# Relativity of Simultaneity Gendankin

Discussion in 'Physics & Math' started by MacM, Feb 3, 2006.

1. ### DaleSpamTANSTAAFLRegistered Senior Member

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I am not exactly sure why you think the location of the inertial observer is important. It doesn't matter if the inertial observer is inside (triangle is rotating around observer) or outside (triangle is rotating next to observer). Either way the distances are observed to be different.

Perhaps by "inside" you actually meant "co-rotating". An observer co-rotating with the interferometer (again, inside or outside the triangle doesn't matter) is not in an inertial frame. The Sagnac interferometer's rest frame is non-inertial.

Remember all of our previous discussion on another thread about rotating reference frames and centrifugal and Coriolis forces? The co-rotating observer can do lots of physics experiments to determine that he is in a non-inertial frame. Projectiles, gyroscopes, and Sagnac interferometers all behave differently than they would in an inertial frame.

-Dale

3. ### 2inquisitiveThe Devil is in the detailsRegistered Senior Member

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That's not the focus of my discussion. I know very well that rotating frames are non-inertial. How many times have I challenged Special Theory being used in rotating frames, from GPS to synchrotrons.

DaleSpam, you stated that Special Relativity 'predicted' the Sagnac effect. Since STR is not applicable in non-inertial rotating frames, I asked how it could predict the Sagnac effect.

Physics Monkey stated quote: "Also, the twin paradox and the Sagnac effect are perfectly explainable with the confines of special relativity. In fact, as has been noted, the explanation of the Sagnac effect doesn't even really need the special theory of relativity so long as the speeds concerned are small."

As I have stated, Special Theory DOES NOT EVEN APPLY to the non-inertial rotating frame where the Sagnac effect is measured. The Sagnac effect is measured in the rest frame of the system, the frame that clould be called 'co-rotating' as DaleSpam has stated.

My observation was, and is, that the speed of light is NOT a constant to all observers. To reiterate, I stated that the distance around the 'co-rotating' frame is the same in each direction, but light travelling in opposite directions will take a different amount of time to complete the trip. A different amount of time for a equal distance in each direction, according to an observer at rest within the system (co-rotating with the system).

5. ### PeteIt's not rocket surgeryRegistered Senior Member

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1) You don't have to directly use a rotating frame to predict the Sagnac effect.

2) Any accelerating frame can be analysed in SRT by considering the momentary comoving inertial frames.

3) I'm also led to believe that something called "curvilinear coordinates" can be used to analyse accelerating frames directly with SR, although I personally don't know how. The speed of light is not necessraily constant in curvlinear systems.

Last edited: Feb 27, 2006

7. ### PeteIt's not rocket surgeryRegistered Senior Member

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This is in accord with SRT, which is built on the foundation of light c being invariant to all non-accelerating observers.

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Pete:

9. ### PeteIt's not rocket surgeryRegistered Senior Member

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Comprehension difficulties?

10. ### CANGASRegistered Senior Member

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Pete:

I am very fluent in standard American English. I have never wasted the time to learn Blithering Gibberish. So, yes, as is often the case, after I have tried to read one of your posts, I realize that you have caused me to enjoy comprehension difficulties.

11. ### DaleSpamTANSTAAFLRegistered Senior Member

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In SR you cannot use accelerating reference frames, but you certainly can have objects that accelerate. There is a rather large body of work in SR dealing with four-momenta and four-accelerations, etc. All of it is done in non-accelerating frames, but definitely with accelerating objects.

Remember from our previous discussion about rotating reference frames that things like the equatorial bulge and stresses in turbine blades could be be derived in both the rotating reference frame and in the inertial frame. In other words, the behavior of the rotating object could be analyzed and described completely in the non-rotating coordinate system. Similarly for SR, it disallows the use of the rotating frame, but you can still analyze rotating objects just fine.

-Dale

12. ### PeteIt's not rocket surgeryRegistered Senior Member

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Which words are you having difficulty with, Mr C?

13. ### 2inquisitiveThe Devil is in the detailsRegistered Senior Member

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DaleSpam, I understand what you are saying and agree with you, up to a point. My reference is to the fact that Special Theory does not predict the 'pseudo' centrifugal force, even though it is real for an observer 'co-moving' with the rotating frame. Special Theory does not predict the different travel times of light measured by an observer co-moving in the rotating frame, even though it is real in that frame.

