The Emission Theory of Androcles
in [Relativity]
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From: Jonah Thomas on 16 Sep 2009 01:42 "Inertial" <relatively(a)rest.com> wrote: > > BTW: If henry's model (whatever it is) predicts phase shift in Sagnac, > then it should predict phase shift here: > > Source (X) and two detectors (D1, D2), equidistant form the source, > with two rays, with wavelength indicated by the < >. So over time we > have: > > D1-----------X-----------D2 > > D1-------<---X--->-------D2 > > D1---<---<---X--->--->---D2 > > D<---<---<---X--->--->--->2 > > The rays will surely arrive at D1 and D2 at the same time and same > speed and same frequency and in phase Yes. > Now look at it in terms of a relatively moving observer 'o' (eg an > observer moving past the device, or the observer is fixed and we put > move the device .. same thing) > > D1-----------------------D2 > .............o............. > > D1-------<---X--->-------D2 > ............o.............. > > D1---<---<---X--->--->---D2 > ...........o............... > > D<---<---<---X--->--->--->2 > ..........o................ > > Look at this from the observer 'o' point of view > > D1-----------------------D2 > .............o............. > > .D1-------<---X--->-------D2 > .............o.............. > > ..D1---<---<---X--->--->---D2 > .............o............... > > ...D<---<---<---X--->--->--->2 > .............o................ > > In the observers frame, the two rays are travelling different speeds > for different path length but over the same time. The frequency of > the rays is different according to the observer. The number of > wavelengths in each path from 'o' to the detectors is different. But the number of turns from the source X to the detectors is the same. > So according to Henry's model, there should be a phase difference at > the detectors. I think he would say that the light heading toward D1 slows down and the light heading toward D2 speeds up, just exactly the amount needed so they reach D1 and D2 at the same time. Same time, same number of turns. No phase difference. > But this is exactly the same set up as at the start, where there is no > phase difference, just seen from someone moving past it. Either there > is a phase difference at the detectors, or there is not. There is not. But this is just like the Sagnac experiment, except it's cut open at the detector and unwrapped. How can we get a phase shift one time and not the other? Wilson claims that in the Sagnac case we have an unambiguous velocity so we can measure the difference in length on the two sides. But clearly the velocity is arbitrary in the unrolled case. But he says that the rotation in the Sagnac case results in an actual change in distance traveled by the wave. This is not a distance that is independent of time and speed. If we look at the actual physical distance that must be crossed, it is, well, the whole way around in both directions. If the number of turns depends on that distance then -- no phase shift. And if it depends on distance and time then the constant number of turns per unit distance is gone. I think probably you're right and my understanding of Wilson's idea does not work. I'll check it again in the morning.
From: Inertial on 16 Sep 2009 03:02 "Jonah Thomas" <jethomas5(a)gmail.com> wrote in message news:20090916011636.1f3b0045.jethomas5(a)gmail.com... > "Inertial" <relatively(a)rest.com> wrote: >> "Jonah Thomas" <jethomas5(a)gmail.com> wrote >> > "Inertial" <relatively(a)rest.com> wrote: >> >> "Jonah Thomas" <jethomas5(a)gmail.com> wrote >> > >> >> > OK, this might not apply to your model, but I have pictures that >> >> > show what the problem is if it does apply. >> >> > >> >> > http://yfrog.com/0xwavecg >> >> > http://yfrog.com/10wavedg >> >> > >> >> >> >So I want to suggest that you talk about maybe "turns". A given >> >> >kind> >of light does x turns per meter, and by stating it that way >> >we> >tend> >to imply that color depends on terms/meter and not >> >> >turns/second.> >Lightspeed can vary with the source, and >> >turns/second> >varies then but> >turns/meter does not. Am I right so >> >far about what> >you're saying?> >> >> >> You're getting close. >> >> >> My definition of wavelength is something like "In the source >> >> >frame, a> photon moves a certain distance in one 'cycle' of its >> >> >intrinsic> oscillation (whatever that may be)". That distance is >> >an> >absolute and> invariant spatial interval....just like the >> >distance> >between the ends> of a rigid rod.. >> >> > >> >> > So, with the model that Inertial and I were using, the photon >> >moves> > forward but doesn't turn. The front of the wave is always >> >the front> > of the wave, and it is in phase with any other >> >front-of-waves it> > happens to meet up with. For it to get out of >> >phase it has to match> > up with something that is not the front of a >> >wave.