Ballistic theory of light demolished
in [Relativity]
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From: Paul B. Andersen on 6 Aug 2010 05:06 On 06.08.2010 00:02, Henry Wilson DSc wrote: > On Thu, 05 Aug 2010 13:41:21 +0200, "Paul B. Andersen"<something(a)somewhere.no> > wrote: > >> On 04.08.2010 23:54, Henry Wilson DSc wrote: >>> On Wed, 04 Aug 2010 15:28:30 +0200, "Paul B. Andersen"<something(a)somewhere.no> >>> wrote: > > >>>> "The X-rays are emitted when the electrons interact with occasional >>>> gas molecules.....whose v>>0 wrt the apparatus frame." >>> >>> Not exactly >>> >>> The updated Wilson Field Theory states that when a charge is accelerated, it >>> reacts back on the applied field causing THAT FIELD to emit an EM quantum. The >>> charge's acceleration vector and its mass determine the direction and amount of >>> energy radiated. >>> >>> The x-rays do not come from the electron at all but from the field. >> >> OK, let's use this theory on a concrete example. >> >> In a synchrotron, the particles are accelerated in RF-cavities >> and guided around the circuit by bending magnets. >> When the particle beam is bent by the magnets, >> synchrotron radiation is emitted. >> >> Your claim is that the energy in this radiation comes >>from the magnetic field in the bending magnets, and not >>from the particles. That means that you claim that >> the particles do not loose kinetic energy in the bends. > > That does not follow at all. The particles don't lose any of their mass, charge > or any other properties when radiation is produced....so obviously it doesn't > originate from THEM. They can still lose kinetic energy due to interaction with > the magnetic field. Ralph, Ralph, Ralph. :-) You are struggling now, aren't you? If the particles loose kinetic energy without radiating, where does the energy go? Come on, Ralph. You are claiming that the re-updated Wilson Field Theory state that charged particles, like electrons, don't emit photons when they loose energy. Maybe you will have to re-update the re-updated Wilson Field Theory? :-) > >> So the questions are: >> >> If the energy in the synchrotron radiation comes from >> the magnetic field in the bending magnets and not from >> the particles, where does the energy put into the >> RF-cavities go when the accelerator is in steady state? >> >> Why is it an upper limit for the energy in the particle >> beam when the particles keep gaining kinetic energy in >> the RF-cavities, and don't loose kinetic energy anywhere? > > Don't rave. The particles simply slow down without radiating. So what? They > speed up again in the electric field. So the energy conservation law is violated. Maybe the re-re-updated Wilson Field Theory state that energy isn't conserved? :-) > > You should ask why, according to the WFT, doesn't this electrostatic > ACCELERATION also cause synchrotron radiation? It probably does...but the gaps > are too small for it to be noticed. Why should I ask when you have given the answer? According to the WFT is synchrotron radiation emitted when charged particles gain kinetic energy in an electric field. BTW, Ralph. Do you know why the next generation high energy accelerators (after LHC) probably will be linear accelerators? Hint: Synchrotron radiation. (They don't use the WFT in the design. :-) ) > >> The energy in the synchrotron radiation is huge in >> a large synchrotron. Why don't we have to put an equal >> amount of energy into the bending magnets? >> Where does the energy lost in the bending magnets come >> from? > > I think you should consider a few possibillities before you ask silly questions > like this. From the energy point of view, it makes no difference whether the > radiation originates from the field or the particle. ....but it makes a huge > difference to the radiation's velocity. You didn't even understand the question, did you? It was: Where does the energy radiated from the bending magnetic field come from? It is NOT fed into the magnets. > >> But when the accelerator is in steady state, the particles >> gain kinetic energy in the RF-cavities, and loose the same >> amount of kinetic energy in the bends. (Thus steady state.) > > good See? The radiated energy is carried from the RF-cavities to the bends as kinetic energy in the particles. The radiated energy is NOT fed into the bending magnetic field, but into the RF-cavities. > >> >> I look forward to see how the re-updated Wilson Field Theory >> explains this. :-) > > I'm glad to see you are trying to learn something new. But you have explained nothing yet. To repeat the probably still unanswered questions: If the synchrotron radiation is emitted from the magnetic bending field, from where does the radiated energy come? It is not fed into the bending magnets. If the particles loose kinetic energy without radiating, where does the lost energy go? -- Paul http://home.c2i.net/pb_andersen/
