The Emission Theory of Androcles
in [Relativity]
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From: Jonah Thomas on 17 Sep 2009 07:04 hw@..(Henry Wilson, DSc) wrote: > Jonah Thomas <jethomas5(a)gmail.com> wrote: > > Ok, I have now worked out why there is confusion about this. > > The fact is, the phase shift already exists at the detector before a > particular photon leaves. The difference originated DURING previous > CHANGES in rotation speed..... as did the different path lengths. So, > at constant rotation speeds, we don't want any new light to change the > status quo. We want the split photons to arrive IN PHASE so the > existing beams remain as they were. > > I'll try to draw this in linear form. > > At constant speed let the broad beams of the two paths be represented > like this(the beams supposedly use coherent light): > > S /\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\ D > S /\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/v D > > They are in phase at the source but out of phase at the detector > because of the different path lengths and the invariant wavelength of > the light used. I drew pictures and found that the way I was thinking of it was wrong. The way you drew the picture was right. The alternative way that Inertial and I were thinking went more like this: S /\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\ D S/\/\/\/\/\/\/\/\/\/\/\ Here the beginning is in phase, and the end is in phase, but the waves are now out of phase at the point D where they both started -- which does not matter. But the reality for our alternative is: S /\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\ D S\/\/\/\/\/\/\/\/\/\/\ They would be in phase at the end and at the spot they started at (if you extend the shorter wave back to that spot), but they are not in phase at the source. Here are the pictures: http://i847.photobucket.com/albums/ab31/jehomas/speedwave4.gif http://i847.photobucket.com/albums/ab31/jehomas/speedwave6.gif Once we assume constant wavelength, it is absurd to have the waves get out of phase at the actual source. Wilson's alternative is the only one that can make sense, unless we find a way to change hidden assumptions I didn't notice I was making. If Wilson's approach doesn't work either then it will probably turn out that it simply does not make sense to have waves with constant wavelength in this circumstance.
From: Jonah Thomas on 17 Sep 2009 07:15 hw@..(Henry Wilson, DSc) wrote: > Jonah Thomas <jethomas5(a)gmail.com> wrote: > >The different velocities will tend to cancel out as they go around > >the circle, from the different inertial observers' points of view. > > > >Except, no, they won't. This model calls for the light in different > >directions to have a constant velocity of c+v or c-v. As it goes > >around the circle it won't change velocity. An observer who is also > >traveling at v would see the c+v light go at c, later it would speed > >up to c+2v and go back to c. > where did you get that idea? Inertial started to look at the situation from the POV of different inertial observers. Try it! Let's do Sagnac with 4 mirrors in a square. At one moment, the way we see the speeds of the light is: c+v +-----<-+ | | c+v | | c+v | | +->-----+ c+v But to an observer who is traveling at v heading to the right, it will look like c+2v +-----<-+ | | c+v | | c+v | | +->-----+ c Or something kind of similar. Right? Not that there's anything wrong with that....
From: Inertial on 17 Sep 2009 09:51 "Jonah Thomas" <jethomas5(a)gmail.com> wrote in message news:20090917070442.32aac7c7.jethomas5(a)gmail.com... > hw@..(Henry Wilson, DSc) wrote: >> Jonah Thomas <jethomas5(a)gmail.com> wrote: >> >> Ok, I have now worked out why there is confusion about this. >> >> The fact is, the phase shift already exists at the detector before a >> particular photon leaves. The difference originated DURING previous >> CHANGES in rotation speed..... as did the different path lengths. So, >> at constant rotation speeds, we don't want any new light to change the >> status quo. We want the split photons to arrive IN PHASE so the >> existing beams remain as they were. >> >> I'll try to draw this in linear form. >> >> At constant speed let the broad beams of the two paths be represented >> like this(the beams supposedly use coherent light): >> >> S /\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\ D >> S /\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/v D >> >> They are in phase at the source but out of phase at the detector >> because of the different path lengths and the invariant wavelength of >> the light used. > > I drew pictures and found that the way I was thinking of it was wrong. Don't let him trick you > The way you drew the picture was right. Nope > The alternative way that > Inertial and I were thinking went more like this: > > S /\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\ > D S/\/\/\/\/\/\/\/\/\/\/\ I don't follow you .. D is behind S there I think you meant at the start (just before rays emitted) we have: S D S D One third of the way thru we have this for the two rays, where s is the stationary point s we marked in the non-rotating frame, and R is the leading edge of the ray (ie the photon/wave/whatever that was first emitted): s S/\/\/\/R D S/s/\/\/R D You can S has been making more photons/waves/whatevers since R .. they come from S's current position, not from s!! Two thirds of the way through we have even more photons/waves/whatevers from S: s S/\/\/\/\/\/\/\/R D S/\/s/\/\/\/\/\/R D At the end we have