> >> >> Yeup >> >> >> >> > But with your model, the front of the wave changes phase as it >> >> > travels. it isn't enough for it to meet another front-of-wave, >> >they> > have to have both traveled the same distance. >> >> >> >> That's what I've been saying .. something must be happening in >> >Henry's> model to make the phase of the two waves change different >> >over the> course of transit, even though they travel for the same >> >time, and are> emitted from the source with the same speed and and >> >frequency .. its> the same ray been split in two. >> > >> > Well, in his model they don't have the same speed. >> >> Yes they do, as emitted from the moving source. its only according to >> a some differently moving observer that the speeds are different > > This is a rotation so it isn't inertial frames. Irrelevant here. Each different inertial observer will measure a different speed for the same ray. > Henry supposes a form of emission theory where one side moves at c+v and > one at c-v and they keep the speed on reflection, and this time it isn't > just an observer effect. What isn't an observer effect? Speed is always observer dependant > He postulates that both sides reach the detector at the same time, and > the detector has moved so that d1 and d2 are in the ratio c+v:c-v. Yes.. this is all standard ballistic sagnac stuff > The distance from the short interval on the ring that a photon was > emitted to the short interval on the ring where the light reaches the > detector can be measured by anybody who doesn't suffer length > contraction. No legnth contraction involved here . we're talking ballistic theory > If everybody agrees that the light reaches the detector > from both sides at the same time, then there isn't a lot of room for > observer effects here. The distances are different, the times are the > same, so the speeds are different. But not according to an inertial observer moving at the same velocity as the detector when the rays arrive >> > They are in phase >> > when they are emitted, >> >> Yes >> >> > but their frequency after they are emitted >> > doesn't have to be the same >> >> Nothing changes it > > If "frequency" is the number of turns per unit time, OK > and they travel > different distances in a unit time, OK > and the number of turns is fixed by > distance OK > then they have different frequencies. And all frames would measure those same two different frequencies. Even a frame that has the same initial velocity as the source at emission point But that contradicts the set up, that we have a single light source with single frequency light at the source. What (other than Henry's wishful thinking) makes the rays have different frequencies? >> > and isn't when the speed is different. >> >> It's all an artefact of who is measuring it .. there is no change in >> the wave itself. its not the case that after emissions one wave >> speeds up and the other slows down. It's just a different measurement >> due to the movement of the observer relative to the source (and the >> rays). There is no change to the rays making their frequency or >> wavelength different (depending on whether the ray is a wave or a >> moving oscillator), its just how they are measured by a relatively >> moving observer. > > If a sagnac device tells you that something is rotating, is that just > due to a relatively moving observer? Eh? >> > He >> > has the "wavelength", the distance it takes for the leading edge to >> > do a complete turn, be constant independent of speed. >> >> Which is not physically possible. The speed is different depending on >> the relative speed of whoever observes it. But the number of turns >> taken isn't .. all inertial observers see the same number of turns >> happening in the same time (we are talking about a non-relativistic >> framework here), but a different speed, so a different wavelength. > > He postulates the distance it takes to do a complete turn to be constant > for a given sort of light, 'Sort of light' ? > independent of speed. So if it is at rest relative to an observer, it can't turn. Because it has to move a given distance to do a turn. But if it is stationary to one observer it is moving to another .. So it must turn. It must turn at different rates at the same time depending on who is looking at it. It is unphysical and impossible > That distance may not > be what you mean by wavelength. Gawd knows .. Henry makes this up as he goes along and changes his mind at each objection. > What you mean by wavelength might not be > constant. Henry has no idea what he means by anything > If the speed is different between the two sides, then all > nonrelativistic observers who count the number of turns will count the > same numbers. > But the two sides will have different numbers of turns in > the same length of time because of the different speeds. > >> Henry's notion is just totally unphysical. > > It may well