swansont Posted February 7, 2014 Posted February 7, 2014 Swanasont said: How is R determined? Reply: R is equal to the shell number cubed or you can just count the rotations of the magnetic field. OK, then. Let's see if your formula accurately predicts the Hydrogen energy levels. The ground state is -13.6 eV. The n=2 level is -3.4 eV. Your equation predicts -1.7 eV. Since R = n3, your formula looks a lot like what physics already predicts, except you have an extra factor of 2. So you need to fix that. More importantly, you need to derive this equation from the underlying physics principles. Just arbitrarily dropping the factor of 2 is an admission that you pulled the equation from … somewhere. Swansont said: How does an electron change direction like that? Where does the angular momentum go (or come from) when it changes direction? Reply: Both the Fine lines and the Hyperfine lines show up on level changes. The energy is accounted for by the capture or release of a photon. The electron changes from one orbit to another that will have either the same or reversed direction of rotation around the nucleus. The slight difference in energy produces the fine line effect. When an electron is heading for the nucleus it only takes a little jolt to make it pass on the other side of the nucleus but the value of the angular momentum could still be the same. A photon only has hbar of angular momentum. An electron in a P state has hbar, and if its excited to a D state is now has 2hbar of orbital angular momentum. If the direction changes, then the difference is 3hbar. Your description violates conservation of angular momentum. Swansont said: There are no wires in a nucleus. How does a proton give you a rotating field, and how come nobody has ever noticed that it has one? Reply: The Handbook of Nuclear Properties of 1997 mentions nuclei with magnetic forces having a rotational character. Unless it was lying, a rotating magnetic field is an accepted concept. So it could be possible for the hydrogen nucleus to have one. It might not be easy to notice. Where, exactly, does it say that, and what, exactly, does it say? If it accounts for the energy structure as you describe, it should be easy to notice. You can't have it both ways. This model has electrons and positrons in the nucleus and photons, electrons, etc consisting of small positive and negative vector like entities that always travel at the speed of light. The large magnetic moment of an electron is accounted for by the currents created by the movement of the entities. The proton should have the extra positron at its center. One possible means of generating a rotating magnetic field is having entities in an off center oblong path around the positron at the center of the proton which would precess because of the fixed speed of the entities. Your digging a deeper hole here, in violating more and more established physics principles. Electrons and photons having structure when none has been observed, positrons being confined inside of protons, etc. Add them to the list of things you need evidence for and new physics to explain. But really, before you pile this any deeper, address the objections that have already been brought up.
Lazarus Posted February 10, 2014 Author Posted February 10, 2014 Swansont said: OK, then. Let's see if your formula accurately predicts the Hydrogen energy levels. The ground state is -13.6 eV. The n=2 level is -3.4 eV. Your equation predicts -1.7 eV. Since R = n3, your formula looks a lot like what physics already predicts, except you have an extra factor of 2.So you need to fix that. More importantly, you need to derive this equation from the underlying physics principles. Just arbitrarily dropping the factor of 2 is an admission that you pulled the equation from … somewhere. Reply: Thank you for pointing out my error. The 2 has no business being in the denominator. This arithmetic stuff gets me in trouble. The only cheating I did was to use the known energy of the lowest hydrogen level. The equation shows how a rotating magnetic field could cause the distinct energy levels of the electron. ------------------------------------------------------------------------------- Swansont said:A photon only has hbar of angular momentum. An electron in a P state has hbar, and if its excited to a D state is now has 2hbar of orbital angular momentum. If the direction changes, then the difference is 3hbar. Your description violates conservation of angular momentum. Let’s save this for entertainment after all the other issues are resolved. --------------------------------------------------------------------------- The Handbook of Nuclear Properties of 1997 Where, exactly, does it say that, and what, exactly, does it say?If it accounts for the energy structure as you describe, it should be easy to notice. You can't have it both ways. The book was in the ASU library. This may be moot now as there is a lot of information on nuclear magnetic moments available. -------------------------------------------------------------------------------- Swansont said: Your digging a deeper hole here, in violating more and more established physics principles. Electrons and photons having structure when none has been observed, positrons being confined inside of protons, etc. Add them to the list of things you need evidence