swansont Posted March 29, 2010 Posted March 29, 2010 So, you confessed it is about light path differentials and not the circle. The light path differences are because of the circular path and motion. Merged post follows: Consecutive posts mergedNow the Earth's rotation around the sun and simultaneously send a clockwise and a counterclockwise light. If the light along the Earth's orbit circle, they returned to the Earth definitely not at the same time. You aren't sending light in a path around the sun. You are sending it around the earth. To send it around the sun, you would have to do an experiment that takes a whole year, and in that time it will circle the earth about 2x10^8 times, and accumulate ~44 seconds of phase. Sagnac from the orbit about the sun is less than a second. In the time it takes for a single trip around the earth, the orbital path is almost a straight line. Very little phase accumulates from the Sagnac effect from this. If the planet’s surface is divided into innumerable equal parts, it also takes them unequal time to go through each part; i.e. clockwise and counterclockwise speeds of light are unequal. If it were true, then the ether wind with velocity of 30 km/s could be detected by Michelson-Morley experiment. But has it ever been detected? If MMX is not sensitive to Sagnac,so you think that MMX can not prove that the principle of constant Light Speed. If the light speed were not isotropic, there would be a phase shift depending on the orientation of the interferometer. The MMX configuration is not sensitive to Sagnac because the encompassed area is essentially zero — the beams retrace their path. Merged post follows: Consecutive posts mergedambros, your discussion is going away from the topic at hand, so I've moved your post to a new thread
vuquta Posted March 29, 2010 Posted March 29, 2010 The light path differences are because of the circular path and motion. 1) Light does not travel in a circle for the ring interferometer. It is a collection of straight lines. http://en.wikipedia.org/wiki/File:Sagnac_interferometer.png 2) The earth is in motion around the sun. If you insist, it is moving relative to the sun although it is virutally absolute at 30km/s. The ring interferometer is in motion relative to the earth at say 30 km/s. You have not explained this. Finally, are you claiming light is isotropic in the vacuum of space which has been verified. I ran into a good modern paper last week showing this astronomically but forgot to save the dm link for later reading. And you are saying light is also isotropic to the earth which we know is moving. How is that?
swansont Posted March 29, 2010 Posted March 29, 2010 1) Light does not travel in a circle for the ring interferometer. It is a collection of straight lines.http://en.wikipedia.org/wiki/File:Sagnac_interferometer.png You can also make one with a coil of optical fiber, which actually is a circle; the actual shape is unimportant. What is important is that it encompass an area. Light going around one way will travel a different distance than light going the other way if there is a rotation of the system, and give you a phase shift. It is the rate of rotation, and the area of the interferometer, that give you the amount of the shift. 2) The earth is in motion around the sun. If you insist, it is moving relative to the sun although it is virutally absolute at 30km/s. The ring interferometer is in motion relative to the earth at say 30 km/s. You have not explained this. The interferometer is NOT in motion relative to the earth. The interferometer in the example is a signal sent around the earth's equator. The motion around the sun is ignored because it closely approximates linear motion in the time it takes to do the experiment, and linear motion does not give you a Sagnac shift (because c is a constant in an inertial frame). I was assuming that the details of the Sagnac effect were understood. Do I need to explain this? Finally, are you claiming light is isotropic in the vacuum of space which has been verified. I ran into a good modern paper last week showing this astronomically but forgot to save the dm link for later reading. And you are saying light is also isotropic to the earth which we know is moving. How is that? It's called the principle of relativity. Motion of the source or receiver does not affect the propagation speed in inertial frames. I know you've heard of it, so I don't understand why you need to ask the question.
Xinwei Huang Posted March 30, 2010 Author Posted March 30, 2010 The light path differences are because of the circular path and motion. Merged post follows: Consecutive posts merged You aren't sending light in a path around the sun. You are sending it around the earth. To send it around the sun, you would have to do an experiment that takes a whole year, and in that time it will circle the earth about 2x10^8 times, and accumulate ~44 seconds of phase. Sagnac from the orbit about the sun is less than a second. In the time it takes for a single trip around the earth, the orbital path is almost a straight line. Very little phase accumulates from the Sagnac effect from this. If the light speed were not isotropic, there would be a phase shift depending on the orientation of the interferometer. The MMX configuration is not sensitive to Sagnac because the encompassed area is essentially zero — the beams retrace their path. Merged post follows: Consecutive posts mergedambros, your discussion is going away from the topic at hand, so I've moved your post to a new thread I have said that the light along the Earth's orbit circle ,it means the Earth around the sun. We do not need to do experiments, do not need to take a whole year. It is easy to calculate that Sagnac effect from the orbit about the sun is 0.628 second. Circle of the earth around the sun is about 942 million kilometers and the distance from China to Japan is about 3,000 km. It is easy to calculate that Sagnac effect from China to Japan is 2000ns. Please note that it is not 0. If it is 0, then the Sagnac effect from the orbit about the sun is 0. Please note that 10000 ÷ 10000 = 1 ≠ 0. If you think 1 is too small so it can be equal to 0, then it multiplying by 10000, the result was 0 instead of 10000.
