Proton Radius Cannot Be Derived From Standard Model
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From: waldofj on 15 May 2010 17:51 On May 12, 8:54 am, PD <thedraperfam...(a)gmail.com> wrote: > On May 11, 6:43 pm, waldofj <wald...(a)verizon.net> wrote: > > > > > On May 10, 4:01 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > On May 10, 2:21 pm, waldofj <wald...(a)verizon.net> wrote: > > > > > > Your statement that the photon is always detected as a particle is > > > > > still factually false. > > > > > I thought that was always the case. Of course I'm thinking in terms of > > > > the Copenhagen interpretation of QM. When is a photon not detected as > > > > a particle? > > > have you heard of wave/particle duality? > > > > Particles do not exhibit interference. > > > sure they dohttp://www.haverford.edu/physics/love/teaching/Physics302PJL2009Recit... > > As I said, one approach is to label photons as "particles" and then > completely revise what is meant by the word "particle" to suit. here is a definition of particle I found on the web: a. A body whose spatial extent and internal motion and structure, if any, are irrelevant in a specific problem. b. An elementary particle I think of elementary as meaning indivisible, or smallest unit. That's what I mean when I call a photon a particle. I'm not revising anything. I guess I just don't know what you mean by "particle". > > > Photons do. > > > they can under the appropriate circumstances. > > > > Just as an example. > > > Light consists of photons, but an assembly of photons has a property > > > called wavelength, which can be directly measured with a grating > > > spectrometer with great precision. > > > you're way off base here. Wavelength is not a property of an assembly > > of photons but a property of each individual photon, as per > > e = hf > > e = hc/lamda > > lamda = hc/e > > This is a measurement of *energy* and then assignment of a wavelength > to that energy. nonsense. I've been trying, unsuccessfully, to find experiments that directly show this relationship. I have found many articles that vaguely refer to such experiments, but not the experiments themselves. Nevertheless the theory is quite clear on this point: this is not just a measurement of energy that a wavelength is being "assigned" to, the ratio of the energy to the wavelength is a fundamental property of each individual photon. > There are devices for *measuring* the wavelength, which do not work so > well for individual photons. I'll go along with that but I thought we were having a theoretical discussion here. Since it's not my line of work I'm not interested in the practical limitations of commercial devices. > > > The wavelength of a single photon > > > is not measurable with a grating spectrometer, > > > sure it ishttp://xmm.esac.esa.int/external/xmm_user_support/documentation/techn... > > And this makes my point, thanks. The resolution of this spectrometer > is contingent on a large number of photons being obtained from the > source. you'll have to show me where it says that. This is what I saw: Nine large format back-illuminated CCDs are operated in single photon counting and frame transfer mode at a temperature of -80 C. For each photon, the position and the energy is measured: the position to determine the high resolution X-ray spectrum as diffracted by the grating module, and the energy and position to separate the contributions from the various overlapping grating orders (and from the in-flight calibration source) and to reduce the background. IOW this instrument is designed to measure individual photons accurately. Of course this is assisted by the fact that these are x- ray photons which pack a lot of energy and so they're comparatively easy to detect. again, I don't know what you mean by resolution. I thought a grating spectrometer works by measuring the angle that light reflects off the grating, so the resolution of the instrument is just a matter of the spatial resolution of the detectors. In the above case that would be the size and spacing of the pixels in the ccd arrays. Since those ccd's are already capable of measuring individual photons I don't see how the resolution (as I have just defined it) is improved by having multiple photons. > > > though the energy or > > > momentum is. > > > since the energy and wavelength are inversely related a direct > > measurement of one is an indirect measure of the other. But think of > > this, how does a grating spectrometer work? How does it measure the > > wavelength of light going through it? Once you've answered that then > > ask why would it be any different if there is just one photon of light > > going through the spectrometer at a time? The only problem > > experimentally is to have a sensitive enough detector. > > No, I'm sorry. There are lots of little Java applets on the web that > simulate the production of a diffraction spectrum from photons > delivered one at a time. It is certainly possible to cut off the > simulation