GETTING RID OF EINSTEIN RELATIVITY
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From: sean on 17 Sep 2007 09:47 On 12 Sep, 15:26, bz <bz+...(a)ch100-5.chem.lsu.edu> wrote: > sean <jaymose...(a)hotmail.com> wrote innews:1189444302.990016.237750(a)d55g2000hsg.googlegroups.com: www.gammarayburst.com http://www.youtube.com/profile?user=jaymoseleygrb > > On 6 Sep, 17:33, bz <bz+...(a)ch100-5.chem.lsu.edu> wrote: > >> sean <jaymose...(a)hotmail.com> wrote > >> innews:1189089230.764471.53320(a)22g2000hsm.googlegroups.com: > > >> > On 30 Aug, 23:55, bz <bz+...(a)ch100-5.chem.lsu.edu> wrote: > >> >> sean <jaymose...(a)hotmail.com> wrote in news:1188512224.511353.237820 > >> >> @i13g2000prf.googlegroups.com: > > >> >> > If you bothered analysing your sean planets sim, youd > >> >> > see that its the only way to have light propagating away > >> >> > from any source at c. > > >> >> One of the most fundamental laws of physics is that an object in > >> >> motion continues in that motion unless acted upon by an outside > >> >> force. > >> > You ignore several things here, First of all if emmision theory > >> > predicts that light always is at c relative to a source > > >> That only applies at the moment of emission. If the source changes its > >> motion after emission, the light does not know or care. > > Maybe in your own personal version of emmision theory. But Im saying > > that if one can model emmision theory as having light propagate away > > from any source always at c relative to any source, then,.... > > one can explain MMx and sagnac > > >> > then it would be impossible for light to be effected by gravity under > >> > these terms. > > >> Once light has left the source, gravity, mirrors, lenses, etc can > >> effect its motion. The 'leaves its source at c' only applies at the > >> moment of emission and then only when measured with equipment that are > >> co-moving with the source and hence in the same inertial frame of > >> reference. If the source suddenly accelerates at right angles and goes > >> flying off, into the night (you have just thrown your flashlight away) > >> the light that has been already emitted from that flashlight does NOT > >> suddenly change its motion in order to maintain a velocity of c wrt the > >> source that it has left far behind. > > How do you know. > > Wheres the proof ? You have none. I discussed this earlier in this > > thread with George and I pointed out that it is technically almost > > impossible to test these claims. At least there is no experimental > > setup > > that exists that can test this clai you make. > > I disagree. > > > If a lab was 10 meters long and a rotating light source on one end > > was observed from the other end. The actual displacement > > due to this `path drag` would be minute. Something like a > > 0.000003 mm deflection over 10 m. > > Let us put a LED (light emitting diode) on the rim of a ultra centrifuge > rotor that is 10 cm in diameter and rotate the disk at 50,000 rpm. The > circumference will be 0.628 meters. The speed of the diode will be 523.599 > meters per second or 1.7 micro c. > > Now, we spin the rotor clock wise for one test and counter clockwise for > another test. We pulse the LED near the point where it is closest to us > (thus traveling at max speed perpendicular to the line of sight) From a > distance of 10 meters we use a long focal length microscope to view the > position of the LED at the moment that it pulses. Our microscope gives us > a 100 to 1 magnification of the image of the LED. We use a ccd camera to > display the image on our computer screen. > > What do I predict and what does your theory predict? > > I predict that the image will appear in the same location, non rotating, > or rotating clockwise or counterclockwise at 50,000 rpm. > > What does your theory predict? Your theory predicts that during the 0.033 > us it take the light to travel from the LED to our tele-microscope 10 > meters away, the light will have moved 0.017 mm in the direction of > rotation. As there are LEDs that are only 20 um in size,http://www.patentstorm.us/patents/6410940-claims.html > so, I calculate that > if your theory were correct that the image would be displaced by 873 > diameters in the direction of travel. In principle this sounds like it could work I get a deflection over 10 meters of about .015mm. so thats pretty close to yours. I wonder if thats large enough to measure over 10 meters? I also wonder about the pulse timing. It seems to me that with the time light travels the 10 meters being about 1/30,000,000 second you would have to have pulse timing thats accurate to 1/30,000,000sec. Maybe one alteration to make it simpler could be this... Use your setup as youve described .....but add a single LED positioned directly above where the other LED is when it pulses and have that on all the time. Then with the imaging scope up very close turn on the rotation and adjust the pulse timing so that the pulse LED turns on just directly below and in line with the other LED. Ie the two LEDs are in line vertically when the pulsed LED is rotating. LED 1 LED 2 Then move the camera back and theoretically if Im right, the rotating LED should slowly move out of line with the non rotating LED the farther back the camera moves from the setup. LED1 LED2 Or not if your prediction is right. This way we know for certain the pulse timing is spot on. Unfortunately the price and tech know-how for this experiment is still probably out of my reach so for the time being this experiment cant be done. Having said all that I realize that Im not sure how to model a source that pulses. Once a source goes OFF it would seem that it no longer exists as a source which means any subsequent motion of an `off` source would not effect the light as it propagates away. So it might be better for the purposes of this experiment to pulse it on and leave it on as it rotates rather than me worrying how to model the motion of a nonexistent source. Hopefully this shouldnt matter for you as it still means the `on` point will be `dragged` or not and the vital theoretical test can still be made > > Youd never be able to measure that. > > Nobody has yet. Therefore no proof exists that tests this possibilty > > one way or the other > >> > Secondly you ignore what I say repeatedly in the post > >> > youve just replied to. > > >> I ignore many things because I am trying to focus your attention on one > >> small point. > > And Im trying to focus your mind on one point. That is that the MMx > > source > > is moving yet the light that leaves the source still has a speed of > > c relative to the moving source. Contrary to what you claim taht it > > isnty > > moving at c in the lab. > > And I have yet to see anything else that works that way. > > The golf ball does not bobble as tiger completes his follow-through. > > It is contrary to all data we have and it is easy to test, as I have just > shown. What I mean above is that the MMx experiment is essentially a source on a rotating table. Except the rotational diameter is large and the interval 1 day. But the rotational speed of the MMx is still fast Dont forget the lab and MMx source are rotating at high speed around the earths axis. If light did not travel away from the source on both paths at c in the source frame as you suggest then one path would be slower than the other and not give a null result. But we observe both paths in MMx to be at the same speed. Which can only mean that light in MMx is propagating *away* from the source on both paths at c. And thats only possible if light was travelling at c only in the source frame. And if light travels at c in the source frame one would have to get this `drag` I predict. In other words you dont have to do your above experiment. MMx has already shown us light drags along with the source. (although drag is georges term. I like to think of it this way.. that light is at c in straight lines only in the source frame) > >> > That emmision theory cannot have light as a particle with mass. > > >> I don't care if light is a wave or a particle at this moment. It > >> doesn't matter. The light, once emitted, goes its merry way. Unlike the > >> teenager, leaving home for the first time, the light does not 'write > >> home' and ask for more money. The light doesn't care what happens to > >> the source after it has left home. > > JUst dont forget that Im argueing here for a wave only emmision model > > to explain the various observed phenomena. So please dont expect me > > to incorporate particle theory in this model. Like for instance.. > > electrons. > > Electrons are sources of photons as they circulate in the synchrotron. > The fact that the electron is a particle is NOT important. > The fact that we KNOW the path that the electron is taking as it emits > those photons IS important. The electrons are moving along a curved path > at a high velocity. But youve just ignored what I said. Electrons dont exist in classical wave theory. One has to describe any stream of electrons in wave terms. So effectively in wave theory a cyclotron isnt modeled as a spiralling beam of particles called electrons. Its modeled as an oscillating magnetic field in a cavity. Hence the source is no longer a rotating beam of particles but a static oscillating cavity source. Ill cover this a bit more below. > The electrons are the SOURCE of the photons. > The source is moving along a curve. > By your theory, the photons are 'tied' to the source and must mirror its > motion. The photons do NOT do that. > This is experimentally verified, as I showed. In my theory there are no photons. It is wave only. It may be that you arent familiar with my other posts but over