 |
 | Home | Contact Us | FAQ | Search & Site Map | Link to Us |
 | Sign In | Join | Other 45 Sites in Network |
Natural Science Forum / Physics / Particle Physics / June 2005Particle Visualization
I am planning to create a sort of tetris like computer game involving particles, and I am trying to create a visual model for the particles. My model for the electron is a bunch of photons enclosed by a neutrino shell, and the positron is enclosed by an anti-neutrino shell. A W boson is a neutrino/anti-neutrino pair, with only one of the shells being filled by photons. The Z bosons have both shells filled equally. An up quark is an anti-neutrino shell filled with photons and has a color charged core. Maybe, not sure yet. A down quark is an up quark bonded with an electron. This will hopefully give a good visualisation of the weak and electromagnetic forces, such as photons coming out of the electron when it moves through an electromagnetic field, photons being emitted when an electron changes energy level, neutrons decaying into protons and W bosons. This to me visualizes the standard model. Is there anything major that i've left out? Also, when an electron changes to a higher energy level, does its speed or mass incease/decrease or stay the same? > I am planning to create a sort of tetris like computer game involving > particles, and I am trying to create a visual model for the particles. > > My model for the electron is a bunch of photons enclosed by a neutrino > shell, and the positron is enclosed by an anti-neutrino shell. Small detail. Electrons and positrons have electric charge. Photons and neutrinos do not. > A W boson is a neutrino/anti-neutrino pair, with only one of the shells > being filled by photons. The Z bosons have both shells filled equally. Small detail. W bosons have electric charge. Neutrinos, antineutrinos, and photons do not. > An up quark is an anti-neutrino shell filled with photons and has a > color charged core. Maybe, not sure yet. Small details. An up quark has electric charge, antineutrinos and photons do not. Moreover, you haven't said exactly *what* the core is that has color charge. Moreover, an antineutrino shell would have a nonzero lepton number, and an up quark does not. > A down quark is an up quark bonded with an electron. > [quoted text clipped - 3 lines] > an electron changes energy level, neutrons decaying into protons and W > bosons. On your last point, I don't see from the above (at all) how a neutron would sensibly decay into a proton and a W. Please explain. > This to me visualizes the standard model. Is there anything major that > i've left out? No, just a bunch of small details. Oh, and you haven't said a thing about the gameplay that would make it an attractive game. > Also, when an electron changes to a higher energy level, does its speed > or mass incease/decrease or stay the same? You mean when it is bound in an atom? Look up the virial theorem. PD I know, that neutrinos and photons and gluons dont have electric charge, But this is the only way I could visualize it. The positive and negative components of the electromagnetic wave(photon) are polarized as they pass through the shell and produce an electric charge whose sign is opposite that of the shell. A neutron decays into a proton, electron and anti-netrino via the weak interaction. The W is the electron and the anti-neutrino or the positron and neutrino. The core of the quark, at the moment, is just a colored sphere. As for the lepton number, I just done a quick check and found that it is B-L that is conserved. If a color core is present, it will produce a change in the baryon number instead of the electron number The gameplay will be like that game bubbles, where you group balls of the same color together. Well this is going to be my first game, and i read somewhere that you should start with a tetris clone. This is just a basic modification of that game. It will be a learning process for me so I am not that bothered wheather people like it or not as long as I learn from the process, but I am well cable of producing this, I have been programming for about 20 years now. I hope I have answered your questions. Remember, I am not trying to describe the actual particles, just create a visual model that will allow me do include the major points of the standard model in the game. I will certainly look up this virial theorem as I have never heard of it before, and I appreciate your input on the subject. > I know, that neutrinos and photons and gluons dont have electric > charge, But this is the only way I could visualize it. > The positive and negative components of the electromagnetic > wave(photon) There are no positively and negatively charged components of an electromagnetic wave. Where did you get that idea? > are polarized as they pass through the shell and produce > an electric charge whose sign is opposite that of the shell. > > A neutron decays into a proton, electron and anti-netrino via the weak > interaction. The W is the electron and the anti-neutrino or the > positron and neutrino. You explained what was in the W already (though I expressed the reasons for my doubts). What you didn't explain is how a W emerges from the transition from a neutron to a proton in your visualization of what each of those things is. In your visualization, what turns into what? > The core of the quark, at the moment, is just a colored sphere. Just checking... what do you mean by color here? What colors are allowed for this mysterious sphere to have? > As for the lepton number, I just done a quick check and found that it > is B-L that is conserved. That is not consistent with experiment. > If a color core is present, it will produce a change in the baryon > number instead of the electron number [quoted text clipped - 12 lines] > describe the actual particles, just create a visual model that will > allow me do include the major points of the standard model in the game. Yes, well, it'll be a standard model of something, though apparently not THE standard model. > I will certainly look up this virial theorem as I have never heard of > it before, and I appreciate your input on the subject. The positive components come from the top part of the wavelength as seen on an oscilloscope, the negative from the bottom. The decay of the neutron happens by the decay of the down quark. The shell of the down quark(neutrino) annihilates with an anti-neutrino and the electric charge is released. Pair production occurs and a neutrino and anti-neutrino are produced.The charges of the up quark bind with the anti-neutrino and the neutrino binds with the released charge from the electron. This produces an up quark and a W-. The colors allowed are rgb, and their anti-colors, as is it used in the Standard Model. > The positive components come from the top part of the wavelength as > seen on an oscilloscope, the negative from the bottom. What do you suppose is being measured on the oscilloscope? That is, what does the vertical axis correspond to (physical quantity, with units), and what does the horizontal axis correspond to (physical quantity, with units)? > The decay of the neutron happens by the decay of the down quark. The > shell of the down quark(neutrino) annihilates with an anti-neutrino and > the electric charge is released. Pair production occurs and a neutrino > and anti-neutrino are produced.The charges of the up quark bind with > the anti-neutrino and the neutrino binds with the released charge from > the electron. This produces an up quark and a W-. I tried to follow this, I really did, but I confess that it is still pretty muddy to me. Too much magic going on, like "Pair production occurs..." What produces the pair?? And in the case of the "released charge from the electron," what electron? I started with a neutron. I'm ending up with a proton and a W. Where did the electron come from? Can you please take this a single step at a time, and be more complete, so I know exactly what changes into what at each step? > The colors allowed are rgb, and their anti-colors, as is it used in the > Standard Model. OK, so the color doesn't reside in the quark, exactly, it resides on a sphere inside the quark, but the sphere isn't a quark, it's something else not yet determined. Have I got that right? PD I was using a sine wave as an example, which has an up phase and a down phase. The axis represent what they really are. (Volts for the vertical and seconds for the horizontal, i think) There are 6 color cores, r+g+b+, r-g-b-, with + corresponding to matter and - corresponding to anti-matter. An electron is a group of photons in a neutrino shell. A positron is a group of photons in an anti-neutrino shell. The photons pass through the neutrino and are trapped within it, The spin of the neutrino, splits the sine wave into an up component and a down component corresponding to positive and negative charge respectively. Negative electric charge is attracted to the up spin of the neutrino, while positive electric charge is attracted to the down spin of the anti-neutrino. An up quark is composed of an electron with a colored core. Therfore, a colored core has -1/3e and an anti-colored core has +1/3e and vice-versa. Yes, these are different from quarks and have not yet been determined, but this is just a visualization for a computer game, not the reality of the situation. Hope this cleared things up. Any more questions or sugestions? A down quark is a combination of an up quark and an electron. A down quark decays into an up quark by an incoming neutrino annihilating with an anti-neutrino, and producing a pair of photons.This means that the shell containing the electron has gone, and a photon or two is released. Between these photons, an annihilation occurs creating a neutrino/anti-neutrino pair, and the rest of the photon(s) are trapped inside one of the neutrinos, creating either the W- or the W+ bosons. A W- boson decays into an electron and an anti-neutrino, while a W+ boson decays into a positron and a neutrino. This decay occurs becase the neutrino/anti-neutrino pair are travelling in opposite directions. What I am basically saying, is that the decay of a down quark into an up quark in a neutron is because of an incoming neutrino causing an annihilation. The mean lifeime of the neutron is 886 seconds, therefore giving the rate at which neutrinos strike the down quark in a