after one photon delivered. I would ask you to try that, > and then determine the wavelength of that photon from the one impact > point on the photosensitive receptor. I don't care about java applets, I refer to the real deal, that x-ray telescope above. > > > Thus, when I measure the wavelength of light, is it a > > > fair statement to say I'm detecting light as a particle? > > > when a photon is detected it is depositing its energy into a single > > atom, bumping electrons into higher orbitals or kicking them out of > > the atom. There are other ways light interacts with matter, being > > split by a beam splitter or reflecting off a diffraction grating or > > some such. These interactions are wave like and I should point out > > during these interactions the light is not being detected, it's not > > until the light deposits its energy into a detector that it is > > detected and this interaction is particle like, one photon at a time. > > So when you're measuring the wavelength of light you're still > > detecting it as a particle, one photon at a time. > > Understood. But a property that is available for the assembly is not > available for the units. You can't say that universally, that's not what QM says about this particular case. > It's like temperature. The temperature of a gas is a property of the > assembly, and you cannot assign a temperature to any one molecule in > the gas. Sure, there's a mathematical relation between the kinetic > energy of the particles and the temperature of the gas, but the > temperature of a single molecule is not even a well-defined quantity. That's certainly true of temperature but I don't see what that has to do with what we're discussing. > > > Light has a litany of properties, including some that are particle- > > > like and some that are wave-like. It is improper to say that when > > > light is exhibiting particle-like properties, that's when it consists > > > of photons. It consists of photons all the time. > > > not according to the Copenhagen interpretation of QM. > > yeah, I know, it's old hat and you always say "keep up" but I'm not so > > sure that keeping up is the best thing to do here. > > In spite of how absurd, illogical, counter-intuitive, and just plain > > insane the Copenhagen interpretation of QM is, I'm not aware of any > > experimental results that contradict it. > > In spite of my own prejudices, I like to go with what works. > > QED works, and that is more along the lines of what I'm talking about. yeah, I've not had the opportunity to study that, and I'll never have the opportunity to study it any more than qualitatively. Do you have any good references? (QED for dummies that is) > > > But photons behave in > > > ways that are unlike a usual particle description. Alternatively, you > > > could say that photons are particles, but that the term "particle" now > > > has to be completely redefined to mean something other than what it > > > did in the 19th century. > > > That's true of all quantum objects, not just photons. > > Yes! Exactly. > > > > > > PD > >
From: PD on 17 May 2010 10:16 On May 15, 4:51 pm, waldofj <wald...(a)verizon.net> wrote: > On May 12, 8:54 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > On May 11, 6:43 pm, waldofj <wald...(a)verizon.net> wrote: > > > > On May 10, 4:01 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > > On May 10, 2:21 pm, waldofj <wald...(a)verizon.net> wrote: > > > > > > > Your statement that the photon is always detected as a particle is > > > > > > still factually false. > > > > > > I thought that was always the case. Of course I'm thinking in terms of > > > > > the Copenhagen interpretation of QM. When is a photon not detected as > > > > > a particle? > > > > have you heard of wave/particle duality? > > > > > Particles do not exhibit interference. > > > > sure they dohttp://www.haverford.edu/physics/love/teaching/Physics302PJL2009Recit... > > > As I said, one approach is to label photons as "particles" and then > > completely revise what is meant by the word "particle" to suit. > > here is a definition of particle I found on the web: > a. A body whose spatial extent and internal motion and structure, if > any, are irrelevant in a specific problem. > b. An elementary particle > > I think of elementary as meaning indivisible, or smallest unit. That's > what I mean when I call a photon a particle. I'm not revising > anything. I guess I just don't know what you mean by "particle". A particle in classical parlance means a bit more than just indivisibility or (for the context at hand) structurelessness. It also means localizability, for instance. The localizability is very much in conflict with the current understanding of a photon. > > > > > > > Photons do. > > > > they can under the appropriate circumstances. > > > > > Just as an example. > > > > Light consists of photons, but an assembly of photons has a property > > > > called wavelength, which can be directly measured with a grating > > > > spectrometer with great