The Emission Theory of Androcles
in [Relativity]
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From: Jonah Thomas on 15 Sep 2009 09:03 "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 > >> > doug <xx(a)xx.com> wrote: > >> > > >> >> Wavelengths are measured with a diffraction grating which is > >only> >> sensitive to wavelength. > >> > > >> > How did you find out that diffraction gratings are only sensitive > >to> > wavelength? I've been looking for information on that and have > >not> > found anything yet. > >> > >> Wikipeida is a good starting point > >> http://en.wikipedia.org/wiki/Diffraction_grating > > > > This one is utterly worthless, they do nothing to distinguish > > effects of wavelength from those of frequency. > > Did you not read the theory of operation? > > The > == > When a plane wave of wavelength λ is incident normally on the > grating, each slit in the grating acts as a point source propagating > in all directions. The light in a particular direction, θ, is made up > of the interfering components from each slit. Generally, the phases of > the waves from different slits will vary from one another, and will > cancel one another out partially or wholly. However, when the path > difference between the light from adjacent slits is equal to the > wavelength, λ, the waves will all be in phase. This occurs at angles > θm which satisfy the relationship dsinθm/λ=|m| where d is the > separation of the slits and m is an integer. Thus, the diffracted > light will have maxima at angles θm given by dsinθm=mλ > == > > It dependant on the distance between waves and the distance between > grating slits > > The frequency doesn't matter Thank you! Only -- it does matter when the waves from different sources reach the detector. Just, at angles where the difference between the distances is equal to a whole multiple of the wavelength, if the speed is equal then the waves will be in phase. And if you know the wavelength and the speed then the frequency is set. I think it follows that waves which travel at different speeds will cancel differently. We have no experience with that from acoustics etc because the medium determines the speed the waves propagate. > >> and here > >> http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/grating.html > > > > Ditto. > > > >> http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/gratcal.html > > > > Likewise. > > > > I haven't seen any analysis at all separating the effects of > > wavelength from frequency at different lightspeeds. There's no > > discussion why it's wavelength that matters instead of frequency. > > Yes there is > > > Of course, at constant > > lightspeed one gives you the other and it isn't really possible to > > separate them. > > Yes .. you can > > Consider the interference in a double slit experiment > > http://www.antonine-education.co.uk/physics_a2/Module_4/Topic_6/derivation_of_young.htm > > Lambda = ws/D where Lambda = wavelength, w = fringe spacing, s = slit > spacing, D = distance from slit to screen. Nothing there at all > related t frequency and speed (except as much as they determine > wavelength) > > Its all about wavelengths and path lengths. Thank you, that's a great link! I notice that the formula you use is an approximation to a more complicated one that also depends only on lambda (or on speed/frequency). I don't see how to get the direct link but it's under Fraunhofer diffraction -> Fraunhofer geometry -> Examine approximation. It uses the simplifying assumption that all the light rays are parallel. It gives a brief explanation about the more complicated case where the single source might be close enough to the grating that the rays are not parallel and the screen may be somewhat close relative to the width of the slit. That one is complicated enough that they didn't draw in much of the diagram. So using the fraunhofer approximation, given light that's the same wavelength but traveling at different speeds that are in phase at the slit, I'd expect them to each self-interfere at the same place. Would the two different interference patterns interfere with each other? No reason to think they would still be in phase at the screen. No, they both have their destructive interference at the same places, and that does not vary by time. So 0+0=0. No, that doesn't make sense. If that was true, how could waves interfere when they are out of phase? If nothing matters but the wavelength and they have the same wavelength, they should all self-interfere at the same exact places. So I'm wrong. I don't understand it yet.
