Kramer Posted October 11, 2015 Posted October 11, 2015 The electron particle must have three electric charges! (A speculation from a layman)The energy of a “static” electron particle is equal: Ee = (-e) * (-e) / (4*pi*ε0* Re) =8.187104679*10^-14 joule …… II-1 “Re” is classic radius derived by Compton wave-length of electron. Am I wrong? My friend S… will say: Yes you are, I see only two electric charges in your formula. And “electron particle” has in fact only one electric charge. Let see the physic’s unity in my formula: Joule = C * C / m *F*m^-1 = C*C / C*(U*m)^-1 *m = C*C*C^-1*U*m /m= C*UThe speculation consist:1- The third electric charge must exists in space in the so called Dirak’s sea, and must have positive sign2- The second charge is associated with the Dirak’s charge and has created Compton electromagnetic wave.3- There is not any annihilation of charges, only a shadow from each other. By three electric charges, two of them that posses different signs create a spherical wave around the first negative charge. Now, shoot for an easy win of green points.
swansont Posted October 11, 2015 Posted October 11, 2015 What is the difference between the two different positively charged particles? (assuming these are distinct particles) Or are you claiming the electron is a composite particle?
ajb Posted October 12, 2015 Posted October 12, 2015 Electromagnetic theory would be very different with three kinds of electric charge. In essence we know about theorires with more than one kind of charge, for example we have Yang-Mills theories.
Kramer Posted October 12, 2015 Author Posted October 12, 2015 SwansonWhat is the difference between the two different positively charged particles? (assuming these are distinct particlesOr are you claiming the electron is a composite particle?----Yes. I think that electron particle is a composite particle. As are all common particles. I say, only, that electric charge must not be confounded as identical “electron particle”. Electric charges are the main components of electron particle. They structure “electron particle”, they are cause of electric static field of “electron particle”, they are the cause of stationary electromagnetic wave and, even this sounds crazy, they posses ability for creating mass, in presence of gravity constant s.c.r.t (G). I am not sure if this is a second own property of “electric charge” or an outsider.Yes, electric charge has a limited dimension. That means it is not zero. AjbElectromagnetic theory would be very different with three kinds of electric charge. In essence we know about theorires with more than one kind of charge, for example we have Yang-Mills theories.----- You know Mr. Ajb, that I am not a physicist. My post is based in data: electron particle posses electric field, posses electromagnetic phenomena even though in static status, exist in space (which has an electric constant) – and has a measurable dimension.Formula gave a precise amount of energy equal with other kind of energies. E = me * C^2. = h*(C / ((2*pi / α) * Re )).For me they are enough for a superficial cognition.This superficial cognition gave me a thought that matter and energy of “common particles” are result “only if interaction” happens between at least two subs. I am not able to discuss about theories that flourish every day, and that became always more and more complicated.
swansont Posted October 12, 2015 Posted October 12, 2015 Swanson What is the difference between the two different positively charged particles? (assuming these are distinct particles Or are you claiming the electron is a composite particle? ----Yes. I think that electron particle is a composite particle. As are all common particles. I say, only, that electric charge must not be confounded as identical “electron particle”. Electric charges are the main components of electron particle. They structure “electron particle”, they are cause of electric static field of “electron particle”, they are the cause of stationary electromagnetic wave and, even this sounds crazy, they posses ability for creating mass, in presence of gravity constant s.c.r.t (G). I am not sure if this is a second own property of “electric charge” or an outsider. What evidence can you present that the electron is composite, or what model can you present that would give predictions of what we should see? My post is based in data: electron particle posses electric field, posses electromagnetic phenomena even though in static status, exist in space (which has an electric constant) – and has a measurable dimension. Formula gave a precise amount of energy equal with other kind of energies. E = me * C^2. = h*(C / ((2*pi / α) * Re )). For me they are enough for a superficial cognition. This superficial cognition gave me a thought that matter and energy of “common particles” are result “only if interaction” happens between at least two subs. I am not able to discuss about theories that flourish every day, and that became always more and more complicated. That's the electron self-energy according to classical physics (hence the name "classical electron radius"). But we know from experiment that the electron radius is not this value. The upper limit is much, much smaller. You are using a relation we know to be wrong.