14. ### 2inquisitiveThe Devil is in the detailsRegistered Senior Member

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DaleSpam, as a followup to my above post, let me say where I believe a problem arises. It is in considering moving frames as 'rest' frames by a so-called observer in the moving frame. Do I believe the speed of light is different in the two different directions? No, I do not. But for an observer co-moving in the rotating frame, that is the only choice he has. The path lengths are the same in both directions, according to his 'rest' position within the frame. From the inertial frame of a distant observer, it is obvious the difference in times is due to the movement of the 'rest' observer while the light is in transit around the triangle. I have often spoke of my distaste for the widespead use of 'rest' frames in all of science to describe the physics as seen by that observer. By considering the moving observer 'at rest', I think inacurate physics can result. I also think these mistakes show up in Special Theory's inertial frames, the so-called 'recipriocity' of the frames. If time really does slow for a frame moving at relativistic speeds, I believe an observer in that frame will see time beating relatively faster in the rest of the universe, not slower. In General Relativity, this is true. An observer located deep in a gravity well will see time passing at a faster rate in the rest of the universe. As I have often said, I much prefer General Theory to Special Theory. That is the main reason, even though I have less of a grasp of the more complicated General Theory.

15. ### DaleSpamTANSTAAFLRegistered Senior Member

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You are correct, because SR does not use rotating frames it would never use something like the centrifugal force. However, even without using the centrifugal force it would still predict all of the real centrifugal effects, like the turbine stress and equatorial bulge.

Similarly for EM, the Sagnac effect is a real effect, and it can be derived just fine in the inertial frame. So SR does predict the different travel times, which is a real effect. Since SR never uses a rotating frame what you would never have in SR is the statement that the distances are the same. That is the equivalent to the centrifugal force, a ficticious force introduced to make Newton's 2nd law work. You would have to introduce a similar ficticious distance metric into a rotating frame to make things work. Rather than do that SR just sticks to inertial frames.

I really don't see what the problem is. A rotating reference frame is just a tool for simplification of certain problem geometries and is never necessary to use. In SR the spacetime geometry is complicated enough that rotating frames are no longer a simplification, so you stick with the inertial frames. What's wrong with that?

-Dale

16. ### kevinalmRegistered Senior Member

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Rindler space, iirc.

17. ### quadraphonicsBloodthirsty BarbarianValued Senior Member

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No, that's not true. Most of the famous postulates and predictions of SR are restricted to inertial frames (constancy of the speed of light being the key one), but that doesn't mean SR can't handle non-inertial frames. As long as you can neglect gravitational effects, SR is in force. To be explicit, SR predicts the following things about non-inertial frames: the speed of light will not generally be constant, and frame forces will generally be evident. While there is no preferred *inertial* rest frame in SR, there is certainly a preference for inertial rest frames over non-inertial ones.

18. ### DaleSpamTANSTAAFLRegistered Senior Member

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I disagree with this statement. Even in the famous "windowless room" the rotating observer knows he is rotating. He can determine the amount of rotation and take it into account. Also, he can easily determine that the path lengths are different by the simple fact that the times are different.

I can understand this complaint. Switching between frames is almost never necessary, and it can lead to significant problems and confusion if done carelessly (as is often the case on this forum). That said, switching frames is occasionally useful for simplifying a particular problem. So although it is rarely necessary it is sometimes convenient. If you prefer to work all of your SR problems from a single frame you should still be able to derive any results you need. After all, the proper time is frame-invariant, and that is the usual reason to switch frames anyway. The only thing that might commonly cause you problems in a single-frame approach is figuring out simultaneity for different observers.

To be honest I really wish I had a better grasp of GR than I do. If you can essentially skip the special theory and go right to the general one then you will always be able to go back and derive any SR result you wish without having to lose sight of the overall general context.

-Dale

19. ### 2inquisitiveThe Devil is in the detailsRegistered Senior Member

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by DaleSpam:
"Similarly for EM, the Sagnac effect is a real effect, and it can be derived just fine in the inertial frame. So SR does predict the different travel times, which is a real effect. Since SR never uses a rotating frame what you would never have in SR is the statement that the distances are the same....
I really don't see what the problem is. A rotating reference frame is just a tool for simplification of certain problem geometries and is never necessary to use. In SR the spacetime geometry is complicated enough that rotating frames are no longer a simplification, so you stick with the inertial frames. What's wrong with that?"
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Remember the Special Theory prediction that rotating frames will slow if we approach them at a relativistic velocity? According to that prediction, the DIFFERENCE in the light travel times will diminish toward zero as we approach the rotating frame. The observer co-rotating with the frame will see no change in the anisotropic light travel times. How do the Lorentz Transforms reconcile that discrepancy? Does the Sagnac effect measured by the co-rotating observer depend on an inertial observer's relative velocity?

20. ### PeteIt's not rocket surgeryRegistered Senior Member

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Can you clarify this statement please? I'm not sure exactly what you mean.

21. ### DaleSpamTANSTAAFLRegistered Senior Member

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Yes. According to SR the relativistic inertial observer will see the rate of rotation decrease as well as the beat frequency of the interference. Both will slow down by a factor of gamma.