the years I have been arguing for a classical wave only emmision model. And incidentally this isnt my theory . Its just observed physics minus the standard model theory assumptions about particles and SR. I suggest NO new laws or assumed variables. Even my idea of `light at c in the source frame only` is actually just me studying what actually was observed in MMx and writing it down. As far as Im aware whatever other theoretical assumptions Lorentz or Einstein made,.. MMX does show us that light HAS TO travel at c in the source frame. And thats all I say. > >> >> The light emitted consists of 'objects'. Once emitted, they must > >> >> travel in straight lines at a constant velocity unless acted upon by > >> >> an outside force. > > >> >> What force would 'tie' the light, once emitted, to the source so > >> >> that it must follow the source? > > >> >> There is no evidence of any such 'tie' between light, once emitted, > >> >> and its source. > >> > You jump to conclusions here. What "evidence" do you have that light > >> > doesnt > >> > travel at c in any direction from a source? > > >> I made no such claim. It DOES travel at c in all directions from where > >> the source was at the moment of emission. But once emitted, it doesn't > >> care where the source goes afterwards. > > This is an odd statement . You say you never claimed that light > > `doesnt travel > > at c relative to a source`. And then you contradict yourself in the > > same > > sentence above and say that in fact it `doesnt travel at c relative to > > the > > source`.?? > > So what do you want to claim? That it does or it doesnt travel at c > > relative > > to the source? > > It travels at c in all directions, as measured in an inertial frame of > reference co-moving with the source at the moment of emission. You can use the term inertial if you want but its a bit of a misleading term to use because it suggests one is measuring the light in a frame other than the rotating MMMx source frame When in fact the truth is that the light is measured in the source frame *only* in MMx. Not some other `inertial` frame. And the source frame is indisputably rotating once a day. So its not inertial contrary to what SR suggests. And incidentally I know what your response willl be.. That the inertial frame approximates the source frame.Ive heard this before and its unnacceptable on two counts. One, that its approximate. In physics this isnt good enough. And two, seeing as its only approximate it would be scientifically more accurate to measure and calculate in a frame that isnt approximnate. And that better frame is the source frame. Its exact , not approximate, Which is what I do in my emmision theory calculations in the exacting source frame. > If the source changes motion or even ceases to exist after the light has > been emitted, the light neither knows nor cares. Its all relative. If your in the source frame,.. the source never moves. Hence the light doesnt have to move either, that is; other than move away from the source at c in straight lines > >> > My evidence that they do is MMx. Notice here that the MMx source is > >> > rotating > >> > around the earths axis yet we also know that the light is travelling > >> > at c on > >> > both paths. This would not be possible if light travelled away from a > >> > rotating > >> > source at different speeds in different directions as you erroneously > >> > suggest.above > > >> You misread my suggestion. > > How did I misread your sugggestion? YOu said... > > > "The light emitted consists of 'objects'. Once emitted, they must > > travel in > > straight lines at a constant velocity unless acted upon by an outside > > force." > > > In other words you say above that once light leaves the MMx source it > > must > > travel in straight lines at c + the speed of earths rotation etc, > > relative > > to the source. > > They travel at c as measured in an inertial frame of reference that was > moving at exactly the same velocity as the source at the instant of > emission. Partly true, but theres more.. It also travels at c relative to the source **after** it has left the source. If it didnt there would be no null result observed. > > THat means that you think that light travels at different speeds away > > from a source relative to that source. > > They travel at c as measured in an inertial frame of reference that was > moving at exactly the same velocity as the source at the instant of > emission. You forgotton something very fundamental here. The MMx experiment measures what light does AFTER it leaves the source. Not just when it leaves, as SR incorrectly assumes. And the MMx measurements indisputably show us that light travels at c on both paths AFTER it leaves the source. And emmision theory predicts that if this is true then a null result should be observed... And guess what... A null result IS observed. !! > Light travels at c as measured in ANY inertial frame of reference, > irrespective of the velocity of that frame of reference wrt the source. > [with the caveat that no inertial frame of reference can ever be defined > so as to move at c with respect to any object that has mass]. I think you are talking about SR here. Primarily Im discussion how emmision theory can also explain both MMx and sagnac . Not SR. As long as emmission theory has the basic rule that light is allowed to travel away from a source at c relative to the source. And I see no reason why it cant , especially considering that this is whats observed in MMx. > >> >> In a cyclotron, particles traveling around a circular path emit > >> >> beams of xrays. > >> >> Those xrays travel in straight lines from the point of emission. > >> >> They do NOT follow, in any way, the subsequent motion of the > >> >> emitting particle. > >> > Two things here.. Firstly,.. whats your proof these xrays > >> > travel in straight lines?? You have none. > > >> You put three lead plates with small holes in them along a straight > >> line leading away from the source. The xrays only reach the detector > >> when all three plates line up with the holes in a straight line. If the > >> xrays were traveling along a curved path, the holes would need to line > >> up along a curved path. > > >> > Show me the part > >> > of an experimental setup that checks the direction relative to the > >> > known > >> > source and the speed of the xray. You have neither nor do both tests > >> > exist. > > >> You have never worked with an xray beam line at a synchrotron. > >> visit a facility such ashttp://www.camd.lsu.edu/andyou will see that > >> they use the synchrotron to generate xray beams. BEAMS of xrays > >> traveling in straight lines. If your idea about the emission following > >> the movements of the source were correct, CAMD would not work as it > >> does. > > >> > This is a fabrication on your part. > > >> I don't need to fabricate anything because I know what I am talking > >> about. > > >> > Secondly,.. you make the assumption that the xrays are produced by > >> > moving particles/ie ...rotating sources like the MMx or sagnac > >> > source. But what proof have you that the sources in a cyclotron > >> > is rotating?? NONE > > >> The xrays are produced when a beam of particles is forced to CHANGE > >> directions. When this is done, they emit photons. This is known as > >> cyclotron radiation. > > >> > You can only assume the xray `sources` are rotating theoretically. > >> > IM not familar enough with the technical setup of a cyclotron > > >> I am. > > >> > but my reference refers to the `particles` not as atoms but as > >> > electrons > > >> Electrons or positrons or protons or heavier particles can be > >> accelerated. When the beams are bent(forced to change direction of > >> motion) photons are emitted. > > >> > As far as Im aware electrons are emr. > > >> You are quite mistaken. The motion of electrons generates electro > >> magnetic radiation but electrons are NOT electro magnetic radiation. > >> But, all of this is beside the point. > > Sorry Im not sure why I said this .Yes the electrons in wave theory > > too , arent the emr but they generate the emr . > > That is ok. We all make mistakes at times. > Wisdom comes from making mistakes and from learning from them. Im not sure if its a big a mistake as youd like to think it is. I was trying to say that electrons in a classical wave theory are waves only . I accidentally used the term emr. I should have said this.. " As far as Im aware electrons are wave only phenomena." > >> The point being that once emitted, light doesn't care what happens to > >> the source. > > >> [snipping the rest of the side issues] > > >> .... > > >> > Lets see if for starters you can supply the lab setup in a cyclotron > >> > that measures the speed and dierection of the observed xrays.. > >> > You cant, because no such test was ever made. > > >> Wrong. The direction is easily determined by three lead plates with > >> holes in them. > > OK fair enough. This is the first time anyonse asked me to define a > > cyclotron in wave only theory so I need to get uo to speed on how they > > are constructed. Its not quite clear in my reference how this is done. > > For instance it shows the defelected `electrons` but not where the > > xrays are generated. I assume the whole cavity generates them or is it > > just a point at where the electron is deflected at the small outlet > > hole? Its very vague my reference. > > Xrays are only emitted where the beam of electrons is bent. Xrays ( or > lower energy emr) can be emitted whenever electrons are made to slow down > suddenly, such as the beam of electrons from the cathode strike the anode > in an x-ray tube or when the electron beam strikes the