swansont Posted March 30, 2010 Posted March 30, 2010 (edited) I have said that the light along the Earth's orbit circle ,it means the Earth around the sun.We do not need to do experiments, do not need to take a whole year. It is easy to calculate that Sagnac effect from the orbit about the sun is 0.628 second. Circle of the earth around the sun is about 942 million kilometers and the distance from China to Japan is about 3,000 km. It is easy to calculate that Sagnac effect from China to Japan is 2000ns. Please note that it is not 0. If it is 0, then the Sagnac effect from the orbit about the sun is 0. Please note that 10000 ÷ 10000 = 1 ≠ 0. If you think 1 is too small so it can be equal to 0, then it multiplying by 10000, the result was 0 instead of 10000. I did not say it was identically zero, I said it was small and can be ignored. In the context of physics experiments, when a << b, we treat a as zero, since it doesn't affect the result. I'd be interested to see your calculation for 0.628 sec; I get about a tenth of a second. edit: got it. factor of 2 pi If the earth did not rotate, what would be the Sagnac signal? The path difference would give you 2*pi of phase over the course of the whole year. But if the experiment takes less time, you get a smaller amount of phase. i.e. if the experiment takes 1 second, you get 1/3.15 x 10^7 of the phase. Merged post follows: Consecutive posts mergedAlso, if the area of the enclosed loop of the light travel were smaller, would you expect the same amount time accumulation? Edited March 30, 2010 by swansont
Xinwei Huang Posted March 31, 2010 Author Posted March 31, 2010 Dear swansont, I would like to correct a misconception that Sagnac effect has nothing to do with the area of the enclosed loop. Have you read Professor Ruyong Wang's paper? http://web.stcloudstate.edu/ruwang/PLA312.pdf http://web.stcloudstate.edu/ruwang/PRL93.pdf His experiments proved that Sagnac effect also exists for linear motion. He proved that Sagnac effect depends on v and L, rather than ω and A. If v and L unchanged, even A = 0, Sagnac effect remain unchanged. I hope that you carefully read his paper. Best Regards Xinwei Huang
swansont Posted March 31, 2010 Posted March 31, 2010 It looks to me like the author showed that if you move an interferometer in a loop, you get a phase shift. Not particularly surprising, IMO. It is true that the path need not be a circle; the existence of Mach-Zehnder interferometers shows this. But this is not strictly the Sagnac effect, which is for a static interferometer in a rotating system, and this is not equivalent to that. So it does not "disprove" the validity of the Sagnac effect in any way, and is not representative of the experiment we are discussing. Now, please answer the questions I asked of you.
Xinwei Huang Posted April 1, 2010 Author Posted April 1, 2010 I'm sorry! Which is your question of the above? Can you repeat your question?
swansont Posted April 1, 2010 Posted April 1, 2010 If the earth did not rotate, what would be the Sagnac signal? Also, if the area of the enclosed loop of the light travel were smaller, would you expect the same amount time accumulation?
vuquta Posted April 2, 2010 Posted April 2, 2010 If the earth did not rotate, what would be the Sagnac signal? Also, if the area of the enclosed loop of the light travel were smaller, would you expect the same amount time accumulation? What does the earth's rotation have to do with anything? The Sagnac effect is seen in GPS with GPS satellites moving relative to the rotation of the earth. So, apparently, it is a relative thing, since the earth is rotating. But, how then do you establish zero rotation?
