The Emission Theory of Androcles
in [Relativity]
|
From: Henry Wilson, DSc on 12 Sep 2009 01:25 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. >> or that >> he is wrong. > >Depends on what about. > >> I do not understand what he is saying, > >I don't think he does either .. he alternates between saying light rays are >moving intrinsic oscillators and that the are waves. He doesn't seem to >know, and uses whichever one lets him argue with someone else > >> and it's quite possible that >> there's a way to look at this which I have missed which gives the >> conclusions he claims. I just don't understand it yet. > >Actually, it appears you do understand quite well. Which is why you also >find that if you follow through the analysis correctly, you get (as every >other physicist has done since Sagnac himself) that a ballistic theory >(where the waves in the non-rotating frame have c+v and c-v velocities) >gives you a zero Sagnac effect. It doesn't >Henry's logic is flawed because he only analyses what a fixed detector in >the non-rotating frame would detect, not the actual moving detector of >Sagnac. He even says that what the moving detector detects is irrelevant to >Saganc !! I mean, really .. how can the actual observed effect that one is >trying to analyze be irrelevant to an analysis? The moving detector receives the same number of waves per second from both rays. There is no doppler shift. That is why the fringe pattern is steady for constant rotation. That does say anythihg about the phasing of the arriving waves. You are incapable of learning anything. Henry Wilson...www.users.bigpond.com/hewn/index.htm Einstein...World's greatest SciFi writer..
From: Jonah Thomas on 12 Sep 2009 02:24 hw@..(Henry Wilson, DSc) wrote: > Jonah Thomas <jethomas5(a)gmail.com> wrote: > >> >OK, let me try this again. I'll put ridiculous numbers on it that > >I> >hope are easy to work with. > >> > > >> >Let's say that our light is 10 Hertz and the path is 1 > >light-second> >long. > > > >10 hertz, one light-second. > > > >> >And then we get it rotating at c/10. > > > >Rotating at c/10. > > > >> Now the light in the forward > >> >direction travels at 1.1c while the light in the back direction > >> >travels at 0.9c. But the apparatus itself is moving at 0.1c, so in > >> >the same time that it previously took for the light to go from > >source> >to target it still goes from source to target in both > >directions. But> >the light in one direction travels 10% farther, > >while the light in> >the other direction travels only 90% as far. > > > >> Right so far. The path lengths are different and the travel times > >are> the same IN THE INERTIAL FRAME. > > > >OK so far. > > > >> >How many cycles have they gone? The same number, 10 cycles. The > >light> >is 10 hertz, so in 1 second they each send out 10 waves. > > > >> Now you are in trouble. You have moved into the rotating frame, > >which> is full of traps. For instance, if you mark the point where a > >> particular wave element is emitted, on the NON rotating frame, that > >> mark moves backwards in the rotating frame. > > > >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? > >guess it would be moving backward in the rotating frame. In the > >nonrotating frame the emitter is moving forward, which is backward > >for the wave that's heading to the back. This is what lets you keep > >the wavelength the same even though the frequency is also the same > >and the speed is different. > > > >I'm not quite clear how I moved into the rotating frame, though. Ten > >times a second a new wave starts. That's frequency, isn't it? That's > >the frequency at the emitter. What's the frequency at the detector? > >If it's true that each spot on each wave reaches the detector from > >both sides in exactly 1 second, and there are 10 of them in each > >second, that would say the frequency at the detector should be 10 per > >second too. Whether you count it from the nonrotating frame or the > >rotating frame either one, that's the frequency at the detector. > > > >> The path lengths are again DIFFERENT just as they were in the > >> nonrotating frame. In the rotating