neutron. Erm, I think I'm starting to get carried away here. Between > > I am planning to create a sort of tetris like computer game involving > > particles, and I am trying to create a visual model for the [quoted text clipped - 6 lines] > Small detail. Electrons and positrons have electric charge. Photons and > neutrinos do not. Photons are not electromagnetic? > > A W boson is a neutrino/anti-neutrino pair, with only one of the > shells [quoted text clipped - 41 lines] > > PD Photons are the force carrier for the electromagnetic force. When they are bound into a shell to form a particle and interact with an other particle, the 2 particles exchange photons, creating the electromagnetic force. Boloney, you cannot divorce electromagnetic field from a photon. > > > I am planning to create a sort of tetris like computer game > involving [quoted text clipped - 10 lines] > > Photons are not electromagnetic? yes but they are not charged. which means there is no trilinear photon coupling. PD [snip] Boloney. A charge is the presence of a field. Show me It's not anything else. > Boloney. > A charge is the presence of a field. > Show me It's not anything else. Nick, there's no showing yon anything because you're not interested. All I can do is give you answers. What you do with them is wholly up to you. I sure am not going to argue with yon. I see that you have learned that they are not charged but can you prove it? You have empty definition. If the photon is electromagnetic how can it not be ectromagnetically charged? > I see that you have learned that they are not charged but can you prove > it? You should have noticed that he has learned not to argue with an idiot. Dirk Vdm > You have empty definition. > > If the photon is electromagnetic how can it not be ectromagnetically > charged? >should have noticed that he has learned not to argue >with an idiot. >Dirk Vdm then you do moron ....ahaha AHAHA ahaha... if the photon is massless, have energy, considered EM field, but no charged, what is energy but not charge ....ahaha AHAHA ahaha... explain it in a few words... and no arithmetics ....ahaha AHAHA ahaha... > >should have noticed that he has learned not to argue >>with an idiot. [quoted text clipped - 13 lines] > > ....ahaha AHAHA ahaha... Why no mathematics? snip----------------------- >> > Small detail. Electrons and positrons have electric charge. Photons >> and [quoted text clipped - 8 lines] > > [snip] If photons are not electromagnetic: a) What are photons? b) From what are the electromagnetic properties of EMR produced? Regards, Monitek (Arden Barker) > snip----------------------- > >> > Small detail. Electrons and positrons have electric charge. Photons [quoted text clipped - 11 lines] > > If photons are not electromagnetic: Reading comprehension problem. Photons are electromagnetic, but they are not charged. > a) What are photons? The carriers of the electromagnetic interaction. The carrier does not need to carry charge itself. The charge identifies the *strength* of the coupling between a (fermion) particle and the appropriate (boson) carrier of the interaction. Think of it as how often the vertex happens. > b) From what are the electromagnetic properties of EMR produced? Too vague. I don't know what you mean by "electromagnetic properties". > Regards, > Monitek (Arden Barker) >> snip----------------------- >> >> > Small detail. Electrons and positrons have electric charge. Photons [quoted text clipped - 21 lines] > the coupling between a (fermion) particle and the appropriate (boson) > carrier of the interaction. Are ther 2 kinds of photon one for positive charge and one for negative charge? Do the photons travel at the speed of light? > Think of it as how often the vertex > happens. I do not understand the relevance of this statement. >> b) From what are the electromagnetic properties of EMR produced? > > Too vague. I don't know what you mean by "electromagnetic properties". One can measure the electrostatic field associated with EMR. One can measure the magnetic field assoiated with EMR. "electromagnetic properties" is a phrase which encompasses both properties. My question is : Where do the electromagnetic properties associated with EMR arise from? If the photon has no charge (I note the word 'need' as used above leaving open the option for a charged photon), then the "electromagnetic properties" must arise from something else. If so what could that something else be? On the other hand if the photon is charged then one does not have to look for something else to derive the electromagnetic properties of EMR. Regards, Monitek (Arden Barker) > >> snip----------------------- > >> >> > Small detail. Electrons and positrons have electric charge. Photons [quoted text clipped - 24 lines] > Are ther 2 kinds of photon one for positive charge and one for negative > charge? No, there is one kind of photon, the same for positive and negative charge. > Do the photons travel at the speed of light? Yes. > > Think of it as how often the vertex > > happens. > > I do not understand the relevance of this statement. I wasn't sure you would. It's a reference to Feynman diagrams. A Feynman diagram is a little schematic of an interaction, where the different parts of the schematic represent terms in an algebraic expression which allows you to calculate the rate at which that interaction occurs. For example, a diagram like a "Y" might have for the three legs an incoming electron, an outgoing electron, and an emitted (or absorbed) photon, and the point where the emission (or absorption) occurs is at the junction of the "Y" -- the vertex. The term corresponding to the vertex is proportional to the charge of the electron. The stronger the charge is, essentially, the more frequently the vertex will occur during an electron's path through spacetime. Note that the charge is carried by the electron in this case, which is codified in the rule that vertices happen on electron lines but never on photon lines. You can't have a "Y" with just photons on the three legs. On the other hand, the statement that the electron has charge means *nothing more* than the fact that it does have vertices where photons are emitted or absorbed. That is, the answer to the question, "What is charge, anyway?" is simply that it is a flag that the particle that bears it participates in such interactions. > >> b) From what are the electromagnetic properties of EMR produced? > > [quoted text clipped - 10 lines] > the other hand if the photon is charged then one does not have to look for > something else to derive the electromagnetic properties of EMR. No, and this is simple to explain. You are under the impression, it appears, that the only sources of electric and magnetic fields are charges. This is not so. If you open a textbook to the chapter on Faraday's law and Maxwell's equations, you will see that there are other sources for electric and magnetic fields; namely, a changing magnetic field is a source of an electric field, and a changing electric field is a source of a magnetic field. This, in fact, is the key thing that lets electric/magnetic fields propagate *together* through space, without carrying charge with them. The fact that the fields are periodic (oscillating) is central to this, because oscillating fields are constantly changing, which is required for them to continue to feed each other. The fact that there are non-charge sources for electric and magnetic fields came as quite a surprise to Faraday, but experiments don't lie. Likewise, the fact that the mathematical solution to propagating, chargeless fields happened to propagate at a speed that was numerically close to the measured speed of light came as quite a surprise to Maxwell, but it too was verified later in experiment. This sequence of stunning revelations completely changed everything in physics. Indeed, of the four fundamental interactions, the electromagnetic interaction is the *only* one that we freely exploit at will. (Some will include gravity in that category, but if we didn't happen to have a fairly huge source of a gravitational interaction nearby, it'd be useless to us.) PD ----- Original Message ----- From: "PD" <TheDraperFamily@gmail.com> Newsgroups: sci.physics.particle Sent: Wednesday, June 01, 2005 2:53 PM Subject: Re: Particle Visualization >> >> snip----------------------- >> >> >> > Small detail. Electrons and positrons have electric charge. [quoted text clipped - 50 lines] > electron. The stronger the charge is, essentially, the more frequently > the vertex will occur during an electron's path through spacetime. Sorry I switched of the idea behind Feynmans aide memoirs the minute he introduced an imaginary exchange particle into the proceedings. > Note that the charge is carried by the electron in this case, which is > codified in the rule that vertices happen on electron lines but never [quoted text clipped - 4 lines] > "What is charge, anyway?" is simply that it is a flag that the particle > that bears it participates in such interactions. The vertices are imaginary. When the rate of change of velocity is large enough then a charged particle causes an electromagnetic disturbance to radiate. >> >> b) From what are the electromagnetic properties of EMR produced? >> > [quoted text clipped - 28 lines] > oscillating fields are constantly changing, which is required for them > to continue to feed each other. Electromagnetic induction is what I would call it. Even that will not work for EMR because a magnetic field can only induce a current in a conductor. Where are the conductors in EMR in vacuo? A conductor is a collection of charges which are free to move. To use electromagnetic induction to propagate EMR you need charged particles. Sorry, your starting assumption in your inductive cycle is totally wrong. You can not start the electromagnetic induction cycle with a magnetic field. The reason being that moving charges are required to create a magnetic field so the changing electric field comes first. Electric fields require charged particles-period. Therefore to create a magnetic field you require charged particles. Where are your charged particles in EMR? > The fact that there are non-charge sources for electric and magnetic > fields came as quite a surprise to Faraday, but experiments don't lie. > Likewise, the fact that the mathematical solution to propagating, > chargeless fields happened to