precision. > > > > you're way off base here. Wavelength is not a property of an assembly > > > of photons but a property of each individual photon, as per > > > e = hf > > > e = hc/lamda > > > lamda = hc/e > > > This is a measurement of *energy* and then assignment of a wavelength > > to that energy. > > nonsense. I've been trying, unsuccessfully, to find experiments that > directly show this relationship. I have found many articles that > vaguely refer to such experiments, but not the experiments themselves. > Nevertheless the theory is quite clear on this point: this is not just > a measurement of energy that a wavelength is being "assigned" to, the > ratio of the energy to the wavelength is a fundamental property of > each individual photon. Then you're going to have to tell me how this relationship is validated directly for single photons by the independent measurement of energy AND wavelength for that photon. > > > There are devices for *measuring* the wavelength, which do not work so > > well for individual photons. > > I'll go along with that but I thought we were having a theoretical > discussion here. Since it's not my line of work I'm not interested in > the practical limitations of commercial devices. It's not a matter of practical limitation. Nor is it a practical limitation that we cannot measure the temperature of an individual molecule in a gas. Rather, it is the case that temperature is not even a well-defined quantity for an individual molecule in a gas. > > > > > The wavelength of a single photon > > > > is not measurable with a grating spectrometer, > > > > sure it ishttp://xmm.esac.esa.int/external/xmm_user_support/documentation/techn... > > > And this makes my point, thanks. The resolution of this spectrometer > > is contingent on a large number of photons being obtained from the > > source. > > you'll have to show me where it says that. This is what I saw: > Nine large format back-illuminated CCDs are operated in single photon > counting and frame transfer mode at a temperature of -80 C. For each > photon, the position and the energy is measured: the position to > determine the high resolution X-ray spectrum as diffracted by the > grating module, and the energy and position to separate the > contributions from the various overlapping grating orders (and from > the in-flight calibration source) and to reduce the background. > > IOW this instrument is designed to measure individual photons > accurately. Position and energy. Where is the wavelength measured? > Of course this is assisted by the fact that these are x- > ray photons which pack a lot of energy and so they're comparatively > easy to detect. > again, I don't know what you mean by resolution. I thought a grating > spectrometer works by measuring the angle that light reflects off the > grating, so the resolution of the instrument is just a matter of the > spatial resolution of the detectors. No, not at all. There is an *inherent* resolution that comes from the uncertainty principle that doesn't have anything to do with the spatial resolution of the detectors. > In the above case that would be > the size and spacing of the pixels in the ccd arrays. Since those > ccd's are already capable of measuring individual photons I don't see > how the resolution (as I have just defined it) is improved by having > multiple photons. > > > > > > > though the energy or > > > > momentum is. > > > > since the energy and wavelength are inversely related a direct > > > measurement of one is an indirect measure of the other. But think of > > > this, how does a grating spectrometer work? How does it measure the > > > wavelength of light going through it? Once you've answered that then > > > ask why would it be any different if there is just one photon of light > > > going through the spectrometer at a time? The only problem > > > experimentally is to have a sensitive enough detector. > > > No, I'm sorry. There are lots of little Java applets on the web that > > simulate the production of a diffraction spectrum from photons > > delivered one at a time. It is certainly possible to cut off the > > simulation after one photon delivered. I would ask you to try that, > > and then determine the wavelength of that photon from the one impact > > point on the photosensitive receptor. > > I don't care about java applets, I refer to the real deal, that x-ray > telescope above. It helps to see the X-ray telescope operating in real time, then. The simulations show how that works for a diffraction grating. You might find it illuminating -- no pun intended. > > > > > > > Thus, when I measure the wavelength of light, is it a > > > > fair statement to say I'm detecting light as a particle? > > > > when a photon is detected it is depositing its energy into a single > > > atom, bumping electrons into higher orbitals or kicking them out of > > > the atom. There are other ways light interacts with matter, being > > > split by a beam splitter or reflecting off a diffraction grating or > > > some such. These interactions are wave like and I should point out > > > during these interactions the light is not being detected, it's not > > > until the light deposits its energy into a detector that it is > > > detected and this interaction is particle like, one photon at a time. > > > So when you're measuring the wavelength of light you're still > > > detecting it as a particle, one photon at a time. > > > Understood. But a property that is available for the assembly is not > > available for the units. > > You can't say that universally, that's not what QM says about this > particular case. > > > It's like temperature. The temperature of a gas is a property of the > > assembly, and you cannot assign a temperature to any one molecule in > > the gas. Sure, there's a mathematical relation between the kinetic > > energy of the particles and the temperature of the gas, but the > > temperature of a single molecule is not even a well-defined quantity. > > That's certainly true of temperature but I don't see what that has to > do with what we're discussing. It has a LOT to do with it. Wavelength is defined for bulk collections of photons, and it is not a well-defined, MEASURABLE quantity for individual photons. > > > > > Light has a litany of properties, including some that are particle- > > > > like and some that are wave-like. It is improper to say that when > > > > light is exhibiting particle-like properties, that's when it consists > > > > of photons. It consists of photons all the time. > > > > not according to the Copenhagen interpretation of QM. > > > yeah, I know, it's old hat and you always say "keep up" but I'm not so > > > sure that keeping up is the best thing to do here. > > > In spite of how absurd, illogical, counter-intuitive, and just plain > > > insane the Copenhagen interpretation of QM is, I'm not aware of any > > > experimental results that contradict it. > > > In spite of my own prejudices, I like to go with what works. > > > QED works, and that is more along the lines of what I'm talking about. > > yeah, I've not had the opportunity to study that, and I'll never have > the opportunity to study it any more than qualitatively. Do you have > any good references? (QED for dummies that is) Feynman's SHORT book on QED is a good start for amateurs. > > > > > But photons behave in > > > > ways that are unlike a usual particle description. Alternatively, you > > > > could say that photons are particles, but that the term "particle" now > > > > has to be completely redefined to mean something other than what it > > > > did in the 19th century. > > > > That's true of all quantum objects, not just photons. > > > Yes! Exactly. > > > > > PD > >
From: spudnik on 19 May 2010 01:34 weren't you the one who was complaining about paradoxical things in QM?... so, you seem to be able to have it both ways, having your wave & your little pizza pie, two, when they were only ever just dual, mathematical representations of one thing; you just don't need to use them, at teh same time, and will probably not be able to in any realistic way. certainly, no-one else has! anyway, arguing with you guys makes me into that "exotic negative mass" stuff, that could build an Einstin-Rosen superbridge to when God-am ever. > The above is correct. The 'particle' portion of the photon can be > considered to be part of the wave itself. The 'particle' portion of > the photon does not have to be a rock of light. thus prove: prove and/or define the most canonical "law of cosines" in trgionometry taht you can; you can define canonical, two. well, I just read the definition of the law, or the supposed outcome of formula, in a large dictionary (of English). thus: I haven't proven that the Bible Code was a hoax; only a hueristical argument about any ring of letters of "all of the letters" ... not the Object or Bunny Rings, neccesarily. however, the biblical topic is "skip codes." --Light: A History! http://wlym.com
From: mpc755 on 19 May 2010 06:08 On May 19, 1:34 am, spudnik <Space...(a)hotmail.com> wrote: > weren't you the one who was complaining > about paradoxical things in QM?... so, > you seem to be able to have it both ways, > having your wave & your little pizza pie, two, > when they were only ever just dual, mathematical representations > of one thing; you just don't need to use them, > at teh same time, and will probably not be able to > in any realistic way. certainly, no-one else has! > de Broglie originated wave-particle duality. In de Broglie wave mechanics, the 'particle' occupies a very small region of the wave. In a double slit experiment the wave enters and exits multiple slits while the 'particle' enters and exits a single slit. The wave creates interference upon exiting the slits which alters the direction the 'particle' travels. Detecting the 'particle' causes decoherence of the associated wave (i.e. turns the wave into chop) and there is no interference. Why is the 'particle' ALWAYS detected exiting a single slit when detectors