frame, photons have the same > >> frequency and because both rays move at c in that frame, the > >elements> that go clockwise doesn't meet their other halves at the > >detector. The> ones that do meet are out of phase. > > > >Um. Pick one that starts at time 0. A wave crest starts in both > >directions. It travels at 1.1c in one direction but in 0.9c in the > >other direction. However, the detector moves, so that the one going > >at 1.1c travels travels 1.1 light seconds while the one going at 0.9c > >travels 0.9 light seconds. So they both take 1 second to reach the > >detector. And at that point they are in phase. Pick any other spot on > >the cycle from wave crest to wave crest and the same thing should > >happen. Pick a spot where the electric field crosses zero going from > >positive to negative. in one direction that travels at 1.1c for 1.1 > >light seconds, in the other direction it travels at 0.9c for 0.9 > >light seconds, taking 1 second for both. It looks to me like they are > >in phase when they reach the detector. I want to try to do this > >entirely in the nonrotating frame first, and never slip into the > >rotating frame. > > > >> That is too difficult for Jerry and Inertial but I expect YOU will > >be> able to understand it. > > > >I'm glad to keep trying but I don't get it yet. > > 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. 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. 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. > >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. > >Frequency at an observer can change depending on the observer's > >velocity. > > > >I notice that the wave which has a fixed wavelength and a fixed > >freqency is different from the path of any particular particle > >composing the wave. The particles can move at a different speed and > >in a different direction. I don't know the implications of that yet. > > > >> >Say you have one wave in each direction starting at time zero, > >> >they should both reach the end at time 0.1 second. Exactly nine > >more> >should reach the end by time 1 second, in both directions. > > > >> See above. You are jumping frames again. You are describing a > >> nonrotating sagnac interferometer. > > > >When it rotates they still create ten new waves per second. It's > >still true that ten new waves arrive at the detector every second. > >Each new wave crest arrives at the same time from both sides. Doesn't > >it? > > 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.
From: Inertial on 12 Sep 2009 02:29 "Henry Wilson, DSc" <hw@..> wrote in message news:vc8ma59oiojm1aoaperkke10ssdam2dsiv(a)4ax.com... > On Sat, 12 Sep 2009 10:47:46 +1000, "Inertial" <relatively(a)rest.com> > wrote: > >>"Henry Wilson, DSc" <hw@..> wrote in message >>news:7sjla55q418ve3t49ftvd1rvfk7v8njuau(a)4ax.com... >>> On Thu, 10 Sep 2009 23:13:26 -0700 (PDT), Jerry > >>>>> Either way, you end up with a null Sagnac result. >>>> >>>>Notice, however, that Henri fantasizes about intrinsic >>>>oscillators that are NOT observed to have the same frequency... >>>>http://mysite.verizon.net/cephalobus_alienus/henri/HWFantasy.htm >>> >>> Poor old Crank. ...has never heard of doppler shift. >> >>Intrinsic oscillators have wavelength doppler shifted, with frequency >>constant. >> >>Waves have frequency doppler shifted, with wavelength constant. >> >>Unless you have SR for things travelling at c and you get both shifted >>(which is what we observe for light) >> >>> Obviously the frequencies >>> are different in the inertial frame >> >>So its not an intrinsic oscillator, its a wave front. And waves fronts >>that >>arrive at a given point at the same time are in phase > > Forget wavefronts. Photons are individual oscillating particles. Then it has the same frequency in every frame of reference. As I said .. you keep changing your story because your 'theory' is self-contradictory .. as soon as you're backed into one corner showing it is wrong working one way, you change it to the exact opposite and the same thing happens again. That's why Jonah can't keep up with you,, you're can't say anything and stick with it .. a clear sogn of your dishonesty.