propagate at a speed that was numerically > close to the measured speed of light came as quite a surprise to > Maxwell, but it too was verified later in experiment. If you have an electrostatic field you have a charged particle responsible for its creation. There are no electric fields which do not have a charged particle as a source. Yes there is no doubt that Maxwells equations were a flash of pure brilliance. The subsequent sweeping of the field source under the carpet was a mistake. > This sequence of stunning revelations completely changed everything in > physics. Indeed, of the four fundamental interactions, the [quoted text clipped - 4 lines] > > PD The weak force has been resolved as electromagnetic. I have modelled the strong force as electromagnetic. The electromagnetic force is electromagnetic. So dont be surprised that there may only be two fundamental forces. Regards, Monitek (Arden Barker) > ----- Original Message ----- > From: "PD" <TheDraperFamily@gmail.com> > Newsgroups: sci.physics.particle > Sent: Wednesday, June 01, 2005 2:53 PM > Subject: Re: Particle Visualization [snip] > >> > Think of it as how often the vertex > >> > happens. [quoted text clipped - 26 lines] > > The vertices are imaginary. I don't know why you say that. Just because virtual particles are *among* those that play in Feynman diagrams, this does not mean that all vertices are associated with virtual particles. Therefore it makes no sense to say that vertices are imaginary. In a sense you are right, though, because the diagrams represent mnemonics on how to write down the calculations for the rates. It so happens that the diagrams are physically suggestive, however. > When the rate of change of velocity is large > enough then a charged particle causes an electromagnetic disturbance to > radiate. And, in your mind, how large is "large enough"? > >> >> b) From what are the electromagnetic properties of EMR produced? > >> > [quoted text clipped - 31 lines] > Electromagnetic induction is what I would call it. Even that will not work > for EMR because a magnetic field can only induce a current in a conductor. That is flat wrong. A changing magnetic field in the evacuated (or air-filled) gap of a dipole magnet will produce a measurable electric field in that gap, which can be tested by the placement of a *stationary* electron or other charged, nonconductive object in that gap. > Where are the conductors in EMR in vacuo? A conductor is a collection of > charges which are free to move. To use electromagnetic induction to > propagate EMR you need charged particles. That is also incorrect. A moving charge can initiate an electromagnetic wave, but the wave will continue to propagate after the charge has been removed or neutralized. > Sorry, your starting assumption in your inductive cycle is totally wrong. > You can not start the electromagnetic induction cycle with a magnetic field. [quoted text clipped - 15 lines] > flash of pure brilliance. The subsequent sweeping of the field source under > the carpet was a mistake. See above. > > This sequence of stunning revelations completely changed everything in > > physics. Indeed, of the four fundamental interactions, the [quoted text clipped - 9 lines] > electromagnetic. So dont be surprised that there may only be two fundamental > forces. The weak force has NOT been resolved as electromagnetic, unless you've done so. I await your demonstration. PD >> ----- Original Message ----- >> From: "PD" <TheDraperFamily@gmail.com> [quoted text clipped - 43 lines] > mnemonics on how to write down the calculations for the rates. It so > happens that the diagrams are physically suggestive, however. I appreciate that Feynman's introduction of the exchange photon produced a mathematical model which fitted the observed data to a T. And cf Maxwell when the maths fit, then the theory which produced the equations, must be right then Feynman's work is powerful juju. However Feynman himself never understood how this vitual exchange particle model matched reality. I venture to suggest that all it means is that they are physically coupled. >> When the rate of change of velocity is large >> enough then a charged particle causes an electromagnetic disturbance to >> radiate. > > And, in your mind, how large is "large enough"? Electrons oscillating at 10 kHz and above emit EMR waves, below that frequency zilch. >> >> >> b) From what are the electromagnetic properties of EMR produced? >> >> > [quoted text clipped - 43 lines] > *stationary* electron or other charged, nonconductive object in that > gap. The only conclusion you can arrive at from both the statement I made above and the following statement you made is that the vacuum *IS A CONDUCTOR*. It has an impedance of 300 ohms. >> Where are the conductors in EMR in vacuo? A conductor is a collection of >> charges which are free to move. To use electromagnetic induction to [quoted text clipped - 3 lines] > wave, but the wave will continue to propagate after the charge has been > removed or neutralized. Which also indicates that the charge measured in the EMR is not due to the original charge. The charge in the EMR induces the magnetic wave front which in turn induces the charge in the next wave front and cancels the charge behind as it collapses and so on ad infinitum. >> Sorry, your starting assumption in your inductive cycle is totally wrong. >> You can not start the electromagnetic induction cycle with a magnetic [quoted text clipped - 25 lines] > > See above. Sorry nobody has shown that the charge field in EMR has no charged particles associated with it. It is an assumption that there are no charged particles in the vacuum. The finesse was to consider only the fields in the equations because it was the fields which did the work. The assumption was "obviously a vacuum has no charged particles in it". >> > This sequence of stunning revelations completely changed everything in >> > physics. Indeed, of the four fundamental interactions, the [quoted text clipped - 15 lines] > > PD Somebody has already got a Nobel for it. http://en.wikibooks.org/wiki/FHSST_Physics_atomic_nucleus_elementary:Forces_of_nature Regards, Monitek (Arden Barker) > >> ----- Original Message ----- > >> From: "PD" <TheDraperFamily@gmail.com> [quoted text clipped - 50 lines] > understood how this vitual exchange particle model matched reality. I > venture to suggest that all it means is that they are physically coupled. Well, that's fine, but QED says more. It says that there is a coupling, and that the coupling is identified as a field with properties consistent with a chargeless, spin-1, massless beast. Moreover, it goes on to say that this is identical with the chargeless, spin-1, massless beast that produces particle-antiparticle pairs in HEP collisions, that induces electron emission in a photocathode, that causes material damage in the vicinity of some radioactive sources, and so on. > >> When the rate of change of velocity is large > >> enough then a charged particle causes an electromagnetic disturbance to [quoted text clipped - 4 lines] > Electrons oscillating at 10 kHz and above emit EMR waves, below that > frequency zilch. I see. So there is no possible way there could be EMR waves of frequency 60 Hz, is that correct? > >> >> >> b) From what are the electromagnetic properties of EMR produced? > >> >> > [quoted text clipped - 47 lines] > and the following statement you made is that the vacuum *IS A CONDUCTOR*. It > has an impedance of 300 ohms. Ah. Perhaps you need to provide a good operational definition of a conductor. I get the feeling you're implementing a definition that is not commonly in use. > >> Where are the conductors in EMR in vacuo? A conductor is a collection of > >> charges which are free to move. To use electromagnetic induction to [quoted text clipped - 8 lines] > in turn induces the charge in the next wave front and cancels the charge > behind as it collapses and so on ad infinitum. That's an interesting proposal. Your next step is to propose an experimental test to distinguish your model from the presently held vacuum model -- i.e. that net or separated charge is present in a vacuum where and when an EM wave passes. > >> Sorry, your starting assumption in your inductive cycle is totally wrong. > >> You can not start the electromagnetic induction cycle with a magnetic [quoted text clipped - 31 lines] > because it was the fields which did the work. The assumption was "obviously > a vacuum has no charged particles in it". Well, I'm not so sure it's simply an assumption. The proof is in an experimental test that would demonstrate or rule out the presence of charged particles in the vacuum in the vicinity of an EM wave. If you'd like your model to be considered for acceptance, propose the test. > >> > This sequence of stunning revelations completely changed everything in > >> > physics. Indeed, of the four fundamental interactions, the [quoted text clipped - 19 lines] > > http://en.wikibooks.org/wiki/FHSST_Physics_atomic_nucleus_elementary:Forces_of_nature No, you are confused. The electromagnetic force and the weak force have been demonstrated to be two different aspects of a combined electroweak force that obeys SU(2) symmetry. The weak has not been subsumed into the electromagnetic, nor the vice versa. The electroweak theory has four gauge bosons, unlike electromagnetism which has but one. PD >> >> ----- Original Message ----- >> >> From: "PD" <TheDraperFamily@gmail.com> [quoted text clipped - 66 lines] > induces electron emission in a photocathode, that causes material > damage in the vicinity of some radioactive sources, and so on. So far so good. How is it that you know this chargeless, massless beast travels at the speed of light between the two particles involed? Or could it be a modification of something that already exists between the two? >> >> When the rate of change of velocity is large >> >> enough then a charged particle causes an electromagnetic disturbance [quoted text clipped - 8 lines] > I see. So there is no possible way there could be EMR waves of > frequency 60 Hz, is that correct? Now if you put it that way I dont see any theoretical reason why that should not be so. >> >> >> >> b) From what are the electromagnetic properties of EMR produced? >> >> >> > [quoted text clipped - 57 lines] > conductor. I get the feeling you're implementing a definition that is > not commonly