The Emission Theory of Androcles
in [Relativity]
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From: Inertial on 11 Sep 2009 20:47 "Henry Wilson, DSc" <hw@..> wrote in message news:7sjla55q418ve3t49ftvd1rvfk7v8njuau(a)4ax.com... > On Thu, 10 Sep 2009 23:13:26 -0700 (PDT), Jerry > <Cephalobus_alienus(a)comcast.net> wrote: > >>On Sep 11, 12:21 am, "Inertial" <relativ...(a)rest.com> wrote: >>> "Inertial" <relativ...(a)rest.com> wrote in message >>> >>> news:00334a99$0$2976$c3e8da3(a)news.astraweb.com... >>> >>> > "Henry Wilson, DSc" <hw@..> wrote in message >>> >>> Let look at Henry's non-relativistic 'BaTH' argument two possible >>> ways... >>> >>> Assume the light rays are a propagating wave. >>> >>> A propagating wave is always observed to have same wavelength by all >>> observers, but their frequency varies. >>> >>> The leading edge of the wave front is always at the same phase (same >>> place >>> in the wave cycle) >>> >>> So if the leading edges of the two ray wave fronts arrive at the >>> detector at >>> the same time, they will be in phase. >>> >>> If the waves also have the same frequency at a detector, then they will >>> remain in phase. >>> >>> This is the case in Sagnac. >>> >>> "BUT" says Henry, "the path lengths are difference, and so the number of >>> wavelengths is different, so the leading edges of the rays are not in >>> phase" >>> >>> For that to be true, that means the position in the cycle of the leading >>> edge of each ray must change over time. >>> >>> So its not a propagating wave, and the rays are instead moving intrinsic >>> oscillators. >>> >>> A moving intrinsic oscillator is always observed to have same frequency >>> by >>> all observers, but its wavelength varies. >>> >>> The position within the cycle of an intrinsic oscillator is determined >>> by >>> time >>> >>> If two same frequency intrinsic oscillators, that started out in phase, >>> arrive at a detector at the same time, they will still be in phase. >>> >>> This is the case in Sagnac. >>> >>> Either way, you end up with a null Sagnac result. >> >>Notice, however, that Henri fantasizes about intrinsic >>oscillators that are NOT observed to have the same frequency... >>http://mysite.verizon.net/cephalobus_alienus/henri/HWFantasy.htm > > Poor old Crank. ...has never heard of doppler shift. Intrinsic oscillators have wavelength doppler shifted, with frequency constant. Waves have frequency doppler shifted, with wavelength constant. Unless you have SR for things travelling at c and you get both shifted (which is what we observe for light) > Obviously the frequencies > are different in the inertial frame So its not an intrinsic oscillator, its a wave front. And waves fronts that arrive at a given point at the same time are in phase
From: Inertial on 11 Sep 2009 20:52 "Henry Wilson, DSc" <hw@..> wrote in message news:3jlla5d3fa6pegus7q6u6vj99b3iu21ogl(a)4ax.com... > On Fri, 11 Sep 2009 04:16:19 -0400, Jonah Thomas <jethomas5(a)gmail.com> > wrote: > >>hw@..(Henry Wilson, DSc) wrote: >>> Jonah Thomas <jethomas5(a)gmail.com> wrote: >>> >hw@..(Henry Wilson, DSc) wrote: >> > >>> This is where the relativist rabble goes wrong. In the NON-ROTATING >>> FRAME the frequencies of the rays are doppler shifted in opposite >>> directions. We are using that frame for our analysis. This is very >>> basic physics ...but clearly too hard for the relativist mentality. >> >>OK, I'll try to stay with the nonrotating frame. I'm trying to >>understand this, it's just easy for me to mess us. >> >>> >The wavelength is the >>> >same because you don't measure wavelength back toward the source when >>> >it emitted the wave, you measure it in the direction of the >>> >wavefront. So in a time interval t units long, one side emits n >>> >cycles at speed c+v and the other side emits n cycles at speed c-v. >>> >Both arrive at the sensors at the same time. During the time for one >>> >wave to pass from the c+v side, one wave will pass from the c-v side >>> >too, slower. I don't see that this gives us a phase shift or a >>> >frequency difference or anything for an interferometer to pick up. >>> >>> ...because you are jumping from one frame to another. If you try to >>> use the rotating frame, there is an imaginary time factor, that I >>> tried to explain before. >>> In the rotating frame, the emission point of a particular element >>> MOVES BACKWARDS. >> >>OK, let me try this again. I'll put ridiculous numbers on it that I hope >>are easy to work with. >> >>Let's say that our light is 10 Hertz and the path is 1 light-second >>long. >> >>And then we get it rotating at c/10. Now the light in the forward >>direction travels at 1.1c while the light in the back direction travels >>at 0.9c. But the apparatus itself is moving at 0.1c, so in the same time >>that it previously took for the light to go from source to target it >>still goes from source to target in both directions. But the light in >>one direction travels 10% farther, while the light in the other >>direction travels only 90% as far. > > Right so far. The path lengths are different and the travel times are the > same > IN THE INERTIAL FRAME. And as the travel times are the same, the leading edges of the two rays arrive at the detector at the same time, and as the phase of a leading edge of a wave never changes, that means no phase shift. No more analysis necessary