Kramer Posted October 13, 2015 Author Posted October 13, 2015 Swanson What evidence can you present that the electron is composite, or what model can you present that would give predictions of what we should see? -----The formula and my interpretation of physics units are not convincing?I insist in this formula because I think is a universal, applicable for all common particles. It fit exact in the case of proton, and maybe in all particles that have Compton wave- length. Seems to me, it is like a pattern. The only variable in this pattern is radius of particle, and indeed the combinations of signs of charges. That's the electron self-energy according to classical physics (hence the name "classical electron radius"). But we know from experiment that the electron radius is not this value. The upper limit is much, much smaller. You are using a relation we know to be wrong. ----- I don’t pretend that have invented or discovered the formula. My interpretations aims to defend nature of matter as particular, based it in the fact that electric charge “e” is a physics concept of something that exist and is exact measurable, as a unity.I think that “electric charge” is the ‘hub of the mater’ of all physics phenomena, together with concept of mass with which is not separable.About “radius” or distance between two “electric charges”, you say is not the value.Seems to me that Radius extracted by Compton wave-length is not a fact convincing.Let see this way:Now I have displayed in my formula that e / (4*pi*e / ( U * d)) = U*d = 1.43994393*10^-9 V*mThis is a universal formula. Divide this value with whatever radius and you will have the second component of energy ----“U”.example: U*d / 1.5346088165*10^-18 = 938264829,8 V for proton.After your concept of electric charge as point with zero dimension, result U = infinite. Which is absurd.That not right. The smallest dimension of electric charge is: “Rmin.” = 1.380543856*10^-36 m.If you don’t want to violate: v < C !With this kind of dimension of electric charges, that “structures” with their movements toward each other the spherical form of common particle, I doubt about results of experiment.
swansont Posted October 13, 2015 Posted October 13, 2015 Swanson What evidence can you present that the electron is composite, or what model can you present that would give predictions of what we should see? -----The formula and my interpretation of physics units are not convincing? No, not in the least. I insist in this formula because I think is a universal, applicable for all common particles. It fit exact in the case of proton, and maybe in all particles that have Compton wave- length. Seems to me, it is like a pattern. The only variable in this pattern is radius of particle, and indeed the combinations of signs of charges. It works for a proton? The proton is ~1876x more massive, so the radius should be correspondingly larger. That will give you a number around 5 x 10^-12m. The actual proton radius (from experiment) is ~5000 times smaller. That's "exact"? So it fails for a proton and an electron. Failure is the only pattern so far. That's the electron self-energy according to classical physics (hence the name "classical electron radius"). But we know from experiment that the electron radius is not this value. The upper limit is much, much smaller. You are using a relation we know to be wrong. ----- I don’t pretend that have invented or discovered the formula. My interpretations aims to defend nature of matter as particular, based it in the fact that electric charge “e” is a physics concept of something that exist and is exact measurable, as a unity. I think that “electric charge” is the ‘hub of the mater’ of all physics phenomena, together with concept of mass with which is not separable. About “radius” or distance between two “electric charges”, you say is not the value. Seems to me that Radius extracted by Compton wave-length is not a fact convincing. Let see this way: Now I have displayed in my formula that e / (4*pi*e / ( U * d)) = U*d = 1.43994393*10^-9 V*m This is a universal formula. Divide this value with whatever radius and you will have the second component of energy ----“U”. example: U*d / 1.5346088165*10^-18 = 938264829,8 V for proton. After your concept of electric charge as point with zero dimension, result U = infinite. Which is absurd. That not right. The smallest dimension of electric charge is: “Rmin.” = 1.380543856*10^-36 m. If you don’t want to violate: v < C ! With this kind of dimension of electric charges, that “structures” with their movements toward each other the spherical form of common particle, I doubt about results of experiment. It's absurd because the formula is wrong. Wrong formulas give wrong answers. Did you consider that?