From: Inertial on 15 Sep 2009 08:58 "Jonah Thomas" <jethomas5(a)gmail.com> wrote in message news:20090915090344.131b4a40.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 >> >> > doug <xx(a)xx.com> wrote: >> >> > >> >> >> Wavelengths are measured with a diffraction grating which is >> >only> >> sensitive to wavelength. >> >> > >> >> > How did you find out that diffraction gratings are only sensitive >> >to> > wavelength? I've been looking for information on that and have >> >not> > found anything yet. >> >> >> >> Wikipeida is a good starting point >> >> http://en.wikipedia.org/wiki/Diffraction_grating >> > >> > This one is utterly worthless, they do nothing to distinguish >> > effects of wavelength from those of frequency. >> >> Did you not read the theory of operation? >> >> The >> == >> When a plane wave of wavelength λ is incident normally on the >> grating, each slit in the grating acts as a point source propagating >> in all directions. The light in a particular direction, θ, is made up >> of the interfering components from each slit. Generally, the phases of >> the waves from different slits will vary from one another, and will >> cancel one another out partially or wholly. However, when the path >> difference between the light from adjacent slits is equal to the >> wavelength, λ, the waves will all be in phase. This occurs at angles >> θm which satisfy the relationship dsinθm/λ=|m| where d is the >> separation of the slits and m is an integer. Thus, the diffracted >> light will have maxima at angles θm given by dsinθm=mλ >> == >> >> It dependant on the distance between waves and the distance between >> grating slits >> >> The frequency doesn't matter > > Thank you! Only -- it does matter when the waves from different sources > reach the detector. Just, at angles where the difference between the > distances is equal to a whole multiple of the wavelength, if the speed > is equal then the waves will be in phase. And if you know the wavelength > and the speed then the frequency is set. > > I think it follows that waves which travel at different speeds will > cancel differently. We have no experience with that from acoustics etc > because the medium determines the speed the waves propagate. > >> >> and here >> >> http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/grating.html >> > >> > Ditto. >> > >> >> http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/gratcal.html >> > >> > Likewise. >> > >> > I haven't seen any analysis at all separating the effects of >> > wavelength from frequency at different lightspeeds. There's no >> > discussion why it's wavelength that matters instead of frequency. >> >> Yes there is >> >> > Of course, at constant >> > lightspeed one gives you the other and it isn't really possible to >> > separate them. >> >> Yes .. you can >> >> Consider the interference in a double slit experiment >> >> http://www.antonine-education.co.uk/physics_a2/Module_4/Topic_6/derivation_of_young.htm >> >> Lambda = ws/D where Lambda = wavelength, w = fringe spacing, s = slit >> spacing, D = distance from slit to screen. Nothing there at all >> related t frequency and speed (except as much as they determine >> wavelength) >> >> Its all about wavelengths and path lengths. > > Thank you, that's a great link! Glad to help. > I notice that the formula you use is an > approximation to a more complicated one that also depends only on lambda > (or on speed/frequency). I don't see how to get the direct link but it's > under Fraunhofer diffraction -> Fraunhofer geometry -> Examine > approximation. It uses the simplifying assumption that all the light > rays are parallel. It gives a brief explanation about the more > complicated case where the single source might be close enough to the > grating that the rays are not parallel and the screen may be somewhat > close relative to the width of the slit. That one is complicated enough > that they didn't draw in much of the diagram. > > So using the fraunhofer approximation, given light that's the same > wavelength but traveling at different speeds that are in phase at the > slit, I'd expect them to each self-interfere at the same place. Would > the two different interference patterns interfere with each other? No > reason to think they would still be in phase at the screen. No, they > both have their destructive interference at the same places, and that > does not vary by time. So 0+0=0. > > No, that doesn't make sense. If that was true, how could waves interfere > when they are out of phase? If nothing matters but the wavelength and > they have the same wavelength, they should all self-interfere at the > same exact places. So I'm wrong. I don't understand it yet. I hope you work it out :)
From: Jonah Thomas on 15 Sep 2009 10:40 hw@..(Henry Wilson, DSc) wrote: > Jonah Thomas <jethomas5(a)gmail.com> wrote: > >hw@..(Henry Wilson, DSc) wrote: > >> >> You are emulating inertial in trying to explain the behavior of > >> >light> by using classical wave thepory....when it has been shown > >> >conclusively> that light is not like that. > >> > >> >What should be used instead? > >> > >> That's the big question. One that says 'wavelength' is absolute and > >> invariant. All you need to do is define what determies this thing > >we> are calling wavelength. > > > >Traditionally people thought of frequency and wavelength and > >wavespeed in terms of concentric circles from a stationary source. > >Just like sound. > > > >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.... > >I found it exciting that emission theories give you that same result > >without having to fudge the lengths or times, or, well, anything. > > > >But you and Androcles both steadfastly maintain that the concentric > >circles are just some sort of useless illusion, and the reality is > >something unrelated. At precisely the time that you'd think the light > >was coming from straight overhead from a star that's moving sideways, > >you have to instead point your telescope off at an angle. So now I'm > >not sure what to think. > > That's called aberration. First Stellar aberration and now mental > aberration.