The Emission Theory of Androcles
in [Relativity]
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From: Inertial on 12 Sep 2009 02:44 "Henry Wilson, DSc" <hw@..> wrote in message news:9ebma5db0utnnuperq27h6v7gphgej2953(a)4ax.com... > On Sat, 12 Sep 2009 13:25:19 +1000, "Inertial" <relatively(a)rest.com> > wrote: > >>"Jonah Thomas" <jethomas5(a)gmail.com> wrote in message >>news:20090911231725.48263503.jethomas5(a)gmail.com... >>> "Inertial" <relatively(a)rest.com> wrote: >>>> >>>> The light source itself is an intrinsic oscillator. > >>> >>> Sure. But that does not mean that I understand Wilson's claims >> >>I don't think anyone does .. not even him. I understand enough of them to >>know that they are mostly inconsistent and unphysical. Which is why he >>seems to change his position all the time, due to his self-conrtadictory >>ideas. > > Of course I understand it. It might have taken a few years but it is quite > straightforward. Shame you can't explain it without changing you mind all the time [snip same old Henry misunderstanding]
From: Jonah Thomas on 12 Sep 2009 12:58 "Androcles" <Headmaster(a)Hogwarts.physics_n> wrote: > "Jonah Thomas" <jethomas5(a)gmail.com> wrote > > hw@..(Henry Wilson, DSc) wrote: > >> 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. > > Don't agree to his senile nonsense mantra. All velocities are frame > dependent > so all wavelengths are frame dependent. > http://www.androcles01.pwp.blueyonder.co.uk/Wave/Relative.gif I'm not sure that you and I ever reached a common understanding about what a wavelength is. I'm pretty sure that part of the problem is that often when physicists mention "wavelength" in actual problems they do not actually measure the wavelength. Often they measure frequency and then assume that wavelength equals velocity/frequency, and sometimes the velocity is assumed. What I mean by wavelength is the physical distance between wavefronts. Ideally you would measure this with a ruler. It is time independent unless it is changing, and if it is changing then at least one of frequency or speed must be changing too. If you can make an instantaneous measurement with a ruler then your relative velocity does not matter. If you suppose that lightspeed can vary, then what does a diffraction grating do? You didn't have much of an answer for that because you acted like you didn't want it to be a wave phenomenon at all which leaves it pretty vague. When I try to think carefully about the traditional wave explanations for a diffraction grating I find them kind of vague too. When c is constant you can measure diffraction and you have one variable. When c is not constant you have two variables -- and a vector if the velocity of the source makes a difference. When c is constant it doesn't matter whether you're measuring a wavelength change or a frequency change -- they have to go together. It doesn't matter which change is happening -- they have to happen together. When c can be different then it makes a difference whether it's frequency that changes or wavelength or both in some ratio. > > 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. > Fuckin' rubbish. You two should be locked in two cells > out of earshot of each other. Well, if wavelength is the distance between wave fronts, then in emission theories it is absolute and frame independent. It doesn't matter what frame you look at it from, you get concentric circles with even spacing. Your frame decides how fast the center of that circle is moving. What does wavelength mean to you?
