Proton Radius Cannot Be Derived From Standard Model
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From: mpc755 on 11 May 2010 19:42 On May 11, 4:52 pm, PD <thedraperfam...(a)gmail.com> wrote: > On May 11, 2:56 pm, mpc755 <mpc...(a)gmail.com> wrote: > > > > > On May 11, 3:09 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > On May 11, 1:30 pm, mpc755 <mpc...(a)gmail.com> wrote: > > > > > On May 11, 1:44 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > On May 11, 10:40 am, mpc755 <mpc...(a)gmail.com> wrote: > > > > > > > On May 11, 11:27 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > > > On May 11, 10:23 am, mpc755 <mpc...(a)gmail.com> wrote: > > > > > > > > > You should also work on answering the following with an answer that > > > > > > > > does not require the future to determine the past. > > > > > > > > Why? The model accounts for all the experimental evidence. It works > > > > > > > just fine. > > > > > > > > > You should also > > > > > > > > work on understanding what experimental evidence is. If a C-60 > > > > > > > > molecule is ALWAYS detected exiting a single slit in a double slit > > > > > > > > experiment this is evidence the C-60 molecule ALWAYS exits a single > > > > > > > > slit. > > > > > > > > No, it isn't. Sorry, it just isn't. It is evidence that the molecule > > > > > > > exits a single slit *when* there is a detector there -- nothing more, > > > > > > > nothing less. The presence or absence of the detector changes that > > > > > > > claim. > > > > > > > The presence of the detector IS the experiment. > > > > > > Nonsense. That's not the experiment at all. Do you not understand the > > > > > double slit experiment? > > > > > In order to determine if the particle exits one or both slits in a > > > > double slit experiment, an experiment is performed. The experiment to > > > > detect if the particle exits a single slit or both slits places > > > > detectors at the exits to the slits. The experiment is called the > > > > 'Detector Experiment'. > > > > That's fine, but that isn't the double slit experiment. And in the > > > double slit experiment, the interesting behavior is what is *observed* > > > when there is no detector at either slit. > > > > > The 'Detector Experiment' is performed over and over again with all > > > > types of particles. After thousands and thousands of executions of the > > > > 'Detector Experiment' with hundreds of different types of particles > > > > the particle is ALWAYS detected exiting a single slit. The 'Detector > > > > Experiment' provides experimental evidence the particle ALWAYS exits a > > > > single slit. > > > > No, it provides experimental evidence that a particle always exits a > > > single slit when there is a detector placed at a slit -- that's all. > > > In order to test whether a particle exits a single slit when there is > > > no detector at the slit, you're going to have to find a way to do it > > > without a detector at the slit. > > > > Don't be a doofus. This should be obvious. > > > It is obvious if you conclude the particle exits both slits when there > > are no detectors at the slit then you are disregarding the > > experimental evidence arrived at from the 'Detector Experiment'. > > No, we are not. The currently accurately predicts what will be > observed when there are detectors at the slits. Hence that data is not > being disregarded. > It also accurately predicts what will be observed when there are no > detectors at the slits. > But the experiment is the detectors at the exits. The 'Detector Experiment' is experimental evidence the particle ALWAYS exits a single slit. > > > > That is what you do when you perform experiments. The arrive at > > conclusions based upon the experiment. In the 'Detector Experiment' > > the particle is ALWAYS detected exiting a single slit so the 'Detector > > Experiment' provides experimental evidence the particle ALWAYS exits a > > single slit. > > No, that would be extrapolating a conclusion beyond what you have data > for. You have data for the class of circumstances in which there is a > detector at the slits. This data does not allow you to conclude what > happens when there is no detector at the slits. You have no data for > that in your "detector experiment". You need a different experiment to > test what happens when there are no detectors at the slits. > > Any bonehead would be able to see that? > > Are you not even able to see what a bonehead can see? > Anyone who has any clue, except for those in a state of delusional denial believing in the Copenhagen interpretation of QM understand placing detectors at the exits to the slits is an experiment. If you conclude something else occur when you do not perform the experiment and you have no evidence of that something else then you are assuming something else occurs which is not supported by the experiment evidence of the 'Detector Experiment'. > > > > Now, you can conclude the particle exits both slits when you do not > > detect it but that is an assumption that is not based upon any > > experimental evidence. The experimental evidence is the particle > > ALWAYS exits a single slit. The reason why ALL of the experimental > > evidence supports the conclusion the particle ALWAYS exits a single > > slit because whenever an experiment is executed in order to determine > > if the particle exits one slit or both slits the particle is ALWAYS > > detected exiting a single slit. > > In experimental science, one learns not to extrapolate findings into > circumstances beyond which you have direct experimental support. You > are guilty of scientific fraud here. > The Copenhagen interpretation of QM is guilty of scientific fraud for assuming the particle exits both slits when not detected. There is no evidence the particle exits both slits. All of the experiment evidence is of the particle ALWAYS exiting a single slit. > > > > > > If you do not perform the 'Detector Experiment' and you assume the > > > > particle exits both slits