for and new physics to explain. But really, before you pile this any deeper, address the objections that have already been brought up. Reply: This is a recounting of objections to this model: The rotating magnetic field that forces the electrons into discrete orbits. Electrons are required to radiate under acceleration. The mysterious constant, .00030394, in the red shift calculation. The zero angular momentum of hydrogen at level S. The electric dipole moment in the nuclei. Neutrino decay. Structure of protons. Small positive and negative vector like entities traveling at c. Heisenberg theory, Pauli theory and this doesn’t agree with current theories. Nobody has complained yet about the explanation of photons passing the sun. The hbar problem. Now the explainations. The rotating magnetic field of the nucleus to produce the discrete orbits is the most important objection. The Rabi Oscillation and the Lamor Precession are intimately involved with Magnetic Resonance Imaging and are just what we would need to force electrons into discrete orbits. However, it requires a magnetic field to occur. So we need to find a magnetic field to satisfy our needs in the nucleus. The 3 quarks for master Mark, the three hump model of the proton and scattering experiments seem to indicate that there are 3 parts to the proton. We know that the proton has a magnetic moment so it is reasonable that each of the 3 parts have a magnetic moment. Then the center part is in a magnetic field and could exhibit Rabi Oscillation and Lamor Precession. Voila, we have the rotating magnetic field we need. Since a photon is much larger than the diameter of an atom the electron makes several complete orbits while a photon is leaving or arriving. It is far fetched to say that there is impossible for there to be a mechanism that prevents a photon from leaving. The constant .00030394 is determined by the shape of the elliptical orbit of the electron and conversely the known energy of the electron and the electron/proton attraction determines the shape of the orbit The number .00030394 is the ratio of the velocity of the electron at the apogee to the velocity of an electron in a circular orbit in the lowest level of hydrogen. The reason that the ratio is very small is that the elliptical orbit is very skinny, not much fatter than a line. There should not be much angular momentum in a pencil shaped ellipse. Thanks to my favorite physicist’s pointing out problems, the model is much improved, so the current model would not have an electric dipole. Since this model has the proton made up of electrons and positrons they can be combined to create a neutrino. The proton consists of 1198 electron/positron pairs plus one positron. The neutron consists of 1200 electron pairs. The proton has 3 distinct parts. The nuclear binding force is replaced by the extra electrons and the sharing of electron pairs. All electrons, positrons and photon consist of many small positive and negative entities traveling at the speed of light. It is not fair or logical to try to discredit a theory by use of an opposing theory, so Heisenberg Uncertainty, Pauli Exclusion and other theories cannot invalidate another theory. I wish someone would complain about the explanation of the failure of classical equations to correctly show the path of a photon passing the sun. The hbar problem can be the last to get resolved.
swansont Posted February 10, 2014 Posted February 10, 2014 Swansont said: OK, then. Let's see if your formula accurately predicts the Hydrogen energy levels. The ground state is -13.6 eV. The n=2 level is -3.4 eV. Your equation predicts -1.7 eV. Since R = n3, your formula looks a lot like what physics already predicts, except you have an extra factor of 2. So you need to fix that. More importantly, you need to derive this equation from the underlying physics principles. Just arbitrarily dropping the factor of 2 is an admission that you pulled the equation from … somewhere. Reply: Thank you for pointing out my error. The 2 has no business being in the denominator. This arithmetic stuff gets me in trouble. The only cheating I did was to use the known energy of the lowest hydrogen level. The equation shows how a rotating magnetic field could cause the distinct energy levels of the electron. ------------------------------------------------------------------------------- Without a derivation of how, exactly, a rotating magnetic field would cause this, it shows nothing. You can derive the hydrogen energy with an assumption of an electrostatic interaction. You are adding in an extra interaction (in an ad-hoc fashion) and getting the same result. This is a recounting of objections to this model: [ Now the explainations. The rotating magnetic field of the nucleus to produce the discrete orbits is the most important objection. The Rabi Oscillation and the Lamor Precession are intimately involved with Magnetic Resonance Imaging and are just what we would need to force electrons into discrete orbits. However, it requires a magnetic field to occur. So we need to find a magnetic field to satisfy our needs in the nucleus. The 3 quarks for master Mark, the three hump model of the proton and scattering experiments seem to indicate that there are 3 parts to the proton. We know that the proton has a magnetic moment so it is reasonable