in use. Quite so. I thought my definition was clear, a region of which contains charged particles which are free to move is a conductor. >> >> Where are the conductors in EMR in vacuo? A conductor is a collection >> >> of [quoted text clipped - 16 lines] > vacuum model -- i.e. that net or separated charge is present in a > vacuum where and when an EM wave passes. Hmmm What I have described is the electromagnetic induction propagation of EMR is that not standard. As for an experiment would a capacitor placed in the path of a standing wave of EMR be sufficient to investigate the presence of charge. Six to ten cm radiation will do nicely. >> >> Sorry, your starting assumption in your inductive cycle is totally >> >> wrong. [quoted text clipped - 79 lines] > > PD So are you saying that the electro weak is an independant force not related to the electromagnetic? Personally I think beta decay is an example of pair creation inside the nucleus. Which means that the neutrino does not exist and the electroweak force does not exist. Regards, Monitek (Arden Barker) [snip] > >> I appreciate that Feynman's introduction of the exchange photon produced > >> a [quoted text clipped - 14 lines] > So far so good. How is it that you know this chargeless, massless beast > travels at the speed of light between the two particles involed? I didn't say that it did. I said it was chargeless, spin-1, and massless. But as it turns out, massless fields have to travel at the speed of light because of a helicity argument. What QED claims is that it is identical to the spin-1, chargeless, massless thing that exhibits the experimental behaviors mentioned, which by the way is *measured* to travel at the speed of light. It's entirely possible that the QED beast is a completely *separate* spin-1, chargeless, massless thing that is involved with electromagnetic interactions, besides a photon, but it's never seemed to be necessary to posit an extraneous object with identical properties. > Or could it > be a modification of something that already exists between the two? Sure it could be. Not much evidence that it needs to be, but it could be. Of course, to suppose that it is begs the question of how such a model could be experimentally distinguishable from a photon model. > >> >> When the rate of change of velocity is large > >> >> enough then a charged particle causes an electromagnetic disturbance [quoted text clipped - 11 lines] > Now if you put it that way I dont see any theoretical reason why that should > not be so. It's not clear from your response whether you think 60 Hz EMR waves are possible or not. If I assume that you are allowing that they are possible, then of course you would also allow that 1 Hz EMR waves are possible, or indeed 0.001 Hz EMR waves are possible. That is, there is no threshold for the production of EMR. Correct? > >> >> >> >> b) From what are the electromagnetic properties of EMR produced? > >> >> >> > [quoted text clipped - 60 lines] > Quite so. I thought my definition was clear, a region of which contains > charged particles which are free to move is a conductor. OK, so let me see if I have this right. If I do NOT put the electron in the gap (i.e. there are no freely moving charged particles), there is no induction, because there is no conductor. But as soon as I put a test charge in there, there is suddenly induction, so there is a potential created *when I put the charge in*? Tell me: If I charge up a Vandegraaf generator, is there no electric field surrounding the generator until I put a test charge in the vicinity of the electrode? Does putting the test charge in the area create the electrostatic field around the vandeGraaf electrode? Let's tweak the scenario. Let's take a rubber rod or any other insulator and charge it up by rubbing it. Now put this charged rod and hold it in place in the gap with a clamp that can measure the force on the rod by means of a pressure sensor at the clamp. Now there are no free charges in the gap. Therefore there is no conductor in the gap. Now change the field. I say there will be a measurable force registered on the pressure sensor. You say what? > >> >> Where are the conductors in EMR in vacuo? A conductor is a collection > >> >> of [quoted text clipped - 21 lines] > the path of a standing wave of EMR be sufficient to investigate the presence > of charge. Six to ten cm radiation will do nicely. I don't understand why this tests for the presence of net or separated charge in the region of EMR. > >> >> Sorry, your starting assumption in your inductive cycle is totally > >> >> wrong. [quoted text clipped - 82 lines] > So are you saying that the electro weak is an independant force not related > to the electromagnetic? Not quite what I said, nor what the Wikipedia article said. What I said is accurate. > Personally I think beta decay is an example of pair creation inside the > nucleus. Which means that the neutrino does not exist and the electroweak > force does not exist. Then you need to explain the following: 1) Why beta decay violates parity and electromagnetic interactions do not. 2) Why electromagnetic interaction rates are orders of magnitude faster than weak interactions, as a class. 3) Why in beta decay, it's the negative electron that gets preferentially ejected from the positive nucleus and not the positive positron (which should be preferentially repelled). 4) Why neutrinos that have traveled long distances have been demonstrated to collide with targets and produce interactions that are only consistent with incident projectiles with neutrino-like properties. Note here the spin of the incident particle is experimentally checkable from the angular distribution of the products, and that a spin 1/2 neutrino is easily distinguishable from a spin 1 photon. 5) Weak isospin selection rules, which are inconsistent with electromagnetic interactions. Perhaps you ought to ask the question why physicists thought that there must be a weak interaction in the first place, and do a little reading. (Hint: why is it called "weak", and what did Fermi have to do with all this?) PD > [snip] >> >> [quoted text clipped - 29 lines] > massless thing that exhibits the experimental behaviors mentioned, > which by the way is *measured* to travel at the speed of light. That interests me have you got a reference for that. I tried to measure the speed of a magnetic field some time ago and got nowhere. The reason I wanted to do that is that I wondered if the speed of a mgnetic field and the speed of an electrostatic field were the same. > It's > entirely possible that the QED beast is a completely *separate* spin-1, [quoted text clipped - 32 lines] > possible, or indeed 0.001 Hz EMR waves are possible. That is, there is > no threshold for the production of EMR. Correct? I would assume that is correct but doesnt the wave length become too long to handle, as the frequency tends to zero then the wavelength tends to infinity. In the case of 0.001 Hz the wave length would be 300,000 x 10^3 kilometres. Or 187.5 million miles! The case we are considering elsewhere where a magnet rotating at 1000 hz would have a EMR wavelength of 186.5 miles so assuming the magnet is the maximum of the magnetic compnent then the maximum of the first electrostatic region would be half a wave length away. So if you give me you location baybe we could make this a joint experiment one day I will rotate the magnet and you will observe the oscilloscope and the next day vice versa. I calculate that 62.1 hz will be a suitable frequency to maximize the electrostatic portion of the EMR. But that is unfortunately close to your mains frequency. Perhaps its better for you to rotate the magnet as our mains is 50 Hz. >> >> >> >> >> b) From what are the electromagnetic properties of EMR >> >> >> >> >> produced? [quoted text clipped - 82 lines] > vicinity of the electrode? Does putting the test charge in the area > create the electrostatic field around the vandeGraaf electrode? Nope magnetic induction works without a conductor in vacuo too. AND generates a current in the vacuum but the vacuum impedance is high at 300 ohms. Now you can try this yourself (using spice software if you like), if you get an air cores choke say of 100 or so turns round a 4" former and you pass through it an alternating current of 500 khz when you monitor the current you will find that it will be low as the impedance is 113104.8 ohms (based on 4" dia x 14" long 100 turn coil which has an inductance of 253.16 microhenries). now if you take a closed loop say a copper ring say 4.5" you will be amazed to see the current rise in the inductor circuit you can take this as an indication of electromagnetic induction working. > Let's tweak the scenario. Let's take a rubber rod or any other > insulator and charge it up by rubbing it. Now put this charged rod and [quoted text clipped - 3 lines] > Now change the field. I say there will be a measurable force registered > on the pressure sensor. You say what? There are no free charges in the gap for sure but the vacuum has been polarized by the charge field. Which means the vacuum is polarized into positive and negative charges but they are not free. The separation of the vacuum charges constitutes a current flow which is known as the displacement current. Just because a charge is not free it does not make it any the less a charge. All I have said is that where a magnetic field is then there are charged particles in motion creating it, wether there are free or associated with others is immaterial. >> >> >> Where are the conductors in EMR in vacuo? A conductor is a >> >> >> collection [quoted text clipped - 31 lines] > I don't understand why this tests for the presence of net or separated > charge in the region of EMR. A standing wave would produce a fixed field which one could measure at ones leasure and actually probe the different regions of the emr field. The capacitor measures the polarised field strength and can be used as a probe. >> >> >> Sorry, your starting assumption in your inductive cycle is totally >> >> >> wrong. [quoted text clipped - 104 lines] > 1) Why beta decay violates parity and electromagnetic interactions do > not. Particle creation of course, in both cases. > 2) Why electromagnetic interaction rates are orders of magnitude faster > than weak interactions, as a class. Expand please. > 3) Why in beta decay, it's the negative