Wang / Sagnac Devices
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From: Androcles on 24 Oct 2009 19:00 "Henry Wilson DSc ." <HW@..> wrote in message news:0ov6e55o8bk6ppnqt55v76k8iluip5t774(a)4ax.com... > On Sat, 24 Oct 2009 10:58:53 +0100, "Androcles" > <Headmaster(a)Hogwarts.physics_p> > wrote: > >> >>"Henry Wilson DSc ." <HW@..> wrote in message >>news:0ug5e5t436u4mi3hu02abv161firlrvsht(a)4ax.com... >>> On Sat, 24 Oct 2009 03:01:44 +0100, "Androcles" > >>>>> >>>>> Einstein...World's greatest SciFi writer.. >>>>and Wilson wants to be better than Einstein, but headless crocodiles >>>>flying kites in thunderstorms doesn't cut the mustard. >>> >>> Why don't you try it yourself...then tell us all about what happens. >> >>You brought it up, it's all yours. I'm content connecting batteries >>across transformers and telling you what happens. > > I'd rather disconnect a battery from a spark coil. Yeah, I'd expect a used VW camper van dealer to graduate to junkyard dog breaking up old cars for their parts, that is just the right occupation for you. Glad it makes you happy.
From: Jonah Thomas on 24 Oct 2009 19:06 HW@..(Henry Wilson DSc). wrote: > Jonah Thomas <jethomas5(a)gmail.com> wrote: > >HW@..(Henry Wilson DSc). wrote: > > > >> Actually, this experiment would not be very difficult. > >> You simply wind two long glass fibres around a former in opposite > >> directions with their ends poking out together. Then you spin the > >> thing at very high speed. Set up a light so that every time the > >ends> pass it, a pulse goes down each one in opposite directions. > > > >No! attach a laser cavity that produces light in two opposited > >directions to the fibers, and have it spin with them. Otherwise, > >which emission theories would predict a difference in lightspeed? > > My version would be more like Fizeau's experiment that Sagnac but it > would be very revealing anyway. > > Your version would also be simple enough, although the rotation speed > would be somewhat limited. > Observing the output might also present a few problems because there > wouldn't be much light involved the time difference between the two > pulses would be very small. > > Let's do the sums. > Say the fibres are 300m. The travel time is around 1us. It isn't impossible to make them 3k for a travel time of 10 us. > Let the ring circumference = 1m. Put the loop that has the laser on a projection that has a 1.5m radius. That gives you about another order of magnitude, if the speed of the laser makes any difference. > If rotation = 30 hz. v = 30 m/s = 10^-7c Do it at 300 hz. That's harder and more dangerous, but -- for science! > So the light speeds relative to the fibres MIGHT be c(1.0000001) and > c(.9999999) in the two directions (according to Einstein) or c (for > BaTh). > > DIFFERENCE in travel times = 2 x 10^-13 seconds. 2 10^-10 > In that time the ring edge moves very little. > > So the pulse output times have to be resolved to less than 10^-13 > seconds. > > This is pretty nearly impossible. Improve 3 orders of magnitude and it's still difficult. Some things can be measured rather more quickly, but I don't know whether this could. Say you had sensors that flipflog depending on which signal comes first. You might get a clear difference between about 50:50 versus 95:5. > It also highlights the basic difficulties associated with measuring > OWLS from a moving source. > This is why Einstein's second postulate has taken so long to refute. > > This also why Andro and I have investigated variable star data and > simulated their brightness curves with BaTh. It is the only conclusive > way to check if OWLS is source dependent.... which it is.
From: Androcles on 24 Oct 2009 19:19 "Henry Wilson DSc ." <HW@..> wrote in message news:o307e5hce7ic1g0mopcv46so92a7ipmjnr(a)4ax.com... > On Sat, 24 Oct 2009 12:43:01 +0100, tominlaguna(a)yahoo.com wrote: >>And, surprise surprise the 2v value appears. It is the same value >>that appears in the Doppler radar formula. So could it be that every >>traffic cop with a radar gun is validating the Ballistic theory with >>each speeding ticket issued? >> >>B. Equation 74 looks vaguely familiar... Oh yes, it's the equation >>for Snell's Law! Tom Roberts claimed that the Ballistic theory >>violated Snell's Law and therefore was not consistent with the Sagnac >>Effect. That appears not to be the case, does it? >> >>I think the bigger issue for Tom Roberts is to find a SRT-consistent >>formulation for the angle of reflection from a moving mirror. I cited >>two different equations for the angle of reflection by mainstream >>physics authors. I know if I do some digging, I will come up with >>other variations. This issue could be interesting to follow. > > This analysis is OK No it isn't. The moving transponder can only emit at c and the photon can only hit the cop at c+v, not c+2v as La Goona Tommy is claiming. You are not paying attention, you've got one eye on the Shiraz and gout on the brain. if light is treated like an elastically bouncing ball. It > is not. > A photon takes time to be emitted and consequently possesses > length....maybe > billions of 'wavelengths'. Oh my gawd... he's back to his headless crocodiles again... That gout must be really troubling you, all the wine barrels are empty.