that each of the 3 parts have a magnetic moment. Then the center part is in a magnetic field and could exhibit Rabi Oscillation and Lamor Precession. Voila, we have the rotating magnetic field we need. Hand-waving is not enough. Show this quantitatively. MRI is entirely consistent with the mainstream QM model. Since a photon is much larger than the diameter of an atom the electron makes several complete orbits while a photon is leaving or arriving. It is far fetched to say that there is impossible for there to be a mechanism that prevents a photon from leaving. Then come up with such a mechanism. "Hasn't been ruled out" (assuming for sake of argument that this is true) is not the same as "evidence for", and the burden of proof lies with you. There should not be much angular momentum in a pencil shaped ellipse. This is science: we quantify things. Angular momentum comes in units of hbar, and you should be able to calculate how much angular momentum a particular orbit has. The S orbital is not pencil-shaped. Since this model has the proton made up of electrons and positrons they can be combined to create a neutrino. The proton consists of 1198 electron/positron pairs plus one positron. The neutron consists of 1200 electron pairs. The proton has 3 distinct parts. The nuclear binding force is replaced by the extra electrons and the sharing of electron pairs. Offered yet again with no corroborating evidence or model. All electrons, positrons and photon consist of many small positive and negative entities traveling at the speed of light. Offered yet again with no corroborating evidence or model. It is not fair or logical to try to discredit a theory by use of an opposing theory, so Heisenberg Uncertainty, Pauli Exclusion and other theories cannot invalidate another theory. HUP and other mainstream theories have lots of evidence to support them. That's why they are mainstream. It is fair, and also in keeping with the rules of the speculations section (and science in general). We don't have to re-prove all of physics every time we make a model or do an experiment. I wish someone would complain about the explanation of the failure of classical equations to correctly show the path of a photon passing the sun. GR addresses this already, and there is experimental confirmation that it does so correctly.
Lazarus Posted February 23, 2014 Author Posted February 23, 2014 Here is the contention of this model. With the assumptions: 1 All matter, electron, protons, photons, etc, all consists of small positive and negative entities that always travel at the speed of light. 2 Protons and neutrons consist of electrons and positrons. 3 Nuclei are held together by electrostatic forces of electrons and positrons. 4 Nuclei have a rotating magnetic field. 5 That Gauss’ electrostatic and electromagnetic Laws, Faraday’s Law and Ampere’s original law are valid. 6 Newton’s Law of gravity is valid. 7 That Newton’s other Laws are corollaries of the electrical laws. This model shows: The reason for discreet orbits of electrons about the nucleus. The relationship of electron orbits to spectral lines and the orbits of moving atoms to the speed of galaxies The reason for slowing of clocks in gravity and at high speeds That physical matter stretches in the direction of motion eliminating the need for time and distance manipulation. That the expansion of space may not be required because the velocity of the source of old light moving away from where the earth is now can account for the Red Shift of light. ------------------------------------------------------------------------- Comments on Radiation The sources of radiation are: Bremsstrahlung radiation shown by Synchrotron radiation or any charge moving through a magnetic field. The reason an electron radiates when affected by a bending magnetic field is that the magnetic moment reacting with the magnetic field provides forces that try to expand the electron causing part of the electron to break free and become a photon. Consider a charged current loop moving through a magnetic field. In addition to the charge causing a change in the direction of motion, each point on the loop receives an outward force. Atomic shell change radiation. The radiation from an electron changing shells is the energy required to offset the difference in the total energy from one shell to the other. Collision radiation from charged particles colliding with atoms. Radiation from the collision can be caused by the Bremsstrahlug radiation and from an electron changing shells. Electrostatic acceleration. It is possible that electrostatic acceleration does not have any unaccounted for energy to produce a photon when the kinetic and latent energy are offset. At this point in the thread: I haven’t proved that this is a valid model of the real world. You haven’t proved that it’s not.
swansont Posted February 23, 2014 Posted February 23, 2014 At this point in the thread: I haven’t proved that this is a valid model of the real world. You haven’t proved that it’s not. Not my job to. You propose the idea, you need to make the model and show how it's supported by experiment. I've asked you to show your model of hydrogen that includes a rotating magnetic field. The existing model works quite well without it. So let's have your model. The rules of speculations require it.
