The Emission Theory of Androcles
in [Relativity]
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From: Inertial on 17 Sep 2009 23:33 "Jonah Thomas" <jethomas5(a)gmail.com> wrote in message news:20090917215927.5bb32949.jethomas5(a)gmail.com... > "Inertial" <relatively(a)rest.com> wrote: >> "Jonah Thomas" <jethomas5(a)gmail.com> wrote >> > "Inertial" <relatively(a)rest.com> wrote: >> >> "Jonah Thomas" <jethomas5(a)gmail.com> wrote > >> >> > 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. >> >> No .. the emitter (as you said) moves with the detector. The path >> from emitter to detector is the SAME LENGTH all the time !! > > Yes, but the path that light must take to get from the emitter to the > detector is different in the two directions. Yes.. my mistake in word .. i should have said 'distance' not 'path' > Because the detector moves away from one light beam and moves toward the > other one. We all agree on that, right? The light is moving toward the detector in both cases. The closing rate between the leading edge of the light ray/photon/wave is the same for both rays. For a circular case (eg fibre optic) the dectector is 2piR away from the leading edge of both rays. Over the time, both rays get close to the detector at the same rate, and arrive at the same time (when the distance is then zero). The detector is moving away from the initial source position as recorded in the inertial frame in one direction, and close to it in the other. But relative to the leading edge of the light (in ballistic analysis) the rays and detector are moving twoard each other at the same rate > In the case of the fast light, it's fast because the emitter is moving > with it. But the detector is moving away from it too. So in my > animation, the path is about 10% longer than it would be if both paths > were the same length. And the slow light has a path that's about 10% > shorter than it would be if both were the same length. That's correct, > isn't it? That's what an inertial frame observer will see .. just like the linear case. When we look from the point of view of observer o who is walking past the experiment, we see: D---------S---------D... ...........o............. ..D------<--S-->------D.. ...........o............. ...D---<-----S----->---D. ...........o............. ....D<--------S-------->D ...........o............. o sees the ray going to the left as moving slower and the ray to the right as faster o sees the distance to the LHS D from him as shorter than the distance to the RHS D when the rays arrive If o wasn't walking past, and we just looked at the light leaving S and arriving at D then we'd not even suggest that there could be a phase difference How can that change because 'o' walked by? If it can't, then how can you get a phase shift in Sagnac? [snip a bit] >> They can't be anything BUT be in phase in a non-relativistic situation >> with constant speeds and same arrival time. > > That's because you assume that the front of the wave stays the same for > both. How can it not? What can change the phase of those photons/rays/waves/whatever? > And that's a reasonable assumption. Indeed it is :) > If you figure that when the light > left the emitter it started at zero and the electric field then climbed > to positive one and then fell to zero, why would the same spot on the > wave later be anything but zero trending positive? > > Still, that is an assumption. And when I drew the picture your way, with > the wave reaching the detector in phase and then I drew it backward to > the sources with constant wavelength and speed 0.9 and 1.1 and distance > 11 and 13, it was out of phase at the source. Its not out of phase at the source (S), its out of phase at 's', which is not the source, but just a place the source passed by. > That does not work. > Agreed? Surely you don't have one wave leave the emitter at some plus > value at the same time the other leaves with some minus value. They leave the emitter in phase and arrive at the detector in phase and it takes the same time for them to do so. The only way for them NOT to arrive in phase is if something happened to the rays in transit > When I draw it in phase at the source, and then I draw the waves with > the same constant wavelength, I get it out of phase at the detector. You are drawing it from the fixed point in the inertial frame. Not from the source > Did I choose some parameter wrong? I tested some things. Yes, the > wavelength is the same. One of them goes 1.1 while the other goes 0.9. > What did I do wrong? Would you like my source code, or would you want to > do it yourself from scratch and see what you get? >> Unless something somehow changes the frequency relative to the >> detector .. but if the frequency is the same and arrival time the same >> (all relative to the detector) then it MUST be in phase. > > Why would the frequency be the same? It is .. Doppler gives that. How can it be different? > The wavelength is the same and the > speed is different. I'd sure expect the frequency to be different. Not at the moving detector. Henry has gotten to you. The detector MOVES. You are assuming this... S D S D s S/'\./'\./R D S/'s./'\./'\.R D s S/'\./'\./'\./'\./D S/'\.s'\./'\./'\./'\./'\./'\.D instead of the unrolled Sagnac case... S D S D s S/'\./'\./'R D S/'s./'\./'R D s S/'\./'\./'\./'\./'\./'\D S/'\./s\./'\./'\./'\./'\D ie you're not looking at the Sagnac scenario > >> > In the model Wilson uses, they do not need to be in phase. >> >> HOW !!!!!! .. if they are emitted in phase they arrive in phase. >> Nothing