are placed at the exits to the slits while the 'particle' is in the slit(s)? Because the 'particle' ALWAYS enters and exits a single slit. A moving C-60 molecule has an associated aether displacement wave. How is a C-60 molecule able to create an interference pattern in a double slit experiment? Because the C-60 molecule ALWAYS enters and exits a single slit and it is the associated aether displacement wave which enters and exits multiple slits. The associated aether displacement wave creates interference upon exiting the slits which alters the direction the C-60 molecule travels. Detecting the C-60 molecule causes decoherence of the associated aether displacement wave and there is no interference. Why is the C-60 molecule ALWAYS detected exiting a single slit when detectors are placed at the exits to the slits while the C-60 molecule is in the slit(s)? Because the C-60 molecule ALWAYS enters and exits a single slit. > anyway, arguing with you guys makes me into > that "exotic negative mass" stuff, > that could build an Einstin-Rosen superbridge > to when God-am ever. > > > The above is correct. The 'particle' portion of the photon can be > > considered to be part of the wave itself. The 'particle' portion of > > the photon does not have to be a rock of light. > > thus prove: > prove and/or define the most canonical "law > of cosines" in trgionometry taht you can; > you can define canonical, two. > > well, I just read the definition > of the law, or the supposed outcome of formula, > in a large dictionary (of English). > > thus: > I haven't proven that the Bible Code was a hoax; > only a hueristical argument about any ring > of letters of "all of the letters" ... not the Object or > Bunny Rings, neccesarily. however, > the biblical topic is "skip codes." > > --Light: A History!http://wlym.com
From: BURT on 19 May 2010 14:19
On May 19, 3:08 am, mpc755 <mpc...(a)gmail.com> wrote: > On May 19, 1:34 am, spudnik <Space...(a)hotmail.com> wrote: > > > weren't you the one who was complaining > > about paradoxical things in QM?... so, > > you seem to be able to have it both ways, > > having your wave & your little pizza pie, two, > > when they were only ever just dual, mathematical representations > > of one thing; you just don't need to use them, > > at teh same time, and will probably not be able to > > in any realistic way. certainly, no-one else has! > > de Broglie originated wave-particle duality. In de Broglie wave > mechanics, the 'particle' occupies a very small region of the wave. In > a double slit experiment the wave enters and exits multiple slits > while the 'particle' enters and exits a single slit. The wave creates > interference upon exiting the slits which alters the direction the > 'particle' travels. Detecting the 'particle' causes decoherence of the > associated wave (i.e. turns the wave into chop) and there is no > interference. > > Why is the 'particle' ALWAYS detected exiting a single slit when > detectors are placed at the exits to the slits while the 'particle' is > in the slit(s)? > > Because the 'particle' ALWAYS enters and exits a single slit. > > A moving C-60 molecule has an associated aether displacement wave. > > How is a C-60 molecule able to create an interference pattern in a > double slit experiment? > > Because the C-60 molecule ALWAYS enters and exits a single slit and it > is the associated aether displacement wave which enters and exits > multiple slits. The associated aether displacement wave creates > interference upon exiting the slits which alters the direction the > C-60 molecule travels. Detecting the C-60 molecule causes decoherence > of the associated aether displacement wave and there is no > interference. > > Why is the C-60 molecule ALWAYS detected exiting a single slit when > detectors are placed at the exits to the slits while the C-60 molecule > is in the slit(s)? > > Because the C-60 molecule ALWAYS enters and exits a single slit. > > > > > anyway, arguing with you guys makes me into > > that "exotic negative mass" stuff, > > that could build an Einstin-Rosen superbridge > > to when God-am ever. > > > > The above is correct. The 'particle' portion of the photon can be > > > considered to be part of the wave itself. The 'particle' portion of > > > the photon does not have to be a rock of light. > > > thus prove: > > prove and/or define the most canonical "law > > of cosines" in trgionometry taht you can; > > you can define canonical, two. > > > well, I just read the definition > > of the law, or the supposed outcome of formula, > > in a large dictionary (of English). > > > thus: > > I haven't proven that the Bible Code was a hoax; > > only a hueristical argument about any ring > > of letters of "all of the letters" ... not the Object or > > Bunny Rings, neccesarily. however, > > the biblical topic is "skip codes." > > > --Light: A History!http://wlym.com- Hide quoted text - > > - Show quoted text - The proton is an infinitely small trio of quarks. It has no radius. Mitch Raemsch |