Lazarus Posted May 30, 2014 Author Posted May 30, 2014 Swansont said: So let's have your model. The rules of speculations require it. --------------------------------------------------------------------------------------- Here is a model of the hydrogen atom, the rest of the elements and the formation of molecules. The assumption is that nuclei have a rotating magnetic field with the time of rotation equal to the period of the hydrogen electron orbit divided by the atomic number to the three halves power. (T = Th/Z^1.5) It is the cause of discrete orbits. Using the equation for circular and elliptical orbits, the approximate size of all atoms can be calculated. The form of the equation is 4Pi^2r^3m = kzq^2t^2. The symbols are: r is semi-major axis of the ellipse, m is the mass of an electron, k is Coulombs Constant, z is the atomic number, q is the charge of an electron and t is the period of the orbit. Molecules are formed by electrons orbiting 2 nuclei. Here are the results of the calculations for some of the elements compared to the McGraw-Hill chart. Z is the atomic number Te-17 is the orbit period of a nucleus with only one electron times 10^17 S1 pm is the semi-major axis for a nucleus with one electron, in picometers. Calc is the calculated semi-major axis of the atom, in picometers. Book is the McGraw-Hill value for the radius of the atom. Diff is the difference. Ratio is the ratio of the book value to the calculated value. Z Te-17 S1 pm Calc Book Diff Ratio Shell 1 1 8.887 37.003 43.844 37.000 -6.844 0.844 2 3.142 26.163 31.000 31.000 -0.000 1.000 Shell 2 3 1.710 21.359 101.234 152.000 50.766 1.501 4 1.111 18.503 87.696 112.000 24.304 1.277 5 0.795 16.550 78.439 85.000 6.561 1.084 6 0.605 15.103 71.583 77.000 5.417 1.076 7 0.480 13.986 66.289 75.000 8.711 1.131 8 0.393 13.083 62.008 73.000 10.992 1.177 9 0.329 12.332 58.449 72.000 13.551 1.232 10 0.281 11.697 55.441 70.000 14.559 1.263 Shell 3 11 0.244 11.154 118.949 186.000 67.051 1.564 12 0.214 10.678 113.872 160.000 46.128 1.405 13 0.190 10.263 109.446 143.000 33.554 1.307 14 0.170 9.891 105.476 118.000 12.524 1.119 15 0.153 9.555 101.896 110.000 8.104 1.080 16 0.139 9.250 98.642 103.000 4.358 1.044 17 0.127 8.975 95.713 99.000 3.287 1.034 18 0.116 8.719 92.979 98.000 5.021 1.054 Shell 4 19 0.107 8.487 160.899 227.000 66.101 1.411 20 0.099 8.273 156.849 197.000 40.151 1.256 31 0.051 6.644 125.953 135.000 9.047 1.072 32 0.049 6.540 123.986 123.000 -0.986 0.992 33 0.047 6.442 122.135 120.000 -2.135 0.983 34 0.045 6.345 120.284 117.000 -3.284 0.973 35 0.043 6.253 118.548 114.000 -4.548 0.962 36 0.041 6.168 116.928 112.000 -4.928 0.958 Shell 5 37 0.039 6.082 180.169 248.000 67.831 1.376 38 0.038 6.003 177.818 215.000 37.182 1.209 49 0.026 5.289 156.664 166.000 9.336 1.060 50 0.025 5.234 155.037 140.000 -15.037 0.903 51 0.024 5.179 153.410 141.000 -12.410 0.919 52 0.024 5.130 151.964 143.000 -8.964 0.941 53 0.023 5.081 150.517 133.000 -17.517 0.884 54 0.022 5.032 149.071 131.000 -18.071 0.879 Shell 6 55 0.022 4.990 212.839 265.000 52.161 1.245 56 0.021 4.947 211.017 222.000 10.983 1.052 81 0.012 4.111 175.348 171.000 -4.348 0.975 82 0.012 4.086 174.307 175.000 0.693 1.004 83 0.012 4.062 173.266 155.000 -18.266 0.895 84 0.012 4.037 172.224 164.000 -8.224 0.952 85 0.011 4.013 171.183 142.000 -29.183 0.830 86 0.011 3.989 170.141 140.000 -30.141 0.823
