The Emission Theory of Androcles
in [Relativity]
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From: Inertial on 13 Sep 2009 01:37 "Henry Wilson, DSc" <hw@..> wrote in message news:igpoa55gccc3161c7t51mn7nrk7ongvi79(a)4ax.com... > On Sun, 13 Sep 2009 09:04:34 +1000, "Inertial" <relatively(a)rest.com> > wrote: > >>"Henry Wilson, DSc" <hw@..> wrote in message >>news:eh3oa5l03f7gt81jodls7fdhmkrleeoeg4(a)4ax.com... >>> On Sat, 12 Sep 2009 16:43:23 +1000, "Inertial" <relatively(a)rest.com> >>> wrote: > >>>>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. >>> >>> Hahahahahahahh! >>> You are moving the start point in the nonrotating frame. >> >>I didn't move any start point. The cylinder is the non-rotating frame. >> >>> Typical false >>> relativist logic! >> >>Typical Henry unable to follow a simple example >> >>> http://www.mathpages.com/rr/s2-07/2-07.htm >>> >>> One path is longer than the other. >> >>Yes it is .. that's what I just showed >> >>> One length of rope is longer than the other >>> no matter how fast you pull it. >> >>That's right .. that's what i just showed. >> >>And the heads of the ropes arrive at the same place at the same time and >>so >>the marks on the rope (the wavelengths) is in phase. > > Are you saying the ropes are flexible? No. I would have used rubber bands if I wanted that. > If one rope is longer than the other, there must be a different number of > turns > in the two. Turns? You mean the marks on the rope I described .. yes there are. Let say we put the marks on the rop at regular intervals from the head of the rope .. then when the ropes meet the detector point at the same time, the marks on the heads line up. > That means there is a phase shift where they meet. No .. it doesn't. The marks on the rope move WITH THE ROPE. > I really think this is too hard for you... Not at all .. you're the only one who is inconsistent and unable to understand basic physics. OK .. you and Androcles.
From: Inertial on 13 Sep 2009 01:38 "Henry Wilson, DSc" <hw@..> wrote in message news:klpoa5l61ni29vjgv428qdvn8u4aoa5sip(a)4ax.com... > On Sun, 13 Sep 2009 09:01:16 +1000, "Inertial" <relatively(a)rest.com> > wrote: > >>"Henry Wilson, DSc" <hw@..> wrote in message >>news:qp2oa5he52gtd4d0tp2c2gof0ntqd0jdoe(a)4ax.com... >>> On Sat, 12 Sep 2009 02:24:47 -0400, Jonah Thomas <jethomas5(a)gmail.com> > >>> One path is longer than the other. The number of turns in each 'path' is >>> different no matter how fast you spin the spring. >>> >>>>Similarly in the other direction. The first backward wave is extruded >>>>slowly, so that at time = 0.1 its forward edge is at 9.1 and its leading >>>>edge is at the emitter at 0.1. Then the second one goes slowly, and at >>>>time 0.2 the forward edge of the first wave is at 8.2 and the end of the >>>>second wave is at 0.2. And by the time we get to 1 second, the leading >>>>edge of the first wave is at 1.0 and the end of the tenth wave is also >>>>at 1.0. Exactly ten waves for that one too. >>>> >>>>If we were wrapping wire around a torus and we had to cover it until we >>>>could quit, then at 1.1c we'd have to cover part of it twice while for >>>>0.9c we wouldn't cover the whole thing. But our wire is sliding along >>>>the torus and it's the first wave of the fast side that covers the same >>>>area twice, and the area that the first wave of the slwo side doesn't >>>>get to, the last wave has backed onto. >>>> >>>>Ten waves both times. No phase shift. >>> >>> No, you don't get the picture at all. >>> >>>>> >I imagine the emitter creating a wave that moves at 1.1c while the >>>>> >emitter itself moves at 0.1c. There are 10 waves present covering the >>>>> >distance around the circle from the emitter to the detector which is >>>>> >in basicly the same place. A new one is being created while the >>>>> >oldest one travels just enough faster than the detector that it is >>>>> >completely consumed by the time the new wave is completely created. >>>>> > >>>>> >Meanwhile, the emitter creates a second wave that moves at 0.9c while >>>>> >the emitter moves away at 0.1c. There are 10 waves present covering >>>>> >the distance around the circle from the emitter backward to the >>>>> >detector. A new wave is being created while the oldest one travels >>>>> >just fast enough into the incoming detector that it is completely >>>>> >consumed by the time the new wave is completely created. >>>>> > >>>>> >Yes, wavelength is absolute and frame independent. If you can look at >>>>> >them, you can measure them. They're in place at any moment of time, >>>>> >ready to be measured. In the absence of length contraction everybody >>>>> >will measure them the same length. >>>>> > >>>>> >And frequency at the source is also absolute and frame independent. >>>>> >You can watch the source create its periodic motion, and everybody >>>>> >gets the same result apart from things like doppler shift which can >>>>> >be easily corrected. >>>>> >>>>> What does 'frequency' mean when applied to light? >>>> >>>>I'm assuming something periodic going on, and in that case the frequency >>>>is the number of times the periodic thing happens per unit time. In my >>>>example that's ten times per second. >>> >>> Nope. Frequency is implied as the 'number of wavecrests' arriving per >>> second. >>> f = hc/lambda. >> >>If light is a little particle, what is a wavecrest? > > A photon is like a'serated bullet' due to its intrinsic oscillation... > whatever > that is. > Its inferred 'frequency' is the number of serations arriving per second. So .. If light is a little particle, what is a wavecrest? >>> I reckon a photon also has an intrinsic oscillation and a natural >>> frequency but >>> that it not the one that we normally detect. >> >>Then what is being detected? Do you have ANY idea? > > Yes. Obviously not one you're able to express.
From: Inertial on 13 Sep 2009 01:41 "Henry Wilson, DSc" <hw@..> wrote in message news:e2soa5958456b2fc0umod81cdg7t2bkmkt(a)4ax.com... > On Sun, 13 Sep 2009 09:02:23 +1000, "Inertial" <relatively(a)rest.com> > wrote: >>So ..that is your argument to the oh-so-obvious result that ballistic >>theory >>cannot give a phase shift because the waves arrive at the same time. > > The fact that the waves arrive at the same time merely tells us the > bleedingly > obvious fact that the fringes DO NOT MOVE during constant rotation. No .. it means that the waves MUST BE IN PHASE because the leading edges of the waves, which are ALWAYS IN PHASE arrive at the detector AT THE SAME TIME.
From: Inertial on 13 Sep 2009 01:43 "Henry Wilson, DSc" <hw@..> wrote in message news:89soa5pqn3ej8av2v35muq42cpm9pahok3(a)4ax.com... > But there's slightly more hope for him than for that other idiot. There is indeed no hope for the other idiot named Henry.
From: Jonah Thomas on 13 Sep 2009 03:30
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. > >> >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. > >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] > >> >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. > >> >> >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. > >> 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? |