The Emission Theory of Androcles
in [Relativity]
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From: Jonah Thomas on 13 Sep 2009 09:36 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. You marked that stationary point on your moving picture. I know what point you're talking about. Why do you care about that point? Doesn't it have purely historical interest? It tells you about the distance that two of the waves travel, and it tells you nothing else. We're clear about the distance they travel, right? One of them travels (c+v)*t and the other travels (c-v)*t where the distance around the path is c*t. Figuring that is all that this point is useful for. Do you have some other use for it? > >> >> 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 It's true that the path that any one wave takes is either ct+vt or ct-vt. And the path is 10 wavelengths long, and you have 10 waves on it at any one time. > >> 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. That would be true if the detector moves but the emitter doesn't. Do you agree it's true if the detector moves but the emitter doesn't? > >> >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 wavelength is the same. the frequency is different and so is the speed. The difference in frequency and the difference in speed exactly cancel out. You needn't get any interference from that. > The frequency of a water wave depends on the speed of one's boat. The > distance between waves does not. Yes, in reality. Although as Androcles points out, if your only way to measure the wavelength of the moving wave is to travel from one wavecrest to the other at some speed with a tapemeasure, and you can't accurately measure your speed and the wave's speed, then you have no way to know what the wavelength actually is. > >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 don't see why you care about the number of waves in that distance. The important thing is the number of waves between the emitter and the detector. That's ten in both directions, all the time. Because the distance between the emitter and the detector is always the same. > 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 was wrong to agree with you about that. The number of waves between the emitter and the detector only changes when the distance between emitter and detector changes, or when c changes. The number of waves between your arbitrary stationary point and the detector at an arbitrary time does change when the rotation speed changes, and only then. > >> >> >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. Agreed. But why do we care how many wavecrests pass a stationary point? > 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. I've tried to keep an open mind and understand what you're saying, but I've reached the point that it is very hard. I'm pretty sure I see how it works, and I don't see how to make your approach work. I believe I see where you make your mistake. I am rapidly losing the ability to listen uncritically because everything you say is compatible with the "mistake" hypothesis and nothing recently brings anything new to the table to change that. However, suppose for a moment that you are wrong on this one point. That means that BaTH is wrong only on the things that require this particular behavior of mirrors. I have the strong impression that essentially none of your work depends on that. So practically everything would carry over to a different emission theory, one that does satisfy the Sagnac experiment. Here is the one I think works: Instead of saying that the speed of light is assigned when the light is emitted and is c+ v*cos(theta) where theta is the angle between the direction of the light and the direction of the source's movement -- instead of that, the velocity of light is cD+vV where D is the direction the light is traveling in the source's frame. That is, once light is emitted then from that point on it behaves as if it is in an aether that is traveling at its source's velocity. This satisfies MM. And it satisfies Sagnac. But for it to work properly for Sagnac it must work inside fiber optics and all the media that the Sagnac effect works. If extinction slows the speed of the fast light to less than c, it must slow the speed of the slow light proportionately. Otherwise we will get a result that is quantitatively different from classical physics which people claim is quantitatively correct within some small limit.
From: Sue... on 13 Sep 2009 09:46 On Sep 13, 9:36 am, Jonah Thomas <jethom...(a)gmail.com> wrote: > > Yes, in reality. Although as Androcles points out, if your only way to > measure the wavelength of the moving wave is to travel from one > wavecrest to the other at some speed with a tapemeasure, and you can't > accurately measure your speed and the wave's speed, then you have no way > to know what the wavelength actually is. There are a lot more instruments based on standing waves than propagating waves. Even the present standard for the speed of light is based on a standing wave. ~Slotted lines~ http://personal.ee.surrey.ac.uk/Personal/D.Jefferies/mwmeas.html http://en.wikipedia.org/wiki/Speed_of_light#Modern_methods Sue...