From: PD on 6 Aug 2010 08:57 On Aug 5, 5:49 pm, ..@..(Henry Wilson DSc) wrote: > On Thu, 5 Aug 2010 15:29:05 -0700 (PDT), PD <thedraperfam...(a)gmail.com> wrote: > >On Aug 5, 5:10 pm, ..@..(Henry Wilson DSc) wrote: > >> On Wed, 4 Aug 2010 15:24:16 -0700 (PDT), PD <thedraperfam...(a)gmail.com> wrote: > >> >On Aug 4, 4:55 pm, ..@..(Henry Wilson DSc) wrote: > >> >> On Wed, 4 Aug 2010 07:13:17 -0700 (PDT), PD <thedraperfam...(a)gmail.com> wrote: > >> >> >On Aug 3, 3:53 pm, ..@..(Henry Wilson DSc) wrote: > >> >> >> On Tue, 03 Aug 2010 22:19:14 +0200, "Paul B. Andersen" <some...(a)somewhere.no> > >> >> >> Hahahahha! > > >> >> >> Is that charged particle moving inertially? > > >> >> >No. You may want to look up what the source of synchrotron radiation > >> >> >is. It's from an accelerated charge. > > >> >> I didn't claim it was moving inertially you dope. I asked YOU a question. > > >> >So why would you ask a non sequitur question about inertially moving > >> >charges in the context of synchrotron radiation? You should KNOW > >> >whether the charge is moving inertially in synchrotron radiation. And > >> >if you don't know, you should at least have the drive to look it up. > > >> Like Paul, you are incapable of thinking before you speak. > > >> Consider a linear accelerator in which a charge is being slowly accelerated. > > >Why would we consider that? We were talking about synchrotron > >radiation in a wiggler, not anything to do with a linear accelerator. > > >> It > >> is obviously not inertial and it appears to intermittently emit an EM quantum. > > >> Q1) How, when and why does it emit an EM quantum? > > >> That process is dependent solely on the particle's acceleration so: > > >No, it's not solely dependent on the particle's acceleration. It > >happens when there IS an acceleration, but the radiation depends on > >other quantities besides the acceleration. Like, the direction the > >particle is going and its charge. > > The particle has NO direction in its own frame. An accelerating particle is not at rest in any inertial frame. > If you were right, radiation > should be emitted equally in all directions. .hence my previous question about > 'inertial movement...' > So I am right. It is dependent solely on the particle's acceleration. No, it doesn't. You could google "angular dependence synchrotron radiation" BEFORE just making something up in your head. You're prone to too many mistakes relying on your own head. > > >> Q2) What does that say about the direction of radiation in a linear vs a > >> circular accelerator? >
From: Paul B. Andersen on 6 Aug 2010 09:31 On 06.08.2010 00:49, Henry Wilson DSc wrote: > On Thu, 5 Aug 2010 15:29:05 -0700 (PDT), PD<thedraperfamily(a)gmail.com> wrote: > >> On Aug 5, 5:10 pm, ..@..(Henry Wilson DSc) wrote: >>> On Wed, 4 Aug 2010 15:24:16 -0700 (PDT), PD<thedraperfam...(a)gmail.com> wrote: >>>> On Aug 4, 4:55 pm, ..@..(Henry Wilson DSc) wrote: >>>>> On Wed, 4 Aug 2010 07:13:17 -0700 (PDT), PD<thedraperfam...(a)gmail.com> wrote: >>>>>> On Aug 3, 3:53 pm, ..@..(Henry Wilson DSc) wrote: >>>>>>> On Tue, 03 Aug 2010 22:19:14 +0200, "Paul B. Andersen"<some...(a)somewhere.no> >>>>>>> Hahahahha! >>> >>>>>>> Is that charged particle moving inertially? >>> >>>>>> No. You may want to look up what the source of synchrotron radiation >>>>>> is. It's from an accelerated charge. >>> >>>>> I didn't claim it was moving inertially you dope. I asked YOU a question. >>> >>>> So why would you ask a non sequitur question about inertially moving >>>> charges in the context of synchrotron radiation? You should KNOW >>>> whether the charge is moving inertially in synchrotron radiation. And >>>> if you don't know, you should at least have the drive to look it up. >>> >>> Like Paul, you are incapable of thinking before you speak. Good grief, Ralph. You asked if the charged particle that is emitting synchrotron radiation was inertial. Either YOU weren't thinking, or you don't know that inertial particles don't radiate. It probably was the former, so why don't you admit that in stead of pretending that your question was a sensible one. >>> >>> Consider a linear accelerator in which a charge is being slowly accelerated. >> >> Why would we consider that? We were talking about synchrotron >> radiation in a wiggler, not anything to do with a linear accelerator. >> >>> It >>> is obviously not inertial and it appears to intermittently emit an EM quantum. >>> >>> Q1) How, when and why does it emit an EM quantum? >>> >>> That process is dependent solely on the particle's acceleration so: >> >> No, it's not solely dependent on the particle's acceleration. It >> happens when there IS an acceleration, but the radiation depends on >> other quantities besides the acceleration. Like, the direction the >> particle is going and its charge. > > The particle has NO direction in its own frame. If you were right, radiation > should be emitted equally in all directions. In the particle's frame of reference, radiation _is_ emitted in all directions, but