s S/\/\/\/\/\/\/\/\/\/\/\/D S/\/\/s/\/\/\/\/\/\/\/\/D The rays arrive at D in phase, they are still in phase at the source S as well. What happens at s doesn't make any difference !!!! > http://i847.photobucket.com/albums/ab31/jehomas/speedwave4.gif > http://i847.photobucket.com/albums/ab31/jehomas/speedwave6.gif You are showing the waves (in the rotating frame point of view) continually emitted a source point that changes distance from D (ie our point s above). There is NO SUCH SOURCE OF WAVES IN SAGNAC !!!! > Once we assume constant wavelength, it is absurd to have the waves get > out of phase at the actual source. Wilson's alternative is the only one > that can make sense, No .. it doesn't > unless we find a way to change hidden assumptions I > didn't notice I was making. > > If Wilson's approach doesn't work either then it will probably turn out > that it simply does not make sense to have waves with constant > wavelength in this circumstance. Why is all this so hard for you and Henry?
From: Jonah Thomas on 17 Sep 2009 12:46 "Inertial" <relatively(a)rest.com> wrote: > "Jonah Thomas" <jethomas5(a)gmail.com> wrote > > hw@..(Henry Wilson, DSc) wrote: > >> At constant speed let the broad beams of the two paths be > >represented> like this(the beams supposedly use coherent light): > >> > >> S /\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\ D > >> S /\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/v D > >> > >> They are in phase at the source but out of phase at the detector > >> because of the different path lengths and the invariant wavelength > >of> the light used. > > > > I drew pictures and found that the way I was thinking of it was > > wrong. > > Don't let him trick you I just did the math. > > The way you drew the picture was right. > > Nope > > > The alternative way that > > Inertial and I were thinking went more like this: > > > > S /\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\ > > D S/\/\/\/\/\/\/\/\/\/\/\ > > I don't follow you .. D is behind S there Yes, because the emitter is moving with the detector. By the time the wavefront which was emitted at D reaches the detector the source has moved behind D on the forward side and ahead of D on the back side. That's why the distances are different. > I think you meant at the start (just before rays emitted) we have: > > S D > S D > > One third of the way thru we have this for the two rays, where s is > the stationary point s we marked in the non-rotating frame, and R is > the leading edge of the ray (ie the photon/wave/whatever that was > first emitted): > > s S/\/\/\/R D > S/s/\/\/R D > > You can S has been making more photons/waves/whatevers since R .. they > come from S's current position, not from s!! Yes, exactly. > Two thirds of the way through we have even more > photons/waves/whatevers from S: > > s S/\/\/\/\/\/\/\/R D > S/\/s/\/\/\/\/\/R D > > > At the end we have > s S/\/\/\/\/\/\/\/\/\/\/\/D > S/\/\/s/\/\/\/\/\/\/\/\/D > > The rays arrive at D in phase, they are still in phase at the source S > as well. > > What happens at s doesn't make any difference !!!! They are in phase at D in the model you use, because you assumed they would stay in phase and required them to do so. In the model Wilson uses, they do not need to be in phase. His model works just fine. Unless I made an arithmetic mistake his model works, it gets a phase shift. I haven't checked whether it is the right phase shift. If you want to argue about whether his assumptions are unreasonable we can, but I don't think it's arguable whether he gets a phase shift or not. > > http://i847.photobucket.com/albums/ab31/jehomas/speedwave4.gif > > http://i847.photobucket.com/albums/ab31/jehomas/speedwave6.gif > > You are showing the waves (in the rotating frame point of view) > continually emitted a source point that changes distance from D (ie > our point s above). There is NO SUCH SOURCE OF WAVES IN SAGNAC !!!! ?? In Sagnac, when the detector moves the source moves with it, right? So one source appears to get farther from D while the other appears to get closer. Because by the time the light that is getting emitted later reaches D, D will be farther away. > > Once we assume constant wavelength, it is absurd to have the waves > > get out of phase at the actual source. Wilson's alternative is the > > only one that can make sense, > > No .. it doesn't It might not, but as I understand it yours makes no sense at all with constant wavelength. You can argue that it can't be constant wavelength, or you can show me how I misunderstood your constant-wavelength model. > > unless we find a way to change hidden assumptions I > > didn't notice I was making. > > > > If Wilson's approach doesn't work either then it will probably turn > > out that it simply does not make sense to have waves with constant > > wavelength in this circumstance. > > Why is all this so hard for you and Henry? Wilson thinks he knows something you don't, but he has trouble explaining it. I'm trying to understand what he's saying and whether he can be right. My previous objection was wrong. I thought I was agreeing with you, and if so you were wrong too. But I may have misunderstood you just like I misunderstood him for so long. Finding that the reason I thought he was wrong was itself wrong doesn't make him right. But it leaves open the possibility. I don't know whether I'll find other objections. And of course, his model can work and still not fit the reality. But I was wrong to say that he couldn't have his Sagnac waves get out of phase. This is progress.