Sensei Posted October 13, 2015 Posted October 13, 2015 (edited) 1- The third electric charge must exists in space in the so called Diraks sea, and must have positive sign 2- The second charge is associated with the Diraks charge and has created Compton electromagnetic wave. 3- There is not any annihilation of charges, only a shadow from each other. By three electric charges, two of them that posses different signs create a spherical wave around the first negative charge. Now, shoot for an easy win of green points. If I recall correctly your older theories, you're postulating existence of sub-particles each with -1e and +1e (and they're not electron/positron). So, to have electron with -1e for outer observer, there is needed -1e -1e +1e sub-particles inside of electron, to have -1e total.. Right? So, positron will have exactly reverse configurations: +1e +1e -1e to have +1e total? Now, what happens during pair-production? Photon has energy 1.022 MeV or more, and 0e charge, after collision there are made electron and positron pair. y + 1.022 MeV -> e+ + e- So photon had to have all these 6 sub-particles inside, right? So far it holds, sort of. But now there is annihilation of newly produced electron-positron: e+ + e- -> y + y (each with 510998.928 eV) If electron has +1-1-1 and positron has -1+1+1, there is no such configuration in which there could be created 2 neutral photons from them.. Possible output configurations are: -1-1-1 = -3e -1-1+1 = -1e -1+1+1 = +1e +1+1+1 = +3e Or I am missing something? It works for a proton? The proton is ~1876x more massive, so the radius should be correspondingly larger. 1836.15x more mass/energy (938.272046/0.510998928 = 1836.15267) Edited October 13, 2015 by Sensei 1
Strange Posted October 13, 2015 Posted October 13, 2015 That sounds a bit like some of the preon models (1), in particular Sundance Bilson-Thompson's (2) work (3). (1) https://en.wikipedia.org/wiki/Preon (2) Best name ever? (3) http://arxiv.org/abs/hep-ph/0503213
Kramer Posted October 14, 2015 Author Posted October 14, 2015 SwansonNo, not in the least. ----- As always. At least to have tried a bit to give for me, where I have flawed. It works for a proton? The proton is ~1876x more massive, so the radius should be correspondingly larger. That will give you a number around 5 x 10^-12m. The actual proton radius (from experiment) is ~5000 times smaller. That's "exact"? ----- And here we differ in reasoning. I say: the more massive a common particle is, the smallest is the radius. This is, in fact, “the cause” why common elementary particles have different Compton wave-length. And I have used the radius of proton wave-length in calculation of energy of proton, or its voltage.So it fails for a proton and an electron. Failure is the only pattern so far. ----Maybe is a failure. But I have impression that you have not read with attention. t's absurd because the formula is wrong. Wrong formulas give wrong answers. Did you consider that? ----- Why is wrong the formula, if it gave right answer? Why you don’t give the dimension of electric charge, or you think that electric charge is only the “ excitation of field” without any dimension? SenseiIf I recall correctly your older theories, you're postulating existence of sub-particles each with -1e and +1e (and they're not electron/positron). So,to have electron with -1e for outer observer, there is needed -1e -1e +1e sub-particles inside of electron, to have -1e total.. Right?-----No. To have a “me” needed (as you say) –1e+-1e +1+e’ subs to have in total: (-1e + 1(photon stationary)) So, positron will have exactly reverse configurations: +1e +1e -1e to have +1e total?-----That right Sensei with above correction. This is my point in this post, but I think you are moving a little fast, creation of photons is another issue. In this post I am speaking about “stationary particles”. Not about interaction, annihilation, creation of photons. If you think that this is an argument against essence of this post, let me know. Now, what happens during pair-production? Photon has energy 1.022 MeV or more, and 0e charge,---- Please explain me (you know I am a lay –man) how photon has obtained 1.022 MeV, when one electron and one positron in, “stationary” situation, create (or they may aren’t able?) “two” photons with overall 1.022 MeV. Or I am out?After this I am ready to debate about this issue.And please remind that the issue isn’t easy: You need to explain what happens with acceleration of particles.