screen on your TV > tube. That is why the voltage on the TV CRT is prevented from exceeding a > certain threshold voltage (somewhere around 30KV if I remember correctly) > to avoid excessive xray production)[but this is again beside the point]. So in my diagram of the cyclotron, radiation comes only from the small point by the exit hole where it says the electron beam exits? The construction is not clear enough yet. My reference is too vague. For instance presumably there is a window material where the `electron beam` exits. Because if it were just an open hole there would be no vacuum. And surely then the window material must then *also slow down* the supposed ` electron beam`in the same way as your anode ..Or maybe the window material *is* the `anode`.?? It doesnt specify this in my reference nor does it specify if the `deflecting` anode is oscillating. Is it? I need more time and info to understand exactly what happens in a cyclotron. > >> The speed is easily determined by measuring the time between the moment > >> that the batch of particles gets forced to change direction of motion > >> and the time the xray pulse hits a target along the beam line. Also, > >> the beam line can contain two detectors, a known distance apart and the > >> time for the pulse to travel between those detectors can be measured. > > You obviously know a good deal about cyclotrons. I admit need to find > > out more about cyclotrons . For instance my reference > > shows the electrons being fired out the small hole? > > The electron beam may start from an electron gun similar to that in your > TV CRT. > > > Or is that the > > xrays. > > The xrays are allowed to exit through a series of small holes so that the > beam is of precise energy and its direction, energy and area of > illumination are known. > > > Its not clear. Any ways in a wave only theory I have to > > explain electrons as waves only so its going to be difficult > > for you to argue that the source is particles. > > It doesn't matter if electrons are waves (they DO have wavelike > properties) or particles (they do have particle like properties) or just > best described by a series of equations and not labeled with 'wave' or > 'particle' at all. It does matter if a wave only model doesnt treat the xray source as a beam of particles. Because then the cyclotron as a whole is treated as one static oscilatting source rather than a series of moving particle sources. And then wave theory predictions are consistent with the observed straight lines of the xrays. Thats why Im trying to find out exactly how the cyclotron works. Because only then will I be able to explain it as an oscillating single wave source. The problem is I havent studied them before and the available description just doesnt say enough. In fact my reference doesnt seem to make a distinction between the electron `beam` and the xray radiation. It almost seems to me from my reference that the two are both the same? Or at least it seems to suggest that the only thing the cyclotron emits are a beam of xrays. No mention of a beam of electrons coming out of the `hole` that I can see. You may be able to clarify this. > > For instance > > I notice that the setup of the cyclotron bears close resemblance > > to the radio transmittter. > > That is no accident. > > > Both have oscillating magnetic fields > > as the `driver` of the radiations and both in the standard model > > describe these as streams of electrons. But wave theory as > > I point out in my "redshift without expansion " describes the > > oscillating field in the transmitter as oscillating magnitic > > fields of the atoms in the wire which produce the emr radiation. > > There are instruments called 'ESR' and 'NMR' (electron spin resonance > spectrometers and Nuclear magnetic resonance spectrometers) that can be > used to precisely determine the properties of the magnetic fields from the > particles in the atom. > > You need to study how these work and what they tell us about the > environment around each atom in a complicated molecule. Please if we could stick to cyclotrons? I can study up and cover these new ESR and NMR points later if youd like, but not now. Dont forget, the points I initailly raise about the MMx source still have yet to be proved incorrect. As it stands, the evidence from MMx show us that light is emitted and travels away at c in the source frame. Which is consistent with the emmision model I outline where I say that if light is at c in the source frame then one can also explain sagnac. So far nobody has supplied viable proof that light isnt at c in the source frame nor viable proof that a model that is based on this cant explain sagnac. Because as Ive said many times... A mathematically correct calculation of sagnac using these parameters does allow emmision theory to explain sagnac. And the reason all other calculations come to the conclusion that emmision theory cant,.. is because they incorrectly assume that light isnt at c in the source frame. You still have yet to prove to me that light does not move away from the source in MMx at c. That was my initial argument in this thread. > If your conjecture were right, these instruments would show much different > results than they do. > > You must be able to explain, for example, why each peak on a 2d 400 MHz > (1H, 1H) COSY-45 spectrum of a particular compound shows up at the exact > location it occurs at. You could ask me to explain every known observation and experiment in physics, and I could spend years asnswering you. But it ignores the fact that so far you have yet to supply any evidence that my initial argument is incorrect. That is .. does light travel at c in the source frame? And so far no one can prove otherwise. But a quick answer to your point above. I did in fact deal briefly a few years back with this point about spectral lines. And its a big subject in itself for just catchphrase answers but a wave only atom, I`ve argued in the past, is a oscillating node in the vacuum/aether with a range of oscilatting frequencies for each atom. THis simple model can easily describe the spctral lines observed. If a wave only atom resonates at `x` nm frequency then it goes without saying that it will emit radiation at that frequency. > Here is a place to start. It will tell you 'how it works'.http://www.cryst.bbk.ac.uk/PPS2/projects/schirra/html/theory.htm > > > No electrons needed. So Id say the cavity in the cyclotron is > > the source and it is the oscillating magnetic field in the > > cavity that generates the xrays. > > Wrong. If you were correct, then my microwave oven would generate xrays. > My ham radio transmitter would generate xrays. We would all be dead > because the cell phone towers would generate xrays. Why would a microwave oven emit xrays? Its source (whattever the mechanism is inside) must obviously be an oscillating magnetic field where the frequency is not that of xrays. In the same way that the radio transmitter oscillates in the radio wave frequencies not xrays. > > Just as the the wire `cavity` > > in the radio transmitter generates the radio waves. > > The 'cavity' does NOT generate the radio waves. Oscillations of current > (electron flow) in the antenna cause electromagnetic radiation to radiate > in the form of radio waves. But youve ignored my request from earlier. Im proposing that a wave only model has to model ALL phenomena as waves . Therefore a wire with a current has to be modeled not as a wire with a stream of electrons but as a wire with atoms that are oscillating their magnetic fields at radio wave frequencies. And these oscillations can be directly explained by the fact that the current is generated by... the wires atoms moving through an oscillating magnetic field in the electric dynamo. I know this is true because I know that electric current is generated by a rotating magnet around (so to speak) a copper wire.Its obvious the atoms in the wire are having their magnetic fields oscillated. One doesnt need electrons to explain electric ac current at least. So please dont insult me by asking me to explain emmision theory by rejecting one of classical wave models most fundamental rules that wave theory allows no particles. Its a bit like a creationist asking a darwinian to explain darwinian theory but only if they (darwinians)accept that the world was created 10,000 years ago ande incorporate that restriction in their explanation. > > I assume > > the cyclotron cavity oscillates at higher frequencies hence the > > higher frequency emmision.Thats whats suggested in my ref anyways. > > Higher frequency, yes. Xray frequency, NO. Are you sure about this? My impression is that somewhere in the setup there has to be an oscillation created that matches the frequency of xrays.Otherwise there could be no xray emmision. And Ive been told that these sort of electric oscilltion frequencies are easily acheived in all sorts of applications. > > As with the transmitter the cavity as a whole is the source. > > No to both. If I fail to hook an antenna to my transmitter, there are no > radio waves emitted. Well, by transmitter I meant both antenaae and power source. But the same then goes for the cyclotron doesnt it? After all if you turn off the oscillating field in the cyclotron you wont get any xrays emitted either. > > And it doesnt move in the lab. It is in the same frame as the > > observor. Hence contrary to what you thought no bent lines in > > the x-radiation are predicted by wave only emmision theory > > because the source isnt moving in the lab > If the microwave accelerators are TURNED OFF, those electrons still in the > beam will continue to circulate until the electrons loose energy and the > beam loses coherence. At the 'bends' xrays would still be emitted. I dont see how that disproves the wave only theory. For instance there must also be a observable decay when a radio transmitter turns off surely. Or what about a lightbulb? THat fades out , yet its emmision can be explained as oscillating magnetic fields in a high resistance filament. > It is NOT the microwaves that produce the xrays. Anymore than it is the > gunpowder that produces the bullet. The gunpowder propels the bullet but > the bullet is a piece of lead that is crimped into the cartridge. > > >> >> He did NOT say 'break apart', he said 'distort'. > >> >> In the MMX, all it takes is moving part of the apparatus by a small > >> >> fraction of the wavelength of light in order to create a detectable > >> >> indication. > > >> >> THAT is why he said 'distort'. Spinning the apparatus would cause > >> >> changes in the instrument that would make it useless for the purpose > >> >> you propose to use it for. > >> > I assume by this that you and george imply that a rotating MM > >> > experiment > >> > physically stretches one arm moreso than another and doesnt give a > >> > null > >> > result? If so please supply a reference for such an observation, > >> > because > >> > I dont believe you have any proof to back up this claim.