swansont Posted May 31, 2014 Posted May 31, 2014 Swansont said: So let's have your model. The rules of speculations require it. --------------------------------------------------------------------------------------- Here is a model of the hydrogen atom, the rest of the elements and the formation of molecules. The assumption is that nuclei have a rotating magnetic field with the time of rotation equal to the period of the hydrogen electron orbit divided by the atomic number to the three halves power. (T = Th/Z^1.5) It is the cause of discrete orbits. Using the equation for circular and elliptical orbits, the approximate size of all atoms can be calculated. The form of the equation is 4Pi^2r^3m = kzq^2t^2. The symbols are: r is semi-major axis of the ellipse, m is the mass of an electron, k is Coulombs Constant, z is the atomic number, q is the charge of an electron and t is the period of the orbit. How do these equations follow from a rotating magnetic field? The orbital parameters should depend on this, right? Why does the field have to be rotating? What is the source of the magnetic field?
Lazarus Posted June 3, 2014 Author Posted June 3, 2014 Swansont said: How do these equations follow from a rotating magnetic field? The orbital parameters should depend on this, right? Why does the field have to be rotating? What is the source of the magnetic field? ---------------------------------------------------------------------- The rotating magnetic field causes the orbits to be discrete. The elliptical orbits have to sync with a multiple of the time of rotation. Otherwise, the orbits would not be stable. A possible cause of the rotation could be that the magnetic field from the 2 end quarks induce a Larmor rotation in the center quark if the 3 quarks are not in a straight line. Some descriptions of the proton put a bend in the proton. The concept might be testable. Line up some hydrogen atoms and apply a strong magnetic field. That should affect the time of rotation causing a change the orbit of the electron which would cause a change in the spectrum. The steps of calculations: Construct orbital equations involving R, V and T. Solve for R, V and T using known values for H, He and He+. Derive the rotational time of an electron. It is proportional to 1/Z^0.5, Helium has a full shell 1 with 2 electrons but hydrogen only has 1 electron. To find the ratio of the size of hydrogen to a full shell, the ratio of sizes between He and He+ can be used. He is smaller than twice the size of He+ so there has to be overlap of the 2 electron orbits of Helium. The ratio of the sizes of He and He+ is 1.1849. Calculate R for an atom with only 1 electron. Get the size of the atom by multiplying R by the outer shell number squared and 1.1849. DERIVATION OF THE EQUATIONS The symbols are: Z = Atomic number 1 Hydrogen M = Mass of an electron 9.109388e-31 kg Q = Charge of an electron -1.602e-19 coulombs R = Semi- major axis 37 picometers Hydrogen E = Kinetic energy of an electron 2.17896e-18 joules Hydrogen K = Coulombs constant 8.9875e9 V = Velocity of electron 2.187e6 m/s Hydrogen T = Time of orbit 1.4806e-16 seconds Hydrogen (Th) S = Shell number 1 Hydrogen EQUATIONS MV^2 / R = Centripetal force KZQ^2 / R^2 = Electrostatic force MV^2/R = KZQ^2 / R^2 Since they are equal R^3MV^2 = KZQ^2 Reduced R^3 = KZQ^2 / MV^2 Semi-major radius cubed R = (KZQ^2 / MV^2)^.33333 Semi-major axis using velocity V = 2PiR/T Velocity is Distance divided by Time 4Pi^2R^2M / RT^2 = KZQ^2 / R^2 Substituting for V in MV^2/R = KZQ^2 / R^2 4Pi^2R^3M = KZQ^2T^2 T^2 = 4Pi^2R^3M / KZQ^2 T squared T = 2PiR^1.5M^.5/K^.5Z^.5Q Time of orbit T = 1.48e-16 / Z^.5 Period of orbit = Period of H divided by Sq root of Z R = (KZQ^2T^2 / 4Pi^2M)^.33333 Substituting for V in R = (KZQ^2 / MV^2)^.33333 R is the semi-major axis of the atom with only 1 electron. To get the semi-major axis of the whole atom multiply by the shell number squared and the He / He+ ratio. r = 1.1849RS^2 Semi-major axis of the whole atom I haven’t found the reference to nuclei with rotating magnetic fields that I stumbled across in the Handbook of Nuclear properties but I did find 3 other references to rotating magnetic fields or oscillating fields in the book. I was surprised