not equally intense in all directions. (If we are talking about a single photon, replace 'intensity in a direction' with 'probability of emission in a direction'). In the frame of reference of the particle, let's call it the S' frame, there is a moving magnetic field. So the velocity vector is still relevant, but here it is the velocity of the B field. The moving magnetic field is perpendicular to the velocity. In the S' frame only an electric field can accelerate the particle. When we transform the moving magnetic field to the S' frame, the electric field is E' = -v*gamma*B, and the direction is perpendicular to the B field and the velocity. So the acceleration is perpendicular to same. The radiation intensity distribution is: I(phi') = Io*cos^2(phi') where phi' is the angle from the velocity vector. Note that for phi' = +/- pi/4, we get I = Io/2, the intensity is half at these angles. This is often called the beam width. So it will radiate almost in all directions, but mostly in both directions along the velocity vector, and nothing perpendicular to it. The radiation diagram will be something like this: (should be two circles) . . * * * * * * * * ----*-----------C-----------*-> v * * * * * * * * * * But we are observing the radiation in the lab frame S, where the particle is moving at the speed v. The intensity at the angle phi = 0 (in the forward direction) will be changed by the square of the Doppler effect: I(0) = Io*(1+v/c)/(1-v/c) equivalently will the intensity in the opposite direction be: I(pi) = Io*(1-v/c)/(1+v/c) (If you think in photons will the energy of each photon be Doppler shifted, and the frequency of arrival of photons will also be Doppler shifted, thus the square.) In addition to this effect will we have aberration: cos(phi) = (cos(phi')+ v/c)/(1 + (v/c)cos(phi')) A beam (or photon) in an off axis direction in S' will be deflected towards the velocity vector in S. It can be shown that the combined effect will lead to a beam width (half intensity) ~= 1/gamma (radians) in the forward direction. For high gammas will we have a very narrow beam along the velocity vector and very little in the backwards direction. Note that this is a prediction of SR, and measurements are in accordance with the prediction. Synchrotron radiation is yet another confirmation of SR. > So I am right. It is dependent solely on the particle's acceleration. No. Observations in the lab frame depend strongly on the particle's velocity in that frame. >>> Q2) What does that say about the direction of radiation in a linear vs a >>> circular accelerator? In any accelerator will we loose a little energy as radiation in the RF-cavities due to the acceleration of the particles. But most of the energy goes into kinetic energy of the particles since the acceleration is along the velocity. But the acceleration of the particles in the bends is much higher than it is in the RF-cavities, and the radiation is _vastly_ more intense. And since the acceleration is perpendicular to the velocity, it doesn't add to the kinetic energy of the particles. The radiation energy is taken from the kinetic energy of the particles, so they loose kinetic energy. In its rest frame, the particle "sees" an electric field E = v*gamma*B. When v ~= c, gamma is thousands and B is very high, that's a gigantic electric accelerating field! That's why the next generation of high energy accelerators probably will be several km long linear accelerators. -- Paul http://home.c2i.net/pb_andersen/
From: Paul B. Andersen on 6 Aug 2010 14:46 On 06.08.2010 15:31, Paul B. Andersen wrote: > > In the particle's frame of reference, radiation _is_ emitted > in all directions, but not equally intense in all directions. > (If we are talking about a single photon, replace 'intensity > in a direction' with 'probability of emission in a direction'). > > In the frame of reference of the particle, let's call it the S' frame, > there is a moving magnetic field. So the velocity vector is still > relevant, but here it is the velocity of the B field. It should be noted that S' is the inertial frame where the particle is _instantly_ at rest. -- Paul http://home.c2i.net/pb_andersen/
From: Henry Wilson DSc on 6 Aug 2010 18:35
On Fri, 06 Aug 2010 11:06:23 +0200, "Paul B. Andersen" <something(a)somewhere.no> wrote: >On 06.08.2010 00:02, Henry Wilson DSc wrote: >> On Thu, 05 Aug 2010 13:41:21 +0200, "Paul B. Andersen"<something(a)somewhere.no> >> wrote: >> >>> >>> Your claim is that the energy in this radiation comes >>>from the magnetic field in the bending magnets, and not >>>from the particles. That means that you claim that >>> the particles do not loose kinetic energy in the bends. >> >> That does not follow at all. The particles don't lose any of their mass, charge >> or any other properties when radiation is produced....so obviously it doesn't >> originate from THEM. They can still lose kinetic energy due to interaction with >> the magnetic