electron that gets > preferentially ejected from the positive nucleus and not the positive > positron (which should be preferentially repelled). Because in this case the positron has annihilated an electron ( part of the e-p pair which I posit resides inside a neutron.) leaving the electron which is outside the neutron. One can envisage this as pair creation either side of the nucleon boundary. > 4) Why neutrinos that have traveled long distances have been > demonstrated to collide with targets and produce interactions that are [quoted text clipped - 3 lines] > and that a spin 1/2 neutrino is easily distinguishable from a spin 1 > photon. Neutrinos are in a mess. They have to have mass to change morphology to explain why we can not find enough neutrinos from the Sun. I do not accept the existance of neutrinos as they are not necessary to explain the physics. > 5) Weak isospin selection rules, which are inconsistent with > electromagnetic interactions. [quoted text clipped - 5 lines] > > PD First of all IIRC it is called weak because it does not occur very often. Fermi was Italian. The suffix ino means small, as Italians are always missing the ends of words to convert Italian to english so whay should have been called a neutralino meaning little neutral one ended up as neutrino. The neutrino was postulated to explain the variation in momentum of the emitted particle in beta decay, the momentum difference being carried by a massless particle. A massless particle has no momentum to carry away 0 * velocity = 0. Regards, Monitek (Arden Barker) Too much to respond to in one post. Am breaking it up into chunks. [snip] > >> So far so good. How is it that you know this chargeless, massless beast > >> travels at the speed of light between the two particles involed? [quoted text clipped - 11 lines] > to do that is that I wondered if the speed of a mgnetic field and the speed > of an electrostatic field were the same. I'm not sure what you're asking. Are you asking if light's speed has ever been measured? Absolutely! Dozens of times. Do you need a reference? Are you asking if the speed of a single photon has been measured? Absolutely. Look up the Advanced Photon Source. Photons are emitted from well-confined bunches of electrons in narrow time windows and they arrive at their targets in narrow time windows, with the time of flight predicted (and fortunately validated) by assuming c. [snip] PD > Too much to respond to in one post. Am breaking it up into chunks. > [quoted text clipped - 31 lines] > > PD The speed of light is the average of the speed of propagation the magnetic field and the electric field. They could travel at different speeds. Has anybody measured the speed of a magnetic field? I assume that the measurement of the magnetic field is easier than the electric. Regards, Monitek (Arden Barker) > > Too much to respond to in one post. Am breaking it up into chunks. > > [quoted text clipped - 34 lines] > The speed of light is the average of the speed of propagation the magnetic > field and the electric field. That is incorrect. The two do not proceed independently. The propagating solution is the solution to the *coupled* Maxwell's equations. You simply do not get a propagating wave solution from electrostatics and another propagating wave solution from magnetostatics. There would be nothing to average. If you don't use the *coupled* equations, you don't get light. PD > They could travel at different speeds. Has > anybody measured the speed of a magnetic field? I assume that the > measurement of the magnetic field is easier than the electric. > > Regards, > Monitek (Arden Barker) Responding to another chunk... [snip] > >> Personally I think beta decay is an example of pair creation inside the > >> nucleus. Which means that the neutrino does not exist and the electroweak [quoted text clipped - 5 lines] > > Particle creation of course, in both cases. If both were due to the same electromagnetic process, they both would conserve the parity quantum number. They don't. You need to look up "parity violation". This is important. You have to do more than account for all particles going in and going out to characterize an interaction. You have to also check that the same "selection rules" apply -- the same patterns of observed behavior. > > 2) Why electromagnetic interaction rates are orders of magnitude faster > > than weak interactions, as a class. > > Expand please. There are many instances of this, but we'll take just one. (If you want more info, download data tables from http://pdg.lbl.gov) The charged pion, which decays via the weak interaction, has a lifetime of 2.6E-8 s. The neutral pion, which decays via the electromagnetic interaction, has a lifetime of 8.4E-17 s, a factor of a BILLION shorter. If the two decayed by the same fundamental interaction, you'd have to explain why the interaction rates are so extremely different. (The fact that whole classes of interactions happen at much slower rates than the electromagnetic interaction was one of the key reasons a new, "weak" interaction was suspected.) > > 3) Why in beta decay, it's the negative electron that gets > > preferentially ejected from the positive nucleus and not the positive [quoted text clipped - 4 lines] > is outside the neutron. One can envisage this as pair creation either side > of the nucleon boundary. You missed my point. If you have an electron and a positron that are created (more on this in a moment) in the nucleus, which has an extremely high density of positive charge, the positron would be preferentially repelled from the positively charged nucleus and the electron would be preferentially attracted to the positively charged nucleus. Instead, you have it the other way around, where the positron is the one that stays behind to eventually annihilate with something in the neutron, and it's the electron that escapes. Why? Secondly, you haven't accounted for where the positron comes from, or where the photons from the annihilation of the positron and the neutron's hidden electron go. The nucleus consists of neutrons and protons; in your model, that's protons and proton-electron bound states. So where does the photon come from to produce the electron-positron pair? More importantly, after the electron is ejected and the positron finds its way into a neutron to annihilate the hidden electron inside, there are two 511 keV photons that emerge from that annihilation. Where do those photons go, and *why aren't they observed at all* in beta decay?? > > 4) Why neutrinos that have traveled long distances have been > > demonstrated to collide with targets and produce interactions that are [quoted text clipped - 7 lines] > explain why we can not find enough neutrinos from the Sun. I do not accept > the existance of neutrinos as they are not necessary to explain the physics. You've missed my point. Accelerator labs produce *beams* of neutrinos, which travel a couple of miles down a beam tunnel, and which cause observable collisions with matter in a target downstream. Their existence is no longer in question. Look up Reines and Cowan, who received a Nobel prize for experimentally establishing the existence of the neutrino. Their experiment was in no way related to beta decay; the confirmation was independent. If you'd like to see a more current experiment, see http://www-boone.fnal.gov/ or more generally, http://www.fnal.gov/pub/inquiring/physics/neutrino/index.html The fact that muon and electron neutrinos have a small mass difference in no way compromises their existence. > > 5) Weak isospin selection rules, which are inconsistent with > > electromagnetic interactions. [quoted text clipped - 7 lines] > > First of all IIRC it is called weak because it does not occur very often. Exactly, which is what distinguishes itself from the electromagnetic interaction, which happens a billion times more frequently. > Fermi was Italian. The suffix ino means small, as Italians are always > missing the ends of words to convert Italian to english so whay should have > been called a neutralino meaning little neutral one ended up as neutrino. > The neutrino was postulated to explain the variation in momentum of the > emitted particle in beta decay, Not just momentum, but spin as well. Note that a spin of 1/2 was required to be accounted for, and an experimentally missed photon (which has spin 1) would not have done the trick. > the momentum difference being carried by a > massless particle. A massless particle has no momentum to carry away 0 * > velocity = 0. m*v is not the correct expression for momentum. That expression for momentum is an approximation that only works even reasonably well for massive objects traveling at velocities much smaller than the speed of light. PD > Responding to another chunk... > [quoted text clipped - 36 lines] > rates than the electromagnetic interaction was one of the key reasons a > new, "weak" interaction was suspected.) In my scheme of things the pion is too small to confine an electron positron pair therefore there is nothing holding up the shell. The negative and positive pion contain an electron and positron respectively, the charge of which sustatins the shell, 284 electron mass IIRC. >> > 3) Why in beta decay, it's the negative electron that gets >> > preferentially ejected from the positive nucleus and not the positive [quoted text clipped - 16 lines] > is the one that stays behind to eventually annihilate with something in > the neutron, and it's the electron that escapes. Why? I said the electron positron pair are created such that the positron is inside the neutron. The neutron already contains a pair so the pair annihilates leaving a proton. The electron leaves with the energy it had when it was created less the energy it lost being pulled back to the proton. Why eject an electron? Because the proton subsurface consists a negative charge which prevents the electron reaching the central positron, remember the charge appears to be fractional in my proposed shell structure for the nucleon so a whole charged particle can not react with the fractional charges because there will always be somne charge left over. > Secondly, you haven't accounted for where the positron comes from, or > where the photons from the annihilation of the positron and the [quoted text clipped - 6 lines] > annihilation. Where do those photons go, and *why aren't they observed > at