The Emission Theory of Androcles
in [Relativity]
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From: Henry Wilson, DSc on 13 Sep 2009 00:08 On Sat, 12 Sep 2009 20:09:28 -0400, Jonah Thomas <jethomas5(a)gmail.com> wrote: >hw@..(Henry Wilson, DSc) wrote: >> Jonah Thomas <jethomas5(a)gmail.com> wrote: >> >hw@..(Henry Wilson, DSc) wrote: >> >> 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. >> >> 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. >> >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. The frequency of a water wave depends on the speed of one's boat. The distance between waves does not. >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. >> >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 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 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. If you can understand the SR 'explanation' you should be able to understand the BaTh one too. There is basically very little difference. >> 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. >> > >> >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. Henry Wilson...www.users.bigpond.com/hewn/index.htm Einstein...World's greatest SciFi writer..
From: Henry Wilson, DSc on 13 Sep 2009 00:10 On Sun, 13 Sep 2009 09:02:23 +1000, "Inertial" <relatively(a)rest.com> wrote: >"Henry Wilson, DSc" <hw@..> wrote in message >news:0q3oa55eebsense5dshgrodrhbu94utal0(a)4ax.com... >> On Sat, 12 Sep 2009 16:35:01 +1000, "Inertial" <relatively(a)rest.com> >> wrote: >> >>>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. >> >> It is bleedingly obvious that the ring rotates during the time the light >> is in >> transit...in BOTH SR AND BATH. > >Yes there is .. derrrr. Is that the comment of an iimbecile? >> It is also bleedingly obvious that you are a clueless troll who is only >> here to >> waste our time. > >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. Henry Wilson...www.users.bigpond.com/hewn/index.htm Einstein...World's greatest SciFi writer..
From: Henry Wilson, DSc on 13 Sep 2009 00:18 On Sat, 12 Sep 2009 23:51:41 +0100, "Androcles" <Headmaster(a)Hogwarts.physics_o> wrote: > >"Henry Wilson, DSc" <hw@..> wrote in message >news:5k4oa5l5d03sf2brfp3c6gi165ebn9h2rd(a)4ax.com... >> On Sat, 12 Sep 2009 15:01:43 -0400, Jonah Thomas <jethomas5(a)gmail.com> >> wrote: >> >>>Fair enough. I'll look at ways for light particles to do interference. >>>If I find something I'll let you know. >> >> As Andro shows in his next post, electrons also diffract. They are >> particles. >> >>>> The emitter is a different molecule. The first molecule has to be >>>> recharged with energy before it can fire off a second pulse (in a >>>> different direction). >>>> It can only send a pulse. Imagine all you want to, there are NO light >>>> waves. Radio waves, yes, but no light waves. It's time you thought >>>> carefully instead of repeating the same old dogma. >>> >>>I'm not disputing you about this. I don't know what the truth is and I >>>have been looking at approaches that seem comfortable and that fit the >>>available facts I know about. I want to try imagining you are right and >>>try looking for ways to make the details work. >> >> Try my model. It seems pretty logical to me. >> Photons are particles that also oscillate intrinsically. When a photon is >> split >> into two and subsequently reunited, the phasing of the two halves >> determines >> the amount of reinforcement or annihilation. Simple, eh? >> >> Diffraction grating angles still operate on the 'intrinsic wavelength' >> principle, on the assumption that photons have a finite cross section. >> >>>Does a continuous laser send waves? It's based on the "principle" that a >>>"charged" molecule can be triggered to "discharge" by light of the >>>frequency it will send. So a single pulse can somehow set off at least >>>two others (being absorbed or partly absorbed by those two?) and you >>>wind up with a cascade that is all in phase but that starts at different >>>times. Does that count as a wave for you? >> >> It's a pulse of photons. Are they all exactly 'in phase' of not? What >> would 'in >> phase' actually mean? >> I would say it means their intrinsic oscillation is coordinated in some >> way by >> the lasing action but that would not necessarily cause them have the same >> intrinsic phase. >> >>>Again I'm not arguing that you're wrong about anything, if you say this >>>is a wave I'm not going to jump up and down and crow that you admitted >>>you were wrong about something. If this special case is a wave then I've >>>learned a little more about your system, and if it isn't a wave then I >>>have more questions. >> >> Have you ever seen a slow motion movie of a falling raindrop? It oscilates >> kind >> of like a dumbell, maybe with a few harmonics. >> >> A photon is obviously a particle that oscillates in some intrinsic way, >> giving >> it both particle and wavelike properties. >> >> Why can nobody else understand this very simple conceot? Am I more >> intelligent >> than everyone else here? >> > >You'll never catch up to me, you lack the mathematics. > http://www.youtube.com/watch?v=cXsvy2tBJlU You didn't compile that. >A photon alternates between electric and magnetic fields >with red representing +ve, blue -ve, yellow north and purple south. > http://www.androcles01.pwp.blueyonder.co.uk/photon.gif That's no good. You need to show it in 3D. > > [hanson writes] >quote/ > Another way to look at photon representation via a sinusoidal EM > parameter display would be by citing/using the **fundamental** > observation that/of > >A collapsing E-field generates an expanding M-field & visa > versa and these first principles / conservation laws say that > > 1) If there is no field of neither M nor E: Nothing happens > 2) If there is a field present but no change: Nothing happens. > 3) If there is a Magnetic Field that starts to collapse, an E field arises. > 4) If M becomes zero, the E will be max+ at pi/2, then > 5) E starts to collapse at p/2 down to 0 at pi while > M rises from 0 at pi/2 to max at pi... ...etc & analog to/till 2pi >/unquote > >It's no good showing youtube to the mild little bunny Jonah, though. He's >only got a toy game-playing computer for his kids and it took a while >before he'd admit he was thus handicapped. > I'd like to see him growl like a kitten for a change, then perhaps he'll >grow to roar like a lion and bite "Inertial's" inert head off. But there's slightly more hope for him than for that other idiot. Henry Wilson...www.users.bigpond.com/hewn/index.htm Einstein...World's greatest SciFi writer..
