The Emission Theory of Androcles
in [Relativity]
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From: Androcles on 13 Sep 2009 03:29 "Jonah Thomas" <jethomas5(a)gmail.com> wrote in message news:20090912211913.77428b0d.jethomas5(a)gmail.com... > "Inertial" <relatively(a)rest.com> wrote: >> "Androcles" <Headmaster(a)Hogwarts.physics_o> wrote > >> > It can only send a pulse. Imagine all you want to, there are NO >> > light waves. >> > Radio waves, yes, but no light waves. >> >> OMG .. Androcles thinks that the set frequency of EM radition we call >> light is a totally different concept from the set EM frequencies we >> call radio waves? He's GOT to be joking. >> >> BAHAHAHA >> >> If not .. I wonder where he thinks the change between being waves and >> being particles happens? > > Androcles has training in electrical engineering, so he quite likely > understands how radio waves are produced. > > He gave the explanation that radio waves are like a tsunami while light > waves are like something else -- a faucet dripping? Rainwaves. > Both water moving, but different forms. Both are water waves. I've always referred to rain as rainwaves. > My guess at where he's heading with that is that light waves are > typically made by atoms, by electrons falling to lower shells or > whatever the wording is. (You can get light without quite that, with > atoms heated to white heat.) Radio waves are made by electrons > accelerated back and forth, typically inside hunks of metal or in > waveguides. A whole lot of electrons moving without much regard to the > atoms they move among. It's different. The electrons that are used to > make radio waves can keep oscillating back and forth as long as you > supply the power. The atoms that make light make one quantum at a time > (or maybe it's two) but they can't do it again until they get their > electrons knocked up again. > > So it's different. So electrons fall to their new levels and if it takes > one cycle for that to happen you get a one-cycle event. If it takes > fifty cycles then you get a 50-cycle event, that's more wavelike than > the short one. How long does it take? I don't know but likely somebody > knows.
From: Inertial on 13 Sep 2009 03:50 "Jonah Thomas" <jethomas5(a)gmail.com> wrote in message news:20090913033001.00ccb381.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: >> >> >hw@..(Henry Wilson, DSc) wrote: >> >> >> Jonah Thomas <jethomas5(a)gmail.com> wrote: >> >> >> >> >> 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. >> > >> >I said back what I understood you to say, and you didn't tell me >> >whether I got it right. >> > >> >The emission point is the point where the first wave we're interested >> >in started out. True or false? >> >> You can look at it that way if you like. BUT THE POINT IS STATIONARY >> IN THE NONROTATING FRAME. > > OK. So you can't mark that point on the rotating apparatus. You can, but its not the point he's referring to > You could, > say, put a rock besice the apparatus where the first wave you care about > starts. Yeup >> >> >> 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. >> > >> >Yes. And still you have ten waves present at a time in each >> >direction, and each of them has the same wavelength. >> >> No you don't. You have 10 + vt/L in one and 10-vt/L in the other. > > Count them as they are produced. At the first wave you make one in each > direction. That's one. At the second wave you make one wave in each > direction. That's two. Three. Four. Five. ... Ten. > > Ten in each direction. Number ten is just finishing its creation as > number one begins to be destroyed by the detector. Yeup >> >> >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. >> > >> >No, the reason you get the same number of turns is that in the >> >forward case by the end the first wave has moved around past the >> >original emission point so that it covers that part twice. The total >> >path is longer but with the first wave (or waves, if the speed is >> >high enough) covering that part twice you still have ten waves >> >covering the whole distance. >> >> They take the same time to get to the detection point (also stationary >> in the nonrotating frame) Not the detector for the Sagnac effect. You have invented another detector and are talking about a difference effect there >> So they meet up at the same instant....but >> their frequencies in the nonR frame are different Yes >> so they are out of >> phase when they meet. No, they don't. Because the leading edge of a wave is always the same phase AND FURTHER, the SAGNAC detector is MOVING in the non-rotating frame, so