all* in beta decay?? The pair separation comes from a random passing EMR and the charge in the neucleus cf SLAC e144. Gamma photons are observed in beta decay there appears to be a delay before their emmision and they may not be 511kev. However, short wave radiation is observed in most beta decays, teh observation of the gamma appears to be atom dependent as well. Can we assume that the morphology of the nucleus be affecting the process? e.g. http://hps.org/publicinformation/ate/q2376.html >> > 4) Why neutrinos that have traveled long distances have been >> > demonstrated to collide with targets and produce interactions that are [quoted text clipped - 19 lines] > > If you'd like to see a more current experiment, see http://www-boone.fnal.gov/ > or more generally, > http://www.fnal.gov/pub/inquiring/physics/neutrino/index.html > > The fact that muon and electron neutrinos have a small mass difference > in no way compromises their existence. Interesting but the solar neutrino problemtells me something doesnt add up. 1% errors trigger physisists to look for new physics, 300% error and the theory still holds good! I dont find it convincing. >> > 5) Weak isospin selection rules, which are inconsistent with >> > electromagnetic interactions. [quoted text clipped - 11 lines] > Exactly, which is what distinguishes itself from the electromagnetic > interaction, which happens a billion times more frequently. Yes but if one is waiting for a passing EMR wave which is powerful enough to cause pair separation and in a particular geometry of pair separation, I am not surprised it is a rare event. >> Fermi was Italian. The suffix ino means small, as Italians are always >> missing the ends of words to convert Italian to english so whay should [quoted text clipped - 6 lines] > required to be accounted for, and an experimentally missed photon > (which has spin 1) would not have done the trick. Didnt the other half spin go with the other particle of the pair? >> the momentum difference being carried by a >> massless particle. A massless particle has no momentum to carry away 0 * [quoted text clipped - 6 lines] > > PD Such as electrons departing after beta decay has occurred? Regards, Monitek (Arden Barker) > > Responding to another chunk... > > [quoted text clipped - 20 lines] > > check that the same "selection rules" apply -- the same patterns of > > observed behavior. You did not address this. > >> > 2) Why electromagnetic interaction rates are orders of magnitude faster > >> > than weak interactions, as a class. [quoted text clipped - 16 lines] > positive pion contain an electron and positron respectively, the charge of > which sustatins the shell, 284 electron mass IIRC. Pions of either variety (neutral or charged) do not decay (at least not very often) into final states with electrons and positrons in them. You should look up the observed decays of the charged and neutral pions. http://pdg.lbl.gov. > >> > 3) Why in beta decay, it's the negative electron that gets > >> > preferentially ejected from the positive nucleus and not the positive [quoted text clipped - 19 lines] > I said the electron positron pair are created such that the positron is > inside the neutron. The neutron already contains a pair A pair of what? Be quite specific please. > so the pair > annihilates leaving a proton. The electron leaves with the energy it had > when it was created less the energy it lost being pulled back to the proton. You haven't addressed the complaint. Why is it the positron stays inside the neutron? In the neutron, you suppose there is both a positive charge (the proton) and a negative charge (the electron). Why is it the positron is drawn to annihilate the electron when it also feels the repulsion of the proton, as well as all the other protons nearby in the nucleus? This is a key point. You can't just say "somehow it finds the electron before it is pushed out of the nucleus". You need to explain at least semiquantitatively why this would be more likely. > Why eject an electron? Because the proton subsurface consists a negative > charge which prevents the electron reaching the central positron, remember [quoted text clipped - 20 lines] > atom dependent as well. Can we assume that the morphology of the nucleus be > affecting the process? No, we cannot. First of all, the gammas which are often (but not always) associated with beta decays are single gammas, not pairs of gammas. Secondly, even if there were random motion of the neutron which causes the 511 keV line to spread a little, it would still show a pronounced peak at that characteristic energy. This is not at all observed, not even in the same ballpark. Third, in your model, the photons would happen with *every* beta-decaying nuclide, and this is also not observed. Don't spend too much time retrofitting your theory to make it work with experiment, unless there is a free parameter already built in to your theory that you can get to fit the data -- if it doesn't agree with experiment, the theory is no good. Period. > e.g. http://hps.org/publicinformation/ate/q2376.html > [quoted text clipped - 31 lines] > 1% errors trigger physisists to look for new physics, 300% error and the > theory still holds good! I dont find it convincing. The solar neutrino problem has been solved. Couple of years ago. You haven't addressed my comments. The experiments I cited have NOTHING to do with solar neutrinos. The accelerators mentioned are neutrino FACTORIES. > >> > 5) Weak isospin selection rules, which are inconsistent with > >> > electromagnetic interactions. [quoted text clipped - 15 lines] > cause pair separation and in a particular geometry of pair separation, I am > not surprised it is a rare event. This does not explain the charged and neutral pion dichotomy. > >> Fermi was Italian. The suffix ino means small, as Italians are always > >> missing the ends of words to convert Italian to english so whay should [quoted text clipped - 8 lines] > > Didnt the other half spin go with the other particle of the pair? No. Look at it this way. The initial particle is a neutron, spin 1/2. Final observed particles are proton, spin 1/2, and an electron, spin 1/2. There is no way you can get a spin 1/2 particle to generate two spin 1/2 particles. Two final state spin 1/2 particles can produce spin 0 or spin 1, but not spin 1/2. Without another particle in the final state, angular momentum is not conserved. Even if you have a photon in the initial state, the photon is spin 1. So there is still a mismatch, with either spin 1/2 or spin 3/2 in initial state, and spin 0 or spin 1 in the final state. You're still missing something. That missing something is the ... neutrino! > >> the momentum difference being carried by a > >> massless particle. A massless particle has no momentum to carry away 0 * [quoted text clipped - 8 lines] > > Such as electrons departing after beta decay has occurred? Those electrons are relativistic. m*v doesn't work for them either. And it sure doesn't work for neutrinos. PD Responding to yet another chunk... [snip] > > Let's tweak the scenario. Let's take a rubber rod or any other > > insulator and charge it up by rubbing it. Now put this charged rod and [quoted text clipped - 12 lines] > charged particles in motion creating it, wether there are free or associated > with others is immaterial. You've missed my point. I'm not interested in measuring polarized charge pulled out of the vacuum. I'm measuring the force on the net material charge that I put on the rubber rod. The charge on the rubber rod did not arise through any change in the magnetic field; it came from the rubbing the rod received. These charges are *stationary* in the gap -- the rod is an insulator and the rod is clamped in place. And yet there is a measurable force on the rod. Because the charges are stationary, that force must be due to an *electric* field, not a magnetic field. Moreover, since the force is horizontal and not radially outward or inward from the gap, it would be difficult to account for such an electrostatic force from redistributed charge on the pole tips. However, to address your explanation, I don't know where you got the idea that displacement current is due to vacuum polarization. In a vacuum-gap capacitor, when the capacitor is charging up, the only motion of charge is from the leads to the plates. There is *no* charge motion in the gap. Nevertheless, there appears a magnetic field circling the gap *as though* there were a current in the gap, and this pseudo-current is called displacement current. Summarizing, if you'll look again at Maxwell's equations, you'll see that there is a possible source of an electric field that is NOT due to any physical electric charge (the field lines do not terminate at any charges), and there is a possible source of a magnetic field that is NOT due to any physical electric current. [snip] PD > Responding to yet another chunk... > [quoted text clipped - 34 lines] > account for such an electrostatic force from redistributed charge on > the pole tips. I would have thought that if you were trying to demonstrate that stationary electrons were NOT affected by a magnetic field then should there be no force on the rod? > However, to address your explanation, I don't know where you got the > idea that displacement current is due to vacuum polarization. In a [quoted text clipped - 3 lines] > circling the gap *as though* there were a current in the gap, and this > pseudo-current is called displacement current. It goes like this. Whatever is between the plates of a capacity is a dielectric ( think about what dielectric means). If you have a vacuum between the plates and the capacitor still charges then the vacuum IS a dielectric and all that being a dielectric means. Secondly the displacement a is not a pseudo-current the magnetic effect can be measured. As I said earlier - must earlier, where you can measure a magnetic field then there are charged particles in motion close by. I stand by that statement and nothing you have said can detract from this. Charging a capacitor creates vacuum polarisation between the plates as the charges separate then there is a real current flowing. The electron moving up say has its magnetic field added to by the positron moving down. > Summarizing, if you'll look again at Maxwell's equations, you'll see > that there is a possible source of an electric field that is NOT