The Emission Theory of Androcles
in [Relativity]
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From: Jerry on 19 Sep 2009 06:11 On Sep 19, 3:18 am, Jonah Thomas <jethom...(a)gmail.com> wrote: > Jerry <Cephalobus_alie...(a)comcast.net> wrote: > > Jonah Thomas <jethom...(a)gmail.com> wrote: > > > Jerry <Cephalobus_alie...(a)comcast.net> wrote: > > > > It appears that the assumption of Ritzian reflections (which I > > > > find utterly bizarre) does result in a fringe displacement in a > > > > mirrored Sagnac apparatus. > > > > I agree that ritzian reflections look bizarre. But are they more > > > bizarre than special relativity? > > > I find special relativity to be utterly natural. > > > One way of looking at SR is that it is a direct expression of the > > geometry of Minkowski space-time. If you understand the geometry, > > all concerns about supposed "paradoxes" vanish. > > Well, if you spend as much time learning about Ritzian theory it might > start to seem utterly natural too. I assure you that SR does not seem > very natural to most people on first examination. > > > > > But here is where the "general argument" falls flat on its face. > > > > Before the invention of fibre optics, it was considered > > > > impossible to achieve a closed light path without the use of > > > > mirrors. Fibre optic gyroscopes represent a mirrorless > > > > implementation of the Sagnac principle, and all versions of > > > > emission theories predict zero fringe displacement given a > > > > mirrorless closed loop. > > > > Do fiber optics work without reflection> > > > Correct. > > Interesting! I had not heard that interpretation before, tell me more? > Post links? > > I did a quick google search on ["fiber optics" reflection] and this is > the first link I found: > > http://electronics.howstuffworks.com/fiber-optic6.htm > > They explained it just the way I usually hear it, about the cladding > around the fibers providing lots and lots of reflections because the > angle of incidence is low enough. Multimode fibres are worthless for long distance transmission of information, and are not used in ring gyros. Look up single mode fibre. No reflections. In any event, Pauli also mentioned use of the Sun as an external light source breaks the closed-loop condition in a Michelson interferometer, and the Ritzian version of emission theory predicts a fringe shift in this circumstance. Miller tried this in 1924 with negative results (at best only about 1 percent of the predicted value, his measurements being completely dominated by noise and systematic error). So Ritzian theory failed here, also. Jerry
From: Inertial on 19 Sep 2009 11:15 "Jonah Thomas" <jethomas5(a)gmail.com> wrote in message news:20090919041330.05e7b3f4.jethomas5(a)gmail.com... > "Inertial" <relatively(a)rest.com> wrote: >> "Jonah Thomas" <jethomas5(a)gmail.com> wrote > >> > http://i847.photobucket.com/albums/ab31/jehomas/speedwave9.gif >> > >> > Here's a picture, with the Sagnac ring unrolled. >> > >> > I set the distance at 10.5 wavelengths. I noticed that when one side >> > goes 9 wavelengths and the other goes 11, at the end they are >> > temporarily in phase. :| >> > >> > Two particles start in opposite directions with different speeds. >> > The wavelength is observably the same. >> >> As I showed you previously, wavelength is always measured from where >> the source is NOW, not where it was. That is how wavelength works .. >> the distance between corresponding point in successive cycles. > > You are thinking about something completely different from what Wilson > is thinking about. Thank god for that .. I wouldn't want a single thought in common with that liar. >> > To have the wavelength be the same and the speed different, the >> > frequency has to shift. >> > And so of course when they meet they are out of phase >> >> No .. they aren't. > > The picture shows two oscillating particles that are at different phases > in their oscillation. > >> > As near as I can tell, this is what Wilson says is happening. >> >> Yes it is, and it shows a fundamental lack of understanding of what >> wavelength is > > But could this be what is happening with light? No > These two particles that > arrive at the same time are going to be the ones that do interference at > that time. And they are clearly not in phase. There is no way for them to be out of phase. > Never mind how well he understands your theories. He has described a > scenario that gives sagnac interference, Nope .. he's described nonsense > and if it deserves to be > refuted then it deserves a demonstration that light cannot behave this > way. Not just that it is incompatible with the way you do things It is incompatible with reality > -- we > already knew it couldn't fit your thinking because if we do things your > way emission theories can't work. They can't. When something doesn't work, it doesn't work. Again.. refer to the linear case: D S D .............s............... D ./'\S/'\. D .............s............... D ./'\./'\S/'\./'\. D .............s............... D/'\./'\./'\S/'\./'\./'\D .............s............... Even if wavelength is dependant on distance, you DO NOT take the distance from s when working out numbers of wavelengths.