From: Inertial on 13 Sep 2009 01:09 "Henry Wilson, DSc" <hw@..> wrote in message news:cupoa5ljmlfmu5o0le74firrmeid8sf497(a)4ax.com... > On Sat, 12 Sep 2009 20:09:28 -0400, Jonah Thomas <jethomas5(a)gmail.com> > wrote: >>OK. The number of wavecrests leaving the source in my example is ten per >>second, in any frame. That's correct > The number of wavecrests that pass any stationary point marked on the > nonrotating ring is NOT ten. Yeup But the detector in Sagnac is NOT a stationary point It moves and so also gets 10 per second > If you can understand the SR 'explanation' you should be able to > understand the > BaTh one too. There is basically very little difference. Except the SR one predicts the observed result and ballistic doesn't. But other than that minor difference ...;) > Androcles is totally confused about Sagnac. He still thinks the detector > is not > rotating with the apparatus. You're such a hypocrite .. that is exactly what your analysis assumes as well. You've even said that how the ray arrive at the detector in Sagnac is not relevant. BAHAHAHAHA
From: Inertial on 13 Sep 2009 01:21
"Jonah Thomas" <jethomas5(a)gmail.com> wrote in message news:20090912204327.00ade286.jethomas5(a)gmail.com... > "Inertial" <relatively(a)rest.com> wrote: >> "Jonah Thomas" <jethomas5(a)gmail.com> wrote > >> > Well, if wavelength is the distance between wave fronts, then in >> > emission theories it is absolute and frame independent. It doesn't >> > matter what frame you look at it from, you get concentric circles >> > with even spacing. Your frame decides how fast the center of that >> > circle is moving. >> >> It depends on whether light is moving intrinsic oscillators, in which >> case wavelength varies (Dopler) for each observer, and frequency >> remains the same. Or if light is a wave, in which case frequency >> varies (Dopler) for each observer, and the wavelength remains the >> same. > > With emission theory wavelength remains the same. If light is a wave .. yes. If it is a moving intrinsic oscillator (as Henry claims) then no. Henry is not consistent between what his theory says about frequency and wavelength and what a photon is. Its probably best to just ignore his ideas of the nature of a photon and just look at a ballistic theory whose waves always leave the source at c wrt the source. In that case, wavelength is the same > Say the source is traveling at speed v in direction V. Yeup > Emit light in direction D at speed c at time 0 and location (0,0,0) Yeup > Then at time t, the light will be at t*(vV+cD) OK .. so the overall velocity of the light, in the observers frame of reference, is the vector sum of the velocity of light relative to the source (cD) and the velocity of the source (vV) > Imagine a spherical wavecrest leaving the source at time 0 and location > (0,0,0). Yeup > At time t that wavecrest will be in a circle with center tvV and radius > ct. Yeup > And without relativity, other inertial observers will also find the > circle to have radius ct. They will only disagree about whare the center > is. They'll all say it is centered on the source. > The same will be true for any other wavecrests, they will all have the > same center, tvV in the frame that says the velocity is vV. They will be > concentric and if they are created at regular intervals the distance > between wavecrests will be the same for everybody. Yeup > Am I wrong? Nope .. that's just fine (assuming ballistic theory and Galilean transforms and Newtonian physics etc). All observers would see concentric spherical wave-crests, centered at the source moving outward from the source at speed c. So the whole arrangement of expanding concentric spherical wave-front 'shells' moves with the source. |