due to [quoted text clipped - 5 lines] > > PD Regards, Monitek (Arden Barker) > > Responding to yet another chunk... > > [quoted text clipped - 38 lines] > electrons were NOT affected by a magnetic field then should there be no > force on the rod? I'm not trying to show that a magnetic field does not affect stationary charges. You already agree that's the case. What I'm trying to show you is that, because there IS a force on stationary charges, it must be an electric field and not a magnetic field. However, as I pointed out elsewhere, the shape of this electric field cannot be accounted for by redistributed charges on the pole tips. > > However, to address your explanation, I don't know where you got the > > idea that displacement current is due to vacuum polarization. In a [quoted text clipped - 6 lines] > It goes like this. Whatever is between the plates of a capacity is a > dielectric ( think about what dielectric means). No, you tell me what you think a dielectric means. It does not mean "that which fills a capacitor". If you think it does, define "paraelectric". > If you have a vacuum > between the plates and the capacitor still charges then the vacuum IS a > dielectric and all that being a dielectric means. Secondly the displacement > a is not a pseudo-current the magnetic effect can be measured. That's why it's called a pseudo-current! Because it generates a real magnetic effect without there being any measurable charge transport in the gap. > As I said > earlier - must earlier, where you can measure a magnetic field then there [quoted text clipped - 3 lines] > a real current flowing. The electron moving up say has its magnetic field > added to by the positron moving down. There are several reasons why this cannot be the case. Let's start with where the supposed vacuum polarization charges go. Do they actually make it to the capacitor plates? If no, then this would result in a dielectric effect, exactly as in a real material, with a measurable dielectric constant > 1. I don't believe that has been measured. If yes, then some of the charge delivered in the leads would be that pulled from the vacuum, as well as that pulled from the battery. That is, there would be more *excess* charge than accountable from the source. Secondly, there would be simple tests to determine whether charge moves in the gap. A phosphorescent screen placed near one plate would register hits from passing charges. I'm pretty sure this has NOT been observed. > > Summarizing, if you'll look again at Maxwell's equations, you'll see > > that there is a possible source of an electric field that is NOT due to > > any physical electric charge (the field lines do not terminate at any > > charges), and there is a possible source of a magnetic field that is > > NOT due to any physical electric current. Have you looked at Maxwell's equations? > > [snip] > > > > PD > > Regards, > Monitek (Arden Barker) >> > Responding to yet another chunk... >> > [quoted text clipped - 51 lines] > However, as I pointed out elsewhere, the shape of this electric field > cannot be accounted for by redistributed charges on the pole tips. Its either induction in the magnet its self or its polarization of the vacuum. If your filed nmorphology does not match that for induction in the magnet then it must be vacuum polarization. >> > However, to address your explanation, I don't know where you got the >> > idea that displacement current is due to vacuum polarization. In a [quoted text clipped - 3 lines] >> > circling the gap *as though* there were a current in the gap, and this >> > pseudo-current is called displacement current. Its my idea that vacuum polarization is responsible for the displacement current. Also it is vacuum polarisation which is the manifestation of the electrostatic field, again my idea. >> It goes like this. Whatever is between the plates of a capacity is a >> dielectric ( think about what dielectric means). > > No, you tell me what you think a dielectric means. It does not mean > "that which fills a capacitor". If you think it does, define > "paraelectric". Dielectric means two electric ie positive and negative electricity. The origin of the thoery stems from polarised molecules. We then have to extend this idea to the vacuum. The simple fact that a capacitor works in vacuum means that the vacuum is a dielectric material. >> If you have a vacuum >> between the plates and the capacitor still charges then the vacuum IS a [quoted text clipped - 5 lines] > magnetic effect without there being any measurable charge transport in > the gap. An effect which is vcreating a magnetic field is a current- not a pseudo-current a real current caused by moving charged particles. If you can measure a magnetic field you care measuring current flow, that after all is the way currents are measured by measuring the magnetic field strength. If you measure a current flow then you are measuring the effects of charged particles moving. To deny that is to say that Faraday's Laws are wrong. >> As I said >> earlier - must earlier, where you can measure a magnetic field then there [quoted text clipped - 15 lines] > is, there would be more *excess* charge than accountable from the > source. Vacuum dielectric constant is defined as 1. The polarised vacuum is polarised by the energy of the charge on the plates of the capacitor. There is no net current flow between the plates via the battery. As the vacuum polarisation takes place there is an effective current flow due to charge separation until the charge reaches equilibrium then the "current flow " stops and the electric field is established and is maintained via the vacuum polarisation. > Secondly, there would be simple tests to determine whether charge moves > in the gap. A phosphorescent screen placed near one plate would > register hits from passing charges. I'm pretty sure this has NOT been > observed. I have some zinc sulphide powder handy will that do? I will have to get the van der graph out though. Is that a fact that only charged particles cause materials to phosphoresce? Might have to wait until the rotating magnet investigation is completed. I have the holder made to fix the magnet in a drill. Would you be happy with a ceramic disc capacitor as a probe to save me making one? As far as the effect has not been seen all tha means is the charge separation is below the minimum for the effect. >> > Summarizing, if you'll look again at Maxwell's equations, you'll see >> > that there is a possible source of an electric field that is NOT due to [quoted text clipped - 3 lines] > > Have you looked at Maxwell's equations? Yes they contain a term for the displacement current. The displacement current is due to charge movement during vacuum polarisation. When the electric field is established the charges stop moving and the magnetic effect ceases. If the equations require a displacement current then the displacement current is real, the charge movement is also real. >> > [snip] >> > >> > PD Regards, Monitek (Arden Barker) > >> > Responding to yet another chunk... > >> > [quoted text clipped - 56 lines] > If your filed nmorphology does not match that for induction in the magnet > then it must be vacuum polarization. The electric field in the gap is such that the electric field lines form closed loops. Explain to me a topology of vacuum polarization that can produce field lines of that shape. (Hint: The electric field of this shape cannot be associated with a scalar potential. Note this is the Maxwell equation that involves the curl of E. What's the curl of a gradient of a scalar potential?) > >> > However, to address your explanation, I don't know where you got the > >> > idea that displacement current is due to vacuum polarization. In a [quoted text clipped - 8 lines] > Also it is vacuum polarisation which is the manifestation of the > electrostatic field, again my idea. Which makes a definite experimental prediction about *real* charge flow in the gap of a capacitor, and therefore that is what you must test. > >> It goes like this. Whatever is between the plates of a capacity is a > >> dielectric ( think about what dielectric means). [quoted text clipped - 7 lines] > this idea to the vacuum. The simple fact that a capacitor works in vacuum > means that the vacuum is a dielectric material. Only if you assume that a capacitor by definition includes a dielectric. I see no reason for that definition. > >> If you have a vacuum > >> between the plates and the capacitor still charges then the vacuum IS a [quoted text clipped - 10 lines] > measure a magnetic field you care measuring current flow, that after all is > the way currents are measured by measuring the magnetic field strength. I would quibble with this. There are many ways to measure current, including measuring the charge delivered by that current over a period of time. Indeed, a coulomb is *defined* as the amount of charge delivered by a current. Presently, physicists do NOT say that the presence of a magnetic field requires a current source -- that's what YOU are saying, and that's what you must prove: that the dE/dt source term for the magnetic field is in fact due to a current. > If > you measure a current flow then you are measuring the effects of charged > particles moving. To deny that is to say that Faraday's Laws are wrong. That's crap. That's not at all what Faraday said. There is a source term for the magnetic field that is dE/dt, and a source term for the electric field that is dB/dt -- these are *in addition to* the current and charge terms. Reread this chapter. > >> As I said > >> earlier - must earlier, where you can measure a magnetic field then there [quoted text clipped - 23 lines] > stops and the electric field is established and is maintained via the vacuum > polarisation. You didn't answer the question. Is the charge in the gap delivered to the plates or no? (If not, WHY NOT?) > > Secondly, there would be simple tests to determine whether charge moves > > in the gap. A phosphorescent screen placed near one plate would [quoted text clipped - 3 lines] > I have some zinc sulphide powder handy will that do? I will have to get the > van der graph out though. You can do the research on what's required for a phosphorescent screen. Got an old computer monitor? (Legal notice: you could kill yourself here if you don't know what you're doing.) > Is that a fact that only charged particles