(notice that > >> > will not be able to supply this as none exists. Youve just fabricated > >> > this evidence) > > >> If you ever spin anything (a car tire for example) you will notice that > >> getting everything 'in dynamic balance' is NOT trivial. Any > >> experimental apparatus consists of many different parts, each with > >> their own mass, density and, in this case, distance from the point > >> about which the apparatus is to be rotated. > > >> The MMX apparatus must maintain all dimensions within fractions of the > >> wavelength of light. > > >> ANY slight imbalance or difference in the distances from the center > >> would produce different torques and invalidate the experiment. Have you > >> ever seen an MMX apparatus? > > >> And one final point. The MMX apparatus IS spinning at a slow rate of > >> spin, 1 revolution per day. > > I dispute these points on several counts. First MMx DID spin the setup > > and it didnt create distortion in the arms. > > WRONG. The movement of the apparatus from one position to another > 'produced fringe shifts of up to 7 fringes' and that is after the > apparatus was allowed to stabilize. You do realize that the entire > apparatus was floating on mercury in order to reduce the stresses of > rotation, don't you????? If the apparatus gave fringe shifts up to 7 then how could they assume a null result? > > So it is possible.Second, > > if this experiment hasnt been tried in your opinion, it doesnt mean > > that > > one can assume that the outcome would be that light wouldnt travel at > > c > > relative to the spinning MMx source,.. which leads to my third point; > > As you point out above. A spinning MMx setup has been tried where > > no distortion occurs and it does show that light from a spinning > > source > > still travels away at c relative to the source. > > Its called the MM experiment. And it spins 1/day. > > And it shows a negative result. Ive heard it called null result. But this means that light was travelling at c on both paths then doesnt it? It seems youve just tacitly admitted this. Yet earlier you suggest that in fact light doesnt travel at c in the source frame. You seem to suggest it only emits at c and then travels at c+-v in the source frame. No offense intended but... clearly MMx refutes this claim of yours.(and others like George) > >> > And secondly assuming the remote chance that it does distort such an > >> > apparatus and not allow us to check what I propose may be observed > >> > (ie whether or not a rotating MMx that isnt distorted does still > >> > have light at c on both paths)...THen how is it that you or george > >> > can say with certainty that a rotating MMx source does not allow > >> > light > >> > to propagate down both arms at c? After all, youve just claimed > >> > that any experiment to test this is impossible. > > >> The math is not subject to the physical limitations of the equipment. > > >> [snipping the rest because it has nothing to do with **the point** that > > >> once light is emitted, subsequent motion of the source can not and does > >> not effect the already emitted light > > This is wrong and unsubstantiated. Ive pointed out many times > > that the MMx results contradict this claim you and others make. > > Because we know that when light leaves the MMx source it still travels > > in all directions at c relative to the source. THis is only possible > > if > > the subsequent motion of the source did effect the already > > transmitted > > light. > > Sean, I have given you the truth, as best I know it. I have suggested > simple experiments that YOU could do to test your ideas. Yes and Id love to be able to perform that test. But until then If you say youve given me the truth then why is it that you seem to be saying two contradictory things about light? On one hand you say light cannot travel at c in the source frame. Yet on the other you admit that MMx shows us that it does travel at c in the source frame. So whats the truth ? Does it or doesnt it? Why cant you or george come clean on this simple fact? It seems to me that you know Im right because you know what MMx shows us. But you also know that if you admit that Im right then that means admitting that emmision theory can explain sagnac. > If your conjecture were correct, the CAMD synchrotron would not work as it > does, the beams of xrays would follow a curve rather than a straight line. Not of the CAMD synchroton was modeled as a single osillating source in the lab frame. As wave theory does model > Neither waves on water nor anything else behaves as you propose for light. I never said waves on water are like light waves. Maxwell made that mistake Not me. > There is no reason to suppose that the light can know or care what happens > to the source of the light once the light has left it. NOTHING else > behaves that way. There is no evidence that light behaves that way. > If the source changes motion or even ceases to exist after the light has > been emitted, the light neither knows nor cares. > In fact, as the 'source' is most often an ion, atom or molecule in an > excited state, > > and the act of emission makes the source 'cease to exist' [in the excited > state] > > then there is no longer a 'reference' to which to tie the light, once it has > been emitted, > > so there is no way for the light to know what it's 'mother' does, once it has > left home. All these above assumptions are made by assuming that the source moves in the source frame after the light has left. But it doesnt.It stays still in the source frame. Thats why light always travels out in straight lines at c in the source frame. And it doesnt need to check back because in the source frame,.. the source always is in the same position no matter how far away the light travels. MMx confirms this. Unfortunately what you you cannot confirm, is that light travels at variable speeds in the source frame. The reason is that any experimental observations show us that in fact light always travels at c in the source frame. I didnt make this up, this is what everyone accepts as being observed in MMx Sean www.gammarayburst.com http://www.youtube.com/profile?user=jaymoseleygrb
From: George Dishman on 17 Sep 2007 14:45 "Henri Wilson" <HW@....> wrote in message news:vqied35c30k174ta1rhp3lscfd41q85ec6(a)4ax.com... > On Thu, 30 Aug 2007 08:18:38 +0100, "George Dishman" > <george(a)briar.demon.co.uk> wrote: >>"Henri Wilson" <HW@....> wrote in message >>news:rs1cd3hh01s0olae1smqq5vuioqflqf23j(a)4ax.com... >>> On Wed, 29 Aug 2007 18:09:16 +0100, "George Dishman" > >>>>Perhaps you weren't following those conversations, but >>>>you are certainly aware that I gave you links to a video >>>>and stills of single-photon experiments when we were >>>>discussing gratings and you know I have repeatedly >>>>pointed out that individual photons are deflected by the >>>>same angle from a grating as predicted by the classical >>>>wave analysis >>> >>> ...and by my photon model >> >>Your equations don't even describe a repetitive >>phenomenon so there is no wavelength in them. >>Maxwell's Equation are second order differential >>so a sine wave is a solution but it propagates at >>a specific speed determined by the coefficients >>and independent of the speed of the source. A >>sine wave with source-dependent speed cannot be >>described that way so you have a major task in >>finding a formulation for that. I don't think >>Ritz ever did it either. > > A photon is not 'a sine wave moving through space'. > A sine wave is a convenient way of representing the field variations.se: > www.users.bigpond.com/hewn/e-field.exe Let's break that into parts to help you along. First note that you have a wavey like thing moving sideways and with an imposed amplitude variation or "envelope". Ignore the envelope and the sideways motion and you have a standing wave. If you send a propagating sine wave against a mirror, that is what you get. Now introduce the motion. To do that you move the mirror to the right slowly. The reflected wave must have the same wavelength, lower velocity and lower frequency (due to Doppler) so without the envelope your diagram shows two "sine waves moving through space". To get the envlope is trickier. You multiply the sine wave going from left to right by the envelope and take a Fourier analysis and you get the Fourier transform of the envelope itself as sidebands on the sine wave which acts as a carrier, but it moves at the speed of the wave, not slowly as you have shown. To solve that you need dispersion, the speed varies with frequency. Do the same to the second wave but note that the envelope is propagating in the wrong direction, left to right while the carrier goes right to left. Once you do the maths for all that, you will find you don't have an intrinsic frequency but two broadband signals with continuous frequencies, not discrete and anisotropic dispersion. In other words, your drawing shows the combination of an infinite number of 'sine waves moving through space' but moving in both directions and at different speeds. So the question is Henry, where is the mirror? George