Xinwei Huang Posted April 2, 2010 Author Posted April 2, 2010 If the earth did not rotate, what would be the Sagnac signal? Also, if the area of the enclosed loop of the light travel were smaller, would you expect the same amount time accumulation? If the earth did not rotate, of course the Sagnac signal did not exist. Also, if the area of the enclosed loop of the light travel were smaller, would you expect the same amount time accumulation? Yes. Professor Wang's experiment tells us that even if the straight line Sagnac effect also exists. Please see pic1.xilu.com/1/3118/6237719/90c6e3416e98735215c8a094f01f5917.jpg
swansont Posted April 2, 2010 Posted April 2, 2010 (edited) What does the earth's rotation have to do with anything? Please review the experiment under discussion. (Sagnac effect from earth rotation vs revolution about the sun) Merged post follows: Consecutive posts mergedIf the earth did not rotate, of course the Sagnac signal did not exist. Also, if the area of the enclosed loop of the light travel were smaller, would you expect the same amount time accumulation? Yes. What about the effect from the sun? Professor Wang's experiment tells us that even if the straight line Sagnac effect also exists. Please see pic1.xilu.com/1/3118/6237719/90c6e3416e98735215c8a094f01f5917.jpg Professor Wang is not measuring the Sagnac effect, he is measuring the Doppler effect. Edited April 2, 2010 by swansont Consecutive posts merged.
vuquta Posted April 2, 2010 Posted April 2, 2010 Please review the experiment under discussion. (Sagnac effect from earth rotation vs revolution about the sun) I am reviewing the discussion. It is also a discussion on the interpretation of MMX and Sagnac. These concepts are in the paper of the OP. So, I asked you why the results of MMX are null but not sagnac. You said it was a result of the circular path. I also showed a sagnac experiment that was a rotating square and hence straight lines paths. Then, you agreed that was not true. If these concepts are not in the domain of this discussion, please tell me and I will find something else to do.
swansont Posted April 2, 2010 Posted April 2, 2010 I am reviewing the discussion. It is also a discussion on the interpretation of MMX and Sagnac. These concepts are in the paper of the OP. So, I asked you why the results of MMX are null but not sagnac. You said it was a result of the circular path. I also showed a sagnac experiment that was a rotating square and hence straight lines paths. Then, you agreed that was not true. If these concepts are not in the domain of this discussion, please tell me and I will find something else to do. For the path around the earth, which is the experiment being discussed, the path IS a circle. In general it does not have to be, it just has to enclose an area, i.e. the light does not return along the same same path as it goes out, at some point in the interferometer.
vuquta Posted April 3, 2010 Posted April 3, 2010 For the path around the earth, which is the experiment being discussed, the path IS a circle. In general it does not have to be, it just has to enclose an area, i.e. the light does not return along the same same path as it goes out, at some point in the interferometer. So, why the negative MMX and positive Sagnac?
swansont Posted April 3, 2010 Posted April 3, 2010 So, why the negative MMX and positive Sagnac? Because the MM interferometer encloses no area, it will not give a Sagnac result. The earth rotates, so a Sagnac interferometer will.
vuquta Posted April 4, 2010 Posted April 4, 2010 Because the MM interferometer encloses no area, it will not give a Sagnac result. The earth rotates, so a Sagnac interferometer will. This does not explain anything. Have you considered something more obvious? Sagnac is measuring two one way light travels. MMX measures round trip. If it is the case that Sagnac measures basically, c+v and c-v because of light path differentials, then MMX would measure (c+v) + (c-v) /2 = c, a constant.
swansont Posted April 4, 2010 Posted April 4, 2010 This does not explain anything. Have you considered something more obvious? Sagnac is measuring two one way light travels. MMX measures round trip. If it is the case that Sagnac measures basically, c+v and c-v because of light path differentials, then MMX would measure (c+v) + (c-v) /2 = c, a constant. One-way speed of light is not part of the OP, and you are not permitted to discuss it here. That is thread hijacking. We discuss two things: the topic of the OP, and the accepted physics that either confirms or refutes it
Xinwei Huang Posted April 6, 2010 Author Posted April 6, 2010 Are we far from the topic? I think we should return to this question. Please see pic1.xilu.com/1/3118/6237719/eb38a9f9bbe7be0a8e3358dcba1e4da0.jpg Can they return to point O at the same time after gyrating a circuit ? Please note that the principle of constant light speed is just a hypothesis. This hypothesis is based on MMX above. According to this hypothesis, the counterclockwise light will return point O earlier than the clockwise light with time discrepancy about 0.02%. But according to the MMX, they can return to point O at the same time after gyrating a circuit. Which is the credible result derived from the hypothesis or the experimental ? Of course, the credible result is derived from the experimental rather than the hypothesis. Many people think that the counterclockwise light will return point O earlier than the clockwise light with time discrepancy about 0.02%. Please note that if the planet's surface is divided into innumerable equal parts, it also takes them unequal time to go through each part; ie clockwise and counterclockwise speeds of light are unequal. If it were true, then the ether wind with velocity of 30 km/s could be detected by Michelson-Morley experiment. But has it ever been detected? Some people try to explain why MMX can not detect the ether wind with velocity of 30 km/s. They think it is reasonable and understandable. However, they did not realize that this is against the theory of relativity. Why? They think that the ether wind with velocity of 30 km/s should exist, but the MMX can not detect it. In other words, MMX is invalid experiment. This is equivalent to say that the MMX can not prove the principle of constant light speed!