From: Sam on 24 Oct 2009 22:55 On Oct 23, 7:46 pm, Jonah Thomas <jethom...(a)gmail.com> wrote: > It concerns me that for some 70 years before we > had fiber optic cables to do the experiment, many people > claimed that Sagnac refuted any ballistic theory, when > they did not. That's a complicated assertion to unpack, because (for example) one might regard quantum electrodynamics in Minkowski spacetime (special relativity) as a "ballistic" or "emission" theory. So, in that sense, the Sagnac effect obviously doesn't refute every possible theory that someone might choose to label as an "emission theory". When people say that the Sagnac effect rules out every ballistic theory it is usually understood that they are referring to a theory in Galilean space and time in which light behaves "more or less" like small material particles in Newtonian mechanics, emitted at the speed c relative to the emitting object, and reflecting off other objects like perfectly elastic particles. The Sagnac effect obviously refutes this class of theories, because they predict zero effect. Sometimes people (like Ritz) postulate "emission theories" with ad hoc special behavior for reflections, stating (e.g.) that the particles continue to move at c relative to the "original source" at the time they were emitted. However, this hypothesis leads again to a prediction of zero Sagnac effect (regardless of which way the original source of light was pointing), so again these theories are refuted by the Sagnac effect. (Ironically, this ad hoc hypothesis is mutually exclusive with the "extinction" argument that has sometimes been invoked to evade the astronomical evidence against ballistic theories, but of course that argument has also been refuted by observations in the x-ray range.) In general, any serious theory that has been called a "ballistic" or "emission" theory by its proponents is refuted by the Sagnac effect.
From: Darwin123 on 25 Oct 2009 00:32
On Oct 22, 2:53 am, Jonah Thomas <jethom...(a)gmail.com> wrote: > Darwin123 <drosen0...(a)yahoo.com> wrote: > > Jonah Thomas <jethom...(a)gmail.com> wrote: > > > Darwin123 <drosen0...(a)yahoo.com> wrote: > So without arguing about it, I don't understand why you can't consider > the detector from an inertial point of view because it's rotating. You can't because of Doppler effect. From the standpoint of the inertial detector, the light that reflects off a mirror undergoes Doppler effect. Consider the emitter is traveling at a constant speed, emitting at a certain frequency as seen by the detector. Note: I am not talking about the rest frame frequency of the emitter. The light reflecting off of the mirror will be at a different frequency than the original emitter. You may not be convinced that light reflecting off a moving mirror has to undergo Doppler shift. If you don't believe that mirrors produce Doppler shift, the Sagnac effect could appear like magic. To convince you, I may need to refer to microscopic details of the mirror surface. When > the question is whether two parts of a split light beam arrive at the > detector at the same time or not, if it's the same time in one frame it > ought to be the same time in all other frames, shouldn't it? However, the two beams may not have the same frequency as seen in a true inertial frame. Even if the beams arrive at the same "time" at the same "place", they can be a different frequencies. The word "beam" convinced you that there is no frequency in the wave. The nodes in one beam are closer together than the nodes in the other beam. So the difference in frequencies is a beat frequency. >If it was a > question when they arrived at two different places then observers might > disagree which was first. We are not measuring difference in arrival time of "photons." We are measuring the beats between two beams of different frequency. At least they are different frequencies as seen in an inertial frame. > But how can they disagree about whether or not > they arrive at the same place and the same time? What is arriving at the same place and time? I think you mean two wave fronts arriving at the same place and time. It is quite possible for a wave front to arrive at a given place without the other wave front arriving. From the POV of the inertial observer the two beams can't have the same frequency. One beam is having its frequency raised by Doppler shift, and the other is having its frequency lowered by Doppler shift. >Don't answer that. I'm > sure there are explanations how it can happen with SR. I really >don't want to know. Sorry I had to tell you. > > My argument was that the cavity itself is accelerating in the > > sense of special relativity. The atoms and electrons of the reflective > > cavity are accelerating. So even if the source and detector are not > > accelerating (the usual case), the Sagnac effect will still occur (the > > always case). > > I don't think so. (Though I could be wrong, and if the experiments have > been done I'll be very interested.) I think the Sagnac effect is due to > the source and detector moving and not the medium. If it's the medium > that moves then you might get some sort of Fizeau effect. Are you calling the reflective material a medium? If so, I agree. If the light wave is traveling as an effenescent wave on a "reflective" surface, then you could technically call it a Fizeau effect. > > Ives's thought experiment could be done. You make a cylindrical mirror > with carefully-spaced holes. You arrange that light shining into the > cylinder through one hole will bounce repeatedly across the inner > surface of the mirror until it leaves by another hole. Position your > stationary source and detector just so, and then spin the mirror. The > spinning mirror will alternately block the emitter (and detector) and > let the light through. Suppose you do that. 1) Take into account Doppler effect. 2) Take into account the diffraction caused by the curvature of the metal. >You should see no Sagnac effect when the light > gets through. Basically, the same experiment that you described has been done with fiber optics. The Sagnac effect was observed. > If you do get an effect I think it will deserve some other > name, The soggy effect? > and there will be a reason that it is not observed during the > usual Sagnac experiment. You could call it anything you want. That won't make it go away. Herbert E. Ives was wrong. > > The reason I predict that will have no effect is that the traditional > Sagnac explanation (the one that says the detector moves so that the > light paths are in proportion c+v:c-v) gives the right answer. >It does > not depend in any way on the movement of the mirrors along the path, but > only on the movement of the detector. If you get the same effect by > moving the mirrors and leaving the detector alone, then this explanation > gives the correct result entirely by accident. I don't follow. > It could be true that > this explanation is not only wrong but has no relation whatsoever to the > correct explanation, Okay, I am lost. I have no idea what you are saying. Maybe it is my lack of comprehension. However, I can't envision what you are talking about. Thanks for talking, though |:-) |