----- after collision there are made electron and positron pair.y + 1.022 MeV -> e+ + e- So photon had to have all these 6 sub-particles inside, right? So far it holds, sort of. But now there is annihilation of newly produced electron-positron: e+ + e- -> y + y (each with 510998.928 eV) If electron has +1-1-1 and positron has -1+1+1, there is no such configuration in which there could be created 2 neutral photons from them..Possible output configurations are:-1-1-1 = -3e-1-1+1 = -1e-1+1+1 = +1e+1+1+1 = +3e 1836.15x more mass/energy (938.272046/0.510998928 = 1836.15267)----- ?? StrangeThat sounds a bit like some of the preon models (1), in particular Sundance Bilson-Thompson's (2) work (3).----- As I know, preons are “bit” of electric charge. Strange you are a great catalog.1) https://en.wikipedia.org/wiki/Preon(2) Best name ever?(3) http://arxiv.org/abs/hep-ph/0503213
swansont Posted October 14, 2015 Posted October 14, 2015 ----- And here we differ in reasoning. I say: the more massive a common particle is, the smallest is the radius. This is, in fact, “the cause” why common elementary particles have different Compton wave-length. And I have used the radius of proton wave-length in calculation of energy of proton, or its voltage. Ah, I misread your equation. So you predict a radius 1836x smaller than the classical electron radius, or ~ 1.5 x 10^-18 m. Still disagrees with experiment. Since the mass is the reason for the Compton wavelength, you are correct. It's the cause. Energy is not the same as voltage. They are different things. It's absurd because the formula is wrong. Wrong formulas give wrong answers. Did you consider that? ----- Why is wrong the formula, if it gave right answer? Why you don’t give the dimension of electric charge, or you think that electric charge is only the “ excitation of field” without any dimension? It doesn't give the right answer. The experimentally determined proton radius is about 0.9 x 10^-15 m. That's only about a factor of 3 smaller than the classical electron radius. You say e / (4*pi*e / ( U * d)) = U*d = 1.43994393*10^-9 V*m but nowhere have you explained what U and d are, or where they come from, or where you got the formula — it just appears out of thin air. But doesn't U*d cancel? e / (4*pi*ε / ( U * d)) = (e*U*d)/(4*pi*ε) so this is the same as saying e/(4*pi*ε) = 1, which is clearly false. Meaning I have no clue how you are getting any of these values. What I suspect what you did was start with the Compton radius (which depends on mass, and thus the mass energy) and did some manipulation and recovered the mass energy. There's no new physics there. But without showing your work, I can't be sure.
Sensei Posted October 14, 2015 Posted October 14, 2015 -----No. To have a me needed (as you say) 1e+-1e +1+e subs to have in total: (-1e + 1(photon stationary)) I don't understand why are you saying 'no'. If we have -1e in center of electron, and photon that has -1e and +1e, then it's exactly like in mine post above. Then why're you saying 'no'? Doesn't make sense to me. Photon has neutral charge 0e My equation was for Q, electric charge inside of electron or positron. Qelectron=-1e-1e+1e=-1e Qpositron=+1e+1e-1e=+1e Then photon has to have Qphoton=+1e-1e=0e That's the only way to not have instant violation of electric charge conservation, within your framework from post #1. -----That right Sensei with above correction. Explain me your correction.. This is my point in this post, but I think you are moving a little fast, But you're moving too slow. After half-year when you're not here, you should have spend time on learning the all leptons, mesons and baryons decays.. You should came up with ready theory that has everything what we already know about them. creation of photons is another issue. In this post I am speaking about stationary particles. Not about interaction, annihilation, creation of photons. Annihilation violated your model instantly. If you think that this is an argument against essence of this post, let me know. I just did in post #8... ---- Please explain me (you know I am a lay man) how photon has obtained 1.022 MeV, when one electron and one positron in, stationary situation, create (or they may arent able?) two photons with overall 1.022 MeV. Or I am out? Photons with high energy can be created by f.e. - cosmic rays, - decay of unstable radioactive isotope - decay of unstable particle f.e. [math]\pi^0\rightarrow\gamma+\gamma[/math] (neutral pion meson decay, it has ~135 MeV/c^2 rest-mass) https://en.wikipedia.org/wiki/Pion#Neutral_pion_decays - fusion in the star f.e. [math]p^+ + D^+ \rightarrow He^3+\gamma+5.49 MeV[/math] - annihilation of mesons or baryons. In other words: they're not from annihilation of electron and positron. At least not stationary electron-positron in our FoR.