(that's a joke) > >> >I'm willing to throw away all the classical > >> >wave stuff if you have something else that works. > >> > >> 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? > >> >But what is it? It > >> >looks to me like you're using stationary waves. > >> > >> Go back to the rope model. No matter how fast the rope is moved > >around> the cylinder and the same number of twists exists between any > >two> points on the cylinder. The two directions of the rope represent > >the> numbers of wavelengths in each path. > >> A photon emitted at one point and moving inside the hollow torus > >moves> much much faster than the rope and experiences virtually the > >same> number of cycles as there are twists. > >> The distance between the two points varies with rope (ring gyro) > >> rotational speed. > >> > >> This is now a pretty clear model. > > > >It isn't at all clear to me, but I'm working on it. > > > >> >> Let's forget about oscillations and frequencies. They are > >totally> >> undefined and you two certainly haven't a clue as to what > >they> >might> imply. Let's just accept the BaTh 'wavelength' > >explanation. It> >works.> The path lengths are different therefore > >each path contains a> >> different number of wavelengths and the rays > >are out of phase when> >> they reunite. End of story. > >> > > >> >If we throw out classical interpretations of wave, frequency, and > >> >wavelength, what do you replace them with? I'm still real unclear > >on> >the details here. It might work for you to throw out all the old > >> >concepts and replace them with new concepts where things work out > >> >right, but I need the concepts. > >> > >> Well work on it....I am. > > > >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. > >That's OK, but when you use the words people use for the traditional > >model people get confused and apply the concepts they have attached > >to those words and they wind up saying you're crazy, confused, lying, > >etc. It might be good to come up with entirely new words. > > I don't take any notice of what those morons say. it only makes me > feel more superior. Still, new words keep people from getting it mixed up with their words that have different meanings. That helps when you find somebody to communicate with. > >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 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. > >> >> 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! > >> My theory works. It says that 'wavelength' (whatever that is) > >remains> constant and 'frequency' (whatever THAT is) is doppler > >shifted in the> nonrotating frame. > >> > >> Form that, we have to speculate on models that might fit. > > > >> >> At constant rotation speed, the fringes do not move. During any > >> >speed> change, they move to a new displacement. > >> > > >> >That's true. But then your task is to explain why they get a phase > >> >change at the very beginning. > >> > >> 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. > 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. > >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. 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! 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. > >> >I say that the light is "in phase" in a sense when it leaves the > >> >emitters -- they both have wave crests at the same times and wave > >> >troughs at the same times etc. But they're in two different places > >> >and a relativist might say that they can't be in phase if they > >aren't> >at the same place and time. > >> > > >> >I say that the light is "in phase" at the moment it reaches the > >> >detectors. But it will not stay in phase for much time or > >distance.> >After all they have different wavelengths. > >> > > >> >Do you agree, in the case where nothing moves but the light which > >> >starts out in phase but which travels at different speeds with no > >> >reflection? > >> > >> No it depends on the model. > > > >?? What is it that you think would vary by model in this simple > >example? > > > I'll think about it. Thank you.
From: Jonah Thomas on 15 Sep 2009 21:40 hw@..(Henry Wilson, DSc) wrote: > Jonah Thomas <jethomas5(a)gmail.com> wrote: > >hw@..(Henry Wilson, DSc) wrote: > >> Jonah Thomas <jethomas5(a)gmail.com> wrote: > >> >But what is it? It > >> >looks to me like you're using stationary waves. > >> > >> Go back to the rope model. No matter how fast the rope is moved > >around> the cylinder and the same number of twists exists between any > >two> points on the cylinder. The two directions of the rope represent > >the> numbers of wavelengths in each path. > >> A photon emitted at one point and moving inside the hollow torus > >moves> much much faster than the rope and experiences virtually the > >same> number of cycles as there are twists. > >> The distance between the two points varies with rope (ring gyro) > >> rotational speed. > >> > >> 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 > >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. 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. 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.
From: Inertial on 15 Sep 2009 21:44
"Jonah Thomas" <jethomas5(a)gmail.com> wrote in message news:20090915214026.1e326db4.jethomas5(a)gmail.com... > hw@..(Henry Wilson, DSc) wrote: >> Jonah Thomas <jethomas5(a)gmail.com> wrote: >> >hw@..(Henry Wilson, DSc) wrote: >> >> Jonah Thomas <jethomas5(a)gmail.com> wrote: > >> >> >But what is it? It >> >> >looks to me like you're using stationary waves. >> >> >> >> Go back to the rope model. No matter how fast the rope is moved >> >around> the cylinder and the same number of twists exists between any >> >two> points on the cylinder. The two directions of the rope represent >> >the> numbers of wavelengths in each path. >> >> A photon emitted at one point and moving inside the hollow torus >> >moves> much much faster than the rope and experiences virtually the >> >same> number of cycles as there are twists. >> >> The distance between the two points varies with rope (ring gyro) >> >> rotational speed. >> >> >> >> 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 > >> >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. 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. > 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. |