to find that this 18 year old book still sells for over $100. Quotes from pages 29, 59 and 81 follow: On page 29: OSCILLATING TRENEDS IN DEFORMED REGIONS Deeper inside major shell regions, where nuclei are deformed, there are oscillating trends superposed on the quadratic variation of the masses. these are reflected in the systematics of mass differences. On page 59: OSCILLATING TRENDS AND DEFORMATION The oscillating trends (section II.F are accounted for by the configuration interaction part (89)) of the mass equation equation. In particular, the depression of the mass surface and the corresponding humps in separation energy lines in deformed regions result from the mixed terms representing mainly neutron-proton correlations. On page 81: INTRODUCTION In specific regions of the nuclear periodic chart, large quadrupole moments ore observed and the low-lying excitations have a rotational character. ------------------------------------------------------------------------------
swansont Posted June 4, 2014 Posted June 4, 2014 There is no magnetic field in your derivation. The first question in my last post is still unanswered.
Lazarus Posted June 4, 2014 Author Posted June 4, 2014 Swanson said: How do these equations follow from a rotating magnetic field? There is no magnetic field in your derivation. The first question in my last post is still unanswered. Reply: The assumption is that nuclei have a rotating magnetic field. The time of rotation of the magnetic field is derived from the known values of hydrogen and that the time of rotation of the magnetic field has to be the same as the lowest possible orbit of the hydrogen electron. Sorry about the oversize print. I never know how the text will come out when it posts.
swansont Posted June 4, 2014 Posted June 4, 2014 Swanson said: How do these equations follow from a rotating magnetic field? There is no magnetic field in your derivation. The first question in my last post is still unanswered. Reply: The assumption is that nuclei have a rotating magnetic field. The time of rotation of the magnetic field is derived from the known values of hydrogen and that the time of rotation of the magnetic field has to be the same as the lowest possible orbit of the hydrogen electron. Sorry about the oversize print. I never know how the text will come out when it posts. None of your derivation includes a magnetic field. There are no predictions you make that include it.
Spyman Posted June 5, 2014 Posted June 5, 2014 (edited) Sorry about the oversize print. I never know how the text will come out when it posts. If you click on the [More Reply Options] button next to the button you get a [Preview Post] button that is very helpful. Edited June 5, 2014 by Spyman
Lazarus Posted June 6, 2014 Author Posted June 6, 2014 Spyman said: If you click on the [More Reply Options] button next to the button you get a [Preview Post] button that is very helpful. Lazurus: Thank you. Swansont said: None of your derivation includes a magnetic field. There are no predictions you make that include it. Reply: The assumption is that there is a rotating magnetic field around nuclei. There seems to be some physical evidence of that. The field doesn’t have to be very strong to affect the orbiting electrons. The strength of the field does not affect the time of the electron orbits as the electron elliptical orbits have to sync with the rotation time. The time of rotation of the magnetic field was calculated from values of hydrogen, centripetal and electrostatic formulas. That is all that is needed to calculate all the orbits of all the elements. The calculations have to give approximations because to get better results it is necessary to track all the electrons and their interactions. No more is necessary to construct a plausible model.