From: Jonah Thomas on 19 Sep 2009 12:02 Jerry <Cephalobus_alienus(a)comcast.net> wrote: > Jonah Thomas <jethom...(a)gmail.com> wrote: > > Jerry <Cephalobus_alie...(a)comcast.net> wrote: > > > Jonah Thomas <jethom...(a)gmail.com> wrote: > > > > Jerry <Cephalobus_alie...(a)comcast.net> wrote: > > > > > But here is where the "general argument" falls flat on its > > > > > face. Before the invention of fibre optics, it was considered > > > > > impossible to achieve a closed light path without the use of > > > > > mirrors. Fibre optic gyroscopes represent a mirrorless > > > > > implementation of the Sagnac principle, and all versions of > > > > > emission theories predict zero fringe displacement given a > > > > > mirrorless closed loop. > > > > > > Do fiber optics work without reflection> > > > > > Correct. > > > > Interesting! I had not heard that interpretation before, tell me > > more? Post links? > Multimode fibres are worthless for long distance transmission of > information, and are not used in ring gyros. Look up single mode > fibre. No reflections. I've looked at multiple sources and found one that mentioned single fiber not having multiple reflections. I want to follow up on that but it's currently a side issue for me. Why do you think that Ritzian light would keep its speed constant when traveling in single-mode fiber? > In any event, Pauli also mentioned use of the Sun as an external > light source breaks the closed-loop condition in a Michelson > interferometer, and the Ritzian version of emission theory > predicts a fringe shift in this circumstance. Miller tried this > in 1924 with negative results (at best only about 1 percent of > the predicted value, his measurements being completely dominated > by noise and systematic error). How would that work? The light from the sun should come at different speeds at dawn and at dusk because of the earth's rotation. So ... that light should then have different speeds in different directions, and those differences would mostly cancel out. How much does the diffraction pattern depend on the speed of the light instead of its wavelength? If different speeds of light that each show no interference give a different interference pattern, then you could see the difference. Is that the argument?
From: Jonah Thomas on 19 Sep 2009 12:14 "Inertial" <relatively(a)rest.com> wrote: > "Jonah Thomas" <jethomas5(a)gmail.com> wrote > > "Inertial" <relatively(a)rest.com> wrote: > >> "Jonah Thomas" <jethomas5(a)gmail.com> wrote > > > >> > http://i847.photobucket.com/albums/ab31/jehomas/speedwave9.gif > >> > > >> > Here's a picture, with the Sagnac ring unrolled. > >> > > >> > I set the distance at 10.5 wavelengths. I noticed that when one > >side> > goes 9 wavelengths and the other goes 11, at the end they are > >> > temporarily in phase. :| > >> > > >> > Two particles start in opposite directions with different speeds. > >> > The wavelength is observably the same. > >> > >> As I showed you previously, wavelength is always measured from > >where> the source is NOW, not where it was. That is how wavelength > >works ..> the distance between corresponding point in successive > >cycles. > > > > You are thinking about something completely different from what > > Wilson is thinking about. > > Thank god for that .. I wouldn't want a single thought in common with > that liar. [sigh] Whatever you think about him personally, he has an idea here that I think deserves some thought. Independent of the source. > >> > To have the wavelength be the same and the speed different, the > >> > frequency has to shift. > >> > And so of course when they meet they are out of phase > >> > >> No .. they aren't. > > > > The picture shows two oscillating particles that are at different > > phases in their oscillation. > > > >> > As near as I can tell, this is what Wilson says is happening. > >> > >> Yes it is, and it shows a fundamental lack of understanding of what > >> wavelength is > > > > But could this be what is happening with light? > > No OK, why not? > > These two particles that > > arrive at the same time are going to be the ones that do > > interference at that time. And they are clearly not in phase. > > There is no way for them to be out of phase. In my picture they are out of phase. You are saying they could not be doing what they do in my picture. What they do in my picture is to oscillate the same amount per unit distance, but with different velocity. Since they travel for different