From: Jonah Thomas on 11 Sep 2009 22:17 Once more dear friends into the breach.... hw@..(Henry Wilson, DSc) wrote: > Jonah Thomas <jethomas5(a)gmail.com> wrote: > >hw@..(Henry Wilson, DSc) wrote: > >> Jonah Thomas <jethomas5(a)gmail.com> wrote: > >> >hw@..(Henry Wilson, DSc) wrote: > > >> In the NON-ROTATING > >> FRAME the frequencies of the rays are doppler shifted in opposite > >> directions. We are using that frame for our analysis. This is very > >> basic physics ...but clearly too hard for the relativist mentality. > > > >OK, I'll try to stay with the nonrotating frame. I'm trying to > >understand this, it's just easy for me to mess us. > >> >The wavelength is the > >> >same because you don't measure wavelength back toward the source > >when> >it emitted the wave, you measure it in the direction of the > >> >wavefront. So in a time interval t units long, one side emits n > >> >cycles at speed c+v and the other side emits n cycles at speed > >c-v.> >Both arrive at the sensors at the same time. During the time > >for one> >wave to pass from the c+v side, one wave will pass from the > >c-v side> >too, slower. I don't see that this gives us a phase shift > >or a> >frequency difference or anything for an interferometer to pick > >up.> > >> ...because you are jumping from one frame to another. If you try to > >> use the rotating frame, there is an imaginary time factor, that I > >> tried to explain before. > >> In the rotating frame, the emission point of a particular element > >> MOVES BACKWARDS. > > > >OK, let me try this again. I'll put ridiculous numbers on it that I > >hope are easy to work with. > > > >Let's say that our light is 10 Hertz and the path is 1 light-second > >long. 10 hertz, one light-second. > >And then we get it rotating at c/10. Rotating at c/10. > Now the light in the forward > >direction travels at 1.1c while the light in the back direction > >travels at 0.9c. But the apparatus itself is moving at 0.1c, so in > >the same time that it previously took for the light to go from source > >to target it still goes from source to target in both directions. But > >the light in one direction travels 10% farther, while the light in > >the other direction travels only 90% as far. > Right so far. The path lengths are different and the travel times are > the same IN THE INERTIAL FRAME. OK so far. > >How many cycles have they gone? The same number, 10 cycles. The light > >is 10 hertz, so in 1 second they each send out 10 waves. > Now you are in trouble. You have moved into the rotating frame, which > is full of traps. For instance, if you mark the point where a > particular wave element is emitted, on the NON rotating frame, that > mark moves backwards in the rotating frame. OK, I see that the emitter moves forward in the nonrotating frame. I guess it would be moving backward in the rotating frame. In the nonrotating frame the emitter is moving forward, which is backward for the wave that's heading to the back. This is what lets you keep the wavelength the same even though the frequency is also the same and the speed is different. I'm not quite clear how I moved into the rotating frame, though. Ten times a second a new wave starts. That's frequency, isn't it? That's the frequency at the emitter. What's the frequency at the detector? If it's true that each spot on each wave reaches the detector from both sides in exactly 1 second, and there are 10 of them in each second, that would say the frequency at the detector should be 10 per second too. Whether you count it from the nonrotating frame or the rotating frame either one, that's the frequency at the detector. > The path lengths are again DIFFERENT just as they were in the > nonrotating frame. In the rotating frame, photons have the same > frequency and because both rays move at c in that frame, the elements > that go clockwise doesn't meet their other halves at the detector. The > ones that do meet are out of phase. Um. Pick one that starts at time 0. A wave crest starts in both directions. It travels at 1.1c in one direction but in 0.9c in the other direction. However, the detector moves, so that the one going at 1.1c travels travels 1.1 light seconds while the one going at 0.9c travels 0.9 light seconds. So they both take 1 second to reach the detector. And at that point they are in phase. Pick any other spot on the cycle from wave crest to wave crest and the same thing should happen. Pick a spot where the electric field crosses zero going from positive to negative. in one direction that travels at 1.1c for 1.1 light seconds, in the other direction it travels at 0.9c for 0.9 light seconds, taking 1 second for both. It looks to me like they are in phase when they reach the detector. I want to try to do this entirely in the nonrotating frame first, and never slip into the rotating frame. > That is too difficult for Jerry and Inertial but I expect YOU will be > able to understand it. I'm glad to keep trying but I don't get it yet. > >But each wave on > >one side is stretched an extra 10% while each wave on the other side > >is just 90% as long. Then we let the waves interfere. They start out > >at the same time. One of them is 10% longer than a lightwave from a > >stationary source, but it also travels 10% faster. The other is 90% > >the length and it travels at 90% of the speed. Won't these interfere > >just exactly like they would if they both started at the same time > >and both were the same length and both traveled at c? This was wrong. The paths that one point travels in either direction are stretched or shrunk. But the waves themselves are not stretched or shrunk, because the emitter moves with the detector. > >I just don't see where the phase shift comes from. I'm missing it. > > Yes > >> I and George Dishman looked at the reflection problem very > >intensely> some years ago. It is not the issue. > >> The point missed by most people is that the emission and detection > >> point of a particular wave element are not the same. SR uses this > >> ...so I can't understand why its followers want to complaiin when I > >> do. > > >I think I accounted for that. That's why in one direction you travel > >10% farther to reach the end and in the other direction you travel > >only 90% as far. Because the detection point has moved 10% since the > >wave left the emission point. > > > >> There is only a small difference between the SR and BaTh > >explanation.