be. I'm not sure I understand what he's talking about. Noone does > But I > understand this much, and so far it looks like a possible way for things > to go. No .. that's the point ... it is NOT possible at all. Its unphysical. Impossible. > I don't understand interference well enough yet, at a minimum I'll have > to sleep on it. But traditional interference looks like it depends only > on wavelength independent of speed and frequency. If you count the > distance traveled in wavelengths, the distance will vary smoothly from > one end of a slit to the other. If the sides of the slit cancel, the > center is the part that's least likely to spread out anyway.... I > started to say more but it might not make sense to me in the morning. So > better not to write it until morning at the earliest. > >> > So if frequency has >> > meaning for him it would absorb all the speed change. >> >> > I think of wavelength as the distance between wave crests, and that >> > needn't have anything to do with the distance it takes for the >> > leading edge to turn. >> >> What wave crests are there? Its a turning object. If you mean the >> distance it travels to make one turn, then that depends on the speed. > > I repeat, he postulates a system where the distance required to make one > turn is independent of speed. And that is physically impossible, unless it travels at the same speed for all observers. >> > Wilson has said that in his model the photon has a >> > definite length, and he appears to picture it like a coil of wire >> > that turns a definite number of times, it is incompressible and the >> > length is completely stable. >> >> So what is its wavelength and what is its frequency and what is its >> phase. These all need to be defined or we can't talk about what >> happens in Sagnac at the detector. > > He doesn't need to define all that, but he does need to define how > interference happens in his system. That might involve wavelength and > frequency and phase, or he might have different concepts that work. Then he has to define it. Light has wave-like behavior, with interference patterns, phase shifts, wavelength and frequency. If his model doesn't have that, it is no use. >> > But if there was value in a model where the photon >> > can be stretched or compressed independent of the fixed distance it >> > takes for the leading edge to turn, that model is available. >> > >> >> Of course, when phase changes at the leading edge like that, what >> >you> have is a moving intrinsic oscillator, and so wavelength varies >> >and> frequency is fixed, and you still end up with them in phase. >> > >> > He's doing something that does not fit your model. I don't know >> > whether it works, but it isn't what you say. >> >> Then what is it? Is it a wave, is it a moving oscillator (eg a >> spinning object) ? > > He talks like it's a spinning object but I haven't yet seen how that's > important. Just to imagine it, I try to imagine a series of spinning > objects. Each of them can spin, and each next one can be a little out of > phase with the previous one. The spin that each one has does not have to > match up with the change in spin for successive particles. (Though > Maxwell's equations might say they do have to match up in some > particular way.) > >> >> > I found that concept alien enough that I simply did not >> >understand> > what you were saying. >> >> > It just did not register. Now the question is whether >> >> > that approach can fit together with the other things we think we >> >> > know, and what has to be changed to fit your model. >> >> > >> >> > I would like it better if we had a model for travel that did not >> >fit> > either of my two pictures. In the one case the leading edge >> >does not> > turn and the Sagnac experiment does not get out of phase. >> >In the> > other case the leading edge turns but another particle >> >following in> > the footsteps of the first, or later wraps of the >> >same photon, would> > give a stationary charge at each spot they >> >traversed until they were> > gone. >> >> > >> >> > There ought to be a third way. >> >> >> >> Or Henry is simply wrong. That's been the concensus for the last >> >few> years. >> > >> > He might easily be wrong. But physics has not advanced by depending >> > on the consensus of people who don't understand what they're >> > deciding about. >> >> Indeed it doesn't .. but Henry has never presented a consistent >> non-contradictory model. And his ballistic analysis of Sagnac is just >> plain wrong. Those are facts, not opinions. > > Are you sure you understand his model? I understand what he is saying .. and its inconsistent nad contradictory. > I'm sure I don't fully understand > it, and I've been giving it the most sympathetic hearing I could manage. > Much harder to understand an idea if you don't give it a sympathetic > hearing. (Of course, even after you do it can still be plain wrong.)