From: Jonah Thomas on 13 Sep 2009 10:45 "Sue..." <suzysewnshow(a)yahoo.com.au> wrote: > Jonah Thomas <jethom...(a)gmail.com> wrote: > > > Yes, in reality. Although as Androcles points out, if your only way > > to measure the wavelength of the moving wave is to travel from one > > wavecrest to the other at some speed with a tapemeasure, and you > > can't accurately measure your speed and the wave's speed, then you > > have no way to know what the wavelength actually is. > > There are a lot more instruments based on standing waves > than propagating waves. Even the present standard for the > speed of light is based on a standing wave. > > ~Slotted lines~ > http://personal.ee.surrey.ac.uk/Personal/D.Jefferies/mwmeas.html > http://en.wikipedia.org/wiki/Speed_of_light#Modern_methods Thank you! Those look like useful links, with a wealth of information buried in the first one like secret treasure. I was interested in the claim that with modern nanosecond-scale timers you can actually measure the delay for light to travel known distances. That might allow some potentially-new experiments.
From: Helmut Wabnig hwabnig on 13 Sep 2009 10:48 On Sun, 13 Sep 2009 10:45:42 -0400, Jonah Thomas <jethomas5(a)gmail.com> wrote: >"Sue..." <suzysewnshow(a)yahoo.com.au> wrote: >> Jonah Thomas <jethom...(a)gmail.com> wrote: >> >> > Yes, in reality. Although as Androcles points out, if your only way >> > to measure the wavelength of the moving wave is to travel from one >> > wavecrest to the other at some speed with a tapemeasure, and you >> > can't accurately measure your speed and the wave's speed, then you >> > have no way to know what the wavelength actually is. >> >> There are a lot more instruments based on standing waves >> than propagating waves. Even the present standard for the >> speed of light is based on a standing wave. >> >> ~Slotted lines~ >> http://personal.ee.surrey.ac.uk/Personal/D.Jefferies/mwmeas.html >> http://en.wikipedia.org/wiki/Speed_of_light#Modern_methods > >Thank you! Those look like useful links, with a wealth of information >buried in the first one like secret treasure. > >I was interested in the claim that with modern nanosecond-scale timers >you can actually measure the delay for light to travel known distances. >That might allow some potentially-new experiments. today we are talking femtoseconds, that is 1000x1000 less than nanos. this is year 2009, not 1909. w.
From: Ahmed Ouahi, Architect on 13 Sep 2009 10:54
Nanosecond-scale http://www.sciencedaily.com/releases/2006/05/060508180735.htm Nanosecond scale carrier dynamics http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6VMT-4GD4SVX-1&_user=10&_rdoc=1&_fmt=&_orig=search&_sort=d&_docanchor=&view=c&_searchStrId=1009926104&_rerunOrigin=google&_acct=C000050221&_version=1&_urlVersion=0&_userid=10&md5=77544af77b9b4993b02c34741721e2e0 nanosecond scale, time-resolved http://www.patentstorm.us/patents/7547508/description.html Photodiodes on Nanosecond Scale http://risorse.dei.polimi.it/spad/1991/IEEE%20EDL%201991.pdf Time Resolved X-Ray Diffraction http://home.comcast.net/~bhoerman/cv/publications/papers/fe290.pdf -- Ahmed Ouahi, Architect Best Regards! "Jonah Thomas" <jethomas5(a)gmail.com> kirjoitti viestiss�:20090913104542.15519570.jethomas5(a)gmail.com... > "Sue..." <suzysewnshow(a)yahoo.com.au> wrote: >> Jonah Thomas <jethom...(a)gmail.com> wrote: >> >> > Yes, in reality. Although as Androcles points out, if your only way >> > to measure the wavelength of the moving wave is to travel from one >> > wavecrest to the other at some speed with a tapemeasure, and you >> > can't accurately measure your speed and the wave's speed, then you >> > have no way to know what the wavelength actually is. >> >> There are a lot more instruments based on standing waves >> than propagating waves. Even the present standard for the >> speed of light is based on a standing wave. >> >> ~Slotted lines~ >> http://personal.ee.surrey.ac.uk/Personal/D.Jefferies/mwmeas.html >> http://en.wikipedia.org/wiki/Speed_of_light#Modern_methods > > Thank you! Those look like useful links, with a wealth of information > buried in the first one like secret treasure. > > I was interested in the claim that with modern nanosecond-scale timers > you can actually measure the delay for light to travel known distances. > That might allow some potentially-new experiments. |