The Emission Theory of Androcles
in [Relativity]
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From: Inertial on 11 Sep 2009 04:41 "Jonah Thomas" <jethomas5(a)gmail.com> wrote in message news:20090911041619.22a7a49d.jethomas5(a)gmail.com... > 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. As SR would agree, also the wavelengths. >> We are using that frame for our analysis. This is very >> basic physics ...but clearly too hard for the relativist mentality. Not at all. SR analysis works very simply and well. No time dilations or length contractinos invovled .. jsut light travelling at c in the inertial frame > 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. True .. messing up isn't hard. >> >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. There's no imaginary time involved. >> In the rotating frame, the emission point of a particular element >> MOVES BACKWARDS. No .. the emission point is fixed in the rotating frame. in the non-rotating frame the emission point (the source) has an instantaneous velocity and moves after the light is emitted. > 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. Yeup .. in the non-rotating frame. > 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. Yeup > But the light in > one direction travels 10% farther, while the light in the other > direction travels only 90% as far. Ok > How many cycles have they gone? The same number, 10 cycles. According to the moving detector, yes. As at the detector (and source) the frequency of both waves is the same > The light is > 10 hertz, so in 1 second they each send out 10 waves. Yeup .. the source has sent out 10 waves in that time. > But each wave on > one side is stretched an extra 10% while each wave on the other side is > just 90% as long. Not if we're talking about waves, no. The wavelength is the same.. count back 10 wavelength from the (in phase) waves that arrive at the detector .. see where you get. Then it will make sense > 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? Yeup .. on phase shift, constructive interference. > I just don't see where the phase shift comes from. I'm missing it. It doesn't .. that's the point > 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. If you manage somehow to change the velocities to be the same as they'd be for light in an aether, you'd be fine >> 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. The detection point when both waves arrive at the same time IS the same. . hence no phase shift >> SR uses this Because in SR the wave fronts do NOT arrive at the detector at the same time >> ...so I can't understand why its followers want to complaiin when I >> do. Because in ballistic theory, the waves arrive at the same time. Hence no phase shift. Its so godamn simple > 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. There's no step missing .. he's bullshitting you. Therer is no speed change ... the apparatus rotates at a constant angular velocity. > 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. Yeup. So no phase shift >> 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. Except that at the detector there is NO frequency difference, the waves arive at the smae speed after the same time with the same frequency. No phase difference > 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? I don't think you're missing all that much. Henry's analysis is wrong.. Everyone (even some of hte other crackpots) recognizes that. Everyone except poor old Henry, who i'm sure MUST know what his mistake is by now, but still claims he's right. That's called lying. >> >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. Its trivially easy to see the phase difference in SR. In the non-rotating frame the light travels at c in opposite directions .. one ray travels a longer distance than the other, so they arrive at different times, so there is a phase shift. > Emission > theory is so much simpler and easier, Its trivially easy to see no phase difference in emission theory in the rotating frame. The light travels at c in opposite directions .. both rays travel the same distance, so they arrive at the same times, so there is no phase shift. Its almost as easy to see no phase difference in emission theory in the non-rotating frame. The light travels a shorter distance, but slower, in one direction and a longer distance, but faster, in the other. Both both rays end up travelling for the same time, so they arrive at the same time, so there is no phase shift. > 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. SR is simpler. Constant speed c. Easy.
From: Inertial on 11 Sep 2009 04:44 "Jonah Thomas" <jethomas5(a)gmail.com> wrote in message news:20090911044249.1fdc25cb.jethomas5(a)gmail.com... > "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. Could be. Henry seems to think that how the waves arrive at the detector doesn't matter for sagnac .. which is complete nonsense, seeing that is where you detect the effect. > 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. No .. they arrive at the same time, with the same frequency, and the same speed .. all relative to the detector. > But no, I got that wrong. Yeup ;):) > The source was moving too, and because of that > the wavelengths are the same. OK > 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. Hope I've helped
From: Jonah Thomas on 11 Sep 2009 04:58 "Inertial" <relatively(a)rest.com> wrote: > "Jonah Thomas" <jethomas5(a)gmail.com> wrote > > "Inertial" <relatively(a)rest.com> wrote: > >> 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. No, the point is that it doesn't forget. Reflect to any angle and it travels at the speed it would have if it was emitted at that angle. > > 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. It doesn't have to average out to c. The forward light travels at c+v and later at c-v. The backward light travels at c-v and later at c+v. So the speeds average out between the two of them, leaving you with a very similar interference pattern compared to the classical case. > > 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 I haven't decided what speed I need at 90 degrees. Since forward and back will both travel an equal distance either way, the differences will disappear regardless of the speed. So I'll save my prediction of that speed until I find some other experiment I have to fudge it to. > > 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 :) > You're a relativist and you're noodging about unrealistic asssumptions? Hey, if Einstein can do it, I can do it. Besides, Ritz did this one first. It's other people who made up a ballistic theory with constant c+v. Ritz did it my way.
From: Jonah Thomas on 11 Sep 2009 05:05 "Inertial" <relatively(a)rest.com> wrote: > "Jonah Thomas" <jethomas5(a)gmail.com> wrote > > 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. > > > > So now my picture is improved. > > Hope I've helped Yes, thank you!
From: Jonah Thomas on 11 Sep 2009 09:03
"Androcles" <Headmaster(a)Hogwarts.physics_n> wrote: > "Jonah Thomas" <jethomas5(a)gmail.com> wrote > I think I was wrong to say anything about SR. SR is a complex concept > that takes a lot of study before students can reliably apply it the > ways their instructors say to. > =========================== > Son, SR cannot be reliably applied to anything. All you can do with it > is write down the answer that some idiot professor expects to see on > an examination paper. The difference between what you said and what I said is that you are more definite and emphatic. > SR is built on two of the madman Einstein's absurd beliefs, > > a) "light is always propagated in empty space with a definite velocity > c" b) the ``time'' required by light to travel from A to B equals the > ``time'' > it requires to travel from B to A. > > Both are ridiculous, but because he wrote it in algebra which YOU > don't understand and the prick calling itself "Inertial" doesn't > understand, the blind lead the blind and the pair of you convince each > other that Einstein must have been a genius because "everyone says > so". I don't understand it well enough to be sure it's bullshit. A whole lot of people have studied it and gone on to use it sometimes. Did a large minority of them figure out it was worthless, but they keep their mouths shut for the sake of their careers? That seems implausible, but it could happen. Did a lot of people see that it's implausible and unintuitive, but they could get adequate results with it so they don't question it? That seems more plausible. If the difference between relativity results versus classical results is seldom important, people could do things whichever way they prefer and then if a reviewer etc notices they didn't use relativity and demands they do so, they can go back and do that and it's unlikely to be any problem beyond the extra effort. > "If EmT works then I may be able to ignore SR entirely. "-- Thomas > If cars have wheels then I may be able to ignore sled skids entirely. But then, sled skids are useful for special purposes. Can't you think of anything better to ignore? Nuclear-powered ice-skates, maybe? |