From: Inertial on 12 Sep 2009 02:35 "Henry Wilson, DSc" <hw@..> wrote in message news:0u8ma55ec24lv1d3urj2sl2qo8bh3rin4u(a)4ax.com... > On Fri, 11 Sep 2009 22:17:47 -0400, Jonah Thomas <jethomas5(a)gmail.com> > wrote: > >>Once more dear friends into the breach.... >> >>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: >>> >>> >> In the NON-ROTATING >>> >> FRAME the frequencies of the rays are doppler shifted in opposite >>> >> directions. We are using that frame for our analysis. This is very >>> >> basic physics ...but clearly too hard for the relativist mentality. >>> > >>> >OK, I'll try to stay with the nonrotating frame. I'm trying to >>> >understand this, it's just easy for me to mess us. >> >>> >> >The wavelength is the >>> >> >same because you don't measure wavelength back toward the source >>> >when> >it emitted the wave, you measure it in the direction of the >>> >> >wavefront. So in a time interval t units long, one side emits n >>> >> >cycles at speed c+v and the other side emits n cycles at speed >>> >c-v.> >Both arrive at the sensors at the same time. During the time >>> >for one> >wave to pass from the c+v side, one wave will pass from the >>> >c-v side> >too, slower. I don't see that this gives us a phase shift >>> >or a> >frequency difference or anything for an interferometer to pick >>> >up.> >>> >> ...because you are jumping from one frame to another. If you try to >>> >> use the rotating frame, there is an imaginary time factor, that I >>> >> tried to explain before. >>> >> In the rotating frame, the emission point of a particular element >>> >> MOVES BACKWARDS. >>> > >>> >OK, let me try this again. I'll put ridiculous numbers on it that I >>> >hope are easy to work with. >>> > >>> >Let's say that our light is 10 Hertz and the path is 1 light-second >>> >long. >> >>10 hertz, one light-second. >> >>> >And then we get it rotating at c/10. >> >>Rotating at c/10. >> >>> Now the light in the forward >>> >direction travels at 1.1c while the light in the back direction >>> >travels at 0.9c. But the apparatus itself is moving at 0.1c, so in >>> >the same time that it previously took for the light to go from source >>> >to target it still goes from source to target in both directions. But >>> >the light in one direction travels 10% farther, while the light in >>> >the other direction travels only 90% as far. >> >>> Right so far. The path lengths are different and the travel times are >>> the same IN THE INERTIAL FRAME. >> >>OK so far. >> >>> >How many cycles have they gone? The same number, 10 cycles. The light >>> >is 10 hertz, so in 1 second they each send out 10 waves. >> >>> Now you are in trouble. You have moved into the rotating frame, which >>> is full of traps. For instance, if you mark the point where a >>> particular wave element is emitted, on the NON rotating frame, that >>> mark moves backwards in the rotating frame. >> >>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. > >>guess it would be moving backward in the rotating frame. In the >>nonrotating frame the emitter is moving forward, which is backward for >>the wave that's heading to the back. This is what lets you keep the >>wavelength the same even though the frequency is also the same and the >>speed is different. >> >>I'm not quite clear how I moved into the rotating frame, though. Ten >>times a second a new wave starts. That's frequency, isn't it? That's the >>frequency at the emitter. What's the frequency at the detector? If it's >>true that each spot on each wave reaches the detector from both sides in >>exactly 1 second, and there are 10 of them in each second, that would >>say the frequency at the detector should be 10 per second too. Whether >>you count it from the nonrotating frame or the rotating frame either >>one, that's the frequency at the detector. >> >>> The path lengths are again DIFFERENT just as they were in the >>> nonrotating frame. In the rotating frame, photons have the same >>> frequency and because both rays move at c in that frame, the elements >>> that go clockwise doesn't meet their other halves at the detector. The >>> ones that do meet are out of phase. >> >>Um. Pick one that starts at time 0. A wave crest starts in both >>directions. It travels at 1.1c in one direction but in 0.9c in the other >>direction. However, the detector moves, so that the one going at 1.1c >>travels travels 1.1 light seconds while the one going at 0.9c travels >>0.9 light seconds. So they both take 1 second to reach the detector. And >>at that point they are in phase. Pick any other spot on the cycle from >>wave crest to wave crest and the same thing should happen. Pick a spot >>where the electric field crosses zero going from positive to negative. >>in one direction that travels at 1.1c for 1.1 light seconds, in the >>other direction it travels at 0.9c for 0.9 light seconds, taking 1 >>second for both. It looks to me like they are in phase when they reach >>the detector. I want to try to do this entirely in the nonrotating frame >>first, and never slip into the rotating frame. >> >>> That is too difficult for Jerry and Inertial but I expect YOU will be >>> able to understand it. >> >>I'm glad to keep trying but I don't get it yet. > > 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). Why? > Now we (and SR) have established that the emission and the detection > points are > separated by the distance vt. Though in SR there are two detection points, due to the difference in time (and therefore phase). There is one detection point in Ballistic theory because the time is the same and so there is not difference in phase. This is so bleedingly obvious, yet for years you've lied about it. [snip henry nonsense]