Kramer Posted October 15, 2015 Author Posted October 15, 2015 Swanson h, I misread your equation. So you predict a radius 1836x smaller than the classical electron radius, or ~ 1.5 x 10^-18 m. Still disagrees with experiment. ----- For sake of clarity, and for help --- may I know experimental value?Since the mass is the reason for the Compton wavelength, you are correct. It's the cause. Energy is not the same as voltage. They are different things.----- Where I have write this mistake?It doesn't give the right answer. The experimentally determined proton radius is about 0.9 x 10^-15 m. That's only about a factorof 3 smaller than the classical electron radius. ---- Not clear for me. I say: classical radius for electron is Re = 2 .8179401*10^-15 m and radius of proton Rp = 1.534698155* 10^-18 m. You say experiments have found:Re = about 0 --- and Rp = 0.9 *10^-15. Right? e / (4*pi*e / ( U * d)) = U*d = 1.43994393*10^-9 V*m but nowhere have you explained what U and d are, or where they come from, or where you got the formula — it just appears out of thin air. But doesn't U*d cancel?e / (4*pi*ε / ( U * d)) = (e*U*d)/(4*pi*ε) so this is the same as saying e/(4*pi*ε) = 1, which is clearly false. Meaning I have no clue how you are getting any of these values. ----- Here is the speculation: E1 = (- e) * [ (-e) / ((4*pi*ε)*1)] = (-e) * U1*1/1= (-e)*U1Let see this part of eq. [ (-e) / ((4*pi*ε)*1)] = (-e) / (+e) / (U1*1) = 1.43994393*10^-9 V*m it is unity potential of two electric charge in 1 m distance.Here U1 is the potential created between (must be) electric charges (-e) and +e in a distance in space =1m ( Which is the unity of distance in system m, kg, s, A). The electric charges are unperceivable because shadow each other, but exist and form stationary Compton wave. The potential of Compton wave interact with perceivable (-e) and determine the electric energy of an perceivable electric charge in a sphere 1m.E1.= (1.43994393*10^-9) V*1m * e / 1m = 2.307044*10^-28 joule. AndEe = (1.43994393*10^-9) V*1m * e / Re = 8.186988333*10^-14 jouleEp = ((1.43994393*10^-9) V*1m * e / Rp = 1.50325604*10^-10 joule What I suspect what you did was start with the Compton radius (which depends on mass, and thus the mass energy) and did some manipulation and recovered the mass energy. There's no new physics there. But without showing your work, I can't be sure. ----- I will try to explain that in elementary common particle all laws of physic (about energy) are applicable with the same value. Even Newton law. At least so seems to me. SenseiI don't understand why are you saying 'no'.If we have -1e in center of electron, and photon that has -1e and +1e, then it's exactly like in mine post above. Then why're you saying 'no'? Doesn't make sense to me. Photon has neutral charge 0eMy equation was for Q, electric charge inside of electron or positron.Qelectron=-1e-1e+1e=-1e----- You say Photon has electric charge 0.But here is the speculation. Photon has two electric charges, unperceivable because have different sign, and seems like they don’t exist in photon, because they shadow each other.Electron has three electric charges: one is perceivable, two are unperceivable because they shadow each other when exist in Compton wave of particle.Qpositron=+1e+1e-1e=+1eThen photon has to haveQphoton=+1e-1e=0e That's the only way to not have instant violation of electric charge conservation, within your framework from post #1. ------ Maybe we say the same but confuse each other. If you are against my speculationslet continue debate.By the way Sensei can you calculate the energy of Compton wave of electron by its length?I want to compare yours with my calculation of this energy which seems to me weird.