cause materials to phosphoresce? Well, pretty much so, but it's irrelevant. If you don't see anything on the screen, then you know for sure you don't have charge flowing through the phosphorescent screen, because moving charges will cause phosphorescence. > Might have to wait until the rotating magnet investigation is completed. I > have the holder made to fix the magnet in a drill. Would you be happy with a > ceramic disc capacitor as a probe to save me making one? Recall that you needed to change the gap so that you can get more than one data point and fit the results to eliminate the pickup in the twisted pair. > As far as the effect has not been seen all tha means is the charge > separation is below the minimum for the effect. And your model should predict what that is. After all, you claim you have a measurable and *real* displacent current, and so you can say that you know exactly how much charge has been moved from one side of the gap to the other. Therefore you know how many vacuum electron and positrons have moved that distance and then you can determine whether that is above or below threshold for seeing it. > >> > Summarizing, if you'll look again at Maxwell's equations, you'll see > >> > that there is a possible source of an electric field that is NOT due to [quoted text clipped - 5 lines] > > Yes they contain a term for the displacement current. They contain a term dE/dt and dB/dt. http://hyperphysics.phy-astr.gsu.edu/hbase/electric/maxeq.html#c3 > The displacement > current is due to charge movement during vacuum polarisation. When the [quoted text clipped - 8 lines] > Regards, > Monitek (Arden Barker) >> >> > Responding to yet another chunk... >> >> > [quoted text clipped - 182 lines] > You didn't answer the question. Is the charge in the gap delivered to > the plates or no? An electron does not flow from the negatively charged plate to the positive plate. The dielectric polarises ie lines its charge up in the direction of the potential gradient. The negative polarisation adjacent to the positive plate causes the electrons to move from the plate by the fact that like charges repel. This leaves a net positive charge on the positive plate. Similarly when EMR passes a capacitor the plates are polarised by the state of vacuum polarisation which is created by the EMR wave. > (If not, WHY NOT?) Why if an electron crossed the gap between the plates it would be a conductor not a capacitor an ideal example of which has infinite resistance. >> > Secondly, there would be simple tests to determine whether charge moves >> > in the gap. A phosphorescent screen placed near one plate would [quoted text clipped - 8 lines] > Got an old computer monitor? (Legal notice: you could kill yourself > here if you don't know what you're doing.) I think the material used for video screens only fluoresces as having a sustained afterglow would be a nuisance. >> Is that a fact that only charged particles cause materials to >> phosphoresce? [quoted text clipped - 3 lines] > through the phosphorescent screen, because moving charges will cause > phosphorescence. Pretty much so is too vague, I really would like a yes/no on this one, because other than using such screens I have never had occasion to condider what they do and why they do it. So is it a fact that only charged particles can cause a phosphorescent material to glow? >> Might have to wait until the rotating magnet investigation is completed. >> I [quoted text clipped - 15 lines] > positrons have moved that distance and then you can determine whether > that is above or below threshold for seeing it. First of all I do not know if the effect of phosphorescence has been observed between capacitor plates or not. I dont know whether it can be seen even if I am right about pair separation being responsible for carrying the electric field. I am looking into it. From the size of the displacement current one can determine the equivalent electron flow and thats about it. Secondly, thats not quite what I said. I said that the displacement current can be measured and has been measured by measuring the magnetic field associated with it - they are synonymous. If you are measuring a real current then real charges are moving to create it. A galvo does not respond to an imaginary current. I personally have not measures the displacement current. As for e-p separation, I wish it was as easy as that. I have determined an approximate figure for the separation distance v charge value of e-p pairs. However, as you know from Maxwells equations that the potential is the sum of all the individual charges and is so in this case. The difference being the value of the charge is a variable and the number of pairs contributing to the charge is unknown. One can make a guess and say that pair separation would be the limit of charge carrying capability after that the vacuum creates pairs and the capacitor becomes a conductor. >> >> > Summarizing, if you'll look again at Maxwell's equations, you'll see >> >> > that there is a possible source of an electric field that is NOT due [quoted text clipped - 21 lines] >> >> > >> >> > PD Regards, Monitek (Arden Barker) > >> "PD" <TheDraperFamily@gmail.com> wrote in message [snip] > >> > There are several reasons why this cannot be the case. > >> > Let's start with where the supposed vacuum polarization charges go. Do [quoted text clipped - 31 lines] > Why if an electron crossed the gap between the plates it would be a > conductor not a capacitor an ideal example of which has infinite resistance. You've defined the displacement as being a *real* current with charge flow. I would suggest that any material where there is a current in the presence of an electric field is a conductor. The usual algebraic formulation of this looks something like this: J = (sigma)*E, where J is the current density, E is the electric field, and (sigma) is the conductivity. You say J is nonzero when E is nonzero, which demands that (sigma) is nonzero. The resistivity is the reciprocal of (sigma). So if you have infinite resistance, what must the conductivity be? If you have nonzero conductivity, how can the resistance be infinite? But algebra aside, let's look at a region of space immediately adjacent to a plate. You say there is charge separation in that region of space *right next* to the plate. What on earth keeps the charge from depositing on the plate. The vacuum gap is nothing like a material dielectric, where there is a surface to the dielectric material beyond which the dielectric's charges cannot move. > >> > Secondly, there would be simple tests to determine whether charge moves > >> > in the gap. A phosphorescent screen placed near one plate would [quoted text clipped - 11 lines] > I think the material used for video screens only fluoresces as having a > sustained afterglow would be a nuisance. Yes, either a fluorescent or a phosphorescent screeen will work. > >> Is that a fact that only charged particles cause materials to > >> phosphoresce? [quoted text clipped - 8 lines] > what they do and why they do it. So is it a fact that only charged particles > can cause a phosphorescent material to glow? Photons, if they are high enough energy, can kick out electrons from the surface of the screen and cause the electron to leave a spot on the screen, but that's a 2nd-order effect. The point here is that moving electrons WOULD be detected. I'm not expecting you to see spots and then having to worry that they are due to something else. I don't expect you to see spots from ANYTHING in the gap, let alone moving charges. > >> Might have to wait until the rotating magnet investigation is completed. > >> I [quoted text clipped - 18 lines] > First of all I do not know if the effect of phosphorescence has been > observed between capacitor plates or not. I dont know whether it can be seen Why wouldn't it be seen? Do the test. Better yet, choose an electron detector of your choice and put that in the gap. > even if I am right about pair separation being responsible for carrying the > electric field. I am looking into it. From the size of the displacement [quoted text clipped - 5 lines] > current then real charges are moving to create it. A galvo does not respond > to an imaginary current. And a galvo is not measuring the magnetic field. > I personally have not measures the displacement > current. Then I suggest you should! > As for e-p separation, I wish it was as easy as that. I have determined an > approximate figure for the separation distance v charge value of e-p pairs. [quoted text clipped - 4 lines] > would be the limit of charge carrying capability after that the vacuum > creates pairs and the capacitor becomes a conductor. Then your model is stuck. At least you could have a free paramater for one and calculate the other in terms of it. > >> >> > Summarizing, if you'll look again at Maxwell's equations, you'll see > >> >> > that there is a possible source of an electric field that is NOT due [quoted text clipped - 24 lines] > Regards, > Monitek (Arden Barker) >> >> "PD" <TheDraperFamily@gmail.com> wrote in message > [snip] [quoted text clipped - 62 lines] > dielectric, where there is a surface to the dielectric material beyond > which the dielectric's charges cannot move. First of all I have not said there is charge flow in between the plates of a capacitor I have said there is charge movement. The charge movement I envisage is that of electron positron pairs in the vacuum separating, but not separating as far as would be required for pair creation. Partially separated pairs will have partial charge and partial mass. Imagine a capacitor with the plates horizontal, the upper plate being negative and the lower plate being positive. The partially separated electron positron pairs whilst they are moving are equivalent to a current flow. In this case the partially separated electron moving upwards would be regarded as a conventional current flow downwards, this would produce a clockwise magnetic field arround the axis of movement. Similarly, the positron moving downwards, creates a clockwise magnetic field round the positron axis of movement. The magnetic fields from both particles are additive. A magnetic field associated with the charging of a capacitor has been measured by Röntgen I am suggesting that this field is a result of movement of charge |
Enable EMail Alerts
Start New Thread
Reply to this Message