From: Inertial on 12 Sep 2009 02:43
"Jonah Thomas" <jethomas5(a)gmail.com> wrote in message news:20090912022447.6fe91124.jethomas5(a)gmail.com... > hw@..(Henry Wilson, DSc) wrote: >> Jonah Thomas <jethomas5(a)gmail.com> wrote: > >> >> >OK, let me try this again. I'll put ridiculous numbers on it that >> >I> >hope are easy to work with. >> >> > >> >> >Let's say that our light is 10 Hertz and the path is 1 >> >light-second> >long. >> > >> >10 hertz, one light-second. >> > >> >> >And then we get it rotating at c/10. >> > >> >Rotating at c/10. >> > >> >> Now the light in the forward >> >> >direction travels at 1.1c while the light in the back direction >> >> >travels at 0.9c. But the apparatus itself is moving at 0.1c, so in >> >> >the same time that it previously took for the light to go from >> >source> >to target it still goes from source to target in both >> >directions. But> >the light in one direction travels 10% farther, >> >while the light in> >the other direction travels only 90% as far. >> > >> >> Right so far. The path lengths are different and the travel times >> >are> the same IN THE INERTIAL FRAME. >> > >> >OK so far. >> > >> >> >How many cycles have they gone? The same number, 10 cycles. The >> >light> >is 10 hertz, so in 1 second they each send out 10 waves. >> > >> >> Now you are in trouble. You have moved into the rotating frame, >> >which> is full of traps. For instance, if you mark the point where a >> >> particular wave element is emitted, on the NON rotating frame, that >> >> mark moves backwards in the rotating frame. >> > >> >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? > >> >guess it would be moving backward in the rotating frame. In the >> >nonrotating frame the emitter is moving forward, which is backward >> >for the wave that's heading to the back. This is what lets you keep >> >the wavelength the same even though the frequency is also the same >> >and the speed is different. >> > >> >I'm not quite clear how I moved into the rotating frame, though. Ten >> >times a second a new wave starts. That's frequency, isn't it? That's >> >the frequency at the emitter. What's the frequency at the detector? >> >If it's true that each spot on each wave reaches the detector from >> >both sides in exactly 1 second, and there are 10 of them in each >> >second, that would say the frequency at the detector should be 10 per >> >second too. Whether you count it from the nonrotating frame or the >> >rotating frame either one, that's the frequency at the detector. >> > >> >> The path lengths are again DIFFERENT just as they were in the >> >> nonrotating frame. In the rotating frame, photons have the same >> >> frequency and because both rays move at c in that frame, the >> >elements> that go clockwise doesn't meet their other halves at the >> >detector. The> ones that do meet are out of phase. >> > >> >Um. Pick one that starts at time 0. A wave crest starts in both >> >directions. It travels at 1.1c in one direction but in 0.9c in the >> >other direction. However, the detector moves, so that the one going >> >at 1.1c travels travels 1.1 light seconds while the one going at 0.9c >> >travels 0.9 light seconds. So they both take 1 second to reach the >> >detector. And at that point they are in phase. Pick any other spot on >> >the cycle from wave crest to wave crest and the same thing should >> >happen. Pick a spot where the electric field crosses zero going from >> >positive to negative. in one direction that travels at 1.1c for 1.1 >> >light seconds, in the other direction it travels at 0.9c for 0.9 >> >light seconds, taking 1 second for both. It looks to me like they are >> >in phase when they reach the detector. I want to try to do this >> >entirely in the nonrotating frame first, and never slip into the >> >rotating frame. >> > >> >> That is too difficult for Jerry and Inertial but I expect YOU will >> >be> able to understand it. >> > >> >I'm glad to keep trying but I don't get it yet. >> >> 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. Better picture. Cylinder. Two marks on the cylinder for start and end point. Put a hole at the start point, inside the cylinder have two two ropes with equidistant marks on them (representing the wavelengths). Pull the ropes from the hole at the start point at two different rates so they take the same time to go around the cylinder in opposite directions and end up at the end point. See how the wavelength stays the same, but the points between which we are measureing the length moves around the cylinder with the rope. When the two ropes reach the end point at the same time, you have the same leading mark on each rope lined up. No phase shift. For an SR-like view, do the same thing, but pull the ropes at the same speed. They arrive at the end point at different times, and the marks on the ropes don't like up. Phase shift. > 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. > > 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. > > 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. Funny how these ballistic explanations .. whether with light as waves, or as moving oscillators, or springs or ropes .. still give no phase shift. But Henry still insists that they do. |