarly, let the length of the diameter be [itex] n_2 [/itex]

then [itex]{\frac{n_1}{n_2}}=π[/itex]

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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  • #7
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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 :smile: - except of course for the cited phrase which should simply be omitted. No gravitational field effect should be introduced in such a derivation!
 
  • #8
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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.
 
  • #9
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Clarification in addition to Mentz114: Your derivation is pure SR, and at a quick glance it looks correct :smile: - 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.
 
  • #10
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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.
 
  • #11
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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.
 
  • #12
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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.
 
  • #13
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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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  • #14
Nugatory
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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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  • #15
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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.

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?

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?
 
  • #16
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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.
 
  • #17
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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
 
  • #18
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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.
 
  • #19
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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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  • #20
A.T.
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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 [Broken]
 
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  • #22
Evo
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This thread will remain open to clear up misconceptions.
 
  • #23
PeterDonis
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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.
 
  • #24
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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.
 
  • #25
Mentz114
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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 [Broken]
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] [Broken]
and find the same ##d\phi## coeffiecient as Dieks.


Phew. I was worried there.

Thanks, Evo.
 
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