The Emission Theory of Androcles
in [Relativity]
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From: Inertial on 12 Sep 2009 18:49 "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: > Try my model. It seems pretty logical to me. > Photons are particles that also oscillate intrinsically. So frequency is absolute and wavelength vcaries with the observer. That's what happens with moving intrinsic oscilaltors. But then you claim that the frequency is different depending on who is looking at it, which is what happens when light is a wave. You can't make up your mind what it is, and end up switching models. [snip] > 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? We understand it .. but you then go saying thing incompatible with your own model. > Am I more intelligent > than everyone else here? I think you confuse 'more' with 'less'
From: Henry Wilson, DSc on 12 Sep 2009 18:49 On Sun, 13 Sep 2009 08:38:49 +1000, "Inertial" <relatively(a)rest.com> wrote: >"Jonah Thomas" <jethomas5(a)gmail.com> wrote in message >news:20090912125853.325f7e11.jethomas5(a)gmail.com... >> "Androcles" <Headmaster(a)Hogwarts.physics_n> wrote: >>> "Jonah Thomas" <jethomas5(a)gmail.com> wrote >>> > hw@..(Henry Wilson, DSc) wrote: >> >>> >> Wavelength is absolute and frame independent in BaTh. >>> > >>> > Yes. Agreed. But each wave is created over the expanse of a >>> > wavelength, it isn't created all at once. >>> >>> Don't agree to his senile nonsense mantra. All velocities are frame >>> dependent >>> so all wavelengths are frame dependent. >>> http://www.androcles01.pwp.blueyonder.co.uk/Wave/Relative.gif >> >> I'm not sure that you and I ever reached a common understanding about >> what a wavelength is. >> >> I'm pretty sure that part of the problem is that often when physicists >> mention "wavelength" in actual problems they do not actually measure the >> wavelength. Often they measure frequency and then assume that wavelength >> equals velocity/frequency, and sometimes the velocity is assumed. >> >> What I mean by wavelength is the physical distance between wavefronts. >> Ideally you would measure this with a ruler. It is time independent >> unless it is changing, and if it is changing then at least one of >> frequency or speed must be changing too. If you can make an >> instantaneous measurement with a ruler then your relative velocity does >> not matter. >> >> If you suppose that lightspeed can vary, then what does a diffraction >> grating do? You didn't have much of an answer for that because you acted >> like you didn't want it to be a wave phenomenon at all which leaves it >> pretty vague. When I try to think carefully about the traditional wave >> explanations for a diffraction grating I find them kind of vague too. >> >> When c is constant you can measure diffraction and you have one >> variable. When c is not constant you have two variables -- and a vector >> if the velocity of the source makes a difference. When c is constant it >> doesn't matter whether you're measuring a wavelength change or a >> frequency change -- they have to go together. It doesn't matter which >> change is happening -- they have to happen together. When c can be >> different then it makes a difference whether it's frequency that changes >> or wavelength or both in some ratio. >> >>> > 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. >>> Fuckin' rubbish. You two should be locked in two cells >>> out of earshot of each other. >> >> 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. Clearly, 'wavelength', whatever that might signify...does not vary with frame. Otherwise BaTh would give the wrong answer....and that is not possible. Wavelength is a fixed distance...(except when a photon accelerates).. It is through experiments like this that we can hypothesize about the structure of a photon. Photons are obviously not just the 'point particles' that relativists claim they are. Henry Wilson...www.users.bigpond.com/hewn/index.htm Einstein...World's greatest SciFi writer..
From: Androcles on 12 Sep 2009 18:51 "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: > >>"Androcles" <Headmaster(a)Hogwarts.physics_o> wrote: >>> "Jonah Thomas" <jethomas5(a)gmail.com> wrote >> >>> > It is time independent >>> > unless it is changing, and if it is changing then at least one of >>> > frequency or speed must be changing too. If you can make an >>> > instantaneous measurement with a ruler then your relative velocity >>> > does not matter. >>> > >>> > If you suppose that lightspeed can vary, then what does a >>> > diffraction grating do? >>> >>> Causes a change in rotation of a photon. >> >>I have found your particle explanation of diffraction to be quite vague. > > Most of what he says is quite vague...but if he would explain it properly > instead of ranting, other people might be able to understand some of it. > >>Unfortunately when I looked in more detail at the classical wave >>explanation of diffraction it turned out to be pretty vague too. >> >>> > When c is constant you can measure diffraction and you have one >>> > variable. When c is not constant you have two variables -- and a >>> > vector if the velocity of the source makes a difference. When c is >>> > constant it doesn't matter whether you're measuring a wavelength >>> > change or a frequency change -- they have to go together. It doesn't >>> > matter which change is happening -- they have to happen together. >>> > When c can be different then it makes a difference whether it's >>> > frequency that changes or wavelength or both in some ratio. >>> > >>> >> > I imagine the emitter creating a wave >>> >>> The emitter is a molecule. 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. >> >>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 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 [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.
From: Inertial on 12 Sep 2009 18:59 "Henry Wilson, DSc" <hw@..> wrote in message news:h32oa5dt4vr3e0m16cgjoqccmdpag1i93f(a)4ax.com... > On Sat, 12 Sep 2009 01:08:39 -0400, Jonah Thomas <jethomas5(a)gmail.com> > wrote: > >>hw@..(Henry Wilson, DSc) wrote: >> >>> Each photon takes the same time to travel the ring but the distances >>> traveled are different. Wavelength is invariant...so there are more >>> waves in one path than the other. >> >>This is one of the places I didn't follow you. Could you maybe show a >>picture of the paths with more waves in one than the other? > > I have. > Here is the reason why one path is longer than the other. > http://www.mathpages.com/rr/s2-07/2-07.htm > (Don't take any notice of the relativistic explanation.) Oh, good heavens no, you might learn something [snip] > There isn't any 'wave' at all. Photons have a spatial periodicity. I > suspect > that this is the result of either a rotation of a +/- pair or the presence > of a > 'standing wave' along its length. Handwaving again. If it is an intrinsic oscillator, then every observer sees it with the same frequency. Just like (say) a weight on a spring bouncing up and down. If you put that on a train (don't you just love trains) the weight on the spring traces out a 'wave' as the train moves.. and the wavelength of that depends on the speed of the observer. Similarly, a standing wave on a moving train has the same frequency for all observers. The standing wave itself also has a fixed wavelength, but the wave generated by the movement of it (like the weight on a spring) has a variable wavelength dependant on the relative speed of the train to the observer. However, the frequency of both is fixed. > Here is my vision of a photon > www.users.bigpond.com/hewn/e-field.exe Yeup .. a moving intrinsic oscillator (the standing wave on a train scenario). It always has a fixed frequency, no matter who sees it, yet Henry says it doesn't.
From: Inertial on 12 Sep 2009 19:01
"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> > 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. > > >>> 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. > >>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. > >>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. I light is a little particle, what is a wavecrest? > 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? |