swansont Posted October 15, 2015 Posted October 15, 2015 Swanson h, I misread your equation. So you predict a radius 1836x smaller than the classical electron radius, or ~ 1.5 x 10^-18 m. Still disagrees with experiment. ----- For sake of clarity, and for help --- may I know experimental value? IOW, you haven't bothered to check your prediction against theory. Gabrielese measured the upper limit as 1 x 10^-18m, limited by they type of experiment he did. This number assumes all experimental uncertainty is due to the electron having a size. http://gabrielse.physics.harvard.edu/gabrielse/overviews/ElectronSubstructure/ElectronSubstructure.html (He also points out that if the electron had structure, the constituent particles would have masses of at least 177 GeV/c^2) There are other experiments that put the limit below 2 x 10^-20 m Since the mass is the reason for the Compton wavelength, you are correct. It's the cause. Energy is not the same as voltage. They are different things. ----- Where I have write this mistake? So you can't even remember what you wrote, or look back to a previous post. Post #10 "I have used the radius of proton wave-length in calculation of energy of proton, or its voltage." It doesn't give the right answer. The experimentally determined proton radius is about 0.9 x 10^-15 m. That's only about a factor of 3 smaller than the classical electron radius. ---- Not clear for me. I say: classical radius for electron is Re = 2 .8179401*10^-15 m and radius of proton Rp = 1.534698155* 10^-18 m. You say experiments have found: Re = about 0 --- and Rp = 0.9 *10^-15. Right? Yes. Your numbers do not reflect what experiment shows. The classical electron radius was calculated using classical physics, which doesn't work on these scales — it's wrong. e / (4*pi*e / ( U * d)) = U*d = 1.43994393*10^-9 V*m but nowhere have you explained what U and d are, or where they come from, or where you got the formula — it just appears out of thin air. But doesn't U*d cancel? e / (4*pi*ε / ( U * d)) = (e*U*d)/(4*pi*ε) so this is the same as saying e/(4*pi*ε) = 1, which is clearly false. Meaning I have no clue how you are getting any of these values. ----- Here is the speculation: E1 = (- e) * [ (-e) / ((4*pi*ε)*1)] = (-e) * U1*1/1= (-e)*U1 Let see this part of eq. [ (-e) / ((4*pi*ε)*1)] = (-e) / (+e) / (U1*1) = 1.43994393*10^-9 V*m it is unity potential of two electric charge in 1 m distance. Here U1 is the potential created between (must be) electric charges (-e) and +e in a distance in space =1m ( Which is the unity of distance in system m, kg, s, A). The electric charges are unperceivable because shadow each other, but exist and form stationary Compton wave. The potential of Compton wave interact with perceivable (-e) and determine the electric energy of an perceivable electric charge in a sphere 1m. E1.= (1.43994393*10^-9) V*1m * e / 1m = 2.307044*10^-28 joule. And Ee = (1.43994393*10^-9) V*1m * e / Re = 8.186988333*10^-14 joule Ep = ((1.43994393*10^-9) V*1m * e / Rp = 1.50325604*10^-10 joule And? What is a Compton wave, much less a stationary Compton wave? But here is the speculation. Photon has two electric charges, unperceivable because have different sign, and seems like they don’t exist in photon, because they shadow each other. Electron has three electric charges: one is perceivable, two are unperceivable because they shadow each other when exist in Compton wave of particle. What is the predicted photon dipole moment, if it's comprised of two charges? What is the predicted electron dipole moment, if it's comprised of three? Why can't I split up an electron (or photon) into its constituent particles? What's the binding energy of the systems?
Sensei Posted October 16, 2015 Posted October 16, 2015 (edited) (He also points out that if the electron had structure, the constituent particles would have masses of at least 177 GeV/c^2) That's up-side-down quantum physics... More mass-energy from decaying smaller mass-energy particle.. ? That's instant violation of energy conservation.. f.e. muon is elementary particle in Standard Model, but the all particles mass-energies it is decaying to, are smaller than rest-mass of muon, 105.66 MeV/c^2. f.e. [math]\mu^-\rightarrow e^- + \bar{V}_e + V_{\mu}[/math] [math]m_e = 0.510998928 MeV/c^2[/math] The all known particles ALWAYS decay to particles with less mass-energy than they have prior decay, to conserve energy.. Edited October 16, 2015 by Sensei
swansont Posted October 16, 2015 Posted October 16, 2015 That's up-side-down quantum physics... More mass-energy from decaying smaller mass-energy particle.. ? That's instant violation of energy conservation.. f.e. muon is elementary particle in Standard Model, but the all particles mass-energies it is decaying to, are smaller than rest-mass of muon, 105.66 MeV/c^2. f.e. [math]\mu^-\rightarrow e^- + \bar{V}_e + V_{\mu}[/math] [math]m_e = 0.510998928 MeV/c^2[/math] The all known particles ALWAYS decay to particles with less mass-energy than they have prior decay, to conserve energy.. That's moot, though, since it wouldn't decay to the constituent particles. The muon you mention, for example, doesn't contain an electron and the neutrinos. The mass of a nucleus or atom is less than the mass of its constituent particles. You have to add energy to break them apart, and the binding energy indicates how much. Decay isn't the relevant process. But the whole exercise is moot, in a way, since there's no indication that the electron is a composite particle.