Lazarus Posted March 21, 2015 Author Posted March 21, 2015 Is this the reason that electrons radiate in a synchrotron? First, I must apologize to GREG H and SWANSONT for pushing my misunderstanding of the effects of momentum change in the production of radiation. In the example of a ball bouncing off of a wall the change of momentum must result in a change of the ball’s kinetic energy because the ball has to lose some energy as a result of the collision and the wall experiences a change in its kinetic energy. The work involved is equivalent to stopping the ball, then shooting it off in the opposite direction. The same effect of energy transfer occurs in a synchrotron when an electron’s path is changed by the magnetic field in the synchrotron. The electron must lose energy and slow down which causes it to radiate to compensate for the energy change. For an electron orbiting a nucleus, the net energy change is zero for a complete cycle. The kinetic energy of the nucleus changes in one direction during one half of a cycle, then changes in the opposite direction during the other half which zeros out the energy changes. Since the photon is much larger than the diameter of the atom, the electron has time to start emitting a photon, then suck it back in as the nucleus returns the kinetic energy from the other half of the cycle.
swansont Posted March 21, 2015 Posted March 21, 2015 Is this the reason that electrons radiate in a synchrotron? First, I must apologize to GREG H and SWANSONT for pushing my misunderstanding of the effects of momentum change in the production of radiation. In the example of a ball bouncing off of a wall the change of momentum must result in a change of the ball’s kinetic energy because the ball has to lose some energy as a result of the collision and the wall experiences a change in its kinetic energy. The work involved is equivalent to stopping the ball, then shooting it off in the opposite direction. The same effect of energy transfer occurs in a synchrotron when an electron’s path is changed by the magnetic field in the synchrotron. The electron must lose energy and slow down which causes it to radiate to compensate for the energy change. For an electron orbiting a nucleus, the net energy change is zero for a complete cycle. The kinetic energy of the nucleus changes in one direction during one half of a cycle, then changes in the opposite direction during the other half which zeros out the energy changes. Since the photon is much larger than the diameter of the atom, the electron has time to start emitting a photon, then suck it back in as the nucleus returns the kinetic energy from the other half of the cycle. But the radiation in a synchrotron is not toward the center, so why would you expect the energy from an atomic electron to go into the nucleus (if it worked this way, which it doesn't)? And how do you know the photon would be much larger than the atom?
Lazarus Posted March 25, 2015 Author Posted March 25, 2015 Swansont said: The radiation in a synchrotron is not toward the center Reply: If an electron approaching from the center changed direction by 180 degrees the radiation would go to the center. Swansont said: How do you know the photon would be much larger than the atom? Reply: The diameter of atoms is about 10^-10 meters. The wave length of visible light is about 10^6 meters. Soft X-Rays are about 10^9 meters. Since a polarizing grate with spacing less than a wave length will block some photons at least one direction across the photon must approximate the wave length. Swansont said: Why would you expect the energy from an atomic electron to go into the nucleus (if it worked this way, which it doesn't)? Reply: A ball bouncing off a wall slows down as it transfers energy to the wall. Similarly, an electron “bouncing” off a synchrotron or an electron making one half of a cycle around a nucleus must transfer energy and slow down. In the synchrotron it appears that there are 2 transfers of energy away from the electron which should be offset by the slowing of the electron. One is the radiation and the other is the reaction to the “bounce” of the electron. One possible solution is that the electron slows enough to compensate for both.
swansont Posted March 26, 2015 Posted March 26, 2015 Swansont said: The radiation in a synchrotron is not toward the center Reply: If an electron approaching from the center changed direction by 180 degrees the radiation would go to the center. I was overly optimistically expecting a physics justification. But how does an electron approach the nucleus from the center? The nucleus is at the center. Swansont said: How do you know the photon would be much larger than the atom? Reply: The diameter of atoms is about 10^-10 meters. The wave length of visible light is about 10^6 meters. Soft X-Rays are about 10^9 meters. Since a polarizing grate with spacing less than a wave length will block some photons at least one direction across the photon must approximate the wave length. What is a "polarizing grate" and what does it have to do with this? Swansont said: Why would you expect the energy from an atomic electron to go into the nucleus (if it worked this way, which it doesn't)? Reply: A ball bouncing off a wall slows down as it transfers energy to the wall. Similarly, an electron “bouncing” off a synchrotron or an electron making one half of a cycle around a nucleus must transfer energy and slow down. In the synchrotron it appears that there are 2 transfers of energy away from the electron which should be offset by the slowing of the electron. One is the radiation and the other is the reaction to the “bounce” of the electron. One possible solution is that the electron slows enough to compensate for both. You can't just throw around the term "synchrotron" and have it mean anything you want. Synchrotron radiation of a high-velocity charge moving in a circle is preferentially emitted along the direction of motion. You can't arbitrarily decide it goes into the nucleus. Since we don't observe this radiation, the obvious conclusion is that the atom doesn't work this way.