happens in between to change that > > Check your assumptions? He isn't using the same assumptions you are. > Are his wrong? Yes > How can you find out if you're so stuck on your own that you > can't even understand his? We know that the light is in phase at the source .. it is a single beam split in two. >> > 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. >> >> There is no phase shift > > My picture could be wrong. Maybe I made a mistake that I haven't found. Yes > If you look at it closely you'll see that it has little bits of wave > evaporating off the front of one wave, and extra little bits magically > appearing at the front of the other wave. There's no reason you'd > predict that would happen, right? But it has to happen if the rest of it > is right. But the rest of it isn't. You have the waves being emitted as if from the fixed point in the inertial frame and not the moving detector > The wavelength is the same, and the frequency at the source is > the same, and the wave is traveling faster. So extra wave appears at the > front. Is that possible? Well, you certainly wouldn't have expected it, > right? Is there maybe something in Maxwells Equations that says it's > impossible? > >> > 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. >> >> Yes .. it most definitely is worth arguing about. >> >> His has a fixed source in the inertial frame and a moving detector, >> with the fixed source emitting two different frequency waves at two >> different speeds. > > It's a moving source, right? Yes >> This is nothing like Sagnac > > I drew the picture straightened out because that was easiest for me. Was > that the problem? Not really. >> >> > http://i847.photobucket.com/albums/ab31/jehomas/speedwave4.gif >> >> > http://i847.photobucket.com/albums/ab31/jehomas/speedwave6.gif >> >> The points on the left side that move left and right (and so get >> further and closer to the destination) correspond to where the source >> WAS when the FIRST photon/wave/ray was emitted. But you show waves >> continually emitted from that location. That is NOT the case. > > No, they correspond to where the source is when it's emitting more. Then it is wrong, as the source is always a fixed distance from the detector. > The > original starting source is elsewhere. > >> The photons/waves/rays are emitted a constant distance from D at all >> times. > > Yes. Would it help if I label D and the original S? Why do I ask, of > course it would. :):) >> >> 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? >> >> Yes .. you don't show that .. you have the distance from source to >> detector changing !!! > > I didn't put the detector in, the detector was where they stop. Clearly > I need to label it better. I'll wait until you label you animation >> > So one source appears to get farther from D while the other appears >> > to get closer. >> >> No .. can't you read that you just contrradicted yourself "detector >> moves with source" "source..get farther..get closer" > > Pick a time when a wave leaves the source in one direction, and another > leaves the opposite direction. OK > The detector is moving away fron one wave > and toward the other. No. It is moving toward both at the same speed relative to the leading edge of the wave/ray/photon. > So later waves will start from farther behind on > one side and closer on the other. > >> The source CANNOT get farther and closer to the detector when ther is >> ONE source and it moves WITH the detector > > Please try to bear with me, I'm not that good at saying this stuff yet. > It's new to me. That's ok :):) >> > Because by the time the light that is getting emitted later >> > reaches D, D will be farther away. >> >> No !!!!! The detector is always THE SAME DISTANCE FORM THE SOURCE. > > It would be easier to show it with a picture, but my current asnimation > software is a little limited for that. I'll extend it or find something > better, but this is a spare-time project.... Yeup .. I understand that >> >> > 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. >> >> It works perfectly with fixed wavelength > > No, when I make it fixed wavelength and make them be in phase at the > detector, they are out of phase at the source. You're not looking at the source. you're looking at some point the source used to be. > I don't consider that > "works perfectly". You can do it if the wavelength isn't fixed. Yes you can .. se my ./'\./'\ example above .. wavelength is fixed. >> > You can argue that it can't be constant wavelength, >> >> I'm not >> >> > or you can show me how I misunderstood your constant-wavelength >> > model. >> >> I don't know what your misunderstanding could possibly be, so I can't >> correct it > > How does your constant-wavelength model work? > > 1. Fixed distance from D to S. yes > 2. Rotates at 0.1c yes > 3. Light in one direction travels at 0.9c, light in the other direction > travels at 1.1c. as seen by a non-rotating observer, yes > 4. The distance traveled is 1.1 times the total fixed distance in one > case, and 0.9 times the fixed distance in the other case. as seen by a non-rotating observer, yes > 5. Both directions have the same fixed wavelength, yes > the same wavelength > they have with no rotation, yes > when both directions travel at c and the > distances are the same. yes > 6. Both light beams are in phase when they leave the source yes > and in phase > when they arrive at the detector. yes > Did I leave something out? Did I add something extra? I can only make > this work for certain particular distances that just happen to match up > the lengths to be a whole number of wavelengths. you seem to think you count wavelengths from where the source was a wave length is the distance between where one cycle starts and the next for the wave/photon/whatever. 