From: Inertial on 16 Sep 2009 03:10 "Jonah Thomas" <jethomas5(a)gmail.com> wrote in message news:20090916014207.1ee0503a.jethomas5(a)gmail.com... > "Inertial" <relatively(a)rest.com> wrote: >> >> BTW: If henry's model (whatever it is) predicts phase shift in Sagnac, >> then it should predict phase shift here: >> >> Source (X) and two detectors (D1, D2), equidistant form the source, >> with two rays, with wavelength indicated by the < >. So over time we >> have: >> >> D1-----------X-----------D2 >> >> D1-------<---X--->-------D2 >> >> D1---<---<---X--->--->---D2 >> >> D<---<---<---X--->--->--->2 >> >> The rays will surely arrive at D1 and D2 at the same time and same >> speed and same frequency and in phase > > Yes. > >> Now look at it in terms of a relatively moving observer 'o' (eg an >> observer moving past the device, or the observer is fixed and we put >> move the device .. same thing) >> >> D1-----------------------D2 >> .............o............. >> >> D1-------<---X--->-------D2 >> ............o.............. >> >> D1---<---<---X--->--->---D2 >> ...........o............... >> >> D<---<---<---X--->--->--->2 >> ..........o................ >> >> Look at this from the observer 'o' point of view >> >> D1-----------------------D2 >> .............o............. >> >> .D1-------<---X--->-------D2 >> .............o.............. >> >> ..D1---<---<---X--->--->---D2 >> .............o............... >> >> ...D<---<---<---X--->--->--->2 >> .............o................ >> >> In the observers frame, the two rays are travelling different speeds >> for different path length but over the same time. The frequency of >> the rays is different according to the observer. The number of >> wavelengths in each path from 'o' to the detectors is different. > > But the number of turns from the source X to the detectors is the same. Just like in Sagnac. But the number of turns fomr o (like the fixed point in the non-rotating frame in sagnac) is not the same >> So according to Henry's model, there should be a phase difference at >> the detectors. > > I think he would say that the light heading toward D1 slows down and the > light heading toward D2 speeds up, just exactly the amount needed so > they reach D1 and D2 at the same time. > > Same time, same number of turns. No phase difference. Just like in Sagnac >> But this is exactly the same set up as at the start, where there is no >> phase difference, just seen from someone moving past it. Either there >> is a phase difference at the detectors, or there is not. > > There is not. Just like in Sagnac > But this is just like the Sagnac experiment, except it's cut open at the > detector and unwrapped. EXACTLY !!! > How can we get a phase shift one time and not the other? EXACTLY !!! > Wilson claims > that in the Sagnac case we have an unambiguous velocity No .. we don't .. its observer dependant like every velocity > so we can > measure the difference in length on the two sides. But clearly the > velocity is arbitrary in the unrolled case. It makes no difference > But he says that the rotation in the Sagnac case results in an actual > change in distance traveled by the wave. How .. why ? > This is not a distance that is > independent of time and speed. If we look at the actual physical > distance that must be crossed, it is, well, the whole way around in both > directions. Just like in my little example above .. according to 'o' there are two different physical distance being crossed > If the number of turns depends on that distance then -- no > phase shift. And if it depends on distance and time then the constant > number of turns per unit distance is gone. > > I think probably you're right and my understanding of Wilson's idea does > not work. I'll check it again in the morning. OK. Have fun. Just refer the same logic back to the diagram above. It has all the important features the same as Sagnac (in a non-relativistic framework). 'o' is analogous to fixed point in the inertial frame at the source that Henry is so fond of. The distance o..D1, and o..D2 are analogous to the different path lengths in Sagnac. The light travels the different distances in the same time at two different speed (c+v and c-v). The lack of phase difference is the same.
From: Henry Wilson, DSc on 16 Sep 2009 05:43 On Tue, 15 Sep 2009 10:40:56 -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: > >> > >> >A moving source would give you a compression, you'd get eccentric >> >circles instead. >> > >> >But you could use that eccentricity to tell who was moving. If there >> >aren't preferred frames then everybody ought to calculate those >> >things as concentric circles. And that's one of the things SR gives >> >you. >> >> It doesn't. It simply says it does by postulate. > >Sure, but it gives a model that does result in the waves moving in >concentric circles independent of frame, and so far it's largely >compatible with experimental evidence. What more would you want? Well, >it would be nice if it made sense. But apart from that.... Wake up. There is NO experimental evidence. >> >> >> >> It was thrown away when the PE effect was discovered....... >> >ironically> by Einstein himself. >> > >> >I don't see that. The PE effect is perfectly compatible with light as >> >waves, isn't it? >> >> NONONONONONONONONO! >> >> That's the problem. >> >> Relativity is incompatible with quantium theory yet idiots like >> inetial and demented dougie still try to defend the nonsense. > >But apart from relativity, there's no particular reason you can't have >light waves and still have atoms that demonstrate the PE effect, right? There is. That's been known for a century >> > >> >It seems to me that you don't have an alternative model, you have a >> >proposal for an alternative model. >> >> Well I have