Kramer Posted October 16, 2015 Author Posted October 16, 2015 SwansonIOW, you haven't bothered to check your prediction against theory.---- A smirk— but anyway. Because I know the only thing of predictions: electron particle has a zero space dimension. My speculation say: the dimension is the main property of an elementary particle, from which depends other properties as mass, energy, frequency etc.. Gabrielese measured the upper limit as 1 x 10^-18m, limited by they type of experiment he did. This number assumes all experimental uncertainty is due to the electron having a size.http://gabrielse.phy...bstructure.html(He also points out that if the electron had structure, the constituent particles would have masses of at least 177 GeV/c^2) There are other experiments that put the limit below 2 x 10^-20 m ---- I am not sure how are convincing the results of dimensions, as I know the scientists doubt. And in my speculation (not theory) the particles are structured by very tiny “subs”, tiny but not zero, in spherical chests created by movements of very fast subs.In this case the only measure are wavelength, and frequency. But I am not sure if experimentations have possibility to measure frequencies let say: fe = 1.2355906*10^20 Hz.Thanks for the link. So you can't even remember what you wrote, or look back to a previous post. Post #10 "I have used the radius of proton wave-length in calculation of energy of proton, or its voltage." ------Another smirk. O come on, Swasont. Don’t pretend that don’t understand. “Calculation of energy or voltage” doesn’t means that they are the same, but that they depends both by radius. Yes. Your numbers do not reflect what experiment shows. The classical electron radius was calculated using classical physics, which doesn't work on these scales — it's wrong. ------ Well. I think that radius is the most important variable in the supposed structure of particles. They have a link with other properties which are measurable with certainty, and that fit.And?What is a Compton wave, much less a stationary Compton wave? ------ In my lay-mans vision about stationary Compton wave length, is the electromagnetic phenomena embedded in a particle. And, --- that electro-“magnetic” phenomena are linked with movement in circle of “electric charge”. And,--- that movement in circle out of concept of radius, is absurd.That the Compton wave length exist, I assume that they must be real. Other ways why they are near every particle in CODATA?What is the predicted photon dipole moment, if it's comprised of two charges? What is the predicted electron dipole moment, if it's comprised of three? ------ I see ! You want That a lay – man to answers on your questions as a physician? For most of your questions as I know, I am not able to answer. But even in your question stand a strong argument against my “radius speculation” like magnetic moment anomaly, may be a specialist can give an answer, or can justify the anomaly with movement in spherical trajectories.And my question for specialist: How explained magnetic moment in an electron with which has zero diameter, and has not any insider movement? Why can't I split up an electron (or photon) into its constituent particles? What's the binding energy of the systems? ------- You tell me if you know. Maybe possession of electron particle ‘me’, in the same time in disposition, with what ever kind of energy and force: Ee = Em =Ef = Eem……..They split only in the moment when exchange their constituent with their antimatter mate, in the process of ….?! Annihilation. Ha?SenseiThat's up-side-down quantum physics... More mass-energy from decaying smaller mass-energy particle.. ? That's instant violation of energy conservation..f.e. muon is elementary particle in Standard Model,but the all particles mass-energies it is decaying to,are smaller than rest-mass of muon, 105.66 MeV/c^2.f.e. The all known particles ALWAYS decay to particles with less mass-energy than they have prior decay, to conserve energy.. ------ As I know, electron and proton are not decaying particles.Their antimatter positron and antiproton, I speculate, are absolute decaying particles, via “gravity attraction” of subs, which is some less than their “electric repelling” of +e charge. They fill space with un-engaged antimatter subs. Sorry I over speculate.