'w' here is the wave length... S S\. S/'\. |-w-| lets look at a couple more cycles, and mark the initial position of the source with 's' S Ss. S/s\. S\.s'\. S/.\s/'\. S\./'s./'\. S/'\./s\./'\. |-w-|-w-|-w-| Notice that after three whole cycles, you can trace three wavelengths back to where the source is NOW (ie where the trailing part of the last cycle emitted was). There is of course, a whole number of wavelengths to the position the source is now (as there must be after a whole number of cycles). There is NOT a whole number of wavelength back to where the source WAS (in some frame of refernce), and no reason why there should. That is what Henry (and now you) wrongly assume in you number-of-wavelengths-different-so-must-be-out-of-phase argument. Oh .. and incase we hear "you are assuming light is a wave", then we can do the same thing with a moving oscillator .. like this S Ss' S/s\. S/'s./' S/'\s/'\. S/'\.s'\./' S/'\./s\./'\. |-w-|-w-|-w-| You still count wavelengths (length travelled in a single cycle) from where the source is NOW to where the leading edge is NOW. [snip] > So anyway, I'll redraw the pictures and make them clearer, and maybe > I'll find a mistake in the process. OK .. I'll see what your new version looks like.
From: Jerry on 18 Sep 2009 07:00 On Sep 17, 9:27 pm, "Inertial" <relativ...(a)rest.com> wrote: > "Androcles" <Headmas...(a)Hogwarts.physics_o> wrote in message > news:_4Asm.141175$I07.118718(a)newsfe04.ams2... > > I would be interested to see your explanation of the phase difference > detected in Sagnac. Androcles has given two completely distinct and incompatible explanations over the years. 1) The standard analysis of Sagnac ignores second-order effects. They do exist, as noted by Paul Andersen in http://home.c2i.net/pb_andersen/pdf/four_mirror_sagnac.pdf By running a gif animation with the rotational velocity a large fraction of the speed of light, Androcles demonstrated that these second order "Coriolis" effects can, in principle, be quite large. 2) My logical challenges to Androcles' second explanation earned me Plonk #5. (DvM has earned more Anrocles Plonks, but I got my plonks with far fewer posts.) Basically, Androcles agrees that no phase difference accumulates in the ring. The phase differences result when c+v and c-v light emerge from the beam splitter and travel to the detector. Jerry
From: Jonah Thomas on 18 Sep 2009 08:41 "Inertial" <relatively(a)rest.com> wrote: > "Jonah Thomas" <jethomas5(a)gmail.com> wrote > > Because the detector moves away from one light beam and moves toward > > the other one. We all agree on that, right? > > The light is moving toward the detector in both cases. > > The closing rate between the leading edge of the light ray/photon/wave > is the same for both rays. > > For a circular case (eg fibre optic) the dectector is 2piR away from > the leading edge of both rays. Over the time, both rays get close to > the detector at the same rate, and arrive at the same time (when the > distance is then zero). > > The detector is moving away from the initial source position as > recorded in the inertial frame in one direction, and close to it in > the other. > > But relative to the leading edge of the light (in ballistic analysis) > the rays and detector are moving twoard each other at the same rate Yes. I found my mistake. > > If you figure that when the light > > left the emitter it started at zero and the electric field then > > climbed to positive one and then fell to zero, why would the same > > spot on the wave later be anything but zero trending positive? > > > > Still, that is an assumption. And when I drew the picture your way, > > with the wave reaching the detector in phase and then I drew it > > backward to the sources with constant wavelength and speed 0.9 and > > 1.1 and distance 11 and 13, it was out of phase at the source. > > Its not out of phase at the source (S), its out of phase at 's', which > is not the source, but just a place the source passed by. I wasn't using 's', I double-counted the distance offset. The result was similar. Oh well. It was fun while it lasted. I tried to find a way for Wilson's idea to be right, and I gave it my best shot. This result fits my original interpretation. The change in speed for the light in the different directions is just enough to make up for the rotation. And without having to deal with the rotation the result is completely symmetrical. It's hard to find anything to work with. > They leave the emitter in phase and arrive at the detector in phase > and it takes the same time for them to do so. > > The only way for them NOT to arrive in phase is if something happened > to the rays in transit It would have to happen differently to the waves that have the different absolute speeds and different directions. I can sort of imagine ways that could happen, but none that follow direction from an emission theory. They'd be ad hoc things added onto the theory to let it fit Sagnac. All in all, I think Wilson would be better off to go with the Ritz emission theory. It treats reflections different, but Wilson's work with double stars probably doesn't involve any reflections so he wouldn't lose much there. And the Ritz theory works with Sagnac with some tiny differences that likely have still not been tested.