field. > >Henry, Henry, V. :-) >You are struggling now, aren't you? >If the particles loose kinetic energy without radiating, >where does the energy go? Gawd! This is elementary physics. Where does the energy go when a rock falls into water? >Come on, Henry. >You are claiming that the re-updated Wilson Field Theory >state that charged particles, like electrons, >don't emit photons when they loose energy. Hahahahhaha! Of course they don't. As with the aforementioned rock, they energise the field and IT emits the radiation ....at c wrt the field electrodes.... Obvious isn't it? >Maybe you will have to re-update >the re-updated Wilson Field Theory? :-) Maybe you will need a brain transplant....get one from a chimp... >>> So the questions are: >>> >>> If the energy in the synchrotron radiation comes from >>> the magnetic field in the bending magnets and not from >>> the particles, where does the energy put into the >>> RF-cavities go when the accelerator is in steady state? >>> >>> Why is it an upper limit for the energy in the particle >>> beam when the particles keep gaining kinetic energy in >>> the RF-cavities, and don't loose kinetic energy anywhere? >> >> Don't rave. The particles simply slow down without radiating. So what? They >> speed up again in the electric field. > >So the energy conservation law is violated. Hahahhaha! No Paul, This is pretty elementary. The energy goes into the field....something like magnetic damping. >Maybe the re-re-updated Wilson Field Theory state >that energy isn't conserved? :-) hahahhahha! my theory involves quite elementary physics. i'm surprised you are having difficulty understanding it. > >> >> You should ask why, according to the WFT, doesn't this electrostatic >> ACCELERATION also cause synchrotron radiation? It probably does...but the gaps >> are too small for it to be noticed. > >Why should I ask when you have given the answer? >According to the WFT is synchrotron radiation emitted when >charged particles gain kinetic energy in an electric field. Well I would be inclined to say that it does not because all the applied energy goes into the particle's increased KE. This is a very different situation from that whereby a charge is deflected by a magnetic field and slowed in the process...(why is it slowed?) However even when a charge is accelerated by a field, there should be a reverse reaction ON that field ...and that could cause radiation emission FROM THE FIELD ITSELF. >BTW, Henry. >Do you know why the next generation high energy accelerators >(after LHC) probably will be linear accelerators? >Hint: Synchrotron radiation. >(They don't use the WFT in the design. :-) ) Of course they do...but they are too ignorant to realise the fact. ...The design is the same. ..... >>> The energy in the synchrotron radiation is huge in >>> a large synchrotron. Why don't we have to put an equal >>> amount of energy into the bending magnets? >>> Where does the energy lost in the bending magnets come >>> from? >> >> I think you should consider a few possibillities before you ask silly questions >> like this. From the energy point of view, it makes no difference whether the >> radiation originates from the field or the particle. ....but it makes a huge >> difference to the radiation's velocity. > >You didn't even understand the question, did you? >It was: >Where does the energy radiated from the bending >magnetic field come from? From the slowing of the particles, of course. That tends to REDUCE the field strength. >It is NOT fed into the magnets. ....maybe some Norwegian physics teachers should be fed to those big white hairy things up north... >>> But when the accelerator is in steady state, the particles >>> gain kinetic energy in the RF-cavities, and loose the same >>> amount of kinetic energy in the bends. (Thus steady state.) >> >> good > >See? Of course I can see something as elemetary as that.. >The radiated energy is carried from the RF-cavities to >the bends as kinetic energy in the particles. >The radiated energy is NOT fed into the bending magnetic >field, but into the RF-cavities. The radiated energy comes from the decelerating particles' reaction on the applied magnetic field. It originates in the FIELD not the particles. >>> I look forward to see how the re-updated Wilson Field Theory >>> explains this. :-) >> >> I'm glad to see you are trying to learn something new. > >But you have explained nothing yet. Paul, why don't you simply acknowledge the fact that I have made the greatest physics discovery since the lever? >To repeat the probably still unanswered questions: > > If the synchrotron radiation is emitted from the magnetic > bending field, from where does the radiated energy come? > It is not fed into the bending magnets. > > If the particles loose kinetic energy without radiating, > where does the lost energy go? The answers are trivial and are given above... Henry Wilson... ........Einstein's Relativity...The religion that worships negative space. |