your assumption is not supported by the > > > > 'Detector Experiment' experimental evidence. > > > > But I'm looking at more experimental evidence than your stupid, > > > restricted "detector experiment" evidence. I'm also including other > > > experiments like the double slit experiment, including the > > > configuration when there is no detector at either slit. > > > > Geez, what an ignoramus. > > > > > > > The experimental > > > > > > evidence associated with the experiment is the particle ALWAYS exits a > > > > > > single slit. > > > > > > No, it isn't. We've just been through that. > > > > > > > Only in the absurd nonsense of the Copenhagen interpretation of QM do > > > > > > you perform an experiment, where the result is the particle ALWAYS > > > > > > exits a single slit, then say the opposite occurs when the experiment > > > > > > is NOT performed, AND say that that supports the experimental > > > > > > evidence. > > > > > > Sorry, that isn't the experiment at all. Perhaps you need to remind > > > > > yourself what the experiment is. > > > > > > > > Time for you to read up on how this is possible. Would you like a > > > > > > > reference to a really good book on this? > > > > > > > > PD > >
From: waldofj on 11 May 2010 19:43 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 do http://www.haverford.edu/physics/love/teaching/Physics302PJL2009Recitation/01%20Matter%20Waves/SingleElectronBuildupOfAnInterferencePatternAJP000117.pdf > 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 > The wavelength of a single photon > is not measurable with a grating spectrometer, sure it is http://xmm.esac.esa.int/external/xmm_user_support/documentation/technical/RGS/ > 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. > 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. > > 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. > 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. > > PD
From: mpc755 on 11 May 2010 23:40 On May 11, 7:43 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... > > > 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 > > > 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... > > > 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. > > > 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. > > > > > 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. > Explain how the following occurs in the Copenhagen interpretation of QM: A C-60 molecule is in the slit(s). Detectors are placed at the exits. The C-60 molecule is detected exiting a single slit. A C-60 molecule is in the slit(s). Detectors are placed and removed from the exits. The C-60 molecule creates an interference pattern in and of itself. In Aether Displacement, the moving C-60 molecule has an associated aether displacement wave. The C-60 molecule is ALWAYS detected exiting a single slit because it ALWAYS enters and exits a single slit. The associated aether displacement wave enters and exits multiple slits. Upon exiting the slits the wave creates interference which alters the direction the C-60 molecule travels. Detecting the C-60 molecule causes decoherence of the associated aether displacement wave (i.e. turns the wave into chop) and there is no interference. If you can't explain the above in terms of the Copenhagen interpretation of QM, without resorting to absurd nonsense like the C-60 molecule enters one slit or multiple slits depending on there being detectors at the exits when it gets there in the future, you might want to reconsider how well it works. > > 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. > > > > > PD > >
From: PD on 12 May 2010 08:54 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. > > > 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. There are devices for *measuring* the wavelength, which do not work so well for individual photons. > > > 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. > > > 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. > > > 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. 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. > > > > > 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. > > > 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 12 May 2010 08:55
On May 11, 10:40 pm, mpc755 <mpc...(a)gmail.com> wrote: > On May 11, 7:43 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... > > > > 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 > > > > 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... > > > > 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. > > > > 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. > > > > 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. > > Explain how the following occurs in the Copenhagen interpretation of > QM: Not to you, no. > > A C-60 molecule is in the slit(s). Detectors are placed at the exits. > The C-60 molecule is detected exiting a single slit. A C-60 molecule > is in the slit(s). Detectors are placed and removed from the exits. > The C-60 molecule creates an interference pattern in and of itself. > > In Aether Displacement, the moving C-60 molecule has an associated > aether displacement wave. The C-60 molecule is ALWAYS detected exiting > a single slit because it ALWAYS enters and exits a single slit. The > associated aether displacement wave enters and exits multiple slits. > Upon exiting the slits the wave creates interference which alters the > direction the C-60 molecule travels. Detecting the C-60 molecule > causes decoherence of the associated aether displacement wave (i.e. > turns the wave into chop) and there is no interference. > > If you can't explain the above in terms of the Copenhagen > interpretation of QM, without resorting to absurd nonsense like the > C-60 molecule enters one slit or multiple slits depending on there > being detectors at the exits when it gets there in the future, you > might want to reconsider how well it works. > > > > 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. > > > > PD > > |