distances when the meet they have rotated different amounts. You are saying this cannot happen with light? You could easily be right. But why not? What is the experiment that shows it does not happen? > Again.. refer to the linear case: > > D S D > ............s............... > > D ./'\S/'\. D > ............s............... > > D ./'\./'\S/'\./'\. D > ............s............... > > D/'\./'\./'\S/'\./'\./'\D > ............s............... > > Even if wavelength is dependant on distance, you DO NOT take the > distance from s when working out numbers of wavelengths. You are talking about how to measure wavelength. Would it help if we use different language here? He is talking about light that "turns". The way circularly polarised light does one complete rotation every now and then. He imagines that the speed of light can be different sometimes. That could of course be completely wrong but I want to consider the possibility until I am convinced it's ruled out. So, imagine that light from the same source but emitted in different directions can have a different speed. Can it still do its rotation in the same distance? I don't see why not. Once we accept that the light can be different speeds, why can't it keep the same distance per turn rather than the same time per term or some combination? Is there some reason this can't happen? You can argue that light has to be constant speed no matter what and therefore it can't happen. But if you accept the possibility that an emission theory could be correct, is there something about emission theories that says that the light emitted from a single source in different directions won't have the same number of turns per unit distance?
From: Jerry on 19 Sep 2009 13:48
On Sep 19, 11:02 am, Jonah Thomas <jethom...(a)gmail.com> wrote: > Jerry <Cephalobus_alie...(a)comcast.net> wrote: > > Jonah Thomas <jethom...(a)gmail.com> wrote: > > > Jerry <Cephalobus_alie...(a)comcast.net> wrote: > > > > Jonah Thomas <jethom...(a)gmail.com> wrote: > > > > > Jerry <Cephalobus_alie...(a)comcast.net> wrote: > > > > > > But here is where the "general argument" falls flat on its > > > > > > face. Before the invention of fibre optics, it was considered > > > > > > impossible to achieve a closed light path without the use of > > > > > > mirrors. Fibre optic gyroscopes represent a mirrorless > > > > > > implementation of the Sagnac principle, and all versions of > > > > > > emission theories predict zero fringe displacement given a > > > > > > mirrorless closed loop. > > > > > > Do fiber optics work without reflection> > > > > > Correct. > > > > Interesting! I had not heard that interpretation before, tell me > > > more? Post links? > > Multimode fibres are worthless for long distance transmission of > > information, and are not used in ring gyros. Look up single mode > > fibre. No reflections. > > I've looked at multiple sources and found one that mentioned single > fiber not having multiple reflections. I want to follow up on that but > it's currently a side issue for me. Why do you think that Ritzian light > would keep its speed constant when traveling in single-mode fiber? > > > In any event, Pauli also mentioned use of the Sun as an external > > light source breaks the closed-loop condition in a Michelson > > interferometer, and the Ritzian version of emission theory > > predicts a fringe shift in this circumstance. Miller tried this > > in 1924 with negative results (at best only about 1 percent of > > the predicted value, his measurements being completely dominated > > by noise and systematic error). > > How would that work? The light from the sun should come at different > speeds at dawn and at dusk because of the earth's rotation. So ... that > light should then have different speeds in different directions, and > those differences would mostly cancel out. How much does the diffraction > pattern depend on the speed of the light instead of its wavelength? If > different speeds of light that each show no interference give a > different interference pattern, then you could see the difference. > > Is that the argument? The Ritzian rules of reflection amount to the statement that, regardless of the number of reflections, light always travels at c with respect to its original emitter. If we consider the Sun to be a "stationary" source, the Ritzian version of emission theory predicts results essentially idential to naive aether theory, with the Sun defining the fixed frame. Jerry |