The Emission Theory of Androcles
in [Relativity]
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From: Jonah Thomas on 12 Sep 2009 20:09 hw@..(Henry Wilson, DSc) wrote: > Jonah Thomas <jethomas5(a)gmail.com> wrote: > >hw@..(Henry Wilson, DSc) wrote: > >> Jonah Thomas <jethomas5(a)gmail.com> wrote: > > >> >OK, I see that the emitter moves forward in the nonrotating frame. > > > >> > >> Nonononno. The emission POINT ...MARKED IN THE NONROTATING > >FRAME...is> at rest in the nonrotating frame. It moves backwards in > >the rotating> frame. > > > >The emission point for you is not where the emitter is at the moment, > >it's where the first wave we're interested in started out. It stays > >in one place in the nonrotating frame, and it moves backward in the > >rotating frame. Do I understand that? > > 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? > >> 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. > >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. 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. > >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. 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 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. OK. The number of wavecrests leaving the source in my example is ten per second, in any frame. > 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?
From: Jonah Thomas on 12 Sep 2009 20:43 "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. Say the source is traveling at speed v in direction V. Emit light in direction D at speed c at time 0 and location (0,0,0) Then at time t, the light will be at t*(vV+cD) Imagine a spherical wavecrest leaving the source at time 0 and location (0,0,0). At time t that wavecrest will be in a circle with center tvV and radius ct. And without relativity, other inertial observers will also find the circle to have radius ct. They will only disagree about whare the center is. 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. Am I wrong?
From: Jonah Thomas on 12 Sep 2009 21:19 "Inertial" <relatively(a)rest.com> wrote: > "Androcles" <Headmaster(a)Hogwarts.physics_o> wrote > > It can only send a pulse. Imagine all you want to, there are NO > > light waves. > > Radio waves, yes, but no light waves. > > OMG .. Androcles thinks that the set frequency of EM radition we call > light is a totally different concept from the set EM frequencies we > call radio waves? He's GOT to be joking. > > BAHAHAHA > > If not .. I wonder where he thinks the change between being waves and > being particles happens? Androcles has training in electrical engineering, so he quite likely understands how radio waves are produced. He gave the explanation that radio waves are like a tsunami while light waves are like something else -- a faucet dripping? Both water moving, but different forms. My guess at where he's heading with that is that light waves are typically made by atoms, by electrons falling to lower shells or whatever the wording is. (You can get light without quite that, with atoms heated to white heat.) Radio waves are made by electrons accelerated back and forth, typically inside hunks of metal or in waveguides. A whole lot of electrons moving without much regard to the atoms they move among. It's different. The electrons that are used to make radio waves can keep oscillating back and forth as long as you supply the power. The atoms that make light make one quantum at a time (or maybe it's two) but they can't do it again until they get their electrons knocked up again. So it's different. So electrons fall to their new levels and if it takes one cycle for that to happen you get a one-cycle event. If it takes fifty cycles then you get a 50-cycle event, that's more wavelike than the short one. How long does it take? I don't know but likely somebody knows.
From: Henry Wilson, DSc on 12 Sep 2009 23:27 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? If one rope is longer than the other, there must be a different number of turns in the two. That means there is a phase shift where they meet. I really think this is too hard for you... Henry Wilson...www.users.bigpond.com/hewn/index.htm Einstein...World's greatest SciFi writer..
From: Henry Wilson, DSc on 12 Sep 2009 23:32
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. >> 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. Henry Wilson...www.users.bigpond.com/hewn/index.htm Einstein...World's greatest SciFi writer.. |