swansont Posted April 6, 2010 Posted April 6, 2010 Are we far from the topic? I think we should return to this question. Please seepic1.xilu.com/1/3118/6237719/eb38a9f9bbe7be0a8e3358dcba1e4da0.jpg If you refuse to address questions asked of you we won't clear up your misconception about the Sagnac effect. It's the etiquette that when one points to an external link, one actually provides valid web address; the server software will turn it into a hyperlink. In any event, I can't connect to xilu.com Can they return to point O at the same time after gyrating a circuit ? Please note that the principle of constant light speed is just a hypothesis. This hypothesis is based on MMX above. According to this hypothesis, the counterclockwise light will return point O earlier than the clockwise light with time discrepancy about 0.02%. But according to the MMX, they can return to point O at the same time after gyrating a circuit. A Michelson interferometer does not enclose an area, and is insensitive to rotations. Which is the credible result derived from the hypothesis or the experimental ? Of course, the credible result is derived from the experimental rather than the hypothesis. Many people think that the counterclockwise light will return point O earlier than the clockwise light with time discrepancy about 0.02%. Please note that if the planet's surface is divided into innumerable equal parts, it also takes them unequal time to go through each part; ie clockwise and counterclockwise speeds of light are unequal. If it were true, then the ether wind with velocity of 30 km/s could be detected by Michelson-Morley experiment. But has it ever been detected? Some people try to explain why MMX can not detect the ether wind with velocity of 30 km/s. They think it is reasonable and understandable. However, they did not realize that this is against the theory of relativity. Why? They think that the ether wind with velocity of 30 km/s should exist, but the MMX can not detect it. In other words, MMX is invalid experiment. This is equivalent to say that the MMX can not prove the principle of constant light speed! Relativity concludes that there is no ether wind, not that one exists but cannot be detected.
vuquta Posted April 7, 2010 Posted April 7, 2010 Are we far from the topic? I think we should return to this question. Please seepic1.xilu.com/1/3118/6237719/eb38a9f9bbe7be0a8e3358dcba1e4da0.jpg Can they return to point O at the same time after gyrating a circuit ? Please note that the principle of constant light speed is just a hypothesis. This hypothesis is based on MMX above. According to this hypothesis, the counterclockwise light will return point O earlier than the clockwise light with time discrepancy about 0.02%. But according to the MMX, they can return to point O at the same time after gyrating a circuit. Which is the credible result derived from the hypothesis or the experimental ? Of course, the credible result is derived from the experimental rather than the hypothesis. Many people think that the counterclockwise light will return point O earlier than the clockwise light with time discrepancy about 0.02%. Please note that if the planet's surface is divided into innumerable equal parts, it also takes them unequal time to go through each part; ie clockwise and counterclockwise speeds of light are unequal. If it were true, then the ether wind with velocity of 30 km/s could be detected by Michelson-Morley experiment. But has it ever been detected? Some people try to explain why MMX can not detect the ether wind with velocity of 30 km/s. They think it is reasonable and understandable. However, they did not realize that this is against the theory of relativity. Why? They think that the ether wind with velocity of 30 km/s should exist, but the MMX can not detect it. In other words, MMX is invalid experiment. This is equivalent to say that the MMX can not prove the principle of constant light speed! Why did you go to all this trouble? There are many in agreement. One such attempt is known as the Emission Hypothesis (or the ballistic theory of light), and was developed partly by Walther Ritz (C&N p.353). According to this theory, light behaves like bullets shot from a gun, its speed with respect to the source being a universal constant and independent of any ether. This idea is consistent with the null results of the Michelson-Morley experiment and many others. http://laser.phys.ualberta.ca/~egerton/specrel3.htm Modern Physics/Michelson-Morley Experiment Walter Ritz's emitter theory (or ballistic theory), was also consistent with the results of the experiment http://en.wikibooks.org/wiki/Modern_Physics:Michelson-Morley_Experiment