The Emission Theory of Androcles
in [Relativity]
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From: Inertial on 11 Sep 2009 03:12 "Jonah Thomas" <jethomas5(a)gmail.com> wrote in message news:20090911022741.14c7c126.jethomas5(a)gmail.com... > "Inertial" <relatively(a)rest.com> wrote: >> "Jonah Thomas" <jethomas5(a)gmail.com> wrote > >> > But there are lots >> > of other versions to try out, and I doubt any of your results depend >> > on reflections. >> >> What other speeds would you think the reflected light would have? >> Given that the mirrors are moving with the same angular velocity as >> the source. If the speed or the rays *does* change at the mirrors, >> then you would get different arrival times (and different arrival >> positions) and so see an effect .. but would it give you the observed >> AMOUNT of effect. > > Here's the way that's obvious to me now, though it might turn out > something else fits better. > > Imagine that the speed of your light has two components. One is c, the > speed that the light travels on its own hook. > > The other is v, the speed that the light travels because of the source's > speed. OK .. so c+v > Imagine that somehow the source speed is always available, and when the > direction of the light changes it winds up traveling at the speed would > have had if it had been emitted in that direction. > > So if the source is traveling in direction V at speed v, and light is > emitted in direction V, it travels at speed c+v. > > But if it then strikes a mirror and reflects into the opposite > direction, then it travels at speed c-v. The c part is now in the > opposite direction but the v part is unchanged and now opposes the > motion instead of adding to it. So if unreflected, it stays at c+v If reflected 180 deg, it changes to c-v (or vice versa) So what happens when the light gets reflected at 90 deg .. does the light then travel at c? But then, it would have forgotten its 'v' part at the next reflection. > So if the light works that way, then in the Sagnac experiment the light > in both directions goes in a complete circle and so all of the velocity > changes will cancel. If you travel at c+v one way and c-v back for the same distance, that doesn't average out to be c. But for relatively small v, it will be close. > Both of them will spend 1/4 of their distance > traveling at c+v and 1/4 at c-v etc. Not really, as you are reflecting at (something near) 90 deg at each mirror, not 180. And I'm not sure yet what your newe reflection theory for light would say about such a reflection > So the diffraction pattern should > be almost the same as the classical and SR case. I think we're getting a bit unrealistic there with how mirrors work :)
From: Inertial on 11 Sep 2009 03:15 "Jerry" <Cephalobus_alienus(a)comcast.net> wrote in message news:1ff7275f-6462-4247-91e0-c0b3caf1c0f8(a)y36g2000yqh.googlegroups.com... > 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 Indeed .. he has some very odd notions at odds with physics, logic and and reality
From: Jonah Thomas on 11 Sep 2009 04:16 hw@..(Henry Wilson, DSc) wrote: > Jonah Thomas <jethomas5(a)gmail.com> wrote: > >hw@..(Henry Wilson, DSc) wrote: > >> Here is the simple explanation of a four mirror Sagnac. > >> http://www.users.bigpond.com/hewn/sagnac.jpg > > > >I see! And that's what Androcles was hinting at too! Very good! I'm > >laughing, that's delightful. > > > >But now I am having trouble with it. The wavelength should stay the > >same, yes. But the frequency stays the same too. > > 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. 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. 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? I just don't see where the phase shift comes from. I'm missing it. But if the light doesn't just travel around the circle at c+v the whole way in one direction and c-v the whole way in the other direction, then it can work fine. And you haven't said anything so far that indicates your theory needs light to stay the same speed after it reflects off a mirror. > 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. 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. > Alternatively, BaTh says the frequencies are doppler shifted > oppositely in the inertial frame and since then travel times are the > same, there is aohase difference when they reunite. If they were traveling at the same speed when they reunited then I'd see the doppler effect. But they're still traveling at different speeds and so I imagine them unrolling in exact overlap. No, wait. They overlap exactly as they cross the finish line, neck and neck at the start and the shorter one goes through slower. But the interference pattern they make past that point? The short, slow wave matched against the long fast one? Is that what I was missing? > >Just as they misinterpreted yours, I think you're misinterpreting > >them. They can have the rays move at c but one of them has to travel > >farther because of the movement of the mirrors etc. > > No, they wriggle out of the problem by claiming that the separation > soeed of the light from the source is indeed c+v. I don't want to talk about SR. It looks like it's real easy to make mistakes with SR, and my intuition is no good there either. Emission theory is so much simpler and easier, and I'm still having trouble with it. I'll try to figure out the easy version first, and only take up SR if emission theory fails.