-Dale

22. ### 2inquisitiveThe Devil is in the detailsRegistered Senior Member

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An observer co-moving with the rotating frame will measure a difference in the clockwise vs. counterclockwise travel time of laser beams projected around the triangle. An observer in an outside inertial frame approaching the rotating triangle at relativistic speed will observer the rotation be be slower and slower as his relative velocity increases wrt the frame. I don't mean the outside observer is accelerating the whole time, just the difference in .866c and .99c for example. At say .9999c relative velocity, STR predicts the triangle will almost stop rotating. However, an observer co-rotating with the frame will still measure the identical Sagnac delay. STR is dependent on no relative velocity to predict the correct Sagnac effect from the distant frame.

An obvious solution to this is what I proposed above. Suppose we add a third frame of reference, one that is not moving relative to the rotating triangle. In this instance, the correct Sagnac effect can be calculated for the rotating triangle and the TIME of the relativistically moving observer can be beating slower. Think of all the rotating frames in the universe. If we use CMB radiation, or the International Celestial Reference Frame, to make our observations from an approximate rest frame, none of the contradictions occur.
Again, the rotating triangle approaching an inertial observer at relativistic speed is not the same as an inertial observer approaching the triangle at a relativistic speed, if time dilation is a fact.

23. ### Physics MonkeySnow Monkey and PhysicistRegistered Senior Member

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I cannot emphasize enough that the special theory is the theory of flat spacetime, and that it remains totally valid locally in the general theory. In order to demonstrate the power and flexibility of the special theory, I will obtain the Sagnac effect using curvilinear spacetime coordinates. Since we're looking at a plane we can use polar coordinates with radius r, angle a, and time t all defined in some inertial frame. Imagine a disk of radius L moving with angular velocity w in this inertial frame. There is an observer on the edge of the disk with various apparatus moving at speed wL with the disk. How can we describe the point of the view of the observer?

Let's introduce the following change of variables r = R, a = A + wT, t = T where uppercase quantities denote variables in the co-rotating frame. This transformation makes simple physical sense: the angle in the inertial frame is just the angle in the rotating frame plus how much the frame has rotated. In the inertial frame, the invariant line element has the form ds^2 = - dt^2 + dr^2 + r^2 da^2. Now the line element is completely invariant and independent of coordinates, and so we can easily obtain its expression in the new rotating coordinates. Its easy to calculate the following: dt = dT, dr = dR, and da = dA + w dT which can now be put together to yield ds^2 = - (1 - R^2 w^2) dt^2 + dR^2 + R^2 dA^2 + 2 R^2 w dA dT .

Now lets examine the physics of this expression. The first part is our old friend the relativistic time dilation. The other interesting part is the term proportional to dA dT which will be responsible for the inhomogenous speed of light. Let me now specialize to the case where w is small which means neglecting higher order terms in w (this situation corresponds to actual experiments for the most part).

The paths of light rays are null lines given by ds^2 = 0, so how do light rays move in our rotating coordinates? Well, if we think about light rays moving around the edge of the disk then dR = 0, and we have something like dT^2 = R^2 dA^2 + 2 R^2 w dA dT. If I divide everything by dT and call the velocity of light the rotating frame C = R dA/dT then 1 = C^2 + 2 w R C. Solving this equation to first order in w gives two solutions given by C = - w R +/- 1, or rearranging things a bit, C = +/- (1 -/+ wR). What is the interpretation of this equation?

The result says that if C > 0 (what the inertial observer would call light moving with the disk rotation) then the speed of light is slightly less than 1 by an amount wR, and if C < 0 then the speed of light is slightly greater than 1 by an amount wR. It's almost magical how easy it is to get these results, and its all special relativity since spacetime is perfectly flat.

Our observer at rest in the rotating frame would conclude that it takes the "fast" light beam only a time T1 = (2&pi;L)/(1+wL) = (2&pi;L)(1-wL) (again expanding to first order in w) to go around while the "slow" light beam takes a time T2 = (2&pi;L)/(1-wL) = (2&pi;L)(1 + wL). The difference in times is T2 - T1 = 4&pi; L^2 w = t2 - t1 (remember our coordinate transformation!) which is precisely the Sagnac effect.

I challenge those of you who may feel uncomfortable with these methods to analyze the whole thing from the point of view of the inertial frame, you'll get all the same answers. It is also easily possible to generalize the result to case of arbitrary angular velocity w in which case the usual time dilation factor also appears. Once more, this is all special relativity since spacetime is completely flat. All that is going on here is the use of curvilinear coordinates in spacetime (not just space!). Depending on your level I can recommend texts where anyone interested can learn how to do calculations of precisely this type. I hope this helps clarify things a bit, and I hope I got all the signs right.