> > >> SR says the rays both move at c and there is a difference in > >distance> and time traveled. BaTh says the travel times are the same, > >the> distances are different but wavelength is the same in both...and > >> therefore there are more waves in one ray than the other. They flow > >in> or out during a speed change. > > > >That's the step I'm missing. It looks to me like the same number of > >waves. > Wavelength is absolute and frame independent in BaTh. Yes. Agreed. But each wave is created over the expanse of a wavelength, it isn't created all at once. I imagine the emitter creating a wave that moves at 1.1c while the emitter itself moves at 0.1c. There are 10 waves present covering the distance around the circle from the emitter to the detector which is in basicly the same place. A new one is being created while the oldest one travels just enough faster than the detector that it is completely consumed by the time the new wave is completely created. Meanwhile, the emitter creates a second wave that moves at 0.9c while the emitter moves away at 0.1c. There are 10 waves present covering the distance around the circle from the emitter backward to the detector. A new wave is being created while the oldest one travels just fast enough into the incoming detector that it is completely consumed by the time the new wave is completely created. Yes, wavelength is absolute and frame independent. If you can look at them, you can measure them. They're in place at any moment of time, ready to be measured. In the absence of length contraction everybody will measure them the same length. And frequency at the source is also absolute and frame independent. You can watch the source create its periodic motion, and everybody gets the same result apart from things like doppler shift which can be easily corrected. Frequency at an observer can change depending on the observer's velocity. I notice that the wave which has a fixed wavelength and a fixed freqency is different from the path of any particular particle composing the wave. The particles can move at a different speed and in a different direction. I don't know the implications of that yet. > >Say you have one wave in each direction starting at time zero, > >they should both reach the end at time 0.1 second. Exactly nine more > >should reach the end by time 1 second, in both directions. > See above. You are jumping frames again. You are describing a > nonrotating sagnac interferometer. When it rotates they still create ten new waves per second. It's still true that ten new waves arrive at the detector every second. Each new wave crest arrives at the same time from both sides. Doesn't it?
From: Inertial on 11 Sep 2009 22:39 "Jonah Thomas" <jethomas5(a)gmail.com> wrote in message news:20090911221747.7f5d8611.jethomas5(a)gmail.com... > Once more dear friends into the breach.... One thing to note. The light source itself is an intrinsic oscillator. The frequency of that oscillator is the same in all frames of reference (if we are assuming a non-relativistic reality) It must be the case as there are particular things happening at fixed events (ie points in space and time). In particular the location and time at which the source begins its next cycle of oscillation. So your claim that in some time, t, the source oscillator has gong through n cycles, is correct in EVERY frame of reference (inertial or not). There is no problem with frame jumping (if you do it) for something that is absolute and the same in all frames.
From: Jonah Thomas on 11 Sep 2009 23:17
"Inertial" <relatively(a)rest.com> wrote: > > The light source itself is an intrinsic oscillator. > > The frequency of that oscillator is the same in all frames of > reference (if we are assuming a non-relativistic reality) Sure. If you want to fudge how many cycles the source goes through in a second, you have to do it with relativistic time contraction or dilation. To change that one you need to be bold, to give up simultaneity etc. > It must be the case as there are particular things happening at fixed > events (ie points in space and time). In particular the location and > time at which the source begins its next cycle of oscillation. > > So your claim that in some time, t, the source oscillator has gong > through n cycles, is correct in EVERY frame of reference (inertial or > not). There is no problem with frame jumping (if you do it) for > something that is absolute and the same in all frames. Sure. But that does not mean that I understand Wilson's claims or that he is wrong. I do not understand what he is saying, and it's quite possible that there's a way to look at this which I have missed which gives the conclusions he claims. I just don't understand it yet. |