the basis of a model. I don't spend all day thinking about >> it. I don't care how many more UNI students are brainwashed by the >> physics establishment. The truth will eventually come out. > >Sure, but your model isn't better than their model until you have >something that works. It does work. It produces the experimentally verified equation for sagnac. Why do you think 'inertial' and Jerry are so desperate to find a flaw in my theory? I am a threat to their whole belief system. Poor old Jerry has backed a loser all his life. Do you think he wants to know about it? >> You're getting close. >> My definition of wavelength is something like "In the source frame, a >> photon moves a certain distance in one 'cycle' of its intrinsic >> oscillation (whatever that may be)". That distance is an absolute and >> invariant spatial interval....just like the distance between the ends >> of a rigid rod.. > >So turns go with distance independent of time. Light can have different >"wavelengths", it can have a different number of turns per unit >distance, but it's distance that matters independent of the time or the >speed. We can stop talking about frequency. Turns, distance, and time >give everything so far. There's polarization, which people interpret as >linear in any direction perpendicular to the direction of travel, or as >circular, or as elliptical. At this point my interpretation of their >interpretation is that the axis of turning can be any direction in 3D. >When it's the direction of travel it's circular polarization. When it's >perpendicular to the direction of travel you get linear polarization in >the third direction. You'll probably want something analogous for your >turns because what they do fits the reality on some level. Any theory about the nature of photons has to explain polarization. Maxwell's wave theory does that but it requires a medium. I believe a photon carries it own little bit of 'aether' ...and that's where the intrinsic oscillation occurs. Fields cannot simply happen in 'nothing'. As I've pointed out many times, space carrying a field must be different from space devoid of fields. What might that difference be? >> >> >> that's the other demo. THe stationary wave is put there purely >> >so> >you> can see the phase difference. >> >> > >> >> >No, this one too. You drew waves that get extended around a >> >circle.> >At any one spot the wave never changes after it gets drawn. >> >Those> >waves are frozen once they are drawn. >> >> >> >> OK. You have to find a model that requires the emitted light to >> >> experience the same number of cycles per path as there are absolute >> >> wavelengths. >> > >> >So, you measure the pathlength and that gives you the number of >> >turns. OK. >> >> Yep. > >OK! >> >> >> >> During any CHANGE in rotational speed, a change also occurs in the >> >> number of wavelengths in each path. They flow out of one and into >> >the> other. >> > >> >Mmmm. You change the rotational speed. The number of turns from the >> >emitter to the detector is unchanged. >> >> No it isn't. The distance 'vt' changes. That's the distance between >> the start and detection points in the inertial frame....according to >> both SR and BaTh. > >The distance from the emitter to the detector never changes. What >changes is the distance from the emitter at time t0 to the detector at >time t1. OK. Yes, these are the emission and detection points (in the inertial frame) for a particular (infinitesimal) element of a ray. You would find this is pretty simple stuff if you would bother to calculate what the distance 'vt' is for different rotational speeds. >> You still haven't understood that mathpages diagram. >> >> >What about the time it takes to >> >get from the emitter to the receiver? The time is the distance >> >divided by the speed. So when it isn't moving the time is d/c. When >> >it's moving at v then the time is >> > >> >t=(d+vt)/(c+v) >> >> In the inertial frame >> 2piR + vt = (c+v)t .........(one ray) >> >> Or 2piR - vt = (c-v)t......(other ray) >> >> So t = 2piR /c > >Yes. t is the same either way. > >> >The distance goes up by the amount the detector turns, and speed goes >> >up by the amount the detector turns. >> > >> >t-vt/(c+v) = d/(c+v) >> >t(c+v) -vt = d >> >ct = d >> >t = d/c >> > >> >The time it takes to get to the detector is independent of v. It >> >takes the same time no matter how fast it spins. >> >> That's correct. THat is easily derived if you use the rotating frame. >> However it isn't as simple as it appears. >> >> >Why would the number of turns it takes to get to the detector be >> >different when the number of turns in that distance is constant and >> >the time it takes to arrive is constant? >> >> That's the big question...and when you answer it, you'll be awarded a >> Nobel prize. Don't forget to mention my name will you. > >[sigh] I almost took you seriously saying you didn't know. Your answer >is that the number of turns is different although it takes the same time >to arrive both ways. > >So, let's pretend for the moment that the light is a series of rotating >particles, and the particles themselves are in phase. One of them leaves >the emitter, then the next one, and then the one after that. Each of >them is at a different part of its rotation cycle when it leaves, so >they will each be at a different part of the cycle when they arrive >anywhere. It is like waves that roll onto the beach, the wave crests do >not stay frozen. If one particle moves at c+v and the other at c-v, and >the first goes a distance d(c+v) and the second at a distance d(c-v), >they will not be in phase at the end because one of them has rotated 2dv >times farther than the other. Even though they start out at the same >place in their rotation they don't end up that way