From: Henry Wilson, DSc on 17 Sep 2009 12:55
On Thu, 17 Sep 2009 03:19:31 -0700 (PDT), Jerry <Cephalobus_alienus(a)comcast.net> wrote: >On Sep 17, 3:44�am, hw@..(Henry Wilson, DSc) wrote: >> On Wed, 16 Sep 2009 23:32:16 -0700 (PDT), Jerry >> <Cephalobus_alie...(a)comcast.net> wrote: >> >On Sep 16, 9:50�pm, hw@..(Henry Wilson, DSc) wrote: >> >> >> A BLOODY RING GYRO MEASURES ABSOLUTE ROTATION ANGLES BY INTEGRATING FRINGE >> >> MOVEMENT DURING A CHANGE IN ROTATIONAL SPEED. THE FRINGE DISPLACEMENT PERFORMS >> >> THE INTEGRATION AUTOMATICALLY. >> >> >Very poor choice of words. Fringe displacement is proportional >> >to rotational speed, and ny integrating the fringe displacement >> >over time, one may obtain the rotation angle. >> >> >The term "fringe movement" implies the first derivative of fringe >> >displacement, and "a change in rotational speed" of course means >> >rotational acceleration. The integral of "fringe movement" is >> >fringe displacement, which of course is proportional to >> >rotational speed. >> >> >By confusing your terminology while yelling, you leave the >> >distinct impression that you are mixed up. >> >> >Which of course you are... >> >> Great try Jerry. >> I realise you didn't want to say straight out that 'inertial' is a complette >> idiot because she is on YOUR side. >> But even YOU must admit that her claim, quote: " there's no speed changing in >> Sagnac. �It rotates at a constant rate" truly epitomises the sad state of the >> relativist mentality. > >Henri, Inertial was discussing the Sagnac EXPERIMENT, when, >completely unannounced, you suddenly started referring to the >capabilities of modern IFOGs. > >In his original experiment, Sagnac typically spun his apparatus >at 1- to 2+ times per second, recording displacements of several >hundredths of a fringe. >http://gallica.bnf.fr/ark:/12148/bpt6k31103.pleinepage.f1410 Is your "1- to 2+ times per second" a constant speed? Both you and inertial don't seem to understand yat the fringes only MOVE during a speed change. The displacement at any CONSTANT speed can only be determined by counting the 'number of fringe widths' a particular fringe moves sideways DURING a speed change. >This is a far cry from modern fibre optic gyros and ring gyros >used in inertial guidance systems, which are capable of detecting >rotational rates down to 0.00001 degree per hour. Please tell Andro why they use lots of turns. Laser systems modulate the signal to get a more sensitive AC output. >Sagnac's original apparatus could not possibly have been used for >monitoring one's heading in an inertial guidance system. So yes, >it is not he, but YOU who is either being massively dishonest >in changing the premises of the discussion, or is simply confused. inertial made the statement that 'ring gyros never change rotation speed'. that was the stupidest thing she has uttered yet and proves she doesn't have a clue what she's talking about. > >Which is it? > >Jerry Henry Wilson...www.users.bigpond.com/hewn/index.htm Einstein...World's greatest SciFi writer.. |