From: bz on 17 Sep 2007 14:02 sean <jaymoseley(a)hotmail.com> wrote in news:1190036876.674031.136830(a)d55g2000hsg.googlegroups.com: > On 12 Sep, 15:26, bz <bz+...(a)ch100-5.chem.lsu.edu> wrote: >> sean <jaymose...(a)hotmail.com> wrote >> innews:1189444302.990016.237750(a)d55g2000hsg.googlegroups.com: > > www.gammarayburst.com > http://www.youtube.com/profile?user=jaymoseleygrb > >> > On 6 Sep, 17:33, bz <bz+...(a)ch100-5.chem.lsu.edu> wrote: >> >> sean <jaymose...(a)hotmail.com> wrote >> >> innews:1189089230.764471.53320(a)22g2000hsm.googlegroups.com: >> >> >> > On 30 Aug, 23:55, bz <bz+...(a)ch100-5.chem.lsu.edu> wrote: >> >> >> sean <jaymose...(a)hotmail.com> wrote in >> >> >> news:1188512224.511353.237820 @i13g2000prf.googlegroups.com: >> ..... >> Let us put a LED (light emitting diode) on the rim of a ultra >> centrifuge rotor that is 10 cm in diameter and rotate the disk at >> 50,000 rpm. The circumference will be 0.628 meters. The speed of the >> diode will be 523.599 meters per second or 1.7 micro c. >> >> Now, we spin the rotor clock wise for one test and counter clockwise >> for another test. We pulse the LED near the point where it is closest >> to us (thus traveling at max speed perpendicular to the line of sight) >> From a distance of 10 meters we use a long focal length microscope to >> view the position of the LED at the moment that it pulses. Our >> microscope gives us a 100 to 1 magnification of the image of the LED. >> We use a ccd camera to display the image on our computer screen. >> >> What do I predict and what does your theory predict? >> >> I predict that the image will appear in the same location, non >> rotating, or rotating clockwise or counterclockwise at 50,000 rpm. >> >> What does your theory predict? Your theory predicts that during the >> 0.033 us it take the light to travel from the LED to our >> tele-microscope 10 meters away, the light will have moved 0.017 mm in >> the direction of rotation. As there are LEDs that are only 20 um in >> size,http://www.patentstorm.us/patents/6410940-claims.html so, I >> calculate that if your theory were correct that the image would be >> displaced by 873 diameters in the direction of travel. > In principle this sounds like it could work > I get a deflection over 10 meters of about .015mm. so thats pretty > close > to yours. You probably used a different speed for light than I did. > I wonder if thats large enough to measure over 10 meters? > I also wonder about the pulse timing. It seems to me that with the > time light travels the 10 meters being about 1/30,000,000 second you > would have to have pulse timing thats accurate to 1/30,000,000sec. > Maybe one alteration to make it simpler could be this... > Use your setup as youve described .....but add > a single LED positioned directly above where the other LED is when it > pulses and have that on all the time. Then with the imaging scope > up very close turn on the rotation and adjust the pulse timing so that > the > pulse LED turns on just directly below and in line with the other LED. > Ie the two LEDs are in line vertically when the pulsed LED is > rotating. > > LED 1 > LED 2 > Then move the camera back and theoretically if Im right, the > rotating LED should slowly move out of line with the non rotating LED > the farther back the camera moves from the setup. > LED1 > LED2 > Or not if your prediction is right. > This way we know for certain the pulse timing is spot on. > Unfortunately the price and tech know-how for this experiment > is still probably out of my reach so for the time being this > experiment > cant be done. The experiment could be done by buying some surplus scientific equipment fairly cheaply. My point was that is was technically possible to test your theory. > Having said all that I realize that Im not sure how to model a source > that pulses. Once a source goes OFF it would seem that it no longer > exists as a source That is a point that I made later. Technically, one an atom has emitted a photon, it no longer exists as a source. It must be excited again before it can re-emit. > which means any subsequent motion of an `off` > source > would not effect the light as it propagates away. Which is what several of us have been trying to tell you. > So it might be better for the purposes of this experiment to pulse it > on > and leave it on as it rotates rather than me worrying how to model > the motion of a nonexistent source. > Hopefully this shouldnt matter for you as it still means the `on` > point > will be `dragged` or not and the vital theoretical test can still be > made > ..... >> And I have yet to see anything else that works that way. >> >> The golf ball does not bobble as tiger completes his follow-through. >> >> It is contrary to all data we have and it is easy to test, as I have >> just shown. > What I mean above is that the MMx experiment is essentially a source > on a rotating table. Except the rotational diameter is large and the > interval 1 day. But the rotational speed of the MMx is still fast > Dont forget the lab and MMx source are rotating at high speed around > the earths axis. If light did not travel away from the source on both > paths at c in the source frame as you suggest then one path would be > slower than the other and not give a null result. But we observe > both paths in MMx to be at the same speed. Which can only mean > that light in MMx is propagating *away* from the source on > both paths at c. Of course it does. But if there were an aether, light traveling one path would have to travel further through the moving aether than light traveling the other path. The different path lengths would make the fringes shift. By the way, the paths perpendicular to the flow of the aether are the ones where the light must travel further and take longer. On the paths with and against the flow of the aether, the upstream and downstream times sum to the same total time. > And thats only possible if light was travelling > at c only in the source frame. And if light travels > at c in the source frame one would have to get this `drag` I > predict. In other words you dont have to do your above experiment. Right. MMX already shows us that the light travel speed is the same in all directions. There is no aether flowing. > MMx has already shown us light drags along with the source. NO! Once the light leaves the source, it travels at a constant velocity with respect to the inertial frame of reference that was co-moving with the source at the instant that the light was emitted. THAT frame of reference continues to move at a constant velocity. What the source does, after emitting the light, is unimportant. > (although drag is georges term. I like to think of it this way.. > that light is at c in straight lines only in the source frame) That is incorrect. Although it is true of the frame that was co-moving with the source at the instant of emission, it is also true that the light will be measured to move at c in ANY inertial frame of reference. If the light source is not in an inertial frame of reference, we just look at the inertial frame of reference that was moving with the same instantaneous velocity as the light source at the moment the light was emitted. That frame continues to move at a constant velocity. The light source does NOT drag it with it. >> >> > That emmision theory cannot have light as a particle with mass. >> >> >> I don't care if light is a wave or a particle at this moment. It >> >> doesn't matter. The light, once emitted, goes its merry way. Unlike >> >> the teenager, leaving home for the first time, the light does not >> >> 'write home' and ask for more money. The light doesn't care what >> >> happens to the source after it has left home. > >> > JUst dont forget that Im argueing here for a wave only emmision model >> > to explain the various observed phenomena. So please dont expect me >> > to incorporate particle theory in this model. Like for instance.. >> > electrons. >> >> Electrons are sources of photons as they circulate in the synchrotron. >> The fact that the electron is a particle is NOT important. >> The fact that we KNOW the path that the electron is taking as it emits >> those photons IS important. The electrons are moving along a curved >> path at a high velocity. > But youve just ignored what I said. Electrons dont exist in classical > wave > theory. One has to describe any stream of electrons in wave terms. Why? > So effectively in wave theory a cyclotron isnt modeled as a spiralling > beam of particles called electrons. Its modeled as an oscillating > magnetic > field in a cavity. NO! The wavelengths of the electrons are MUCH TOO SHORT to resonate in the cavities involved. You need to study how the cyclotrons actually work. The model of pushing batches of particles around works much better and easier than trying to consider the wave nature of the electrons. > Hence the source is no longer a rotating beam > of particles but a static oscillating cavity