From: Jonah Thomas on 12 Sep 2009 15:01 "Androcles" <Headmaster(a)Hogwarts.physics_o> wrote: > "Jonah Thomas" <jethomas5(a)gmail.com> wrote > > It is time independent > > unless it is changing, and if it is changing then at least one of > > frequency or speed must be changing too. If you can make an > > instantaneous measurement with a ruler then your relative velocity > > does not matter. > > > > If you suppose that lightspeed can vary, then what does a > > diffraction grating do? > > Causes a change in rotation of a photon. I have found your particle explanation of diffraction to be quite vague. Unfortunately when I looked in more detail at the classical wave explanation of diffraction it turned out to be pretty vague too. > > When c is constant you can measure diffraction and you have one > > variable. When c is not constant you have two variables -- and a > > vector if the velocity of the source makes a difference. When c is > > constant it doesn't matter whether you're measuring a wavelength > > change or a frequency change -- they have to go together. It doesn't > > matter which change is happening -- they have to happen together. > > When c can be different then it makes a difference whether it's > > frequency that changes or wavelength or both in some ratio. > > > >> > I imagine the emitter creating a wave > > The emitter is a molecule. It can only send a pulse. Imagine all you > want to, > there are NO light waves. Radio waves, yes, but no light waves. It's > time you thought carefully instead of repeating the same old dogma. Fair enough. I'll look at ways for light particles to do interference. If I find something I'll let you know. > The emitter is a different molecule. The first molecule has to be > recharged with energy before it can fire off a second pulse (in a > different direction). > It can only send a pulse. Imagine all you want to, there are NO light > waves. Radio waves, yes, but no light waves. It's time you thought > carefully instead of repeating the same old dogma. I'm not disputing you about this. I don't know what the truth is and I have been looking at approaches that seem comfortable and that fit the available facts I know about. I want to try imagining you are right and try looking for ways to make the details work. Does a continuous laser send waves? It's based on the "principle" that a "charged" molecule can be triggered to "discharge" by light of the frequency it will send. So a single pulse can somehow set off at least two others (being absorbed or partly absorbed by those two?) and you wind up with a cascade that is all in phase but that starts at different times. Does that count as a wave for you? Again I'm not arguing that you're wrong about anything, if you say this is a wave I'm not going to jump up and down and crow that you admitted you were wrong about something. If this special case is a wave then I've learned a little more about your system, and if it isn't a wave then I have more questions.
From: Henry Wilson, DSc on 12 Sep 2009 16:56 On Sat, 12 Sep 2009 01:08:39 -0400, Jonah Thomas <jethomas5(a)gmail.com> wrote: >hw@..(Henry Wilson, DSc) wrote: > >> Each photon takes the same time to travel the ring but the distances >> traveled are different. Wavelength is invariant...so there are more >> waves in one path than the other. > >This is one of the places I didn't follow you. Could you maybe show a >picture of the paths with more waves in one than the other? I have. Here is the reason why one path is longer than the other. http://www.mathpages.com/rr/s2-07/2-07.htm (Don't take any notice of the relativistic explanation.) >Did Androcles give a picture of that you could point to? > > >> What does this tell us about the nature of light? > >I'm real unclear about that too. Every other theory gives us a picture >where you can imagine a bunch of particles moving in all directions, or >you can imagine a wave moving in the same directions. Emission theory >gives us particles moving from the source, or waves moving from a center >that moves. For the first time we get a theory where in many directions >the hypothetical particles are going in a different direction from the >hypothetical waves. I haven't worked out the implications of that but >they look profound. There isn't any 'wave' at all. Photons have a spatial periodicity. I suspect that this is the result of either a rotation of a +/- pair or the presence of a 'standing wave' along its length. Here is my vision of a photon www.users.bigpond.com/hewn/e-field.exe ...only much fatser an maybe with millions (billions) of cycles along its length. The intrinsic wavelength is set by the distance between peaks. It is an absolute and invariant length. >I imagine particles that spin or something, and when they do >cancellation they sort of disappear. What happens to the energy they >were carrying? I imagine particles streaming out in all directions, the >large majority of them destined never to interact with anything much, >they head off to the ends of the universe. Billions of years of stars >putting out light particles and the majority of them never interact with >anything. How much of the mass of the universe would be tied up in that >by now? Not that there's anything wrong with that, not like it can't >happen, it just seems strange to think of it. I agree, The majority of EM in the universe is in transit...going somewhere....but physics knows absolutely nothing about the structure of traveling light. Everything we know about it has been gained at the instant of emission or absorption. Henry Wilson...www.users.bigpond.com/hewn/index.htm Einstein...World's greatest SciFi writer..