they are the SAME frequency at the detector. Gees. . we've gone thru thing many many times. And you keep making the same mistake. The Saganc efefct happens AT THE DETECTOR. The detector IS MOVING. >> This is terribly simple really. It is .. yet after many many years, you still can't get it. > At what time do the beginning of the first wave in each direction reach > the detector? > > At what time do the beginning of the second wave in each direction reach > the detector? > >> The frequency of a water wave depends on the speed of one's boat. The >> distance between waves does not. > > Agreed. Because the water does not move with the boat (ie the speed of waves leaving a boat is not always the same wrt the boat). With light it is always c. >> >In the backward case, by the time the start of the first wave >> >has reached the sensor, the last wave has been created entirely >> >behind the original emission point. So you get ten waves with the >> >same wavelength even though the total distance the first wave covers >> >is only nine wavelengths. >> >> No, you haven't gotten it yet. > > [sigh] I know .. its frustrating, isn't it [snip a bit] >> I think the point you are missing is that this is a continuing >> process. The number of waves in each path only changes when the >> rotation speed changes. > > I agree. And every successive wave of the two rays arrives at the same time and in phase. This is soooo simple [snip a bit] >> >OK. The number of wavecrests leaving the source in my example is ten >> >per second, in any frame. >> >> The number of wavecrests that pass any stationary point marked on the >> nonrotating ring is NOT ten. > > Yes. But why count the number that pass a stationary point EXACTLY !!!! > when the > detector is moving? We've been telling Henry this for years. His response .. what happens at the detector in Sagnac is not relevant !!! > Isn't it wavecrests that pass the detector that > count? It sure is. Can you appreciate the frustration the rest of us haev with Henry. He's been told this SOOO many times, and he still either a) doesn't get it or b) deliberately lies about it for god-only-knows-what reason. >> If you can understand the SR 'explanation' you should be able to >> understand the BaTh one too. There is basically very little >> difference. > > The difference I see is that the SR explanation has the speed of light > constant in both directions. So their waves are out of phase when they > meet. Yes .. it means the leading edges of the waves don't arrive at the detector at the same time. Hence a phase shift. Soooo simple >> >> 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. > > Agreed, no doppler shift. To get the phase different you'd have them get > out of phase by a constant amount And leading edges of a wave are always in phase. So only way to get them out of phase is for the mto arrive at different times. NOTE: A difference in frequency at the detector will result in a continually changing phase difference that starts in phase (when rays first arrive) and then shifts continually > and then they would all arrive at the > same speed but one side would be slow consistently by that constant > amount. But your moving picture does not show that. It shows them > arriving at the same time, every time. And so in phase >> >> >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. >> > >> >At this point in my imagination Androcles is saying the pathlengths >> >are history. Why do the pathlengths matter? >> >> Androcles is totally confused about Sagnac. He still thinks the >> detector is not rotating with the apparatus. > > His pictures don't show the detector standing still. Yeup .. he's not THAT confused about it. He seems to think its a coriolis effect, but the coriolis with (relatively) VERY fast light compare to rotation rate, the difference that could possibly make is minimal. > At this point we agree about most of the facts. The only thing I don't > understand is why you say the waves in the different directions are out > of phase. Yeup. That's the killer > You show each wave arriving at the detector at the same time. > How are they out of phase? He'll probably go back to light being a moving oscillator, so the wave front phase changes all the time. But oscillators have the same frequency in every frame of reference, so those the phase of the oscillators would not be zero (unless the time was a multiple of the period), they would both be at the same point in their oscillation cycle, and still in phase. You're fighting a losing battle trying to get a consistent and physically realistic answer from Henry.