From: Jonah Thomas on 18 Sep 2009 08:45 "Inertial" <relatively(a)rest.com> wrote: > "Jonah Thomas" <jethomas5(a)gmail.com> wrote > > Because the detector moves away from one light beam and moves toward > > the other one. We all agree on that, right? > > The light is moving toward the detector in both cases. > > The closing rate between the leading edge of the light ray/photon/wave > is the same for both rays. > > For a circular case (eg fibre optic) the dectector is 2piR away from > the leading edge of both rays. Over the time, both rays get close to > the detector at the same rate, and arrive at the same time (when the > distance is then zero). > > The detector is moving away from the initial source position as > recorded in the inertial frame in one direction, and close to it in > the other. > > But relative to the leading edge of the light (in ballistic analysis) > the rays and detector are moving twoard each other at the same rate Yes. I found my mistake. > > If you figure that when the light > > left the emitter it started at zero and the electric field then > > climbed to positive one and then fell to zero, why would the same > > spot on the wave later be anything but zero trending positive? > > > > Still, that is an assumption. And when I drew the picture your way, > > with the wave reaching the detector in phase and then I drew it > > backward to the sources with constant wavelength and speed 0.9 and > > 1.1 and distance 11 and 13, it was out of phase at the source. > > Its not out of phase at the source (S), its out of phase at 's', which > is not the source, but just a place the source passed by. I wasn't using 's', I double-counted the distance offset. The result was similar. http://i847.photobucket.com/albums/ab31/jehomas/speedwave7.gif http://i847.photobucket.com/albums/ab31/jehomas/speedwave8.gif Oh well. It was fun while it lasted. I tried to find a way for Wilson's idea to be right, and I gave it my best shot. This result fits my original interpretation. The change in speed for the light in the different directions is just enough to make up for the rotation. And without having to deal with the rotation the result is completely symmetrical. It's hard to find anything to work with. > They leave the emitter in phase and arrive at the detector in phase > and it takes the same time for them to do so. > > The only way for them NOT to arrive in phase is if something happened > to the rays in transit It would have to happen differently to the waves that have the different absolute speeds and different directions. I can sort of imagine ways that could happen, but none that follow direction from an emission theory. They'd be ad hoc things added onto the theory to let it fit Sagnac. All in all, I think Wilson would be better off to go with the Ritz emission theory. It treats reflections different, but Wilson's work with double stars probably doesn't involve any reflections so he wouldn't lose much there. And the Ritz theory works with Sagnac with some tiny differences that likely have still not been tested.
From: Jonah Thomas on 18 Sep 2009 14:45
Jerry <Cephalobus_alienus(a)comcast.net> wrote: > On Sep 17, 9:27 pm, "Inertial" <relativ...(a)rest.com> wrote: > > "Androcles" <Headmas...(a)Hogwarts.physics_o> wrote in message > > news:_4Asm.141175$I07.118718(a)newsfe04.ams2... > > > > I would be interested to see your explanation of the phase > > difference detected in Sagnac. > > Androcles has given two completely distinct and incompatible > explanations over the years. > > 1) The standard analysis of Sagnac ignores second-order effects. > They do exist, as noted by Paul Andersen in > http://home.c2i.net/pb_andersen/pdf/four_mirror_sagnac.pdf > By running a gif animation with the rotational velocity a > large fraction of the speed of light, Androcles demonstrated > that these second order "Coriolis" effects can, in principle, > be quite large. But there's no particular reason to expect that second order effects would come out the same as the first-order effects predicted by classical theory and SR, the effects that have been observed. So that is not very useful unless it turns out that the effects predicted match the effects seen. > 2) My logical challenges to Androcles' second explanation earned > me Plonk #5. (DvM has earned more Anrocles Plonks, but I got my > plonks with far fewer posts.) Basically, Androcles agrees that > no phase difference accumulates in the ring. The phase > differences result when c+v and c-v light emerge from the beam > splitter and travel to the detector. That ought to be testable. Change the distance to the detector and the phase difference from that cause should change too, shouldn't it? It would be great if that was true, we would have a reliable plentiful source of bi-speed light to experiment with. Once again, it looks to me like the Ritz form is best so far, everybody seems to agree that it fits the Sagnac results, it is designed so that it will, so you don't have to come up with strange reasons for it to do so. |