This rules out any conceptually coherent ballistic theory of light propagation, according to which the speed of light is the vector sum of the velocity of the source plus a vector of magnitude c. Ironically, the original Michelson-Morley experiment was consistent with the ballistic theory, but inconsistent with the naïve ether theory, whereas the Sagnac effect is consistent with the naïve ether theory but inconsistent with the ballistic theory. Of course, both results are consistent with fully relativistic theories of Lorentz and Einstein, since according to both theories light is propagated at a speed independent of the state of motion of the source. http://www.mathpages.com/rr/s2-07/2-07.htm Now, this might seem to suggest that light is a disturbance in a material medium in which the objects A,B,C just happen to be at rest, but this is ruled out by the fact that it applies regardless of the state of (uniform) motion of those objects. Naturally this implies that the flash of light propagates isotropically with respect to the inertial rest coordinates of object D as well. http://www.mathpages.com/rr/s1-04/1-04.htm
Xinwei Huang Posted April 7, 2010 Author Posted April 7, 2010 If you refuse to address questions asked of you we won't clear up your misconception about the Sagnac effect.It's the etiquette that when one points to an external link, one actually provides valid web address; the server software will turn it into a hyperlink. In any event, I can't connect to xilu.com A Michelson interferometer does not enclose an area, and is insensitive to rotations. Relativity concludes that there is no ether wind, not that one exists but cannot be detected. Dear swansont and vuquta, The link pic1.xilu.com/1/3118/6237719/eb38a9f9bbe7be0a8e3358dcba1e4da0.jpg is failed. Please see pic1.xilu.com/1/3118/6237719/d6c698c33153555f2d41488b9769ec82.jpg Please seriously consider, if you think that the counterclockwise light will return point O earlier than the clockwise light with time discrepancy about 0.02%,so the planet's surface is divided into innumerable equal parts, it also takes them unequal time to go through each part; ie clockwise and counterclockwise speeds of light are unequal. If it were true, then the ether wind with velocity of 30 km / s could be detected by Michelson-Morley experiment. Do you think so?
swansont Posted April 7, 2010 Posted April 7, 2010 Please seriously consider, if you think that the counterclockwise light will return point O earlier than the clockwise light with time discrepancy about 0.02%,so the planet's surface is divided into innumerable equal parts, it also takes them unequal time to go through each part; ie clockwise and counterclockwise speeds of light are unequal. If it were true, then the ether wind with velocity of 30 km / s could be detected by Michelson-Morley experiment. Do you think so? Where do you get 0.02%? Sagnac for the earth is just over 200 ns, for a travel time of about 0.13 sec, which is 1.5 parts in 10^6. If you divide the path into 200 sections, each one would contribute 1 ns to the delay. As the sections get smaller, the contribution from each gets smaller. This, incidentally, is why the contribution from the sun is so small — it's a tiny fraction of an orbit during that 0.13 second. But that's all irrelevant, because the phase accumulation from the Sagnac effect depends on the area of the interferometer, too, and a Michelson interferometer encloses no area. It will not measure rotation effects. It is sensitive to linear motion, if light has an unequal speed of propagation due to the ether. But we get a null result. Ergo, no ether.
vuquta Posted April 7, 2010 Posted April 7, 2010 Dear swansont and vuquta, The link pic1.xilu.com/1/3118/6237719/eb38a9f9bbe7be0a8e3358dcba1e4da0.jpg is failed. Please see pic1.xilu.com/1/3118/6237719/d6c698c33153555f2d41488b9769ec82.jpg Please seriously consider, if you think that the counterclockwise light will return point O earlier than the clockwise light with time discrepancy about 0.02%,so the planet's surface is divided into innumerable equal parts, it also takes them unequal time to go through each part; ie clockwise and counterclockwise speeds of light are unequal. If it were true, then the ether wind with velocity of 30 km / s could be detected by Michelson-Morley experiment. Do you think so? What is aether wind? How does it have a speed? What is it? Since MMX has null results, how do you compare your logic to MMX? Are you claiming MMX is wrong or are you claiming the two directions are wrong. I am just not seeing a logical argument.
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