From: Inertial on 13 Sep 2009 04:03 "Jonah Thomas" <jethomas5(a)gmail.com> wrote in message news:20090913034536.6ff1dded.jethomas5(a)gmail.com... > "Inertial" <relatively(a)rest.com> wrote: >> "Jonah Thomas" <jethomas5(a)gmail.com> wrote >> > "Inertial" <relatively(a)rest.com> wrote: >> >> "Androcles" <Headmaster(a)Hogwarts.physics_o> wrote >> > >> >> > It can only send a pulse. Imagine all you want to, there are NO >> >> > light waves. >> >> > Radio waves, yes, but no light waves. >> >> >> >> OMG .. Androcles thinks that the set frequency of EM radition we >> >call> light is a totally different concept from the set EM >> >frequencies we> call radio waves? He's GOT to be joking. >> >> >> >> BAHAHAHA >> >> >> >> If not .. I wonder where he thinks the change between being waves >> >and> being particles happens? >> > >> > Androcles has training in electrical engineering, so he quite likely >> > understands how radio waves are produced. >> >> Who knows what he knows >> >> Radio waves are just EM radiation .. the distinction between different >> names 'types' of EMR is arbitrary .. just different sets of >> frequencies. > > I agree and I expect Androcles agrees with your idea -- that it's all > EMR and the frequencies are different. Androcles would not agree with - me on principle. He rarely agrees with anyone. I agree with him when he's right (sometimes he is), I agree with Henry when he is right (sometimes he is). They are both, however, very often wrong :) [snip a bit] >> You can make light and other EMR waves in a number of different ways. >> There is no fundamental difference between them EXCEPT frequency. > > I think it's the nonfundamental difference he's talking about. You can > make radio waves with a radio transmitter, and they are waves. The > electrons move back and forth continuously and they generate a > continuous wave. Yeup. > But visible light is made by individual atoms that each make a small > quantum of light that starts and ends quickly. No continuing wave. No .. there are multiple ways of making both visible light and radio waves Eg. light from a lightbulb is due to the heat in the filament causing motion. That's why it has a broad spectrum of radiation (not just a number of discrete frequencies that you get with electrons changing orbitals .. eg like light from sodium atoms) > He's talking about how they are made and the difference between them > because of how they are made. No .. he's just being a smartarse :) > I may have misunderstood him, but this is what I think he's saying, not > that there's a fundamental difference but that there's a difference in > practice. Not all EM from atoms is visible light. Not all EM from moving electrons is radio waves. The method of production has little to do with the waves/rays/light/radiation themselves, other than the frequencies and intensity they have. The waves are still just EMR / photons.
From: Androcles on 13 Sep 2009 04:29 "Jonah Thomas" <jethomas5(a)gmail.com> wrote in message news:20090913034536.6ff1dded.jethomas5(a)gmail.com... > "Inertial" <relatively(a)rest.com> wrote: >> "Jonah Thomas" <jethomas5(a)gmail.com> wrote >> > "Inertial" <relatively(a)rest.com> wrote: >> >> "Androcles" <Headmaster(a)Hogwarts.physics_o> wrote >> > >> >> > It can only send a pulse. Imagine all you want to, there are NO >> >> > light waves. >> >> > Radio waves, yes, but no light waves. >> >> >> >> OMG .. Androcles thinks that the set frequency of EM radition we >> >call> light is a totally different concept from the set EM >> >frequencies we> call radio waves? He's GOT to be joking. >> >> >> >> BAHAHAHA >> >> >> >> If not .. I wonder where he thinks the change between being waves >> >and> being particles happens? >> > >> > Androcles has training in electrical engineering, so he quite likely >> > understands how radio waves are produced. >> >> Who knows what he knows >> >> Radio waves are just EM radiation .. the distinction between different >> names 'types' of EMR is arbitrary .. just different sets of >> frequencies. > > I agree and I expect Androcles agrees with your idea -- that it's all > EMR and the frequencies are different. I will never agree with the cackling idiot. I wonder where he thinks the change between being tsumanis and being rain happens? Rain is just water. Tsunamis are just water. The distinction between different names 'types' of water is arbitrary. "BAHAHAHA" Killfile the troll.