From: Henry Wilson, DSc on 12 Sep 2009 17:09
On Sat, 12 Sep 2009 02:24:47 -0400, Jonah Thomas <jethomas5(a)gmail.com> wrote: >hw@..(Henry Wilson, DSc) wrote: >> Jonah Thomas <jethomas5(a)gmail.com> wrote: > >> > >> >OK, I see that the emitter moves forward in the nonrotating frame. >> >> >> Nonononno. The emission POINT ...MARKED IN THE NONROTATING FRAME...is >> at rest in the nonrotating frame. It moves backwards in the rotating >> frame. > >The emission point for you is not where the emitter is at the moment, >it's where the first wave we're interested in started out. It stays in >one place in the nonrotating frame, and it moves backward in the >rotating frame. Do I understand that? There are two rotations, the ring is rotating and the photons are rotating around the ring. Here it is in the nonrotating frame. the distance between the emission and detection points is vt....where t is the travel time around the ring. http://www.mathpages.com/rr/s2-07/2-07.htm If you can't understand that you shouldn't be here. >> Here's a very simple way to look at it. >> >> Imagine a photon as being like a long tightly wound coil spring. (The >> coils always remain touching). >> Now we (and SR) have established that the emission and the detection >> points are separated by the distance vt. Wrap the spring loosely >> around a cylinder so that it can be rotated around it. Mark two points >> on the cylinder to represent the above two points for a particular >> turn. Spin the spring clockwise around the cylinder. No matter how >> fast you do that, the number of turns between the two fixed points >> remains the same. In the anticlockwise direction, the number of turns >> between the two points is different from that of the first because the >> distance from the emission point and the detection point is >> different.....but again independent of spin rate. Changing the >> distance between points is equivalent to changing a ring gyro's >> rotation speed. > >I think I see that picture now. > >My thought was that if the path is 10 wavelengths long, then you're >going to have 10 waves on it at a time. I still think that. One path is longer than the other. Have another look at it. http://www.mathpages.com/rr/s2-07/2-07.htm Even SR gets that right. It's simple stuff. >So try it out. You're wrapping the spring around a torus, and the >emitter -- the point you're extruding the spring from -- is traveling at >0.1 c. So in 0.1 seconds you have one whole wave wrapped around the >torus and if we mark the starting spot now the leading point on the >spring is at 1.1 and the trailing point is at 0.1 where the emitter is. >At the 0.2 second mark there is a second wave wrapped around the torus. >the leading edge of the first wave is at 2.2 and the trailing edge of >the second wave is at 0.2. At time 0.3 we have a third wave around the >torus, the leading edge of the first wave is at 3.3 and the trailing >edge of the third wave is at 0.3. > >At time 0.9 the nineth wave has its end at 0.9 and the leading edge of >the first wave is at 9.9. >At time 1 second the tenth wave has its end at 1.0 and the first wave is >also at 1.0. The first wave has entirely passed its emission point. >There is not an eleventh wave, there are ten of them. One path is longer than the other. The number of turns in each 'path' is different no matter how fast you spin the spring. >Similarly in the other direction. The first backward wave is extruded >slowly, so that at time = 0.1 its forward edge is at 9.1 and its leading >edge is at the emitter at 0.1. Then the second one goes slowly, and at >time 0.2 the forward edge of the first wave is at 8.2 and the end of the >second wave is at 0.2. And by the time we get to 1 second, the leading >edge of the first wave is at 1.0 and the end of the tenth wave is also >at 1.0. Exactly ten waves for that one too. > >If we were wrapping wire around a torus and we had to cover it until we >could quit, then at 1.1c we'd have to cover part of it twice while for >0.9c we wouldn't cover the whole thing. But our wire is sliding along >the torus and it's the first wave of the fast side that covers the same >area twice, and the area that the first wave of the slwo side doesn't >get to, the last wave has backed onto. > >Ten waves both times. No phase shift. No, you don't get the picture at all. >> >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. >> >> What does 'frequency' mean when applied to light? > >I'm assuming something periodic going on, and in that case the frequency >is the number of times the periodic thing happens per unit time. In my >example that's ten times per second. Nope. Frequency is implied as the 'number of wavecrests' arriving per second. f = hc/lambda. I reckon a photon also has an intrinsic oscillation and a natural frequency but that it not the one that we normally detect. >> What is not appreciated is that for a constant rotation speed there is >> a constant fringe DISPLACEMENT but no fringe MOVEMENT. > >Yes. > >> At constant speed, the same number of waves arrives at the detector >> each second from BOTH rays.....there is NO doppler shift of >> frequency....that's why the fringe pattern is stable. BUT the phasing >> of the two is different because the number of wavelengths in each path >> is different. > >That's the part I don't understand, why the number of wavelengths is >different. Because the pathlengths are different. If you didn't keep reverting to the rotating frame you would understand that. http://www.mathpages.com/rr/s2-07/2-07.htm Henry Wilson...www.users.bigpond.com/hewn/index.htm Einstein...World's greatest SciFi writer.. |