source. The beams are actually observable. Batches of electrons in the beams are observable. There is nothing to indicate a 'static oscillating cavity source. > Ill cover this a bit more below. >> The electrons are the SOURCE of the photons. >> The source is moving along a curve. >> By your theory, the photons are 'tied' to the source and must mirror >> its motion. The photons do NOT do that. >> This is experimentally verified, as I showed. > In my theory there are no photons. It is wave only. It may be that > you arent familiar with my other posts but over the years I have been > arguing > for a classical wave only emmision model. And incidentally this isnt > my > theory . Its just observed physics minus the standard model theory > assumptions > about particles and SR. I suggest NO new laws or assumed variables. A 'classical wave only model fails to explain the particle like properties of photons, electrons, neutrons, etc. > Even my idea of `light at c in the source frame only` is actually > just me studying what actually was observed in MMx and writing it > down. > As far as Im aware whatever other theoretical assumptions Lorentz or > Einstein > made,.. MMX does show us that light HAS TO travel at c in the source > frame. If you would recognize that 'the source frame' means 'an inertial frame of reference that was co-moving with the source at the instant of emission' then we would have no reason to disagree with you. That means, however, that you must recognize that what happens to the source after it emits the light has no effect on the light already emitted. > And thats all I say. ..... >> It travels at c in all directions, as measured in an inertial frame of >> reference co-moving with the source at the moment of emission. > You can use the term inertial if you want but its a bit of a > misleading > term to use because it suggests one is measuring the light in a frame > other than the rotating MMMx source frame One MUST use an inertial frame of reference. > When in fact the truth is that the light is measured in the source > frame > *only* in MMx. > Not some other `inertial` frame. And the source frame > is > indisputably rotating once a day. Such a source frame is NOT an inertial frame of reference. > So its not inertial contrary to what > SR suggests. It can only approximate an inertial frame of reference over a short period of time or you can use an inertial frame of reference that is co-moving with the source for a short period of time, until the source and that inertial frame of reference have diverged 'significantly' in position or velocity. > And incidentally I know what your response willl be.. > That the inertial frame approximates the source frame.Ive heard this > before > and its unnacceptable on two counts. One, that its approximate. > In physics this isnt good enough. You don't understand experimental science. ALL experiments are 'only approximate'. Our test equipment has limited accuracy. And even if our test equipment were infinitely precise, the laws of physics prevent knowing anything 'exactly. > And two, seeing as its only > approximate > it would be scientifically more accurate to measure and calculate in > a > frame that isnt approximnate. And that better frame is the source > frame. NO. It is scientifically more accurate to measure what can be measured. Calculate what the theory says should be measured and examine the discrepancies between what is predicted and what is observed. That is what science does. The theory is based on inertial frames of reference, so that is how the calculations of predicted values are to be made. Those values are then compared with what is actually measured. Repeated experiments give slightly different answers and the error bars for the data are calculated. If the data with the error bars shows values different than predicted then the theory must be revised. That is how science works. > Its exact , not approximate, Which is what I do in my emmision theory > calculations in the exacting source frame. >> If the source changes motion or even ceases to exist after the light >> has been emitted, the light neither knows nor cares. > Its all relative. If your in the source frame,.. the source never > moves. Your mistake is that you don't recognize that the rule 'light moves at c with respect to the source' ONLY works if the source continues to be stationary in the inertial frame of reference that was co-moving with the source at the time of emission. > Hence the light doesnt have to move either, that is; other than move > away > from the source at c in straight lines The light moves at c in straight lines relative to the location that the source was in at the time of emission, in an inertial frame of reference that was co-moving with the source at the instant of emission. If the source leaves that position, the light neither knows nor cares. .... >> They travel at c as measured in an inertial frame of reference that was >> moving at exactly the same velocity as the source at the instant of >> emission. > Partly true, but theres more.. It also travels at c relative to the > source > **after** it has left the source. If it didnt there would be no null > result observed. >> > THat means that you think that light travels at different speeds away >> > from a source relative to that source. >> >> They travel at c as measured in an inertial frame of reference that was >> moving at exactly the same velocity as the source at the instant of >> emission. > You forgotton something very fundamental here. The MMx experiment > measures > what light does AFTER it leaves the source. Not just when it leaves, > as > SR incorrectly assumes. The entire MMX apparatus approximates an inertial frame of reference during the time of travel of the light from the source, through the beam splitter, to the various mirrors. > And the MMx measurements indisputably show us > that light travels at c on both paths AFTER it leaves the source. Yes. The entire apparatus approximates an inertial frame of reference. > And emmision theory predicts that if this is true then a null > result should be observed... And guess what... A null result > IS observed. !! The important point of MMX was NOT that the observation MATCHED what was predicted. Science does NOT work that way. An experiment where the outcome matches the predictions is only mildly interesting. It just adds support to the current theory. A theory can not be proved. Many theories can predict the same results and one can not choose between the theories on the basis of experiments where the results agree with the theories. What was earthshaking about MMX is that it did NOT produce the predicted results. The 'currently accepted theory' of the day was the aether theory. The aether theory predicted that MMX would produce a positive result. When MMX failed to observe the aether drift, science had to toss out the aether and formulate a new theory. It formulated many different theories. Several of which are currently usable. You have a fundamental misunderstanding of science. You are trying to PROVE your theory. Theories can NOT be proved. Only disproved. You are wasting your time and the time of those who are trying to help you see the problems with your approach. >> Light travels at c as measured in ANY inertial frame of reference, >> irrespective of the velocity of that frame of reference wrt the source. >> [with the caveat that no inertial frame of reference can ever be >> defined so as to move at c with respect to any object that has mass]. > I think you are talking about SR here. Primarily Im discussion how > emmision > theory can also explain both MMx and sagnac . Not SR. SR is perfectly happy explaining MMX an sagnac. > As long as emmission theory has the basic rule that light is allowed > to > travel away from a source at c relative to the source. And I see no > reason > why it cant , especially considering that this is whats observed in > MMx. first of all, no one that I know of, besides you, claims that the emitted light continues to travel at c wrt the source, no matter what the source does after the light has been emitted. Second, the fact that your strange twist on the emission theory seems consistent with MMX does nothing useful. Experiments do NOT prove theories. Any theory that is consistent with MMX is just that, consistent with MMX. Theories must stand other tests. Your theory FAILS many tests because it provides NO way to tie the emitted light to the source. It would be NICE if your theory were correct because we could then have faster than light signaling. How? take a laser diode and wiggle it side to side, keeping the laser beam parallel at all times. If your theory were correct, the entire beam would have to wiggle side to side, instantly, along its entire length. At a distant site, that wiggle could be detected and it would provide an instant method of communications, faster than light. ..... >> That is ok. We all make mistakes at times. >> Wisdom comes from making mistakes and from learning from them. > Im not sure if its a big a mistake as youd like to think it is. > I was trying to say that electrons in a classical wave theory are > waves > only . I accidentally used the term emr. I should have said this.. > " As far as Im aware electrons are wave only phenomena." Electrons seldom behave like waves, and when they do, the wavelength is very short. That is why electron microscopes let us look at very small things. We are using electrons as waves. If you look at the image from an electron microscope, you will get some idea of the 'wavelength' of the electrons. It must be much smaller than any object imaged. .... >> Xrays are only emitted where the beam of