swansont Posted March 26, 2015 Posted March 26, 2015 I just ran the numbers, and using the critical frequency from http://en.wikipedia.org/wiki/Synchrotron_radiationfor hydrogen (assuming a Bohr orbit) the wavelength is 3.5 x 10-11 m. i.e. smaller than the atom. (It also corresponds to a 35 keV photon, which is a pretty hefty x-ray.)
xyzt Posted March 26, 2015 Posted March 26, 2015 Reply: If an electron approaching from the center changed direction by 180 degrees the radiation would go to the center. Electron does not "approach from the center". You cannot make up sh!t and throw it around trying to see if it sticks to the wall. 1
Lazarus Posted March 30, 2015 Author Posted March 30, 2015 Swansont said: I just ran the numbers, and using the critical frequency from http://en.wikipedia....otron_radiationfor hydrogen (assuming a Bohr orbit) the wavelength is 3.5 x 10-11 m. i.e. smaller than the atom. (It also corresponds to a 35 keV photon, which is a pretty hefty x-ray.) Reply: Yes, an electron with that energy would blaze its way out of an atom. In a static atom the energy levels do not reach that high. The escape energy (kinetic energy) of the lowest level of hydrogen is 13.6 eV or a wave length of about 9*10^-8 meters. More importantly, the escape energy for a uranium nucleus with only one electron is about 12,500 eV or 10^-10 meters compared to the size of the uranium atom of about 3*10^-10 meters. So you are right that at the extreme end of the periodic chart the wave length could be smaller. The normal configuration is to have 2 electrons in the lowest level which would mean an escape energy of about 3800 eV or 3*10^-10 which comes closer. It might well be that uranium atom with only one electron would be unstable. xyzt said: Electron does not "approach from the center". You cannot make up sh!t and throw it around trying to see if it sticks to the wall. Reply: Perhaps you should read the description of the hypothetical experiment again. -1
swansont Posted March 30, 2015 Posted March 30, 2015 Swansont said: I just ran the numbers, and using the critical frequency from http://en.wikipedia....otron_radiationfor hydrogen (assuming a Bohr orbit) the wavelength is 3.5 x 10-11 m. i.e. smaller than the atom. (It also corresponds to a 35 keV photon, which is a pretty hefty x-ray.) Reply: Yes, an electron with that energy would blaze its way out of an atom. In a static atom the energy levels do not reach that high. The escape energy (kinetic energy) of the lowest level of hydrogen is 13.6 eV or a wave length of about 9*10^-8 meters. More importantly, the escape energy for a uranium nucleus with only one electron is about 12,500 eV or 10^-10 meters compared to the size of the uranium atom of about 3*10^-10 meters. So you are right that at the extreme end of the periodic chart the wave length could be smaller. The normal configuration is to have 2 electrons in the lowest level which would mean an escape energy of about 3800 eV or 3*10^-10 which comes closer. It might well be that uranium atom with only one electron would be unstable. You appear to have completely missed the point. The cyclotron frequency is the frequency of the emitted radiation. Nothing to do with escape energy. The electron would not "blaze its way out of the atom", it would very quickly spiral in. Which it doesn't do. Further, the cyclotron frequency would increase with increasing Z, as this decreases r and increases v. Hydrogen is not an extreme case. It's the opposite. The bottom line is that we don't see this radiation, and you have provided no evidence that this is going on. Along with the lack of evidence for any of your other claims.
Phi for All Posted March 30, 2015 Posted March 30, 2015 ! Moderator Note OK, seven pages and I don't see any supportive evidence, though it's been requested several times. This is a requirement in Speculations, since anybody can have an unsupported idea. Mainstream science has been tested using the same requirements, so if you want to offer something better, you need to support it as much as you can. Please don't re-open this topic without some supportive evidence. Thread closed.
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