because they are >rotating at different speeds, proportional to their speed and >proportional to the distance they cover. > >OK, I can see it that way. The detector doesn't care much about the element itself. It is receiving 'waves' while the element is in transit from E to D. The number of waves counted is:pathlength/wavelength. >> >It looks to me like when we assume that the speed of the light in the >> >two directions is c+v and c-v and that speed stays constant at c+v >> >and at c-v the whole distance, we should get no interference. But >> >when the speed of light is the vector sum of cD+vV where V is a unit >> >vector in the direction of the source and v is the speed of the >> >source, and D is the direction that will give us a vector sum in the >> >direction we're interested in, then we get precisely the amount of >> >interference we'd expect by classical or by SR methods, the amount >> >that is experimentally observed. >> >> Yes, the classical explanation is indeed very attractive...except that >> an aether does not exist and there is no obvious explanation as to why >> the rays should move at c+v and c-v wrt the source. > >If the frequency is constant then it goes one way. If the wavelength is >constant it goes differently. As I said, it it not the frequency of the particular element that matters. It is the number of waves counted while the element is in transit. >How do we choose between those? Well, I have assumed that the particles >are in phase when they leave the emitter. The two that leave in two >opposite directions and at different speeds are at the same angle in >their rotation. If one of them rotates slower than the other and travels >slower, then when the time comes to send the second particles the slow >ones will be closer together. So if you look at the distance between the >waves, it will be closer for the slow side! Even though we said >"wavelength" was the same! On the slow side the particles are moving >slower and rotating slower so they each rotate once in the same >distance, but the apparent wavelength will be shorter! No it wont. It is an absolute distance. You are using the wrong model. >So number of turns is something distinct from traditional wavelength. >You have to give it a new name or people will confuse it with wavelength >and they will misunderstand. There has never been a proper definition of eitIher 'wavelength' OR 'frequency' in the case of light. Wave theories appeared to explain diffraction and quit a few other properties of light but quantum mechanics put an end to all that. In reality nobody knows a bloody thing about the nature of a photon or about light in transit. Physics is very much in its infancy.... and Einstein hasn't helped one iota. Henry Wilson...www.users.bigpond.com/hewn/index.htm Einstein...World's greatest SciFi writer..
From: Henry Wilson, DSc on 16 Sep 2009 05:57
On Tue, 15 Sep 2009 21:40:26 -0400, Jonah Thomas <jethomas5(a)gmail.com> wrote: >hw@..(Henry Wilson, DSc) wrote: >> Jonah Thomas <jethomas5(a)gmail.com> wrote: >> >> >> >> This is now a pretty clear model. >> > >> >It isn't at all clear to me, but I'm working on it. > >OK, this might not apply to your model, but I have pictures that show >what the problem is if it does apply. > >http://yfrog.com/0xwavecg >http://yfrog.com/10wavedg I discussed those two possibilities with Paul several years ago....the 'frozen Norwegian snake' model or the 'warm wriggling Australian' one. I don't really think light behaves like either although my 'rayphases' demo uses that principle. >> >So I want to suggest that you talk about maybe "turns". A given kind >> >of light does x turns per meter, and by stating it that way we tend >> >to imply that color depends on terms/meter and not turns/second. >> >Lightspeed can vary with the source, and turns/second varies then but >> >turns/meter does not. Am I right so far about what you're saying? >> >> You're getting close. >> My definition of wavelength is something like "In the source frame, a >> photon moves a certain distance in one 'cycle' of its intrinsic >> oscillation (whatever that may be)". That distance is an absolute and >> invariant spatial interval....just like the distance between the ends >> of a rigid rod.. > >So, with the model that Inertial and I were using, the photon moves >forward but doesn't turn. The front of the wave is always the front of >the wave, and it is in phase with any other front-of-waves it happens to >meet up with. For it to get out of phase it has to match up with >something that is not the front of a wave. You are regarding the photon as a simple oscillator. You cannot do this. >But with your model, the front of the wave changes phase as it travels. >it isn't enough for it to meet another front-of-wave, they have to have >both traveled the same distance. No, they travel for the same time with different speeds and over diffrent distances . You need a model that conforms to that. >I found that concept alien enough that I simply did not understand what >you were saying. It just did not register. Now the question is whether >that approach can fit together with the other things we think we know, >and what has to be changed to fit your model. > >I would like it better if we had a model for travel that did not fit >either of my two pictures. In the one case the leading edge does not >turn and the Sagnac experiment does not get out of phase. In the other >case the leading edge turns but another particle following in the >footsteps of the first, or later wraps of the same photon, would give a >stationary charge at each spot they traversed until they were gone. > >There ought to be a third way. Yes. We haven't yet found it. Henry Wilson...www.users.bigpond.com/hewn/index.htm Einstein...World's greatest SciFi writer.. |