electrons is bent. Xrays ( or >> lower energy emr) can be emitted whenever electrons are made to slow >> down suddenly, such as the beam of electrons from the cathode strike >> the anode in an x-ray tube or when the electron beam strikes the screen >> on your TV tube. That is why the voltage on the TV CRT is prevented >> from exceeding a certain threshold voltage (somewhere around 30KV if I >> remember correctly) to avoid excessive xray production)[but this is >> again beside the point]. > So in my diagram of the cyclotron, radiation comes only from > the small point by the exit hole where it says the electron beam > exits? The electron beam does NOT exit. Its path is bent and it continues to circulate around the cyclotron ring. > The construction is not clear enough yet. My reference is too vague. > For instance presumably there is a window material where the `electron > beam`exits. The beam would only be allowed to exit if one wanted a beam of electrons. If one wants xrays, you just bend the beam. > Because if it were just an open hole there would be no > vacuum. There are many articles on how cyclotrons work. Google is your friend. ..... I snipped continued misconceptions about how the cyclotron works. > I need more time and info to understand exactly what happens in a > cyclotron. YES. ..... >> It doesn't matter if electrons are waves (they DO have wavelike >> properties) or particles (they do have particle like properties) or >> just best described by a series of equations and not labeled with >> 'wave' or 'particle' at all. > It does matter if a wave only model doesnt treat the xray > source as a beam of particles. Because then the cyclotron as a > whole is treated as one static oscilatting source rather than a series > of moving particle sources. You need to look at how the cyclotron works. You have linear acceleration sections and you have beam bending sections. The xrays are emitted in the beam bending sections. > And then wave theory predictions > are consistent with the observed straight lines of the xrays. The electrons are the source of the xrays. The electrons are forced to make a turn. If the xrays were 'tied' to the source, they would ALSO have to make a turn. They don't. > Thats why Im trying to find out exactly how the cyclotron works. It is no secret. > Because only then will I be able to explain it as an oscillating > single wave source. The problem is I havent studied them before and > the > available description just doesnt say enough. In fact my reference > doesnt seem to make a distinction between the electron `beam` > and the xray radiation. It almost seems to me from my reference that > the two are both the same? They are quite distinct. What is your source, anyway? > Or at least it seems to suggest that > the only thing the cyclotron emits are a beam of xrays. That is correct. > No mention of a beam of electrons coming out of the `hole` that > I can see. You may be able to clarify this. That is correct. The electrons circulate around the cyclotron's 'race track'. Wherever they must make a turn, xrays are emitted, in a straight line. Kind of like the sound of squealing tires as the race cars skid around the turns on one of those tracks that have multiple turns and straight away sections. The electrons don't leave the track any more than the race cars do. ..... > >> You need to study how these work and what they tell us about the >> environment around each atom in a complicated molecule. > Please if we could stick to cyclotrons? I can study up and > cover these new ESR and NMR points later if youd like, but not now. > Dont forget, the points I initailly raise about the MMx source still > have > yet to be proved incorrect. As it stands, the evidence from MMx > show us that light is emitted and travels away at c in the source > frame. MMX showed that there was no aether. You really need to understand how science works. > Which is consistent with the emmision model I outline where I say that > if light is at c in the source frame then one can also explain sagnac. > So far nobody has supplied viable proof that light isnt at c in the > source frame nor viable proof that a model that is based on this cant > explain sagnac. Because as Ive said many times... A mathematically > correct calculation of sagnac using these parameters does allow > emmision theory to explain sagnac. And the reason all other > calculations come to the conclusion that emmision theory cant,.. > is because they incorrectly assume that light isnt at c > in the source frame. You still have yet to prove to me that > light does not move away from the source in MMx at c. > That was my initial argument in this thread. >> If your conjecture were right, these instruments would show much >> different results than they do. >> >> You must be able to explain, for example, why each peak on a 2d 400 MHz >> (1H, 1H) COSY-45 spectrum of a particular compound shows up at the >> exact location it occurs at. > You could ask me to explain every known observation and experiment > in physics, and I could spend years asnswering you. Your theory, to be successful, must do exactly that. > But it ignores > the fact that so far you have yet to supply any evidence that my > initial > argument is incorrect. I have shown several times and ways that your ideas are wrong. ANY experiment that is contrary to your theory disproves your theory. You have been shown many. > That is .. does light travel at > c in the source frame? And so far no one can prove otherwise. I have shown several reasons and ways to test, as well as consequences that would be contrary to what we know IF your theory were correct. > But a quick answer to your point above. I did in fact deal > briefly a few years back with this point about spectral > lines. And its a big subject in itself for just catchphrase answers > but a wave only atom, I`ve argued in the past, is a oscillating node > in the vacuum/aether with a range of oscilatting frequencies for each > atom. > THis simple model can easily describe the spctral lines > observed. If a wave only atom resonates at `x` nm frequency then > it goes without saying that it will emit radiation at that frequency. An atom can only emit energy when it changes from a higher energy state to a lower one. >> Here is a place to start. It will tell you 'how it >> works'.http://www.cryst.bbk.ac.uk/PPS2/projects/schirra/html/theory.htm >> >> > No electrons needed. So Id say the cavity in the cyclotron is >> > the source and it is the oscillating magnetic field in the >> > cavity that generates the xrays. >> >> Wrong. If you were correct, then my microwave oven would generate >> xrays. My ham radio transmitter would generate xrays. We would all be >> dead because the cell phone towers would generate xrays. > Why would a microwave oven emit xrays? Its source (whattever > the mechanism is inside) must obviously be an oscillating magnetic > field > where the frequency is not that of xrays. The cyclotron uses microwave energy to accelerate the electrons around the 'racetrack', but it produces xrays. The cavities are the same order of magnitude in size as the cavities in my microwave oven. If your theory were correct, both would produce xrays. > In the same way that the radio transmitter oscillates in the > radio wave frequencies not xrays. You claimed that oscillating magnetic fields generate xrays. Well, my radio antenna has oscillating magnetic fields. Why shouldn't it produce xrays? The syncrotron does NOT have oscillating magnetic fields at xray frequencies, why should it produce xrays and my antenna not produce them? >> > Just as the the wire `cavity` >> > in the radio transmitter generates the radio waves. >> >> The 'cavity' does NOT generate the radio waves. Oscillations of current >> (electron flow) in the antenna cause electromagnetic radiation to >> radiate in the form of radio waves. > But youve ignored my request from earlier. Im proposing that > a wave only model has to model ALL phenomena as waves . Therefore > a wire with a current has to be modeled not as a wire with > a stream of electrons but as a wire with atoms that are oscillating > their > magnetic fields at radio wave frequencies. If it worked the way you suggest, my radio transmitter would put out xrays. That model does not work. Atoms are much too massive to oscillate at the frequencies involved. Your theory is not viable. > And these oscillations can be directly explained by the fact that > the current is generated by... the wires atoms moving through > an oscillating magnetic field in the electric dynamo. > I know this is true because I know that electric current > is generated by a rotating magnet around (so to speak) a > copper wire.Its obvious the atoms in the wire are having their > magnetic fields oscillated. One doesnt need electrons to explain > electric ac current at least. So please dont insult me by > asking me to explain emmision theory by rejecting one > of classical wave models most fundamental rules that wave > theory allows no particles. Its a bit like a creationist > asking a darwinian to explain darwinian theory but only > if they (darwinians)accept that the world was created 10,000 years ago > ande incorporate that restriction in their explanation. You have yet to give me a wave explanation for the particle like properties of the electron. >> > I assume >> > the cyclotron cavity oscillates at higher frequencies hence the >> > higher frequency emmision.Thats whats suggested in my ref anyways. >> >> Higher frequency, yes. Xray frequency, NO. > Are you sure about this? My impression is that somewhere > in the setup there has to be an oscillation created that matches the > frequency > of