From: Androcles on 13 Sep 2009 04:21
"Jonah Thomas" <jethomas5(a)gmail.com> wrote in message news:20090913033001.00ccb381.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: >> >> >hw@..(Henry Wilson, DSc) wrote: >> >> >> Jonah Thomas <jethomas5(a)gmail.com> wrote: >> >> >> >> >> 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. >> > >> >I said back what I understood you to say, and you didn't tell me >> >whether I got it right. >> > >> >The emission point is the point where the first wave we're interested >> >in started out. True or false? >> >> You can look at it that way if you like. BUT THE POINT IS STATIONARY >> IN THE NONROTATING FRAME. > > OK. So you can't mark that point on the rotating apparatus. You could, > say, put a rock besice the apparatus where the first wave you care about > starts. > >> >> >> 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. >> > >> >Yes. And still you have ten waves present at a time in each >> >direction, and each of them has the same wavelength. >> >> No you don't. You have 10 + vt/L in one and 10-vt/L in the other. > > Count them as they are produced. At the first wave you make one in each > direction. That's one. At the second wave you make one wave in each > direction. That's two. Three. Four. Five. ... Ten. > > Ten in each direction. Number ten is just finishing its creation as > number one begins to be destroyed by the detector. You've by now realised Wilson can't count. I warned you the old chap is senile. I've seen dementia in others, Wilson's is slowly worsening as expected. C'est la vie. >> >> >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. >> > >> >No, the reason you get the same number of turns is that in the >> >forward case by the end the first wave has moved around past the >> >original emission point so that it covers that part twice. The total >> >path is longer but with the first wave (or waves, if the speed is >> >high enough) covering that part twice you still have ten waves >> >covering the whole distance. >> >> They take the same time to get to the detection point (also stationary >> in the nonrotating frame) So they meet up at the same instant....but >> their frequencies in the nonR frame are different so they are out of >> phase when they meet. >> >> This is terribly simple really. > > At what time do the beginning of the first wave in each direction reach > the detector? > > At what time do the beginning of the second wave in each direction reach > the detector? > >> The frequency of a water wave depends on the speed of one's boat. The >> distance between waves does not. > > Agreed. Only Wilson and Thomas can be in two places at once. What's the wavelength here: http://www.20days.co.uk/2008events/lanzarote-surfing.jpg Observation disproves your dogma. > >> >In the backward case, by the time the start of the first wave >> >has reached the sensor, the last wave has been created entirely >> >behind the original emission point. So you get ten waves with the >> >same wavelength even though the total distance the first wave covers >> >is only nine wavelengths. >> >> No, you haven't gotten it yet. > > [sigh] You are not going to ever get it, Wilson contradicts himself. He has senile dementia. >> >> >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. >> > >> >OK, what could I do to get the picture? Could you draw a picture of >> >what happens? I could try to draw it my way, and we could compare. >> >> I have. >> www.users.bigpond.com/hewn/ringgyro.exe >> >> Both the circles of dots are moving at c wrt the source (rotating >> frame). The blue and green dots represent a particular photon wave >> element. When they meet, one path is clearly longer than the other. >> The sine wave shows the nbmber of wavelengths in each path. > > I played that on one of the kids' computers. Your picture is just like > the one I would have made except you have more waves around the circle. > And it shows just what mine would have shown. The blue and green dots > start out at the same time and place and they reach the detector at the > same time and place. How do you figure they are out of phase? There are > more waves on the fast side than on the slow side if you count from the > nonrotating point where they first started. But why do you care about > that point? The emitter is rotating around the circle just like the > detector is. So count the number of waves on each side from the last one > the emitter has made to the last one the detector has consumed. If you > count them that way you get the same number in each direction. > >> I think the point you are missing is that this is a continuing >> process. The number of waves in each path only changes when the >> rotation speed changes. > > I agree. [sigh] Hopeless... >> >> >> >I imagine the emitter creatinu 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. >> > >> >OK. The number of wavecrests leaving the source in my example is ten >> >per second, in any frame. >> >> The number of wavecrests that pass any stationary point marked on the >> nonrotating ring is NOT ten. > > Yes. But why count the number that pass a stationary point when the > detector is moving? Isn't it wavecrests that pass the detector that > count? > >> If you can understand the SR 'explanation' you should be able to >> understand the BaTh one too. There is basically very little >> difference. > > The difference I see is that the SR explanation has the speed of light > constant in both directions. So their waves are out of phase when they > meet. The speed of light is c in the rotating frame. Therefore it HAS to be c+v and c-v in the nonrotating frame. Even SR says that. "But the ray moves relatively to the initial point of k, when measured in the stationary system, with the velocity c-v" >> >> 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. > > Agreed, no doppler shift. To get the phase different you'd have them get > out of phase by a constant amount and then they would all arrive at the > same speed but one side would be slow consistently by that constant > amount. But your moving picture does not show that. It shows them > arriving at the same time, every time. > >> >> >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. >> > >> >At this point in my imagination Androcles is saying the pathlengths >> >are history. Why do the pathlengths matter? >> >> Androcles is totally confused about Sagnac. He still thinks the >> detector is not rotating with the apparatus. > > His pictures don't show the detector standing still. > > At this point we agree about most of the facts. The only thing I don't > understand is why you say the waves in the different directions are out > of phase. You show each wave arriving at the detector at the same time. > How are they out of phase? Because he's senile, can't count and is totally confused. |