From: Jonah Thomas on 11 Sep 2009 04:42 "Inertial" <relatively(a)rest.com> wrote: > "Jonah Thomas" <jethomas5(a)gmail.com> wrote > > hw@..(Henry Wilson, DSc) wrote: > >> Here is the simple explanation of a four mirror Sagnac. > >> http://www.users.bigpond.com/hewn/sagnac.jpg > > > > I see! And that's what Androcles was hinting at too! Very good! I'm > > laughing, that's delightful. > > The diagram is correct for the ballistic analysis, not for SR though > > The path lengths are fine too. Different lengths. > > The times for the two rays to travel is the same > > The two rays arrive at the destination at the same time > > And as far as the moving detector is concerned, they arrive with the > same frequency and the same speed I think that's what I was missing. I figured that they arrive at the detector at the same time but at different speeds and they would exactly overlap. But the detector is moving and that would make them out of phase after all. But no, I got that wrong. The source was moving too, and because of that the wavelengths are the same. When v = c/10 the slow side isn't squeezed into 90% of the length, because the source keeps moving away. How strange. The frequency is the same, 10 hertz. The wavelength is the same, 0.1 light-second. But the speeds are different because the source and target are both moving. I thought frequency * wavelength had to equal velocity? No, for emission theory it doesn't have to. Source and target travel at the same speed so there's no doppler effect and no redshift. Wavelength stays the same. Frequency stays the same. Velocity varies with the speed of target and source. > > But now I am having trouble with it. The wavelength should stay the > > same, yes. But the frequency stays the same too. > > At the detector, yes, because it is moving. > > If you look at point S' as a fixed point in the inertial frame, then > no, due to doppler shift, the speeds and frequencies are different. Yes, my picture of it was wrong. Wavelength is the same. Frequency is different. It's the fixed point that does it. So now my picture is improved. > But the detector is NOT fixed.. it is moving and has an instantaneous > tangent velocity when it reaches point S', and relative to it, the > rays have the same speed and same frequency. > > > 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. > > That's correct > > > 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. > > It doesn't .. this has been explained countless times to Henry. > > > It looks to me like the version that has the light emitted at some > > speed, call it c+v, and it stays that speed despite any reflections > > or refractions, is probably not compatible with Sagnac. > > It isn't > > > But there are lots > > of other versions to try out, and I doubt any of your results depend > > on reflections. > > What other speeds would you think the reflected light would have? > Given that the mirrors are moving with the same angular velocity as > the source. If the speed or the rays *does* change at the mirrors, > then you would get different arrival times (and different arrival > positions) and so see an effect .. but would it give you the observed > AMOUNT of effect. > > >> >> > and at the same time shown why Sagnac refutes SR. > >> >> > >> >> Which, of course, is also incorrect, as Sagnac does not in any > >way> >refute SR> and never has. It shows that the light speed, in > >the> >inertial frame of> reference, is not affected by the speed of > >the> >source, and so is consistent> with both SR and aether theories, > >but> >not with ballistic / emission> theories. > >> > > >> >Agreed. Relativity was carefully designed to provide classical > >> >results under classical conditions, and there is nothing here that > >> >would result in a relativistic difference. > >> > >> Sagnac refutes SR because it requires that the rays move at c+v and > >> c-v wrt the source. > > > > Just as they misinterpreted yours, > > We don't :) > > > I think you're misinterpreting them. > > He is misinterpreting both :) > > > They can have the rays move at c but one of them has to travel > > farther because of the movement of the mirrors etc. > > That's correct. > >
From: Jonah Thomas on 11 Sep 2009 04:46
"Androcles" <Headmaster(a)Hogwarts.physics_n> wrote: > "Jonah Thomas" <jethomas5(a)gmail.com> wrote > > "Androcles" <Headmaster(a)Hogwarts.physics_n> wrote: > > > >> You can depend on an old 33 RPM record playing turntable not > >> to depend on SR/GR, but the strobe light on the side used to verify > >> its speed by illuminating regularly spaced marks on the rim meshes > >> perfectly with Sagnac. > > > > It turned out I had seen that, I didn't realise that was the one you > > were presenting again until I saw it again. > > > > So what is the take-home message here? Something about strobes.... > > Something about the sum of two traveling wave making a standing wave > > or a slow traveling wave.... > > Simple, isn't it? > If you DEFINE wavelength = speed/frequency then increasing speed > has to increase the wavelength. You can't change the 50Hz (60 Hz USA) > frequency of the strobe light, so by your definition the wavelength > changes. The way I see it, when you increase speed you increase the frequency of the dots going past. You don't change the frequency of the strobe that provides the standing wave, but you increase the frequency that matters. So the wavelength can stay the same. > Yet that is ridiculous, nobody is repainting the marks on the side of > the turntable, so your definition must be wrong or the distance > between marks isn't the wavelength. > Yet the teeth around a gear look awfully like a travelling wave to me. > http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/MechSagnac.gif > > So let's see if you can think. You tell me what the take-home message > is. You're trying to trick me, see if I fall for it? |