xrays. I am sure that your impression is wrong. > Otherwise there could be no xray emmision. It is the bending of the beam of electrons that causes the electrons to emit xrays. > And Ive been told that these sort of electric oscilltion frequencies > are > easily acheived in all sorts of applications. You have also been told that much of what you have been thinking is wrong. >> > As with the transmitter the cavity as a whole is the source. >> >> No to both. If I fail to hook an antenna to my transmitter, there are >> no radio waves emitted. > Well, by transmitter I meant both antenaae and power source. > But the same then goes for the cyclotron doesnt it? > After all if you turn off the oscillating field in the cyclotron > you wont get any xrays emitted either. That is incorrect. The xrays will continue to be emitted as long as the beam of electrons continues to circulate. Eventually, the electrons will 'run out of gas' and stop, but it won't happen instantly. ..... >> If the microwave accelerators are TURNED OFF, those electrons still in >> the beam will continue to circulate until the electrons loose energy >> and the beam loses coherence. At the 'bends' xrays would still be >> emitted. > I dont see how that disproves the wave only theory. For instance > there must also be a observable decay when a radio transmitter turns > off > surely. A transmitter can be designed so as to turn off very rapidly. Radars send out very short pulses of microwaves and then listen for the echos. > Or what about a lightbulb? THat fades out It takes time for the hot filament to cool down. > , yet its emmision > can be explained as oscillating magnetic fields in a high resistance > filament. NO the emission is due to the high temperature of the filament. ..... > If the apparatus gave fringe shifts up to 7 then how could they assume > a null result? Because the number of fringe shifts should have been much larger. Have you bothered to read the experiment that you have spent so much time claiming as a proof of your theory? ..... >> >> And it shows a negative result. > Ive heard it called null result. But this means that light was > travelling at c on both paths then doesnt it? Of course it means light travelled at c on both paths. The point is that if the aether theory had been correct, light would have taken longer to travel one path than the other when the apparatus was oriented in the proper direction. The important thing about the experiment to science was that the results were different from the predicted results. > It seems youve just tacitly admitted this. > Yet earlier you suggest that in fact light doesnt travel at > c in the source frame. You seem to suggest it only emits at c and then > travels at c+-v in the source frame. You keep confusing the 'inertial frame co-moving with the source' and 'the source frame'. At the moment of emission, they are the same. Once the light has been emitted, the first one keeps moving at the same velocity, the emitted light only travels at c with respect to the first. What happens to the source after emission is unimportant to the light that has already been emitted. > No offense intended but... > clearly MMx refutes this claim of yours.(and others like George) NO. MMX refuted aether. SR is consistent with MMX. If it were not, SR would not have been a success. ...... >> >> Sean, I have given you the truth, as best I know it. I have suggested >> simple experiments that YOU could do to test your ideas. > Yes and Id love to be able to perform that test. But until then If you > say > youve given me the truth then why is it that you seem to be saying two > contradictory things about light? I said NOTHING contradictory. It only seems contradictory to you because you keep confusing 'the source frame' with 'the inertial frame that was co-moving with the source at the instant the light was emitted'. ..... >> There is no reason to suppose that the light can know or care what >> happens to the source of the light once the light has left it. NOTHING >> else behaves that way. There is no evidence that light behaves that >> way. If the source changes motion or even ceases to exist after the >> light has been emitted, the light neither knows nor cares. > >> In fact, as the 'source' is most often an ion, atom or molecule in an >> excited state, >> >> and the act of emission makes the source 'cease to exist' [in the >> excited state] >> >> then there is no longer a 'reference' to which to tie the light, once >> it has been emitted, >> >> so there is no way for the light to know what it's 'mother' does, once >> it has left home. > All these above assumptions are made by assuming that the source moves > in the source frame after the light has left. But it doesnt.It stays > still in the source frame. Thats why light always travels out in > straight > lines at c in the source frame. And it doesnt need to check > back because in the source frame,.. the source always is in the same > position no matter how far away the light travels. Only if the source is in inertial motion. If the source moves any other way AFTER the light has been emitted, your theory insists that ALL the light ever emitted by that source must also instantly move in the same way. ..... -- bz please pardon my infinite ignorance, the set-of-things-I-do-not-know is an infinite set. bz+spr(a)ch100-5.chem.lsu.edu remove ch100-5 to avoid spam trap
From: George Dishman on 17 Sep 2007 15:50 "bz" <bz+spr(a)ch100-5.chem.lsu.edu> wrote in message news:Xns99AE849FAE9DEWQAHBGMXSZHVspammote(a)130.39.198.139... > sean <jaymoseley(a)hotmail.com> wrote in > news:1190036876.674031.136830(a)d55g2000hsg.googlegroups.com: .... >> Having said all that I realize that Im not sure how to model a source >> that pulses. Once a source goes OFF it would seem that it no longer >> exists as a source > > That is a point that I made later. Technically, one an atom has emitted a > photon, it no longer exists as a source. It must be excited again before > it can re-emit. > >> which means any subsequent motion of an `off` source >> would not effect the light as it propagates away. Whoopee! > Which is what several of us have been trying to tell you. Thanks Bob, you seem to have finally got across to Sean what I have trying to tell him for months. >> So it might be better for the purposes of this experiment to pulse it on >> and leave it on as it rotates rather than me worrying how to model >> the motion of a nonexistent source. No! That's the whole point Sean, after the light leaves and the source is switch off, how can it affect the light that has already been emitted? Look again at my diagram: http://www.georgedishman.f2s.com/Sean/Sean_Planets.html The light was shown as a dot moving away from each planet because it came from a hand torch which you pointed at Sirius, switched on for a fraction of a second and then switched off. So then you lay it on the ground, hit it with a hammer several times, jump on the bits, dissolve them in acid, bury the goo in a landfill site and never see it again. How the (insert your favourite expletive here) can the remains of that torch, "the source", possibly affect the motion of the flash of light which is by now leaving the solar system? It doesn't make any sense. George
From: bz on 17 Sep 2007 17:24
"George Dishman" <george(a)briar.demon.co.uk> wrote in news:LO6dnW5Q6YqDRHPb4p2dnAA(a)pipex.net: > > "bz" <bz+spr(a)ch100-5.chem.lsu.edu> wrote in message > news:Xns99AE849FAE9DEWQAHBGMXSZHVspammote(a)130.39.198.139... >> sean <jaymoseley(a)hotmail.com> wrote in >> news:1190036876.674031.136830(a)d55g2000hsg.googlegroups.com: > ... >>> Having said all that I realize that Im not sure how to model a source >>> that pulses. Once a source goes OFF it would seem that it no longer >>> exists as a source >> >> That is a point that I made later. Technically, once an atom has >> emitted a photon, it no longer exists as a source. It must be excited >> again before it can re-emit. >> >>> which means any subsequent motion of an `off` source >>> would not effect the light as it propagates away. > > Whoopee! > >> Which is what several of us have been trying to tell you. > > Thanks Bob, you seem to have finally got across to > Sean what I have trying to tell him for months. > >>> So it might be better for the purposes of this experiment to pulse it >>> on and leave it on as it rotates rather than me worrying how to model >>> the motion of a nonexistent source. > > No! That's the whole point Sean, after the light > leaves and the source is switch off, how can it > affect the light that has already been emitted? > > Look again at my diagram: > > http://www.georgedishman.f2s.com/Sean/Sean_Planets.html > > The light was shown as a dot moving away from each > planet because it came from a hand torch which you > pointed at Sirius, switched on for a fraction of > a second and then switched off. So then you lay it > on the ground, hit it with a hammer several times, > jump on the bits, dissolve them in acid, bury the > goo in a landfill site and never see it again. How > the (insert your favourite expletive here) can the > remains of that torch, "the source", possibly affect > the motion of the flash of light which is by now > leaving the solar system? It doesn't make any sense. > > George Not to mention the fact that it would allow faster than light signaling! Just wiggle your source laser side to side and the whole laser beam instantly follows it, according to Sean's interpretation of 'light leaves the source at c in the sources frame of reference' ! I don't think so!!!! -- bz please pardon my infinite ignorance, the set-of-things-I-do-not-know is an infinite set. bz+nanae(a)ch100-5.chem.lsu.edu |