swansont Posted October 17, 2015 Posted October 17, 2015 Swanson IOW, you haven't bothered to check your prediction against theory. ---- A smirk— but anyway. Because I know the only thing of predictions: electron particle has a zero space dimension. My speculation say: the dimension is the main property of an elementary particle, from which depends other properties as mass, energy, frequency etc.. Predictions are one thing, experimental results are another. If it disagrees with experiment, it's wrong. Your prediction disagrees with experiment. It's wrong. ---- I am not sure how are convincing the results of dimensions, as I know the scientists doubt. And in my speculation (not theory) the particles are structured by very tiny “subs”, tiny but not zero, in spherical chests created by movements of very fast subs. In this case the only measure are wavelength, and frequency. But I am not sure if experimentations have possibility to measure frequencies let say: fe = 1.2355906*10^20 Hz. Thanks for the link. Which scientists doubt these results? So you can't even remember what you wrote, or look back to a previous post. Post #10 "I have used the radius of proton wave-length in calculation of energy of proton, or its voltage." ------Another smirk. O come on, Swasont. Don’t pretend that don’t understand. “Calculation of energy or voltage” doesn’t means that they are the same, but that they depends both by radius. Voltage makes no sense in this context, so no, I don't understand. But I know what you did; just because the units match if you take an energy and divide by charge does not mean the quantity is meaningful. Yes. Your numbers do not reflect what experiment shows. The classical electron radius was calculated using classical physics, which doesn't work on these scales — it's wrong. ------ Well. I think that radius is the most important variable in the supposed structure of particles. They have a link with other properties which are measurable with certainty, and that fit. Radii are measurable with certainty, and your results don't fit. And? What is a Compton wave, much less a stationary Compton wave? ------ In my lay-mans vision about stationary Compton wave length, is the electromagnetic phenomena embedded in a particle. And, --- that electro-“magnetic” phenomena are linked with movement in circle of “electric charge”. And,--- that movement in circle out of concept of radius, is absurd. That the Compton wave length exist, I assume that they must be real. Other ways why they are near every particle in CODATA? The Compton wavelength is a calculable quantity from a formula. It's the wavelength of a photon whose energy is the same as the rest mass energy of the particle (h/mc) making it a useful number for calculations. It does not imply that there is a phenomenon known as a Compton wave What is the predicted photon dipole moment, if it's comprised of two charges? What is the predicted electron dipole moment, if it's comprised of three? ------ I see ! You want That a lay – man to answers on your questions as a physician? For most of your questions as I know, I am not able to answer. But even in your question stand a strong argument against my “radius speculation” like magnetic moment anomaly, may be a specialist can give an answer, or can justify the anomaly with movement in spherical trajectories. And my question for specialist: How explained magnetic moment in an electron with which has zero diameter, and has not any insider movement? Yes, absolutely. I want you to be able to discuss science if you're posting on a science board. It is what is expected of you. Why can't I split up an electron (or photon) into its constituent particles? What's the binding energy of the systems? ------- You tell me if you know. Maybe possession of electron particle ‘me’, in the same time in disposition, with what ever kind of energy and force: Ee = Em =Ef = Eem…….. They split only in the moment when exchange their constituent with their antimatter mate, in the process of ….?! Annihilation. Ha? My answer is that it is not a composite particle. What is your answer, since you claim it is? That's the issue, and what you are expected to address.
Phi for All Posted October 17, 2015 Posted October 17, 2015 I see ! You want That a lay – man to answers on your questions as a physician? For most of your questions as I know, I am not able to answer. ! Moderator Note Kramer, you make it so difficult to discuss science with you. You make claims you can't support, and when asked you fall back on the fact that you're not a scientist. Yet you continue to make your bold, anti-mainstream claims, asserting them in a way that demands evidence, which you don't provide. Discussions like this eventually circle back to the same crossroads; provide some supportive evidence, or start making assumptions and guesses that may not have a foundation in reality. And we don't go down that latter road here. It lacks all rigor. You continue to use the same process in your inquiries, so your speculative thread closures are on a par with others who challenge mainstream science without providing evidence to support their reasoning. I would hope you post here instead of some of the more loosely run forums where wild guesswork is allowed and even encouraged, because we ask you for a more rigorous approach. We challenge you to find that which will provide a reason for our membership to think your ideas reflect reality better than one of our current explanations. No more hand-waving, no more Arguments from Incredulity, and no more "I know I'm right but I don't have to show evidence because I'm a layman". Without supportive evidence, I have to close this speculation. Don't open this one again until you're prepared to be more rigorous.
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